Jun-ichi Igusa (Auth.) - Theta Functions

xeRn

let J). denote the hypercube in Rn defined by -A.~Xi~A. for 1 ~i~n. If we take A. sufficiently large, the support of

7

§ 2. Plancherel Theorem for R"

to I;. can be extended to RR as an infinitely differentiable periodic function with 2 AZR as its period group. Let S denote a partial sum of its Fourier expansion. If S contains sufficiently many terms, we have IS(x)I~2M and also lexp(n x tx) 4>(x)- S(x)IP~tp"/2 llP for every x in I;.. We observe that cp(x)=exp( -n xtx) S(x) is an element of~. Moreover we have

J 14>(x)-cp(x)IP dx+ J Icp(x)IP dx ~t·pR/211P, J exp(-pnxtx)dx+(2M)P. J

114>-cpll~=

~

~-~

I;.

exp(-pnxtx)dx

R"-I;.

hence 114>-cpllpb for all b in RR is dense in IJ'(RR). The following is the Plancherel theorem for RR:

~

spanned over C

Theorem 2. There exists a unitary mapping 4> --+ 4>* of I3 (Rn) to itself satisfying (4)*)* (x) = 4>( -x) for every 4> in L2(Rn) such that if 4> is also contained in E(Rn), we have 4>* (x) =

J4>(y) e(xty) dy.

R"

Proof For every 4> in~, we define 4>* by the formula in the theorem. Then, for every 4>b(X) = exp( -nxtx)e(btx), we have

4>: (x) = exp( -nbtb) 4>ib(X) , and hence (4):)* (x) = 4> _ b(x) = 4>b ( - x). Consequently, the correspondence 4> --+ 4>* gives a C-linear bijection of ~ to itself. Furthermore we have (4)b' 4>b') = 2-"/2 exp( -(n/2)(b-ll)t(b-b'») = (4):, 4>;)

for every b, b' in C". Therefore the above C-linear bijection of ~ to itself preserves the norm. Since ~ is dense in L2 (Rn), this can be extended uniquely to a unitary mapping of L2 (R") to itself, also denoted by 4> --+ 4>*. We have (4)*)*(x)=4>(-x) for every 4> in I3(R"). This proves the first part. We shall prove the second part. Suppose that 4> is an arbitrary element of L2 (R"). Then we have (4)*,4>:)=(4>, 4>b) and 4>b(Y) = (4):)* ( - y), hence (4)*,4>:)= 4>(y)( 4>:(x)e(x ty)dx)dy.

J

R"

J

R"

-

8

I. Theta Functions from an Analytic Viewpoint

If ~ also belongs to L!(Rn), we can apply the Fubini theorem to the above repeated integral. In this way, we see that the difference of ~* and the function of x defined as the integral of ~(y)e(xty) over R n with respect to dy is orthogonal to every ~:, hence to ~. Since ~ is dense in L2 (Rn), this gives the integral representation of ~* (x). q.e.d. The unitary mapping in Theorem 1 is called the Fourier transformation of L2 (Rn) to itself.

§ 3. The Group A(X) We shall start by recalling the Pontrjagin duality theory for commutative locally compact groups. Suppose that X is an arbitrary commutative locally compact group. The law of composition in X will be written additively. Let X* denote the commutative group of characters of X, which are continuous homomorphisms of X to C;. We shall use q to denote the multiplicative group of complex numbers of absolute value 1. The law of composition in X* will be written also additively. With the topology of uniform convergence on compact subsets of X, X* becomes a locally compact group. We call X* the dual of X. If x, x* are elements of X, X*, the value of x* at x will be denoted by (x, x*) instead of x*(x). Let X, Y denote commutative locally compact groups and IX a continuous homomorphism of X to Y. Let X*, y* denote the duals of X, Y and x, y* general elements of X, Y*. Then the formula (x IX, y*) = (x, y* IX*)

defines a continuous homomorphism IX* of y* to X*. We call IX* the dual of IX. We may say that the class of all commutative locally compact groups forms a "category" and the asterisk gives a "contravariant functor" of this category to itself. The important fact is that there exists a bicontinuous isomorphism of X to (X*)* for every X which is "natural" in the sense that if IX is as above, it commutes with IX and (IX*)*. Because of this fact, we usually identify X with (X*)*. Also we put (x*, x) = (x, x*) for every x, x* in X, X*. Suppose that X is a finite dimensional vector space over R. With the standard topology, X forms a locally compact group. Furthermore X* can be identified with the dual space of X. In fact, there exists an Rvalued R-bilinear form [x, x*] on X x X* satisfying (x, x*) =e([x, x*])

for every x, x* in X, X*. The bilinear form [x, x*] is unique and nondegenerate. If the dimension of X is n, with respect to any R-base of X, we will get a bicontinuous isomorphism of X to Rn. If x is mapped to

§ 3. The Group A(X)

9

(Xl'" x n), the Lebesgue measure dXl ... dXn on Rn determines a Haar measure dx on X via this isomorphism. Similarly we get a Haar measure dx* on X*. We note that dx, dx* are not intrinsic. We shall consider the case where the isomorphisms of X, X* to Rn are defined with respect to dual bases. In this case, if x, x* are mapped to (Xl'" Xn), (xt ... X;), we have n [x,x*]= LXjxj=xtx*. Therefore, if we put

j=l

J

4J*(x*)= 4J(x) (x, x*) dx x

for every 4J in I!(X) n L2 (X), the Plancherel theorem for Rn implies that 4J -+ 4J* can be extended uniquely to a unitary mapping, also denoted by 4J-+4J*, of L2(X) to L2(X*). Moreover we have (4J*)*(x)=4J(-x) for every 4J in L2(X). As a consequence, if dx, dx* are arbitrary Haar measures on X, X*, the mapping 4J-+ 4J* can be defined in the same way, and it is a "similitude." The Haar measures dx, dx* on X, X* which give rise to a unitary mapping are called dual measures. We shall assume, in the following, that dx, dx* are dual measures. Let R, R* denote the regular representations of X, X*. We recall that R is defined as (R(u) 4J)(x) = 4J(x + u) for u, x in X and 4J in JJ(X). We put

(S(u*) 4J)(x)=(R*(u*) 4J*)*( - x) for every u* in X*. We observe that S is the unitary representation of X* in L2 (X) which is equivalent to the regular representation of X* by Fourier transformation. We shall denote by A(X) the subgroup of the unitary group Aut(L2(X)) generated by the images of Rand S. We observe that, although the norm in L2 (X) depends on the choice of dx, Aut(L2(X)), R, S are all intrinsic. Therefore A(X) is an intrinsically defined subgroup of Aut(L2(X)). We shall determine the structure of A(X). We need the following lemma: Lemma 3. Let u, u* denote elements of X, X*. If we identify them with the functions (u, ), ( ,u*) on X*, X, we have

(R(u) 4J)* =( -u) 4J*, for every 4J in L2 (X).

(u* 4J)*=R*(u*) 4J*

Proof. If we take 4J from L(X), the vector space over C of continuous functions on X with compact supports, the two identities follow immediately from the integral representation of the Fourier transformation. By continuity, they hold for every 4J in L2 (X). q.e.d.

10

1. Theta Functions from an Analytic Viewpoint

As a consequence of Lemma 3, we get S(u*) cI>=u* cI>, i.e., (S(u*) cI>)(x) = (x) , for every u* in X* and cI> in L2 (X). Therefore, if we introduce an element U(u, u*, t) of Aut(L2(X)) depending on u, u*, t in X, X*, C~ as we have

(U(u, u*, t) cI>)(x) = t (x + u), U(u, u*, t)=t S(u*)R(u).

Also we have R(u)S(u*)cI>=R(u)(u* cI»= = for every cI> in J3(X). In this way, we get the following commutation relation: R(u) S(u*)=
Theorem 3. Let A(X) denote the product space X x X* x Cr; introduce a law of composition in A(X) as (u I , ut, t 1)(U2, uI, t 2)= (u 1 + U2' ut + uI,
Then A(X) becomes a locally compact group. Furthermore, if we put (U(u, u*, t) cI>)(x)=t(x+u) for every (u, u*, t) in A(X), cI> in J3(X), and x in X, we get a unitary representation U of A(X) in L2(X). The image group U(A(X)) is A(X) and U gives a bicontinuous isomorphism of A(X) to A(X). Proof In view of our previous observations, what is left to be proved is that U gives a bicontinuous mapping of the locally compact space A(X) to the subgroup A(X) of Aut(L2(X)). Since we have U(u, u*, t)= t S(u*) R(u), the mapping U is certainly continuous. We shall show that

11

§ 4. The Irreducibility of U

if U(U, U*, t) is close to the identity in Aut(L2(X»), (u, u*, t) is close to (0, 0, 1) in A(X). We take an element ~ of L(X) satisfying II ~ II = 1. Then if we have II U(u, u*, t) ~ - ~ II < 21-, there exists a point x of X satisfying ~(x+u)~(x)*O. In that case, u is the difference of two points of the support of~. Therefore, if the support of ~ is small, u is close to 0 in X. If we pass to the representation of A(X) in If (X*) which is equivalent to U by Fourier transformation, the roles of u and u* interchange. This follows from Lemma 3. Therefore u* is close to 0 in X*. Since U is continuous, U (u, u*, 1) is close to the identity in Aut(L2(X»); hence t in U(u, u*, t) U(u, u*, 1)-1 =t· id is close to 1 in q. This proves the assertion. Finally, if U(u, u*, t) is close to U(uo, u~, to) in Aut(L2(X»), U(u, u*, t) U(uo, u~, t o)-l= U(u-uo, u*-u~, tt(

1)

is close to the identity in Aut(L2(X»). By what we have shown, this implies that (u, u*, t) is close to (uo, u~, to) in A(X). q.e.d. We shall denote a general element of X x X* by w. We see that U(w, t) is contained in the center of A(X) if and only if w=O. We usually identify t with (0, t) and U(O, t). Then q becomes the center of A(X).

Moreover, the quotient group A(X)/C~ is bicontinuously isomorphic to X x X*. Sometimes, the two-step nilpotent group A(X) is called the Heisenberg group, and its unitary representation U is called the Schrodinger representation.

§ 4. The Irreducibility of U In order to make the proof transparent, we shall recall basic (elementary) facts on group representations. Suppose that G is a locally compact group and dg a (right invariant) Haar measure on G. Then L(G) forms an associative ring by the convolution-product:

We call L(G) the group ring of G. If U is a unitary representation of G in a Hilbert space Hand t/I an element of L( G), the formula (U(t/I)a, b)= J(U(g) a, b) t/I(g) dg G

defines a bounded C-linear mapping U(t/I) of H to itself. In fact, for any fixed a, the complex conjugate ofthe right side defines a C-linear mapping T of H to C. By applying the Schwarz inequality, we get IT(b)l~

JI(U(g)a, b)llt/I(g)1 dg~ 11~llllbllllt/llll'

G

12

I. Theta Functions from an Analytic Viewpoint

According to a well-known elementary theorem, there exists one and only one element a' of H satisfying T(b)=(b, a') for every b. If we put a' = V(",) a, we see that V(",) is a C-linear mapping of H to itself. Furthermore, we have I V(",) all ~ 11"'111 Iiall for every a. This explains the definition of V ("'). The point is that the correspondence'" -+ V(",) gives a ring homomorphism of L( G) to the nng of bounded C-linear mappings of H to itself. In other words, we have

"'z

for every "'1> in L( G). The verification is straightforward. We shall apply these facts to G = A(X), H = J3 (X), and to the unitary representation V of A(X) in LZ(X) defined in Theorem 3. Let du, du* denote dual measures on X, X* and put dw=du du* for w=(u, u*). Let dt denote the Haar measure on Cr normalized by the condition that the total measure is 1. And finally put dg=dwdt for g=(w,t). This is a bi-invariant measure on A(X). If cjJ is an element of L(X x X*), we define cjJ# as cjJ#(w, t)=cjJ(w) t- 1 ; if'" is an element of L(A(X)), we define ",b as

"'b(w)= S ",(w, t) t dt. c; Then '"' and bare C-linear mappings of L(X x X*) to L(A(X)) and of L(A(X)) to L(X x X*) satisfying

(cjJ#)b = cjJ,

V(",b)#) = V(",).

Therefore, if we define V(cjJ) as V(cjJ#), the image under V of L(A(X)) is the same as that of L(X x X*). Furthermore, if we put

k(x,x+u)= ScjJ(u,u*)(x,u*)du*, x*

for every tP in L(X) we have

(V(cjJ) tP)(x) = S k(x,y)tP(y)dy. x

Also, by the Planche rei theorem for R n we get

S Ik(x, x+uW dx= S IcjJ(u, u*W du*.

x

x*

If we integrate both sides with respect to du and apply the Fubini theorem, we see that the correspondence cjJ -+ k gives a C-linear mapping of L(X x X*) to LZ(X x X) which preserves the LZ-norms with respect to dw and dx dy. Therefore it can be extended uniquely to a norm-preserving

§ 4. The Irreducibility of U

13

mapping, say T, of LZ(X x X*) to LZ(X x X). We shall show that Tis surjective, hence unitary. We take k from L(X x X) and put

J

l/J(u, u*)= k(x, x+u) (x, -u*) dx x

for every (u, u*) in X x X*. Then, similarly as above, we see that the correspondence k -+ l/J gives rise to a norm-preserving mapping, say T*, of LZ(X x X) to LZ(X x X*). If we take l/J and k from L(X x X*) and L(X x X), by the Fubini theorem we get (Tl/J, k)=(l/J, T* k). By continuity, this formula is valid for every l/J in LZ(X x X*) and k in ]} (X x X). Since T is norm-preserving, its image is closed. Hence, if T is not surjective, there exists at least one k =F 0 in LZ (X x X) which is orthogonal to the image of T. Then we get T* k = O. Since T* is norm-preserving, we get k = O. But this is a contradiction. We observe that if k is an arbitrary element of IJ(X x X), for every iP in LZ(X) (K iP)(x) = k(x, y) iP(y) dy x

J

is defined except for those x belonging to a set of measure 0, and KiP is in LZ (X). In fact, by the Fubini theorem the function k(x, ) on X belongs to LZ(X), hence (KiP)(x) = (k(x, ), ~ is defined, except for those x belonging to a set of measure 0; by the Schwarz inequality we get I(KiP)(x)l~ Ilk(x, )lllIiPll, hence IIKiPl1 ~ IlklllliPll. The C-linear mapping K of LZ(X) to itself is called the Hilbert-Schmidt operator on X with kernel k. If K 1 , K z are Hilbert-Schmidt operators with kernels kl> kz, the product Kl K z is the Hilbert-Schmidt operator with

k3 (x, y) =

Jkl (x, z) k z (z, y) dz

x

as its kernel. In fact, similarly as above, k 3(x, y)=(k1(x, ), k z ( ,y») is defined except for those x, y belonging to sets of measure 0, and I k311 ~ Ilkll1llkzll; hence by the Fubini theorem we get

J J = Jk 3(x, y)iP(y) dy.

(Kl K z iP)(x) = k 1(x, z)( kz(z, y) iP(y) dy) dz x x x

The set of Hilbert-Schmidt operators on X forms an associative ring and a Hilbert space with (Kl' K z )=(k1 , k z ) as its scalar product. We have thus obtained the following theorem:

Theorem 4. Let U denote the unitary representation of A(X) in IJ (X) defined as (U(w, t) iP)(x) = t(x, u*) iP(x+u)

14

I. Theta Functions from an Analytic Viewpoint

for every (w, t)=((u, u*), t) in A(X) and

(
§ 5. Induced Representations In order to have a better perspective, we shall explain the concept of induced representations and Mackey's theorem. Let G denote a locally compact group, K a closed subgroup, and X a character of K, i.e., a continuous homomorphism of K to C;. Suppose that the homogeneous space K\G has an invariant measure. We shall denote the (right invariant) Haar measures on G, K by dg, dk and an invariant measure on K\G by dg. We can normalize one of these measures so that we have

St/I(g)dg= S (Jt/I(kg)dk)dg G

KIG

K

for every t/I in L(G). Let He denote the vector space over C of those continuous functions t/I on G satisfying

t/I(k g}= X(k) t/I(g)

§ 5. Induced Representations

15

for every k in K and g in G and with the property that the continuous functions on K\ G well dermed by g= Kg-I 1/1 (g) 1have compact supports. Then He forms a pre-Hilbert space with (1/11,1/12)=

J

1/11(g)I/I2(g)dg

K\G

as its scalar product. Let H denote its completion. And finally, for every s, gin G and 1/1 in He' put (Ux(s) I/I)(g) = I/I(g s).

Then Ux(s)I/I is in He and Ux extends to a unitary representation of G in H, also denoted by Ux' This is called the representation of G induced by the character X of K.

As our first example, we take G = A (X), K = X* x C; embedded in G as (x*, t)-(O, x*, t), and X(x*, t)=t. In this case, we can identify X with K\G by x-g=K(x,O, 1). Furthermore, if dX,dx*,dt are Haar measures on X, X*, C;, we can take dg=dx dx* dt, dk=dx* dt, dg=dx. Let
1/1 (x, x*, t)=t
Then we get a norm-preserving bijection T of L(X) to He satisfying (T-lI/I)(x)=I/I(x, 0, 1). Moreover, if
((T- 1 Uis) T)
Therefore, passing to the completions L2 (X) and H of L(X) and He and to the corresponding extension of T, we see that the unitary representation U in Theorem 3 is unitary equivalent to the representation of A(X) induced by the character X(x*, t)=t of X* xC;.

In the second example, we need a certain lattice in X x X*. Let A denote a lattice in X, i.e., a discrete subgroup of X with compact quotient. Let A* denote the "annihilator" of A in X*, i.e., the set of elements x* of X* satisfying (~, x*) = 1 for every ~ in A. Then A* forms a lattice in X* and A is the annihilator of A* in X. Let A' denote a lattice in X* contained in A*; then A' is a subgroup of A* with finite index. Put W=X x X* and L=A x A'. Then L is a lattice in W. We take G=A(X), K=Lx C;, and X(w, t)=t. If we put L*=(A')* x A*,

the centralizer of K in G is L* x Cr. We observe that A*/A' can be identified with the dual of (A')*/ A by the restriction of (x, x*) to L*. In

16

1. Theta Functions from an Analytic Viewpoint

particular, we have

We take the Haar measure dx on X/A for which the total measure of X/A is 1; we normalize a Haar measure dx on X so that we have

J

JcP(x)dx=

x

X/A

(IcP(x+e»)dx ~eA

for every cP in L(X). In the same way, we normalize Haar measures dx*, dx* on X*/A', X*. We put dw=dxdx* for w=(x,x*) and dw=dxdx*. Then, for any Haar measure dt on Cl', we can take dg=dw dt, dg=dw, and as dk the Haar measure on K normalized by the condition that its restriction to Ci is dt. Let L(W, L) denote the vector space over C of continuous functions () on W satisfying (}(w+m)=(e, -x*) (}(w) for every w=(x, x*) in Wand m=(e, e*) in L. Then L(W, L) forms a pre-Hilbert space with «(}h (}2)=

J (}1(W) (}2(W) dw

W/L

as its scalar product; let H(W,L) denote its completion. Let (), '" denote functions on W, A(X); consider the correspondence () -+ t/I defined by t/I(w, t)=t(}(w).

Then we get a norm-preserving bijection

11

of L(W, L) to He satisfying

(11- 1 t/I)(w)=t/I(w, 1). We shall denote the unique extension of 11 to a

unitary mapping of H(W, L) to H also by T1• We observe that, for every () in L(W, L), we have

((11- 1 Ux(u, u*, t) 11) (})(x, x*)= (x, u*) t (}(x+u, x* +u*). We shall decompose Ux into a sum of irreducible representations. First we shall settle a special case:

Lemma 4. In the case where A' = A*, Ux is unitary equivalent to U; hence Ux is irreducible. Proof Let cP denote an element of L(X) and put (T2 cP)(w) =

I

(u (w, 1) cP)(~)

~EA

= ~(e,x*)cP(x+e) ~EA

17

§ 5. Induced Representations

for w= (x, x*). Then T2 II> is in L(W, L) and T2 is norm-preserving. On the other hand, let 0 denote an element of L( W, L) and put (T2*O)(x)=

J O(x,x*)dx*.

X·IA'

We shall show that T2* 0 is in If{X) and Tl norm-preserving. At any rate, TlO is a measurable function on X. (In fact, it is contained in LOO(X).) For a fixed x, O(x, ) is in L(X*/A'). Moreover (Tl O)(x+~)=

J

O(x, x*)<~, -x*) dx*

X·IA'

is the coefficient of <~, x*) in the Fourier expansion of O(x, ). Since A' = A*, we have A = (A')*; hence the orthonormal system «~, x*) )~eA in If (X* / A') is complete. Therefore we have

J

X.IA'

hence

IO(x, x*W dx* =

L I(T2* O)(x + ~W;

~eA

JI(Tl O)(xW dx= J (L I(T2* O)(x+~W) dx XIA = J IO(x, x*W dx dx*.

X

~eA

WIL

This shows that T2* is norm-preserving. If II> and 0 are as above, we have (T2 11>, 0)=(4), T2* 0). If we extend T2 and Tl to norm-preserving mappings of i}(X) to H(W, L) and H(W, L) to L2(X), by continuity, this formula is valid for every II> in L2(X) and 0 in H(W,L). Then, as in §4, we conclude that T2 is unitary and T2- 1 = T2*. Furthermore, for every 0 in T2(L(X)) we have ((T2 U(u, u*, t) T2- 1) O)(x, x*)=
11 T2 gives a unitary equivalence of Ux and

U.

q.e.d.

Going back to the general case, we recall that L x C; and L* x C; commute elementwise. Therefore, for every m = (~, ~*) in L*, the mapping ifJ(w, t)-+ ifJ(m, l)-l(W, t)) keeps He invariant. If we pass to L(W, L) by 11, we get a mapping Veo of the form (Veo O)(w) =

<-~,x*-~*) O(w-m)

for w=(x, x*). We observe that Vc.o is norm-preserving, depends only on m+L; and Therefore, if we convert A (A')*/A) = (L*/L) x C; into a locally compact group in the same way as A(X) by using (~, ~*) -+ -~, ~*), the correspondence (m+L,t)-+tVeo gives rise to its unitary representation in H(W,L).

<

18

I. Theta Functions from an Analytic Viewpoint

We recall a simplest spectral theorem: Suppose that H is a Hilbert space, Y a finite commutative group, V a unitary representation of Y in H; for every y. in Y·, define Hy* as the set of those ifJ in H satisfying V(y) ifJ = (y, y.) ifJ

for every y in Y. Then Hy* forms a closed Y-invariant subspace of H; Hyt , HYi for yr =1= y! are orthogonal; and H= Ef) H y •• y'eY'

In fact, if yr =1= y!, there exists an element y of Y satisfying (y, yt) =1= (y, YD. Let ifJl' ifJ2 denote elements of Hyt, Hy !; then

(ifJt> ifJ2) = (V(y) ifJt> V(y) ifJ2) = (y, yr - yD(ifJl' ifJ2), hence (ifJt> ifJ2)=O. Moreover, for every ifJ in H, p,.(ifJ) = l/card(Y)· L (y, - y.) V(y) ifJ yeY

is in H y .; and, by the" orthogonality of characters," we have

ifJ= L

y*eY'

p,.(ifJ)·

This proves the assertion. Let A denote a character of A./A', i.e., a character of A. taking the value 1 on A'; let A"" denote its extension to a character of W = X x X· trivial on X. Consider the set HJ. of those 0 in H(W, L) satisfying VCI) O=A(~·)O

for every w=(O, ~.) in L •. Then HJ. forms a closed A(X)-invariant subspace of H(W, L); HJ., HJ.' for A=I=A.' are orthogonal; and H(W, L)= Ef)HJ.. J.

We observe that HJ. for A=O coincides with H(W, A x A.) as vector spaces and their norms differ only by a constant factor. Therefore the subrepresentation in HJ. for A=O is irreducible by Lemma 4. We shall show that the subrepresentations in HJ.' HJ.' are unitary equivalent. We may assume that A.' = O. Then the multiplication by A"" maps HJ. to Ho , and it gives such a 'unitary equivalence. We shall state some of our results as follows:

Proposition 1. Let L = A X A' denote a lattice in W = X x X* such that for every (~, ~.) in L; consider the representation Ux of A(X) induced by the character X of K=LxCl defined by X(w,t)=t. Then Ux (~, ~.)= 1

§ 5. Induced Representations

19

decomposes, up to unitary equivalence, into a finite sum of the unitary representation U of A(X) in L2(X); its multiplicity is equal to the square root of the index of K in the centralizer of K. In the course of the above proof, we have encountered a group which is quite similar to A(X). We observe that A(X) can be defined for any locally compact commutative group X. Also, the unitary representation U can be defmed in general by the same prescription. According to Mackey's theorem, any unitary representation of A(X) under which every element t of q is mapped to the scalar multiplication by t decomposes into a sum of irreducible representations all unitary equivalent to U. Except for the finiteness, Proposition 1 follows immediately from this theorem. In order to obtain some insight to Mackey's theorem, we shall discuss the simplest case where X is finite: Proposition 2. Let X denote a finite commutative group and U * a unitary representation of the compact group A(X)=X x X* x q in a finite dimensional Hilbert space H such that U* (0,0, t) is the scalar multiplication by t for every t in Ct. Then U* is unitary equivalent to a finite sum of U. In particular, U is irreducible and the multiplicity of U in U* is equal to dim(H)/card(X).

Proof. We can consider X as a subgroup of A(X) by x --+ (x, 0, 1). For every x* in X*, consider the set H,.. of those ljJ in H satisfying U* (x) ljJ =
Furthermore U*(u, u*, t) gives a unitary mapping of Hx' to Hx*+U*' In particular, we have dim (H) = card (X*) dim(Ho). Choose an element ljJ of Ho satisfying IIljJll = 1 and put

ljJ,..= U*(O, x*, l)ljJ for every x* in X*. Then ljJ,.. is in H,.. and

U*(u, u*, t) ljJx.=t
H' = $ CljJx' x*eX*

is an A(X)-invariant subspace of H. Denote by T the C-linear mapping of L2(X) to H' defined by Tx*=T< ,x*)=ljJx' for every x* in X*.

20

I. Theta Functions from an Analytic Viewpoint

If the Haar measure on X is normalized by the condition that the total measure is 1, T is unitary. Moreover, if U' denotes the subrepresentation of U* in H', we have T- 1 U'(u, u*, t)T= U(u, u*, t). Therefore. U' and U are unitary equivalent. Let H" denote the orthogonal complement of H' in H and apply the same argument to H" instead of H. After a finite number of steps, we will get the desired decomposition of U*. q.e.d.

§ 6. The Group Sp(X) In § 1 we have defined Aut (G) as a topological group for any locally compact group G. In the special case where G is a finite dimensional vector space, say Y, over R, Aut(Y) is usually denoted by GL(Y). If the dimension of Y is m, GL(Y) is bicontinuously isomorphic to GLm(R)= GL(m, R). We go back to X, X* as before. Consider the following "multiplicative alternating form" on X x X*:

(x, x*), (y, y*») ~ (x, y*) ( - y, x*). Then the set of those (T in GL(X x X*) which keep this alternating form invariant forms a closed subgroup of GL(X x X*). We call this the symplectic group of X and denote it by Sp(X). We shall denote general elements of X, X* by x, x* and for every (T in GL(X x X*) we put

(x, x*) (T=(x a + x* y, x {3 + x* <5). Then a, {3, y, <5 are continuous homomorphisms of X, X, X*, X* to X, X*, X, X*. For the sake of simplicity, we say that (T is "composed of" a, {3, y, <5; we write

We shall denote the identity element of any group by "1". Then (T is contained in Sp(X) if and only if ( a {3) (

Y <5

<5* - {3*) = 1, -y* a*

i.e., if and only if a{3*={3rx*, y<5*=<5y*, a<5*-{3y*=1. Since gg'=1 is equivalent to g' g = 1 in any group, the above condition can be written as a* y=y* a, {3* <5=<5* {3, a* <5-y* {3= 1. If we map X, X* to Rn with respect to dual bases and identify x, x* with their images, we will have (x, x*) =e(x tx*). Therefore Sp(X) consists of those (T in GL 2n (R) satisfying

§ 6. The Group Sp(X)

21

Hence Sp(X) becomes the usual real symplectic group SP2n(R) defined in § 1. We go back to the intrinsic notation and denote by P(X) the closed subgroup of Sp (X) defined by y = o. Also we shall denote by Q (X) the open subset of Sp(X) defined by the condition that l' is an isomorphism. Although Q(X) is not a subgroup, we have Q(X)-l=Q(X) and P(X)Q(X)=Q(X)P(X)=Q(X). The following two lemmas will be used later: Lemma 5. For any 0"0 in Q(X), we have Q(X)=P(X) 0"0 P(X). Proof. Clearly, it is enough to show that for some 0"0 in Q(X) we have Q(X)=P(X) 0"0 P(X). We take a bicontinuous isomorphism Yo of X* to X and put

_(0Yo

0"0 -

_(y~)-l)

0

.

Then, for any given 0" in Q(X), we try to solve the equation 0"=0"10"00"2 in the "unknown" 0"1>0"2 in Sp(X) of the following forms:

If 0" is composed of a,

p, y, b, the equation 0"=0"10"00"2 can be written as

( a f3\ = (Pl Yo y b J b1 Yo

Pl Yo P2 -al(y~)-l) b1 Yo P2 .

We observe that a=Pl Yo, y=b1Yo, b=b 1Yo P2 can be solved uniquely in Pl, b1, P2. The condition that 0"1' 0"2 are in Sp(X) can be expressed as a 1 bt = 1, al Pt = Pl at, P! = P2. In particular, we have al =(bt}-l, and hence 0"1' 0"2 are already determined. Therefore we have only to show that the equations P= Pl Yo P2 -al(y~)-l,

al f3t = Pl at,

P! = P2

are automatically satisfied. If we express Pl, P2, al in terms of a, p, y, b and Yo, the right side of the first equation becomes ay-l b _(y*)-l. If we multiply 1'* from the right, the equation becomes P y* = a y -1 b y* -1. But this is a consequence of y b* = b y* and a b* - P y* = 1. The other two can be verified in a similar way. q.e.d. Lemma 6. We have Sp(X)=Q(X)2. Proof. Let 0" denote an arbitrary element of Sp(X) composed of a, p, y, b. The problem is to show that 0" is contained in Q(X)2. Since we have P(X)Q(X)P(X)=Q(X), we can replace 0" by any element of P(X) 0" P(X). We shall use SP2n(R) instead of Sp(X). Let 0"1,0"2 denote elements of SP2n(R) with" P" = "y" =0. Then1:he "y" in 0"10"0"2 is b1 y a2 ,

22

1. Theta Functions from an Analytic Viewpoint

in which c5 1 is the" c5" for suitably, we get

(J 1

and

a2

is the" a" for

(J 2'

If we choose (J 1, (J 2

in which y' is an element of GL(n', R) for some n' ~ n. By the previous remark, we may assume that y itself has this form. Then the condition ta y = ty a implies that a is of the form (

a'

0)

* (x: '

in which a' is of degree n'. Furthermore we haveta' y' =ty' a' and det(a") =1=0. On the other hand, for any real symmetric matrix p of degree n (J3=

(In

0)

-p In

is in SP2n(R), and the "y" in (J3"1(J is pa+y. Therefore we have only to find a p for which det(p)=I=O and det(pa+y)=I=O, or det(a+p-l y)=l=O. We may assume that n'~ 1. Choose a real number 11.=1=0 satisfying det(A y' -a')=I=O and put

p=

(Al ,-a'(y')-l n

o

0 I n _ n,

)-1.

Then p is symmetric and det(p)=I=O. Moreover we have det(a+p-l y)= det(Ay')det(a")=I=O. q.e.d. A non-empty subset C of a vector space over R is called a convex set if, for every Xl' X2 in C and AI, 11.2 ~ 0, )'1 + 11.2 = 1, Al Xl + 11.2 X2 is in C. The convex closure of a given (non-empty) set F is the smallest convex set which contains F; it is the intersection of all convex sets containing F. For the sake of completeness, we shall give a proof for the following elementary lemma: Lemma 7. If F is a compact subset of a finite dimensional vector space over R, the convex closure of F is a closed set.

Proof. Let E denote a vector space over R. By a simplex in E, we understand a "non-degenerate simplex." We observe that the convex closure of a finite subset J of E is the union of simplices with vertices in J. This can be proved, e.g., by an induction on card (J). Consequently, the convex closure of a set F is the union of all simplices with vertices in F. After recalling this, assume that E is finite dimensional and F compact. Let X denote a limit point of the convex closure of F. Then there exists a sequence (xp)p in the convex closure of F such that xp --+ X as p --+ 00. We have seen that each- xp is contained in a simplex with its

§ 6. The Group Sp(X)

23

vertices in F. Since the number of vertices is at most equal to dim (E) + 1, by passing to a subsequence we may assume that it is a fixed number, say n. We have n xp= L Api Ypi' i=l

in which Ypi are in F and Api ~ 0, ApI + ... + Apn = 1 for all p, i. Since F and [0,1] are compact, by passing to subsequences we may assume that Ypi ~ Yi' Api ~ Ai as p ~ 00 for all i. Then we have n

X=

LAiYi,

i= 1

in which Yi are in F and Ai~O, .,1.1 + ... +An= 1. This shows that x is in the convex closure of F. q.e.d.

Lemma 8. Let G denote a topological group and K a compact subgroup of G such that the quotient space G/K is homeomorphic to a convex set C in a finite dimensional vector space over R. Then every compact subgroup of G is conjugate by an inner automorphism of G to a subgroup of K. If K is also a Lie group, it is a maximal compact subgroupof G. Proof We let G act on C from the left via the homeomorphism between G/K and C. Let Xo denote the point of C which corresponds to K. Then K is the stabilizer subgroup of G at Xo' Let K' denote an arbitrary compact subgroup of G and dk' the Haar measure on K' normalized by the condition that the total measure is 1. For every open neighborhood V of the identity in K', choose a finite subset J of K' satisfying V J = K'. Then, for a suitable choice of J, we have

J(k' . x o) dk' = lim (card (J))-l . L k'· Xo'

K'

V

k'eJ

If we denote this point (of the ambient space of C) by Xl' therefore, it is a limit point of the convex closure of the compact set K' . Xo. By Lemma 7 the convex closure is closed. Since K' . X o is a subset of the convex set C, Xl is contained in C. Moreover we have k'· Xl =Xl for every k' in K'. Since G acts transitively on C, there exists an element g of G satisfying g. XO=Xl' Then g-l K' g is contained in the stabilizer subgroup of G at Xo, which is K. Let K' denote a compact subgroup of G which contains K. By what we have shown, there exists an element g of G satisfying g-l K' gcK. We also have g-1 Kgcg- l K' g, hence g-1 KgcK. Let KO denote the connected component of the identity in K. Then we have g-1 KO gcKo. If K is a Lie group, this implies g-1 KO g=Ko. Also, if K is a Lie group, the quotient group K/Ko is finite. Therefore we have g-l Kg=K; hence K'cgKg- 1 =K, hence K'=K. q.e.d.

24

I. Theta Functions from an Analytic Viewpoint

We shall apply Lemma 8 to determine maximal compact subgroups of G=SP2n(R). We take K=SP2n(R)n0 2n (R) and as C the Siegel upperhalf space of degree n denoted by 6 n • We have to show that G/K is homeomorphic to C. We recall that 6 n is an open subset of the vector space over C, hence over R, of symmetric elements of Mn(C); a symmetric matrix 't in Mn(C) belongs to 6 n if and only if Im('t) is positive-definite. We observe that 6 n is a convex cone in its ambient space. (We have encountered 6 n already in Theorem 1.) If (1 is an element of SP2n(R) composed of ex, p, 1', lJ and 't a point of 6 n , then det(y't+lJ)=I=O and (1.

't = (ex 't + P)(y 't +lJ)-1

is a point of 6 n • The classical proof is as follows: Let E denote the standard alternating matrix of degree 2n; E is the element of SP2n(R) composed of 0, 1n, -In' o. If WI> W2 are elements of Mn(C),

(1/20

t( ) (-) :~ E :~ = (1/2iWwl W2 -tW2 WI)

is a hermitian matrix of degree n; call it h. Then the first observation is that if h is positive-definite, then det(w2)=I=O. Otherwise there exists an element v=l=O of M n,I(C) satisfying W2 v=O; by multiplying tv and v from the left and the right, we get tv h v= 0; but this is a contradiction. Secondly, ifw is an element of Mn(C), we have

t(~) E (~) =tw-w, t(~) E (~) =tw-w. Thirdly, if (1 is an element of SP2n(R), we have t(1E(1=E. In fact, by definition we have (1 E t(1=E; take the inverses of both sides and multiply t(1 and (1 from the left and the right; finally replace E- l by - E. Therefore, if (1 is in SP2n(R) and 't in 6 n , we have Im('t)=(1/2i) tGJ E

GJ

=(1/2i) tGJ t(1 E(1

GJ P) .

= (1/2 i) t(ex 't + f3\ E (ex 't + Y't+lJI Y't+t5

Since Im('t) is positive-definite, we have det(y't+t5)=I=O. Put 't'= (ex 't+ P)(y-r+lJ)-I. Then, by an argument similar to the one above, we get

P)

t(ex 't + f3\ (ex 't + t't -'t = l' 't + lJ lEy 't + d =t(y 't +lJW't' - 't')(y 't +lJ).

§6. The Group Sp(X)

25

Since r is symmetric, r' is also symmetric. Once we know this, by the above calculation we get Im(r)=t(y r +15) Im(r/)(y r +b). Since Im(r) is positive-definite, Im(r/) is also positive-definite. Therefore r' is a point of 6 n • We observe that 6 n has a complex structure as an open subset of the vector space over C of symmetric elements of Mn(C)' Moreover, for every a in SP2n(R) the mapping of 6 n to itself defined by r -+ a· r is holomorphic. If at> a2 are elements of SP2n(R), we have a 1 · (a2 . r)= a1 a2' rand 12n · r=r. Therefore SP2n(R) acts biholomorphically on 6/1' Furthermore i In is a point of 6 n and the stabilizer subgroup of SP2n(R) at i In consists of all elements of M2n(R) of the form

in which y, b are elements of Mn(R) satisfying y tb=b ty and y ty+b tb = In. An element of SP2n(R) has this form if and only if it commutes with E. Therefore the stabilizer subgroup of SP2n(R) at i In is K = SP2n(R)n02n(R).lfy,b are elements of Mn(R), we have ytb=bty and y ty + b tb = In if and only if y i + 15 is contained in Un, the unitary group of degree n. Since the correspondence

(~ ~y) -+ Y i+b is compatible with multiplications, we get a bicontinuous isomorphism of K to Un. We shall show that SP2n(R) acts transitively on 6 n • Let r denote an arbitrary point of 6 n and put x=Re(r), y=Im(r). Since y is positivedefinite, there exists an element 15 of GLn(R) satisfying tb y b = In. Put a

= (tb- 1 Xb) 0 b'

Then a is in SP2n (R) and a . i In = r. This proves the assertion. We have thus obtained the following theorem: Theorem 5. Let 6 n denote the Siegel upper-half space of degree n. Then SP2n(R) acts transitively and biholomorphically on 6 n as r-+a'r=(ar+{))(yr+b)-1 with K=SP2n(R)n0 2n (R) as the stabilizer subgroup at i1 n • The correspondence a -+ y i + b gives rise to a bicontinuous isomorphism of K to Un; K is a maximal compact subgroup of SP2n(R); all maximal compact subgroups of SP2n(R) are conjugate by inner automorphisms of SP2n(R).

26

I. Theta Functions from an Analytic Viewpoint

We shall show that SP2n(R) is connected: If X, Yare topological spaces and f a surjective open mapping of X to Y such that Y and all fibers f-l(y) are connected, X is connected. Since we have U1 =C~, U1 is connected. Suppose that n~2 and assume that Un _ 1 is connected. We embed Un_ 1 in Un as

and denote the image group also by Un_i. Then the quotient space UjUn _ 1 can be identified with the space of n-th column vectors of elements of Un, which is a (2n-1)-sphere. Hence, by what we have recalled above, Un is connected. Since K=SP2n(R)n0 2n (R) is homeomorphic to Un, it is connected. Since the quotient space SP2n(R)/K is homeomorphic to 6 n and since 6 n is connected as a convex set, by the same reason as above, SP2n(R) is connected. We shall show that SP2n(R) is contained in SL2n(R). Since we have a Eta=E for every a in SP2n(R), we get det(a? = 1. The connectedness of SP2n(R) implies, therefore, that det(a) = 1. § 7. The Group B(X) We take the normalizer of the group A (X) in Aut(L2 (X») and denote it by B(X). If s is an arbitrary element of B(X), we get a bicontinuous automorphism (w, t) -+ (w', t') of A(X) as S-l

U(w, t) s= U(w', t').

The element of Aut (A (X») so obtained keeps Cr elementwise invariant. Let B(X) denote the subgroup of Aut(A(X») consisting of all elements with this property. Every element s of B (X) is determined by its effect on (X x X*) x 1. Suppose that w is a general element of X x X* and put (w, 1)s=(wa,f(w»). Then a is an element of GL(XxX*) and f a continuous mapping of XxX* to Ct. We shall write s=(a,f). If (a,f) and (a', 1') are elements of B(X), we have (a,f) (a',f') = (a a',f") ,

in which f"(w) = f(w)f'(w a). Therefore we have (1, 1)= 1 and (a,f)-l = (a-t, 1'), in which I' (w)= f(w a-i)-i. Suppose that s=(a,f) is an arbitrary element of B(X). Since we have (Wi' t1 ) (W2' t 2)= (W2' t 2) (Wi' t 1 ),

by applying s to both sides we see that
§ 7. The Group B(X)

27

that u is composed of 0(, p, y, ~, i.e., put

u=(~ ~. Then, by applying s to (WI' 1) (W2' 1)=(w1+W2' (u 1, u!», we get (u 1, u!> f(w 1+W2)=(U1 O(+ut y, u 2 P+u!~) f(w 1)f(w 2).

We observe that if we put fa(w) = (uO(, 2- 1 U P) (uP, u* y) (u* y, 2- 1 u* ~),

we get one such f The verification is straightforward. Then f and fa differ by an element of (X x X*)* =X* x X. In other words, there exist m, m* in X, X* satisfying f(w) = fa(w) (u, m*) ( -m, u*)

for every w = (u, u*) in X x X*. The sign-change of m is made because of (m, m*), 1)-1(W, 1)(m, m*), 1)=(w, (u, m*) (-m, u*».

Conversely, suppose that u is an arbitrary element of Sp(X) and f a continuous mapping of X x X* to Cf defmed as above in terms of fa and (m,m*) in XxX*. Then (w.t)s=(wu,f(w)t) defines an element s of B(X). Consequently the correspondence s -+ u gives a surjective homomorphism, which is obviously continuous, of B(X) to Sp(X) such that the kernel is bicontinuously isomorphic to X x X*. The fact is that the correspondence u -+ (u, fa) defines a continuous homomorphism of Sp(X) to B(X). This needs verification, but again it is straightforward. Therefore B(X) is bicontinuously isomorphic to a semidirect product of X x X* by Sp(X). We have spent enough time already for the group B(X). We shall discuss the relation between the three basic groups A(X), B(X), and Sp(X). We recall that every s in B(X) determines an element s=(u,f) of B(X) as

S-l

U(w, t) s= U(w u,f(w) t).

If we put n(s)=u, we get a homomorphism n ofB(X) to Sp(X). In general, if a locally compact group A is contained in a topological group B as a normal subgroup, by associating to each element of B the restriction to A of the corresponding inner automorphism, we get a continuous homomorphism of B to Aut (A). Therefore the homomorphism B(X) -+ B(X) defmed by s -+ s is continuous; hence the homomorphism n is continuous. We shall show that A(X) is the kernel of n. It is clear that A(X) is contained in the kernel of n. Let s denote an element of the kernel of n.

28

I. Theta Functions from an Analytic Viewpoint

Then the previous argument shows that s differs from U(m, m*), 1) by an element of the centralizer of A (X) in B(X). We have seen in § 4 that the centralizer of A (X) in Aut(L2(X)) is Ct. Therefore s is contained in A(X). We have shown that 11: gives rise to a continuous injective homomorphism B (X)/A (X) - Sp (X). The fact is that this is surjective and bicontinuous. We observe that 11: is surjective if and only if the homomorphism B(X) - B(X) is surjective. The surjectivity of B(X) - B(X) is an immediate consequence of Mackey's theorem that we have explained in § 5. However, since we did not prove this theorem, we shall proceed differently. It will be enough to show that there exists an open subset of Sp (X) which generates Sp (X) and on which the inverse mapping is defIned and continuous. Since the open subset Q(X) of Sp(X) generates Sp(X) by Lemma 6, it is a candidate for such an open subset. We shall fIrst construct a cross-section for 11: over P(X). We recall that P(X) is the closed subgroup ofSp(X) defIned by y=O. Let u denote an element of Sp(X) composed of oc, /3, 0, (j. Then for any function ([> on X we put (p(u) ([»(x)=loclt (xoc, 2- 1 x/3) ([>(xoc). We see that p(u) maps L2(X) to itself and is norm-preserving. Furthermore we have (i) p(u) p(u')=p(u u'), p(1)= 1; (ii)

U(w, t) p(u)=p(u) U(w u, fa(w) t)

for every u, u' in P(X) and (w, t) in A(X). Therefore p is a homomorphism of P(X) to B(X) satisfying 1I:(p(u))=u for every u in P(X). We shall show that p is continuous. We have only to show that p(u) ([> depends continuously on u at u = 1 for every ([> in any fIxed dense subset of L2 (X). We use L(X) as this subset and take any ([>*0 from L(X). If V is a compact neighborhood of the identity in P(X), for every u in V composed of oc, /3, 0, (j the support of ([>(x oc) as a function of x is contained in a compact subset, say C, of X depending only on ([> and V. Moreover ([>(xoc) as a function of (x, u) is uniformly continuous on the product X x V. Therefore, for any positive real number 8, if we take V sufficiently small, the L2-norm of ([>(x oc)- ([>(x) as a function ofx becomes less than 8/3 for every u in V. On the other hand, (x oc, 2- 1 x /3) as a function of (x, u) is uniformly continuous on the product C x V. Therefore, if we pass to a smaller V, the uniform norm of (x oc, 2 -1 x /3) -1 as a function of x in C becomes less than 8/311([>11 for every u in V. We may also assume that 11-locl- t l <8/311([>11 for every u in V. Then, by writing (p(u) ([>)(x)-([>(x) as the sum of three successive differences of

§7. The Group B(X)

29

and by evaluating their L2-norms for 0' in V, we get IIp(O') cP-cPll <8/3 +8/3 +8/3 =8. This proves the continuity of p. We shall construct a cross-section for 11: over Q(X). This will settle the continuity problem. We take an element 0' of Q(X). We recall that 0' is an element ofSp(X) composed of IX, p, y, (j, in which y is an isomorphism of X* to X. We define Iyl with respect to dual measures dx, dx* on X, X*; we have ly*I=IYI. For every cP in L(X) we put

J

(r(O') cP){x) = Iylt cP(XIX+X* y)fa(x, x*) dx*. X*

If we have a special 0', say 0'0' for which 1X=(j=O and y=Yo' we get

(r(O'o) cP}(x) = IYol-t cP*( - X(y~)-l). In particular, we have Ilr(O'o) cPll = II cP* II = IIcPll for every cP in L(X); r(O'o) can be extended uniquely to an element of Aut(L2 (X»). We also have (iii)

(iv) (v)

P(O'l) r(O') = r(O'l 0'),

r(O') p(0'2)=r(0' 0'2);

U(w, t) r(O')=r(O') U(w O',ja(w) t);

(r(O') cP1, cP 2)=(cP1, r(O'-l) cP 2)

for every 0'1' 0'2 in P(X), 0' in Q(X), (w, t) in A(X), and cP1, cP 2 in L(X). The first two should be understood as identities of" operators" on L(X). The verifications are all straightforward. As an example, we shall give the details for the first identity in (iii): If 0'1 is composed of 1X 1,Pl'O,(jl'O'l0' is composed of 1X1 1X+Pl y, 1X1 P + Pl (j, (jl y, (jl (j. Therefore, for every cP in L(X) we have

J = l(jl ylt JcP(x 1X11X +(x Pl + x* (jl) y) fal (x, x*)

(r(O'l 0') cP) (x) = l(jl ylt cP(X(1X11X+ Pl y)+ x* (jl y) fal a(x, x*) dx* x*

X*

. fa (x 1X1' X Pl + x* (jl) dx*.

We have used the fact that "O'-+(O',ja)" gives a cross-section for the homomorphism B(X)-+ Sp(X). We have fal (x, X*)=(XIX 1, 2- 1x Pl); if we put y* =X Pl +x* (jl' we have dy* = l(jll dx* and l(jll = 11X11- 1. Therefore we get

J

(r(O'l 0') cP) (x) = 1IX1it (x 1X1' 2- 1 x Pl) Iylt cP(x 1(1) IX + y* y) x' . fa (x 1X1' y*) dy* =(P(O'l) r(O') cP) (x).

30

I. Theta Functions from an Analytic Viewpoint

By Lemma 5 every
Theorem 6. The normalizer B(X) of A (X) in Aut(L2 (X») is connected and locally compact. Moreover the homomorphism 1t of B(X) to Sp(X) defined as S-l U(w, t) s= U(W1t(s), *) is continuous, surjective, and open. Therefore the quotient group B (X)jA (X) is bicontinuously isomorphic to Sp(X).

Remark. In general, an extension of a Lie group by a Lie group is a Lie group. The proof becomes trivial if the extension has a local crosssection. Therefore B(X) is a Lie group. Moreover B(X) is unimodular in the sense that its Haar measure is bi-invariant. An outline of the proof is as follows: Let ds denote a Haar measure on B(X) and define L1 (s) as d(s-l)= L1 (s) ds. Then L1 gives a continuous homomorphism of B(X) to and the kernel of L1 is the largest unimodular normal subgroup of B (X). Since A(X) is a unimodular normal subgroup of B(X), we get L1 = 1 on A(X). except for Since there is no continuous homomorphism of Sp(X) to the trivial one, we get L1 = 1 on B(X). We shall proceed to the determination of all maximal compact subgroups of B(X). We shall first obtain preliminary information. Let K denote a compact subgroup of B(X). By replacing K by KCf, we shall assume that K contains Cf. We -see (either directly or using Lemma 8)

R:

R:

§8. Fock Representation

31

that et is the unique maximal compact subgroup of A(X); hence KnA(X)=et. Moreover 1t(K) is a compact subgroup ofSp(X). In view of Theorem 5, it is expected, therefore, that a maximal compact subgroup ofB(X) is an extension of et by Un' In order to prove this, the fact that the extension splits, and other important information, we shall introduce another realization of A (X).

§ 8. Fock Representation We shall denote a general point ofC" by z=(z1 ... zn)' Let x=(x1... x n), Y = (Y1 ... Yn) denote the real and the imaginary parts of z; put dx = dx1... dxn, dy=dY1 ... dYn, dz=dxdy, and fmally dJ.L(z)=exp( -1tz'z)dz. Then dJ.L(z) is a measure on C" for which the total measure of C" is 1. Let t/I denote a holomorphic function on en and t/I(z)=

Lap zP P

its Taylor expansion. The letter p stands for an n-tuple of non-negative integers (P1 ... pJ and zP for the monomial zit ... z:n. Since the power series is uniformly convergent on every compact subset of en, if we denote the maximum oflz11, ... , IZnl by Izloo' we have

J1t/I(zW dJ.L(z) = lim J

en

r_ 00 Izloo;lir

1t/I(z)1 2 dJ.L(z)

We observe that the integral of zPzq for p+q is O. On the other hand, for any non-negative integer m and positive real number a, we have

Jxm exp( -ax) dx=m!/am+1. 00

Therefore we get

o

J1t/I(zW dJ.L(z) = L la I2 enJzPzP exp( -1t z 'Z) dz p

en

P

=

L lapI2 1t- 1p1 p!. P

The absolute value Ipi stands for the sum P1 + ... + Pn and the factorial p! for the product P1 ! ... Pn!. Let Hn denote the set of holomorphic functions t/I on en for which the above integrals are convergent. Then Hn forms a pre-Hilbert space with (t/l1' t/l2)=

Jt/l1(Z) t/l2(Z) dJ.LJz)

en

32

I. Theta Functions from an Analytic Viewpoint

as its scalar product. Furthermore, if we put 4>p(Z)=(p!)-t(nt z)P,

we get an orthonormal system (4)p)p in Hn. We shall show that Hn is a Hilbert space and the orthonormal system (4)p)p complete: If 1/1 is an element of Hn , since 4>p(z) is zP up to a constant factor, we can write its Taylor expansion as I/I(z) =

L cp 4>p(z). p

Then we have cp=(1/1, 4>p) for every p and

111/111 =(L ICp 12 )t < 00. p

Conversely, suppose that (cp)p is a sequence of complex numbers such that the sum ofalllcpl2 is finite. Consider the formal power series 1/1 (z) defined as above. Then by the Schwarz inequality we get

L ICpll4>p(z)1 ;£(L ICp1 )t exp(n/2) z z). 2

p

t

p

This shows that the formal power series converges absolutely and uniformly on every bounded subset of en. Hence it represents a holomorphic function on en of which it is the Taylor expansion. This completes the proof. In the course of the above proof, we have shown that for every 1/1 in Hn we have 11/1 (z)l;£ exp(n/2) z tz) II 1/111. Consequently, the norm convergence in Hn implies the uniform convergence on every bounded subset of en. We observe that for any ( in C" the holomorphicfunction z --+ exp(nz t,) on en admits the following Taylor expansion: exp(n z t,)= L 4>p(z) 4>p('). p

We can regard this as the Fourier expansion of the function z --+ exp(n z t,) in Hn with respect to th~ orthonormal system (4)p)p. Consequently, if ~e denote by (1/1 (z), exp(n z t(») the scalar product ofl/l in Hn andz --+ exp(n z t(), we have (1/1 (z), exp(n z tf)) = L (1/1, 4>p) 4>p((). p

We can regard the right side as the Fourier expansion of 1/1, evaluated at (, with respect to the same orthonormal system. Since we have seen that the norm convergence in H~ implies the pointwise convergence on

§ 8. F ock Representation

33

en, this is equal to 1/1 ((). Therefore exp(n z It) is the "reproducing kernel" for Hn in the sense that I/I(z)= Jexp(nzlt)I/I(Odjl(O en

for every 1/1 in Hn. We shall consider the product

en x e[

and put

(V(e, t) 1/1) (z)= t· exp( -n(z+e/2) Ic+(n i/2) Im(e Ie))· I/I(z +e)

for every (e, t) in en x e[ and z in en. The function 1/1 on en can be arbitrary. If 1/1 is holomorphic on en, V(e, t) 1/1 is also holomorphic on en; if we take 1/1 from H n , we get II V(c, t) 1/111 = 111/111. We also have V(c 1 ,

t 1)

V(c 2 , t 2 ) = V(c 1 + c 2 , e( - 1m (c 1 ) IRe (c2))

for every (c 1 , t 1 ), (c 2 , t 2 ) in

en x e[, and

tl

t2)

V(O, 1)= 1. Therefore, if we put

V((u, u*), t)= V( -u i+u*, t),

we get a homomorphism V of A (Rn) to Aut(Hn) such that every t in e[ is mapped to the scalar multiplication by t. We shall show that there exists a unitary mapping I of IJ (Rn) to Hn satisfying IU((u, u*), t)= V((u, u*), t) I

for every ((u, u*), t) in A (Rn). Since the unitary representation U is irreducible, such an I is unique up to an element of er. We shall see that I is the unitary mapping which maps the complete orthonormal system of normalized Hermite functions on R n to the complete orthonormal system (cP p)p. We put k(x, z) = 2 n/4 exp( - n x IX) e(x IZ) exp((n/2) z IZ) for every x in R n and z in en. Then k( ,z) is an L 2 -function on Rn. In fact we have Jk(x, z) k(x, () dx=exp(n zIt) Rn

for every z, ( in en. We have seen that this is the reproducing kernel for Hn. On the other hand, although k(x, ) is a holomorphic function on en, it does not belong to Hn. Therefore we have to proceed cautiously. Lemma 9. For a fixed x in R n , we expand the holomorphic function k(x, ) on

en into

its Taylor series as k(x, z)=

L hp(x) cPp(z). p

Then (hp)p forms a complete orthonormal system in L2 (Rn). Moreover, for a fixed z in en, this is the Fourier expansion of k( ,z) with respect to this orthonormal system.

34

I. Theta Functions from an Analytic Viewpoint

Proof. Since k(x, z), tPp(z), and dx decompose into n-fold products of the similarly defmed functions and measures for n= 1, e.g., as k(x, z) =kn(x, z)=k l (Xl' Zl)'" kl (Xn' Zn)'

we have only to prove the lemma in the case where n= 1. Since we have we get

k(x, z)=2* exp(n x 2) exp( -2n(x-i Z/2)2) , hp(x)=2*/(2i)P(P !)t nP/ 2 . exp(nx2) dP exp( - 2n x 2)!dxP

for p =0, 1, .... In particular exp(n x 2) hp(x) is a polynomial in X of degree pwith 2* (2 i)P n P/ 2/(p !)t . x P as its highest term. Since, up to a constant factor, exp( -nx 2) hp(x) is the

p-th derivative of exp ( - 2n x 2 ), by integration by parts we get

Jhp(x)exp( -nx 2)x

Q

R

J

dx=const. exp( -2nx 2). dP xq/dx p · dx R

for every p, q ~ O. Since this is 0 for q = 0, 1, ... , p -1, we get (h p ' hq ) = 0 for q < p, hence for every q =t= p. In the case where q = p, by making the "const." explicit we get (h p, hp)= 1. Therefore (hp)p forms an orthonormal system in IJ (R). Furthermore we have

J

(k( ,z), hp)=const. exp(n z2/2+2n i z x)· dP exp( -2n x 2)!dxp · dx R

=const. exp(n z2/2) zp·

Jexp( - 2n x 2+ 2n i x z) dx

R

=const. zp.

Again, by making the "const." explicit we get (k( ,z), hp)=tPp(z). Therefore k( ,z)=

L tPp(z) hp p

is the Fourier expansion of k( ,z) with respect to the orthonormal system (hp)p in L2 (R). It remains to show that (hp)p is complete. Suppose that an element f/> of L2(R) is orthogonal to every hp. Then f/> is orthogonal to

k( ,z) for every zinC. According to Lemma 2, the set of k( ,z) for every

z in in

e spans a dense subspace of L2(R). Therefore we get f/>=O.

q.e.d.

We are ready to settle our problem: For every f/> in L2(Rn) and z

en, put

(I f/>)(z) =

Jk(x, z) f/>(x) dx.

Rn

Then by the Schwarz inequality we get

1(1 f/>)(z) I;;;; exp(n/2) z tz) II f/> II.

§8. Fock Representation

35

In particular, the above integral is uniformly convergent with respect to the parameter z on every bounded subset ofC". Hence [cJI is a holomorphic function on en. By Lemma 9 we have

(IcJI) (z)= L cPp(z} (h p,
We may regard this as the Taylor expansion of [cJI. Then, as we have seen in the beginning, we get

J1([cJI} (zW dJ.L(z) = L I(hp, 4'>W·

en

p

Since (hp}p forms a complete orthonormal system in IJ (Rn), the right side is IlcJl112. Therefore [is a norm-preserving mapping of L2(Rn) to Hno Since we have [h p= cP pfor every p and since (cPp}p forms a complete orthonormal system in Hn, [is surjective, hence unitary. The verification of [U(u, u*), t) = V( -ui+u*, t} [is straightforward. We have thus obtained the following theorem:

Theorem 7. Let Hn denote the set of holomorphic functions ifJ on C" for which II ifJll =( lifJ(z}1 2 exp( -n z I Z) dz)t

J

en

are finite. Then Hnforms a Hilbert space. Moreover, ifwe put (V(u, u*), t) ifJ) (z)= t· exp( -n(z +c/2} Ie +(n i/2) Im(c 'C})· ifJ(z + c} with c = - u i + u* for every (u, u*), t) in A (R") and ifJ in Hn, we get a unitary representation V of A (Rn) in Hn which is equivalent to the unitary representation U in L2 (Rn). The equivalence is given by the unitary mapping [ of L2(Rn} to Hn defined by (IcJI)(z)= Jk(x,z}cJI(x}dx, in which

Rn

k(x, z}=2n/4 exp( -n x IX} e(x IZ } exp(n/2} z IZ).

The unitary representation V in Theorem 7 is called the Fock representation of A (Rn). We shall use the corresponding realization of

A(Rn} to construct a maximal compact subgroup of B(Rn}: Let p denote an element of Un and put

(M(p) ifJ) (z}=ifJ(z p) for every ifJ in Hn and zinC". Then M(p} is a norm-preserving mapping of Hn to itself satisfying M(p} M(p'}=M(P p'} for every p, p' in Un, and M(ln}=1. Therefore M gives a homomorphism of Un to Aut(Hn}, which is clearly injective. We shall show that M is continuous. We have only to show that M(p} ifJ depends continuously on p for every ifJ in a fixed subset of Hn which spans a dense subspace of Hn. We observe that the [mite

36

I. Theta Functions from an Analytic Viewpoint

dimensional subspace of homogeneous polynomials in Zl' ... , Zn of any ftxed degree is M(p)-invariant for every p in Un. Therefore we have only to observe that the corresponding subrepresentation is continuous and the union of all such subspaces spans a dense subspace of Hn. On the other hand, for every p in Un and (c, t) in C" x C:, we have

M(P)-l V(c, t) M(P)= V(c p, t'), in which

t' = t· e(a-) Im(c(1n - p tp) tc)).

Therefore, if we put

M1(p)=J- 1 M(p)J,

MI(P) is an element of Aut(L2(Rn)) and it normalizes A(Rn). In this way, we get a continuous homomorphism MI of Un to B(Rn). Since M is injective and Un compact, MI gives a bicontinuous isomorphism of Un to the image group MI (Un). Furthermore, if we decompose p into its real and imaginary parts as p = -1 i + b and put (1

we have

=

(b1 -1)b'

MI (p)-l U(w, t) MI (p)= U(w (1, t')

for every (w, t) in A(Rn) with t' as above. Hence, in the notation of Theorem 6, we have n(M1 (p))=(1 for every p in Un. As we have seen in §6, elements of the form (1 form the maximal compact subgroup SP2n(R) n 02n(R) of SP2n(R). Therefore, by what we have said at the end of § 7, if we put

K =MI(U)CX o

n

l'

we get a maximal compact subgroup Ko ofB(Rn), which is bicontinuously isomorphic to the product Un X

C: .

§ 9. The Set ~(X) We shall determine maximal compact subgroups ofB(X) and, at the same time, characterize functions on R n of the form e(!x1: t x+c t x) in which (1:, c) are in 6 n x C". We shall start by proving the following lemma: Lemma 10. Consider the representation M' of Un in the ring of polynomials C[Z]=C[Zl' ... , zJ defined by (M'(P) p)(z)=P(z p)

for every p in Un and P in C [z]. Then the subrepresentation Md of M' in the vector space C [Z]d of homogeneous polynomials of degree d is irreducible for every d.

§9. The Set

37

~(X)

Proof Since C [Z]d is one dimensional, and hence Md irreducible, for n = 1 or d = 0, we shall assume that n;E; 2 and d;E; 1. We observe that, as a closed subspace of Hn , C [Z]d has the structure of a Hilbert space and Md is a unitary representation of Un in this Hilbert space. Therefore, as we have seen in § 4, we have only to show that any C-linear mapping T of C [Z]d to itself which commutes with every Md(P) is a scalar multiplication. If we take a diagonal matrix in Un as p, ™d(P) = Mip) T for every such p implies that T P = cp P with cp in C for every P of the form P(z)=zP=zf 1 ••• z:",

Ipl=Pl +"·+Pn=d.

We have only to show, therefore, that Cp=C(dO ... O) for every p. At any rate, we have pk=l:O, Pk+l ="'=Pn=O for some k;E;1. Since Cp=C(dO ... O) is trivially true for k = 1, we can apply an induction on k assuming that k;E; 2. Defme P' by the condition that p;=pJor i=l:k-1, k and P~-l =Pk-l +Pk' p~=O. We have only to show that cp=c p" We embed U2 in Un by letting U2 act in an obvious manner on the two dimensional subspace of cn defined by Z/ =0 for i =I: k -1, k and trivially on its orthogonal complement. Then we write down the condition that TMd(P) P' = Md(P) TP' for every P in U2 and P'(z)=zp'. This will give cp=c p" Since the verification is really in the two dimensional case, we shall give its detail assuming that n=2: If Pij is the (i,j)-coefficient of P for 1 ~i,j~2, the condition implies that

Since P11 P21 =1:0 in general, we get c(ab)=c(dO)'

q.e.d.

Lemma 11. Let u denote an element of SP2n(R) composed of a, p, ,)" () with det(')')=1:0 and (r, c) a point of6 n x en. Then,for we have

4>(x)=e(! x -r tx + c tx)

(r(u) 4>)(x) = Idet(')')I-t det(i- 1(-r +,),-1 {»)-t e( _! c(-r +,),-1 (»-1 tc)

in which

. e(! x -r' tx + c' tx), -r' =(a -r + P)(')' -r + (»-1,

c' =C(')' -r+ {»-1;

the sign of det(i- 1(-r+')'-1 {»)t is determined by the condition that det(i- 1(-r + ,),-1 (»)t tends to det (Iro (-r»)t >0 as Re(-r) ~ _,),-1 (). Proof We have defined r(u) 4> in § 7 for every 4> in L(X) as

J

(r(u) 4» (x) = 11'1"1 4> (x a+ x* ')') fa (x, x*) dx*

x·

and observed that r(u) can be extended uniquely to an element of Aut(L2 (X»). We also have (r(u) 4>1' 4>2)=(cP1,r(u- 1) 4>2) for every 4>1'

38

I. Theta Functions from an Analytic Viewpoint

<1>2 in L2(X). If we take <1>1 from Ll(X) n L2 (X) and <1>2 from L(X), r(a- 1 ) <1>2

admits the integral representation and the Fubini theorem can be applied to (<1>1' r(a- 1 ) <1>2). Therefore, as in the second part of the proof of Theorem 2, we see that the integral representation for r(a) <1> remains valid even when <1> is in Ll(X) n 13 (X). In particular, we can take as <1> the function specified in the lemma. Then, by using the facts that IX tfl, y-l b are symmetric and IXtb-pty=ln' we get (r(a) <1» (x)=lyl-t S<1> (y) fAx, (y-xlX)y-l)dy x = Idet(y)l- t e(! X (IX y-l) tx) Se(! y(-r +y-l b) ty +(c-x ty-lrY) dy. K"

By applying Theorem 1 to this integral, we see that it is equal to det(i-l(r + y-l b))-t e( -!(c - x ty-l) (r + y-l b)-l t(c -

X

ty-l )).

On the other hand, we have

If we take the transposes of both sides and multiply t(yr + b)-l from the left and ta- 1 from the right, we get t(YL+b)-l('r In)=(L'1n)

(_~ -~),

hence t(y L+b)-l = -L' y+lX. By using this, we can transform the expression for (r(a) <1» (x) into the form stated in the lemma. q.e.d. In the following theorem, we shall assume that case where dim (X) = 1 will be discussed later.

dim(X)=n~2;

the

Theorem 8. Let K denote a maximal compact subgroup ofB(X). Then there exists a uniquely determined one dimensional K-invariant subspace of13 (X). It is of the form C<1> with <1> (x) = e(q (x)); q is a quadratic polynomial on X satisfying q (0) = 0 such that its degree 2 component has a positivedefinite imaginary part. The point <1> is characterized among all points ofC<1> by the condition that <1>(0) = 1. Conversely,for every such <1> there exists a uniquely determined maximal compact subgroup K of B(X) which keeps the subspace C<1> invariant. The normal subgroup ofK which keeps <1> invariant is bicontinuously isomorphic to Un and K decomposes into the direct product of this subgroup and Ct. The group B (X) acts transitively on the set of all maximal compact subgroups of B(X) via inner automorphisms. This gives rise to a transitive and biholomorphic action ofB(X) on the complex manifold of all quadratic polynomials on X of the above-described nature.

§9. The Set ~(X)

39

Proof We shall identify X, X* with R n with respect to dual bases. We keep the notations in § 8, e.g., Ko =M1 (Un) C(. We shall determine all one dimensional Ko-invariant subspaces of 13 (Rn). If we pass to Hn by the unitary mapping I of LZ (Rn) to H n , they are mapped to one dimensional M(Un)-invariant subspaces of Hn. Let Ct/J denote one of such spaces. Then we have t/J(z p)= X(p) t/J(z)

with X(p) in C for every p in Un and z in en. Consequently, if t/J d denotes the homogeneous component oft/J of degree d, Ct/Jd is an Md(p)-invariant subspace of C [ZJd for every p. Since Md is irreducible by Lemma 10, this implies that t/Jd=O unless dime C[ZJd= 1. Since we have assumed that n ~ 2, dime C [z Jd = 1 if and only if d = O. Hence we get t/J = t/J (0). Therefore, if we put tPo(x)=exp( -nx 'x),

CtPo is the only one dimensional Ko-invariant subspace of LZ (Rn). Furthermore MI (Un) is the stabilizer subgroup of Ko at tP o . Let s denote an arbitrary element of B(Rn) and put S-1 U(w,t)s= U(w O",f(w) t) for every (w, t)=((u, u*), t) in A (Rn). We have seen in § 7 that f(w) can be written in the form f,,(w) e(u tm* - m tu*) for some m, m* in Rn. We shall show that if we take e(! x or tx+ C tx) with (or, c) in 6 n x en as tP(x), we have (s· tP)(x) = const. e(! x or' tx + cltx) , in which

or' = 0". or =(a or+ P)(y or+b)-1 c' =c(y or+b)-1 +mor' +m*.

First of all, since we have Sp (X) = Q (X)Z by Lemma 6, there exist 0"1' 0" Z in Q(Rn) satisfying n(s)=0"=0"1 O"z. Then sand r(0"1)r(O"Z) differ by an element of the kernel of n, which is A (Rn). Therefore we get s = U ((m o , m6), to) r(O" 1) r(O" z) for some ((m o , m6), to) in A (Rn). If we calculate S-1 U(w, t)s by using some of the formulas in § 7, we get

U(w O",f,,(w) t')

with

t' =e(u tm6 - mo tu*) t;

hence mo = m, m6 = m*. On the other hand, if O"j is composed of ()( j' Pj' Yj' 15 j for j = 1, 2, we have yor+b=(Y1(O"z . or)+b1) (yz or+b z)· Therefore, if we calculate (r(0"1)· (r(O"z)· tP)) (x) by using Lemma 11, we get const. e(t x (0" . or)'x+ c(y or + 15)-1 tx); by applying U((m, m*), to) to this, we get (s· tP) (x) = const. e(! x or' tx + c' tx) with (or', c') as above. We let B(Rn) act on 6 n x en by (*). Then (or', c') depends continuously on (s, (or, c)) and the correspondence (or, c) ~ (or" c') gives a biholomorphic

40

I. Theta Functions from an Analytic Viewpoint

mapping of 6 n x c n to itself. Moreover, since SP2n(R) acts transitively on 6 n as 't -+ 0' • 't, n is surjective, and R n 't + Rn = cn for every 't in 6 n, B(R") acts transitively on 6 n x en. We shall show that Ko is the stabilizer subgroup of B(Rn) at (i In' 0). We have seen that Ko is contained in the stabilizer subgroup of B(Rn) at (i 1n, 0). Suppose that s is an arbitrary element ofthis stabilizer subgroup. Then n(s) is contained in the stabilizer subgroup of SP2n(R) at i In' which is the image group of Ko under n. Hence, by multiplying a suitable element of Ko to s, we may assume for our purpose that n(s)= 1. Then we get s= U(m, m*), t) for some (m, m*), t) in A (Rn). Since we have mi+m*=O, we get m=m*=O. Therefore sis in C1. This proves the assertion. The rest follows from Lemma 8. q.e.d. In the case where n = 1, we have a similar but slightly different situation: Let K denote a maximal compact subgroup of B(X). Then the one dimensional K-invariant subspace of L2(X) is not unique; it is of the form C ~ qJ with qJ as in Theorem 8 and ~ a polynomial function of degree d on X uniquely determined up to a constant factor by K for d = 0, 1, .... This is the only change that we have to make. In other words, the decomposition of L2 (X) into irreducible K-invariant subspaces takes place by the same pattern. In the case where n = 1, K is commutative, and hence these subspaces all become one dimensional. We shall denote by ~(X) the set of functions qJ on X of the form e(q(x»), in which q is a quadratic polynomial on X such that its degree 2 component has a positive-definite imaginary part; ~ (X) may be called the set of "Gaussian densities". We observe that the correspondence q -+ qJ gives an open mapping of the space of such quadratic polynomials to the subset ~(X) of L2(X). We shall show that ~(X) is contained in the Schwartz space of X; the definition of the Schwartz space is as follows: Let P denote an arbitrary polynomial function on X and D a translation-invariant differential operator on X. Then a Schwartz function qJ on X is an infmitely differentiable function on X such that

liP· DqJll 00 = sup IP(x) (DqJ) (x)1 < 00 xeX

for every such P and D. Let [I'(X) denote the vector space over C of all Schwartz functions on X. If qJ is in [I'(X), p. DqJ is also in [I'(X) for every P and D. Let (~, Dj)j denote a rmite set of pairs of polynomial functions and translation-invariant differential operators on X. We take the set of all elements qJ of [I' (X) satisfying II ~ . Dj qJ 1100 < 1 for every j as a neighborhood of 0 in [I' (X). The set of all such neighborhoods can be taken as a base of open neighborhoods of 0 in [I' (X). In this way, [I' (X) becomes a locally convex topological vector space over C. This is the Schwartz space of X.

§9. The Set ~(X)

41

We shall show that ~(X) is a subset of 9'(X). More generally, for any given (P, D) and any compact set C of quadratic polynomials q on X such that the degree 2 components have positive-dermite imaginary parts, liP· D4>111X> for 4>(x)=e(q(x)) is bounded when q varies in C. The proof is as follows: We identify X with R n and put r(x)=(x l )2 + ... + (xn)2)t

for x = (Xl'" x n). We observe that P(x) (D4» (x) can be written in the form in which 1l (x, q) is a polynomial in Xl'"'' xn and in the coefficients of q. Since q varies in the compact set C, if d is the degree of ll(x, q) in Xl' ... , x n , we have

1l(x, q) 4>(x),

11l(x, q)1 ~const. max(l, r(xt)

for every x in R and q in C. On the other hand, if we write q(x)= !x't'tx+ctx+q(O), the smallest eigenvalue of Im('t') has a positive minimum and c, q (0) are bounded when q varies in C. Hence there exist A. > 0 and Jl ~ 0 satisfying n

14>(x)1 ~const. exp( -A.r(x)2 + w(x)) for every x in Rn and q in C. Therefore we have only to make the final observation that sup t1 exp( -A. t 2+Jl t)< 00. t~l

We recall that 9'(X) is a dense subspace of I!'(X) for every p. The proof is as follows: Let 4> denote an arbitrary Schwartz function on X. Then 4> is, at any rate, a continuous function on X. We identify X with Rn and use the function r(x) as before. Choose A. from R satisfying A.> nip and put In the case where p= 00, for every e in R~, the points x of X satisfying 14>(x)1 ~e form a closed subset of the solid sphere ofradius (M/e)lO,; hence it is compact. Therefore 4> is in LIX>(X). In the case where l~p 14>(x)iP dx~MP r(x)-).P dx~const. rn-l-).p dt< 00.

J

r(x)~l

J

r(x)~l

J

1

Since the integral of 14>(x)iP over the solid sphere of radius 1 is clearly rmite, we get 114>11 P < 00. Therefore 4> is in I!' (X). Since inrmitely differentiable functions on X with compact supports are all Schwartz functions on X, 9' (X) is dense in I!' (X) for every p. We have seen that every element s ofB(X) keeps ~(X) invariant. We shall show that s also keeps 9' (X) invariant. Since s can be written in the form U(w, t) r(ul ) r(u 2 ), we may assume that s is either U(w, t) or r(u). It is clear that U(w, t) keeps 9'(X) invariant. We recall that r(u) can be

42

I. Theta Functions from an Analytic Viewpoint

written in the formp(o-l) r(o-o) P(0-2) for any given 0-0 in Q(X). If an element 0- of Sp (X) is composed of a, p, 0, f>, we have (p(o-) tl» (x) = lal t (x a) for every tl> in L2(X). Therefore, if tl> is in 9"(X), p(o-) tl> is also in 9"(X). If 0- 0 is composed of 0, Po, Yo, 0, we have (r(o-o) tl>)(x)=IYol- t tl>*( _X(y~)-l) for every tl> in L2 (X). We take dual bases of X, X* and identify X, X* with Rn. Then we can take Yo = - In and we get r (0- 0 ) tl> = tl>* for every tl> in L2 (X). Let p* denote a polynomial function and D* a translationinvariant differential operator on X*. Then there exist a polynomial function P and a translation-invariant differential operator D on X satisfying (D(Ptl»)* =p* . D* tl>* for every tl> in 9" (X). This can be proved by a repeated application of (1/2n i) atl>* /axp= f(yP tl>(y») e(x ty) dy Rn

(-2n i x p) tl>* (x) =

J(atl>/aYp)e(x ty)dy

Rn

for 1 ;£p;£n valid for every tl> in 9"(X). Since we have IIP* . D* tl>* IL.,;£ IID(Ptl»111 < 00, tl>* is in 9"(X*). Therefore every s in B(X) keeps 9"(X) invariant.

§ 10. The Discrete Subgroup 1;. We shall consider the situation which has been discussed as the "second example" in § 5: We take a lattice A in X and denote by A* the annihilator of A in X*. For the sake of simplicity, we take A x A* itself as L. Then L is its own annihilator in W = X x X* with respect to (x, x*), (y, y*») -+
<- y, x*).

Hence LxI forms a subgroup of A (X); its normalizer in A(X) is Lx Ct. Let U(L) denote the image of LxI under the unitary representation U of A(X) in L2(X); take the normalizer, say N L , of U(L) in B(X). Let s denote an element ofB(X) and (o-,f) its image in B(X),i.e., puts- 1 U(w, t)s = U (w 0-, J(w) t) for every (w, t) in A(X). Then s is contained in N L if and only if Lo- = Land J = 1 on L. We observe that the set of elements of Sp (X) which keep L invariant forms a discrete subgroup of Sp (X). On the other hand, we know that there exist m, m* in X, X* satisfying

<

J(w) =J.,.(w) u, m*)

<- m, u*)

§ 10. The Discrete Subgroup Ii

43

for every w=(u, u*) in W. Suppose that (1 is composed of IX, p, y, D. Then L(1=L implies that the correspondence e --+
the series is uniformly convergent on every compact subset of X. Proof We choose a Z-base of A and identify A with zn; by R-linearity we identify X with Rn. We take a sufficiently large ro such that for the solid sphere, say C, of radius r 0/2 we have X = C + A. We can restrict x to C. Then, for every y in X satisfying r(y)~ro, we have r(y)lr (x + y);;;; r(y)/(r(y) -r(x));;;; 2. Choose M from R satisfying

II rm ([J II

00 ;;;;

M

for m=O, n+ 1. This is the only place where the assumption that ([J is a Schwartz function is used. At any rate, for r( e) ~ r0 and m = n + 1, we have

r(et I([J (x + e)1 =(r(e)/r(x + e))m Ir(x + et ([J(x+ e)1 ;;;;2m M. Let N denote the number of lattice points in 2 C, i.e., the number of elements e of A satisfying r(e);;;;r o' Then we have

L 1([J(x+e)I;;;;NM + ~EA

L

1([J(x+e)1

~EA, r(~)?;ro

;;;;NM +2m M· ~EA,

L

r(e)-m.

r@?;ro

We observe that this series and the integral 00

ftn-1-mdt

ro

converge for the same values of m and the integral does converge for m>n, hence for m=n+1. q.e.d.

I. Theta Functions from an Analytic Viewpoint

44

The above proof shows that if iP is restricted to a subset of Y (X) for X = R n on which II rm iP II 00 ~ M for m = 0, n + 1, the convergence of the series in Lemma 12 is uniform also in iP. An example of such a subset is as follows: Let q(x)=! x T tx + c tx+q(O) denote a quadratic polynomial on R n with T in 6 n and iP(x)=e(q(x)). Restrict T by the condition that the n eigenvalues of Im(T) stay away from O. As for c and q(O), assume that their imaginary parts are bounded. Then the argument in §9 about cg(X) being contained in Y(X) shows that we have IIrdiPlloo~ const. (depending on d) for every d. The identity in the following lemma is called the Poisson formula: Lemma 13. Let iP denote an arbitrary Schwartz function on X. Then we have

L iP(e) = L iP*(e*)·

~EA

~EA.

Proof. We have seen in § 9 that iP* is a Schwartz function on X*. It follows from Lemma 12 that F
L iP(x+e) ~EA

defines a continuous function p

e

L

F(x)=

a(e*) (x, e*);

~EA.

we have

a(e*)= JF(x) (x, -e*) dx X/A

= JiP(x) (x, -e*) dx, X

and this is nothing else than iP* ( - e*). In this way, we get the following "pointwise" identity:

L iP(x+e)= L iP*(-e*) (x, e*); ~EA

~'EA*

this becomes the formula in the lemma for x=O.

q.e.d.

We shall obtain a set of generators for the group SP2n(Z). We shall first prove the following simple lemma:

§10. The Discrete Subgroup ll.

Lemma 14. Suppose that an element ring A has the following form:

(1=

(

(X'

(Xl 2

y'

Y12

*

o

(1

P'

of SP2n(A) for any coefficient

*)

* *

(X"

*

II 0

0

45

,

()"

in which (X', p', y', ()' are of degree n' and (X", ()" of degree n" = n - n'. Then (1

'=((X' P') y' ()'

is in SP2n,(A), and (X12 =Y12 =0, (X" t{)" = In'" The set of elements such as forms a subgroup of SP2n(A), and the correspondence (1-+ (1' gives a surjective homomorphism of this subgroup to SP2n,(A); the corresponding extension splits.

(1

Proof We recall that an element of M 2n (A) composed of (x, p, y, () is contained in SP2n (A)ifand only if (X tp, Yt{) are symmetric and (X t{)_ p ty = In' For the above (1, (X t{) - P ty = In implies that (x" t{)" = In"; hence y t{) is symmetric if and only if Y12 =0 and y't{)' symmetric. Hence, in addition to (X" t{)" = In'" (X t{) - P ty = In implies that (X12 =0 and (Xtt{)' - P' ty' = In" Moreover, since (X tp is symmetric, (X'tp' is symmetric. Therefore (1' is in SP2n,(A). The rest is clear. q.e.d. We shall introduce a notation (which has become necessary already in the above proof). If (1' and (1" are elements of M2n'(A) and M2n',(A) composed of (x', p', y', ()' and (x", P", y", {)", we put

0 0)

(X' P' o (X" 0 P" ( (1' EB (1" = y' 0 ()' 0 o

y"

0

.

()"

This is an element of M 2n (A) for n=n' +n". If (1', (1" are in SP2n,(A), SP2n,,(A), (1'EB(1" is in SP2n(A).

Lemma 15. The group SP2n(Z) is generated by elements of the forms

Proof We shall keep some of the notation used in Lemma 14. For the sake of simplicity, we call an element of SP2n(A) with "y"=O or "P"=O an element of type P or tP. We shall use P' instead of P for

46

I. Theta Functions from an Analytic Viewpoint

SP2n,(A). If 0"' is an element of type p' or tp', 0"' Ef> 12n" is an element of type P or tP. If we multiply E to an element of type P or tp from both sides, we get an element of type tp or P. Moreover we have

Therefore, if we denote by r the subgroup of SP2n(A) which is generated by elements of types P and tp, r is also generated by E and elements of type P only. After these remarks, we shall prove the lemma. Let 0" denote an arbitrary element of SP2n(Z) composed of IX, P, Y, band Yn , bn the last row vectors in Y, b. If x=(xl ... xJ is an element of K" for any field K with an absolute value, we shall denote by Ix 100 the maximum ofl xli, ... , Ixnl. We shall introduce three operations which are right multiplications by elements of r: (i) If Yn=l=O, pass from 0" to 0" E; (ii) if bn=l= 0, pass from 0" to tu- l ( 0" 0

~)

so that the new bn becomes (0 ... 0 d) for d ~ 1 ; (iii) if bn is of the form (0 ... 0 d) for d ~ 1, pass from 0" to

so that Inew Ynloo ~d/2. We observe that the positive integer d in (ii) is the greatest common divisor of the n coefficients of bn • If we apply (i), (ii), (iii), in this order, to any given 0" with Yn=l=O, we get Inew Ynloo ~ IYnI00/2 . Consequently, a repeated application of (i), (ii), (iii) will produce an element 0"1 for which" Yn" is O. Then" bn" for 0"1 is different from O. By applying (ii), we get an element 0"2 for which "Yn" is 0 and" bn" is (0 ... Od). Then d is necessarily a positive unit of Z; hence d = 1. In the case where n= 1,0"2 is an element of type P, hence an element of r. In the case where n ~ 2, Lemma 14 shows that 0"2 is the product of 0"' Ef> 12 and an element of type P. If we apply an induction on n, we see that 0"' is contained in the subgroup r' of Sp2n' (Z) for n' = n -1 generated by elements of types P' and tp'. Therefore 0"' Ef> 12 is contained in the subgroup r generated by elements of types P and tP. q.e.d.

§ 10. The Discrete Subgroup Ii

47

Now we are ready to prove the following theorem: Theorem 9. Let iP denote an arbitrary Schwartz function on X such that the function F on B(X) defined by F(s)=

L (siP) (~) ~eA

is different from the constant O. Then the stabilizer subgroup Ii ofNL at F, which consists of those So in N L satisfying

F(so s)=F(s) for every sin B(X), is a discrete subgroup ofB(X), and it does not depend on iP. The homomorphism 1t restricted to Ii has U(L) as its kernel and the subgroup ofSp(X) consisting of those 0' which keep L invariant as its image. Moreover N L decomposes into the direct product of Ii and Ct. Proof First of all, the function F = P is certainly different from the constant 0 for any R~ -valued Schwartz function iP on X. (Such a iP can be found even in ~(X).) Therefore Ii is well defined. We observe that the intersection of Ii and Ct consists of the identity element of B(X). Since we have seen that the quotient group NJCt is discrete, Ii is a discrete subgroup ofNL, hence ofB(X). Moreover Ii contains U(L). We take dual bases of A and A*, i.e., Z-bases ~1"'" ~n and ~t, ... ,~: of A and A* satisfying

for every x = (Xl'" xn) and x* = (xt. .. x:) in Rn. With respect to such dual bases, we identify A, A* with zn and X, X* with Rn. Then an element 0' ofSp(X) keeps L invariant if and only if 0' is in SP2n(Z), Put 0'0=E. Then we have LO'o=L and fao(~,~*)=<-~,~*>=l for every (~,~*) in L; hence r(O'o) is in N L. Since 9'(X) is invariant under every sin B(X) and since r(O'o) is the Fourier transformation, we get F(r(O'o)s)=F(s) by Lemma 13. Therefore r(O'o) is in Ii. In general, if s is a square matrix, we shall denote by So the row vector whose p-th coefficient is the p-th diagonal coefficient of s for every p. Let 0'1 denote an element of SP2n(Z) composed of at:, p, 0, lJ. Then we have <~at:, 2- 1 ~ p>=<~, 2- 1 (at:P*)0>

for every

~

in A. Also we have det(at:)=

±l. Put

Sl = U((O, -2- 1 (at:P*)0), l)p(O'l)' Then Sl is in NL and (Sl siP) (~)=(siP) (~at:) for every ~ in A. Since at: keeps A invariant, we have F(Sl s)=F(s). Therefore 81 is in Ii.

48

1. Theta Functions from an Analytic Viewpoint

Let r denote the subgroup of I;. which is generated by U(L), r(a o), and elements of the form Sl. We observe that r does not depend on F. We shall show that I;. =r. Let S2 denote an arbitrary element of NL and put n(s2)=a 2. Then a 2 is in SP2n(Z). On the other hand, we have n(r)=SP2n(Z) by Lemma 15. Therefore, by multiplying a suitable element of r to S2' we may assume that a 2 = 12n . Then we have S2 = U(w, t) for some (w, t) in A(X). Since S2 is in N L , W is in L; hence F(S2 s)= t· F(s). We have assumed so far that S2 is an element of NL and have modified S2 by an element of r. If S2 is contained in 1;., we have t· F(s)=F(s). Since F is not the constant 0, we get t = 1; hence S2 is in U (L). Therefore we have I;. =r and also NL =1;.. et. q.e.d.

Remark. In general, a discrete subgroup r of a unimodular group G is called a lattice if the (total) measure of the homogeneous space G/r is finite. In the case where r is normal in G, e.g., in the commutative case, this means that G/r is compact. We shall point out that by a result of Siegel our group I;. is a lattice in B (X): First of all, we have A (X) n I;. = U(L), and U(L) is a lattice in A(X). Hence the homogeneous space A (X) rdrL has finite measure. On the other hand, by Siegel's theorem SP2n(Z) is a lattice in SP2n(R). Hence B(X)/A(X) I;. has finite measure. Putting these together, we see that B(X)/I;. has fmite measure. We can modify the above argument by passing to the quotient of B (X) by its maximal compact subgroup. The necessary theorem of Siegel will then take the classical form that the quotient of 6 n by SP2n(Z) has finite measure. We take dual bases of A, A* and identify A, A* with zn and X, X* with Rn. Let a l and a denote elements of Q(Rn), i.e., elements of SP2n(R) composed of al , Pl' Yl' 01 and a, p, y, 0 with det(Yl)=!=O and det(y)=!=O; let tP denote an arbitrary element of~(Rn). Then by Lemma 11 we get

for some (Tl' el ) in 6 n x en. We shall fix a l and tP; hence we may assume that the "const." is 1. Let m, m* denote elements of Rn, t an element of et, and put s= U((m, m*), t) r(a). We shall regard m, m*, t, a as variables. Then elements of the form s r(a l ) form an open subset of B(X) depending on al and by Lemma 6 and others B(X) is the union of such open sets for all choices of a l . We observe that (for a fixed a l )sr(al ) determines (m,m*,t,a) uniquely. In fact, we can consider (m, m*, t, a) as local coordinates of sr(a l ) valid in the above open subset of B(X). Moreover, applying Lemma 11 again,

§10. The Discrete Subgroup Ii

we get

F(s r(0"1») =

49

L (s r(O"l) 4» (~) ~ezn

=t 'ldet(y)l-t det(i-l('t'l +y-1 15»-t

·e(-! C1('t'1 + y-l 15)-1 ' Cl)

· L e(!(~+m)'t't(~+m)+ct(~+m)+m*t~), ~ezn

in which 't'=(eX't'1 +f3)(y't'l +15)-1,

c=C1(Y't'1 +15)-1.

This shows that F(sr(0"1») depends real-analytically on the local coordinates (m, m*, t, 0") of sr(O"l)' Therefore F is a (complex-valued) realanalytic function on B(X). Since we have F(so s)=F(s) for every So in Ii. and s in B(X) by Theorem 9, F is really a function on the homogeneous space Ii. \B(X). We call Fa theta function on B(X) and regard the above explicit form of F(sr(0"1») as a local expression of the theta function F. On the other hand, for every (m m*) in R 2n and ('t', z) in <=>n x en, we put

0mm*('t', z)=

L e(!(~+m) 't' t(~+m)+(~+m)t(z+m*»).

~eZ"

This infmite series, called the theta series, converges absolutely and uniformly on every compact subset of R 2n X <=>n X en. In fact, by the remark after the proof of Lemma 12, the convergence is absolute and uniform with respect to (m m*), 't', c) provided that (m m*), c are bounded and the n positive eigenvalues ofIm('t') stay away from O. Therefore, for any fixed (m m*), 0mm*('t', z) is a holomorphic function on <=>n X and we have

en,

F(sr(0"1»)=t ·ldet(y)l- t det(i- 1('t'1 +y-1 15»)-t

· e( -! C1('t'1 +y-1 15)-1 tCl -m 'm*) Om m*('t', c).

We observe that F(sr(0"1») and 0mm.('t', c) differ only by an "elementary factor." We call 0mm*('t', ) a classical theta function, or simply a theta function, of characteristic (m m*) and of modulus 't'. It is possible to show that the invariance of F under Ii. implies the well-known complicated functional equation for the classical theta function. However we shall not go through the details here. We merely point out that the identity (0.5) below is nothing else than the invariance of F under U(w,l) for w=(~, ~*) in L. We shall give five identities concerning the classical theta functions: We have (0.1) 0mm'('t', -z)=O -m _m'('t', z); the verification is trivial. If ~, (0.2)

~*

are in

zn, we have

50

I. Theta Functions from an Analytic Viewpoint

the verification is again trivial. If u, u* are in R n, we have (0.3)

0mm.(L, z+u L+u*)=e( -tu L'u-u '(z+u*»)

this follows immediately from

He +m) L'(e +m)+(e +m)'(z+u L+U* +m*) =

-t u L 'u-u '(z+u*+m*)

+t(e + m+u) L'(e + m+u)+(e + m+u)'(z+m* +u*). We put m=m* =0 and replace u, u* by m, m* in (0.3); then we get (0.4)

0mm.(L, z)=e(t m Ltm+m'(z+m*») OOO(L, z+mL+m*)

valid for every m, m* in Rn. We put (0.3) and (0.2) together; then we get (0.5)

0mm,(L, z+ e L+ e*)=e( -t e L'e - e 'z) e(m 'e* - e 'm*) 0mm.(L, z)

valid for every e, e* in

zn. These identities will be used later.

Chapter II

Theta Functions from a Geometric Viewpoint § 1. Hodge Decomposition Theorem for a Toms Let X denote a fmite dimensional vector space over Rand L a lattice in X. (We shall see that X will play the role of W=X x X* in Chapter I.) Then L spans X over R. Otherwise, let X' denote the subspace of X spanned by Lover R. Then the quotient group X/X' is a vector space over R of a positive dimension. On the other hand, it is a quotient group ofthe compact group X/L; hence X/X' is compact. But this is a contradiction. Let 1t denote the canonical homomorphism of X to X/L. Then (X, 1t) is a covering space, in fact the universal covering space, of X/L. Hence X/L inherits the real-analytic structure of X. If the dimension of X is m, X/L is bi-analytically isomorphic to (R/Zr; it is called a real torus or just a torus. Ifwe start from a vector space over C instead of a vector space over R, we will get a complex manifold, called a complex torus. We shall prove the existence of a "simple" covering (with a partition of unity subordinate to the covering) and the Hodge decomposition theorem both in the special case oftori; the proofs are exceptionally simple in this case. If M is a set and (Ui)ieI a set of subsets U; of M indexed by I such that

U U;=M,

ieI

we say that (Ui)ieI is a covering of M; we shall also write (Ui)i instead of (Ui)ieI· Let (U;)ieI and (tj)jeJ denote two coverings of M. If there exists a mapping of I to J denoted by i -+ ii such that U; is contained in tj; for every i, we say that (U;)i is a refinement of (tj)j. If, in particular, the mapping i -+ ii is injective and U; = tj; for every i, we say that (U;)i is a subcovering of (tj)j. A covering (Ui)i of a topological space M is called locally finite if, for any compact subset C of M, the set of those i's for which en U; =I=!!> is finite. If M is compact, a locally finite covering is necessarily finite. A covering (Ui)i of a topological space M in which all U; are open is called an open covering. Suppose that M is an m dimensional (infinitely) differentiable manifold. An open -covering (U;)i of M is called

52

II. Theta Functions from a Geometric Viewpoint

simple if every non-empty intersection l!;, f""'I ••• f""'I l!;p is diffeomorphic to an open convex subset of Rm. Let (l!;)iel denote a locally finite open covering of M. If there exists a set (Pi)iel of infmitely differentiable, nonnegative real-valued functions Pi on M indexed by I such that the support of Pi is contained in Ui for every i and

LPi=l,

iel

we say that (Pi)i is a partition of unity subordinate to (l!;)i'

Lemma 1. Let M =X/L denote a torus and (lj)j an open covering of M. Then there exists a finite simple refinement (l!;)i of (Jij)j and a partition of unity (Pi)i subordinate to (l!;)i' Proof. We define a Euclidean distance d(x, y) of two points x, y of X by using the coordinates of x, y. Let d denote the distance from 0 to the complement of 0 in L, i. e., the minimum of d (0, for +0 in L. For any e in R~ and a in X, we shall denote by u,,(a) the interior of the sphere of radius e centered at a. Since the metric is translation-invariant, we have u,,(a+a')= u,,(a)+a' for every a' in X. For every a, we choose e~d/4 such that n(U.(a)) is contained in some lj. Then, for any distinct rr in L, we have (u,,(a)+e)f""'I(U.(a)+rr)=~; also we have either

e)

e

e,

U.(a)f""'I(u",(a')+e)=~

or

U.(a)f""'I(u",(a')+rr)=~.

Since (n(u,,/2 (a)))a forms an open covering of M and M is compact, we can fmd a finite subcovering, say (n(U. i/2(a i)))ieI' We put l!;=n(U.i(a i)) for every i in 1. It is clear by defmition that (l!;)i forms a finite simple refinement of(lj)j' We shall construct a partition of unity (Pi)iel subordinate to (l!;)iel' Consider the function P•. a on U.(a) defined by

P•. a(x)=exp( -l/(l-(d(x, a)/e)2)) for every x in u,,(a). If we put (n.p •. J(n(x))=P •. a(x), we get a welldefined function n. P•. a on n(u,,(a)). If we assign the value 0 to every point of M -n(u,,(a)), we will get an infmitely differentiable function, say 4>•. a' on M. By construction, 4> •• a is positive at every point of n(U.(a)) and 0 elsewhere. Since (l!;')i for U; = n (u"i/ 2 (a i)) forms an open covering of M, the sum of 4>.i/2.ai for all i is positive at every point of M. Therefore

Pi= 4>.i/2.a/(L 4>.i/2.ai) iel is an infmitely differentiable, non-negative real-valued function on M with the closure of l!;' as its support. It is clear that (Pi)i forms a partition of unity subordinate to (l!;)i' q.e.d.

§1. Hodge Decomposition Theorem for a Torus

53

Suppose that M is an m dimensional differentiable manifold and ¢ a differential form of degree p on M. Let Xl' ... , Xm denote local coordinates of M valid in some open subset U. Then ¢ admits a local expression of the following form: A-(X)= 'f'

"L.

A. . (x)dx.'-1 If'll ••• lp

il<···
/\.·./\dx., I-p

in which /\ denotes an exterior product; the coefficients ¢i, ... ip are (C-valued) functions on U; and the expression is unique. Unless otherwise stated, we shall assume that ¢ is infinitely differentiable, i. e., all ¢i, ... ip for all choices of U are infinitely differentiable. We shall denote by iP(M) the vector space over C of differential forms of degree p on M and by .fF(M) the direct sum of all .fFP(M). We have .fFP(M)=O for p>m and, for every ¢ in .fFP(M) and t/! in .fFq(M), their exterior product ¢ /\ t/! is in .fFp+q(M). If ¢ is in .fFP(M), by using its local expression, we put (d¢)(x)=

L

(d¢i, ... i)(X)/\dx i, /\ ... /\dx ip '

il<"'
in which

m

A-. ..• lp . )(x)= "(8 A-. (d 'PI-! i..J "fill'"

.

I-p

/8x.)(x) dx.l l

i=l

for every iI, ... , i p. Then d¢ is a well-defined element of .fFP+ I (M) with the above (d ¢ )(x) as a local expression. By linearity, we define d ¢ for every ¢ in .fF(M); we have d2 =0 in the sense that d(d¢)=O for every ¢ in .fF(M). Moreover, if ¢ is in .fFP(M) and t/! in .fF(M), we have d(¢ /\ t/!)=d¢ /\ t/!+( -l)p ¢ /\dt/!.

If M "" is another differentiable manifold and n an (infinitely) differentiable mapping of M"" to M, we get a C-linear mapping n* of .fF(M) to .fF(M"") which is degree-preserving. We recall that if xf, ... denote local coordinates of M"" valid in some open subset U"" of n-I(U), at every point x"" of U"" we have

x:,

(n* ¢)(x"")=

in which

L

it<···
¢i, ... ip (n(x""») dxdn (x ""») /\

dxi(n(x""») =d(Xi n)(x"") =

... /\ dxip(n(x""»),

L (8(Xi n)/8 x;)(x"") dx; j

for every i. If¢, t/! arein.fF(M), we have n*(¢ /\ t/!)=n* ¢ /\ n* t/!. Moreover n* commutes with d in the sense that n*(d ¢)=d(n* ¢) for every ¢ in .fF(M).

We shall apply what we have said to M =X/L, M"" =X, and n the canonical homomorphism. We take an R-base of X and identify X with Rm. Let Xl' ... , Xm denote the coordinates of a general point x of X. Then

Chapter II

Theta Functions from a Geometric Viewpoint § 1. Hodge Decomposition Theorem for a Toms Let X denote a fmite dimensional vector space over Rand L a lattice in X. (We shall see that X will play the role of W=X x X* in Chapter I.) Then L spans X over R. Otherwise, let X' denote the subspace of X spanned by Lover R. Then the quotient group X/X' is a vector space over R of a positive dimension. On the other hand, it is a quotient group ofthe compact group X/L; hence X/X' is compact. But this is a contradiction. Let 1t denote the canonical homomorphism of X to X/L. Then (X, 1t) is a covering space, in fact the universal covering space, of X/L. Hence X/L inherits the real-analytic structure of X. If the dimension of X is m, X/L is bi-analytically isomorphic to (R/Zr; it is called a real torus or just a torus. Ifwe start from a vector space over C instead of a vector space over R, we will get a complex manifold, called a complex torus. We shall prove the existence of a "simple" covering (with a partition of unity subordinate to the covering) and the Hodge decomposition theorem both in the special case oftori; the proofs are exceptionally simple in this case. If M is a set and (Ui)ieI a set of subsets U; of M indexed by I such that

U U;=M,

ieI

we say that (Ui)ieI is a covering of M; we shall also write (Ui)i instead of (Ui)ieI· Let (U;)ieI and (tj)jeJ denote two coverings of M. If there exists a mapping of I to J denoted by i -+ ii such that U; is contained in tj; for every i, we say that (U;)i is a refinement of (tj)j. If, in particular, the mapping i -+ ii is injective and U; = tj; for every i, we say that (U;)i is a subcovering of (tj)j. A covering (Ui)i of a topological space M is called locally finite if, for any compact subset C of M, the set of those i's for which en U; =I=!!> is finite. If M is compact, a locally finite covering is necessarily finite. A covering (Ui)i of a topological space M in which all U; are open is called an open covering. Suppose that M is an m dimensional (infinitely) differentiable manifold. An open covering (U;)i of M is called

§ 1. Hodge Decomposition Theorem for a Torus

55

polynomial function p* on X there exists a translation-invariant differential operator D on X satisfying

p*(e*)a(e*)= J(D f)(x) e( -x te*) dp.(x). M

This can be proved by a repeated application of (2ni e;)a(e*)=

J(of/oxp)(x) e( -x te*) dp.(x),

M

in which e; is the p-th coefficient of e* for 1 ~ p ~ m. In particular, we have sup IP*(e*) a(e*)1 < 00.

~eL"

Furthermore any translation-invariant differential operator D on X commutes with the summation over L*, i.e., (Df)(x) is obtained by simply replacing e(x te*) by D(e(x te*»). Conversely, if we have a set (a(e*»)~eL" of complex numbers a(e*) satisfying the above "growth condition ", the sum of a(e*) e(x te*) over L* certainly defines an infinitely differentiable periodic function on X with L as a period group. After this remark, we take n* cjJ as f and consider the following series: (1/2n)2.

L

(a (e*)/q (e*») e(x te*).

~eL,,-O

Since q(x) is positive-definite and since we have excluded e* =0 in the summation, the series is well defmed; by what we have said, it represents an infmitely differentiable periodic function, say g, on X with L as a period group. Hence there exists an element l/I of §"O(M) satisfying n* l/I=g; l/I is uniquely determined by g, hence by cjJ. If we put GcjJ=l/I, G is a C-linear mapping of §"O(M) to itself. The verification of LJGcjJ= GLJcjJ=cjJ-HcjJ, HGcjJ = GHcjJ=O is straightforward. q.e.d. Since H is a projection, §"P(M) decomposes into the direct sum of its kernel and image; in the theory of algebras, this is called the Peirce decomposition of §"P(M) relative to the idempotent endomorphism H. At any rate, if we denote by $"P(M) and JilfP(M) the kernel and the image, we have §"P(M) = JilfP(M) EB $"P(M).

On the other hand, we have H LJ = LJH = 0, i. e.,

HLJcjJ=LJHcjJ=O for every cjJ in §"P(M); this follows from the defmition (and also from the proof of Theorem 1). Therefore LJ maps JilfP(M) to 0 and $"P(M) to itself. Since we have HG=GH=O by Theorem 1, G has the same property. The other formula in Theorem 1 implies that LJGcjJ=GLJcjJ=cjJ for every

56

II. Theta Functions from a Geometric Viewpoint

4> in JP(M). We shall show that JffP(M) is the kernel of LI. We decompose an arbitrary element 4> of .?FP(M) into 4>1 +4>2 with 4>1 in JffP(M) and 4>2 in %P(M), i.e., 4>1 =H4>l and H4>2=0; then we get Ll4>=Ll4>2. If 4> is in the kernel of LI, we have Ll4>2 =0, hence GLl4>2 =4>2 =0. This proves the assertion. For the same reason, JffP(M) is the kernel of G. We extend LI, G, H by linearity to F(M); we shall denote by Jff(M), %(M) the direct sums of JffP(M), %P(M) respectively for all p. Then F(M) becomes the direct sum of Jff(M) and %(M), and both LI and G map Jff(M) to 0 and %(M) to itself. Moreover LI and G are the inverses of each other on %(M). We shall show that any C-linear mapping TofF(M) to itselfwhich commutes with LI commutes with G and H. Ifwe represent a general element 4>=4>1 +4>2 of F(M) with 4>1 in Jff(M) and 4>2 in %(M) by the row vector (4)1 4>2)' Tgives rise to a matrix with four entries 7;j as

We observe that Tcommutes with LI if and only if 7;.2 =0, T21 =0, T22 A= AT22 • Then T certainly commutes with G and H. We shall pass to a complex torus: Let Z denote a vector space over C of dimension g, hence a vector space over R of dimension m = 2 g, L a lattice in Z, and Z/L the corresponding complex torus. Then what we have shown so far applies to the "underlying" real torus M of Z/L. We take a C-base of Z and identify Z with O. Let Zl' ••• , Zg denote the coordinates of a general point Z of Z. Then the differentials dz 1, ... , dzg , dz1, ... ,dzg determine translation-invariant elements m1' ... , mg , w 1, ... ,Wg of F1(Z/L) satisfying n*(mi)=dzi , n*(wi)=dzi for l~i~g. If U is an open subset of M, every 4> in F(U) can be written uniquely in the form

in which ,/,.(p,q)=",/,. . "I' l.J 'Pl.!

...

..• l.pdt ... Jq

m·'1 A···Am·lp AW.Jl A···AW.Jq

with 4> i l •.• ip,jl ••• jq in FO(U); the summation extends over the set of indices satisfying i1 <··· into 4>(P,q) is intrinsic, i.e., it does not depend on the identification of Z with cg. We say that 4> is of type (p, q) if 4> = 4>(p,q). The elements of F(U) of type (p, q) form its subspace F(P,q)(U) over C, and F(U) decomposes into the direct sum of all F(P,q)(U).

§ 1. Hodge Decomposition Theorem for a Torus

57

Iffis an element of §"O(U), we have

df =d'f +d"f, in which

g

g

d'f = L cPiOJi'

d"f= L!/I/jjj j_l

i=l

with cPi,!/Ij in §"O(U); we put cPi=O;f, !/Ij=o'/ffor l~i,j~g. If cP is an element of §"(P,q)(U), by replacing eachf=cPil ... ip,it ... jq by d'fwe get an element d' cP of §"(P+l,q)(U); similarly we get an element d" cP of §"(p,q+l)(U). We extend d', d" by linearity to §"(U); we have d=d'+d" in the sense that dcP=d' cP+d" cP for every cP in §"(U). We observe that d', d" are intrinsic. Moreover d2 =0 gives rise to (d')2 =0, (d")2 =0, d'd" +

d"d'=O. There are two more C-linear mappings, denoted by ~', ~", of §"(U) to itself defined as follows: For the sake of simplicity, we put 0 1 =OJ·II /\ ... /\ OJ ip'

0 1a =OJ·'1 /\ ... /\ OJ·14 -1 /\ OJ·'«+1 /\ ... /\ OJ·'p

for 1 ~a~p; also we put and derme 02b similarly as 0la for mapfOl /\ O2 to

l~b~q. q

p

2· L(-1)a(0;:f)01a/\02' a=l

2· L

Then we require that

~',~"

( _1)P+b (olb f) 0 1 /\ 02b

b=l

for every fin §"O(U). Since every cP in §"(U) can be written uniquely as a sum of terms such as f 0 1 /\ O2 , this prescription and the linearity derme ~', ~".It follows from the definition that (~y =0, W')2 =O,~'~" +~"~' =0; ~', ~" map §"(p,q)(U) to §"(P-l,q)(U~ §"(p,q-l)(U). We note that ~', ~" are not intrinsic. Now we derme a C-linear mapping Ll of §"(U) to itself as g

Ll (fOl /\ ( 2 )= -4· L (0; o;'f) 0 1 /\ O2 ; i=l

then we have

Ll =2(d'~' +~' d')=2(d"~" +~" d''). This can be verified without any difficulty by applying the three" operators" to f 0 1 /\ O2 , If we denote the real and imaginary parts of Zi by X 2i _1 and X 2i for 1 ~i~g, we get g

m

-4· L o;o;'f= - L(oYf i=l

j=l

58

II. Theta Functions from a Geometric Viewpoint

for m=2g. Therefore the above ,1 becomes the previous ,1 relative to the quadratic form q (x) = (X 1)2 + ... + (X m )2 ; hence Theorem 1 can be applied. Since (d')2=0, we have d' ,1 =2d'(d' {)' +{)' d')=2d' {)' d' =L1d';

we see similarly that d", {)', {)" commute with ,1. Therefore d', d", {)', {)" commute with G and H. On the other hand, by definition, ,1 maps !JF(P,q)(ZjL) to itself. If we denote by T the projection of !JF(ZjL) to !JF(p,q)(ZjL), this means that T commutes with ,1. Then T commutes with G and H; hence G and H map !JF(p,q)(ZjL) to itself.

§ 2. Theta Function of a Positive Divisor We shall explain the concept of positive divisors of a complex manifold. We observe that if V is a connected complex manifold andfa holomorphic function on V different from the constant 0, the open subset, say VI' of V consisting of those points z where f(z)=!=O is dense in V. After this remark, we take a complex manifold M; every non-empty open subset V of M is itself a complex manifold; we shall denote by l!JM(V), or simply by l!J(V), the ring of holomorphic functions on V. If V is a (non-empty) open subset of V, we have the restriction homomorphism l!J(V) ~ l!J(V); we shall denote the image in l!J(V) of an elementf of l!J(V) also by fbut adding, e.g., "on V". An elementf of l!J(V) is a zero divisor of l!J(V) if and only if it is the constant on some connected component of V; hence, iffis not a zero divisor, it is not a zero divisor on any open subset Vof V. We shall consider a set (Vi,h))iEI of pairs (Vi,h) indexed by I, in which (Vi)i forms an open covering of M andh in l!J(V;) for every i; for the sake of simplicity, we shall use (Ui,h)iEI or (Vi,h)i instead of (Vi,h))iEI' Also, for any covering (Vi)i of M, we put

°

U.H ... Zp . =u.11 (l ... (lu.Zp

provided that this intersection is not empty. We shall assume that (U;,h)i satisfies the following two conditions:

(i) h is not a zero divisor of l!J(U;) for every i; (ii) there exists a unit fij of l!J (Vi) satisfying .f; = hj jj on each Vij. Suppose that (~, gj)j is a similarly defined set. If there exists a unit hij of l!J(U; (l~) satisfying h = hij gj on every non-empty intersection U;(l Vj, we say that (U;,h)i and (Vj,g)j are equivalent. This is an equivalence relation in the class of all sets of the form (Vi' h)i' If D is the equivalence class of (Vi,h)i' we say that D is the positive divisor of M with (U;, h)i as a representative; also we say that "J; =0" is a local equation

§ 2. Theta Function of a Positive Divisor

59

for D in U;. We observe that every positive divisor of M has a representative (V;, !;); in which V; =l= Vj for every i =l= j; the class of all such representatives of all positive divisors of M is a set. Therefore, the class of all positive divisors of M is also a set. Suppose that (V;); is a refinement of (Vj)j under i ~ j; and D the positive divisor of M with (Vj, gj)j as a representative. Then, if we denote by!; the restriction of gj; to V;, we get another representative (V;,!;); of D. On the other hand, any finite number of open coverings of M has a common open refinement, i. e., an open covering which is a refinement. Therefore, if D, D' are positive divisors of M, we can find their representatives of the form (V;,!;);, (U;,.t;');. Then (U;,!;!;'); defmes a positive divisor D+D' of M; clearly D+D' depends only on D, D'. Under this addition, the set of positive divisors of M forms an additive monoid. Since this monoid satisfies the" cancellation law," i. e., D + D" = D' + D" implies D = D', it can be embedded into the group consisting of expressions of the form D - D' with an obvious identification; this is called the divisor group of M. The identity element of the divisor group is a "positive" divisor; for any open covering (V;); of M, it has (V;, 1); as a representative. If D has a representative of the form (M,f), we shall write D = (f); fis unique up to a unit of (()(M). The "second Cousin problem" asks whether every positive divisor of M is ofthe form (f). If M is compact and connected, we have (()(M)=C; this is an immediate consequence of the maximum modulus principle (for one complex variable). Therefore, if M is a complex torus with a positive divisor D =l= 0, in which case we are primarily interested, the answer to the second Cousin problem is negative. In order to modify the problem in a sensible manner, suppose that M is a connected complex manifold; let (M*, n) denote a covering manifold of M. Then every positive divisor D of M determines a positive divisor n*(D) of M*; if (V;,!;); is a representative of D, (n- 1 (V;),!;o n); is a representative of n* (D). A modified problem asks whether n* (D) is of the form (F) for some F in (() (M*) if(M*, n) is the universal covering manifold ofM. We shall examine the formal aspect of the problem. Suppose in general that M is a complex manifold and r a group of biholomorphic mappings of M to itself. Then r acts in a natural manner on the monoid of positive divisors of M as follows: Let y denote an element of rand D a positive divisor of M; choose a representative (U;,!;); of D. Then (y-i (U;),!; ° y); defines a positive divisor y* (D) of M; y* (D) depends only on y and D; and we have y*(D+D')=y*(D)+y*(D'), (Yi Y2)*(D)= yi(yf(D)), l*(D)=D for every y, Yi' Y2 and D, D'. Suppose that M is connected and let r denote its fundamental group. Then, by using a realization of r as the group of homotopy classes of closed curves with

°

60

II. Theta Functions from a Geometric Viewpoint

reference to a fixed point of M, we can let r act on the universal covering manifold M* of M; the action (y, z)-+y(z) is biholomorphic and 1toy=1t for every y. Therefore, if D is a positive divisor of M, we have y* (1t* (D») = 1t* (D). Consequently, if we have 1t* (D) = (F), for every y in r there exists a unit uy of 19(M*) satisfying F(y(z») =F(z) uy(z)

at every point z of M*. The unit uy is an "automorphy factor" in the sense that uYI Y2 (z) = uYI (yz (z») un (z) for every Y1' yz in r. Conversely, if F is an element of 19(M*) different from 0 with the above property, there exists a positive divisor D of M satisfying 1t*(D) = (F); clearly D is unique. Although the existence-proof is straightforward, for the sake of completeness we shall give its details: We take an open covering ([1;); of M, in which every [1; is connected and simply connected. Choose a connected component, say [1; 0' of 1t- 1 ([1;) for each i. Then 1t maps [1;0 biholomorphically to [1;; and 1t- 1 ([1;) is the disjoint union of y([1;o) for all y in r. We defme/; by /;(1t(z») = F(z) for every z in [1;0. Then/; is an element of 19(UJ different from O. If [1;; is defmed, the intersection y-1(U;0)n U;o is not empty for some y in 'r; 1t maps the disjoint union of all such intersections biholomorphically to U;j; and, for every z in y-1(U;0)n U;o we have /;(1t(z») = F(y(z») =F(z) Uy(z) = .fj(1t(z») uy(z).

This shows that ([1;,/;); defmes a positive divisor D of M satisfying (F). Let Z denote a fmite dimensional vector space over C and L a lattice in Z; we shall take the complex torus ZjL as M, Z as M*, and the canonical homomorphism as 1t; we can take L as r.

1t* (D) =

DefmitioD. A thetafunction on Z relative to L is a holomorphicfunction () on Z different from the constant 0 satisfying (}(z+~) = (}(z) e(Q~(z)+c~),

in which Q~ is a C-linear form on Z and c~ an element ofCJor every z in Z and ~ in L. The theta function () is a function of the form F with u~(z)=e(Q~(z)+c~)

as an automorphy factor. Therefore, by what we have shown, there exists a unique positive divisor D of ZjL satisfying 1t* (D) = «(}). We observe that the set of theta functions on Z relative to L forms a multiplicative monoid under the ordinary multiplication with the constant 1 as its

§ 2. Theta Function of a Positive Divisor

61

identity element. We shall determine the group of units of this monoid. We take a C-base of Z and identify Z with cg. Let (J, (J' denote theta functions on Z satisfying (J(J' = 1; then iP log (J/8zJJz j for 1 ~ i, j~g are holomorphiq periodic functions on Z with L as a period group; hence they are all constants. By integration, we get (J(z)=e(P(z)),

in which P is a polynomial function of degree 2 on Z. Conversely, every such function is a theta function on Z relative to any lattice in Z. Therefore, if we call it a trivial theta function on Z, the group of units consists of trivial theta functions; this group is independent of the lattice L. Now we can state and prove the following important theorem: Theorem 2. Let (J denote a theta function on Z relative to Land D the positive divisor of Z/L satisfying n*(D)=((J). Then the correspondence (J -+ D gives a surjective homomorphism of the multiplicative monoid of theta functions on Z relative to L to the additive monoid of positive divisors of Z/L; the kernel of this homomorphism is the group of trivial theta functions on Z. Proof The main point of the theorem is the surjectivity of the homomorphism, i.e., the existence of (J satisfying n* (D) = ((J) for any given D; the rest is clear by what we have said. We shall start by making some simple remarks: In general, if U is a complex manifold andfan infinitely differentiable function on U, i.e., an element of g-O(U), we have df =d'f +d"j, in which d'j, d"fare in g-(1,O)(U), g-(O,l)(U). Since d is "real", i.e., depends only on the underlying real manifold of U, the complex conjugate of df is dj; hence the complex conjugates of d'j, d"f are dill, d'j Moreover, we have d"f = 0 if and only iff is in (!) (U). The reason is that if we express d"f = 0 in terms of local coordinates, we get the "Cauchy-Riemann differential equations" (for one complex variable) in each coordinate. Suppose that U is connected and simply connected. Then, for every element
II. Theta Functions from a Geometric Viewpoint

62

on ~jk' In particular, by taking k= i, we get J;Jji= 1 on ~j; hence the assumption in (ii) that k is a unit of (i)(Vi) is redundant. We shall go back to the complex torus M =Z/L. As we have remarked before (in the general case), we can replace (Vi)i in the representative (~,J;)i of D by anyone ofits refmements. Therefore, we shall assume by Lemma 1 that (VJiEI is a fmite simple covering with a partition of unity (Pi)iEl subordinate to (~)i' We choose a connected component, say U;o, ofn- 1 (U;) for each i; let Vi~ denote the image of ~o under the translation z --+ z + ~ for every ~ in L. Then n- 1 (Vi) is the disjoint union of ~ ~ for all~, and n maps each U;~ biholomorphically to Vi' Since Vj k is connected and simply connected, gjk=(1/2ni) log.fjk is an element of (i)(Vjk) unique modulo Z. Therefore Im(gjk) is well defined, and we have 1m (gij)+ 1m (gjk) + Im(gki) =0 on ~jk' We observe that Im(gjJPi can be extended to Vj as an infmitely differentiable function by simply assigning the value 0 to every point of ~-Vji" We shall denote the so-extended function also by Im(gjJPi; and we put gj= l)m(gjJ PiiEI

Then gj is in $'O(V); and we have

I =I

gj-gk=

(Im(gjJ-Im(gki)) Pi

iEI

1m (gjk)Pi = 1m (gjk)

ieI

on ~k' Put ¢j=2i. d' gj;

then ¢j is in

$'(1, 0) (Vj ).

Since gjk is in (i)(Vjk ), we have d" gjk=O; hence

¢j-¢k=2i· d'(gj-gk)=2i· d' Im(gjk)=d'(gjk-gjk)=dg jk ·

Since we also have d"(dgjk)=d" d' gjk= -d' d" gjk=O, we get d" ¢ .=d" ¢k on ~k' Therefore d" ¢j is the restriction to Vj of an element of $'(1, l)(M). Since ¢j is of type (1,0), we have 15" ¢j=O; hence LJ¢j=2(d" 15" +15" d") ¢j=2J" (d" ¢j)'

Therefore LJ ¢ j is the restriction to Vj of an element of $'(1, O)(M); we shall denote this element by ljJ. Since we have H 15" = 15" H = 0, we get H ljJ = O. Therefore, if we apply Theorem 1 to ljJ, we get ljJ=LJGljJ; put ¢j=¢j-GljJ

§ 2. Theta Function of a Positive Divisor

63

on ~. Then c/Jj is an element of g;;(1,0)(Uj) satisfying Llc/Jj=O such that c/Jj-c/J~ = c/Jj -c/Jk =dgjk

on ~k. Since we have d(dgjk)=O, we get dc/Jj=dc/J~ on Ujk" Therefore dc/Jj is the restriction to Uj of an element, say oc, of g;;2(M). Since c/Jj= 2i· d' gj' we have d' c/Jj=O, hence d' t/J =d' LI c/Jj=LI d' c/Jj=O; hence d' Gt/J = Gd't/J=O. Therefore we get d'c/Jj=dic/Jj-d'Gt/J=O; hence oc=dc/Jj=d"c/Jj on ~. Since c/Jj is oftype (1, 0), this shows that oc is oftype (1, 1). Moreover we have LI oc=LI d" c/Jj=d" LI c/Jj=O, hence oc=H oc. Therefore, if we identify Z with e g, we have g oc = L oc jk Wj 1\ Wk, j,k=1

in which oc jk are in e (and Wj' Wk are as in § 1). Let Z denote a point of Ui ~ and put g

c/Ji~(z)=c/J;(n(z))+

L ocjk(Z-e)kdzj; j,k=1

then c/Ji~ is an element of g;;(1,0)(U;~) satisfying dc/Ji~=O. Since U;~ is connected and simply connected, there exists an element hi~ of (!)(U;~) satisfying dhi~=c/Ji~; put Oi~(Z)= /;(n(z)) e( - hi~(Z))

for every Z in (:+1=)

U;~.

Then, if Z is in

Oi~(Z)=Oj,,(z)e (

Uj ,,, we have

U;~n

±

k,I=1

OCkIZk(e-il)l+const.).

In fact, if we apply" d log" to both sides, take the difference, and use the defmitions of gij' hi~' Oi~' (=#=) will be reduced to g

L

ockl(e -il), dzk=O; k,I=1 but this follows from the definition of c/Ji~. In the same way, if z is in U;o, hence z+~ in Ui~' we have Oi~(Z+ ~)= 0iO(Z) e(const.).

dgiAn(z)) -c/Ji~(Z)+ c/Jj.,(z)-

We choose an index io and extend 0ioO to the entire manifold Z by a repeated application of ( =#= ). Since Z is connected and simply connected, we get a well-defined element 0 of (!)(Z); clearly 0 is different from o. If Z is in U;~, we have

O(Z)=Oi~(z)e ( -.

±

J,k=1

OCjkZjek-tconst.).

64

II. Theta Functions from a Geometric Viewpoint

In fact, this is true by definition in U;o 0' i. e., for every z in Uio o' Suppose that it is true in Ui~ and let Uj~ denote any "immediate neighbor" of U;~, i.e., any Uj~ which overlaps U;~. Then we see quite easily by using (*) and the induction assumption that it is true also in Uj~' Therefore it is true in general. In particular, if z is in Uio 0' hence z + ~ in Uio ~, we have O(z)=0ioO(z)e(const.) and also

O(Z+~)=Oio~(Z+~) e ( Therefore, since we have

.

±()(jkZj~k+const.).

J,k=l

Oio~(Z+~)=OioO(z)e(const.),

O(z+~)=O(z) e (- .

±

J,k=l

we get

()(jk Zj ~k+const.) .

By the principle of analytic continuation, this remains true for every z in Z. Therefore 0 is a theta function on Z relative to L. Since the restrictions of 0 and/; 0 n to Ui~ differ only by a unit of (9(Ui~) for every i and ~, we have (O)=n*(D). q.e.d.

§ 3. The Automorphy Factor

u~(z)

We shall obtain necessary conditions satisfied by the automorphy factor u~(z)=e(Q~(z)+c~) of any theta function on Z relative to L. In § 4 we shall show that these conditions are sufficient for u~(z) to be the automorphy factor of some theta function on Z relative to L. We shall introduce some terminology: Let Z denote a vector space over C. A mapping (z, w) --+ Q(z, w) of Z x Z to C is called a quasi-hermitian form on Z x Z if it is C-linear in z, R-linear in w, and if

AQ(z, w)= 1/2i· (Q(z, w)-Q(w, z)) is in R for every z, w in Z. A quasi-hermitian form H(z, w) on Z x Z is called a hermitian form if

H(iz, w)+H(z, iw)=O for every z, w in Z. We shall show that if Q is quasi-hermitian,

H(z, w)= 1/2i· (Q(iz, w)-Q(z, iw)) is hermitian. It is clear that H(z, w) is C-linear in z and R-linear in w. Moreover, if we put AH(z, w)= 1/2i· (H(z, w)-H(w, z)), we have

AH(z, w)=t· (AQ(z, w)+ AQ(iz, iw)); hence AH(z, w) is in R for every z, w in Z. Therefore H is quasi-hermitian; and clearly H(iz, w)+H(z, iw)=O. We have shown also that AH(z, w)=

65

§3. The Automorphy Factor u~(z)

AH(i z, i w) for every z, w in Z. We shall show that if Q is quasi-hermitian, S(z, w)= 1/2i· (Q(iz, w)+Q(z, iw»)

is symmetric and C-bilinear in z, w. We have S(z, w)-S(w, z)=Aa(iz, w)+Aa(z, iw) =i(Aa(z, w)-Aa(iz, iw»);

hence S(z, w)-S(w, z) is in Rn iR=O. Therefore S(z, w) is symmetric in z, w, and C-linear in z, hence C-bilinear in z, w. Moreover we have Q(z, w)=H(z, w)+S(z, w).

Conversely, if H(z, w) is hermitian and S(z, w) symmetric and C-bilinear in z, w, their sum Q(z, w) is quasi-hermitian. We call Hand S the hermitian and symmetric parts of Q; we shall write H=Her(Q) and S=Sym(Q). Since we have Aa(z, W)=AH(Z, w), we shall drop the subscripts Q, Hand denote this simply by A (z, w). Then we have H(z, w)=A(iz, w)+iA(z, w),

A(iz, w)=A(iw,z),

andA(z, w)= -A(w, z); hence H(w, z)is the complex conjugate of H(z, w). The set of quasi-hermitian forms on Z x Z forms a vector space over R and "Her" gives the projection of this space to the subspace over R of hermitian forms on Z x Z. If H is a hermitian form such that H(z, z)~O for every z in Z, we write H~O; we write Hl~H2 if HI-H2~O. The set of hermitian forms H on Z x Z satisfying H ~ 0 forms a convex cone. Lemma 2. Let H denote a hermitian form on Z x Z with 1m (H) = A; suppose that H~O. Then a point Zo ofZ satisfies A(zo, w)=Ofor every w in Z ifand only ifH(zo, zo)=O.

Proof. Let Zo and Z~ denote the subsets of Z defined respectively by A(zo, w)=O for every w in Z and H(z~, z~)=O; the problem is to show that Zo = Z~. Since A (z, w) is R-linear in z, Zo is clearly a subspace over R. Since A is alternating and A(iz, w) symmetric in z, w, for every zo, w in Zo, Z we have A(izo, w)=A(iw, zo)= -A(zo, iw)=O;

hence izo is in Zo; and Zo is a subspace over C. We observe that A, hence also H, vanishes on Zo x Z and Z x Zo' Therefore Zo is contained in Z~. We shall prove the converse: Suppose that z~, w, A. are in Z~, Z, R; then we have H(iw+A.z~, iw+A.z~)=H(w, w)+2A. ·A(z~, w)~O.

This implies that

A(z~,

w)=O; hence z~ is in ZOo

q.e.d.

66

II. Theta Functions from a Geometric Viewpoint

Suppose that Z is finite dimensional and L a lattice in Z as in §§ 1-2; let H denote a hermitian form on Z x Z. We say that H is a Riemannform relative to L if A=Im(H) is Z-valued on Lx L and H~O. The Riemann form H is non-degenerate if and only if it is positive-definite, i. e., H (z, z) > 0 for every z=l=O in Z. By Lemma 2, this is the case if and only if A is nondegenerate. More generally, suppose that X is a fmite dimensional vector space over R, L a lattice in X, and A an alternating form on X x X which is Z-valued on Lx L. Then there exists an R-bilinear form B on X x X which is Z-valued on Lx L and satisfies A (x, y)=B(x, y)-B(y, x) for every x, y in X. In fact, if ~l' Xl' ... ,xm' YI' · .. ,Ym in R we have

in which A(~i' define B as

~)= -A(~j' ~J

B

... ,

~m

form a Z-base of L, for every

are all in Z. Therefore we have only to

CtIXi~i' it/i~i) =i~/(~i' OXih

On the other hand, we recall that a continuous mapping f of a locally compact commutative group Y to Ct such that (YI' Y2) -+ f(YI

+ Y2)/f(YI) f(Y2)

is a bicharacter of Y, i.e., a character of Yin YI and Y2' is called a second degree character of Y. If a mapping l/t of L to Ct has the property that X(~) = l/t (~) e(!· B(~, ~))

is a character of L, it is certainly a second degree character of L; we say that such a l/t is associated with A. We observe that if C is a symmetric Z-valued form on Lx L, ~ -+ e(!· C(~, ~)) is a character of L. Since an arbitrary" B" differs from the above B by such a C, the property of l/t being associated with A does not depend on the choice of B. Let Xo denote the set of those Xo in X satisfying A(xo, y)=O for every Y in X. Then Xo is a subspace of X over R; and A vanishes on Xo x X and X x Xo. Therefore there exists a unique alternating form, say A*, on (XjXo) x (XjXo) such that A is the reciprocal image of A* under the canonical homomorphism X -+XjXo .

Lemma 3. Put Lo = L n X 0; then Lo and LjLo are lattices in X ° and XjX o . If LI is a supplement of Lo in L and Xl the subspace of X spanned by LI over R, we get an R-linear isomorphism XI-+XjX O mapping LI

§3. The Automorphy Factor u~(z)

67

to LILo: the restriction Al of A to Xl X Xl is non-degenerate and it is the reciprocal image of A* under the isomorphism Xl -+XIXo' Proof. By deftnition, LILo is Z-free; hence there exists a supplement Ll of Lo in L. (We have only to choose a Z-base of LILo, replace each element of the base by a representative in L, and take as Ll the subgroup of L generated by the representatives; every supplement Ll can be obtained in this way.) Let Xl denote the subspace of X spanned by Ll over R. Then Ll is a lattice in Xl' We shall show that the restriction Al of A to Xl X Xl is non-degenerate. Since Al is Z-valued on Ll XLI' if Al is degenerate, the restriction of Al to Ll X Ll is degenerate; i.e., there exists at least one e l =4=0 in Ll satisfying A l (e l ,lh)=A(el ,'71)=0 for every '71 in L l • Since we also have A(el , '70)=0 for every '70 in L o , we get A(e l , '7) = 0 for every '7 in Ll + Lo = L. Since L spans X over R, this implies that e l is in LlnXO=O; but this is a contradiction. We shall show that Lo spans Xo over R. By deftnition, Lo spans a subspace X~ of Xo over R. Since L spans X over R, an arbitrary element Xo of Xo can be written in the form Xl +X~ with Xl in Xl and x~ in X~; we have Al (Xl' Yl)=A(x l , Yl)=A(x o, Yl)=O for every Yl in Xl' Since Al is non-degenerate, we get Xl =0; hence xo=x~, and hence Xo=X~. The rest is clear. q.e.d. We observe that A* and Al are Z-valued on (LILo) x (LILo) and Ll x L l . Suppose that a second degree character 1/1 of L is associated with A. We say that 1/1 is strongly associated with A if I/I(e + eo)=I/I(e) for every e in L and eo in Lo. In this case, 1/1 is the reciprocal image of a unique mapping 1/1* of LILo to C 1 under L-+LILo. We shall show that 1/1* is a second degree character of LILo associated with A*. Let 1/11 denote the restriction of 1/1 to L l . Then 1/11 is the reciprocal image of 1/1* under Ll-+LILo. Therefore we have only to show that 1/11 is a second degree character of Ll associated with AI' Let Bl denote an R-bilinear form on Xl XXI which is Z-valued on Ll XLI and satisftes Al(Xl'Yl)= Bl (Xl' Yl) - Bl (Yl' Xl); let B denote the reciprocal image of Bl under the product of X -+XIXo and XIXo-+Xl , Then B is an R-bilinear form on X x X which is Z-valued on Lx Land satisftes A(x, y)=B(x, y)B(y, x). Since 1/1 is associated with A, X(e)=I/I(e) e(!. B(e, e») is a character of L. If Xl denotes the restriction of X to Ll> we get Xl(el) = 1/11 (el) . e(!· Bl (el' el») for every el in L l . This proves the assertion.

Theorem 3. Let u~ (z) denote the automorphy factor of a theta function e on Z relative to L. Then there exist a quasi-hermitian form Q on Z x Z, a C-linear form I on Z, and a second degree character 1/1 of L satisfying u~(z) =e(1/2i) Q(z, e) +(1/4i) Q(e, e)+ l(e») 1/1 (e)

58

II. Theta Functions from a Geometric Viewpoint

e

for every z in Z and in L. The triple (Q, I, "') is uniquely determined by H = Her(Q) is a Riemann form relative to L; and", is strongly associated with A=Im(H). u~(z);

Proof If we write down the condition that u~(z)=e(Q~(z)+c~) is an automorphy factor, i.e., uN.,(z)=U~(Z+I1) u.,(z), we get Q~+.,(z)+c~+.,== Q~(Z+I1)+C~+Q.,(z)+C., mod 1

for every e,11 in L. This implies that

Q~(z)

is Z-linear in

eand

c~+.,==Q~(I1)+c~+C., mod 1,

e,

hence Q~ (11) == Q., (e) mod 1, for every 11 in L. Since L spans Z over R, for any fixed z, Q~(z) has a unique R-linear extension, say Qw(z), to Z; put

Q(z, w)=2i· Qw(z). Then Q(z, w) is C-linear in z and R-linear in w. Moreover we have

A(z, w)= 1/2i· (Q(z, w)-Q(w, z»)=Qw(z)-Qz(w). Since Q~(I1)==Qi~)mod1, A(e,l1) is in Z for every e,11 in L; hence A(z, w) is in R for every z, win Z. Therefore Q(z, w) is a quasi-hermitian form on ZX Z. As for c~, we put Then we have

d~=c~-t· Q~(e). d~+.,==t· A(e, 11)+d~+d.,mod 1

e.

for every e,11 in L. This implies that Im(d~) is Z-linear in Let r(z) denote the R-linear extension of Im(d~) to Z and I(z) the C-linear form on Zwith r(z) as its imaginary part, i.e.,

I(z)=r(i z)+ i r(z).

e

Then d~-I(~) is in R for every in L. On the other hand, let B(z, w) denote an R-bilinear form on ZX Z which is Z-valued on Lx Land satisfies A(z, w)=B(z, w)-B(w, z). Then

is a character of L. Therefore

is a second degree character of L associated with A. Putting these together, we get the first part of the theorem.

§3. The Automorphy Factor u~(z)

69

We shall show that Q, l,I/I are uniquely determined by u~(z). Since d log(u~(z)) = n . Q (dz, e), the restriction of Q to Z x L, hence Q itself, is uniquely determined by u~(z). On the other hand, if a C-linear form f on Z is R-valued, f(iz)=if(z) shows that f(z)=O. Since L spans Z over R, any R-linear form on Z is R-valued if it is R-valued on L. Therefore, if (Q, 1', 1/1') is another triple derived from u~(z), by applying what we have said to f = l-l', we get 1= l'; hence 1/1 = 1/1'. We shall show that H=Her(Q) is a Riemann form relative to L. We have seen that A = Im(H) is Z-valued on LxL. Therefore we have only to show that H~O. Let (Jo denote a trivial theta function on Z; , it is of the form (Jo (z) =const. e(1/4i) So(z, z) + lo(z)), in which So is a symmetric C-bilinear form on Z x Z and lo a C-linear form on Z. We get rid of the (unimportant) constant factor by assuming that (Jo takes the value 1 at z=O. We observe that "Q", "l", "1/1" for the product (Jo (J are Q + So, 1+ lo, 1/1. In particular, "1/1" and "H" (=Her(Q)) are unchanged. We say that (J is normalized if Sym(Q)=O and l=O. Since So and lo in (Jo are arbitrary, there exists one and only one (Jo such that (Jo (J is normalized. Lemma 4. If a theta function (J on Z relative to L is normalized, 1(J(z)1 exp( -nI2· H(z, z)) is a continuous periodic function on Z with L as a period group; hence . 1(J(z)1 ~ M . exp(nl2 . H(z, z))

for some constant M ~ O. Proof Since the continuity of the function is obvious, we shall prove its periodicity. Since (J is normalized, we have 1(J(z+ e)1 = 1(J(z)1 exp(n(Re(H(z, e))+!· H(e, e))) for every z in Z and e in L. This is equivalent to the periodicity in question. q.e.d. We go back to the proof of H~O. Since .. H" is invariant under the normalization-process, we shall assume that (J is normalized. If z, Zo are in Z and A in C, obviously we have

H(Z+A zo, Z+A zo)=H(zo, zo) IAI2 +O(A), in which O(A) denotes a quantity such that o (A)/A. is bounded as A--+OO, i. e., IAI-+- 00. We observe that as a function of A, (J(z + Azo) is holomorphic on C. If we have H (zo, zo) < 0 at some zo, by Lemma 4 and by the above relation, this holomorphic function tends to 0 as A--+ 00. Therefore, by the maximum modulus principle, it is the constant 0; hence (J(z)=O. Since z is arbitrary, we get (J=O; but this is a contradiction.

70

II. Theta Functions from a Geometric Viewpoint

Finally, we shall show that t/I is strongly associated with A. For the same reason as above, we shall assume that e is normalized. As in the proof of Lemma 2, let Zo denote the subspace of Z over C consisting of those Zo which satisfy A(zo, w)=o for every w in Z; then A and H vanish on Zo x Z and Z x Zo; hence, for every z in Z, Zo in Zo, and A. in C, we have H(z+A.zo, z+A.zo)=H(z, z). Therefore by Lemma4 we get le(z +A.zo)1 ~M . exp(n/2. H(z, z)); hence, for fIxed z, zo, e(z + A. zo) is a bounded holomorphic function on C. According to Liouville's theorem, it is a constant; hence e(z+zo)=e(z). On the other hand, we have

e(z +~) = e(z) e(1/2 i) H(z, ~) + (1/4 i) H(~, ~)) t/I(~) for every zin Zand ~ inL. If we replace eby~+~o for any ~o in Lo=LnZo, we get t/I(~ +~o)= t/I(~). Therefore t/I is strongly associated with A. q.e.d. If (Q, I, t/I) is the triple derived from the automorphy factor u~(z) of a theta function e, we say that e is of type (Q, I, t/I). We observe that e is normalized if and only if it is of type (H, 0, t/I) in which H = Her (H). Also, in the notations of the above proof, if e, H, and t/I are the reciprocal images of e*, H*, and t/I* under Z ~ Z/ Zo and L ~ L/L o , e* is a theta function (on Z/Zo relative to L/Lo) of type (H*, 0, t/I*). Clearly, every theta function of type (H*, 0, t/I*) can be obtained in this way. We observe that H* is non-degenerate.

§ 4. The Vector Space L(Q, I, t/I) We shall prove the converse of Theorem 3 with a quantitative supplement. Let Z/L denote a complex torus; suppose that Q is a quasihermitian form on Z x Z, I a C-linear form on Z, and t/I a second degree character of L such that H=Her(Q) is a Riemann form relative to L and t/I strongly associated with A = 1m (H). Then the converse ofTheorem 3 will just be the existence of a theta function (on Z relative to L) of type (Q, I, t/I). Let L(Q, t, t/I) denote the union of the set of theta functions of type (Q, I, t/I) and the constant 0. Then L(Q, I, t/I) forms a vector space over C; and the existence ofa theta function is equivalent to L(Q, t, t/I)=1=O, i. e., dime L(Q, I, t/I);;;; 1. We shall give a formula for this dimension involving only A and L. Again, let X denote a fInite dimensional vector space over R, L a lattice in X, and A an alternating form on X x X which is Z-valued on Lx L. Then the set L* of those x in X satisfying A(x, 11)=0 mod 1 for every 11 in L forms a closed subgroup of X. Since A is Z-valued on

§4. The Vector Space L(Q, /, l/I)

71

Lx L, L* contains L. Moreover the subspace Xo of X consisting of those Xo which satisfy A(xo, y)=O for every y in X is certainly contained in L*. Proposition 1. The group index [L*: L + Xo] is finite, and it is the square of an integer. Proof. If we use the notations in the proof of Lemma 3, we have L* = (L1)* + X o , in which (L 1)* consists of those Xl in Xl satisfying Al (Xl' '11)=0 mod 1 for every '11 in L 1. This implies that [L*: L + Xo] = [(L 1 )*: L 1 ].

Therefore we have only to show that [L*:L] is the square of an integer if A is non-degenerate. This is a well-known fact; but for the sake of completeness, we shall give the proof. Lemma 5. Let L denote a free Z-module offinite rank and A a Z-valued non-degenerate alternating form on L x L. Then there exist a Z-base ~1' ···'~2g of L and a set of positive integers e 1 , .•• ,eg , each e i dividing e i + 1 for 1 ~i
Proof. We shall exclude the trivial case where L=O and apply an induction on the rank of L. Since A is non-degenerate, for any ~ =F 0 in L there exists at least one '1 in L satisfying A(~, '1)=F0. By replacing '1 by -'1 if necessary, we may assume that A(e, '1»0. Consider the set of positive integers of the form A(~, '1) for ~,'1 in L. We have seen that this set is not empty; hence it contains a smallest element, say e1 • Let ~1,'11 denote elements of L satisfying A(~1,'11)=el. Then the set 1: of elements ~' of L satisfying A(~l' ~')=A('11> e')=0 forms a submodule of L; clearly (Ze1 +Z'11)n1:=0; we shall show that L=(Z~l

+Z'11)+1:.

Consider the homomorphism L--+Z defined by ~--+A(~l' ~). Then the image of L is a submodule of Z which contains e1 as the smallest positive integer; hence the image is e1 Z. For the same reason, the image of L under the homomorphism L-+Z defined by ~-+A(e,'11) is e1 Z. Therefore, for every ~ in L, ~'=~-A(~,'11)/el·el-A(~1,e)/el ·'11

is contained in 1:. This gives the direct sum decomposition of L. We observe that the restriction A' of A to 1: x.£ is non-degenerate. Since

72

II. Theta Functions from a Geometric Viewpoint

otherwise the proof is already finished, we shall assume that E =1= O. Then, for every q in Z and ~', r( in E, we have

A(q ~1 +~', 1]1 +I]')=q e1+ A'K, 1]'). Consequently, if A'(~', 1]') is not a multiple of e1, for ~ =q ~l + ~', I] =1]1 +1]' with a suitable q, we have O
L*= L(Zei1~i+Zei1~g+i); i=1

hence [L*:LJ=(e 1 ... eg )2.

q.e.d.

The Z-base ~1> ... , ~2g of L in Lemma 5 is called a canonical base of L relative to A. Since e1,el,e2,e2, ... ,eg,eg are the elementary divisors of the matrix (A(~i'~j»)i,j' they, hence also e1 , ... ,eg , depend only on Land A. The square root of [L*:LJ, which is equal to e1 ... eg , is called the Pfaffian of A. In the general case, the square root of [L*: L + X oJ is the Pfaffian of Al on LI x L I , which is also the Pfaffian of A * on (LjLo) x (LIL o); it is called the reduced Pfaffian of A relative to L. Theorem 4. Let ZIL denote a complex torus and (Q, I, l/I) a triple which satisfies the conditions in Theorem 3. Then the dimension of the vector space L(Q, I, l/I) over C is equal to the reduced Pfaffian of A= Im(Her(Q») relative to L.

Proof We recall that the multiplication by a suitable trivial theta function on Z gives a C-linear isomorphism of L(Q, I, l/I) to L(H, 0, l/I) in which H = Her (Q). We recall also that, in the notations of § 3, "taking the reciprocal image under Z-+ ZjZo" gives a C-linear isomorphism of L(H*, 0, l/I*) to L(H, 0, l/I). Since Zo is the subspace of Z over C consisting of those Zo which satisfy A(zo, w)=O for every w in Z, the reduced Pfaffian of A relative to L is the Pfaffian of A * = 1m (H*) on (Lj Lo) x (II Lo). Therefore we have only to prove the theorem in the case where H is non-degenerate. After t his reduction, we shall find a suitable symmetric C-bilinear form S on Zx Z and examine L(H +S, 0, l/I) instead of L(H, 0, l/I); these vector spaces are isomorphic under the multiplication of eo(z)= e(lj4 i) S (z, z»). We choose a canonical base ~l>'''' ~2g of L relative to A and denote by X' and X" the subspaces of Z over R spanned by ~l> ... ,~g and ~g+I""'~2g; X' and X" are both of dimension g. We shall show that there exists one. and only one S such that Q = H + S

§4. The Vector Space L(Q, /, l{!)

73

vanishes on Z x X". We first observe that if Y is a subspace of Z over R such that A vanishes on Y x Y, we have Y n i Y = 0. Otherwise, there exist z, w4=O in Y satisfying Z= iw; we have H(z, z)=A(iz, z)=A(z, w)=O. Since H is positive-definite, we get z=O; but this is a contradiction. Since A vanishes on X" x X", we can take X" as Y; hence X" + iX" is a subspace of Z over R of dimension 2 g; and hence X" + i X" = Z. Therefore, if a C-bilinear form on Z x Z vanishes on X" x X", it vanishes on Z x Z; clearly, this implies the uniqueness of S. We shall prove its existence. We take ell ~g+l' ... , e;l ~2g as a C-base of Z and identify Z with cg. Then we have H(z, w)=zhtw, in which h is a hermitian positive-definite matrix of degree g. Since A vanishes on X" x X", H is R-valued on X" x X"; hence the coefficients of h are in R. Consequently S(z, w)= -zhtw is a symmetric C-bilinear form on Zx Z and Q=H +S vanishes on ZxX". We shall examine the properties of Q. Since Q vanishes on Zx X", for every (w, x") in Zx X", we have Q(x", w)=Q(x", w)-Q(w, x")= 2iA(x", w). Therefore, for every z, win Zwe have

g

=2i· L zA: 1A(~g+i' w). i=l

Hence, if we put

2g W=

LYi~i

i=l

with Y1' ... , Y2g in R, we get

g

Q(z, w)= -2i· L ZiYi' i=l

For 1 ~ i ~ g, let

('til'"

't ig)

denote the coordinates of ~ i' i. e., put g

~i=

then we have

L 'tij( ej

1

~g+);

j=l

Q(~i' ~)= -2i't ij ·

Since A vanishes on X' x X', H, hence also Q = H + S, is symmetric on X' x X'. Therefore 't=('ti)i,j is a symmetric matrix. On the other hand, since we have Q(z, w)=z ht(w-w)= -2iz-h tIm (w) ,

74

II. Theta Functions from a Geometric Viewpoint

we get

Re(Q(z, w))=2Im(z h tIm(w))=2Im(z) h tIm(w).

Consequently, we have Re(Q(z, z))>o for every z in Z with Im(zHO, i. e., for every z in Z - X", hence a fortiori for every z:} in X'. This is equivalent to saying that

°

Im(T)=t· Re(Q(~i'

OL

is positive-definite. We have thus shown that T is a point of 6 g, the Siegel upper-half space of degree g. We note also that if we put z, w= ~1' ... , ~g in Re(Q(z, w))=2Im(z) h tIm (w), we get Im(T)= Im(T) h Im(T); this implies that h=Im(T)-1. We shall examine the special properties of the automorphy factor u~(z) for L(Q, 0, t/I), i. e., of every O:}O in L(Q, 0, t/I). Put 2g 2g ) g ( B i~1Xi~i' i~/i~i =i~1eixiYg+i' Then B is R-bilinear on Zx Z, Z-valued on L x L, and A(z, w)=B(z, w)B(w, z); hence is a character of L. Moreover we have u~(z)=e((1/2i) Q(z, ~)+(1/4i) Q(~, ~)+(1/2) B(~, ~)) X(~).

Since Q vanishes on ZxX" and B on X" xX", we have u~(z)=X(~) for every ~ in LnX". We choose m'=(m1 .. , mg), m"=(mg+1 ... m2g ) from Rg so that we have X

(.~ ni~i) =e (.f (ming+i-nimg+i)) ,=1 ,=1

for every n1 , ••• ,n2g in Z; m',m" are unique modZ g. Let 0 denote an element of L(Q, 0, t/I) and take g

as

~;

then we have

g

L:Pi~g+i= L:Piei(ei1~g+i) i=1 i=1

O(z+ ~)=O(z) X(~)=O(z) e(m' tp) for every P=(P1 ... Pg) in zg. Hence, if we denote by e the diagonal matrix of degree g with ei as its i-th diagonal coefficient,

01 (z)=O(z) e( - m' e- 1 t z) is a holomorphic function on Z, which is periodic with zg e as a period group. Therefore 01(z) becomes a holomorphic function of e(zt!e1 ), ••• , e(zg/eg), and it can be expanded into a "Laurent series." which converges absolutely and uniformly on every compact subset of (C x )g. Since this

§4. The Vector Space L(Q, /, t/I)

75

is the Fourier series of the restriction of 01 to K", by which 01 is uniquely determined, we call it the Fourier expansion of 01 , If we multiply e(m' e- 1 t z) to both sides, we get O(z)=

c(r) e(r+m'e- 1) tz);

L ,eZ B e- 1

the series converges absolutely and uniformly on every bounded subset ofCg. We shall examine the coefficients c(r). Take g

as

e; then we have

g

LPie i = L Pi7:ij(e;1eg+ j) i=1 i,j=1

O(z+e)=O(z) e( -p tZ -!P7: tp _ p tm")

for every P.=(P1 ... p,) in zg. By the uniquen~ss ofthe Fourier expansion, the coeffiClents of e\(r+m' e- 1) tz) on both SIdes are equal; hence c(r) e(r+m' e- 1) 7: tp)=c(r+ p) e(

-! p7: tp -

p w")

for every r in zg e- 1 and p in zg. Consequently, if we put Om'm,,(7:, z)= L

we have

peZa

O(z)=

e(!(p+ m') 7: t(p+m')+(p+m')'z+m"»), L

,modza

const. O,+m'e-1 ",,,(7:, z),

in which r runs over a complete set of representatives of Z g e- 1 /Z g ; the "const." depends on 7: but not on z. We recall that, except for a minor notational difference, 0m'm,,(7:, z) is the theta series in Chap. I, § 10. We have seen that 0m'm,,(7:, ) is a holomorphic function on cg and satisfies 0m'm,,(7:, z+n' 7:+n")=Om'm,,(r, z) e( -tn' 7: tn' -n' tz ) e(m' tn" -n'lm")

for every n', n" in zg. Therefore each O,+m'e-1m,,(7:, ) is an element of L(Q, 0, 1/1) and, again by the uniqueness of the Fourier expansion, the [Zg e- 1 :Zg]=e1 ... eg of these under the summation sign are linearly independent over C. Hence we have dime L(Q, 0, 1/I)=e1 ... ego q.e.d. If 0i is a theta function (on Z relative to L and) of type (Q;, I;, I/Ii) for i = 1, 2, clearly 0102 is a theta function of type (Q1 + Q2' 11 + 12 ,1/111/12)' We observe also that the set of Riemann forms (on Z x Z relative to L) forms an additive monoid with the constant 0 as its identity element.

Corollary. Let 0 denote a theta function of type (Q, I, 1/1) and put H = Her (Q). Then 0 - H gives a surjective homomorphism of the multiplicative monoid of theta functions to the additive monoid of Riemann forms with the group of trivial theta functions as its kernel.

76

II. Theta Functions from a Geometric Viewpoint

We have only to show that H =0 only if 8 is trivial. Since H is invariant under normalization, we may assume that 8 is normalized. Then what we have to prove is that H =0 only if 8 is a constant. We observe that a second degree character (of L) is strongly associated with if and only if it is 1; also L(O, 0, 1) = C. This implies the assertion. A theta function 8 on Z relative to L can very well be a theta function (on Z) relative to a larger lattice. In fact, if 8 is of type (Q, I, 0/) and H =Her(Q) degenerate, this is always the case. We shall show that if H is non-degenerate, this does not happen for any "general" 8 in L(Q, I, 0/). We shall prove the following simple lemma:

°

Lemma 6. If 8 is a theta function of type (Q, I, 0/) and a in Z, then 81(z)=8(z+a) is a theta function of type (Q, 11> 0/1)' in which 11(z)=/(z)+ 1/2i· Q(z, a),

0/1(~)=0/(~)e(A(a, ~)).

Proof Since 8 is a theta function of type (Q, I, 0/), we have 8(z +~) = 8(z) e(1/2 i) Q (z, ~) + (1/4 i) Q (~, ~) + I(~)) o/(~)

for every z in Z and ~ in L. By replacing z by z + a in this, we see that the automorphy factor of 81 has e(1/2i) Q(a,~)) as an extra factor. We have Q(a,z)=Q(z,a)+2iA(a,z); Q(z, a) is C-linear in z and e(A(a,~)) gives a character of L. This implies the lemma. q.e.d. Theorem 5. Let 8 denote a theta function of type (Q, 1,0/) and A = 1m (Her (Q)). Then every element a of Z for which 8 and 81 defined by 81(z)=8(z+a) differ by a unit of (!)(Z) belongs to the subgroup L* of Z defined by A(z,~) == mod 1 for every ~ in L. If A is non-degenerate, there exists a finite number of subspaces of L(Q, I, 0/) of smaller dimensions such that if 8 does not belong to their union, every such a is contained

°

in L. Proof If 8 and 81 differ by a unit of (!)(Z), this unit is necessarily a trivial theta function. Therefore it is of the form

80 (z) = const. e(1/4 i) So (z, z) + 10 (z)). By Lemma 6, 81 is of type (Q, 11,0/1); and 80 8 is of type (Q + So, 1+ 10 ,0/). Since 81 =8 0 8, we get So=O, 11 =1+/0' 0/1 =0/; hence 10(z)=1/2i·Q(z,a), and e(A(a, ~)) = 1 for every ~ in L. This proves the first part. Assume that A is non-degenerate; then L*/L is a finite group. For every 8=l=0 in L(Q, I, 0/), consider the set L* of elements of Z such as a; then L* forms a subgroup of L* containing L; and 8 becomes a theta function relative to L*. Suppose that 8 is of type (Q*, 1*,0/*) relative to L*. Since Q* and Q have the same restriction to ZX L, we have Q* =Q; similarly 1* =1. As for tjJ*, we have 0/* =0/ on L; moreover,

§4. The Vector Space L(Q, I, 1/1)

77

since 1/1* is associated with A, if B* is an R-bilinear form on ZX Z which is Z-valued on L* x L* and satisfies A(z, w) = B* (z, w) - B* (w, z) for every z, w in Z, X* (a) = 1/1* (a) e(t· B* (a, a») is a character of L*. After observing these, we consider all subgroups L* of L* containing L but different from L for which B* as above can be found; the number of such subgroups is fmite. We consider all characters X* of L* which extend the character X of L defmed by

X(e) = 1/1 (e) e(t· B*(e, e»); the number of such extensions is finite. For each X*, we derme 1/1* by the above formula and consider L(Q, I, 1/1*) relative to L*. We observe that (L*)*, defined similarly as L*, is contained in L*; hence the Pfaffian of A on L* x L* is smaller than the Pfaffian of A on L x L. (In fact, they differ by the factor [L* :L].) Therefore by Theorem 4 we get dimcL(Q, I, I/I*)
Remark. We shall examine the case where Her(Q) in L(Q, I, 1/1) is non-degenerate, Le., the case where L*/L is a finite group. We shall show that V=L(Q, 1,1/1) can be considered as an irreducible module of a certain extension of Cf by L*/L: We choose a canonical base 1 , ••• , 2g of Lrelative to A=Im(Her(Q») and define an R-bilinear form B on Z x Z as

e

2g

2g

i_1

i_1

B ( Lxie i , Lyie i

e

)

g

= LeixiYg+i; i_1

then we have A(z, w)=B(z, w)-B(w, z) for every z, w in Z. Consequently

x(a)=I/I(a) e(t. B(a, a») for a in L defines a character of L. We extend X to a character of L*; denoting the extension also by X, we define I/I(a) for every a in L* by the above formula; finally, we put ua(z)=e(1/2i) Q(z, a)+(1/4i) Q(a, a)+ I(a») 1/1 (a). Then ua(z) for a=e in L coincides with our previous u~(z). We take (a, t) arbitrarily from L* x Cf and, for every e in V=L(Q, I, 1/1), we put (U(a, t) 0) (z)= t· Ua(Z)-l e(z+a); then we get (U(a, t) e) (z + e)=e(A(a, e») u~(z) (U(a, t) e) (z) = u~(z) (U(a, t-) e) (z)

78

II. Theta Functions from a Geometric Viewpoint

e

for every in L; hence U(a, t)() is in V. By Theorem 5, we have U(a, t)()=() for every () in V if and only if (a, t) is in LxI; moreover, we have

1, 1)

U(a t (U(a

2,t 2)())= U(a1+a2,e( -B(a1,a2») t1 t 2)()

for every a1 , a 2 in L* and t 1 , t2 in Cf. We shall incorporate some observations in Chap. I, § 5. If we put

it1 Z(ei 1ei>/it1 Z ei, i, y* = it1 Z(ei 1e +i)/it1 Z e + Y=

g

g

we have L*/L= Y x Y*. Moreover, if y, y* are elements of L Z(ei 1

L Z(ei eg+J, 1

i

e

i ),

B(y, y*) mod 1 depends only on the images of y, y* in

i

Y, Y*. If we denote them also by y, y* and put (y, y*)=e( -B(y, y*») ,

this converts y* into the dual of Y. Finally, if we put U(y, y*), t)= U(y+ y*, t),

the new U gives a representation of A(Y)=(Yx Y*) x Cf in v. As we shall see in § 5, V can be converted in a natural manner into a Hilbert space; and U becomes a unitary representation of A(Y) in V. By Theorem 4, we have dimdV)=e1 ... eg=card(Y). Therefore, by Chap. I, Proposition 2, the unitary representation U is irreducible.

§ 5. A Change of the Canonical Base Let H denote a non-degenerate Riemann form on Z x Z relative to Land IjJ a second degree character of L associated with A=Im(H); let 1 , ... , 2g denote a canonical base of L relative to A and X" the subspace of Z over R spanned by g+ 1, ... , 2g . Then, as we have seen in §4, there exists a unique symmetric C-bilinear form 8 on ZX Z such that Q=H +8 vanishes on ZX X"; we have also found a particular C-base of L(Q, 0, 1jJ). We shall examine the way how this particular base depends on the choice of the canonical base. We have introduced coordinates in Z with respect to e1 1 eg + 1 , ... , e;-1 2g : If Z is a point of Z, its coordinates (Z1 ... Zg) are defined by

e

e

e

e

e

g

Z=

L zi(ei 1 eg+J.

i=1.

§5. A Change of the Canonical Base

79

We have seen that the matrix r with the coordinates of ~i as its i-th row vector for 1 ~ i ~ g represents a point of 6 g' Furthermore, if we do not distinguish z from (Zl ... Zg), we have

H(z, w)=z h tw,

S(z, w)= -z h tw

for h=Im(r)-l. Conversely, if r is an arbitrary point of 6 g, the row vectors of r and the diagonal matrix e (with ei as its i-th diagonal coefficient) are linearly independent over R; hence these 2g vectors generate a lattice in e g. Moreover the hermitian form H(z, w)=z Im(r)-l tw on eg x eg defines a non-degenerate Riemann form relative to this lattice: Clearly H is positive-definite. If A = 1m (H), we have A(xr,yr)=O,

A(xr,ye)=xety,

A(xe,ye)=O

for every x, y in Rg; hence A is Z-valued on the product of the lattice in question. Suppose that H is an arbitrary Riemann form on Z x Z relative to Land rjJ a second degree character of L strongly associated with A = Im(H). Then, for every (}1'(}2 in L(H,O,rjJ), (}1(Z)(}2(z)exp(-nH(z,z)) is a continuous periodic function on Z with L as a period group. This follows from Lemma 4. Therefore we can derme the scalar product ((}l' (}2) of (}l' (}2 as the integral of this function over Z/L with respect to a Haar measure on Z/L; we shall use the Haar measure normalized by the condition that the total measure is 1. Then L(H, 0, rjJ) becomes a finite dimensional Hilbert space over C. If L(Q, I, rjJ) is arbitrarily given, there exists a unique trivial theta function (}o satisfying (}o (0) = 1 such that the multiplication by (}o maps L(Q, I, rjJ) to L(Her(Q), 0, rjJ); if we put S=Sym(Q), we have (}o(z)=e( -(1/4i) S(z, z)-l(z)). We shall convert L(Q, I, rjJ) into a Hilbert space so that the mapping becomes unitary. For instance, if we are in the situation that we have recalled in the beginning, the element of L(H, 0, rjJ) which corresponds to an element () of L(Q, 0, rjJ) is

(}(z) e( -(1/4i) S(z, z)) = (}(z) exp(n/2. z Im(r)-l tz). Moreover we have Re(z Im(r)-l tz)_ H(z, z)= -2Im(z) Im(r)-l tlm(z). Therefore, for every (}1' (}2 in L(Q, 0, rjJ) we have ((}1'(}2)=

J

R2g/Z2g

(}l(Z) (}2(z)exp (-2nxlm(r)tx)dxdy,

80

II. Theta Functions from a Geometric Viewpoint

in which z=xr+ye and dx=dx 1 ... dxg, dy=dY1'" dYg. We recall that if we put . 8m'm,,(r, z)=

L e(t(P+ m') r t(p+m')+(p+m') t(z + mil)),

PEZg

the particular base of L(Q, 0, ljJ) consists of 8r+m'e-'m,,(r, ), in which r runs over a complete set of representatives of Z g e- 1/Z g• Lemma 7. The particular C-base of L(Q, 0, ljJ) consists of orthogonal vectors of the same squared length det(2Im(r))-t. Proof. Suppose that r, s are elements of zg e- 1 and denote by 81 ,8 2 the corresponding elements of L(Q, 0, ljJ). If p, q are the summation indices for the theta series 81 ,8 2 , (0 1 , ( 2 ) is an integral with respect to dx dy of a double series with p, q as summation indices. Since this

double series is uniformly convergent on the domain of integration, which has finite measure, the integral and the summation can be interchanged. (This is also a good case where the Fubini theorem can be applied.) Therefore (0 1 , ( 2 ) becomes a series of integrals with p, q as summation indices. In each integral, the integrand is a product of e((p+r-q-s) e ty) and a factor which does not depend on y. The integral of the first factor over Rg/zg with respect to d y is 1 for p + r = q + s and 0 otherwise. Consequently, if r$s mod 1, we get (8 1 , ( 2 )=0. This proves the orthogonality of distinct elements of the particular base. In the case where r=s mod 1, we have 01 =02 ; denote this by O. Then, if we put n'=r+m' e- 1, we have (0,0)=

L

J exp( -2n:(x+ p+ n') Im(r) t(x + p+ n')) dx

PEZg Rg/Zg

=

Jexp(-2n:x Im(r)tx)dx.

Rg

By Chap. I, Theorem 1, this last integral is equal to det (2 1m (r)tt.

q.e.d.

For the sake of completeness, we shall prove two well-known simple lemmas; the first of these can be considered as a generalization of the maximum modulus principle: Lemma 8. Let f1' ... , fN denote holomorphic functions on a connected open subset U of cn. If the continuous non-negative function N

¢(z)=

L 1.t;(z)1 2

on U attains its maximum at some point of U, fl' ... ,fN are all constants. Proof. Suppose that ¢ attains its maximum at some point a of U; by applying the translation z -+ z + a, we may assume that a = O. If

§5. A Change of the Canonical Base

81

we take a sufficiently small p >0, the polycylinder defined by IZll ~ p, ... , IZnl~P is contained in U. For each i, let

J;(z) =

I

aip zP

p

denote the Taylor expansion of J;, in which P=(P1 ... Pn) and zP=

Zf' ... z~n; then, by assumption, we have N

N

cP(O)= IlaiOI2~cP(z)= IJ;(z)J;(z) i=1 i=1 for every z in U. If we integrate both sides over the above polycylinder with respect to a Haar measure on en, we get N

I

i=l

N

laiOl2~

I I

i=l

laip l2 p2Iplj(p + 1),

p

in which Ipi = PI + ... + Pn and p+ 1 =(Pl + 1) ... (Pn + 1). This implies that aip=O for every p=I=O, hence J;(z) = aiO for 1 ~ i~N. Since U is connected, this relation holds everywhere on U. q.e.d. Lemma 9. Let X denote a non-empty set, K a field, and fl' ... ,fN mappings of X to K which are linearly independent over K. Then there exist N elements Xl' ... , XN of X satisfying

det( JiI". (x.)) . . = det( JiI".(X·))1 <. ·
Proof Since the lemma is true for N = 1, we shall assume that N ~ 2 and apply an induction on N: We choose N -1 elements Xl' ... , XN - l of X satisfying d = det (J; (X))1 ;;;i,j
dfN(x)+

I

diJ;(x)=O

for some d1 , ••• , dN -1 which are independent of X; but this is a contradiction. q.e.d. After these preliminaries, we go back to the situation that we had in the beginning. If ~~, ... , ~~g form a Z-base of L, there exists a unique element (J of GL 2g (Z) such that the column vector with ~t· as its i-th entry is obtained from the column vector with ~i as its i-th entry by the left multiplication by (J. Moreover ~~, ... , ~~g form a canonical base of L relative to A if and only if (J

(oe) (-eoe)! 0 -e -0 . (J=

82

II. Theta Functions from a Geometric Viewpoint

If u is composed of submatrices IX, p, 1', J of degree g, this is the case if and only if (I. e Ip, I' e IJ are symmetric and (I. e IJ - Pe II' = e. In this

case, everything that we have defined and proved using ~l' ... , ~2g can be defmed and proved using ~t, ~fg. In fact, we have only to put "sharp" everywhere. In order to avoid confusion, we shall use (z) and (z#), just for a moment, to denote coordinates of a point z of Z. Then, there exists an element v of GLg(C), which is independent of z, such that (z#)=(z) v; if we express the relation between the two canonical bases in terms of their coordinates, we get

... ,

i.e.,

.#=«(I..+pe)v,

e=(y.+Je)v;

hence I' • + J e has an inverse, and (z#)=(z)(yo+J e)-l e,

.# = «(1..+ p e)(y. +J e)-l e.

Since the relation between. and .# is symmetric, this defines a biholomorphic mapping of 6 g to itself. On. the other hand, if we express H(z, z) by using (z) and (z#), we get v Im(.#)-l 'V=Im(.)-l.

Let () and ()# denote the elements of L(Q, 0, rfJ) and L(Q #,0, rfJ) which correspond to the same element of L(H, 0, rfJ). Then, using the complex conjugate of the above relation between Im(.#)-l and Im(.)-l, we get ()# (z)= ()(z) e({1/4 i) (S# (z, z) - S(z, z)))

= ()(z) e(t(z#) Im(v- 1) Im(.)-l I(Z)). Since we have Im(v- I )=Im(e- 1 (y.+Je))=e- 1 y Im(.) and (z#)= (z)(y.+Je)-le, we get ()# (z)= ()(z) e(t(z#) e- 1 I' I(Z)) =e(z) e(t(z) (I'. +J e)-l y t(z)).

Therefore, if we denote (z), (z#) simply by z, z# and put (m')# = n', (m") # =n", we have ()r+n' e- 1 n"(.#' z#) = e(tz(y. + J e)-l y'z)

L

smodi

Crs ()s+m' e-1

m"(.' z),

in which the coefficients Crs for r, s in Z g e- I are independent of z; we shall see that they are holomorphic functions on 6 g.

§S. A Change ofthe Canonical Base

83

Let 'to denote an arbitrary point of e g • If we apply Lemma 9 to the set of N=[Zg e- 1 : Z~ functions 0s+m'e-'m"('t o , ) on x=cg indexed by s mod 1, we see that there exist N elements Xl' ••• , X N of X satisfying det(0s+m'e-'m"('t o , xj))•./=t=O. We take 't, Zl' ••• , ZN in some respective neighborhoods of 'to, Xl' ••• , x N and replace Z in the above relation involving Cr. by Zl' ••• , ZN; we can uniquely solve this system of N linear equations in the unknown Cr.' Since 't* and z* depend holomorphically on 't, Z, each Cr. becomes a quotient of holomorphic functions ofT, Zl' ••• , ZN such that the denominator does not vanish at 't='to, Zl =X 1 , ••• , ZN=XN , Therefore cr. is a holomorphic function ofT, Zl' ••• , ZN in possibly smaller neighborhoods. Since Cr. depends only on 't, it is a holomorphic function on some neighborhood of 'to. Since 'to is arbitrary, Cr. is a holomorphic function on e g • On the other hand, if 0 and 0* are elements of L(Q, 0, I/t) and L(Q*,O, I/t) which correspond to the same element of L(H, 0, I/t), 0 - 0* gives a unitary mapping of L(Q, 0, I/t) to L(Q*, 0, I/t); this follows immediately from the definition. Therefore, by applying Lemma 7 to the relation involving Cr.' we get br ,r2 det(2Im('t*))-t =

L

,mod1

cr "

Cr2 ,

det(2 1m ('t))-t,

in which br ,r2 = 1 for r1 == r2 mod 1 and br ,r2 =0 otherwise. Since we have

v Im('t*)-l tv = 1m ('t)-l, we get det(Im('t*))-t Idet(v)1 = det(1m ('t)tt.

Putting these together, we get

L

,mod1

cr"

Cr2 ,

Idet (v)1 = br ,r2'

Since e g is connected and simply connected, the square roots of det(v) = det(Y't+be)-le) give rise to two holomorphic functions on e g • We choose one of them and put ur,=det(y 't+b e)-l e)t Cr.'

Then, by what we have shown, (u r ,)", forms a unitary matrix. Since the entries of this matrix are holomorphic functions on e g , Lemma 8 shows that they are all constants. We shall examine the relation between m=(m'm") and m* = (n'n"). If ~ is an arbitrary element of L, we have 2g

2g

~= LPi~i= LPt~t i=l

i_1

84

II. Theta Functions from a Geometric Viewpoint

for some Pl"'" P2g and p!, ... , pfg in Z. We put p' = (PI ... Pg), p" = (Pg+I ... P2g) and q'=(Pt ... p:), q"=(P:+I ... pfg). Then we have 1jJ(~)= x(~) e(!. B(~, ~))= x* (0 e(!· B* (~, ~)).

We recall that x(~)=e(m'

Ip" - p' 1m"),

B(~, ~)=p'

e Ip";

x* (~), B* (~,~) are similar. Therefore we get

We recall that the column vector with ~i* as its i-th entry is obtained from the column vector with ~i as its i-th entry by the left multiplication by CT; hence (P' p")=(q' q") CT. In the above congruence relation, we replace p', p" by q' a + q" y, q' {J + q" {j. If (a e I{J)O denotes the row vector with the i-th diagonal coefficient of a e I{J as its i-th coefficient, since a e I {J is symmetric, we have

similarly we have

! q" y e l{jlq" =!(y e 1{j)O Iq" mod 1. Therefore, by using a e I{j - {J e Iy = e we get n'=m' l{j -m" Iy+!(y e 1{j)O'

n"= -m' I{J +m" la+!(a e t{J)o mod l.

We have thus obtained the following theorem:

Theorem 6. Let (r, z) denote an arbitrary point of 6 g x cg and a, {J, y, (j elements of Mg(Z) satisfying

(ay (J){j ( -e0 0e) I(rxy

(J)

{j =

( 0

e)

-e 0 .

Then, for every m', m" in Rg, we have 8r+n'e- ' n"('*' z*)=e(!z(y, + in which

(j e)-I

y tz) det(e- I (y, +

.smod L 1 urs 8s+m'e- ' m"("

(,*, z*)=((a, +

(j e»)t

z),

(J e) (y, +{j e)-I e, z(y, +(j e)-I e),

(n'n")=(mlm,,)I(_; -:) +!((yel{j)o(ael{J)o); and (urs)r,s is a constant unitary matrix of degree det(e).

§5. A Change of the Canonical Base

85

In the special case where e= 19, we have the following simpler situation: Corollary. Let (or, z) denote an arbitrary point of 6 g x cg and u an element of SP2g(Z) composed of IX, (3, y, b. Then, for every m=(m'm") in R 2 g, we have Omll (or * ,z*)=e(t z(y 't' + b)-l Y tz) det(y 't' +b)t . U Om('t', z), in which

('t'*, Z*)=((IX 't' + (3)(y 't' +b)-l, z(y 't' +b)-l), m* =mu- 1 +t((y tb)o(lXt{3)o);

and u is an element of C{ independent of 't', z.

Chapter III

Graded Rings of Theta Functions § 1. Graded Rings We shall collect several lemmas on graded rings, some of which will become necessary only in later chapters. If S is a commutative ring (with the identity) containing a sequence of additive subgroups So, Sl' ... such that S is their direct sum and SpSqcSpH for every p, q, we say that S is a graded ring over So; we observe that So is a subring of Sand Sk an So-module. If M is an S-module containing a sequence of additive subgroups M o, M I , ... such that M is their direct sum and SpMqcMp+q for every p, q, we say that M is a graded S-module with Mk as its homogeneous component of degree k. An element of Mk is called a homogeneous element of M of degree k; every element x of M can be written uniquely as XO+x I + ... with Xk in M k, in which xk=O for almost all k; we call x k the homogeneous part of x of degree k. A graded S-module N contained in M is called a graded submodule if it is a submodule and Nk=N nMk for k=O, 1, .... We observe that S itself is a graded S-module. If K is a commutative ring and K [x] = K [Xl' ... ,xn] is the ring of polynomials in n letters Xl' ... ' xn with coefficients in K, K [x] can be considered as a graded ring over K with K[X]I=Kxl+···+Kx n ; a graded submodule of K[x] is called a homogeneous ideal of K[x]. A graded ring R contained in S is called a graded subring if it is a subring and a graded submodule. For any positive integer d, we shall denote by S(d) the subring of S generated by Skd; S(d) is the direct sum of So, Sd' ... and SpdSqdcS(P+q)d for every p, q. Therefore S(d) can be considered as a graded ring over So with Skd as its homogeneous component of degree k. In general, let A denote a commutative ring and M an A-module. We recall that M is called an A-module of finite type, or just a finite A-module, if every element of M can be written as a linear combination of a fixed finite number of elements of M with coefficients in A. Let B denote a commutative ring which contains A as a sub ring. Then B is called a ring of finite type, or finitely generated, over A if every element of B can be written as a polynomial in a fixed finite number of elements

87

§ 1. Graded Rings

of B with coefficients in A. There are two finiteness conditions for the graded ring S: One is the "componentwise finiteness" requiring that Sk be a fmite So-module for every k. Another is the "ring-fmiteness" requiring that S be finitely generated over So. The ring-fmiteness implies the componentwise fmiteness; but the converse is false. If S satisfies the componentwise finiteness and So is a field, we shall denote by dim(SJ the dimension of the vector space Sk over So. We observe that the additive structure of S is determined by the following formal power series:

which is called the "generating function" of S. We shall be concerned primarily about the ring-fmiteness of graded integral domains over So=c. Let S + denote the ideal of S generated by Sl' S2' ... , i. e., put S+=Sl +S2+····

Then, if S+ is a fmite S-module, S is fmitely generated over So; more precisely, we have the following lemma: Lemma 1. Let Xl' .•. , xn denote homogeneous elements of S such that S + = S Xl + ... + S Xn. Then we have

Proof. We have only to show that every element of Sk is contained in So [X] =So[X p .•• ,xJ for k=O, 1, .... Since this is trivially true for k = 0, we shall assume that k ~ 1 and apply an induction on k. We observe that if mi is the degree of Xi' we have mi>O. Let a denote an arbitrary element of Sk. Since a is in S+, we have a=b1 Xl + ... +bnxn with bi in S. Let b; denote the homogeneous part of bi of degree k-mi ; we take b;=O for k-mi
By the induction assumption, all b; are in So [x], hence a is in So [x]. q.e.d. Lemma 2. Let (c ij) for 1 ~ i ~ m, 1 ~j ~ n denote a set of m n integers; consider the following set of m homogeneous linear equations: n

'~::CijXj=O

j=l

in the unknown non-negative integers of all solutions is finitely generated.

Xl' ... ,

Xn. Then the additive monoid

88

III. Graded Rings of Theta Functions

Proof We say that an element of any additive monoid is "minimal" if it can not be written as a sum of two elements of the monoid which are different from O. We observe that any monoid of non-negative integer n-tuples is fmitely generated if and only if the set of minimal elements is fmite. After this remark, we shall settle the case where m = 1. We change our notation and write the given equation in the form P

q

i_I

j_1

"£ . .e~o x~ e'! x'!, Il = " £..oJJ

in which e;, e'j are all positive integers. We shall exclude the trivial case where p=O or q=O. Let x=(x' x") for x' =(x~ ... x~),

x" =(x~ .. , x~)

denote an arbitrary element of the monoid. If, for some i, x;.is larger than ... ,e~ and, for some j, x'j not smaller than e;, x becomes the sum of the following two elements: e~,

(x~

'" x; -e'j .,. x~x~ ... x'j -e; ... x~),

(0 ...

e'j .,. 0

o. ..

e; ... 0);

these are elements of the monoid and different from O. Let le'l = le'l"" denote the maximum of e~, ... , e~; let le"l, Ix'l, Ix"l be similar. Then, as we have seen, if x=(x' x") is a minimal element, Ix'I>le"l implies that Ix"l < Ie' I. We observe that the set of elements x of the monoid satisfying either Ix'i ~ le"l or Ix"l ~ le'l is bounded (in Rp+q), hence fmite. Therefore the set of minimal elements is fmite. We shall consider the case where m~2 and apply an induction on m. Consider the additive monoid of solutions of the m-th equation and let ~k=(~k1 ... ~k,,) for 1 ~k~r denote a finite set of generators. Then every element x of the original monoid can be written as

with non-negative integers YI' ... , Yr such that

±

j=1

eijxj=

±(± Cij~kj)

k= I

j= I

Yk=O

for l~i
L ~k'1ik k=1 for 1 ~ i ~ N form a set of

gene~ators

of the original monoid. q.e.d.

§ 1. Graded Rings

89

Lemma 3. Suppose that a graded ring S is finitely generated over So. Then S(d) is also finitely generated over So: moreover, for a suitable d we have S(d)=SO[Sd].

Proof We shall exclude the trivial case where S+ =0. By assumption, there exists a finite number of elements Xl' ... 'X n of S such that S = So [Xl' ... , Xn]. Since S is generated over So by the homogeneous parts of Xl"'" X n , we may assume (by changing the notation) that each Xi is homogeneous of degree mi;;;; 1. Consider the following equation: n

LmjPj-dq=O

j_l

in the unknown non-negative integers Pl' ... ,Pn,q and denote by M(d) the additive monoid of all solutions; M(d) is rmitely generated by Lemma 2. Let (PH ... Pin qJ for 1 ~ i~N denote a finite set of generators of M(d) and put n Y j-- TIxPiJ' j' j=l

then we have S(d) = So [Yl' ... , YN]' This proves the first part. We shall prove the second part. Let m denote the least common multiple of ml , ... , mn ; consider the finite subset E of M(m) consisting of those (Pl'" Pnq) in which O~Pj
(q-times)

for q = 1, 2, .... Since this is trivially true for q = 1, we shall assume that q;;;;2 and apply an induction on q. Let (Pl'" Pnq) denote an arbitrary element of M q • For eachj, we write Pj=mlmj · ej+Pj, in which ej is an integer and O~pj
q' = 11m· L mjPj. j=l

Then we have

n

qQ= Lej+q', j=l

hence q' is an integer. Therefore

(P~

... p~ q') is an element of E, hence + ..~ + en;;;; Q. Hence we can

q' ~ Q. Since q;;;; 2, this implies that el

90

III. Graded Rings of Theta Functions

choose (in many ways) non-negative integers et, ... ,e: satisfying for 1 ~j ~ n such that n qQ= LeT+Q; j_l put Pj=mlmj · eT +p'j.

eT~ej

Then (Pi ... Pnq) decomposes into the sum of the following two elements: (p~ ... p~ 1),

which are respectively in M q _ l and Mi' We have thus shown that Mq is contained in M q_ l +Ml =Ml + .. ·+Ml . This implies that M q= Ml + .. ·+Ml ; and the induction is complete. It follows from what we have shown that S(4)=SO[S4]' q.e.d. According to the above proof, the positive integer d such that (S(4»1 =S4 generates S(d) over So can not be taken arbitrarily large; it has to be a multiple of "m". This limitation lies in the nature of things as the following example shows: Example. Consider the ring of polynomials S=K[x,y] in two letters x, y of respective degrees 2, 3; the commutative ring K is arbitrary. Then Sk is the free K-module generated by x P ~ in which 2p+3q=k; we have So=K, Sl =0, S2=Kx, S3=Ky, ... and, in general, S2k=Kxk+Kxk- 3 y2+ "', S2k+l =S2k-2Y' In the notations of the proof of Lemma 3, "m" = 6 and "E" consists of 0 only; hence d can be any (positive) multiple of 6. We shall show that this (sufficient) condition is also necessary: If d is odd, say d=2k+ 1, we have S4=Sd_3Y and Su=Kxd+ ... ; hence K[SJ does not contain Su' Therefore, if S(d)=K[Sd]' d has to be even, say d=2k; moreover, is contained in (S4)3 = L Kx 3 (k-p-q-r) y2(p+q+r),

t

hence k = P + q + r for some non-negative integers p, q, r. Since we also have k'!?,3p, 3q, 3r, we get k=3p=3q=3r. Therefore d=2k is a multiple of 6. In general, if A is an integral domain, we shall denote by F(A) the field of quotients of A. Suppose that S is a graded integral domain over So. An element ~ of F=F(S) which can be written as the quotient alb of an element a of Sp+k by an element b+O of Sp for some p, k is called a homogeneous element of F of degree k. Let Fk=Fk(S) denote the set of all such elements; then Fo is a field and Fk an Fo-module. Moreover, the subring of F generated by Fo, Fl , ... is a graded ring over Fo with Fk as its homogeneous component of degree k; also Sk = S n F1 for k=O, 1, ....

§ 1. Graded Rings

91

Lemma 4. Let S denote a graded integral domain over So. Then, for every positive integer d, we have Fo (S) = Fo (S(d)}. Moreover, if =1= 0 is a homogeneous element of F(S} of the smallest positive degree, is transcendental over Fo(S} and F(S}=Fo(S)(e}.

e e

Proof Since Fo(S} contains FO(S(d)}, we shall prove the converse. Let 1'/ denote an arbitrary element of Fo(S}; we have 1'/=a/b for some a,b in Sk' in which b=l=O. Since abd-l, bd are in Sdk and 1'/=abd- l /b d,

1'/ is in FO(S(d)}; this proves the first part. We shall prove the second part. Let d denote the degree of we write any given integer k as k=de+k', in which e is an integer and o~ k' < d. Then every element 1'/ of Fk(S} is of the form 1'/ = 1'/' with 1'/' in Fk,(S}; hence, if we have 1'/=1=0, we get k'=O, i.e., k=de. Therefore, if we have Fk(S}=I=O, we get k=de and Fk(S} = Fo (S) ee. This implies that F(S}=Fo(S)(e}. We shall show that is transcendental over Fo(S}, If is algebraic over Fo(S}, we have

e;

ee

e

en+oc i

e

n- I

e

+ ... +ocn=O

for some OC I , ••• , oc n in Fo (S). Since en, OC I en-I, ••• , OCn are homogeneous elements of F of degrees n d, (n -1) d, ... ,0, we get =0; but this is a contradiction. q.e.d.

e

If A is a subring of B, in which B is a commutative ring (with the identity), by the normalization of A in B, or simply the B-normalization of A, we understand the set of elements of B which are integral over A; the B-normalization of A is a subring of B containing A; if B is a subring of C and A* the B-normalization of A, the C-normalizations of A and A* are the same. If B is the total ring of quotients, i. e., the ring of quotients of elements of A by non-zero divisors of A, the B-normalization A * of A is called the normalization of A; A is called normal if A * = A. We shall use this terminology only in the well-known case where the rings have no zero divisors different from O.

Lemma 5. Let A denote an integral domain, A* its normalization, and t a letter. Then the normalization of A[t] is A*[t]. Proof If we put K=F(A}, we have F(A[t]}=K(t}. Since K[t] is a unique factorization ring, it is normal. Therefore the normalization of A[t] is a subring of K [t] containing A* [t]. Let f(t} denote an element of K [t] which is integral over A [t]; we have only to show that f(t} is in A * [t]. There exists a monic polynomial P(y}= yn + PI (t) yn-I + ... + Pn(t}

92

III. Graded Rings of Theta Functions

with coefficients Pi(t) in A [t] satisfying P(j(t))=O. Choose a positive integer m which is larger than the degrees of P1 (t), ... , Pn(t), and f(t); let x denote another letter and express P(x + t"') as a polynomial in x:

P(x + t"') =(x + tm)n + P1(t)(X+ t"')n-1 + ... + Pn(t) =Xn+q1 (t) x n - 1 + ... +qn(t). Then qi(t) are all in A [t] and qn(t) is a monic polynomial (of degree m n); moreover, f(t) - t'" is a root of the above monic polynomial in x. Therefore, t"'-f(t) divides qn(t) in K[t]. Since qn(t) is a monic polynomial with coefficients in A, all roots of qn(t) (in some algebraic closure of K) are integral over A. Since t'" - f(t) divides qn(t), its roots are among the roots of qn(t). Since t'" - f(t) is a monic polynomial (of degree m), therefore, the coefficients of t'" - f(t), hence those of f(t), are integral over A. We have thus shown that f(t) is in A* [t]. q.e.d.

Lemma 6. Let S denote a graded integral domain and S* its normalization. Then S* is a graded ring over the Fo-normalization of So with S* n Fk as its homogeneous component of degree k for k = 0, 1, .... Proof. Let cP denote the mapping of S to SEt] dermed by

in which ak is in Sk for every k and t a letter. Then cp is an injective homomorphism of S to S [t]; hence it can be extended uniquely to an injective homomorphism of F=F(S) to F(S[t])=F(t). We can see quite easily that an element ~ of F is contained in Fk if and only if CP@= ~ t. Suppose that ~ is an arbitrary element of S*; then we have ~n+c1 ~n-1

+ ... +cn=O

with Ci in S; by applying cp, we get cp(~)n+cp(c1) cp(~)n-1

+ ... +CP(cn)=O.

Since cp(cJ are in SEt] and cp@ in F(S[t]), this shows that normalization of S [tJ. Therefore, by Lemma 5 we have cp(~)=

cp(~)

is in the

L ~k t

k:5;O

with ~k in S*. We shall show that ~=~0+~1 + .... Since ~ is in F, it is of the form a/b with a,b in S andb=l=O; weputa=aO +a1+ "', b=bo +b 1+ ... with ak, bk in Sk' Then ~=a/b implies that cp(~) cp (b) = cp (a), i.e.,

(L ~kt)( L bkt)= L akt; k:5;O

1:5;0

1:5;0

§ 1. Graded Rings

93

by replacing t by 1, we get ~ = ~o + ~l + .... We shall show that ~k is in Fk • Assume that ~k is in Fk for every k>p; we have only to show that ~p is in F;,. Let q denote the largest index for which bq=l=O; then ~ b = a implies that ~pbq+~p+lbq_l +···=ap+q. By applying
This shows that

~

+ clO ~n-l + ... +cnO =0.

is in S6; hence S* n Fo is contained in S6.

q.e.d.

A commutative ring A is called a noetherian ring if every ideal of A is a finite A -module; in this case, every A-submodule of a finite A -module is a finite A-module. We recall that if A is a normal noetherian integral domain and F a finite separable extension of F(A), the F-normalization of A is a fmite A-module. (We refer to van der Waerden's Algebra, Springer for the proof.) Let K denote a normal noetherian integral domain and B an integral domain which is finitely generated over K. Suppose that there exists a finite number of elements YI' ... ' Yd of B which are algebraically independent over F(K) such that B is integral over A = K [YI' ... , Yd] and F = F(B) separably algebraic over F(A). Then the normalization B* of B is a finite B-module. The proof is as follows: Clearly, A is a noetherian integral domain and F a fmite separable extension of F(A); moreover, a repeated application of Lemma 5 shows that A is norma1. Therefore the F-normalization, say A", of A is a finite A-module. Since B is the B-normalization of A, the F-normalizations of A and B are same; hence A" = B*. Therefore B* is a finite A-module, hence a fortiori a finite B-module. Lemma 7. Suppose that an integral domain B is finitely generated over a field K of characteristic O. Then the normalization B* of B is a finite B-module; hence B* is finitely generated over K.

Proof By assumption, there exists a finite number of elements, say of B such that B = K [Xl' ... , xn]. Let d denote the transcendence degree of F(B)=K(Xl' ... , xn) over K. According to our previous observation, we have only to show that there exist d linear combinations n Yi= L cijXj (1 ~i~id) Xl' ... , X n,

j=l

94

III. Graded Rings of Theta Functions

of Xl' ..• , xn with coefficients cij in K which are algebraically independent over K such that B is integral over K[Yl' ... ,y.,]. If we have d=n, we can simply take Yi = Xi for 1 ~ i ~ d. Therefore we shall assume that d < n and apply an induction on n - d. In the course of the proof, it is permissible for us to apply any non-singular linear transformation with coefficients in K to Xl' •.. , X n • Since Xl' ... , Xn are algebraically dependent over K, there exists a polynomial P=I=O in n letters with coefficients in K satisfying P(x)=O for x=(xl ... xn)' Suppose that P is of degree m and let p", denote the homogeneous part of P of degree m. Since K is an infInite fIeld, there exists an element a=(a l ... aJ of K n satisfying p"'(a)=l=O. (This is well known; it can be proved easily by an induction on n.) After a non-singular linear transformation with coefficients in K, we may assume that al = ... =an_ l =0, an=l. Then xn is integral over K[x l , ... , X n_l ]. By the induction assumption, there exist d linear combinations Yl' ... , Yd of Xl' ..• , x n _ l with coefficients in K which are algebraically independent over K such that K[x l , ... ,xn_tJ is integral over K[Yl' ""Yd]. Since xn is integral over K[xl, ... ,Xn_l ], it is integral over K[Yl""'Yd]; hence B=K[x l , ... , xJ is integral over K[Yl> ""Yd]' q.e.d. The following lemma is a conditional converse of (the fIrst part of) Lemma 3: Lemma 8. Let S denote a graded integral domain over a field K of characteristic 0 such that S(d) is finitely generated over K for some d. Then S itself is finitely generated over K.

Proof. We shall exclude the trivial case where S = K. Then by Lemma 4 we have for some homogeneous element e=l=O; if the degree of e is q, we have e=a/b with a in Sp+q and b=l=O in Sp for some p. Put R=S(d)[a,b]; then we have S(d)c:.Rc:.S, and F(R) = F(S). Since we have (Sk)dc:.Skdc:.S(d) for every k, S is integral over S(d), hence over R. Therefore, if R* is the normalization of R, we have Rc:.Sc:.R*. By assumption, S(d) is fInitely generated over K; hence R is fInitely generated over K. Therefore R* is a fInite R-module by Lemma 7. Since R is noetherian, S is also a fmite R-module; hence S is fmitely generated over K. q.e.d.

§ 2. Algebraic and Integral Dependence Let S denote a graded integral domain over a fIeld K; then an estimate for the transcendence degree of Fo(S) over K can be obtained from the asymptotic behavior of dim(SJ as k -+00. In order to formulate this statement concisely, we shall introduce some notations. If c/J and '" are R-valued functions on a set X such that 1c/J(x)l~c·",(x) at every

95

§2. Algebraic and Integral Dependence

x in X for some positive constant c, we write --1/1 on X, we write -<1/1 on X. In the case where X is a sequence in R tending to 00, we also write
as k~oo. Proof The set of monomials xi' ... X~d in which Pl' ... , Pd are nonnegative integers satisfying mlPl +···+mdPd=k forms a K-base of R k • Therefore dim(RJ is equal to the number of such monomials, which is equal to the coefficient ofrk in the power series expansion of1 we shall express this fact as

IIi

I

(1-("");

i_l

dim(Rk)=Coef· tk (1/fJ (1-("")). Then, obviously, we have i.e.,

Coef.t"'k(1/(1-(",)")~dim(RmJ~Coef·t"'k(1/(1-t)~,


for k=O, 1, .... This implies that dim(Rmk)>-
q.e.d.

Proposition 1. Let S denote a graded integral domain over a field K satisfying the componentwise finiteness condition. Then

dim (SJ-
as k ~ 00 implies that any n + 2 homogeneous elements of S are algebraically dependent over K. Proof For 1 ~ i ~ d = n + 2, let Xj denote a homogeneous element of S of degree mj; we shall show that Xl' ... , Xd are algebraically dependent over K. Suppose that they are algebraically independent over K; then we have mj?;,1 for 1~i~d. Therefore, by applying Lemma9 to R= K[x l , ... ,x,J, we get dim (SmJ>-k"+1 as k ~ 00. On the other hand, by assumption we have dim (SmJ-«m k)">-
k~oo.

We thus have a contradiction.

q.e.d.

96

III. Graded Rings of Theta Functions

Remark. If a graded integral domain S over a field K is as in Proposition 1, i. e., if it is componentwise finite and satisfies the "growth condition ", dim (Sk)--O; by Proposition 1, the n + 2 elements Xi for ~ i ~ n + 1 are algebraically dependent over K; then, by applying Lemma 4 to the graded subring K [xo, ... , xn+ IJ of S, we see that ~ 1, ... , ~n+ 1 are algebraically dependent over K. We shall show that if ~1> ••• , ~n are algebraically independent over K, the degree of I'/=~n+l over K(~l' ... , ~n) is bounded. We write ~i=X;/XO for 1~i~n as above; this time we express 1'/ as Y/Yo with Yo4'O,y in Spq for some q > 0. This is certainly possible, and we can regard p but not q as fixed. Let k, N denote positive integers and, for z = Yo or y, consider the set of monomials

°

in which eo + ... + en + q e = k and e < N. If these monomials are linearly independent over K, since they are elements of Spb we get

as k ~ 00; this implies that N· kn/n! -- ... , ~n)' In fact, if the degree of 1'/ over F=K(~l' ... , ~n) attains its maximum, by the existence of a "primitive element", we have F(I'/, ()=F(I'/) for every ( in Fo(S), hence Fo(S)=F(I'/)=K(~l' ... , ~n' 1'/). We go back to Lemma 8: A graded integral domain S over a field K of characteristic is finitely generated over K if (and only if) SId) is finitely generated over K for some d. We shall examine this condition:

°

°

Lemma 10. Let Xl"'" Xn 4' denote elements of Sd for some d ~ 1 such that SId) is integral over R=K[x 1 , ••• ,xJ and Fo(R)=Fo(S(d»). Then SId) is finitely generated over K.

§2. Algebraic and Integral Dependence

97

Proof. We shall use a similar argument as the one used in the proof of Lemma 8: If we have n = 0, we get R = K, hence SId) = K; we shall exclude this trivial case. Then we have F(R) = Fo (R) (Xl) = Fo (S(d») (Xl) = F(S(d»).

Since SId) is integral over R, the normalization R* of R contains SId). Since R* is a fmite R-module by Lemma 7 and R noetherian, SId) is also a finite R-module; hence SId) is fmitely generated over K. q.e.d. We shall examine the condition that SId) is integral over R. We shall use the following well-known lemma: Let A denote a commutative ring (with the identity 1) and COl' ••• , CON a finite number of elements of A satisfying N CO i = Laijcoj

e,

e

j_1

with aij in A for 1 ~ i, j ~ N .. suppose that at least one of the COl' ••• , CON is not a zero divisor of A. Then is integral over the subring of A generated by aij (and 1). We recall that one way to prove elementary properties of normalization, e.g., those stated before Lemma 5, is to use this lemma. For the sake of completeness, we shall give a proof of the lemma: By regarding A as a Z-algebra, we shall express the subring of A generated by aij as Z [aij] = Z [au, a12 , ••• ,aNN]. By assumption, we have

e

N

L(Oije-aij)COj=O

j_l

(1~i~N),

in which Oii= 1 and 0ij=O for i*'j. Let f(t) denote the characteristic polynomial of the matrix (aij)i,j; then we have f(e)=det(oij -aij)iJ, hence f( e) coj = 0 for every j. Since not all COl' ... ,CON are zero divisors of A, we get f
e

e

Lemma 11. Let R = K [Xl' ... , xJ denote a graded integral domain over a field K of characteristic 0, in which Xl; ... , Xn 0 are in R l , and

*'

e

98

III. Graded Rings of Theta Functions

an element of Fk(R) for some k~O; suppose that ~/x~ is integral over K[x 1/x j, ... ,xJxJ for 1~i~n. Then ~ is integral over R. Proof Let S denote the normalization of R. Then S is a graded ring by Lemma 6; and S is a fInite R-module by Lemma 7. Therefore, there exists a fmite number of homogeneous elements ~1' ... , ~m of S of degrees d1 , ••• ,dm such that S=R~l +···+R~m' Take an integer d satisfying d ~ d1 , ••• , dm and denote the products of monomials in Xl' ... , Xn of degree d-d j and ~jfor 1~i~m by '11' ... , '1q • Then every homogeneous element of S of degree at least equal to d is a linear combination of '11' ... , '1q with coefficients in R. On the other hand, let l:j denote the Then, by the multiplicative submonoid of R consisting of 1, Xj, compatibility of localization and normalization, Sl:j-l is the normalization of Rl:j- 1 ; hence (Sl:j-l)O is the normalization of (Rl:j- 1)o. We observe that (Sl:j-l)O is the subring of Fo(R) consisting of a/xi with a in Se for some e~O; also (Rl:j- 1)o=K[x1/x j, ... ,xJxJ and its fIeld of quotients is Fo(R). Since ~/x~ is an element of Fo(R) and integral over (Rl:j- 1)o, it is in (Sl:j-l)O' Therefore Xii~ is contained in S for some ej~O. Take an integer e at least equal to

x;, ....

n

L(ej-1)+1 j=l

and denote by Yl' ... , Yp the monomials in Xl' ... 'Xn of degree e. Then ~ Yj' hence also ~ Yj '1j' are all contained in S. Therefore every ~ Yj'1j is a linear combination of '11' ... , '1q with coefficients in R k+ e =Rk Yl + .. ·+Rkyp·

Hence, if we denote the N = p q products Yj'1j by w 1 '

... ,

wN ' we have

N

~ Wj=

L aij Wj

j=l

with aij in Rk for 1 ~ i, j over R. q.e.d.

~ N. As we have seen, this implies

that ~ is integral

We shall examine the condition that an element ~ of the fIeld of quotients F of an integral domain R is integral over R. Let p denote a prime ideal of R; then l:=R-p is a multiplicative submonoid of R containing 1 but not O. Put Rp=Rl:-l; Rp is the subring of F consisting of alb with a in Rand b in R - p. If ~ is integral over R, ~ is certainly integral over Rp for every p. We shall show that if ~ is integral over Rp for every maximal ideal p of R, ~ is integral over R. Let a denote the set of elements b of R such that b ~ is integral over R; then a is an ideal of R. We have only to show that a contains 1. If this is false, a is contained in a maximal ideal p of R. Since ~ is integral over R p , we have ~m+atlbl . ~,-,,-1

+ ... + am/bm=0

§3. Weierstrass Preparation Theorem

99

for some al , ... , am in Rand bl , ... , bm in R - p; put b = bi ... bm and multiply bm to the above equation. In this way, we get an equation showing that be is integral over R. Then b is in n, hence in p; but this is a contradiction. We shall rephrase the condition that is integral over RI' for a prime ideal p of R. We shall exclude the trivial case where =0 and put '" = lie; let t denote a letter. Then, if is not integral over RI" for every element f(t) of R [t] satisfying f(",)=O, f(O) is contained in p. The proof is a triviality: Put f(t)=amrm+···+aIt+a O and assume that ao=f(O) is not in p; then em+aIla o · m - I + ... +am/ao=O shows that is integral over RI'.

e

e

e

e

e

§ 3. Weierstrass Preparation Theorem We shall review the Weierstrass preparation theorem and its immediate consequences. First we shall fIX our notations. Suppose that U is a complex manifold; for every non-empty open subset Vof U, we have defined (l)(V); it is the ring, more precisely the C-algebra, of holomorphic functions on V. If V' is a non-empty open subset of V, the restriction of functions gives a C-algebra homomorphism (9 (V)~ (l)(V'); if V" is a non-empty subset of V', the product of (l)(V)~(l)(V') and (l)(V')~(l)(V") is (l)(V)~(l)(V"). Therefore, if we consider (l)(V) for every open set V containing a given point a of U, we can take the direct limit

If f is an element of (l)(V), its image in (l)a is called the germ of fat a. Since the homomorphism (l)(V)~(l)a is injective if V is connected, it is customary to use the same notation for f and its germ (even if V is not connected). The classical proof of the Weierstrass preparation theorem makes use of the following facts: , Let PI denote a positive real number and Wi the subset of C defmed by \W\
L (w )e=(1/2n i) J i

i=1

Bd(W,)

=

we. d log F(w)

J we+I(F'IF)(w)dlJ,

R/Z

100

III. Graded Rings of Theta Functions

in which F' denotes the derivative of F and w=Ple(O). This is well known in function theory. Let K denote a field of characteristic 0 and Wl' ... 'Wm letters; then the power sums wr+···+w:;' for l;;;;!e;;;;!m and the elementary symmetric functions Wl + ... + wm ' ••• , Wl ... Wm generate the same subring of K[w l , ... , w.J over K. This follows from the "Newton formulas." Suppose that U is a connected complex manifold and f an element of l!J(U); if f =0 on a non-empty open subset of U, then f =0. In particular, if g is a continuous function on U, which is different from the constant 0, f(z)=O for every z in U satisfying g(z)9=O implies thatf=O. Theorem 1. Let V denote a complex manifold, Wan open subset of C defined by Iwl

(g- fh)(z, w)= L b;(z) w; ;=0

with bo , ... , bm _ l in l!J(V).

Proof. We shall first prove the uniqueness of h, hence also of bo, ... ,bm _ l • Suppose that k is an element of l!J(U) which is similar to h. Then f(h-k) is an element of l!J(V)[w] of degree at most m-l. Hence, for every z in V, f(z, )(h-k)(z, ) is either the constant 0 on W or has at most m -1 zeros in W. Since f(z, ) has exactly m zeros in W, we have the first alternative, hence f(h-k)=O. Since f is not the constant 0 on any connected component of U, we get h-k=O. This proves the uniqueness of h. Let a denote an arbitrary point of V. We shall show that there exist an open subset »'1 of W defmed by Iwl
J

lId(W,)

we·dlogf(z,w).

§3. Weierstrass Preparation Theorem

101

According to the "holomorphicity of a function defmed by an integral" in Chap. I, §1, the right side defmes an element of (!)(lI;.). In the special case where e=O, this function is Z-valued and takes the value m at a. Since lI;. is connected, it is the constant m on lI;.. We have seen that all power sums of Wi> ••• , Wm are in lD(VJ; hence their elementary symmetric functions are also in (!)(lI;.). In other words, if we put m p(z, w)= (w-wi)=wm+a1(z) wm - 1 + ... +am(z),

n

i=l

the m functions a1, ... , am on lI;. are in (!)(lI;.). Moreover, for every (z, w) in U1 = lI;. X ~ satisfying f(z, w)=+=O, by the Cauchy theorem for one complex variable, we have p(z, w)/f(z, w)=(1/2n i)

J

(p(z, w)/f(z, w») d log(w-w).

Bd(Wl)

For a similar reason as above, the right side defmes an element u of (!)(U1); and we have p(z, w)=u(z, w)f(z, w) for every (z, w) in U1 satisfying f(z, w)=+=O; hence p=uf on U1. Since f(z, ) and p(z, ) have the same zeros in ~, u(z, ) is a unit of (!)(~); hence u is a unit of (!)(U1). Let g denote an arbitrary element of (!)(U); then k(z, w)=(1/2n i)

J

(g(z, w)/p(z, w») d log(w-w)

Bd(Wl)

defines an element k of (!)(U1); moreover, we have (g - p k)(z, w)

=(1/2 n i) If we write

J

(g(z, w)/p(z, w»)(p(z, w)- p(z, w») d log(w -w).

Bd(Wl)

m-1

p(z, w)-p(z, w)=(w-w)·

L ci(z, w) Wi,

i=O

each ci(z, w) is an element of (!)(lI;.) [w]. Therefore bi(z) = (1/2 n i)

J

(g(z, w)/p(z, w») cj(z, w) dw

Bd(W1)

defines an element bi of (!)(lI;.), and (g-pk)(z,w)=

m-1

L b;(z)wi

i=O

for every (z, w) in U1. If we put h=uk, h is also an element of (!)(U1), and pk=fh on U1. We recall that the element h of (!)(U1) is uniquely determined by the condition that g- fh is an element of (!)(lI;.)[w] of degree at most m-l. Since P1 can be replaced by any larger number less than P, h can be extended uniquely to an element of lD(lI;. x W) satisfying the above condition. If we denote lI;. by v,. and h by ha , again

102

III. Graded Rings of Theta Functions

by the uniqueness, we have ha = hb on v"b x W for v"b = v" n Vi,. Therefore we get an element h of (lI(U) such that g-fh is an element of (lI(V)[w] of degree at most m -1. q.e.d. In the course of the above proof, we have shown that f differs from an element p of (lI(Jt;.)[w] of the form w"'+ lower terms by a unit of (lI(U1). Actually, since p is uniquely determined by f, it can be extended to U. In other words, under the same hypothesis as in Theorem 1, there exists a unique element p of (lI(V)[w] of the form w"'+ lower terms which differs from f by a unit of (lI(U). We will get another proof of this fact if we take as g in Theorem 1 the element of (lI(U) defined by g(z, w)=wm • We observe that if a and b are points of complex manifolds of the same dimension, (lIa and (lib are isomorphic as C-algebras. Therefore we have only to consider (lI = (lIo for the complex manifold en. If U is an open neighborhood of 0 and f an element of (lI(U), the Taylor

expansion of f centered at 0 depends only on the germ of f at O. In this way, we get an isomorphism of (lI to the C-algebra of convergent power series with coefficients in C in the coordinates Zl"'" Zn of a general point Z of en. Let f +0 denote an element of (lI; then we can rewrite its Taylor expansion as f(z) = fm(z) + fm+l(Z)+"',

in which h is a homogeneous polynomial of degree d and fm+O. We call fm the leading form and m the leading degree of f. We observe that (lI - 0 forms a multiplicative monoid, i. e., (lI is an integral domain, and taking the leading degrees gives a homomorphism of this monoid to the additive monoid of non-negative integers; moreover, the set oF of elements with positive leading degrees forms the ideal of non-units of (lI. We shall assume that n ~ 2; we shall show that if f + 0 is an element of (lI, there exists an open neighborhood of 0 in en for which f satisfies the condition of Theorem 1: We choose an element a+O of en satisfying fm(a)+O. Since GLn(C) acts transitively on en-o, after a non-singular linear transformation in cn, we may assume that a=(O ... 0 1). Then we have f(O, zn) = const. z;:' + higher terms

with the "const." +O. In the following, we shall use only this property of f. We put z' = (Zl ... Zn_l)' Let r, p denote positive real numbers and V, W the open polycylinders in en-I, C defined by Iz'loo
§3. Weierstrass Preparation Theorem

103

for every w in W, hence f(O, w)=I=O for every w in W -0; then, by making r smaller if necessary, we may assume that f does not vanish at any point of Vx Bd(W}. Then, as we have seen in the proof of Theorem 1, for every z' in V, f(z', } has m zeros in W. This proves the assertion. If g is an arbitrary element of ~, we may assume that it is holomorphic on the above V x W. Then we can apply Theorem 1 to f, g on V x W. As we have seen, there exists an element p of ~(V)[w] of the form wm+ lower terms which differs from f by a unit of ~(V x W). Since the m zeros of f(O, } in Ware all at 0, we have p(O, w}=wm. If we denote the "~" and ".F" for cn-l by ~' and .F', we get the following Weierstrass preparation theorem:

Corollary. Iff =1=0 is an element of ~ of leading degree m, after a nonsingular linear transformation in cn, we have fm (z) = const. in which the" const." =1= 0;

if f

z: + lower terms,

is an element of ~ such that

f(O, zn)=const.

z: + higher terms

m-l with the "const."=I=O, we have ~=~f + L ~' z~; ;=0

moreover, f can be written uniquely as the product of a unit of ~ and a "Weierstrass polynomial" p of degree m, which is an element of ~' [zn] of the form ..111..111 1 <:n +a1<:n - + ... +am with a1, ... , am in .F'.

If we agree to take ~' =C and .F' =0 for n= 1, the above statement is true also in this case; moreover, .F = ~ Zl and all ideals of ~ are powers of oF, i.e., ~ is a discrete valuation ring. In the general case, i.e., for every n~ 1, ~ is a noetherian unique factorization ring. We shall show that ~ is a noetherian ring. We shall assume that n~2 and ll:pply an induction on n. Let f denote an ideal of ~; we shall exclude the trivial case where f =0. Let f =1=0 denote an element of f. After a non-singular linear transformation in en and mUltiplying a unit, we may assume that f is a Weierstrass polynomial of degree m. Consider the following ~' -module: m-l fn L ~'z~; ;=0

this is an ~' -submodule of a finite ~' -module. Since ~' is noetherian by the induction assumption, the above ~'-module is fmite; hence it is generated over ~' by a finite number of elements, say it, f2' .... The corollary of Theorem 1 shows that f is generated over ~ by f, it, f2' ... ; hence f is a finite ~-module.

104

III. Graded Rings of Theta Functions

Suppose in general that A is an integral domain. An element of A is called irreducible if it is not a unit and if it can not be written as a product of two non-units of A. If every element of A -0 can be written uniquely up to units as a product of a finite number of irreducible elements, we say that A is a unique factorization ring. In this case, for a letter t, A [t] is also a unique factorization ring. After recalling this, we shall show that @ is a unique factorization ring. If an element f of @-O is a product of two non-units, the leading degrees of factors are smaller than the leading degree of f It follows from this fact that every non-unit of @-O can be written as a product of a fmite number of irreducible elements. Therefore we have only to prove the uniqueness. We shall assume that n ~ 2 and apply an induction on n. If the product of a finite number of elements f, g, ... of @ differs from a Weierstrass polynomial by a unit, its Taylor expansion contains a power of Zn with non-zero coefficient; clearly, this property is inherited by the factors f, g, ... ; hence f, g, ... differ from Weierstrass polynomials by units. On the other hand, the product of a fmite number of Weierstrass polynomials is certainly a Weierstrass polynomial; moreover, if a Weierstrass polynomial divides another Weierstrass polynomial in @, the quotient is necessarily a Weierstrass polynomial; this follows from the corollary of Theorem 1. After observing these, let f denote an irreducible element and g, h elements of @-O such that f divides g h but not g; we have only to show that f divides h. After a non-singular linear transformation in en, f g h differs from a Weierstrass polynomial by a unit. As we have observed, this implies that f, g, h are, up to units, all Weierstrass polynomials; hence we may assume that they are Weierstrass polynomials. Then f is irreducible and f divides g h but not g, all in @' [zJ. Since @'[zn] is a unique factorization ring by the induction assumption, this implies that f divides h in @'[zn], hence in @. We shall show that if U is a connected complex manifold, the ring @(U) is normal. First of all, since U is connected, for every a in U, the homomorphism @(U)--+@a is injective; hence @(U) is an integral domain. Let ~ denote an element of F(@(U)) which is integral over @(U); write ~=flg with f,g in @(U) and g=l=O. Then ~ is an element of F(@a)which is integral over @a. Since @a is a unique factorization ring, it is normal; hence ~ is in @a. In other words, there exists an element ha of@a satisfying f = g ha in @a for every a in U. We observe that the set (ha)a defines an element h of @(U) satisfying h(a)=ha(a) for every a in U; and we have ~= flg=h in F(@(U)). We shall examine the definition of positive divisors given in Chap. II, § 2. We fix a complex manifold U. Let D denote a positive divisor (of U) with (~,ni as a representative. If we denote by Xi the set of points Z of ~ which satisfy .t;(Z) =0, Xi isa closed subset of~; and XJ1Uij=

§3. Weierstrass Preparation Theorem

105

Xj n Uji for every Uij" Therefore, if we denote by X the union of all Xi' X is a closed subset of U satisfying X n U;=Xi for every i; moreover, X does not depend on the choice of the representative (U;'!;)i of D. We call X the support of D. If a point a of U is contained in U;, the principal ideal (fJa!; is defmed; and it does not depend on the index i nor on the choice of the representative (U;'!;)i of D. We call (fJa!; the "ideal of D" at a. The following theorem is basic: Theorem 2. For any positive divisor D, there exists a unique positive divisor Dred with the following properties: Dred has the same support as D; D - Dred is a positive divisor; if D' is a positive divisor which has the same support as D, D' - D red is a positive divisor.

Proof. Since the uniqueness of Dred is clear, we shall prove the existence. Let a denote a point of U and (fJaf the ideal of D at a; then we have - ret ... I.e, uf -J1 t ,

in which f1' ... ,J; are irreducible elements of (fJa satisfying (fJa!; =t- (fJaJj for i =t-j and e1 , ••• ,et positive integers for some t ~ 0; and u is a unit of (fJa' In the case where t = 0, we take Ua= U - X and fa = 1. In the case where t~ 1, we define Ua and fa as follows: We take local coordinates (Z1 ... zJ of U valid in some open neighborhood of a such that It ... J; differs from a Weierstrass polynomial by a unit of (fJa; then we can assume that It, ... ,J; are all Weierstrass polynomials. If we take connected open neighborhoods V, W of a' =(a1 ... an_ 1), an in cn-1, C sufficiently small, we can identify V x W with an open neighborhood, say Ua' of a; we may assume that It, ... ,J; are elements of (fJ(V) [znJ. By making V, W smaller if necessary, we may assume that f =0 is a local equation for D in Ua' i.e., (fJJ is the ideal of D at c for every c in Ua' and u a unit of (fJ(Ua). Then Nt ... J;e, = 0 is a local equation for D in Ua; we put fa =It ... J;. By making V still smaller, we may assume that, for every z' =(Z1 ... Zn_1) in V, all zeros of f(z', ) are in W. We shall show that {Ua,fa)a is a representative of a positive divisor. The key point is the fact that (fJcfa is "reduced" at every point c of Ua in the sense that (fJc/(fJcfa has no nilpotent element different from O. We shall exclude the trivial case where fa= 1. Let d denote the discriminant of the polynomial fa in zn; then d is an element of (fJ(V). Since fa has no multiple factor in (fJa' [zJ, we have d=t-O as an element of (fJa" Suppose that (fJcfa is not reduced at some point c=(c 1 '" cn) of Ua' Then the germ of fa at c contains an irreducible element of (fJc as a multiple factor. This implies that, for every z' in a small open neighborhood of c'=(c1 ... cn _ 1 ) in V,fa(z', ) has a multiple zero. Then d will vanish on this neighborhood of c'. Since V is connected, we get d=O as an element of (fJ(V), hence a fortiori as an element of (fJa'; but this is not the case. By what we have shown, if c is an arbitrary point

106

III. Graded Rings of Theta Functions

of Uab , we have {9c fa = {9c!;,; hence (Ua,fJa is a representative of a positive divisor. If we take this as Dred , it is easy to verify that Dred has the required properties. q.e.d. The above proof shows that if we write the ideal of D at any point a as {9aN' ... f/', in which ft, ... ,it are irreducible elements of {9a satisfying {9a1; {9ah for i j and e1, ... , et positive integers, the ideal of Dred at a is {9af1 ... it· We shall show that, by changing the notation, if D1, D2 are positive divisors with {9a f1' {9af2 as the ideals at a and fa the greatest common divisor of f1,f2 in {9a' {9afa can be considered as the ideal at a of a positive divisor D. The proof is similar to that of Theorem 2. We have only to show that fa is the greatest common divisor of It,f2 at every point in some open neighborhood of a. After dividing by fa' we may assume that fa = 1. In other words, we have only to show that if f1,f2 are relatively prime in {9a' they are relatively prime in {9b for every b in some open neighborhood of a. If we take a suitable local coordinates valid in some open neighborhood of a, we may assume that f1,f2 are Weierstrass polynomials; moreover, by using the same notation as in the proof of Theorem 2, we may assume that It, f2 are elements of (9(V) [znJ and, for every z' in V, the zeros of f1(Z', ),f2(Z', ) are all in W. Let d denote the resultant of f1' f2; d is an element of (9(V), and d,*,O as an element of {9a'; hence d,*,O as an element of {9b' for every b' in V. Thus f1' f2 are relatively prime in {9b' [znJ; hence, for every b in V x W, f1,f2 are relatively prime in {9b' We say that a positive divisor is irreducible if it can not be written as a sum of positive divisors different from O. Let D1, D2 denote irreducible divisors with {9aft, {9af2 as the ideals at a such that f1,f2 are not relatively prime in {9a for at least one a,' then we have D1 =D 2. In fact, under the assumption, there exists a positive divisor D'*'O such that D1 - D, D2 - D are both positive; hence D1 = D = D2. Suppose that D, D' are positive divisors and D' irreducible,' then D - D' is a positive divisor (f and only if the support of D contains the support of D'. Since the only if-part is clear, we shall prove the if-part. Suppose that the support of D contains the support of D' but D-D' not positive; let {9af, {9af' denote the ideals of D, D' at a. Then f,f' are relatively prime in {9a; otherwise, there exists a positive divisor D1 0 such that D - D1, D' - D1 are both positive; hence D'=D1' and D-D' positive; but this is not the case. Once we know that f,f' are relatively prime in {9a' by converting them into Weierstrass polynomials and taking the resultant etc., we can find a point b satisfying f'(b)=O but f(b),*,O. This is a contradiction. As a consequence, if a positive divisor can be expressed in the form

'*'

'*'

'*'

t

D= LeiD i , ~

i= 1

§4. Geometric Lemmas

107

in which D 1 , ••• ,Dt are irreducible (and distinct) and e1 , ••• ,et positive integers, these are uniquely determined by D; moreover, we have t

D red =

LD i=l

i•

We say that a positive divisor D is reduced if D=D red • If D is reduced, the restriction of D to any non-empty open subset is reduced. Suppose that a positive divisor D is reduced; then there exists a point b of the support of D such that if (lib f is the ideal of D at b and (Zl ... Z,,) local coordinates valid in some open neighborhood of b, we have gradb(f)=(ofloZl ... oflozn)(b)=t=O.

We first observe that the property gradb(f)=t=O does not depend on the choice of the local coordinates. We shall use the same notation as in the proof of Theorem 2; we may assume that f = fa. Choose a point b' =(b1 ••• bn_ 1 ) of V such that d(b')=t=O; let bn denote any zero of f(b', ) and put b=(b1 ••• bn). Then bn is a simple root of f(b', zn)=O, hence (oflozn)(b)=t=O. § 4. Geometric Lemmas

We need two lemmas on algebraic varieties over C. These lemmas are "elementary" in the sense that they are more or less at the level of van der Waerden's Einfiihrung in die algebraische Geometrie, Springer. Nevertheless, if we try to give complete proofs (assuming only standard materials in the same author's Algebra), it will take us out of the main subject of this book. Therefore, we shall just" explain" the lemmas; the explanation can be made into proofs by any reader who is familiar with that level of algebraic geometry. We refer to Weil [20], Appendix for complete proofs. We fix an algebraically closed field O. Let (mid denote a set of polynomials P; in n letters 11, ... , T,. with coefficients in O. Then the subset X of on consisting of those points a=(a1 ••• a,,) which satisfy p;(a)=O for every i in I is called a Zariski closed set in on defined by (miel' Let ~ denote the ideal of 0 [11, ... , TJ generated by (miel' Since the ring 0[11, ... , TJ is noetherian, or, by the Hilbert "Basissatz", ~ has a finite ideal base, say (FjteJ; and X can equally be dermed by (Fj)jeJ' Let ~(X) denote the set of all polynomials P in 0[1;', ... , TJ which vanish at every point of X. According to the Hilbert "Nullstellensatz", ~(X) is the "root" of ~. Moreover, the correspondence X ~~(X) gives an inclusion-reversing bijection of the set of Zariski closed sets in on to the set of "reduced" ideals of 0[11, ... , TJ. (The root of an ideal a of a commutative ring A consists of all elements of A

108

III. Graded Rings of Theta Functions

which are nilpotent mod a; it is denoted by r(a); the ideal a is called reduced if r(a)=a.) The intersection of an arbitrary number of Zariski closed sets is a Zariski closed set; also, the union of a fmite number of Zariski closed sets is a Zariski closed set. Therefore, the set of all Zariski closed sets in fl' can be taken as the set of "closed sets"; this is called the Zariski topology in OR. Let X denote a Zariski closed set in OR. We say that X is an irreducible affine variety, or just a variety, if ~(X) is a prime ideal. In this case, the transcendence degree of the field of quotients of O[~, ... , T,,]/~(X) over 0 is called the dimension of X; it will be denoted by dim (X). Every Zariski closed set X can be represented uniquely as the union of a fmite number of varieties Xl' ... , X t none containing another. The maximum of dim (XJ for 1;:;;'; i;:;;'; t is called the dimension of X; it will be denoted again by dim(X). Let X denote a Zariski closed set in fl' and K a subfield of O. If ~(X) has an ideal base contained in K[~, ... , TJ, we say that K is a field of definition of X or X is defined over K. In this case, if we put ~K(X)=~(X)nK[~,

we have

... , TJ,

O[~, ... , T,,]/~(X)=(K[7;., ... , TJ/~K(X))®KO.

Suppose that X is a variety, i.e., ~(X) is a prime ideal; let ~i denote the residue class of T; mod ~ (X) for 1;:;;'; i;:;;'; t. Then the transcendence degree of K(~)=K(~l' ... , ~J over K is equal to that of O(~) over 0, which is equal to'dim(X) by definition. A point x=(x 1 ••• xJ of X is called a generic point of X over K if the homomorphism K[~]-+K[x] over K mapping ~i to Xi for 1;:;;'; i;:;;'; n, which exists because X is a point of X, is an isomorphism. This imples that the transcendence degree of K(x) over K is equal to dim(X). Conversely, if x is a point of X such that the transcendence degree of K(x) over K is equal to dim (X), x is necessarily a generic point of X over K. If the transcendence degree of 0 over K is at least equal to dim(X), X has a generic point over K. Suppose that a point x of fl' and a subfield K of 0 are given. Then, in the case where the characteristic of 0 is 0, x can be considered as a generic point over K of a variety defined over K if and only if K is "maximally algebraic" in K(x), i.e., if and only if every element of K(x) which is algebraic over K is contained in K. Let X denote a variety in OR of dimension d and (/DieI a finite ideal base of ~(X). Then we say that a point a of X is a simple point of X, or that a is simple on X, if the rank of the matrix

§4. Geometric Lemmas

109

evaluated at a=(al ... a") is n-d. The property of a being simple on X does not depend on the choice of the ideal base. If X is defmed over K and x a generic point of X over K, x is simple on X. If we extract square matrices of degree n - d from the above matrix and take their determinants, we get a finite set of polynomials in .0 [7;., ... , TJ; this defmes a Zariski closed set, say Y, in .0"; and X.=X - Y is precisely the set of all simple points of X.1f a is simple on X, X is "locally at a a complete intersection of n-d hypersurfaces". A more precise, purely algebraic statement is as follows: Let Fl , ... , F,.-d denote elements of~(X) such that

rank(oFJo1j)(a));,j=n-d. Then for every P in ~(X) there exists an element Q of .o[~, ... , TJ satisfying Q (a) =t=0 such that PQ is contained in the ideal of .o[~, ... , TJ generated by Fl , ... , F,.-d' From now on, we take C as .0; then .0" = C" has two topolo'gies, i. e., the usual topology and the Zariski topology. If we do not explicitly mention the topology, we shall be talking about the usual topology; we observe that a Zariski closed set is closed. Suppose that X is a Zariski closed set in C" and (In; a finite ideal base of ~(X); let K denote the extension field of Q obtained by adjoining all coefficients of all If to Q; then K is a countable subfield of C over which X is defmed. (If A is a countable (commutative) ring and t a letter, A [t] is countable; any homomorphic image of a countable ring is countable; the total ring of quotients of a countable ring is countable. We have only to apply these obvious facts repeatedly.) Since the composite field of a fmite number of countable fields is countable, we can find a countable common field of definition of any fmite number of Zariski closed sets. Therefore we shall make an agreement that a (common) field of defmition is a countable subfield of C. Since C is not countable, the transcendence degree of C over any field of defmition is infinite; hence, we can always find generic points of varieties over their fields of definition.

Proposition 2. Let X denote an irreducible affine variety in C" defined over K; then the set of generic points of X over K is dense in X. We shall try to give some idea how the proof goes: In the special case where X = C", K is an arbitrary countable subfield of C. Since K [~, ... , TJ is countable, we can convert this. set into a sequence, say Po = 0, ~, 11, .... Let X; denote the Zariski closed set defined by If; then 00

C"-UX; is precisely the set of all generic points of C" over K. The rest follows from an elementary theorem in general topology (known as "the Baire category theorem ").

III. Graded Rings of Theta Functions

110

In the general case, as we have seen in the proof of Lemma 7, we can fmd d = dim (X) linear combinations n

'1i=~>ijej j=1

(1~i~d)

of el' ... , en with coefficients cij in K such that K [e] is integral over K ['11' ... , '1d]' Let a denote an arbitrary point of X and put bi = L cij aj j

for 1 ~ i ~ d. Then, by the special case that we have treated, we can find a sequence (y(v»)v of generic points y(v) of C' over K which converges to b=(b1 ... bd) as v-+oo. According to the "positivity of multiplicity" in the elementary case (which can be reduced to the "continuity of roots" of an algebraic equation as functions of the coefficients), we can find a sequence (x(v»)v of points xcv) of X satisfying

for 1~i~d which converges to a as v-+oo. Since the transcendence degree of K(x(v») over K is d, xcv) is a generic point of X over K for every v. If X is an irreducible affine variety in en of dimension d and X. the set of simple points of X, at every point a of X., we can find n - d elements F1 , ... , F" -d of .:J (X) and an open neighborhood V of a in C" such that rank (oFJo1j)(x))j,j = n-d for every x=(x1

•••

x n) in V and V nX consists of all x satisfying F1 (x) = ... =F,,_d(X)=O.

'DJ-is follows from the fact that X is locally at a a complete intersection of F1 =0, ... ,F,,_d=O. Therefore X. can be considered as a submanifold of C" of complex dimension d.

Proposition 3. If X is an irreducible affine variety in nected open subset of X.

en, X.

is a con-

Again we shall try to give the idea of the proof. The main point is the connectedness of X •. In the case where X is a curve, i. e., d = dim (X) = 1, this is intuitively clear and can be proved easily. The case where d~2 can be reduced to this case as follows: Suppose that a, b are two points of X.; we want to show that a, b are in the same connected component of X •. Let K denote a field of defmition of X. Since X. is locally connected (as a submanifold of C"), by Proposition 2 we can replace a by a generic point x of X over K; also, we can replace b by a generic point y of X over K(x); x,y are called "independent generic points" of X over K. Let (~Jj,k for 1~j
§4. Geometric Lemmas

111

the subspace L of the vector space M(d -1, n; C) over e defmed by n

n

k_1

k_1

L UjkXk = L UjkYk=O

(l~j
then L is an irreducible atfme variety defined over K(x, y). Let u=(Ujk)j,k denote a generic point of Lover K(x, y); then U is a generic point of M(d -1, n; C) over K. Consider the subspace M of en defmed by n

L u jk 1k=0

(l~j
k-1

then M is a "generic subspace" of en of dimension n - d + lover K passing through x, y. Put C=X nM. According to "Bertini's theorems", C is an (irreducible) curve and C.=CnX•. Therefore x,y are points of C.; and we have remarked that C. is connected. Hence x, y belong to the same connected component of X •. We shall go back to the end of § 2, change the notation accordingly, and translate the criterion of integral dependence into language involving the usual topology of en: Let X denote an irreducible atfme variety in C" and put e[l;., ... , TJ/~(X)=e[X1' ... , xJ=R; let a=(a1 ••• aJ denote a point of X and p the image of ~(a) in R; then the correspondence a --+ p gives a bijection of X to the set of all maximal ideals of R. We shall examine the condition that an element ~ of F(R) = C(x) is integral over Rp. We write ~=P(x)/Q(x) with P(x), Q(x) in Rand Q(x) =1=0; let Y denote any Zariski closed set in C" which is strictly contained in X and contains all points a of X satisfying Q(a)=O. We shall show that if ~(a(v»=P(dv»)jQ(a(v» (1 ~v< (0) are bounded for any sequence (dV»v in X - Y which converges to a as V--+OO, ~ is integral over Rp. The proof is as follows: Suppose that ~ is not integral over Rp; then we have ~=I=O, hence '1= lie is defined. Let X* denote the irreducible atfme variety in C"+l with e [x, '1] as its coordinate ring. If t is a letter and ~ the kernel of the homomorphism e[l;., ... , T", t] --+ e [Xl' ... , x n' '1] over e defmed by T;--+ Xi for 1~i~n and t --+ '1, this means that X* is the Zariski closed set in C"+l satisfying ~(X*)=~. We shall show that (a, 0) is a point of X*. If (a, 0) is not a point of X*, there exists an element f(t) of R[t] satisfying f('1) =0 but f(O) not in p; this will imply that ~ is integral over Rp. Let y* denote the set of points of X* whose projections to X belong to Y; then y* is a Zariski closed set in C"+l strictly contained in X*. We take a common field of definition K of X* and y*; by Proposition 2 we can find a sequence (a(V), bv)v of generic points (a(V), by) of X* over K which converges

112

III. Graded Rings of Theta Functions

to (a, 0) as v --+ 00; this sequence is necessarily contained in X* - y*. Since we have Q(x)-P(x) 11=0, we get Q(a(V»)_P(a(v») bv=O for 1 ~v< 00. We observe that (alV»)v is a sequence in X - Y which converges to a as v --+ 00 such that ~ (al V») = l/b v for 1 ~ v < 00 are unbounded. This proves the assertion.

§ 5. Automorphic Forms and Projective Embeddings In general, let U denote a connected complex manifold; suppose that a group r acts on U as a group of biholomorphic mappings. Let p denote a holomorphic automorphy factor relative to r. This means that (p(q, ))... is a set of units of (!)(U) satisfying p(q q', z)= p(q, q' . z) p(q', z)

for every q, q' in rand z in U; we have p(l, z)= 1, p(q, Z)-1 = p(q-1, q. z)

for every q in rand z in U. By using such an automorphy factor, we can let r act on the vector space (!)(U) over C as (q. f)(z) = p(q-1, Z)-1 f(q-1 . z).

We observe that, for every integer k, l is also a holomorphic automorphy factor relative to r. Hence, by using pk, we can let r act on the same vector space. Let A(T)k denote the set of r-invariant elements of (!)(U), i. e., the set of holomorphic functions f on U satisfying f(q· z)=p(q, Z)k f(z)

for every q in rand z in U. Then A(T)k forms a subspace of (!)(U) over C. LetA(T) denote the subring of the ring (!)(U) generated by A(T)o,A(Th, ... , i.e., put A(T)= L A(T)k· k~O

Since U is connected by assumption, the ring (!)(U), hence also its subring A(T), is an integral domain. We shall give a sufficient condition under which A(T) forms a graded ring. Let roo denote the set of those q in r satisfying p(q, )= 1. Then, with q, q' in roo' q q' and q-1 are both in roo; hence roo is a subgroup of r. Lemma 12. If the index of roo in r is infinite, A(T) is a graded ring. Proof. Suppose that q', q" are elements of r satisfyingp(q', )= p(q", ). Then, if we put q=q"(q')-1, we have p(q", z)=p(q, q'. z) p(q', z) for every z in U. By cancelling p(q",z)=p(q',z), we see that q is in roo; the converse is also true. Therefore we have p(q', )=p(q", ) if and only if roo q' = roo q".

§5. Automorphic Forms and Projective Embeddings

113

Suppose that we have an element h of A(T)k for each k satisfying We shall show that fo+'" + fN=O implies h=O for every k. Let u denote an arbitrary element of r; then we have O~k~N.

N

N

L h(u' z)= L p(u, Z)k h(z)=O

k=O k=O for every z in U. Since the index of roo in r is inftnite by assumption, we can fmd N + 1 elements u O' u 1 ' ... , UN of r satisfying roo u;=t=roo u j ' i.e., p(u;, ) =t=p (u j , ), for every i=t=j. Then the function det(p(u;, r);,k=

±

n(p(u;,

)-p(uj ,

))

i
on U is different from the constant O. Let X denote the set of points of U where this function vanishes; then, for every z in U - X, the following system of linear equations: N

L p(u;,zr h(z)=O

(O~i~N)

k=O has only the trivial solution fo (z) = '" =fN(Z) =0. This implies that h=O for every k. q.e.d. We shall assume that A(T) is a graded ring; we shall show that A(T) is normal, i.e., integrally closed in its field of quotients F(A(T»). By Lemma 6, the normalization, say S, of A(T) is a graded ring. Therefore we have only to show that Sk is contained in A(T)k for every k~O. We have shown in §3 that the ring (!)(U) is normal. Since A(T) is a subring of (!)(U), S is contained in (!)(U). Therefore, if f is an element of Sk,f is holomorphic on U. On the other hand, f can be written in the form alb with a in A(T)k+p and b=t=O in A(T)p for some p~O. This implies that f(u· z)= p(u, Z)k f(z) for every u in rand z in U. Hence f is contained in A(T)k' We keep the assumption that A(T) is a graded ring. Suppose that r is contained as a subgroup in a group r* which acts on U as a group of biholomorphic mappings. Suppose also that there exists a holomorphic automorphy factor p* relative to r* which restricts to p, i.e., satisfying p*(u, )=p(u, ) for every u in r. Then, by using (p*)k for a given k, we can let r* act on the vector space (!)(U) as before. If u* is in r*, by defmition we have u* . A(T)k=A(u* r(U*)-l)k' Therefore, if we apply u* to each A(T)k by using (P*)k for k=O, 1, ... , we get a degree-preserving isomorphism of the graded ring A(T) to the ring A(u* r(u*)-l) provided that this is also a graded ring. In the special case where r is a normal subgroup of r*, r* acts on A(T) as a group of degree-preserving automorphism.

114

III. Graded Rings of Theta Functions

We shall discuss two special cases. Let Z/L denote a complex torus and u~(z) =e(1/2 i) Q(z, ~) +(1/4 i) Q(~, ~)+ l(~») I/I(~), in which z is in Z and ~ in L, the automorphy factor in Chap. II, Theorem 3. Then we can take Z as U, L as r, and u~(z) as p(~, z), where the action of Lon Z is defmed as ~. z=z+~; we have A (L)k = L(k Q, k l, I/I k)

for every k~O. If the index of LrTJ in L is finite, LrTJ is a lattice in Z. Then, by the uniqueness (in "Theorem 3") applied to u~(z) for ~ in L rTJ , we get Q=O, l=O. Since 1/1 is strongly associated with Im(Her(Q»)=O, we get 1/1=1; hence LrTJ=L, A(L)k=C for every k~O. Except for this case, A(L) is a graded ring over A(L)o=C. We shall pass to the second example. If (1 is an element of SPzg(R) with (y t5) as its bottom row and 1: a point of the Siegel upper-half space U=($g'

p«(1, 1:)=det(y 1: +t5)

defines a holomorphic automorphy factor p relative to SPzg(R). We shall consider restrictions of p to certain subgroups of SPzg(R). If 1 is a positive integer, we shall denote by I;(l) the subset of SPZg(Z) consisting of those (1 which satisfy the congruence: (1 == 1z gmod l. Since I; (l) is the kernel of the homomorphism SPZg(Z) - SPzg(Z/l Z) defined by (1- (1 mod I, it is a normal subgroup of SPZg(Z) of finite index. We call I;(l) the principal congruence group (of degree g and) of levell; we have I;(l)=SPZg(Z). A subgroup r ofSPzg(R) which contains I;(l) for some 1 as a subgroup of fmite index is called a modular group. If r is a modular group, we define the ring A(r) by the general procedure. In the special case where g= 1, for our later purposes, we have to modify the definition of A(r). By assumption, r contains 11 (l) for some I. Since 11 (l) is normal in 11(1), if f is in A(r)k' for every (1 in 1}(1), (1·f is in A(I1(l))k· Let lo denote the smallest positive integer such that (1. f is invariant under 1:-1:+10; clearly 10 divides I; moreover, if we put t=e(1:/10), «(1.f)(1:) can be expanded into a Laurent series as «(1. f)(1:) =

Lap t

P,

peZ

and it converges for 0 < Itl < 1. The additional condition is that, for every (1, this is a power series in t, i.e., ap=O for all p
§5. Automorphic Forms and Projective Embeddings

115

if-part. Suppose that 1'*,0. Choose u, v from GLg(R) such that if we put

° 0) (IX

(u

we have

tu-l

l'

1'* =

° 0) = (IX*

fJ\ (V

JJ

tv - l

1'*

f3*)

J*'

(~ ~), det(y') *,0.

Since 1'* tJ* is symmetric, we have

J*= (Jo'

*)

J" .

Hence, if we put

we have

det(y T+ J)=c . det(T' +1" -1 J');

the constant c*,O is the product of the determinants of u, v, 1", J". If the degree of 1" is g', the mapping 6 g ~ 6 g, defined by T~ T' is surjective. Since det(T' +1" -1 J') is not a constant function of T', it is not a constant function of T. This proves the only if-part. It follows from what we have shown that roo consists of those (J in r for which l' = and det (J) = 1. We shall show that A(r) is a graded ring. By assumption, r contains 1;(1) for some I; then r contains

°

*'

for every k in IZ. Since we have roo (Jk roo (Jk' for k*,k', the index of is infinite. Therefore, by Lemma 12, A(r) is a graded ring. In the case where g = 1, the modified A(r) is contained in the A(r) defined by the general procedure. Since this is a graded ring and the modification is componentwise, the modified A(r) is also a graded ring. We introduced the ring of "automorphic forms" abstractly and gave two examples. In the following we shall discuss the general pattern of "projective embeddings" by automorphic forms. Let U, W denote connected complex manifolds; by an embedding of U in Wwe understand a holomorphic mapping f of U to W which gives rise to a biholomorphic mapping of U to a locally closed submanifold of W. Once we know that a holomorphic mapping f of U to W is injective and the image locally closed, f will be an embedding if it is "immersive ", or "differentially injective", in the following sense: Let a denote an arbitrary point of U and (Z1 ••• zn) local coordinates of U valid in some open neighborhood of a; similarly, let (WI ... W N ) denote local coor-dinates of Wvalid in some

roo in r

116

III. Graded Rings of Theta Functions

open neighborhood of f(a). Then .t;(z) = Wi (J(z)) for 1 ~ i ~ N are defined and hoI omorphic on some open neighborhood of a. The condition is that rank (ofJoz)(a))i,j = n. Geometrically this means that the differential df of f gives a C-linear injection of the tangent space of V at a to the tangent space of W at f(a). We shall be concerned exclusively about the case where W is a complex projective space. In this case, f is called a projective embedding of V. We shall recall the definition of a projective space and projective varieties. Let Q denote an algebraically closed field and n a non-negative integer. We let QX act on Qn+1_0 as (x o ... xn)--+(txO'" tx n)

(tEQ X ) .

Let P"(Q) denote the corresponding quotient set and n the canonical mapping of Qn+1 -0 to P"(Q); we call P"(Q) the n dimensional projective space. A subset X of P"(Q) is called aZariski closed set in P"(Q) ifn-1(X)u 0 is a Zariski closed set in Qn+1. We observe that a Zariski closed set X* in Qn+ 1 is of the form n -1 (X) u 0 if and only if X* is defmed by a set of homogeneous polynomials in n + 1 letters To, ... , T" with coefficients in Q, i.e., if and only if ~(X*) is a homogeneous ideal of Q[To, ... , T,,]. In this case, the dimension of X is defined as dim(X)=dim(X*)-l. We say that X is an irreducible projective variety, or just a projective variety, if ~(X*) is a prime ideal; we say that a point x=n(x o '" x n) of X is a simple point of X, or x is simple on X, if (xo .. , x n ) is a simple point of X*; this depends only on x. Let U; denote the complement of the hyperplane in P"(Q) defmed by 7;=0. Since this "hyperplane" is a Zariski closed set, in fact a projective variety, Vi is a Zariski open set in P"(Q); moreover, we have n

P"(Q)=

U U;.

If we map Qn to Qi X 1 x Qn-i in an obvious manner and then map this to P"(Q) by n, we get a bijection of Qn to Vi' If we identify Vi with Qn through this bijection, for every Zariski closed set X in P"(Q), Xi=X n U; is a Zariski closed set in Qn; and we have n

X=UX i· If X is a projective variety, Xi is either empty or a variety of the same dimension; and a point x of Xi is simple on X if and only if it is simple on Xi'

§5. Automorphic Forms and Projective Embeddings

117

If we take C as Q, P"(Q)= P,.(C) has two topologies, i.e., the usual quotient topology and the Zariski topology. We shall not specify the topology if we are talking about the usual topology. We observe that P,.(C) is compact; also, it can be considered as a complex manifold with ~=cn as a coordinate neighborhood; this is the n dimensional complex projective space. Every Zariski closed set X in P,.(C) is closed, hence compact. If X is a projective variety of dimension d, the set Xs of simple points of X forms a submanifold of P"(C) of complex dimension d. Furthermore X. is an open connected subset of X; this follows immediately from Proposition 3. Let U denote a connected complex manifold, r a group acting on U as a group of biholomorphic mappings, and p a holomorphic automorphy factor relative to r; then we can define A(r) by the general procedure. Suppose that A(r)k is a finite dimensional vector space over C and let 10 J1' ." IN denote a C-base of A(r)k' Then we can consider the holomorphic mapping U ~ CN + 1 defined by z~(Jo(z)I1(z) ... IN(z)),

If 10 J1' ... I N do not vanish simultaneously at any point Z of U, by composing with n, we get a holomorphic mapping U ~.&(C). We observe that, for every (j in r, z and (j • z have the same image in .&(C). Therefore, if r\ U denotes the quotient space of U by r, we get a welldefined continuous mapping

I: r\u ~PN(C)' We say that the action of r on U is properly discontinuous if, for every compact subset C of U, the set of all (j's in r satisfying (j. C n C::j::~ is finite. In this case, the quotient space r\ U is a locally compact Hausdorff space. If, moreover, r acts freely, i. e., without fixed points, on U, r\ U becomes a complex manifold and U its covering manifold. In this case, if the image is locally closed and f, dI both injective, I gives a projective embedding of r\ u. We observe that the change of the C-base Io,ft, ... IN of A(r)k amounts to applying a "projective transformation" in '&(C), which is a biholomorphic mapping of '&(C) to itself. Therefore we sometimes say that A(r)k gives a projective embedding of r\ u. We recall that if U can be considered as the coset space G/K of a topological group G by a compact subgroup K, the action of any discrete subgroup r of G on G/K is properly discontinuous. In fact, if C is a compact subset of G/K, its inverse image, say C", under the canonical mapping G--+- G/K is also compact. Then C" (C")-1 is compact, hence r n C" (C") -1 is finite. Therefore we have only to observe that every (j in r satisfying (j. CnC::j::~ is contained in rnC"(C")-l. We shall show that the two examples fall into this pattern. In the first case, we

118

III. Graded Rings of Theta Functions

can take G=Z, K=O, and r=L; this is an obvious case. In the second case, Chap. I, TheoremS shows that we can take G=SP2g(R), K= SP2g(R)n02g (R), and r=any discrete subgroup of Sp2g(R). Therefore, if r is a modular group acting freely on 6 g, the quotient space r\ 6 g is a complex manifold with 6 g as its universal covering manifold. The fmal remark is that if fo, h' ... ,fN are holomorphic functions on some open neighborhood of a point a=(al ... aJ of cn and fo(a)=j:O, we have rank ((0 (fJ fo)/oZj)(a»)i,j= rank(/;(a)(ojJozj)(a»)i,j-l. The verification is straightforward. We shall use this fact in verifying the injectivity of df.

§ 6. Polarized Abelian Varieties Let Z/L denote a complex torus; consider the additive monoid of Riemann forms on Z x Z relative to L. We say that Z/L is an abelian variety if this monoid contains at least one positive-definite hermitian form. In this case, consider the submonoid of all non-degenerate Riemann forms (on Z x Z relative to L). Two elements ofthis submonoid are called "commensurable" if their suitable positive integer multiples are equal; and each commensurability class is called a polarization of Z/L. If Z/L satisfies a certain general condition, neither the monoid of all Riemann forms nor the submonoid of non-degenerate Riemann forms is fmitely generated; hence there exist infinitely many polarizations. We shall show that every polarization contains a unique Riemann form of which all others (in the polarization) are positive integer multiples. Let H denote an arbitrary element of the polarization; we choose a canonical base "','2g of L relative to A=lm(H). Then there exist positive integers el> ... , eg , each ei dividing ei+l for 1 ~i
'1'

A

(~lXi 'i> i~/i 'i) = itlei(Xi YHi-Yi XHi)'

It is clear that ell H is an element of the polarization and all others are positive integer multiples of ell H. We call this unique Riemann form the minimal element of the polarization. If we choose a polarization of Z/L, this becomes what we call a polarized abelian variety. A polarized abelian variety is, therefore, the pair of Z/L and a polarization. Let H denote the minimal element of the polarization and Q a quasi-hermitian form on Z x Z with H as its hermitian part; let I denote a C-linear form on Z and", a second degree character of L associated with A = 1m (H). We shall consider the subring

§ 6. Polarized Abelian Varieties

119

S = S (Q, I, l/I) of the ring (9 (Z) generated by

Sk=L(kQ, k I, l/I k) for k = 0, 1, .... We call S the ring of theta functions of the polarized abelian variety Z/L, i.e., of (Z/L, (kH)k> 0)' We have seen in § 5 that S is a normal graded integral domain over C with Sk as its homogeneous component of degree k. Let Pf(A)=PfdA) denote the Pfaffian of A relative to L; then, by Chap. II, Theorem 4, we have dim(SJ= Pf(A) kg for k = 1, 2, ... , in which g = dimdZ). In particular, S satisfies the componentwise finiteness condition. We shall see later that S is ring-finite over C. Let (Z/L,(kH)k>o) and (Z*/L*,(kH*)k>o) denote two polarized abelian varieties; let v denote a C-linear isomorphism of Z to Z*. We say that v is an isomorphism of (Z/L, (kH)k> 0) to (Z* /L*, (kH*)k>o) if v maps L surjectively, hence bijectively, to L* and H* to H. In this case v gives rise to a holomorphic isomorphism of Z/L to Z* /L*; conversely, every holomorphic isomorphism of Z/L to Z* /L* comes from a C-linear isomorphism v of Z to Z* mapping L bijectively to L*. Therefore we can say that an isomorphism of polarized abelian varieties is a holomorphic isomorphism of the underlying complex tori which is compatible with the given polarizations. On the other hand, we say that (Z/L, (kH)k> 0) and (Z* /L*, (kH*)k> 0) are of the same type ifthere exists an isomorphism of L to L* mapping (the restriction to L* x L* of) Im(H*) to 1m (H). Clearly, isomorphic polarized abelian varieties are of the same type. Let C(J denote the class of all polarized abelian varieties of the same given type; consider the totality of isomorphism classes of polarized abelian varieties in C(J. We shall show that this "totality" is a set and that there exists a "naturally defined" bijection of this set to a certain quotient space of the Siegel upper-half space. Let (Z/L, (kH)k::'O) denote an arbitrary polarized abelian variety in C(J; choose a canonical base ~l' •.. , ~Zg of L relative to A=Im(H); then we get g positive integers el, ... ,eg as before. Let e denote the diagonal matrix of degree g with ei as its i-th diagonal coefficient for 1 ~ i ~ g; then e does not depend on the choice of the canonical base nor on the polarized abelian variety (Z/L, (kH)k>o); in fact, it depends only on C(J. As in Chap. II, § 4, we introduce coordinates in Z with respect t.o ell ~g+l' ... , e;:l ~Zg. We have seen that the matrix T with the coordinates of ~i as its i-th row vector for 1 ~ i ~ g is a point of 6 g. Consider the subgroup G of GLzg(R) consisting of those (J in Mzg(R) which satisfy (J

e) (-e0 0e)t (0 -e- 0 . (J=

120

III. Graded Rings of Theta Functions

Suppose that an element (1 of G is composed of oc, p, y,~; and put T# =(ocT+pe)(YT+~e)-l e. We have seen in Chap. II, § 5 that T# is a point of 6 g and G acts biholomorphicallyon 6 g as T--+T#. Put GZ =GnM2g (Z). Then the image ofT in the quotient space Gz \ 6 g does not depend on the choice of the canonical base el> ... , e2g; hence it depends only on (Z/L, (kH)k> 0)' Moreover two polarized abelian varieties in ~ have the same image in Gz \ 6 g if and only if they are isomorphic; and every point of Gz \ 6 g comes from a polarized abelian variety in ~. All these are contained in Chap. II, § 5. We shall show that G is conjugate to Sp2g(R) by an inner automorphism of GL 2g (R). Let (1 denote an arbitrary element of M 2g (R); then, clearly, (1 is contained in G if and only if

(1*=(~g ~rl(1(~g ~) is contained in SP2g(R). This proves the assertion. We observe that if (1 and (1* are composed of oc, p, y, ~ and oc*, p*, y*, ~*, we have oc*=oc,

p*=pe,

y*=e- 1 y,

~*=e-l~e.

At any rate, the correspondence (1--+ (1* gives an isomorphism of G to Sp2g(R). We observe that ifT is an arbitrary point of 6 g and (1 an element of G, the above definition of T# can. be rewritten as T# =(1*' T, in which (1* . T is the familiar action of SP2g(R) on 6 g. Consequently, the isomorphism of G to Sp2g(R) is compatible with their actions on 6 g. Let r denote the image of Gz in SP2g(R); then, by what we have said, the quotient space Gz \ 6 g can be identified with the quotient space r\ 6 g. We shall show that r is a modular, group; we have only to show that r contains .fg(eg) as a subgroup of finite index. Let I denote any positive integer multiple of ego If (1* is in .fg(l), oc =oc*, p= P* e- 1, y =ey*, ~=e~* e- 1 are all in Mg(Z); hence (1 is in Gz . This shows that r contains .fg(l). Let I' denote a positive integer and Gz(l/) the normal subgroup of Gz defined by (1= 12g mod 1'; then the index of Gz(l/) in Gz is certainly finite. We observe that the image of GzW) in Sp2g(R) is contained in .fg(l); hence [r:.fg(l)] ~ [G z : GzW)] < 00.

Therefore r is a modular group. We go back to the ring of theta functions ofa polarized abelian variety. We observe that there are infinitely many rings of theta functions for the

§ 6. Polarized Abelian Varieties

121

given polarization; nevertheless, they are all isomorphic under degreepreserving isomorphisms. This follows from the following theorem: Theorem 3. Let Z/L denote an abelian variety and consider the graded ring of theta functions:

in which H = Her(Q) is non-degenerate. Then S is uniquely determined by H up to a degree-preserving isomorphism over C. Proof Let ~b ••• , form B on Z x Z as

~2g

denote a Z-base of L and define an R-bilinear 2g

2g

)

B i~IXi~;' i~/i~i =i~/Ri,OXiYj, (

in which A=Im(H); then I/Io(~) =e (1· B(~, ~))

is a second degree character of L associated with A. We have only to show that there exists a degree-preserving isomorphism CI( of S to L(k H, 0, I/Ii) over C. Let a denote a point of Z and, for every () in

L

k;;;O

L(Q, I, 1/1), define ()1 as

()1(Z)=()(z+a);

then, by Chap. II, Lemma 6,

()1

is in L(Q, ib 1/11), in which

11(Z)=I(z)+ 1/2i· Q(z, a),

1/11(~)=I/I(~) e(A(a, ~)).

Since A is non-degenerate, for any character X of L, there exists a point a of Z satisfying x(~)=e(A(a, ~)) for every ~ in L. In particular, this is true for the character x=I/IoN; thus we get 1/11 =1/10' Let ()o denote the trivial theta function on Z defined by ()o(z)=e( -(1/4i) S(z, z)-11(z)),

in which S = Sym (Q) and 11 as above. Then, for every ()1 in L( Q, Ib 1/10), ()2 = ()o ()1 is in L(H, 0, 1/1 0)' After observing these things, we put (CI(k

())(z) =

()o (Z)k

()(z + a)

for every () in Sk = L(k Q, k I, I/Ik). Then Cl(k () is in L(k H, 0, I/Ii) and Cl(k gives a C-linear isomorphism of Sk to L(kH, 0, I/Ii); moreover, if (), ()' are in Sk, Sk" we have Cl(k+ k' (() ()') = (CI(k ()) (CI(k' ()').

Therefore CI(=(CI(k)k
k;;;O

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III. Graded Rings of Theta Functions

We shall be interested in the "structure constants" of the ring of theta functions of a polarized abelian variety; roughly speaking, these are the "theta constants" to which the next two chapters will be devoted. In order to get some feeling about the ring of theta functions, we shall examine the special case where g = 1. First we shall determine the monoid of all Riemann forms on Z x Z relative to L. We choose a Z-base ';1,';2 of L and introduce a coordinate in Z via the C-linear isomorphism C ~ Z defined by z ~ Z';2. If we identify a point of Z with its coordinate, we get ';2 = 1. Since ';1,';2 are linearly independent over R, ';1 is not in R. By replacing ';1 by its negative if necessary, we may assume that ';1 has a positive imaginary part, i.e., ';1 is in 6 1; we put ';1 =T. Let H denote an arbitrary hermitian form on CxC; then h=H(l, l) is in R and we have H(z,w)=hzw for every z, w in C. We have H(z, z)~O for every z in C if and only if h~O; and A=Im(H) is Z-valued on Lx L if and only if h· Im(T) is in Z. We have thus proved the following fact: Every complex torus ofcomplex dimension 1

is an abelian variety with only one polarization; it is isomorphic to (CIL, (kH)k> 0) in which L=ZT+Z for some Tin 6 1 and H(z, w)= (111m (T») zw

for every z, w in C. Moreover, if we map (CIL, (kH)k> 0) to the image point

of T in 11 (1)\ 61> we get a bijection of the set of all isomorphism classes of complex tori of complex dimension 1 to 11(1)\61 . By using the above notations, we put A=Im(H); then we have

A(x' T+X", y' T+Y")=X' y"-y' x" for every x', x", y', y" in R. Therefore the second degree character 1/1 of L defined by

I/I(PT+q)=( -l)pq

for every p, q in Z is associated with A. Moreover the quasi-hermitian form Q on C x C defined by

Q(z, w)=(l/Im(T»)z(w-w) has H as its hermitian part. We shall determine the ring of theta functions S=S(Q, 0, 1/1). We observe that St consists of hoiomorphic functions 0 on C satisfying O(Z+pT+q)=e( _k(p2 T/2+pz») O(z) for every p, q in Z. Since the functions z ~ O( - z) satisfies the same condition, it is in St. Therefore, if we denote by and Sk the subspaces of Sk consisting of even and odd functions in Sk, we have the direct sum decomposition:

st

§ 6. Polarized Abelian Varieties

123

On the other hand, since Pf(A) = 1, we get dim(Sk)=k for k=1,2, .... In particular Sl is generated over C by one element; and we know what this element is. For the convenience of the reader, we shall copy the definition of the familiar theta series and their properties (in the case where the dimension is 1). We put 9 ij (z)= L e{(r/2) (p+ i/2f +(p+ i/2) (z+ j/2)) peZ

for every i,j in Z (and Z in C); then we have (1)

9ij( -z)=( _l)ij 9ij(z),

(2)

9;+2 iz)=9ij(z),

(3)

9 ij (z) =e(i2 r/8 + i(z/2 + j/4)) 9 00 {z +(i r + j)/2) ,

(4)

9 ij (z+ p r +q)=( -l);q-j p e( - p2 r/2- p z) 9 ij (z)

9;j+2(Z)=( _l)i 9ij(z) ,

for every p, q in Z. In fact, in the notation of Chap. I, § 10, we have 9ij(z) = fJ;/2j/2 (r, z);

and (1), (2), (3), (4) follow respectively from (fJ.1)-(fJ.2), (fJ.2), (fJ.4), (fJ.5). By (2) we may assume that i, j = 0, 1; then (1) is equivalent to saying that 9ij(z) is an even function for (ij) =(00), (01), (10) and an odd function for (ij) =(11). We go back to Sk' We see by (4) that 9 00 is in Sl; hence Sl

=st =C9 00 '

We observe that 9 00 has only one zero (of order 1) in C/L. In fact, if we recall Theorem 2 and the corollary of Theorem 4 both in Chap. II, this follows from the fact that the Riemann form of 9 00 is minimal; it is also an easy exercise in function theory. We see by (3) that every 9ij has only one zero (of order 1) in CjL. Since 9 11 is an odd function, it has the image of 0 in C/L as its zero; hence the zeros of 900,901,910 in C/L are respectively the images of (r + 1)/2, r/2, i. On the other hand, we see by (4) that the (9;)2 are all in S2' Since the zeros of 9;j are different, any two of the (9ij)2 are linearly independent over C. Since we have dim(S2)=2, we get We shall determine Sk for every k"?; 3. We shall show that C [S 2] is isomorphic over C to the ring of polynomials in two letters with coefficients in C. We know that C[S2] =C[(9 00 )2, (9 11 )2] is a graded integral domain over C and (9 00 )2, (9 11 )2 are elements of degree 2. Therefore, if (9 00 )2, (9 11 )2 are algebraically dependent over C, there exists a homogeneous polynomials F =1= 0 in two letters with- coefficients in C satisfying

III. Graded Rings of Theta Functions

124

F(.9 00 )2, (.9 11 )2)=0. On the other hand, since C is algebraically closed, F is a product of linear factors; hence a linear combination of (.9 00 )2 and (.9 11 )2 with coefficients in C has to vanish. But this can not be the case. As a consequence, we have dim(S2Y')=p+1 for every p~O with the understanding that (S2)0=C. Suppose that k is even, say k=2p; then Sip contains (S2Y and Sip contains (S2)p-2 .900 .901 .910 .9 11 , If we compute the sum of the dimensions of these two subspaces over C, we get (p + 1) + (p-2+ 1)=2p=dim(S2p); hence we have Sip = (S2)P,

Sip = (S2)p-2 .900 .901 .9 10 .9 11

for every p ~ 2. Suppose next that k is odd, say k = 2 p + 1. Then, in the same way as above, we have

Sip+1 =(S2y- 1 .901 .9 10 .9 11

Sip+1 = (S2Y' .900 ,

for every p ~ 1. In particular, we have

S=C[x,Y,z], in which x=.9 oo , y=(.9 11 )2, z=.9 01 .9 10 .9 11 , We shall determine the "ideal of relations" between x, y, z. If we introduce a ring of polynomials C [11, T2, T3] in three letters 11, T2, T3 and consider the surjective homomorphism C [11, T2 , T3 ] -. C [x, y, z] = S over C defined by 11 -. x, T2 -. y, T3 -. z, the kernel is the ideal of relations in question. Since (.9 00)2, (.9 11 )2 form a C-base of S2, we have (.9 01 )2 = K' (.9 oo f + K(.9 11 )2

(5) for some we get (6)

K,

Ie' in C. If we evaluate both sides at z+(r+ 1)/2, by using (3)

(.910)2 = K(.9 00 )2 - K' (.9 11 )2.

Therefore, if we put p= T32-(K 112 -K' T2)(K'.112+K T2) T2, we have P(x, y, z)=O; hence we get a surjective homomorphism C [11, T2, T3]/(P) -. S, in which (P) is the principal ideal C [11, T2, T3] P. If we compute the generating functions of these two graded rings over C, we get (1-t 6 )/(1-t)(1-t2)(1-t 3) for C[11, T2, T3]/(P) and

1+

k~1kt'=1+t.d t~/) /dt =

1+t/(1-t)2

for S. Since they are equal, the above homomorphism is an isomorphism. We have thus shown that Z2=(KX 2-K' y) (K' X2 +KY) Y

§ 7. Projective Embeddings

125

is the only relation between x, y, z in the sense that (P) is the ideal of relations between x, y, z. We shall make the constants K, K' explicit and fmd their

relations. If we evaluate both sides of (5) at z+t and use (3) again, we get (7)

by evaluating (5), (6), (7) at 0, we get K

= .9 10 (0)2/.9 00 (Of,

K'

=.9 01 (0)2 /.9 00 (0?

K2+(K'f=1.

We shall see later that K2 + (K'f = 1 is the only relation between K and K' as functions on 6 1 .

§ 7. Projective Embeddings We shall prove a theorem on projective embeddings of a polarized abelian variety. We need several lemmas:

Lemma 13. Let Z/L denote a complex torus and () an element of L(Q, 1, tjJ). Then,for any m points al> ... , am ofZ satisfying a l + ... +arn=O,

n (}(z+ai) m

e(z)=

i=l

defines an element e of L(m Q, m 1, tjJrn). Proof Let u~(z) denote the automorphy factor of L(Q, I, tjJ) and write it as e(Q~(z)+c~); then we have e(z+~)=e(z)·

n u~(z+a;) rn

i=l

=e(z)e

ttl (Q~(z+ai)+c~))

= e(z) e(m(Q~(z)+c~)).

We have only to observe, therefore, that e(m(Q~(z)+c~))=u~(zt is the automorphy factor of L(mQ, m 1, tjJm). q.e.d.

Lemma 14. Let Z/L denote a complex torus and a a given point of z. Then there exists an element e of L(mQ, m I, tjJm) satisfying e(a)=l=O provided that m ~ 2. Proof Let () denote a theta function of type (Q, 1, tjJ) and define e as in Lemma 13. Let X denote the support of the positive divisor «(}); it is the set of points z of Z satisfying (}(z) =0. Then X is a closed subset of Z

126

III. Graded Rings of Theta Functions

and it does not contain any non-empty open subset of Z. Consequently, we can choose ai so that a + ai is not contained in X for 1 ~ i < m. Also, by replacing one of the a10 ... , a...-1, say a1, by another point in a small open neighborhood of a1 if necessary, we can assume that a+am=a(a 1 + ... +am-1) is not contained in X. Then we have e(a)=I=O. q.e.d. We recall that if H is a hermitian form on Z x Z, H ~ 0 means H(z, z) ~ 0 for every z in Z; and H1~H2 means H1-H2~0. If S is a symmetric matrix in Mn(R), we get a hermitian form on en x en as H(z, w)=zstw; hence S ~ 0 and Sl ~ S2 are defined. Without passing to en, S ~ 0 can be defined as xstx~o for every x in Rn. The following lemma will be used in the proof of Lemma 16:

Lemma 15. If S10 S2 are symmetric matrices in Mn(R) satisfying Sl ~S2~0, we have tr(Si)~tr(S~). Proof. We recall that if S is positive-definite, it can be written as v tv for some v in GLn(R). Therefore, for every S2~0 we have tr(SS2)= tr(v tv S2) =trCv S2 v) ~ O. For every Sl ~O and A. > 0, S = Sl + A.1nis positivedefinite; by what we have shown, we have tr(SS2)= tr(Sl S2)+A. tr(S2)~0; hence tr(SlS2)~0. On the other hand, for a, b, c in R with c~O, a + 2 b A. + cA.2 is a monotone increasing function of A. ~ 0 if and only if b~O. Therefore, for every S10 S2~0,

tr(Sl +A.S2)2)=tr(S~)+2A. tr(SlS2)+A.2 tr(S~) is a monotone increasing function of A.~O; hence tr(Sl +S2)2)~tr(S~). Except for the notation, this is our lemma. q.e.d. In the following lemma, by Riemann forms we understand Riemann forms on Z x Z relative to L:

Lemma 16. For any Riemann form Ho, the number of Riemann forms H satisfying Ho ~ H is finite. Proof. We take a Z-base e1, ... , e2g of L and identify Z with R 2g as 2g Z= LXiei-+X=(X1 ... X2g)' i=l

Then, to the multiplication by i in Z corresponds in R 2g the right multiplication by an element J of M 2g (R); we have J2= -1 2g . Moreover, if we associate with H the alternating matrixE in M 2g (Z) as E= (A(e;. e))i,j' in which A = Im(H), for z-+x, w-+ y we have H(z, w)=A(iz, w)+iA(z, w) =(xJ) E ty+ixE ty;

§ 7. Projective Embeddings

127

hence JE=t(JE) and JE~O. Since J is non-singular, if we put

q(E) = tr(JE) t(JE») for every alternating matrix E in M 2g (R), we get a positive-defInite quadratic form q on this subspace of M 2g (R). Therefore the number of "integer points" E satisfying q(E)~const. is fInite. We observe that the R-bilinear form H defIned by an alternating matrix E in M 2g (R) as above is a Riemann form if and only if J E = '(J E), J E ~ 0, and E an integer point. Suppose that Eo is associated with Ho; then we have Ho~H if and only if JEo~JE. By Lemma 15, this implies that q(Eo)~q(E). Therefore the number of E's, hence the number of H's, is fInite. q.e.d.

As an application of Lemma 16, we shall show that every positive divisor D of the complex torus Z/L can be written uniquely in the form t

D= ~>iDi> i=1

in which Dl> ... , Dt are irreducible divisors, i. e., minimal elements of the additive monoid of positive divisors, and el> ... , et positive integers. Suppose that Di> D2 are positive divisors of Z/L satisfying D=D1+D2 , Di> D2 +0; let H, Hi> H2 denote the Riemann forms of theta functions of D, Di>D 2. Then we have H=H 1+H2' H1, H2+O. By Lemma 16, the number of Riemann forms H' satisfying H~H' is fInite. Therefore, we can not keep on decomposing D indermitely; hence we can represent D as stated; the uniqueness has been proved in the general case in § 3. In the following lemma, when we say a triple (Q, I, 1/1), it is understood that the vector space L(Q, I, 1/1) for the complex torus Z/L contains a theta function, i.e., Q, I, 1/1 satisfy the conditions in Chap. II, Theorem 3:

Lemma 17. Let (Q, I, 1/1) denote a triple in which Her(Q) is non-degenerate. Then it can be written as

i.e., Q=Q1 +Q2' 1=11 +12' 1/1=1/111/12' in which Her(Q1) is non-degenerate and such that every () + 0 in L(Q1, 11> 1/11) determines a reduced divisor ofZ / L. Proof Suppose that Hi> ... , Ht are hermitian forms on Z x Z satisHi~O for 1~i~t. Then, for any positive real numbers e1' ... ,eo L ei Hi is positive-dermite if and only if L Hi is positive-dermite. In fact,

fying i

i

for any point z of Z, Lei Hi(z, z)=O, Hi(z, z)=O for every i, and i

L Hi(z, z)=O are all equivalent. After this remark, we shall prove the i

lemma. By Lemma 16, if we put H = Her(Q),-the number of Riemann

III. Graded Rings of Theta Functions

128

forms H' satisfying H"?;,H' is finite. Consequently, H can be written as

L ei Hi> in which ei is a positive integer and Hi a minimal Riemann form. i Since H is non-degenerate, by what we have remarked, L Hi is nondegenerate. If L Hi can be written as Lei Hi, in which ei is a positive integer not all 1 and Hi a minimal Riemann form, we replace L Hi by L Hi. i

i

j

i

j

After a finite number of steps, we will get a non-degenerate Riemann form H' satisfying H"?;,H', which is "reduced" in the sense that, for no Riemann form H"=t=O, we have H' "?;,2H". Let (Q2,1 2, 1/12) denote a triple in which Her (Q2) = H - H'; such a triple exists by the corollary of Chap. II, Theorem 4. Put Q1 =Q-Q2' 11 =1-12' 1/11 =1/111/12; then (Q1,1l> 1/11) is a triple in our sense and Her(Q1)=H'. Moreover, since H' is reduced, every O=t=O in L(Ql> 11, 1/11) determines a reduced divisor of Z/L. q.e.d. We say that a positive divisor D of Z/L is non-degenerate if the theta function of D has a non-degenerate Riemann form. We observe that D is non-degenerate if and only if the group of translations in Z/L which keep D invariant is finite. In fact, if D is degenerate, the remark at the end of Chap. II, § 3 shows that the group in question is infinite. If D is nondegenerate, Chap. II, Theorems 2 and 5 show that the group is finite. Lemma 18. Let 0 denote a theta function on Z which determines a non-degenerate, reduced divisor D of ZjL; let (Zl ... Zg) denote coordinates in Z. Then

for all a satisfying O(a)=O spcm the entire space O. Proof. Since the lemma is trivially true for g = 1, we shall assume that g"?;,2. If the lemma is false, there exists an element C=(C1 ... cg)=t=O of c g such that g L Ci(OO/OZi) (b) =0 i=l

for every b satisfying O(b)=O. After a non-singular linear transformation in 0, we may assume that c=(1 0 ... 0). Decompose D into irreducible divisors Dl> ... , Dt and, for each i, let Ai denote the imaginary part of the Riemann form of the theta function of Di ; then A = L Ai is the imaginary i

part of the Riemann form of the theta function of D. We shall show that for at least one index i and an element ~ of L we have Ai(c, ~)=t=0. Otherwise, we have A(c, ~)=O for every ~ in L, hence A is degenerate. This contradicts the assumption that D is non-degenerate. Let a=(a1 ... ag) denote a point of Z such that n(a) for n:Z--+ZjL is contained in the support of Vi but not in the support of Dj for j =t= i.

§ 7. Projective Embeddings

129

Then, since D is reduced, as we have seen in § 3, for a suitable choice of a, we get grada(O)=l=O. Since the first coordinate of grada(O) is 0, after a non-singular linear transformation in Z2, ••• , Zg' which does not affect c=(1 0 ... 0), we may assume that grada(O)=(O ... O 1). Then, by the Weierstrass preparation theorem, we get

in which u is a unit of (9a and f a holomorphic function on some open neighborhood of a'=(a1 ... ag_1) satisfying f(a')=O. Let V, W denote open neighborhoods of a', ag in cg- 1, C such that u is a unit of (9 (V x W) and f an element of (9(V); by making V smaller if necessary, we may assume that bg=ag-f(b') is in W for every b' =(b1... bg_ 1) in V. Then we have O(b)=O for b=(b1... bg); hence

(fJO/fJz1) (b)=u(b) (fJf/fJz 1) (b)=O. Since u(b)=l=O, we get (fJf/fJz 1) (b) = (fJf/fJz 1) (b') =0; hence f depends only on Z2,"" Zg_1' Therefore, for every small tine, D j and its image 1Y; under the translation: Z1 -+ Z1 +t, Zj-+ Zj (1
Theorem 4. Let S denote the ring of theta functions of a polarized abelian variety (Z/L, (kH)k> 0)' Then,for every m~ 3, Sm gives a projective embedding of Z/L. Proof Let S=S(Q, I, 1/1) and decompose (Q, I, 1/1) into (Q1, 11, 1/11)+ (Q2, 12 , 1/12) as in Lemma 17; we recall that Her(Q1) is non-degenerate and reduced. We shall first show that L(m Q1> m 11> I/Ii) gives a projective embedding of Z/L. If N + 1 is the dimension of this vector space over C, by Lemma 14, it gives rise to a holomorphic mapping, say f, of Z/L to ]1(C). For this, it is sufficient that m~2. Since f is continuous and Z/L compact, the image of Z/L under f is a closed subset of PN(C). We shall show that f is injective. Let a,b denote points of Z such that n(a-b)=l=O. By Chap. II, Theorems 2 and 5, there exists an element 0 of L(Q1, i1, I/It> such that the identity is the only translation in Z/L which keeps the divisor D of Z/L satisfying n*(D)=(O) invariant. We know that D is reduced. Since the translation by n(a-b) maps D to D' =l=D, there exists an irreducible divisor D1 whose support is contained in the support of D but not in the support of D'. Otherwise, as we have seen in § 3, D' - D

130

III. Graded Rings of Theta Functions

becomes a positive divisor. Since the Riemann form of the theta function of D' -D is 0, we get D' -D=O; but this is not the case. Choose a point c of Z such that n(c) is contained in the support of Dl but not in the support of D'; then we have 9(c)=0 but 9(c-a+b)=I=O. Put a1=c-a and choose a2' ... , am so that al + ... + am = o. We say that the choice of a2, ... , am is "general" relative to b if 9(a2 + b) ... 9(am+ b) =1= O. As in the proof of Lemma 14, we can make such a choice provided that m~3. If we derme 8 as in Lemma 13, we get 8(a)=O but 8 (b) =1=0. Since 8 is an element of L(mQl' mil' I/In f is injective. We shall show that df is injective at any given point a of Z. By Lemma 18, there exist g points b1, ... ,bg of Z satisfying 9(b i)=O such that gradb,(9) for l;;;;;i;;;;;g are linearly independent over C. Let bo denote any point of Z satisfying 9(b o)=I=O. We define g+ 1 elements 8 i of L(mQ1, mil' I/Ii) as

n 9(z+ai), m

8 i (Z)=

j=l

in which ail = -a+b;, ail + ... +aim=O for O;;;;;i;;;;;g; we may assume that the choice of ai2 , ... , aim is general relative to a. Then we have 8 0 (a) =1= 0 and 8 i (a)=0 for 1 ;;;;;i;;;;;g; hence /;=8;/8 0 is a holomorphic function on some open neighborhood ofa satisfying/;(a)=Ofor 1 ;;;;;i;;;;;g. Moreover, since we have l
for l;;;;;i;;;;;g, they are linearly independent over C; hence df is injective at a. Therefore f gives a projective embedding of ZjL. We shall show that L(mQ, m l, I/Im) gives a projective embedding of ZjL. We observe that L(mQl,mll,I/Ii)L(mQ2,mI2,I/I~) is contained in L(mQ, m I, I/Im); and we have seen that L(mQb mil' I/Ii) gives a projective embedding of ZjL. On the other hand, by Lemma 14, for any given point a of Z, there exists an element of L(m Q2, m 12, I/I~) which does not vanish at a. By putting these together, we see that L(m Q, m 1, I/Im) gives a projective embedding of ZjL. q.e.d. We shall give the usual form of Theorem 4. Suppose that (Ql, 11, 1/11) and (Q2, 12, 1/12) are triples for ZjL satisfying Her(Q1)=Her(Q2); then there exists a C-linear isomorphism of L(Q1' 110 1/11) to L(Q2, 12, 1/12). More precisely, as we have shown in the non-degenerate case, there exist a trivial theta function 90 on Z and a point a of Z such that the isomorphism is given by 9(z) ---+ 9 0 (z) 9(z + a).

Therefore, if L(Ql, 11, 1/11) gives a projective embedding of ZjL, L(Q2, 12,1/12) also gives a projective embedding of ZjL. Consequently, if(Q', 1',1/1') is a

131

§ 7. Projective Embeddings

triple for Z/L such that Her(Q')=mH, in which m~3 and H a non-degenerate Riemann form (on Z x Z relative to L), L(Q', I', l/I') gives a projective embedding ofZ/L. We change our notation and assume that L(Q, I, l/I) gives a projective embedding of Z/L; then H = Her(Q) is non-degenerate. Let 0 , 1 , ••• , ON denote a C-base of L(Q, I, l/I); then the mapping

°°

z --+ (0 0 (z) 0l(Z) ... ON(Z))

gives rise to a projective embedding, say f, of Z/L. We shall show that the image X of Z/L under f is a projective variety in &(C) of dimension g=dimdZ). Consider the ring of polynomials C[To , .•. , TN] in N + 1 letters To, ... , TN and denote by ~ the kernel of the homomorphism C [To, ... , TN] --+ R = C [° 0 , ••• , ON] over C defined by 11 --+ 0i for 0 ~ i ~ N. Since R is a graded integral domain over C, ~ is a homogeneous prime ideal of C [To, ... , TN]' A homogeneous polynomial P is contained in ~ if and only if P(x) = 0 for every x in X. Therefore, if Y denotes the projective variety defined by ~, it is the smallest Zariski closed set in &(C) which contains X. We shall show that X = Y. If we put A = Im(H), we have dim(RJ~Pf(A)

kg

for k = 1, 2, .... Therefore, by Proposition 1, any g + 2 homogeneous elements of R are algebraically dependent over C; this implies that d=dim(Y)~g. Let Ys denote the set of simple points of Y; then Ys is a connected dense open subset of Y, and a submanifold of PN(C) of complex dimension d; moreover, Y - Ys is a Zariski closed set in PN(C). We have explained the idea of the proof in § 4. Since Y is the smallest Zarlski closed set containing X, Y- Ys does not contain X; hence X n Ys=!=S'. Let x denote an arbitrary point of X n Ys and take an open neighborhood V of x in PN(C) satisfying Y n V = Ys n V; since Ys is open in Y, this is possible. Then, since X is contained in Y, we have

(X n Ys)n V=X n Vc:Ysn V. We recall that X and Ys are submanifolds of PN(C) of complex dimensions g and d satisfying d~g. Therefore, we get d=g; also, if we make V smaller, we get X n V= Ysn v. This shows that X n Ys is open in Ys. On the other hand, since X is compact, X n Ys is closed in Ys. Since X n Ys is not empty and Ys connected, we get X n Ys= Ys; hence Ysc:X. Since Ys is dense in Y, we get Y=X. This completes the proof. We shall mention the case where g= 1. We have seen in § 6 that, up to an isomorphism, there is only one ring S to consider; we have S=C[x,y,z] in which

132

III. Graded Rings of Theta Functions

is the only relation between x=8 oo , y=(8 11 )2, z=8 01 810 8 11 , We have shown also that form a C-base of S3' If we express the relation between x, y, z in terms of xo, Xl' X2, we get

This defines an irreducible cubic curve Y in ~(C) which is "smooth" in the sense that Y= y.. By the general observation, z - (xo (z) Xl (z) X2 (z)) gives rise to a biholomorphic mapping of Z/L to Y. It is an instructive exercise to prove this fact, together with S(3)=C[S3]' without using general theorems.

§ 8. The Field of Abelian Functions We shall start by recalling the definition of meromorphic functions on a complex manifold U. Let f denote a holomorphic function on some dense open subset U' of U. Suppose that for any given point a of U there exist an open neighborhood V of a and two elements g, h of C/(V), in which h is not a zero divisor of (!) (V), satisfying hf = g on U' n V. Then we say that f is a meromorphic function on U. We observe that the intersection of a finite number of dense open subsets of U is a dense open subset of U. Therefore we can identify two meromorphic functions on U if they coincide on a dense open subset of U. We observe also that an element h of (!)(V) is not a zero divisor of (!)(V) if and only if h=l=O as an element of (!)b for every b in V. We see that the set of all meromorphic functions on U forms a ring, in fact a C-algebra, which contains (!)(U) as a subring. Moreover, if U is connected, we get a field, which contains F((!)(U)) as a subfield. We shall assume that U is connected.- Let f denote a meromorphic function on U different from O. We shall show that, in the definition of f, we can assume that g, h are relatively prime in (!)b for every b in V. We recall that g, h are elements of (!)(V) satisfying hf = g on U' n V. Since f=l=O, neither h nor g is a zero divisor of (!)(V). Let k denote a greatest common divisor of g, h in (!)a; by making V smaller if necessary, we may assume that g/k, h/k are in (!) (V). Therefore, from the beginning, we can assume that g, h are relatively prime in (!)a' As we have seen in § 3, if we make V still smaller, g, h are relatively prime in (!)b for every b in V. This proves the assertion. We shall define two positive divisors of U. We shall write v", ga' ha instead of V, g, h. Then we have haf=ga, hbf=gb on U' n v"b; hence

§ 8. The Field of Abelian Functions

133

on U' n Yab. Since U' n Yab is a dense open subset of Yab, we have ha gb=hb ga on Yab. Since ga and ha are relatively prime in @c for every c in Yab, we get gb=uga, hb=uha for some u in @(Yab)' Since gb and hb are relatively prime in @c for every c in Yab, U is a unit of @(Yab)' Therefore both (Ya, gala and (Ya, ha)" can be considered as representatives of positive divisors of U. Bya similar argument as above, we see that they depend only on f; hence we shall denote them by (f)o and (f)oo' Clearly, f can be extended to the complement of the support of (f)oo as a holomorphic function; and f is in @(U) if and only if (f)oo =0. We take as U the complex torus Z/L which admits at least one polarization. The meromorphic functions on Z/L are called abelian functions. In the special case where g=dimc(Z)= 1, they are called elliptic functions. We shall prove the following theorem:

Theorem 5. Let S denote the ring of theta functions of a polarized abelian variety (Z/L, (kH)k> 0); then Fo(S) is the field ofall abelianfunctions on Z/L. Moreover S is finitely generated over C. Proof. Since every element of Fo(S) is a meromorphic function on Z/L, we shall prove, conversely, that every meromorphic function f on Z/L is contained in Fo(S). We may assume thatf ,*0; also we shall identify f with f 0 n. Let ()' denote the theta function of (f)oo and suppose that it is of type (Q', I', 1/1'); then f()' is also a theta function of type (Q', 1', 1/1'). Let S=S(Q, I, 1/1) and put (Q*, 1*, 1/1*) = d(Q, I, 1/1) + (Q', I', 1/1'») for any d;;;;3; then Sd' L(dQ', dI', (I/I')d) is contained in L(Q*, 1*, 1/1*). Let ()o, ()t> ••• and ()t, ()r, ... denote C-bases of Sd and L(Q*, 1*, 1/1*); then

(()t(z) ()r(z) ... )

z-+ (()o(z) ()1(Z) ... ),

give rise to projective embeddings of Z/L. Let X, X* denote the corresponding images and put R=C[Sd] =C[()o, ()1' ... ];R*=C[()t,()r, ... ]. If ()" ,*0 is an element of L(dQ', dl', (I/I,)d), ()i ()" is in L(Q*, 1*, 1/1*); hence ()i ()" =

L cij ()j j

with cij in C. If we associate ()i ()" to ()i for i = 0, 1, ... , we get an injective homomorphism of R to R* over C. Although this homomorphism depends on ()", it gives rise to the inclusion of Fo(R) = C«()1/()0 , ()2/()0' ... ) in Fo(R*)=C«()tI()t, ()!I()t, ... ), which is intrinsic. Let K denote a countable subfield of C which contains, besides cij' the coefficients of finite ideal bases ofthe homogeneous prime ideals for X, X*. Let Y denote the Zariski open subset of X which we denoted by X 0 in § 5; in short, we may say that Y is defined by () 0 O. We observe that .y is the image of the set of

'*

134

III. Graded Rings of Theta Functions

points z of Z satisfying 00 (z) =1= 0 under the projective embedding of Z/L. We define y* similarly. Then Y, y* are irreducible affine varieties defined over K. Let y* =(yt y! ... ) denote a generic point of y* over K and put Yi=(C;o+ ~>ij yj)/(coo+ ~>Oj yj) j*O j*O

for i = 1, 2, ... ; then y = (Yl Y2 ... ) is a generic point of Y over K. In fact, theisomorphismK(Ot/O~, O!/O~, ... ) --+ K(y*) over K defined byOj/O~ --+ yl gives rise to the isomorphism K(OdOo, O2/0 0 , •.. )--+K(y) over K dermed by 0i/OO --+ Yi. Therefore, K(y*) is a finite algebraic extension of K(y); we shall show that K(y*)=K(y). Consider any conjugate field of K(y*) in Cover K(y) and denote the image of y* by y** =(yt* y!* ... ); then y** is also a generic point of y* over K. If we denote by z*, z** the points of Z which are mapped to y*, y** under the biholomorphic mapping Z/L--+ X*, we have yj = OJ (z*)/O~ (z*),

yj* = OJ (z**)jO~ (z**)

for every j; hence Oi(Z*)/OO (z*) = Yi = Oi{Z**)/Oo(z**)

for every i. This implies that y* = y**. Since K(y*) has no other conjugate than itself over K(y), we have K(y*)=K(y). Consequently, every yj can be written in the form Fj(Yl> Yl, ... )/GAYl> Y2, ... ), Gj(Yl, Y2, ... )=1=0 with Fj, Gj in the ring of polynomials K[J;, T2 , ... J. This implies that OJ/O~=Fj(OdOo,

O2/0 0 ,

.•• )/Gj(OdOo,

O2/0 0 ,

••. )

for every j; hence Fo (R) = Fo (R *). Since the meromorphic function jean be written as jOO(O')d/OO(O')d, it is in Fo(R*). Since R is contained in S(d), Fo(R) is contained in Fo(S(d»)=Fo(S); hence j is contained in Fo(S). We shall show that S is finitely generated over C. By Lemma 8, we have only to show that S(d) is finitely generated over C. Since we have shown that Fo (R) = Fo (S) = Fo (S(d») for R = C [Sd], by Lemma 10, S(d) is finitely generated over C if S(d) is integral over R. Let 0 denote an arbitrary element of Skd for any k~O. By Lemma 11, 0 is integral over R if O/O~ is integral over for every i. Let Y= Xi denote the Zariski open subset of X defined by 0i =1= 0; take any point y of Y and a sequence (y(V»)v in Y which converges to y as v --+ 00. We choose a sequence (z(V»)v in Z such that each z(v) is mapped to y(V) under the biholomorphic mapping Z/L--+ X. We may assume that (z(V»)v converges to a point z of Z as v --+ 00; then z is mapped to y. We observe that 0i(Z(V») =1=0 for every v and 0i(Z)=I=O. Therefore (O/O~)

(y(V») = o(Z(v»)j0i (Z(v»)k

(1 ~ v < (0)

§ 8. The Field of Abelian Functions

135

are bounded. Since y is an arbitrary point of Y and (y(V»v an arbitrary sequence in Y which converges to y as v -+00, in view of the results in §§ 2 and 4, O/e't is integral over Ri • q.e.d.

Remark. It follows from Theorem 5 and Lemma 3 that for a suitable d we have S(d)=C[Sd]. Since S is normal, SId), hence also C[Sd], will then be normal. On the other hand, if Sd gives a projective embedding of Z/L, the image is smooth in the sense that every point is simple (cf. Weil [20], Appendix, Theorem 8). Therefore, by a theorem of Zariski, the image is a "normal variety," and hence SId) and C[Sd] coincide except possibly for homogeneous components of small degrees. In other words, we have (S(d»k=C[Sd]k for almost all k. We shall give a rather peculiar proof to this fact in Chap. IV, § 5 in the special case where d=Omod 4.

Chapter IV

Equations Defining Abelian Varieties § 1. Theta Relations (Classical Forms) We shall start this chapter by proving "theta relations," i.e., relations between theta functions. More precisely, we shall be interested in polynomial relations between 0rn(-r, z), 0m(r, 0) with constant coefficients. From the" labyrinth" of theta relations, we shall select just two, which are themselves not unrelated: The first one is called" Riemann's theta formula" and the second one the" addition formula." Their shortest proofs depend on the following lemma: Lemma 1. Let L denote a discrete commutative group and L l , L2 two subgroups of finite indices; let iP denote an Ll-junction on L, i. e., a C-valued function on L such that the sum ofliP(~)1 over L is convergent. Then we have [L:LlJ' LiP(~)=L (LX(I1+n iP (l1+n), ~eLl

x.,

'I eL 2

in which X runs over the dual of LIL land (, over a complete set of representatives of LIL 2 . Proof We observe that if 11 runs over L2 and, over a complete set of representatives of LIL 2, ~=11+' runs over L. Since iP is an V-function on L (and X runs over a finite set), the right side of the proposed identity is an absolutely convergent series. If we fix ~ = 11 and take the summation over the dual of LIL l , by the orthogonality of characters (of the finite commutative group LIL l ), we get [L:LlJ' iP(~) if ~ is contained in Ll and 0 otherwise. q.e.d.

+,

We recall that the theta series 0m(r, z) has been defined as 0m(r, z)= L

e(! (p + m') r t(P + m') +(p + m'f(z + m")) ,

peZ g

in which m=(m' m") with m', m" in Rg, r in 6 g, and z in cg. Unless otherwise stated, the notation m=(m' m") will be kept throughout; if m', m" have complicated forms, we shall write Oem' m"J (r, z)instead ofOm'm,,(r, z).

§ 1. Theta Relations (Classical Forms)

137

Theorem 1. Let mh m2, m3' m4 denote elements ofR2gand Zl' Z2, Z3, Z4

points of Cg ; let

1 1 1 1) 1 1 -1 -1 ( T=! 1 -1 1 -1 (1 = 12g , 19), 1 -1 -1

1

and put (nl n2 n3 n4) = (ml m2 m3 m4) T, (Wl W2 W3 W4) = (Zl Z2 Z3 Z4) T.

Then we have

0mlr, Zl) ... 0m4(r, Z4)= 1/2g· L e( -2m! 'a") 0nl+a(r, Wl)'" a

0n4+a (r,

W4),

in which a=(o' a") runs over a complete set of representatives of! z 2g/z2 g • Proof Suppose that the «1" in the matrix T stands for ln for any positive integer n; then T is a symmetric orthogonal matrix of degree 4n. Therefore, if Uh U2, U3' U4 are elements of en and (Vl V2 V3 V4)= (Ul U2 U3 U4) T, the relation between u's and v's is symmetric, i.e., (Ul U2 U3 U4)=(Vl V2 V3 V4) T. Moreover we have 4

4

,.=1

P_l

L 'V,. V,. = L 'Up up,

4

4

,.=1

P=1

LV,. 'V,. = L uptup;

we can polarize these quadratic relations to convert them into bilinear relations. After this remark, we take n = g and replace Lh L 2 , L in Lemma 1 by Z4 g, Z4g T, Z4g + Z4g T. Moreover, we put 4

4'(e)= IT e(!(e,.+m~) 't'(e,.+m~+(e,.+m~'(z,.+m~») ,.=1

for every e =(el e2 e3 e4) in L and try to make the identity in Lemma 1 explicit. We observe that the right multiplication by T permutes Lh L2 and keeps L invariant. Moreover, if p runs over a complete set of representatives of Zg/2Zg, !(P p p p) runs over a complete set of representatives of LIL1 • In particular, we have [L:L 1] = [L:L 2] =2g • We take a" from! zg and put x(e)=e(2el 'a")

e

for every in L. If a" runs over a complete set of representatives of ! Zg/Zg, X runs over the dual of LIL 1 • Finally, if a' runs over a complete set of representatives of! Zg/Zg, '=(20' 000) runs over a complete set of representatives of LIL 2. This follows from the fact that it is a set of 2g elements of L which are incongruent mod L2 .'

138

IV. Equations Defining Abelian Varieties

We have seen (in Chap. I) that tJ> is an L1-function on L; the left side of the identity is clearly 2g • Om! (r, Zl) ... Om. (r, Z4). We shall compute the right side: If we put '1 = (ql q2 q3 q4), the right side becomes L e(2{ql +2a') ta") tJ>{ql + 2a', q2' q3, q4)'

L a

~eL2

t

in which a runs over a complete set of representatives of Z2 g/Z2 g • For the sake of simplicity, we put (rl r2 r3 r4)={ql +2a' q2 q3 q4); then this can be written as L L e{2ql ta")e

(

t

4

4

)

L{r,,+m~)Tt{r,,+m~)+ L{r,,+m~r{z,,+m~) .

a ~eL2 ,,= 1 ,,= 1 Since we have tr{AB)=tr{BA) for any two matrices A, B such that the products AB, BA are defined, we get

4

4

L (ra+m~) Tt{r,,+m~)= L tr({ra+m~) Tt{ra+m~)) a=l a=l 4

= L tr(Tt{ra+m~){ra+m~)). a=l By the definition of L 2 , (Pi P2 P3 P4)={ql q2 q3 q4) T runs over Z4 g ; and (rl r2 r3 r4) T={Pl +a' P2 +a' P3 +a' P4 +a').

Therefore, by the remark that we have made in the beginning, we get 4

4

L t{ra+m~){r,,+m~)= L t{pp+a' +np) (pp+a' +np); a=l P=l

similarly, we get 4

4

L (ra+m~t{za+m~)= L (pp+a' +np)t{wp+np). a=l P=l

Therefore, the right side becomes equal to L

L e{2ql ta") e

a PccEZg

(t ±(Pa+n~+a') Tt{Pa+n~+a') a:=1

+ atl{Pa+n~+a')t{Wa+n~)). Since

wa+n~ =wa+n~ +a" -a",

L

L e

a p"eZ g

(2 q

1

ta" -

this can be written as

i (p,,+n~+a')

ta")

,,=1 4

. TI e(t{Pa+n~+a')

T

t{p,,+n~+a')

§ 1. Theta Relations (Classical Forms)

139

Since LPa=2Qi and Ln~=2m~, the first exponential factor is equal to a

a

e( -2m~ ta"); hence the entire expression is Le( -2m~ ta") 0nl+a(r, Wi)'" 0n4+a(r, W4)' q.e.d. /I Before we pass to the addition formula, we remark that each summand e( - 2m~ taft) 0nt+a(-r, Wi) ... 0n4+a(-r, W4) in Riemann's formula depends only on a "mod Z2g. In fact, if we replace a by a+b for any b in Z2 g, the exponential factor is multiplied by e( -2m~ tb") while, in view of Chap. I, § 10, (0.2), the other factor is multiplied by e(n~ + a') tb" + ... +(n4 + a') tb")=e(2m~

let

Theorem 2. Let m1> m2 denote elements ofR2 g and

tb"). Z1> Z2

points ofCg ;

(1 1)

T="21 1 -1 (1 = 12g, 19), and put (ni n2)=(mi m2) T,

Then we have

(Wi W2)=(Zl Z2) T.

0ml (-r, Zl) Omz(-r, Z2)= 1/2g 'L e( - 2m~ taft) /I"

. 0[2n~ nl. + a"] (t-r, W1) 0[2n2 n2 + a"] (t-r, W2), in which a" runs over a complete set of representatives of t zg/zg. Proof. Suppose that the "1" in T stands for in; then T is a symmetric matrix satisfying T2 =t 12n . Therefore, if Uh U2 are elements of Cn and (V1 V2)=(U1 U2) T, we have (U1 u2)=2(V1 V2) T. Moreover we have 2 2 2 2 L tVa Va=t L 'Up Up, L VIZ'Va=t L Up tup; a=l P=l a_1 P_1 we also have the corresponding bilinear relations. We take n=g and replace L1> L 2, L in Lemma 1 by Z2 g, Z2g T, Z2g + Z2g T. Moreover, we put 2 4>@=

ne(!(~IZ+m~ -rt(~IZ+m~)+(~a+m~t(Za+m~»)

a=l for every ~ =(~1 ~2) in L and try to make the identity in Lemma 1 explicit. Since T- 1=2T is a Z-matrix, we have Z2 g T- 1 cZ2g, hence L=L 2. If p runs over a complete set of representatives of Zg/2Zg, p) runs over a complete set of representatives of L/L1 ; hence [L:LtJ =2g. Also, if a" runs over a complete set of representatives of t zg/zg,

t(P

x(~)=e(2~1

runs over the dual of L/L 1 •

ta")

140

IV. Equations Defining Abelian Varieties

Now, the left side of the identity is 2' . Om, (r, Z1) Om2 (r, Z2). If we put ,,=(q1 q2), the right side becomes

L L e(2q1 ta") lP(ql> q2)= L L e(2q1 ta") a" "eL2

a" "eL2

. e (tlXt(qlX+m;J't't(qlX+m;J+

IXt(qlX+m;Jt(ZIX+m~)),

in which a" runs over a complete set of representatives of t zg/zg. We can proceed exactly in the same way as in the proof of Theorem 1; we have made necessary remarks already. In this way, we get

~ ~ge(2q1 ta")e (!lXt1(p.. +2n~)'t't(pIX+2n;J p ..

+ IXt1 in which (q1 q2)=(P1 P2) T. Since written as

~ ~g e (2 q

1

P..

(p1X+2n~)t(wlX+n~)),

wlX+n~ =wlX+n~ +a" -a",

this can be

taft - IXt (P1X+ 2n;J ta")

.1X=1 ne(!(P1X + 2n~) 't' t(P1X + 2 n~) + (PIX + 2n;J t(WIX + n~ + a")). 2

Since L PIX = 2q1 and L n~ = m~, the first exponential factor is equal to IX IX e( -2m~ ta"); hence the entire expression is Le( -2m~ taft) O[2n~n~+a"] (f't', w1)O[2n2 n2+a"] (f't', W2). q.e.d. a" We remark that each summand on the right side of the addition formula depends only on a" mod Z'. In fact, if we replace a" by a" + b" for any b" in zg, the exponential factor is multiplied by e( -2m~ tb") and the other factor by e(2nl tb" +2n2 tb")=e(2m~ tb"). We shall make a few comments on Theorems 1 and 2. Suppose that r is a positive integer and r ml == ... == r m4 == 0 mod 1, i.e., mod Z2,; then we have r n1 == ... == r n4 mod 1. Therefore, if one of them is a Z-vector, all four become Z-vectors. This is the first comment. If we take we get We keep a,b arbitrary but assume that 2m==Omod 1; put

§ 1. Theta Relations (Classical Forms)

141

Then, by Theorem 1, we get

8 m (t, Z)= 1/2g • Le( -2m' In") n

. 02m+n(7:, WI) On+a(7:, W2) 0n+b(7:, W3) 0n_a_b(7:, W4); hence, by using Chap. I, § 10, (0.2), we get

8 m (7:, Z)= 1/2g • L e( -2(m" n" -n" m"») 8 n (7:, W), n

in which n runs over a complete set of representatives of t Z2 g/Z2 g • This (usually under the additional assumption that 2a=2b=0 mod 1) is the classical form of Riemann's formula. If we take we get In the corresponding formula, we put U3=U4=0 and replace UI' U2 by U, v; then, by omitting 7:, we have

Om,(U+v) 0m2(u-v) Om3(0) Om4(0) = 1/2g

•

L e(- 2ml 'a") On,+a(u) 0n2+a(u) 0n3+a(v) On4 +a (v) , a

t

in which a runs over a complete set of representatives of Z2 g/Z2 g• This is the classical addition formula for theta functions. In this respect, Theorem 2 is more a "transformation formula" (of degree 2) than an addition formula. (The word "transformation" means, in the modem terminology, an isogeny, i.e., a covering of an abelian variety by another.) We shall derive the following corollary from Theorem 2, which will play an important role in § 2: Corollary. Let m', n' denote elements ofRg; then we have

° .°

Om' 0(7:, z) On' 0(7:, z)= L [t(m' +n')+d 0] (27:, 2z) a'

[t(m' -n')+a' 0] (27:,0),

in which d runs over a complete set of representatives of t zg/zg. We replace ml' m2 by (t(m' +n')+a' 0), and Zl' Z2 by 2z, O. Then, we get

°

(t(m' -n')+d 0), 7: by 27:,

[t(m' +n')+a' 0] (27:, 2z) O[t(m'-n')+d 0] (27:,0)

= 1/2g • L e( -(m' +n' +2d) 'a") O[m' +2d a"] (7:, z) O[n' a"] (7:, z), all

-

142

IV. Equations Defining Abelian Varieties

t

in which a" runs over a complete set of representatives of zgjzg. Since 2a' is in zg, by definition we have O[m'+2a'a"J (T,z)=O[m'a"J (T,Z). Therefore, by taking the summation with respect to a' over a complete set of representatives of zgjzg, we get the corollary.

t

§ 2. A New Formalism Let ZjL denote an abelian variety and L(Q, I, t/J) a familiar vector space of theta functions on Z in which H = Her(Q) is non-degenerate. We know, by Chap. III, Theorem 3, that the graded ring L L(kQ, kl, t/Jk) k;;;O

is uniquely determined by H up to a degree-preserving isomorphism over C; the isomorphism is given componentwise by a translation in Z followed by the mUltiplication of a trivial theta function on Z. Actually, we have seen that L(Q, I, t/J) possesses a more significant uniqueness characteristic. Because of its fundamental nature, we shall recall its details. Let el, ... , e2g denote a canonical base of L relative to A; then there exist g positive integers el , ... , eg each ei dividing ei+l such that we have

A

(~l Xi ei' i~/i ei) = itl ei(Xi Yg+i - Yi Xg+J

for every Xl' ... , X2g' Yb ... , Yzg in R; we define, as before, an R-bilinear form B on Z x Z as B

(

2g

2g

)

g

i~lxiei' i~/iei =i~/iXiYg+i.

Consider the finite commutative group i=l

and define the compact group A (X) = (X x X*) x C1' as in Chap. I, § 5. Then A (X) has an irreducible unitary representation U in L2 (X) defined as

(U ((u, u*), t)
§ 2. A New Formalism

143

We shall recall the representation of A(X) in L(Q, I, ljJ). If we put x(a)= ljJ (a) e(!· B(a, a»)

for every a in L, we get a character X of L; we extend X to a character of L* (denoted also by X); then we extend ljJ to a second degree character of L* by the above formula; fmally, we put Ua(Z) = e((1/2 i) Q(z, a)+(1/4i) Q(a, a)+I(a») ljJ(a)

for every a in L*. Let (a, t) denote an element of L* x of L(Q, I, ljJ); put

Cr and () an element

(U(a, t) ()) (z)=t· Ua(Z)-l ()(z+a).

Then U(a, t) () is in L(Q, 1, ljJ), and we have U(ab t 1 ) U(a2' t 2)= U(a 1 +a2' e( -B(ab a2») tl t 2);

moreover, U(a, t) depends only on the image of (a, t) in (L*/L) x C[ ; we shall denote it also by (a, t). Let e denote the diagonal matrix of degree g with ei as its i-th diagonal coefficient; and, for x=(x 1 ••• x g), Y=(Yl ... Yg) with Xi' Yi in ei 1 Z, put (x, y) =e( -x e ty). We observe that (x, y) depends only on the images of x, Y in X; if we denote them also by x, y, this gives an identification of X with X*. We take an arbitrary element (x, x*) of X x X* ( = X x X) and map it to the element g

g

a= 2:Xi~i+ 2:Xr~g+i i=l

i=l

of L*/L; then U(a, t) is defined as above. This gives the irreducible unitary representation of A(X) in L(Q, 1, ljJ) under which every element t of C[ is mapped to the scalar multiplication by t. . We can replace (Q, 1, ljJ) by any other triple as long as H =Her(Q) remains the same. The triple which gives rise to the classical theta functions is the one in which Q vanishes on Z x (2: R ~g+ i) and 1= 0; we i shall assume also that

for every x = (Xl ••. x g), x* =(xf. .. xi) in Rg, or at least in X. If we map Z to cg with respect to the C-base ell ~g+1' ... , e;:l ~2g of Z, as we have seen in Chap. II, § 4, the theta functions ()r 0(., ), in which r runs over a complete set ofrepresentatives of zg e- 1 /zg, fOTm a C-base of L(Q, 0, ljJ);

144

IV. Equations Defining Abelian Varieties

we recall that the coefficients Tij of T are determined by

for l~i~g. Moreover, for every ((u,u*),t) in A(X) and () in L(Q,O,t/I), we have (U((u, u*), t) ()) (z)=t· e(z tu+! u T tu+u e tu*) ()(z+u T+U* e).

Since ()r o( T, ) depends only on r mod 1 and zg e- 1 jzg = X, we can consider r as an element of X. Then, by Chap. I, § 10, (().2), (().3), we get U((u,u*),

1) ()xo(T, )=<-x-u,u*) ()x+uO(T, )

<

for every x in X. We observe that x* = ,x*) for all x* in X* form a C-base of L2(X); if we choose ()=l=OfromL(Q, 0, t/I) satisfying U(x, 0, 1)()=() for every x in X, which is unique up to a scalar factor in C x , the correspondence x* -+ U(O, x*, 1) () gives rise to an isomorphism of L2(X) to L(Q, 0, t/I) as A (X)-modules. This is the way Chap. I, Proposition 2 was proved. Let bx denote the" delta function" on X at x, i.e., the function on X satisfying bx(X) = 1, bx(Y)=O for y=l=x; then we have b x = 1jcard(X)·

L <-x, x*) x*.

x*eX*

Consequently, the element of L(Q, 0, t/I) which corresponds to bx is 1jcard(X)·

L

<-x,x*)U(O,x*,l)().

x*eX*

Because of U((u, 0), 1) ()x 0(T, )=()x+uo(T, ), we can take ()=

L ()y 0(T,

);

yeX

then, because of U((O,u*),l)()yo(T, )~<-y,u*)()yo(T,), the above element becomes ()_xo(T, ). We have thus shown that the isomorphism of L2(X) to L(Q, 0, t/I) as A(X)-modules maps b x to ()-xo(T, ) for every x in X. We have been using ()m(T, ) for the function Z=(Zl'" Zg) -+ ()m(T, z) on cg. We have learned that elements of L2(X) can be interpreted as theta functions. We shall try to express the multiplication of theta functions in the language of elementary spaces such as L2 (X). If (), ()' are elements of L(P Q, 0, t/lP), L(q Q, 0, t/lq), their product () ()' is an element of L((P+q) Q, 0, t/lPH). We put g

Xn = EEl (2n ei)-l ZjZ i~l

§ 2. A New Formalism

145

and consider the group Gn=A(X,J for n=O, 1, .... Then, as in the special case of n=O, we have an isomorphism of v" =L2 (X,J to L(2n Q, 0, 1/1 2") as Gn-modules; the isomorphism is unique up to a scalar multiplication; we can normalize the scalar factor so that the delta function Ox on Xn at x is mapped to the theta function Z=(Zl'" Zg)- 0_xo(2n T,

2n z)

on eg• The reason we have 2n T, 2n Z instead of T, Z is as follows: The canonical base of L relative to A is also a canonical base of L relative to 2nA; but we have 2n e instead of e; hence the mapping of Z to eg relative to 2nA differs 'from the mapping of Z to eg relative to A which we have ftxed by the factor of 2n. We shall change our notation: For every x in X n, we shall denote by Qn(x) the delta function Ox on Xn at x. Also we put

eh ... ,e2g

qn(x)=0_xo(2n T, 0);

°

we observe that qn(x) is a theta constant; in fact, it is the value at of the theta function which corresponds to Qn(x). By Chap. I, § 10, (0.1), we have

qn( -x)=qn(x). Moreover, if we replace m', n', T, Z in the corollary of Theorem 2.by x, y, 2n T, 0, we get

qn(x) qn(y) = L qn+l(! (x+ y)+r) qn+l (!
in which r runs over! zgjzg. We observe that if we put

u=!
v=!(x-y)+r,

we have u+v=x, U-V= y; and, ifr runs over! zgjzg, (u, v) runs over the set of solutions ofu+v=x, u-v=y. Therefore we can write

qn(x) qn(y) =

L

qn+l(U) qn+l(V).

U+V=X, U-V=Y

After that remark, we define a multiplica~ion: v" x v,,- v,,+l so that it becomes the usual multiplication: L(2n Q, 0,1/1 2") x L(2n Q, 0, 1/1 2") _ L(2n + 1 Q, 0, 1/12"+ ')

under the above isomorphisms. Then, by replacing m', n', T, Z in the corollary of Theorem 2 by - x, - y, 2" T, 2" z, we get u+v=x, u-v=y

In § 3 we shall investigate the properties of the sequence (v,,)n equipped with this multiplication.

146

IV. Equations Defining Abelian Varieties

§ 3. Theta Relations (Under the New Formalism) We take an algebraically closed field a of characteristic different from 2, put Xn= Ea (2n el)-;-1 Z/Z

,

I_I

for n=O, 1, ... , and consider the set v" of all a-valued functions on Xn • We recall that eh ... , e, are positive integers, each ei dividing el+l for 1~i
Also, for every positive integer r, we shall denote by Kr the kernel of the endomorphism of Xoo dermed by x-+rx; we have K2 =tZ'/Z', K4 =! Z'/Z' etc.; and the mUltiplication by 2 gives a surjective homomorphism of Xn+l to Xn with K2 as its kernel. Suppose that we have a set of elements qn(x) of a indexed by nand x in Xn satisfying qn( -x)=qn(x) and

Let Qn(x) denote the delta function on Xn at x and define a multiplication: v" x v" -+ v,,+1 as

Because qn+l( -V)=qn+l(V), the multiplication is commutative. Let f,f' denote arbitrary elements of.v" and x a point of X n + 1 ; then we have the following formula: (1)

(f-f')(x)=

L

f(x+y)f'(x-y)qn+l(Y)·

yex+Xn

In fact, since we have f

we get

= L f(u) Qn(u), ueXn

f' = L f' (v) Qn(v), veX"

f·f' = Lf(x+ y)f'(x-y) Qn+l(X) qn+l(Y)'

in which x, y are in Xn+l andx+ y,x-yin X n. Sincex+ y=x- y mod X n , for a fixed x, the condition is simply that y is in x + Xn; this proves the formula.

§ 3. Theta Relations (Under the New Formalism)

147

We shall assume that Xo contains K 2 , i.e., that e1 , hence all ei' is even; equivalently, we could have assumed that n ~ 1. Then, for every x, y in Xn+ 1 satisfying x == y mod Xn and for every r in K 2 , we have Qn(x+ y+r)· Qn(x- y+r)=

L Qn+l(U) qn+l(V),

in which u+v=x+y+r, u-v=x-y+r. This implies that 2u=2x, hence u=x+r+s for some s in K 2 , and V= y+s; the converse is also true. Therefore, we can write Qn(x+ y+r)· Qn(x- y+r)=

L Qn+l(x+r+s) qn+l(Y+S).

seK2

Let K~ denote the commutative group of homomorphisms of K2 to Q x and A an element of K~; then we have A(r)=A(r+s)A(s), and hence

L A(r) Qn(x+ y+r)· Qn(x- y+r) TeK2

(2)

= ( L A(r) Qn+l(x+r») ( L A(r) qn+l(y+r»). reK2

feK2

For the sake of simplicity, we shall denote the first factor (on the right side) by Qn+l(A, x) and the second factor by qn+l(A, y). If u, v are other elements of Xn+l satisfying u==v mod X n, we have

L A(r) Qn(u+v + r)· Qn(u-v+r)= Qn+l(A, u) qn+l(A, v).

reK2

We can replace Qn by qn and the" dot-multiplication" by the ordinary multiplication (of elements of Q) and we get

L A(r) qn(u+v+r) qn(u-v+r)=qn+l(A, u) qn+l(A, v).

(2')

reK2

If x, y, u, v are elements of Xn+l satisfying x==y==u==v mod X n, we have

( L A(r) Qn(x+y+r)· Qn(x-y+r») (L (3)

A(r) qn(u+v+r)qn(u-v+r»)

=( L A(r) Qn(x+v+r)· Qn(x-v+r») reK2

(L

A(r) qn(u+ y+r) qn(u- y+r»).

reK2

In fact, the left side is equal to (Qn+l(A, x) qn+l(A, y») (qn+l(A, u) qn+l(A, v») and the right side to (Qn+l(A, x) qn+l(A, v») (qn+l(A, u) qn+l(A, y»); hence they are equal. If we take the summation with respect to A over K~, we get

L Qn(x+ y+r)· Qn(x- y+r) qn(u+v+r) qn(u-v+r)

(4)

reK2

=

L Qn(x+v+r)· Qn(x-v+r) qn(u+ y+r) qn(u- y+r).

reK2

148

IV. Equations Defining Abelian Varieties

We can replace Qn by qn and the dot-multiplication by the ordinary multiplication; in this way, we get new formulas (3'), (4') from (3), (4). For instance, we have L qn(x+ y+r) qn(x- y+r) qn(u+v+r) qn(u-v+r) (4')

re K 2

= L qn(x+v+r)qn(x-v+r)qn(u+y+r)qn(u-y+r). reK2

We can give a more symmetric form to (4'): Let a, b, c, d denote elements of Xn satisfying a+b+c+d= -2z for another element z of Xn; then we have

L~~+~~~+~~~+~~~+~

(4")

re K 2

= L qn(a+z+r) qn(b+z+r) qn(c+z+r) qn(d+z+r). reK 2

In fact, if x, y, u, v are elements of Xn+l satisfying x=y=u=v mod X n, a=x+y, b=x-y, c=u+v, d=u-v are elements of Xn satisfying a+b+ c+d= -2z for z= -x-u also in Xn • Conversely, if a, b, c, d, z are elements of Xn satisfying a+b+c+d= -2z, we choose an element x of Xn+l satisfying 2x=a+b and put y= -x+a, U= -x-z, V= -u+c; then x, y, u, v are elements of Xn+l satisfying x=y=u=v mod Xn, and x+y=a, x-y=b, u+v=c, u-v=d. We have a+z= -u+y, b+z= -u-y, c+z= -x+v, d+z= -x-v. Since qn is an even function on Xn , we get (4"). Let a, b denote arbitrary elements of Xn; then we have qn(a) qn(b) =

L U+V=G,

u- v=b

qn+l(U) qn+l(V)

in which x+y=u, x-y=V, u+v=a, u-·v=b. This implies that 2x=a, 2y=b; the converse is also true. Therefore, we can write qn(2x)qn(2y)=( L qn+2(x+r))( L Qn+2(y+r)), re K 2

reK2

in which x, yare arbitrary elements of Xn + 1 • By taking qn(2x)=

X=

y, we get

± L qn+2(x+r); re K 2

if we use the above formula again, we see that the sign does not depend on x. We recall that, in the given sequence of .Q-valued functions qn on Xn , we can make arbitrary sign changes qn -+ ± qn for n = 0, 1, ... (without affecting the defining properties). Therefore, by "adjusting" the signs

§ 3. Theta Relations (Under the New Formalism)

149

successively for n = 2, 3, ... , we may assume that

qn(2x) = L qn+2(x+r)

(5)

,eK2

for every x in XII+1 for n=O, 1, .... We shall show that if a=c and qn(x)=O_xo(2n"t,0), this adjustment is unnecessary since the Eq.(5) is automatically true. We observe that if q runs over zg and rover t zg/zg, p=2q+2r runs over zg. Therefore, for every x in X n+1, we have

L 0H,o(2n+2"t, 0)= Le(2n- 1(2q+2r+2x)"tt(2q+2r+2x») r

~q

=

L e(2n- 1(p+2x) "tt(p+2X»)=02xO(2n "t, 0); p

this proves the assertion. We have so far used only the formal property: u+v=x, u-v=y

of the sequence of a-valued even functions qn on Xn for n=O, 1, .... In particular, the trivial case where qn =0 for all n has not been excluded. We shall make a more restrictive assumption on (qn)n' We shall assume the existence of a graded ring S over a with the following property: There exists a sequence of a-linear isomorphisms of v" to S2n which are compatible with the multiplications: v" x v,,-+ v,,+1 for n=O, 1, .... A graded ring such as S may not be unique; we are interested in the relation between any two such rings. In order to analyze this problem, we shall introduce a particular ring which is "universal". Consider the ring fIl of polynomials in Qn(x) for all n, x with coefficients in a, in which Qn(x) are regarded as "letters". In other words, fIl is the symmetric algebra of Vo EiH'i E9 ... over a. We convert fIl into a graded ring over a by assigning 2n as its degree to each element of v". If J,f' are elements of v" and f f' the product of J, f' in fIl, it is a homogeneous element of degree 211+ 1; hence f f' - f· f' is also a homogeneous element (of fIl) of degree 2n + 1. Therefore, if we take the quotient of fIl by the ideal generated by f f' - f l' for all J,f' in v" for n = 0, 1, ... , we get a graded ring R over a; this is the universal ring. We shall show that there exists a degree-preserving homomorphism 0(: R -+ S over a. Byassumption, there exist a-linear mappings v" -+ S for n = 0, 1, ... ; they give rise to a degree-preserving homomorphism fIl-+ S over a. Also by assumption, the image offf' - f· f' in S is for all J, f' in v" for n = 0, 1, ... ; hence we get a homomorphism 0(: R -+ S over a, which is degree-preserving. We call 0( the canonical homomorphism of R to S. In order to examine the properties of IX, we shall use the following two lemmas:

°

150

IV. Equations Defining Abelian Varieties

Lemma 2. Let A, x denote arbitrary elements of K!, X n + l ; suppose that the multiplication: v" x v" -4 v,,+l is "strongly non-degenerate ", i.e., f -f' = for J,f' in v" implies that f = or f' = 0. Then there exists an element y of x + K4 such that

°

°

qn+I(A, y)=

L A(r) qn+l(y+r)=j=O.

reK2

of

Proof For every A, z in K!, X n, we have defined an element Qn(A, z)

v" as

L A(r) Qn(z+r);

Qn(A, z)=

reK2

since the Qn(z')'s for all z' in Xn form an Q-base of v" , we have Qn(A, z)=I=O. In particular, since the multiplication is strongly non-degenerate, we have Qn(O, 2x)· Qn(A, 0)=1=0. According to the definition, we have

Qn(O, 2x)· Qn(A, 0)=

L

Qn(2x+r)· A(S) Qn(s)

r. SeK2

in which y+v=2x+r, y-v=s and r, s in K 2 • This implies that y=x+r', v= y+s with 2r' =r+s; conversely, if y=x+r', V= y+s with r', s in K 4 , K 2 , by putting r=2r' +s, we get y+v=2x+r, y-V=S. Therefore, we can write

Qn(O, 2x)· Qn(A, 0)=

L

Qn+I(Y) (

L

Qn+I(Y) qn+I(A, y).

yex+K.

=

yex+K.

L A(S) qn+I(Y+S))

seK2

Since the Qn+I(Y')'S for all y' in Xn+l form an Q-base of v,,+l and Qn(O, 2x)· Qn(A, 0)=1=0, we have qn+I(A, y)=I=O for at least one yin x+ K 4 . q.e.d. Lemma 3. Let R denote a graded ring which is generated by RI over Ro; let M denote a graded R-module and a finite R-module. Then

Mk,=O for any irifinite sequence (ki)i implies that Mk=Ofor almost all k. Proof Since M is a graded R-module and a finite R-module, we have M = R Xl + ... + R x n , in which Xl' ... , Xn are homogeneous elements of M, i. e., elements of Mdt' ... , Md n for some db ... , dn • Since the case where n = 0, i. e., M = 0, is trivial, we shall first assume that n = 1; let J denote the kernel of the surjective homomorphism R -4 M defined by a -4 a X (for X = Xl' d = d l ). This homomorphism gives rise to an isomorphism of

§ 3. Theta Relations (Under the New Formalism)

151

R,JJk to Mk+d for k=O, 1, ... ; and we have Rki-d=Jki-d,

hence (Rlti-dcJ, for every ki?,d. Put e=ki-d for any ki>d; then we have (Rl)k=(R1)k-e(R1)ecJ for all k?,e. Since R is generated by Rl over Ro , we have (R1t=R k; hence Rk=Jk , i.e., Mk+d=O, for all k?,e. We shall assume that n?, 2 and apply an induction on n. Put M' = R Xl + ... +RXR_l and Mil =M/M'; then M', Mil are graded R-modules generated over R by less than n homogeneous elements; and we have M,JM,,=M;: for k = 0, 1, .... Clearly, we have Mk = if and only if M" = M;: = 0. Therefore, we have M"i = M~ =0 for i = 1, 2, .... By the induction assumption, this implies that M,,=M;:=O, hence Mk=O, for almost all k. q.e.d.

°

The assumption that R is generated by Rl is quite essential in Lemma 3: If R is the ring of polynomials in one letter of degree d> 1 with coefficients in any commutative ring (with the identity) and M =R considered as an R-module, we have Mk=O for all k not divisible by d; yet Mkd=l=O for k=O, 1, .... We are ready to prove the following theorem: Theorem 3. Let O(=(O(k)k denote the canonical homomorphism of the universal ring R to S,' then the Q-linear mapping,' O(k: Rk~Sk is bijective for k=2R, n=O, 1, .... If Xo contains K4 and S an integral domain, R is generated by Rl over Q. If, moreover, S is a finite R-module via the homomorphism 0(, O(k is bijective for almost all k. Proof. We shall show that 0(2n: R 2n ~ S2n is bijective. By definition, 0(2n maps the image of v" in R 2n surjectively to S2n; moreover, we have dim(v,,)=dim(S2n). Therefore, we have only to show that the image of v" in R 2 n is R 2 n. We observe that every element of R can be written in the form f= L const. h f2 ... J.,

in which the" const." is in Q and /; in v"i' i.e., in the image of v"i in R, for l~i~r. If we have ni=nj for i=l=j, /;jj is in v"i+ l ' Therefore we may assume that ni =1= nj for i =1= j. On the other hand, every positive integer k has a unique diadic representation: k=2Rl +2R,+ ... +2Rr , in which n 1>n 2 > .. ·>nr ?,0. Therefore, if f is an element of R 2n, by what we have said, we get r = 1; hence f is in v", i.e., in the image of v" in R. We shall show that if Xo contains K4 and S is an integral domain, R is generated by Rl over Q. Since R is generated by the images Rb R 2 ,

IV. Equations Defining Abelian Varieties

152

R 4 , ••• of Yo, Vi> V2, ... , we have only to show that (R 2n)2=R 2n +l for n=O, 1,2, .... We observe that R 2n+l is spanned over Q by Qn+l(2,x)=

L 2(r) Qn+l(x+r)

reK2

for all 2 in K! and x in

Xn+l.

On the other hand, (R2n)2 contains

L 2(r) Qn(x+ y+r)· Qn(x- y+r)=Qn+l(2, x) qn+l(2, y)

reK2

for every 2 in K! and x, y in Xn+l satisfying x=y mod X n. Since S is an integral domain, the multiplication: v" x v" -+ v" + 1 is strongly nondegenerate; also Xn contains K 4 • Therefore we have only to incorporate Lemma 2. If, besides the above assumptions, S is a finite R-module, CXk is bijective for almost all k. The proof is as follows: Let J denote the kernel of cx; then J is a homogeneous ideal of R, i.e., an ideal of R and a graded R-module. Since R is generated by Rl over Q, R is a noetherian ring; we also have J2 n = for n = 0, 1, .... By Lemma 3, this implies that Jk = 0, i. e., ('4 is injective, for almost all k. We have not used the assumption that S is a finite R-module. Let R' denote the image of R under cx; then R' and S, hence also M =S/R', are graded R-modules; by assumption, M is a finite R-module; also we have M2n=0 for n=O, 1, .... Again by Lemma 3, this implies that Mk=O, i.e., CXk is surjective, for almost all k. q.e.d.

°

Corollary. Under the assumptions of Theorem 3, we have Q[Sl]k=Sk for almost all k. Since R is generated by Rl over Q and CXl: Rl -+ Sl bijective, Q [Sl] is the image of R under cx and Q [Sl]k the image of Rk under CXk. Since CXk is bijective for almost all k, we have Q[StJk=Sk for almost all k.

§ 4. The Ideal of Relations We take the same setup as in § 3. We shall assume that Xo contains K 4 , i. e., e1 , ... , eg are multiples of 4, and S is an integral domain. If V is a vector space over Q, we shall denote the symmetric algebra of V by 9"[V] instead of Q[V]; if VI' V 2, ... form an Q-base of V, 9"[V] is isomorphic over Q to the ring of polynomials in Vb V2' ... (considered as letters) with coefficients in Q. Iff, f' are elements of 9"[V], we shall denote their product ff' sometimes by f 0 f'. The problem is to analyze the kernel of the degree-preserving homomorphism 9"[Sl] -+ S over Q; by Theorem 3, this problem is more or less reduced to the analysis of the kernel I of the surjective degree-preserving homomorphism

9"[Vo] -+ Rover

Q.

153

§ 4. The Ideal of Relations

We have a long way to go; we shall start by proving the following lemma, which is needed in the proof of Lemma 5:

Lemma 4. Let al"'" a4 denote elements of Xn+ 1 satisfying L a; ==

omodXn ;

then we have

;

n qn+l(al +a;+1- 2b)Qn+3(b), 3

L Qn+l(al +r)· .... Qn+l(a4 +r)= L reK2

b ;=1

in which b runs over the set of all solutions of4b=La;. i

Proof We have

Qn+1(a 1+ r)· Qn+1(a 2 +r)= L Qn+ 2 (x) qn+2(Y)'

in which x+y=a1+r, x-y=a2+r. Choose an element C1 of Xn+2 satisfying 2 C1 = a1 + a2; then we have x = C1 + rl> Y = a1 + r - C1 - r1 for some r1 in K 2; conversely, this implies that x+y=a1+r, x-y=a2+r. Therefore, we can write Qn+1(a1 +r)· Qn+1 (a2 +r)= L Qn+2(C1 +r1) qn+2(a1 +r- c1- r1)' rl eK2

By repeating the same argument, we get Qn+1 (a3 +r)· Qn+1(a4 +r) = L Qn+2(a- c1+r2) qn+2(a3 +r-a+c1- r2), r2eK2

Qn+2(C1 +r1)· Qn+2(a- c1+r2) = L Qn+3(XO+S1) qn+3(C1 +r1- x O- s1), 'l e K2

qn+2(a1 +r- c1- r1) qn+2(a3 +r-a+c1- r2) = L qn+3(YO+S2)qn+3(a1+ r - c1- r1-Yo-S2), '2 eK 2

in which 2a=La;, 2xO=a+r1+r2, 2YO=a1+a3-a.,....r1-r2=a-a2i

a4 - r1- r2' On the other hand, we have qn+1(2x)= L qn+3(X+t) teK2

for every x in X n + 2 • Putting these together, we get L Qn+1(a1 +r)· .... Qn+1(a4 +r) reK2

n qn+1(a1 +ai+1- a - r)Qn+3(XO+S), 3

= L

r,seK2 i=l

in which 2xo=a+r; obviously, this can be written as in the statement of the lemma. q.e.d.

154

IV. Equations Defining Abelian Varieties

We have so far used only K!; in the following lemmas, we need Kt and K~; these are the groups of homomorphisms of K4 and Ks to lY. We observe that the homomorphism 4: Ks -+ K2 defined by r" -+ 4 r" is surjective and has K4 as its kernel. Moreover, the restriction homomorphism K~ -+ Kt is surjective and has 4* K! as its kernel, in which 4*: K! -+ K~ is defined as (4* A)(r") = A(4r") for every A in K! and r" in Ks; we have 4* K! =4K~, and 4K~ is the kernel of2: K~ -+ K~; if A" is an extension of A to K s , we have 4* A=4A". In the same way, 2* K! = 2 Kt is the kernel of the restriction homomorphism Kt -+ K! etc. If A', x' are elements of Kt, X n + 1 , we put

L A'(r')Qn+1(x'+r');

Q~+l(A',X')=

r' eK 4

we define we put

q~+l(A',X')

similarly. Also, if A", x" are elements of K~, X n+2 ,

Q~+2 (A", x") =

L

r"eKs

A"(r") Qn+2(X" + r");

we define q~+2 (A", x") similarly. They have a certain periodicity property; e.g., for every r' in K 4 , we have Q~+l (A.',

x' - r') = A.'(r') Q~+l (A', x').

On the other hand, if A', x" are elements of Kt, X n + 2 , we have q~+l (A',

2 x") =q~+3 (2* A.', x").

In fact, we have

L A.'(r') qn+1 (2x" + r') =lj2 L A.'(2r") qn+1(2x" +2r").

q~+l (A.', 2 x") =

r'eK4

g •

r"eKs

Since we have

qn+1(2(x" + r")) =

L qn+3(X" +r" +s),

seK2

this is equal to

L

A'(2r") qn+3(X" +r")=q~+3(2* A.', x").

r"eKa

The following lemma can be considered as a refinement of Lemma 2: Lemma 5. Let A~, A2, A~ and a~, a2, a~ denote arbitrary elements of K! and X n+1; then there exist elements A', r" of K!, Ks such that

n q~+1(A;+2A.', a;±r")=l=O, 3

;=1

in which the indicated signs can be chosen arbitrarily.

155

§ 4. The Ideal of Relations

2A.~

Proof Let A~ denote an extension of A~ + A2 + A~ to Ks; then we have = 2* (A~ + A2 + A~). If we put Ai'+ 1 = 2* Ai - A~ for 1 ~ i ~ 3, we have

We choose an element ai' of Xn+2 satisfying 2a;'=ai for 1~i~3; then we choose an element aT of X,,+3 satisfying 2aT=a~+d2+a3. Put at+1 = ai' -aT for 1 ~i~3; then we have 4

~a*-O i ..;i-·

i=1

By Lemma 4, we have Q~+3(A~,

aT)· .... Q~+3(A4' al)

n q,,+3(ai' +r{' +r;'~1-2b)Qn+s(b), 3

= LA~(r{') ... A4(r4) L

;=1

in which the frrst summation is taken over (KS)4 modulo the "diagonal" of (K 2t and the second summation over the set of solutions of 4b= L ri'· Since L Ai' =0, A~(ri').·· A4(r4),and also qn+3(ai' +ri' +r;~1 -2b), i

i

L ri', are invariant under rt - ri' +r(l ~i~4) for any r in K 2. Therefore, ;

we can take the first summation over (KS)4 provided that we multiply by 1/2g• We can transform it further into 1/24g • L LA"(4b-Lr;")A.~(r{')···Qn+s(b), l"eKl

i

L

in which r;" runs over Ks and b over K 32 • By using Ai' =0 again, we can write the product A"(4b- LrnA~(r{') ... A4(r4) as ; ;

A"(4b)(A~ + A")(2 r{')

n(A;'+1 - A")(r{' + r;~1)· 3

;=1

Therefore we get 3

1/24g • LA"(4b)(A~ +A.")(2r")n q~+3(Ai~1-A", ai' -2b)Qn+s(b), i=1

in which A" runs over Kt, r" over K s , and b over K 32 • On the other hand, we have L (A~ +A")(2r")=23g or 0 r"eK8

IV. Equations Defining Abelian Varieties

156

according as 2 (A~ + A") = 0 or =1= 0; and 2 (A~ + A") = 0 if and only if A~ + A" is in 4K~=4* K~. In this way, we get

nq~+3(2* ..1.;+4* A, a;' -2b) Qn+s(b), 3

=

1/2g • L(A~ +4* ..1.)( -4b)

i=1

in which A runs over K~ and b over K 32 . For the same reason as in the last part of the proof of Lemma 2, there exist elements A, b of K~, K32 such that 3 q"n+3 (2* X+4* A" d' -2b)=I=O • I.

n

Since K 32 is contained in Xn + 3, a;' - 2 b is in Xn + 2 for 1 ~ i ~ 3. Therefore q~+3(2* ..1.;+4* A, a;' -2b) can be replaced by q~+1(A;+2* A, a;-4b); we take as A' an extension of A to K4 and r" = - 4 b. Since q~ + 1(A; + 2..1.', a; - r") differs from q~+1(A; + 2..1.', a; +r") just by the factor of (A; + 2 ..1.')(2 r")=1= 0, we can replace r" by - r" in anyone of the three factors. q.e.d. Lemma 6. Let . 1.1, ..1.2 denote arbitrary elements of K! and a', b' elements of Xn+1 satisfying a' =b' mod Xn; denoting the restriction of A; to K2 also by A;, define an element T(A1' . 1.2; a', b') of 9"[v,,]2 as

Then, for every r', s' in K 4, we have T(Al, . 1.2; a' - r', b' -s')=A1(r' +s') A2(r' -s') T(A1' . 1.2 ; a', b'); moreover, the image of T(A1' . 1.2 ; a', b') in R 2n+l is and also L (..1.1 + A2)(r') qn (..1.1' a' + b' + r') qn (..1.2' a' - b' + r')

r' eK 4

Proof The periodicity property of T relative to K4 follows from that of Q relative to K 2" The image of T(A1' . 1.2 ; a', b') in R 2n+l is L A1(r' +s) A2(r' +t) Qn(a' +b' +r' +s)· Qn(a' -b' +r' +t),

in which r' runs over K4 and s, t over K 2. The pair (r' + s, r' + t) determines the triple (r',s,t) up to (r',s,t)->.(r'+r,s+r,t+r) for some r in K 2. We

§ 4. The Ideal of Relations

157

can write r' +s, r' +t as s'+t', s' -t' with some s', t' in K 4 , and conversely. The pair (s', t') is unique up to (s', t') --+ (s' +r, t' + r) for some r in K 2 • Therefore, for any Q-valued function 4> on (K4)2, we have

L 4>(r'+s,r'+t)= L4>(s'+t',s'-t');

r' ,s, t

s', t'

in particular, we can write the above element of R 2 n+l as L(A.~ +A.z)(s')(A.~ -A.z)(t') Qn(a' +s' +b' +t'). Qn(a' +s' -b' -t')

= L (A.~ + A.z)(s')(A.~ - A.z)(t') L Qn+l(a'+s'+r)qn+l(b'+t'+r).

reK1

Since we have (A.~ +A.z)(s')(A.~ -A.z)(t')=(A.~ +A.z)(s' +r)(A.~ -A.z)(t' +r),

this is equal to 2gQ~+1(A.~+A.Z,a')q~+1(A.~-A.z,b'). By replacing Q by q, we also get the last formula in the lemma. q. e. d.

K:

Let A.~, A.z, A.3 denote arbitrary elements of and a', b', c' elements of Xn+l satisfyinga'=b'=c' mod Xn; then we get an element of 9'[v,,]2 as q~+l(A.~ -A.z +2A3' c') T(A.~,

A.z; a', b')

-q~+l(A.~ -A.z, b') T(A.; +A.3, A.z -A.3;

a', c').

Since we have to deal with such an element in Lemma 8, we shall denote it by U(A.;, A.z, A.3; a', b', c'); this element has the following obvious property: Lemma 7. Let A.;, ... , A,~ denote arbitrary elements of K: and a', b', c', d' elements of Xn+l satisfying a'=b'=c'=d'modXn; then we have q~+l(A.; -A.z +2A,4,

d') U(A.;, A.z, A.3; a', b', c')

=ifn+l(A.~ -A.z +2A3' c') U(A.~,

A,z, A.4; a', b', d')

+q~+l(A.~ -A.z, b') U(A.~ +A.4, A.z -A,4, A.3 -A.4;

a', d', c').

We are ready to prove the following principal lemma: Lemma 8. Let W denote the subspace of 9'[v,,]2 spanned over Q by

for all flo ... ,f4 in v,,-l (n~l); then W contains U(A.;,A,z,A.3;a',b',c') for aUA.;, A.z, A.3 in K: and a',b',c' in Xn+l satisfying a'=b'=c' mod X n.

158

IV. Equations Defining Abelian Varieties

Proof Since the proof is quite involved, for the simplicity of printing, we shall assume that n = 1. (Actually, this assumption does not affect the generality.) Let a, b, c, d denote arbitrary elements of Xo; then we can find elements x, y, u, v of Xl satisfying x+y=a,

x-y=b,

u+v=c,

u-v=d.

Choose an element z of Xl satisfying -2z=a+c; then we have (- y-z)+(x+z)=b,

(- y-z)-(x+z)=c,

(-v-z)+(u +z)=d,

(-v-z)-(u +z)=a.

By definition, we have Qo(a)· Qo(b)= L QI(x+r) ql(y+r)

= 1/2g • L QI(A, x) ql(A, y); AEKi

we have similar expressions for Qo(c)· Qo(d), Qo(b)· Qo(c), Qo(d)· Qo(a). In this way, we see that W contains the following element:

-(L QI(A, - y- z) ql(A, x+ z))o{L QI(A, -v-z) ql(A, u+z)). A

A

We observe that x, y, u, v, z are elements of Xl satisfying x==y,

u==vmodXo ,

x+y+u+v+2z=O;

and they can be taken arbitrarily subject to these conditions. We choose an element a' of X 2 satisfying2a' =X +u and putb' =x-a',c' = - y-z-a'; then we get elements a', b', c' of X 2 satisfying a' + b' = x,

a' - b' = u,

a' + c' =

- y - z,

a' - c' =

- v - z.

Since ql(A, y)=ql(A, - y) etc., the above expression can be written as L Ql(A 1 , a' + b')o Ql (A2' a' - b') ql (Ab a' + c' + z) ql(A2, a' - c' + z) - L Ql(A 1 , a' +C')oQl{A2' a' - c') ql(A b a' +b' +Z)ql(A 2 , a' -b' +z),

in which AI' A2 run over K!. We observe that a', b', c', z are elements of X 2 satisfying a' == b' == c' mod Xl' b' - c' == z mod X 0; and they can be taken arbitrarily subject to these conditions. We replace a',b',c',z by a'+r',b'+s',c'+t',z-r' for r',s',t' in K 4 ; then we take A~, A~, A~ arbitrarily from K! and apply L r', s'. t'EK4

A~(r')A~(s')A.~(t')."

§ 4. The Ideal of Relations

159

In this way, we get an element of W of the form A - B, in which A=

I( I A~ (r') A~(S') Ql (A1o a' +b' +r' +s')

0

Ql (A2' a' -b' +r' -s'))

r',5'

(IA~(t') ql(A1o a' +c' +z+t') ql(A2, a' -c' +z-t')), t'

B=

I( I A~(r') A~(t') Ql (Al' a' +c' +r' + t')

0

Ql (A2' a' -c' +r' -t'))

r', t'

(I A~(S') ql(A1o a' +b' +z+s') Ql(A2, a' -b' +z-s')); s'

Al, A2 run over K! and r', s', t' over K 4 . We shall show that A=O unless A10 A2 satisfy certain relations. If we fIx Al, A2, the term

A~, A~, A~,

under the fIrst summation in A can be written as

I

A~ (r') A~ (s') A~ (t') Al (rl + r3) A2 (r2 + r4)

Ql(a' +b' +r' +s' +r1) 0 Ql(a' -b' +r' -s' +r2) Ql(a' +c' +z+t' +r3) Ql(a' -c' +z-t' +r4),

in which r', s', t' run over K4 and r1 , ..• , r4 over K 2 . If r'o, sO are elements of K4 satisfying r'o=sO mod K2 and to an element of K 2 , we can take to after replacing r', s', t' the partial summation with respect to ro, by r' +ro, s' +so, t' +to and r10 r2 , r3 , r4 by

So,

We observe that each term is simply multiplied by

moreover, the sum of such factors is different from 0 if and only if A~+A~+2*Al=0 and A~=A~=Al+A2 on K 2 . Let Ji,~ denote an extension of Ai to K4 and put Ji,~ = A~ - Ji,~; then we have

for some Ji,~ in K!; also we have Al=Ji,~, A2=Ji,~ on K 2 . By processing B in the same way, we get A~ +A~ +2* Al =0, and A~ =A~ =Ai +A2 on K 2 . We recall that A~, A~, A~ are arbitrary elements of K!. Therefore, we can take Ji,~, Ji,~, Ji,~ arbitrarily from K! and defIne A~, A~, A~ as above; then, as we have seen, in A the contribution comes only from the pair (}'l, A2) such that Ai = f1~, A2 = f12 on K 2 . Moreover, in B we get 2*Ai=2(Ji,~+Ji,'3)' hence Ai=Ji,~+Ji,~ on K 2 ;-also A2=Ji,~+Ji,~ on K 2 ·

IV. Equations Defining Abelian Varieties

160

We choose A.~, ,1.2, ,1.3 as above and replace /1; by A.; for 1 ~ i ~ 3; then we get

A=( I

A.~ (r' +s') ,1.2 (r' -s') Ql (A.~, a' +b' +r' +s') 0

Ql (,1.2, a' -b' +r' -s'))

r',5'

(I(A.~ - ,1.2 + 2A.3)(t') ql (A.~, a' + c' + z+ t') ql (,1.2, a' - c' + z- t')) , I'

B= (

I

r',t'

(A.~ + A.3)(r' + t')(A.2 - A.3)(r' - t')

Ql(A.~ +,1.3' a' + c' +r' + t') 0

Ql (,1.2 +,1.3' a' -c' +r' - t')) (I (A.~ - A.2)(S') ql (A.~ + A.3, a' + b' + z + s') ql (,1.2 + ,1.3, a' - b' + z - s')). S'

As in the proof of Lemma 6, the first factor in A can be written as

I

A.~(r' +s)A.2(r' +t)Ql(A.~, a' +b' +r' +s) 0

r', s. t

Ql(A.2, a' -b' +r' +t);

by the periodicity of Q, this is equal to 2 2g T(A.~, ,1.2; a', b'). By the periodicity of q, the second factor in A can be written as I(A.~ +,1.2 +2A.3)(t') ql(A.~, a' +c' +z+t') ql (,1.2 , a'-c' +z+t'); I'

since A.~ =A.~ +2A.3 on K 2 , we can replace A.~ by A.~ +2,1.3 in ql; hence this is equal to 2g q2 (A.~ + ,1.2 + 2,1.3, a' + z) q2 (A.~ - ,1.2 + 2A.3, c'). In the same way, we see that the first factor in B is 2 2g T(A.~ + ,1.3' ,1.2 - A.3; a', c') and the second factor 2g q2 (A.~ + ,1.2 + 2,1.3, a' + z) q2 (A.~ - ,1.2, b'). Therefore W contains A - B = 23g q2 (A.~ + ,1.2 + 2,1.3, a' + z) U(A.~, ,1.2, ,1.3; a', b', c');

we can get rid of the factor q2 (A.~ + ,1.2 + 2A.3 , a' + z) as follows: Since we can take z=b' -c',

q2 (A.~ + A.2 + 2A.3, a' + b' - c') U(A~, ,1.2, ,1.3; a', b', c') is in W for every A.~, ,1.2, ,1.3 in K! and a', b', c' in X 2 satisfying a' == b' == c' mod Xl. We take A.~ from K! and d' from a' +Xl = ... =c' +Xl . If we multiply q2 (A.~ +,1.2 + 2A.~, a' +b' -d') q2 (A.~ +,1.2 +2,1.3 -2A.~, a' +d' -c')to the identity in Lemma 7 for n = 1, we get 1(.

U(A.~,

,1.2, ,1.3; a', b', c')

= 1(1 q2 (A.~ + ,1.2 + 2A.~, a' + b' - d') U(A.~, ,1.2, A.~; a', b', d') + .

1(2

q2 (A.~ + ,1.2 + 2,1.3 - 2 A.~, a' + d' - c')

U(A.~

+ A.~, ,1.2 -

A.~,

,1.3 -

A.~;

a', d', c'),

§ 4. The Ideal of Relations

in which

161

"1> "2 are in Q and "=q2(A,1 +A,2 +2A,~, a' +b'-d') q2 (A,1 + A,2 + lA,3 - 2A,~, d - c' +d')

q2 (A,1- A,2 + lA,~, d').

By what we have shown, the right side of the above identity is in W. According to Lemma 5, for any given A,i, A,2, A,3 in K! and d, b', c', d' in X 2 satisfying d=b'=c'=d' mod Xl> there exists an element A,~ of K! for which ,,+0 provided that we modify d' by a suitable element of Ks. Therefore U(A,1,A,2,A,3; a',b',c') is in W. q.e.d.

Lemma 9. Let V denote a vector space over Q and l' a subspace of 9'[V]2; put (ab)*=ab+1' for every a, b in V; and let W denote the subspace of 9'[9'[V]2/1']2 spanned over Q by (ab)*(cd)*-(ac)*(bd)* for all a, b, c, d in V. Then we have

Proof. Let x, y denote elements of 9' [V]2 and put x* = x + 1', y*=y+l'; then xy+9'[V]21' depends only on x*, y*; and (x*,y*)-+ x y+9'[V]21' is symmetric and Q-bilinear. Therefore it gives rise to an Q-linear mapping of 9'[9'[V]2/1']2 to 9'[V]4/9'[V]21', under which all (ab)*(cd)*-(ac)*(bd)* are mapped to 0; hence we get an Q-linear mapping {3 of 9'[9'[V]2/1']2/W to 9'[V]4I9'[V]21'. Conversely (a, b, c, d) -+ (a b)* (c d)* + W is symmetric and Q-multilinear. Therefore it gives rise to an Q-linear mapping of 9' [V]4 to 9' [9' [V] 2/1'] 2/W, under which 9'[V]21' is mapped to 0; hence we get an Q-linear mapping (X of 9'[V]4I9'[V]21' to 9'[9'[V]2/1']2/W. Lete, 1/ denote the images of (a, b, c, d) in 9'[V]4I9'[V]21', 9'[9'[V]2/1']2/W under the obvious

mappings; then we have e=abcd+9'[V]21',

1/=(ab)*(cd)*+ W,

hence (X (e) = 1/, {3(1/)=e. Since elements such as e,1/ span the above two vector spaces over Q, (X, {3 are the inverses of each other. q.e.d. We shall derme two sequences of ideals. We temporarily revoke the standard notation and denote by R(n) the universal ring for the sequence, v", v,,+1' ... ; we have a surjective degree-preserving homomorphism 9'[v,,] -+ R(n)

over Q; let l(n) denote the kernel and put J(n)=9'[v,,] 1~). Then we have J1n)=I~n), and R(O)=R, 1(0)=1; we-put J(O)=J.

J(n)cl(n),

162

IV. Equations Defining Abelian Varieties

Lemma 10. We have

Ji") = I!:)

9'[v,,]4iJi") =9'[v,,+tJ2/I~"+l), for n=O, 1, ....

Proof. For the same reason as before, we shall assume that n=O. We are supposed to show that

In Lemma 9, we take Vo, 12 as V, I'; then we get 9'[VO]4/J4 =9'[V1]2/W ,

in which W is as in LemmaS for n=l. We have WcI~l) and 9'[Yt]2/I~l)=R~)=R4; we also have J4 cI4 and 9'[Vo]4/I4=R4. Let W' denote the subspace of W spanned by U(A~, A2, A;; a', b', e') for all A~, A2, A; in Kt and a', b', e' in X 2 satisfying a'==b' =e' mod Xl' If we can show that we will get W'= W=I~l) and J4 =I4 . We shall show that 9'[Yt]2 is spanned by T's. If A is an element of K~, the restriction of 2* A to K2 is 0; hence, by using the periodicity of Q relative to K 2 , we get

L T(A~ + 2* A, A2; a', b') = L (A~ + A2 + 2* A)(r') Ql (A~, a' + b' + r') Ql (A2' a' - b' + r') A,r' = 2g L (A~ + A2)(r) Ql (A~, a' + b' + r) Ql (A2' a' - b' + r)

AeK!

0

0

reK2

=22gQl(A~,

a' +b') 0 Ql(A2, a' -b').

We recall that, for every x, y in Xl> there exist a', b' in X 2 satisfying x=a' +b', y=a' -b'; also that the Ql(A, x)'s for all A, x in K~, Xl span Yt. Therefore, T's span 9'[Yt]2' We shall show that T(A~, A2; a', b') is in W' if q2(A~ -A2, b')=O. By assumption, we have U(A~,

A2, A;; a', b', e') =

K • T(A~,

A2; a', b'),

in which K = q2 (A~ - A2 + 2A.;, e'). By Lemma 5, we have K =1= 0 for some A; , e'; therefore T(A~, A2; a', b') is in W'. We fIx A.', a' in Kt, X 2 and consider the set of all T(A~, A2; a', b') in which A~ + A2 = A.' and b' == a' mod Xl; we shall show that it spans a vector space over Q such that its image in 9'[Yt]2/W' is of dimension at most one. Suppose that an element T(Al' A~; a', b') of the set is not

§ 4. The Ideal of Relations

163

in W'; then we have Q2(A1-A2, b')*O. We observe that an arbitrary element of the set can be written as T(AI + A3, A2 - A3; a', c') for some A3 in Kt and c' in a'+X1 • Since W' contains U(AI ,A2,A3; a', b', c') and Q2 (AI - A2, b') 0, it is a constant multiple of T(AI' A2; a', b') modulo W'; this proves the assertion. Now, by Lemma 6, we have

*

T(A'1, A'· 2, a' - r' , b') = (A'1 + A'2) (r') T(A'1, A."2, a', b')

for every r' in K 4 • We let A.' run over Kt and a' over a complete set of representatives of X 2 /K 4 ; for each pair (A.', a'), we pick a T(Al' A2; a', b') which is not in W', if any, in which Al + A2 = A' and b' == a' mod Xl' Then, by what we have shown, the set of all such T's spans 9'[J';]2 modulo W'. We observe that the number of elements in this set is at most equal to card (Kt) card (X2/K 4) = card (X2) = dim(V2) = dim (R4). q.e.d. We shall engage in the so-called "diagram-chasing" in the proof of the next theorem. This is a simple device to visualize the relations of many modules and homomorphisms; we shall recall a few basic terms. A "diagram" is a non-empty set of a-modules and a-homomorphisms, i.e., a-linear mappings; a-homomorphisms are indicated by "arrows". A "sequence" is a special diagram consisting of a-modules Mh M 2 , M 3, .. , and a-homomorphisms M 1 -+M2 , M2 -+M3' .... If the products of a-homomorphisms in all sequences from M to N in the given diagram are same, we say that the diagram is "commutative". A sequence M 1 -+M2 -+M3 -+··· is called "exact" if Im(M1 -+M2)= Ker(M2 -+M3 ), Im(M2 -+M3 )=···, in which "1m" denotes the image and "Ker" the kernel. A sequence of the form 0-+ M' -+ M -+ M" -+ is exact if and only if M' -+ M is injective, M -+ M" surjective, and Im(M' -+ M)= Ker(M -+ M"). Let M' -+ M -+ M" denote a sequence; if M' -+ M, M -+ M" are both injective or surjective, the product M' -+ M" is injective or surjective; if M' -+ M" is injective, M' -+ M is injective; if M' -+ M" is surjective, M -+ M" is surjective. Since a is a field, the tensorization is a "covariant exact functor", i.e., if -+ M' -+ M -+ M" -+0 is exact, for any a-module N, 0-+ M' ® N -+ M ® N -+ M" ® N -+ 0, 0-+ N ® M' -+ N ® M -+ N ® M" -+ are both exact.

°

°

°

Theorem 4. We have 9'[VO]2n+1_212 =I2n+l

for n=O, 1,2, ... and 9'[Vo]k_2 12=Ik for almost all k. Proof We observe that the first part can be written as J2n+1=I2n+l for n=O, 1, .,. and the second part as Jk=Ik for almost all k. If we apply Lemma 3 to the graded ring 9'[Vo] and the graded 9'[VoJ-module IjJ, therefore, the second part follows from the. first part. Since the first

164

IV. Equations Defining Abelian Varieties

part is trivially true for n=O, we shall assume that n!?; 1 and apply an induction on n. In particular, we have J2n=12n; this implies that J2n+l =9'[VO]2n+I_2 12 =9'[VO]2n9'[VO]2n_212

=9'[VO]2n J2n=9'[VO]2n12n.

On the other hand, by Lemma 10, we have 9'[v,,_1]21r-1) =Jin - 1)=1lr- 1); hence the sequence 0-+9'[v.._l]21~n-l) -+9'[v,,-1]4 -+ Rlr-1)-+0

is exact; we can identify R~-l) with R 2n+l. We tensorize the exact sequence 0 -+ 12n-+ 9' [VO]2n -+ R 2n-+ 0 with itself, and we get a commutative diagram of the following form:

o

0

t

t

0

t

0-+ /.!, -+ L -+ /.!,' -+ 0

t

t

t

t

t

t

t o

0

0-+ M' -+ M -+ Mil -+ 0 0-+ N' -+ N -+ N"-+O

t

t

0

with exact horizontal and vertical sequences, in which L=12n ®9'[VO]2n,

M'=9'[VO]2n ® 12n,

M =9'[VO]2n ® 9'[VO]2n,

Nil =R 2n ® R 2n.

This gives rise to an exact sequence 0 -+ L + M' -+ M -+ Nil -+ 0, in which L+ M' is the submodule of M generated by the images of L, M' in M and M -+ Nil the product of M -+ Mil, Mil -+ Nil. This sequence can be embedded in the following commutative diagram: 0

0

0

O-+A'

-+A

-+A"

-+0

O-+L+M'-+M

-+N"

-+0

t

t

t

t

t

t

t

t

t

0-+ J2n+1 -+9'[VO]2n+1-+ R 2n+1-+0

t

0

t

_0

t

0

§ 4. The Ideal of Relations

165

with exact vertical sequences, in which M = 9"[VoJ2n ® 9"[VoJ2n ~ 9"[VoJ2n+l is the Q-homomorphism defined by a ® b~ab, and L + M' ~J2n+l, N" ~ R 2n+' are similar; the top sequence may not be exact at A", i.e., A ~ A" may not be surjective; and the bottom sequence may not be exact at 9"[VoJ2n+l. We observe that if one of them is exact, the other is also exact; hence we have J2n+l = 12n+l if and only if A ~ A" is surjective. The point of the proof is to embed A ~ A" in a different diagram. We observe that the Q-homomorphism 9"[9"[VoJ2n-I]2 ~ R 2n can be factored as 9"[9" [VoJ2n-,h ~9"[R2n-IJ2 ~ R 2 n and also as 9"[9"[VoJ2n-I]2~9"[VoJ2n~R2n. (Since 9"[VoJ~R is a ring homomorphism, the two factorings indeed give the same obvious homomorphism of 9"[9"[VO]2n-I]2 to R2n.) We put

P= 9"[9"[VoJ2n-I]2 ® 9"[9"[VoJ2n-I]2,

Q=9"[R 2n-IJ2 ® 9"[R2n-1J2'

and derme B, C by the following exact sequences:

o~ B ~ P~9"[9"[VoJ2n-I]4 ~O O~C~Q~9"[R2n-IJ4

~O,

in which P ~ 9" [9" [VOht-l]4 is the Q-homomorphism defined by a®b~ab, and Q~9"[R2n-,J4 is similar. We can embed these and

in a commutative diagram; in this way, we get a sequence B ~ C ---? A". Also, we can embed the first sequence (involving B) and

O---?A ---?M ---?9"[VoJ2n+I---?O

t

t

t

o---? A" ~ N" ---? R 2n+'

~0

in a commutative diagram; in this way, we get a sequence B ---? A ---? A". Since P ~ Q ~ Nil and P ---? M ---? N" make up a commutative diagram, B ~ C ---? A" and B ---? A ---? A" also make up a commutative diagram. The problem is to show that A ---? A" is surjective; we have only to show, therefore, that B ~ C, C ---? A" are both surjective. We shall settle the surjectivity of B ----* C. Since the Q-homomorphism 4>: 9"[VoJ2n-l ---? R 2n-, is surjective, we can take one of its "liftings ", t/!: R 2n-,---?9"[VO]2n-1; rjJ is an Q-homomorphism satisfying 4> 0 t/! = the identity mapping of R 2n-l. Then t/! determines unique liftings of 9"[9"[VoJ2n-l]p---?9"[R2n-1J p for every p, in particular for p=2, 4, and of P----*Q; in this way, we get a lifting C~B of B----* C; hence B~ Cis surjective.

166

IV. Equations Defining Abelian Varieties

We shall settle the surjectivity of C --+ A". We process the exact sequence exactly as above. Then we get an exact sequence of the form

°

--+ l~n-l) ® y[v,,-1]2 + y[v,,-1]2 ® 1lJ.'-I)

--+y[v,,-1]2 ® y[v,,-1]2 --+ R 2n ® R 2n--+ 0; as above, this can be embedded in a commutative diagram; and it tells that Ker(y[v,,_1]2 ® y[v,,-1]2 --+y[v,,-1]4) --+ Ker(R 2n ® R 2n--+ R 2n+') is surjective if and only if the sequence y[v,,-1]2 l~n-l) --+ y[v,,-1]4 --+ R 2n+. is exact. Since this sequence has been shown to be exact, we have the surjectivity. We can replace v,,-l by R 2n-. and we get the surjectivity of C --+ A". q.e.d. Corollary. Let 1 (S) denote the kernel of the degree-preserving homomorphism Y[SI]--+S over Q; then we have Y[SI]k_21(S)z=I(S)kfor almost all k. We can identify v", R 2n, S2n for n=O, 1, ... via the Q-linear mapping and the canonical homomorphism R --+ S; then we get Y [Vo] = Y[Sd and 1 cl(S), 12n+.=I(S)zn+' for n=O, 1, .... This implies that Y[SI]2n+'_21(S)z=I(S)zn+. for n=O, 1, ... ; the rest is the same as in the proof of the second part of Theorem 4.

v" --+ R

§ 5. Quadratic Equations Defining Abelian Varieties We shall determine quadratic relations between elements of Sl in S. In the notation of the corollary of Theorem 4, we have Y[VO]=Y[SI] and 12 =1(S)z; hence we have only to make elements of 12 explicit. We shall proceed as in the proof of Lemma 10 where we have settled a similar problem for I~l); in fact, the present case is much simpler. We shall introduce similar notations: Let 2 denote an arbitrary element of K! and a, b elements of Xl satisfying a=b mod Xo; define an element T(2; a, b) of Y[VO]2 as T(2; a, b)=

I

2(r) Qo(a+b+r) 0 Qo(a-b+r).

Then, for every r, s in K 2 , we have T(2; a+r, b+~)=2(r+s) T(2; a, b);

§ 5. Quadratic Equations Defining Abelian Varieties

167

this is clear. Moreover, as we have seen in § 3, the image of T(A; a, b) in R2 is QI (A., a) ql ()" b); and also

L A(r) qo (a + b + r) qo(a- b + r) =ql (A., a) ql (A., b).

rEK2

If A., a, b are as above and c in a+Xo=b+Xo, we put U (A.; a, b, c)= ql (A., c) T(A.; a, b)- ql (A., b) T(A; a, c); this is an element of 12 • We shall prove the following theorem:

Theorem 5. The subspace 12 of .9'[VO]2 is spanned over Q by ql (A, d) U(A.; a, b, c)

for all A. in

K~ and a, b, c, d in Xl satisfying a=b=c=d mod Xo. The of Qo(x)oQo(Y) in ql (A., d) U(A.; a, b, c) are linear combinations of qo(U) qo(vLfor U, V in Xo with coefficients in Z. Proof We shall prove the second part. We put

co~fficients

t(A; a, b)=

L A(r) qo(a+b+r) qo(a-b+r);

reK2

then t(A.; a, b) is a linear combination of qo(U) qo(V) with coefficients in Z, and t(A.; a, b)=ql (A, a) ql (A., b); moreover, we have

ql(A., d) U(A.; a, b, c)=t(A.; c, d) T(A.; a, b)-t(A.; b, d) T(A.; a, c). If we recall the definition of the T's, we see that the coefficients of Qo(x)oQo(y) in ql (}o, d) U(}.; a, b, c) have the required form. Let 1', I" denote the subspaces of 12 spanned respectively by elements such as ql (A., d) U(A.; a, b, c), U(A.; a, b, c); then we have I' cI". According to Lemma 2, for any given A., a in K~, Xl' there exists an element d of a+Xo such that ql(A.,d)::j::O; hence I'=I". We shall show that 1'=12 . We observe that the T's span.9'[VoJ2. In fact, for any given x, y in Xo, we can find a, b in Xl satisfying a+b=x, a-b= y; and we have Qo(x) 0 Qo(y) = 1/28 • L T(A.; a, b). AeKl

Therefore, we will get I' =12 if we have dim (.9' [VoJ21 I') ~ dim(R2). We observe that if ql (A., b)=O, T(A.; a, b) is in 1'. In fact, ql (A., b) =0 implies that U(A.; a, b, C)=ql(A., c) T(A; a, b); again by Lemma 2, we can find c in a+ Xo such that ql(A., c)::j::O; hence T(A; a, b) is in 1'. We shall show that, for any given A., a, the T(A; a, b)'s for all b in a + X 0 span a vector space over Q such that its image in .9'[Vo]211' is of dimension at most one. Suppose that T(A.; a, b) for b in a + Xo is not in 1'; then we have ql (A., b)::j::O. Let c

168

IV. Equations Defining Abelian Varieties

denote an arbitrary element of a+ Xo; then T{2; a, c) is in QT{2; a, b)+ 1'. This proves the assertion. Consider the set of all pairs (2, a) in which 2 runs over K! and a over a complete set of representatives of XdK 2 ; choose b from a + X 0, if any, such that T{2; a, b) is not in 1'. Then, by what we have said, the set of all such T's spans 9"[VO]2 modulo 1'; and the number of elements in this set is at most equal to card (Xd = dim (R2)' q.e.d. We go back to the ring of "genuine" theta functions: We take as S the graded normal integral domain over Q = C such that m'

m' runs over zg /zg for the diagonal matrix e with e1 , .•• , eg as its diagonal coefficients. We have assumed that e==O mod 4; we shall show that S is afinite R-module via the canonical homomorphism R --+ S. According to our previous notation, the image of R in S is C [SI]. As we have shown in Chap. III, Theorem 5, S is finitely generated over C and integral over C [SI]; therefore S is certainly a finite C [SI]-module. In particular, by the corollary of Theorem 3, we have C [SI]k = Sk for almost all k. We change our notation and denote by S the graded ring of theta functions of an arbitrary polarized abelian variety; let d denote a positive integer satisfying d == 0 mod 4. Then, by Chap. III, Theorem 3, the above result can be applied to S(d) instead of S. Hence we have {S(d»)k=C[Sd]k for almost all k; this is the "peculiar proof" mentioned at the end of Chapter III. We shall prove an important consequence of Theorem 5 as Theorem 6; we need two lemmas: e- 1

Lemma 11. Let ("t", a) denote an arbitrary point of 6 g x exists at least one m' in 1zg such that Om' 0 ( "t", a) =1= O.

cg,"

then there

Proof This is a consequence of Chap. III, Theorem 3 and Lemma 14. For the convenience of the reader, we shall give a complete proof. In the formula of Theorem 2, we take ml = m2 = (m' 0), m' in 1zg, ZI = Z, Z2 = a; then we get n1 =(m' 0), n2 =0, WI =l(z+a), W2 =l(z-a). Hence we have

Om'O("t", z) Om'O("t", a) =

1/2g 'Le( -2m' I r") Oor,,(h, l(z + a)) OOr,,(h, l(z-a)), r"

in which r" runs over 1zg/zg. If we apply runs over 1zg/zg, we get

OoH"t", !(z+a)) 0o(!"t", !(z- a)) =

L to both sides, in which m' m'

L Om'O("t", z) Om'O("t", a). m'

§ 5. Quadratic Equations Defining Abelian Varieties

169

t

Therefore, if we have ern' 0 (r, a) =0 for every m' in zg, the right side, hence also the left side, is O. Since z is an arbitrary point of e g , we get eo(tT, )=0; but this is not the case. q.e.d.

Lemma 12. Let Z/L and Z* /L* denote complex tori, and Z/L-+ Z* /L* a holomorphic mapping. Then it can be lifted to a holomorphic mapping Z -+ Z*, and every lifting is an affine linear mapping. Proof Since Z is connected and simply connected, the holomorphic mapping Z/L-+ Z* /L* can be lifted to a holomorphic mappingf: Z -+ Z*. If ~ is an arbitrary element of L, for every z in Z, f(z+~)-f(z)=~*

is contained in L *. Since z -+ ~* gives a continuous mapping Z -+ L* and since L* is discrete, ~* does not depend on z. We introduce coordinates Z1' Z2, ... and zf, zf, ... in Z and Z*, and put (z;* f)(z)=.(;(z) = I; (Z1' Z2' ... ) for i = 1, 2, .... Then all partial derivatives 0/;/0 Zj are holomorphic periodic functions on Z with L as a period group; hence they are constants. By integration, we see that each I; (z) is a polynomial in Z1> Z2, ... of degree one; hencefis an affine linear mapping. q.e.d.

Theorem 6. If the diagonal matrix e of degree g satisfies e==O mod 2, we get a holomorphic mapping of6g to ~(c)for N + 1 = det (e) given by T -+

(ern' 0 (T, O»)rn' ,

in which m' runs over zg e- 1/zg. Moreover, if two points T, T * of 6 g have the same image in ~(C) and e==O mod 4, there exists an element (J' of M 2g (Z) composed of rx, p, y, 0 satisfying

(J'

e)t (0-e 0e)

0 ( -e 0

(J'=

such that

for every m' in

zg e- 1 /Zg, in which

(T*, z*) =(rxT + pe)(y-r +0 e)-1 e, z(y T+0 e)-1 e).

t

Proof If e == 0 mod 2, zg/zg is contained in zg e- 1/zg. Therefore, by Lemma 11 there exists at least one m' in zg e- 1 /zg such that ern' O(T, 0)=1=0, in which T is an arbitrary point of 6 g ; this implies the first part. If e==O mod 4, by Chap. III, Theorem 4, we get a projective embedding of eg/(zg T + zg e) given by

170

IV. Equations Defining Abelian Varieties

let X denote the image. If we consider em' 0 (" ) as letters, we denote them by To, ... , TN' We define the ring S for, and denote by I (S) the kernel of SI'[Sl] -+ S; then, by the corollary of Theorem 4, we have SI'[Sl]k_2 I(Sh = I(S)k

for almost all k; moreover, we have SI'[Sl] =C[To ,"" TN]. Every element of I(S)k vanishes at every point of X; the converse is also true. Let x denote a point of &(q not in X; then there exist elements L, P of SI'[Sl]l' I(S)p for some p such that L(x) =1=0, P(x) =1=0. If we take a sufficiently large integer q, PLq is in I(S)k=SI'[Sdk_21(Sh for k=p+q; then we get

PU=2, AiF;, i

in which Ai are in SI'[Sl]k_2 and F; in 12 =1(Sh. This implies that

2, Ai (x) F; (x) =1= 0, i

hence F;(x) =1=0 for at least one i. We have thus shown that X consists of points of &(C) which satisfy F(x)=O for every F in 12 , We can restrict F to any subset of 12 which spans 12 over C; e.g., we can take the set of all "ql (A., d) U(.1; a, b, c)" in Theorem 5 as such a set. We recall that each ql (A., d) U(.1; a, b, c) is a quadratic polynomial in To, ... , TN whose coefficients are themselves quadratic polynomials in em' 0 (,,0) with coefficients in Z. After this crucial observation, we take another point of 6 g which has the same image as , in &(C); let X* denote, as X, the image of cg/(zg + zg e) in &(C). Then, up to a constant factor, we have the same set of quadratic polynomials in To, ... , TN which define X, X* in &(C); hence X =X*. We observe that the "zeros" in

,*

,*

cg/(zg, + zg e),

Cg/(zg

,* + zg e)

are mapped to the same point of X = X*. Therefore we can take the lifting of under which 0 is mapped to O. Then by Lemma 12 we get a non-singular linear transformation z-+ z* =zv of Cqo itself satisfying (zg, + zg e)v= zg + zg e; furthermore, there exists a non-vanishing function ton C g, which depends on " such that

,*

,*

em,o('*' z*)=t(z) em,o(', z) for every m' in zg e- /zg and z, or z*, in C g. Since, for every z in eg, we have em'o("z)=I=O for at least one m', t is a non-vanishing holomorphic function on eg. Since t is also the quotient of theta functions, it is a trivial l

171

§ 5. Quadratic Equations Defming Abelian Varieties

theta function; if we change the signs ofm', z, by using Chap. I, § 10, (0.1), we get t( -z)=t(z); hence t(z)=const. e{!· zstz) for some s=ts in Mg(C). Moreover, there exists an element (1ofGL 2g (Z) composed of oc, p, y, 0 such that

r:) =(; ~ G)

v;

this implies that v=(y-r+oe)-l e and -r* =(oc-r+pe)v=(oc-r+pe)(y-r+oe)-l e.

If we replace z* by z* +pe, i.e., z by z+(py)-r+poe, for any pin by using Chap. I, § 10, (0.5), we get

zg,

(y-r +oe) s=y. This implies that (y-r+oe)st(y-r+oe)=y-rty+yeto; hence yeto is symmetric. Similarly, if we replace z* by z* +p-r*, i.e., z by z+(pCl.}-r+ppe,

19=(oc-(oc-r+pe)s)tv-l.

we get

Since s=ve- l y, this is equal to (oc t(y-r +0 e}-(oc-r + p e)ty) e- l =(oc e to - p e ty) e- l

;

hence IXeto-pety=e. We can rewrite 19=(IX-(IX-r+pe)s) tv-l as

octv- l = 19+-r* V-I stv- l ; hence octv- l -r* =-r* +-r* V-I stv-l -r* is symmetric. Also, we can rewrite -r* =(oc-r+pe)v as etp= _-rtoc+tv- l -r*; hence ocetp= -oc-rtOC+octV-I-r* is symmetric. Therefore (1 satisfies the required condition. q.e.d. We can continue our discussion and give a characterization of the element (1 in Theorem 6. However, we shall postpone this analysis until the next chapter; we shall derive a corollary of Theorem 6: Let r denote a positive integer and m = (m' m") an element of Z2g r- l ; then, by replacing p by r p + q in

Om (-r, r z) =

L e(t(p + m') -r t(P + m') + (p + m') t(r Z + m"») ,

peV

we get

0m(-r, rz)=

L

e(q+mVm") O[(q+m')r- l 0] (r2 -r, r2 z).

qmodr

We can solve this system of linear equations; and we get

Oem' r- l 0] (r2 -r, r2 z)= 1/r'· L e( -m,tm,,) 0m(-r, rz), m"

172

IV. Equations Defining Abelian Varieties

in which mil runs over zg r- 1jzg. If r is even, we can apply Theorem 6 to the holomorphic mapping of 6 g to .f!v(C) for N + 1 =r2g defined as r2 T --+ (8 em' r- 1OJ (r2

T,

O))m"

in which m' runs over zg r- 1jr zg. We have e=r 2 1g; hence the element (J is in SP2g(Z) and (r 2 T)#=r 2(IXT+p)(}'T+c5)-1, (r 2 z)#=r 2 z(}'T+c5)-1 etc. By changing the notation slightly, we can state the following result: Corollary. If r is an even positive integer, we have a holomorphic mapping of 6 g to .f!v(C)for N + 1 = r2g defined as

T --+ (8m (T, O))m, in which m runs over a complete set of representatives of Z2 gr-1jZ2 g. Moreover, ifT, T# have the same image in .f!v(C), there exists an element (J ofSp2g(Z) composed of IX, P, y, c5 such that

Om(T*, Z#) = con st. e(tz(YT+c5)-l ytz) 0m(T, z) for every m, in which

(T#, z#) =((OCT + PHy T+ c5)-1, z(y T+ c5)-1).

Chapter V

Graded Rings of Theta Constants § 1. Theta Constants Let z -+ Om ('r, z) denote the familiar theta function on C'; we recall that m=(m' mil) is in R 2g and 't' in 6 g • If m is in Q2 g, we call Om('t', O) a theta constant. Since in this chapter we shall mainly consider theta constants rather than theta functions, we make an agreement that Om means the function 't' -+ Om('t', 0) on 6 g ; accordingly, we shall write Om(r,O) =Om('t'). By an abuse of language, we call Om also a theta constant. Let eb ••• , eg denote positive integers each ej dividing ej + 1 for 1~ i < g and e the diagonal matrix of degree g withe! as its i-th diagonal coefficient; we have denoted by G the subgroup of GL 2g (R) consisting of

such that a e' p, y e 'b are symmetric and a e 'b -

Pe 'y = e. This implies that

-e'p e- 1 ) . eta e- 1 ' hence we can equally assume that 'ae- 1 y, 'pe- 1 b are symmetric and 'ae- 1 b-'y e- 1 p=e- 1 • The subgroup Gz =GnM2g (Z)ofGhasappeared in Chap. II, Theorem 6 and also in Chap. IV, Theorem 6; it is sometimes called the "paramodular group" relative to e. If e is a scalar matrix, we get G=Sp2g(R), GZ =Sp2g(Z); in the general case, we have

G=(lo 0)e SP2g (R) (l0 0)-1 e . g

g

The following lemma will be used in § 3: Lemma 1. Let m, n denote elements of R 2 g such that Om ('t', z) = t(z) On('t', z) for a trivial theta function t; then we have m=n mod 1 and t(z)=e(n"(m-n)").

v. Graded Rings of Theta Constants

174

Proof. This is a consequence of the proof of Chap. II, Theorem 4. For the convenience of the reader, we shall give another proof: Write t(z)= eH·zs'z+b'z+c) for some S=IS in Mg(C), b in C g, and c in C. Apply the translation z-+z+p for any p in zg; then, by using Chap. I, § 10, (0.5), we get s=O and m'==b+n' mod 1; in particular, b is in Rg. Similarly, by applying the translation z -+ z + p -r, we get m" == - b -r + n" mod 1; this implies that b· Im(-r) =0, hence b=O. The rest follows from Chap. I, § 10, (0.2). q.e.d. If 2m==0 mod 1, by Chap. I, § 10, (0.1), (0.2), we have

hence, in the case where 2m==0, 2m' Im"==t mod 1, we get Om=O. We shall show that, in every other case, we have Om =1=0. We need the following simple lemma: Lemma 2. Let K denote a field and x, y elements of Kn satisfying y, i.e., Xi Xj= Yi Yj for every i, j; then we have X= ±y.

IX X=ly

Proof. Since the lemma is trivially true for n = 1, we shall assume that and apply an induction on n; we shall exclude the trivial case where x=O. By changing indices, we may assume that Xl =1=0. If we denote by x*, y* the elements of Kn-l obtained from x, y by dropping Xn, Yn, by the induction assumption, we have x* = ± Y*; then we get Xl Xn = Yl Yn = ±Xl Yn' hence Xn= ±Yn with the same sign as in x*= ±y*. q.e.d. n~2

Theorem 1. We have Om=O, i.e., Om is the constant 0, 2m==0, 2mllm"==tmod 1.

if and only if

Proof. Take a positive integer r such that rm==O mod 1; then Om (2 r2 -r) becomes a series in e( L: eij -rij) for ell, egg running over z.

e12, ... ,

i~j

Moreover the coefficient of e(L:eij-rij) is the sum of e((p+m')lm") for all p satisfying i~j eii =r2 (Pi + mi)2,

eij= 2r2(pi + m;)(pj + mj)

for 1 ~ i ~ g, 1 ~ i < j ~ g; and we have Om = 0 if and only if all such coefficients are O. Consequently, we have Om =0 if and only if, for every p in zg, the sum of e((q+m')lm") for all q in zg satisfying I(q+m')(q+m')= l(p+m')(p+m') is O. By Lemma 2, this means that q+m'= -(p+m') has a solution q in zg for every p in zg and e((p+ m')lm") +e( -(p + m')lm")=O holds. The condition can be rewritten as 2 m' == 0 mod 1, e (2 m' 1m") = - 1, and e(2p 1m") = 1 for every pin z!, i.e., 2m==0, 2ml l m" ==!mod 1. q.e.d.

§ 1. Theta Constants

175

In the special case where g= 1, we have a more defmite result. We have shown in Chap. III, § 6 that Oo(T, ) has only one zero in C!(ZT+ Z), and it has (T+ 1)/2 as a representative. Since Om(T) and Oo(T, m' T+m") differ only by an exponential factor, we have 0m(T)=O if and only if m' == m" ==! mod 1. Therefore, in the case where g = 1, all Om which are not the constant 0 have no zeros in 6 g • We shall tum our attention to the "transformation law" of Om under the paramodular group. We recall that if a is an element of G composed of oc, p, y, (j, for every Tin 6 g ,

a· T=(OC T+ Pe)(y T+(j e)-l e is also in 6 g ; and, in this way, G acts transitively on 6 g •

Theorem 2. For any a in G and m in R 2 g, we put (j - ocY) +i(yet(j)o(ocetp)o), n=m t( _p
.L K(a)xy Oy+m'e-

1

m,,(T),

y

in which x, yare in zg e-1/Zg and, if we put K(a)=(K(a)xy)x,y, then K(a) is a unitary matrix of degree det(e) depending only on a with the same sign ambiguity as that of det(e-1(YT+(je))t. Proof We shall denote the "n" for m=O by b and write n=a+b; i.e., we put a' = m' t (j - m" t y , a" = _m,tp+m"t oc ,

hence


We also put V=(YT+(je)-le, s=ve-1y, T*=a'T, and z*=zv. If we take m = 0 in Chap. II, Theorem 6 and denote the unitary matrix (Urs)r, s by K(a), we get

Ox+b' e- 1b" (T*, z*)=e(tz s tz) det(v-1)t L K(a)xy OYO(T, z); - y

176

V. Graded Rings of Theta Constants

we observe that IC(O) depends only on 0' and it has the same sign ambiguity as that of det(v- 1)t. In this identity, we replace z by m' e-1..r + m", hence z* by a' e- 1 .* +a"; then, by Chap. I, § 10, (0.3), 0 [x + b' e- 1 b"] (.*, z*) is equal to

and Oyo(', z) equal to

Therefore, we have only to show that cPm(a) is congruent modulo 1 to a' e- 1 tb" +!a' e- 1 .* e-lta' +a' e- 1 ta" -!m' e- 1 • e- 1 'm' -m' e- 1 'm" +!(m' e- 1 .+m") s'(m' e- 1 .+m").

We shall show that they are actually equal; by what we have said, it is enough to show that

for z=m'e- 1 .+m", hence z*=zv=a'e- 1 .*+a". By s=ve- 1 y, this is surely true if a' = m' e- lt v- 1 e - z ty; the right side is equal to

and this is a' by definition.

q.e.d.

If we take e = 19 in Theorem 2, we get the following simpler situation:

Corollary. For any 0"

0'

in Sp2g(R) and m in R 2g, we put

m =m 0'-1 +!(y t<5)o (IX tP)O) ,

cPm(O') = -!(m"<5/Jlm' -2m"f3y tm" +m" ty lX'm") +!(m' t<5 - m" ty) '(IX '(3)0; then, for every

0'

in SP2g(Z), we have

in which IC(O'f is an element of

C; depending only on 0'.

The above corollary is called the "transformation formula" of theta constants. We observe that the formula is explicit except for the constant IC(O'). (In the case where det(Y)=FO, it can be expressed by a (multiple) Gaussian sum.)

§ 2. Some Properties of K(q)2

177

§ 2. Some Properties of rc(a)Z We shall examine the constant ,,«1) in the corollary of Theorem 2; we have to introduce certain subgroups ofSp2g(Z). For our later purpose, we shall consider, more generally, the corresponding subgroups of Gz . We have seen that G is conjugate to SP2g(R) by an inner automorphism of GL 2g (R); hence G is a subgroup of SL2g (R) and Gz a subgroup of SL2g (Z). For every positive integer I, we have defmed Gz(l) as the kernel of the homomorphism Gz -+ SL2g (Zjl Z) which maps every (1 in Gz to (1 mod I; Gz(l) is a normal subgroup of Gz of finite index. We shall denote by D the diagonal matrix of degree 2g composed of e, 0, 0, e, i.e., we put

D=(~ ~); as before, we shall denote by E the element of SP2g(Z) composed of 0, 19 , -1g, O. We have DE =ED; and G consists of all (1 in M 2g (R) satisfying a (DE) t(1 = DE. Lemma 3. Let Gz (e) denote the set of all a in Gz of the form 12g + DS with S in M 2g (Z); then Gz(e) is a normal subgroup of Gz of finite index. Proof If (1,0"' are elements of Gz(e) and (1= 12g +DS, a'=1 2g +DS', we have (1 (1' = 12g + D(S + S' + SDS');

hence aO"' is in Gz(e). For every a in G, we have (1-1=(DE)t(1(DE)-1. Therefore, if (1 is an element of Gz(e) and (1= 12g +DS, we have (1-1= 12g +D(E tSE- 1);

hence (1-1 is in Gz(e). We have shown that Gz(e) is a subgroup of Gz . If (11 is in Gz and (1= 12g +DS in Gz(e), we have all (1(11 = 12g+D(Et(11E-1 S(11)' . Since this is an element of Gz(e), Gz(e) is normal in Gz . Since Gz(e) contains Gz(eg), the index of Gz(e) in Gz is finite. q.e.d. As before, for any square matrix s, we shall denote by So the row vector whose i-th coefficient is the i-th diagonal coefficient of s. If s is in Mn(Z) and tsss mod 2, for every u in Mm,n(Z), we have

(ustu)ossotu mod 2.

Lemma 4. Let Gz (e,2e) denote the set of all (1=1 2g +DS in Gz(e) satisfying (SE)osOmod 2; then, if esO mod 2, Gz(e, 2e) is a normal subgroup of Gz of finite index.

178

V. Graded Rings of Theta Constants

Proof If we expand (J'(DE)t(J'=DE for (J'= 12g +DS in Gz(e), we get SE-t(SE)+SDEtS=O. Therefore, if e=O mod 2, we have SE=t(SE) mod 2; furthermore, (J'--+ SE mod 2 gives a homomorphism of Gz(e) to the additive group of symmetric elements in M 2g (Z/2Z). In particular, (J'--+(SE)o mod 2 is a homomorphism of Gz(e) to (Z/2Z?g. Since Gz(e, 2e) is the kernel of this homomorphism, it is a normal subgroup of Gz(e); we shall show that Gz(e, 2e) is normal in Gz . We have seen that, for every (J'1 in Gz and (J'= 12g +DS in Gz(e), the" S" for (J'lt (J' (J't is Et(J't E- l S (J'1; since we have if (J' is in Gz(e, 2e), (J'11 (J'(J't is also in Gz(e, 2 e). Since Gz(e, 2e) contains Gz (2e g), the index of Gz(e, 2e) in Gz is finite. q.e.d. If 1 is an arbitrary positive integer and e=11g, we have G=Sp2g(R), GZ =SP2g(Z); also Gz(e) consists of all (J' in SP2g(Z) which satisfy (J'= 12g mod I; hence Gz(e) = I; (I). If 1 is even, we have e=Omod2; hence Gz (e,2e) is a normal subgroup of Gz =I;(1); we shall denote it by I;(l, 21). If we write an element (J' of I; (I) as

(J'=(;

~)=12g+le

;),

a)

we have

-b SE= ( -d c .

Therefore, (J' is in I; (I, 21) if and only if bo = Co = 0 mod 2, i. e., if and only if

(ex tf3)o =(')1 t<5)o =0 mod 21; we define a subset I;(l, 21) of I; (I) by this condition for every 1. If 1 is odd, we have I;(I, 21)=I;(l)nI;(1, 2). We shall show that I;(1, 2), hence also I;(l, 21), is a subgroup of I;(1). If (J' is an element of Sp2g(R) composed of ex, {3, ')I, <5 and m an element of R 2g, we put (J"

m=m (J'-1 +a((J'),

a((J')=!((')I t<5)o (ex t{3)o);

clearly, we have 12g · m =m. If n is another element of R 2g, we have (J'·m-(J'·n=(m-n)(J'-t; therefore, if (J' is in I;(1) and m=nmod1, we have (J' . m = (J' . n mod 1. Lemma 5. If (J', (J" are elements of I;(1) and m an element of R 2g, we have

(J'(J'" m=(J" ((J'" m) mod 1; hence I;(1) acts on the set R 2 g/Z 2 g.

§ 2. Some Properties of K(O-j2

179

Proof We have u u'· m - u· (u'· m)= a(u u') - a (u') u- 1 - a(u). Put u u' = u" and suppose that u, u', and u" are composed of a, f3 etc., a', f3' etc., and a", f3" etc. Then the" first component" of the above vector is

!(y"I(jIl)O -(y' W)ot(j +(a'If3')oty -(y t(j)o). We shall show that this is in zg. We have

y" =y a' +(j y',

(j" =y f3' +(j (j';

therefore, by using the condition that u' is in I;(I), we get

(y"I(jIl)O

=(a

lt

f3')oly + (y' t(j')ot(j +(y 1(j)O mod 2;

hence the first component is in zg. In the same way, we see that the second component is in zg. q.e.d. We have shown that I;(I) acts on the set R 2gjZ2 g ; we observe that the stabilizer subgroup of I;(I) at 0 consists of those u in I;(I) which satisfy a(u)=O mod 1, i.e., (y 1(j)O =(a l f3)o =0 mod 2. Since this is precisely the definition of I;(I, 2), it is a subgroup of I;(I). Although I; (I, 21) is a subgroup of I;(I) for every I, if I is odd, it is not normal in I;(I); in fact, it is not normal even in I;(I). We observe that the element E of I;(I) is contained in I;(I, 2). On the other hand, for every u in Sp2g(R), we have lu=E- 1 u- 1 E. Therefore Sp2g(R), I; (1), and I;(1,21) for every I are invariant under u -" tu. In particular, I;(l, 21) can equally be defined as a subset of I;(l) by Lemma 6. Both I;(I) and I;(2) are generated by elements of the forms

Proof For I; (1), this has been proved as Chapter I, Lemma 15; more precisely, in the proof of that lemma. Therefore, we have only to consider I;(2). Suppose that p is odd and q even; then, for some even k, we have Iq+kpi ~Ipl. Since pis odd andq+kpeven, we have Iq+kpi
180

V. Graded Rings of Theta Constants

(i) If ')Ig =FO, pass from

(1

to (1

so that Inew <5gg l ~lrgloo; (ii) pass from (1 to (1

(1g

o

P) 19

(U~1 ~)

so that the new <5g becomes (0 ... 0 d) for d ~ 1; (iii) if <5g is of the form (0 ... 0 d) for d ~ 1, pass from (1 to (1

(1p 0)19 g

so that Inew ')Igloo
U=(1v

g- 1

0)

±1

for some v in 2Zg- 1, in which ±1=the sign of <5gg . Then the new <5gg is positive and, for a suitable v, the new <5gj for i =F g becomes less than <5gg in absolute value. If we still have <5gj =FO for some i=Fg, we take = U

(10

g_ 1

tV) 1

for some v in 2 Zg-1 such that all coefficients of v are 0 except for Then the new <5gg is <5gg +k<5gj ; hence, for a suitable k, we get

Vj

= k.

Inew <5gg l ~1<5gd < <5gg . A repeated application ofthese two operations will give the operation (ii). Since d is the greatest common divisor of the coefficients of the old <5g, we have d~lold <5ggl. The existence of (iii) is clear. If we apply (i), (ii), (iii) in this order to any given (1 with ')I g=F 0, we get

Inew ')Igloo < l')Igloo· The rest is the same as in the proof of "Lemma 15."

q.e.d.

§ 2. Some Properties of "(0")2

181

After these preliminaries, we shall start examining the constant K(O"); we shall review a part of Chap. III, § 5: If 0" is an element of SP2 g(R) with y, b as its bottom row and -r a point of 6 g, for every integer k, p(O", -r)= det (y -r + bY' defines a holomorphic automorphy factor relative to SP2g (R); hence we can let Sp2g(R) act on the vector space of holomorphic functions on 6 g as (0". f)(-r) = p(0"-1, -r)-1 f(0"-1. -r). We take 0" from r,(1) and write down 0"-1. f for 2k

f= nO".ml; i=1 then, by the corollary of Theorem 2, we get 2k

(2k) 2k

0"-1)] O".m,=K(0")2k e i~1q,ml(0") ig 0ml· In the following lemma, we shall use this fact for k = 1 and m1 = m2: Lemma 7. If 0" is an element of r,(1) composed of (X, K(0")2 = KrO"-1)2 =det (b).

p, 0,

b, we have

Proof We shall show that K(O"f =det(b) for every 0" in r,(l) composed of (X, p, 0, b. By the corollary of Theorem 2, we have

If we compute O".o(O"·-r) directly, by using (PP)t(Prx):=(rxtP)otpmod2, we get O".o(O"·-r)=Oo(-r); hence K(0")2det(b)=1. Since det(b)=±l, we can write this as K(0")2 =det(b). We shall show that K2 gives a character of r,(1,2), i.e., a homomorphism of r,(1, 2) to Ct. If 0" is in r,(1,2), we have 0"·0:=0 mod 1; hence Oa.O=OO. This implies that

If we compute (0"0"')-1.(0 0 )2 for 0", 0"' in r,(l,2), we get K(0"0"')2= K(0")2 K(u'f. We shall show that K(0")2=K(tO"-1)2 for every 0" in r,(1) composed of (X, p, 0, b. We have to"=E- 10"-1 E. We shall apply E, 0"-1, E- 1 in this order to (0 0 )2: Since E is in r,(l, 2), we have

We solve the equation 0". m=O in m, and we get

182

V. Graded Rings of Theta Constants

since m'=O and "y"=O for (J, we have l/Jrn((J)=O. Therefore we get

(J-1 . (0 0 )2 = (J-1 . (0 00 , m)2 = K((Jf (Orn)2. We solve the equation E· n=m in n, and we get

n=( -m" m') = (!(a IP)O b 0); since n" = 0 and "b" = 0 for E, we have l/Jn (E) = O. Therefore we get

E- 1 . (Ornf =E- 1 . (OE'nf =K(E)2 (On)2. Since K2 gives a character of Fg(1, 2), we have K(E)2 K(E-1)2 = 1; hence

I(J. (Oof = K((J)2 (Onf. On the other hand, we have 1(J-1 . n = 0; since n" = 0 and I(J-\ we have l/Jn(I(J-1)=0; therefore we get

"/3" =

0 for

I(J . (Oof = (,(J-1)-1 . (O'a -1. n)2 = K('(J-1f (On)2 . Putting these together, we get K((J)2=K('(J-1f.

q.e.d.

Theorem 3. We have K((J)8 = 1 for every (J in Fg(1); K((J)2 =e(!· tr((J -1 2g )) = ( _l}ttr(~-1g) for every (J in Fg(2) composed of a, of Fg(1, 2).

/3,

y, b; and K2 gives a character

Proof The last part has been proved (in the proof of Lemma 7); we shall prove the first part. If 2m=Omod1, we have e(8l/Jm((J))=1 for every (J in Fg(1); hence (J-1 . (Oa' rn)8 = K((J)8 (Om)8. By LemmaS, we have (J(J"m=(J'((J"m) mod 1 for every (J, (J' in Fg(1) and m in R 2 g; by Chap. I, § 10, (0·2), if 2m=0 mod 1, (Omf depends only on m mod 1. Therefore we get

((J (J')-1 . (0 0000" m)8 = K((J (J')8 (Om)8 =((J,)-1. ((J-1. (0 00 , (a"m»)8)=K((J)8 K((J')8 (Om)8. Since Om+O, e.g., for m=O, we get K((J(J')8=K((J)8 K((J')8; therefore K8 is a character of Fg(1). By Lemma 7, for every (J in Fg(1) composed of a, /3, 0, b, we have K((J)4=K('(J-1)4=1, hence K((J)8=K(I(J-1)8=1. By Lemma 6, such (J and 1(J-1 generate Fg(1); hence K8 = 1 on Fg(1). In order to prove the second part, we make the following observation: For every odd p, we put X(P)=( _1)t(p-1). Then X(P)= 1 for p= 1 mod4, and X(P)= -1 for p= -1 mod4; hence x(p)=pmod4, and X(Pp')=X(P)X(P') for every odd p, p'. For every (J

§ 3. Holomorphic Mappings by Theta Constants

183

in .fg(2) composed of IY., {3, y, 15, we put

we shall show that IjJ is a character of .fg(2) satisfying IjJ (a) = ljJea- 1). If a' is an element of .fg(2) composed of IY.', {3', y', b' and 15" the "15" for a a', we have b"=y{3'+bb'. Therefore we get g

bi'i= Lbijbji=biibiimod4, j=1

hence ljJ(aa')=IjJ(a) ljJ(a'). We observe that the "15" for la- 1 is Since IY. lb-{3ty=lg, we get

IY..

g

L lY.ijbij=lY.iibii= 1 mod4;

j=1

hence lY.ii=b ii mod 4. Therefore we have ljJ(a)=IjJ(fa- 1)=e(k· tr(a-1 2g»).

What remains to be shown is that the two characters 1(2 and IjJ of .fg(2) are the same. Let a denote an element of .fg(2) composed of IY., {3, 0, b. Then, by Lemma 7, we have l(a)2 =det(b); also

=nbii=ljJ(a) mod4. g

det(b)

i=1

Since l(a)2 and ljJ(a) are ± 1, we get l(a)2 = IjJ (a). By Lemma 7 and by what we have shown, we get l((fa- 1)2=I((a)2=IjJ(a)=ljJea- 1). By Lemma 6, .fg(2) is generated by such a and la- 1 ; hence 1(2=1jJ on .fg(2). q.e.d.

§ 3. Holomorphic Mappings by Theta Constants We shall give a characterization of the element a of Gz discussed in Chap. IV, Theorem 6:

,#

denote arbitrary points of 6 g ; then, in the case Lemma 8. Let " 0 mod 4, we have

where e

=

Oxo('#)= t· Ox 0 (,) with t =t= 0 independent of x for every x in for some a in Gz(e, 2 e).

zg e- 1/Z g if and only if, # =

a .,

Proof Suppose that we have Oxo(,#)=t· Oxo(') for every x in

zg e- 1/Z g ;

then, by Chap. IV, Theorem 6, there exists an element a of Gz composed of iX, {3, y, 15 with the following properties: If we put

184

v. Graded Rings of Theta Constants

v=(y,+oe)-I e, z* =zv, s=v e-Iy, we have T* =a· ,=(cu+pe)v and Oxo(r*, z*)=const. eH z stz) Oxo(T, z) for every x in zg e-I/Zg, in which the "const." is independent of x. In particular, OO(T*,Z*) and 00(T,Z) differ by a trivial theta function. In this relation, we apply the translation z* -+ z* +x e, i.e., z -+ z+(xy) T+x b e, for any x in Zge- 1 ; then, by Chap. I, §10, (0·2), (0·3), OoxAT*,Z*)= Oo(T*,Z*) and Oxyx~e(T,Z) differ by a trivial theta function. Hence the same thing holds for the pair 00 (" z) and Oxyx~e(T, z). According to Lemma 1, this implies that x y == x 0 e == 0 mod 1. Since x is an arbitrary element of Zge-l, we get y=ec with c in Mg(Z). Similarly, by applying the translation z* -+z* +x,*, i.e., z-+Z+(xex)T+xpe, for any x in Zge-l, we get xex==x, xpe==Omod1; hence ex=l g +ea, p=eb'e- I with a, b' in Mg(Z). We observe that the relation between T, T* is symmetric; also that the "ex" for a-I is etoe- I. Hence e toe- 1 =lg+e td, i.e., 0 = 19 + e d, with d in Mg(Z). We shall examine the second translation more closely: If we use Oo(T*,z*)=const.eHzstz) 00 (" z) and Chap. I, §10, (0.2), (0.3) in full, we get a rather complicated congruence relation mod 1; by using ,* =tvt(ex T+ p e), tv =ex-(ex T+ p e) s, this can be simplified; and we get !xexetptx==O mod l. This implies that e- 1 exe tpe- 1 ==Omod1, (e- 1 exe tpe- 1}0 ==0 mod 2. If we substitute ex=l g +ea, p=eb'e- I in the first congruence equation, we get e- lt b'==Omod1; hence tb'=etb, i.e., p=eb, with b in Mg(Z); and, from the second, we get b o == 0 mod 2. We have thus shown that a is in Gz(e) and bo==Omod2; we shall show that a is in Gz(e, 2e); we have only to show that co==O mod 2. We examine the first translation in the same way as we have examined the second; then, by using (yr+oe)s=y, we get !xy etotx==O mod 1; this implies that Co == 0 mod 2. We shall prove the converse. We recall that, for every a in G composed of ex, p, y, 0, a* =

(~: ~:) = c_ex y e~ ~ e) i

is in Sp2g(R); moreover, for every T, z in 6 g,

cg, we have

T* = (ex T+ p e)(y T+0 e)-l e=(ex*, + p*)(y* T+b*)-l, z* =z(y T+0 e)-I e = z(y* T+ b*)-I.

§ 3. Holomorphic Mappings by Theta Constants

185

Suppose that a is in Gz(e, 2e); put cx= 19 +ea, p=e b, y=e c, <5 = I g +ed. Then we have cx*=lg +ea, p*=ebe, y*=c, <5*=lg+de; in particular, a* is in Fg(l, 2). Moreover, for any x in zg e- 1, if we put m=a*' (x 0), we get m'=x+tco=x, hence, by Chap. I, § 10, (0.2), we get Oa*.(XO)(r#)=e(xtm") Oxo(T#)=Oxo(r#); also, we have

cPx 0 (a*) = -t(xe) bt(xe)=O mod 1.

Therefore, by the corollary of Theorem 2, we get Ox o(T#)= ,,(a*) det(y* T+ <5*)! Ox o(T); hence we have only to put t=,,(a*)det(Y*T+<5*)!.

q.e.d.

We shall show that, in the case where a is in Gz(e, 2 e), if we choose the sign of det(Y*T+<5*)!=det(e- 1 (YT+<5e»)! suitably, we actually get Ox o(a· T) =det(e- 1 (y T+<5 e»)! Ox O(T) for every x in Z g e- 1/Zg and T in 6 g • By using the same notation as in the above proof, we have tc=c mod 4. Let P denote a symmetric matrix of degree g satisfying P=C mod 4; we can take pji=pij=Cij for l~i~j~g; let a1 denote the element of Fg(I) composed of I g , 0, p, 19 • Then, because of co=O mod 2, a1 is in Fg(I, 2); moreover, a2 =a1"1 a* is in Fg(4). Therefore, by Theorem 3, we have ,,(a*)2=,,(a1)2 ,,(a2)2=,,(a1)2; by Lemma 7, this is equal to "ral1)2 =det(lg)= 1; hence ,,(a*) = ± 1. We can include this sign in det(Y*T+<5*)!. We shall show that the element a of Gz(e, 2e) satisfying T# =a' T is unique, i.e., Gz (e,2e) acts freely on 6 g , for e=O mod 4. We recall that G acts transitively on 6, with compact stabilizer subgroups. In general, let G denote a topological group, K a compact subgroup, and r a discrete subgroup of G; then, if r has no element of finite order other than 1, it acts freely on G/K. In fact, if a K is an arbitrary point of G/K, the stabilizer subgroup of r at aK is rnaKa- 1. Since this is a finite subgroup of r, by assumption it consists of 1 only. In view of this, we have only to prove.the following lemma:

Lemma 9. Let T denote an element of Mn(Z) different from 1n, satisfying T = 1n mod d for some integer d ~ 3; then T has an infinite order, i.e., Tk=F 1n for k= 1, 2, .... Proof Suppose first that d is a power of 2; then we can write T=ln+2as with S in Mn(Z), S$Omod2 for some integer a~2. This

v. Graded Rings of Theta Constants

186

implies that T2 = I n +2a+1 S 1,

SI =S +2a- 1S 2=S mod2;

and, for every b ~ 0, T2b = I n+2a+bS b,

Sb=S mod2.

Let k denote a positive integer; write k = 2b k' with an odd k'. Then we have Tk=(ln+2a+bSbY"

in which

= I n +2a+b(k' Sb+ R),

=

hence k' Sb + R k' S:$ 0 mod 2. If d is not a power of 2, it has an odd prime factor, say p; we can write T = in + pa S with S in Mn (Z), S:$ 0 mod p for some a ~ 1. This implies that TP= I n +pa+1S mod pa+2

even for a = 1; the rest is the same as in the case where p = 2. q.e.d. We shall use the following lemma, which really belongs to Chap. III, § 7, in the proof of Lemma 11:

Lemma 10. For every min R 2 g and 't" in 6 g , the divisorojCg/(zgt+Zg) determined by Om(t, ) is non-degenerate and reduced. Proof LetJ denote the element of M 2g (R) defined by i(tlg)=(tlg)!J; also, let E denote the standard alternating matrix of degree 2g, i.e., the element of M2g (Z) composed of 0, 19, -lg, O. By Riemann forms, we understand Riemann forms on Cg x Cg relative to zg t + zg. Let H denote the Riemann form of Om(t, ); then, for every x=(x' x"), y=(y' y") in R2 g, we have H(x' t+x", y' t+ y")=x(JE) ty+ixEty.

In particular, H is non-degenerate, hence the divisor of Cg/(zg t + zg) determined by Om(t, ) is non-degenerate; we shall show that it is reduced. We have only to show that, for no Riemann form H' =1=0, we have H~2H'. If we have H~2H' for some H'=I=O, H-2H'=H" is also a Riemann form; and the values ofIm(H'), Im(H") at (x' t+x", y't+y") are of the forms x Elty, x E" ty, in which E', E" are alternating matrices in M2g (Z). We have E=2E'+E"; JE', JE" are symmetric and JE', J E" ~ 0; and J (E' + E") is positive-definite. Let T denote an element ofGL 2g (R) such that T(J(E'+E,,»)tT=1 2g and T(JE,)tT=adiagonal matrix; let AI> ... , A2g denote the diagonal coefficients; then we have Ai~O for 1~i~2g. Since T(,JE)tT is the diagonal matrix with

187

§ 3. Holomorphic Mappings by Theta Constants

1 + A1, ... , 1 + A2 g as its diagonal coefficients, we have

n(1 +A;); 2g

det(JE)=det(T)-2

;=1

similarly, we have det(J(E' +E")}=det(T)-2. On the other hand, we have det (E' + E") ~ 1 = det (E). By putting them together, we get n(l+A;)~l; hence A;=O for all i, i.e., T(JEYT=O. This implies that ;

E' =0; hence H' =0. But this is not the case.

q.e.d.

We observe that the theta series for Orn(r, z) can be differentiated term-by-term by Z1' ..• ,Zg,rll>r12, ... , rgg any number of times; in this way, we get 13 2 Om (r, z)/i3 Zj 13 Zk = 2n i 130m(r, z)ji3rjk

(j +k)

=4nii30m (r,z)/i3rjj

(j=k).

If C=(Cjk)j,k is an arbitrary symmetric matrix in Mg(C), we can put them together as g

L

J,k=l

Cjk 13 2 Om (r, Z)/i3 Zj 13 Zk =4n i L Cjk 130m(r, Z)/i3rjk' j;;;k

Since this becomes a heat equation, i.e., a parabolic differential equation which governs the conduction of "heat", for a suitable choice of C (and the corresponding restriction of r), we call it a heat equation; this will play the role of a "turntable" in the proof of Lemma 11. We need another observation: Let k denote a positive integer; consider the vector space over C of all holomorphic functions e on cg satisfying for every p, q. in zg. Then, by Chap. II, Theorem 4, its dimension is kg; by Chap. I, § 10, (O.S), it contains all products Oml(r, z+a 1) .. · OmJr, z+ak),

in which m1, ... , mk are in R 2 g, L m;=O mod 1 and a1, ... , ak in Cg, La;=O. i

i

Let r denote a positive integer; then, for every m in Z2g r- 1, Z ~ Orn(r, r z) belongs to the above vector space for k = r2; this follows again from Chap. I, § 10, (O.S). We have remarked in Chap. IV, § S (after the proof of Theorem 6) that z~ Orn(r, r z) with m running over a complete set of representatives of Z2g r- 1/Z 2g and Z ~ Ox 0 (r2 r, r2 z) with x running over zg r- 2/Z g are related by an element ofGL(kg, C) whose coefficients are in the field of k-th roots of 1. In particular, they span the same vector space over C; and it coincides with the vector space of all e's for k=r2.

V. Graded Rings of Theta Constants

188

Lemma 11. Let -r denote an arbitrary point of 6 g ; then, in the case where e == 0 mod 4, we have rank (Ox o(-r)(oOx o/ihjk)(-r»)x=tg(g+ 1)+ 1,

in which x runs over

zg

e-1/Zg•

Proof It is enough to prove the lemma in the special case where e=4· I g • Then, because of the linear relation between Ox o (4-r) and Om(-r) for 2m==0 mod 1, we have only to prove that rank (Om (-r)(O Om/o-rjJ (-r»)m =t g(g + 1)+ 1 for every -r in 6 g ; where m runs over a complete set of representatives of Z2g 2- 1/Z 2g. Suppose that we have rank (Om (-r)(OOm/O-rjk)(-r»)m
for all m, i.e., for all m in the above set of representatives. We define a symmetric matrix C of degree g as C=(Cjk)j,k' We shall show that c=I=O. If c=O, we get AOm(-r)=O for all m. On the other hand, by Chap. IV, Lemma 11, we have On(-r) =1=0 for some n among the representatives. Then A 01l(-r)=0 implies that A=O; but this is not the case. By the heat equation, the above relation can be transformed into g

L Cjk(02 0m(-r, z)/OZj OZk)(-r, 0)=4n i A 0m(-r, 0); j,k=l

we can replace 0m(-r, z) by 0m(-r, 2z) provided that at the same time we replace A by 4;,. Since m runs over a complete set of representatives of Z2g 2- 1/Z 2g, by the previous observation, we can replace 0m(-r, 2z) by 8(z) = 0mt(-r, z+a1) ... 0m4(-r, z+a4),

in whichml> ... , m4 are in R 2 g, Lnlj==O modI andal> ... , a4in Cg,Laj=O. j

j

We take ml=···=m4=0, and choose al=a, a2=b so that we have Oo(-r,a)=Oo(-r,b)=O; then we choose a3, a4 satisfying Laj=O general j

enough so that 0o(-r, a3) 0o(-r, a4)=I=O. As we have remarked in the proof of Chap. III, Lemma 14, this is possible. In this way, we get grada(Oo(-r, »)ctgradb(Oo(-r, »)=0 for every a, b in satisfying 0o(-r, a)=Oo(-r, b)=O; where grada(Oo(-r, ») is as in Chap. III, Lemma 18, i.e., itis the row vector with (oOo(-r,z)/ozj)(-r,a) Cg

§ 4. The Classical Reduction Theory

189

as its j-th coefficient for 1 ~j ~ g. By Lemma 10, the divisor of cg/(Zg-c+Zg) determined by 0o(-c, ) is non-degenerate and reduced; hence, by "Lemma 18", the grada(Oo(-c, »,s for all a in c g satisfying 0o(-c, a)=O span Cg over C. Therefore, by the above identity, we get c=O; but this is a contradiction. q.e.d. By putting Lemmas 8, 9, and 11 together (and using some observation in Chap. III, § 5), we get the following theorem:

Theorem4. If e=Omod4, the quotient space Gz (e,2e)\6 g is a complex manifold with 6 g as its universal covering man!fold; the holomorphic mapping of6g to &(C)for N+l=det(e) defined by -c --+ (Ox o (-c»)x, in which x runs over zg e-1/Zg , gives rise to an injective mapping [) of Gz(e, 2e)\ 6 g ; moreover, [) maps a small open neighborhood of every point of Gz(e, 2e)\ 6 g biholomorphically to a submanifold of & (C).

If r is an even positive integer, we have a similar result for I;(r2, 2r2) (instead of Gz(e, 2 e») and for the holomorphic mapping of 6 g to &(C) (N + 1 =r2g ) given by in which m runs over a complete set of representatives of Z2g r- 1/Z 2g ; we call this the corollary of Theorem 4. We shall be concerned with the global image of Gz(e, 2e)\ 6 g under [) and also with its closure in & (C). In order to clarify these, we shall follow an entirely different line of thought.

§ 4. The Classical Reduction Theory We shall review a certain aspect of the reduction theory for GLn(Z) and SP2n(Z); we shall use the notation introduced in Chap. III, §2. Let Xn denote the set of symmetric matrices in Mn(R); Xn is a vector space over R of dimension !n(n+ 1). Let ~n denote the subset of Xn consisting of positive-definite matrices; ~n is an open convex cone in Xn • We let GLn(R) act on Xn as then GLn(R) keeps ~n invariant; and the action on ~n is transitive. In fact, if (x, y) is an arbitrary point of Xn x ~n' there exists an element u of GL n(R) such that u· x = a diagonal matrix and u· y = In. If y is an element of ~n' we write y>O; if y is an element of the closure of ~n in X n , we write y~O as before. We observe that y~O implies that ~ yt~~o for every ~ in Rn. Since every y in Xn can be diagonalized, the converse is also true. We shall use the following simple lemma:

190

V. Graded Rings of Theta Constants

Lemma 12. Let K denote a field, y a symmetric matrix in Mn(K) in which the n' -th principal minor, i.e., the submatrix of y obtained by cross'ing out its i-th row and column for all i > n', is non-singular. Then the equation

t(

In' Y= 0

W)

In"

0)

(y' (In' W) 0 y* 0 In"

in the unknown y', y*, win Mn,(K), Mn,,(K), Mn'.n,,(K) is uniquely solvable; moreover, y' is the n' -th principal minor of y and y* is symmetric. Proof If we carry out the multiplication on the right side, we get

(

y'

y'w ) twy'w+y* '

twy'

hence the unique choice for y' is the n' -th principal minor of y. Since y' is in GLn,(K) by assumption, the equation is uniquely solvable in w and y*; and y* is clearly symmetric. q.e.d. Suppose that y is a symmetric matrix in M" (K) of rank n'. By applying the same permutation to rows and columns of y, we may assume that the n'-th principal minor of Y is non-singular; then we have y*=O in Lemma 12. Consequently, if y is a symmetric matrix in Mn(K), there exists an element u of SLn(K) satisfying uy

t

U=

(Y'0 0)0 '

in which y' is non-singular. We shall take R as K. Ify is in ~n' by applying Lemma 12 successively starting from n' = 1, we get in which u is a triangular matrix with 1 on the diagonal, 0 below the diagonal and a is a diagonal matrix with positive diagonal coefficients; we call this the Jacobi decomposition of y. If we denote the (i, j)-coefficient of u by uij and the (i, i)-coefficient of a by ai> we have i-1

Yij=ai uij+ Lap U pi U pj ,

hence

p=1 i-I

Yi=Yii=ai+ L ap(up;)2~ai' p=1

for

l~i~j~n.

Since a1 ... an=det(a)=det(y), we get det(Y)~Yl'" y,,;

§ 4. The Classical Reduction Theory

191

this is called "Hadamard's theorem." The following lemma is called "Hermite's theorem": for

Lemma 13. Let y denote a point of '-l3n and m(y) the minimum of ~ =1= 0 in zn; then we have

~ yt~

m(y)n--<~t~ on Rn, ~ y t~ for ~ =1=0 in zn attains its minimum m(y), say at u 1 ; and the coefficients of U 1 are relatively prime. There exists, therefore, an element u of GLn(Z) with Ul as its first row vector. By replacing y by U· y, we may assume that Yl = m (y). If we apply Lemma 12 by taking n' = 1, for e = (~' ~") in R n with ~' in R, we get ~ yt~= Y1(~' +CtW)2+~" y*tC.

We choose C=I=O from zn-l so that we get Cy*t~"=m(y*); we choose ~' from Z so that ~* =~' + C tw is at most equal to t in absolute value. Then we have hence Yl ~!' m(y*). By the induction assumption, we have m(y*)"-I-< det(y*) on '-l3n-l' This implies that m(y)n= Yi~(1)n-l Ylm(y*)"-I-
on '-l3n.

q.e.d.

If we apply the induction slightly more carefully, we will get m(y)" ~(1}!n(n-l) det(y) for every y in '-l3n. We say that a point y of '-l3n is Minlwwski reduced, or simply Mreduced, if it has the following two properties: (M.1) For every ~=(~1'" en) in zn in which ~k' ... , ~n are relatively prime for some k, we have ~ yt~~Yk; (M.2) Yk.k+l~O for 1 ~k
192

v. Graded Rings of Theta Constants

Let Yo denote an arbitrary point of ~n; then ~ Yo t~ for ~ +0 in zn attains its minimum, say at UI. Consider the set of all ~ in zn such that UI' ~ can be included in a Z-base of zn; then ~ Yot~ attains its minimum, say at U2. By replacing U2 by - U2 if necessary, we may assume that UI Yo tU2 ~ O. By continuing this process, we can fmd a Z-base U1 , ••• , Un of zn satisfying UkYOtUk+I~O and ~ Yot~~UkYotUk for every ~ in zn such that U1 , ••• , Uk-I' ~ can be included in a Z-base of zn. Let U denote the matrix with Uk as its k-th row for 1 ~k~n; then U is an element of GLn(Z), and y=U· Yo is M-reduced. If Y is M-reduced, we have 0
IYijl ~tyi

for 1 ~i
for every y in 9tn • The following theorem, due to Minkowski, is a generalization of this fact:

Theorem 5. We have det(Y)>-
limYI .. · Yn/det(y) = yeS

00.

If we have YI>-< ... >--
on S; but this is a contradiction. There exists, therefore, an index k satisfying 1 ~k-< ... >-
§ 4. The Classical Reduction Theory

193

We choose e" *0 from zrr-k so that we get e" y* te" =m(y*); we choose e' from Zk so that each coefficient of e* = e' + e" tw is at most equal to t in absolute value. Then, by (M.l) for k+l instead of k, we have Yk+l~eyte; hence, by using the Hermite theorem, we get Yk+l ~e* y'te* + m(Y*)-
on S. Since ykiYk+l--+0 on S, we can drop Yk from this relation. Clearly, if Y is in 9trr , y' is in 9tk • Therefore, by the induction assumption, we have Yl ... Yk-
--
~rr'

we

y,,/al ... a,,= Yl ... y,,/det(y)>-
on 9t" by Theorem 5; hence Yi>-
Uij = Yij/ai - L aJai · Upi Upj p=1

for i ... , ui,,-<1 on 9t,. provided that up,p+l> ••• ,u pn -<1 on 9tn for 1 ~ p < i; therefore, we have only to use an induction on i. Corollary 2. Let e=(el ... en) denote an element of R"; then we have

" el eyte>-< LYi i=1

on R"x 9tn • Let d denote the diagonal matrix with di = Y! as its i-th diagonal coefficient for 1 ~ i ~ n; then the relation to be proved becomes

194

V. Graded Rings of Theta Constants

If we replace ~ d by ~, this becomes equivalent to ~ y* t ~ X in which y*=d- 1 yd- 1 • Since we have

~ t~

on Rn x 9l n,

ly;t}1 = IYijlldi dj~IYijIIYi~-! and det(y*) = det (y)fYl ... YnX1 on 9ln, y* belongs to a compact subset of ~n' Therefore its eigenvalues belong to a compact subset of R~; hence ~ y* t~x~ t~ on Rn x 9ln. Let -r denote a point of 6n=Xn+i~n' u an element of SP2n(R) composed of rx., {3, y, b, and U· -r = (rx. -r + (3)(y -r + b)-I; then, as we have seen in Chap. I, § 6, we have Im(u, -r)=t(y-r+b)-l Im(-r)(y:r+b)":l. Therefore we have det(Im(u· -r»)~det(Im(-r») if and only if Idet(y -r+b)1 ~l. We say that -r=x+iy is Siegel reduced, or S-reduced, if it has the following three properties: (S.l) det(Im(u· -r»)~det(Im(-r») for every u in r=SP2n(Z); (S.2) y is M-reduced; (S.3) IXijl~-!for l~i,j~n. We shall denote by O:n the set of S-reduced points. Since 9ln is closed in ~n' O:n is closed in 6 n ; we shall see later that O:n is closed in the ambient space of 6 n , which is the vector space over C of all symmetric matrices in Mn(C). By modifying our previous notation, we shall denote by roo the subgroup of r consisting of those elements u with "y" = 0; an element of roo gives a transformation in 6 n of the form

in which u is in GLn(Z) and s rise to x --+ u Xtu + s, y --+ u Y tu 6 n, roo' -ro contains a point -r -r also satisfies (S.l), hence -r is

a symmetric matrix in Mn(Z). This gives for -r = x + i y. Therefore, for every -r 0 in satisfying (S.2), (S.3); if -ro satisfies (S.l), S-reduced.

Lemma 14. Let -r denote an arbitrary point of 6 n: then, for any e in R~, the set of real numbers of the form det(Im(-r*») for -r* in r·-r satisfying det(Im(-r*»)~e is finite. Proof Put -r=x+iy and -r*=x*+iy*. For our purpose, we can restrict -r* to the subset of r· -r defined by the condition that (y*)-l is M-reduced. In fact, it is enough to show that the set of y* for -r* in r·-r such that (y*)-l is M-reduced and det(y*)~e is finite. By Theorem 5, we have

195

§ 4. The Classical Reduction Theory

on this set. Write 1."* =(a 1." +P)(}"t +15)-1; then we have

(y*)-1 =(Y 't +15) y- 1I(y:r +15)=(, x +15) y-lt(y X +15)+y yly. Let Yk,l5 k denote the k-th rows in y, 15 for

l~k~n;

then we have

(y*);l =(Yk X + 15J y-l/(Yk x+l5 k)+Yk ytYk' Since Yk and 15k are not both 0, we get (y*);l~Yk y lYk»1

for yk=l=O and

(y*);1=l5 k y- 1/I5k»1

for Yk=O. By putting this and (y*)11 ... (y*);1-<1 together, we get (Y*)k"l>-<1 on the set of y* that we have defined. Then the above expression for (y*)k"1 shows that Yk, 15k can have only finite numbers of possibilities; and hence the set itself is finite. q.e.d. As an application of Lemma 14, we shall show that SP2,,(Z) . (j" = 6". Let 'to denote an arbitrary point of 6,,; we have only to show that r· 'to for r=Sp2,,(Z) contains an S-reduced point. By Lemma 14, det(Im(O"' 'to») for 0" in r attains its maximum, say at 0"0' Put 1."=0"0' 1."0; then 1." satisfies (S.l). By applying an element of roo to 1.", we may assume that 1." also satisfies (S.2), (S.3); then 1." is S-reduced.

Lemma 15. If 't=x+iy is a point of (j", we have Y1~3t/2; and y-1 belongs to a subset of ~" bounded in ~". Proof. By using the notation in Chap. I, § 10, we put 0"= ( _

~ ~) $12,,_2;

this is an element of r, and we have Y1." + 15 = ( - 0't1

*) .

1,,_1 '

hence Idet(Y1."+I5)I=I1."ll~l by (S.l). On the other hand, by (S.3) we have IRe(1."l)I~!; hence Y1=Im(1."1)~(1-(!)2)t=3t/2. Let y=tU({U denote the Jacobi decomposition of y=Im('t); then, by Corollary 1 of Theorem 5, we have uij-<1 for i-
V. Graded Rings of Theta Constants

196

subset of (R~r which is bounded in Rn. Hence y-l =u- 1 a- 1Iu- 1 varies in a subset of ~n bounded in ~n' q.e.d. Lemma 16. The subset ~n of 6 n is closed in the ambient space of 6 n; moreover,for any c in R~, the subset ~n(c) of ~n defined by det(Im(r»)~c is compact unless it is empty. Proof. We identify the ambient space of 6 n with eN for N =tn(n+ 1). If y=Im(r), by Lemma 15, we have det(Y)>--l

on ~n; therefore, the closure of ~n in eN is contained in 6 n • Since ~n is closed in 6 n , it is closed in eN. Suppose that ~n(c) is not empty; then l-
~n-h

7:

= (7:'

there exists AO in R~ such that

0)

0 iA

is in ~n for every A~AO' Proof. Put

7:' =

x' + i y' and X=

7: = X

0)

X' (0 0 '

+ i Y; then we have

(Y' 0)

y= 0 A ;

hence 7: satisfies (S.3). Moreover, 7: satisfies (S.2), i.e., Y satisfies (M.1), (M.2), for every A~Yn_l' In fact, since y' satisfies (M.2), Y also satisfies (M.2). Let ~=(~1'" ~n) denote an element of zn in which ~k' ••• , ~n are

relatively prime; put ~'=(~1 ... ~n-l)' If ~n=O, we have 1 ~k-- ... >-(y#);1, and det(y#)-l n) y-l/(Yn X +<>n)+Yn Y'Yn'

§ 5. Modular Forms

197

if we omit a fmite number of terms of S, we get Yn=O and bn=(O ... 0 ± 1). According to Chap. I, Lemma 14, this implies that

*)

0 13' * +1 *

1:1.' q

(

*

= y' - 0 b'

o

0 0

*'

±1

in which a', 13', y', b' form an element q' of T' = Sp2 n_ 2(Z). Since .' is in ~n-1' we have Idet(y'.'+b')I~l by (S.l). Since we have det(y.+b)= ±det(y'.' +b'), we get det(y#)~det(y); butthisis a contradiction. q.e.d. Lemma 18. If we put "n = sup tr(y-1) IYn

for .=x+iy in ~n' we have "1~···~"n
for every A.~A.o. This implies that tr(y'-1)~"n' hence "n-1~"n.

q.e.d.

Remark. Let Y; denote the submatrix of y obtained by crossing out its i-th row and column; then, by using the Hadamard theorem, we get tr(y-1)= on

~n'

n

n

;=1

;=1

L det(Y;)/det(y)~cn L 1/y;~2n cJ3t

in which en is the Minkowski constant. In particular, we have

"1 ~ 2/3t , "2 ~ 16/3t .

§ 5. Modular Forms We shall prove some basic properties of modular forms relative to SP2n(Z), Let r denote a modular group (contained in SP2n(R)) and defme the graded ring A(r) as in Chap. III, § 5; an element f of A(r)k is called a modular form of degree n and of weight k relative to r; it is a holomorphic function on 6 n satisfying f(q· .)=det(y.+b)k f(.)

for every q in r with (y b) as its bottom row; this definition has to be modified, as in Chap. III, § 5, if n = 1. In the special case where r = SP2n(Z), we shall omit "relative to r."

198

V. Graded Rings of Theta Constants

Let L denote a lattice in the vector space Xn of symmetric matrices in Mn(R); then the subset L* of Xn defined by tr(s s*)=O mod 1 for every s in L is also a lattice in Xn. For instance, if L is the lattice of symmetric matrices in Mn(Z), L* consists of "half-integer matrices," i.e., elements s* = (SO)i, j of Xn with sf, 2 in Z for all i, j. If a holomorphic function f on 6 n is invariant under the translation -r -+ -r + s for every s in a lattice L, f can be expanded into a Fourier series:

So

L a (s*) e(tr(s* -r)).

f(-r)=

s*eJ..

This series converges absolutely and uniformly on every compact subset of 6 n; and the coefficient a(s*) is given by

a(s*) = l/m(L) . e( - i . tr(s* y))

J

xmodL

f(x + i y) e( - tr(s* x)) dx,

in which m(L) is the total measure of the quotient by L; for the sake of simplicity, we shall also use s to denote elements of L*. The following lemma is called "Koecher's theorem":

Lemma 19. Suppose that f is a holomorphic function on 6 n which is invariant under the transformation -r -+ u-r'u+s for every u in SLn(Z) and s='s in Mn(Z). Then, in the case where n;;;;2, the Fourier expansion off takes the form f(-r)=

L a(s) e(tr(s-r)), s$:O

in which s runs over the set of half-integer matrices satisfying s;;;; O. Proof. We have seen that the Fourier expansion of f has the above form in which s runs over the lattice of half-integer matrices; we shall show that a(s)=O unless s;;;;O. By the invariance of f under -r-+u-r'u, we have a(ustu)=a(s) for every u in SLn(Z), Suppose that a(s)+O for some half-integer matrix s for which there exists an element ~ of Rn satisfying ~ s'~ <0. Since Q" is dense in R n, we may assume that ~ is in Qn. By multiplying to ~ a common denominator of its coefficients, we may assume that ~ is in zn; we may also assume that the coefficients of ~ are relatively prime; then, there exists an element u of SLn(Z) with ~ as its first row. By passing from a(s) to a(us'u), we may assume that Sl <0. Let y denote a diagonal matrix in Mn(R) with positive diagonal coefficients; then the Fourier series for f is absolutely convergent at i y. Since a(s)+O, this implies that

L exp( -2-n tr(s' y))< 00,

§ 5. Modular Forms

199

in which the summation is taken over the set of all s of the form tv s v for v in SLn(Z); this is not the summation over SLn(Z). We take as v the triangular matrix with 1 on the diagonal and 0 off the diagonal except at (1,2). In this way, we see that L exp( -2n(L S; Yi+(S1 k 2 +2s12 k+s 2) Y2)),

keZ

;*2

in which Si = Sii' Yi = Yii etc., is a partial sum of the above convergent series; but, because of S1 <0, this series is divergent. We thus have a contradiction. q.e.d. We observe that if f is a modular form of degree n and of any weight, it satisfies the condition of the lemma; hence, in the case where n~2, its Fourier expansion has the form stated in the lemma. On the other hand, by our definition in Chap. III, § 5, this is true (trivially) for n = 1. We shall consider Fourier series of the form f(T)=

L a(s) e(tr(sr:»), s~O

in which S runs over the subset defined by s~O of a lattice in ~n; we shall denote this lattice by L instead of L*. For the sake of simplicity, we· shall assume that dL is contained in Mn(Z) for some positive integer d; also, we shall assume that the Fourier series is absolutely convergent at every point of 6 n. If f is a modular form of degree n and of any weight, as we have seen, the above conditions are satisfied with d = 2. We shall examine the behavior ofthe Fourier series when or approaches a "boundary point" of 6 n. For this purpose, we shall recall the concept of dominant series: Suppose that we have a series L F; of functions Fi on a set X and a i

convergent series L M; of non-negative real numbers M;, in which i runs ;

over the same set, say 1. If we have

suplF;(x)I~Mi x

except for a fmite number of, i's in I, we say that

for almost all, i.e.,

L M; is a dominant series i

for L F;; and we write ;

LF;~LM;.

ieI

In this case, for every x in X,

ieI

L F;(x) converges absolutely; and the i

convergence is uniform in x. If all F; depend on a parameter, say A, and sup IF; (x, .,1.)1 ~Mi for almost all i, any limit process with respect to A x,A

commutes with the summation, i. e., we have lim L F;(x, .,1.)= L lim F;(x, A)

A-Ao i

i A.... Ao·

200

V. Graded Rings of Theta Constants

for every x in X provided that lim Fi (x, A.) as A. -do exists; if, moreover F;(x, A.) approaches lim F;(x, A.) uniformly in x for each i, the above

convergence is uniform in x. We have seen that if s, yare elements of X" satisfying s, y~o, we have tr(sy)~O: Since every y~O in X" is a limit ofa sequence in 'lJ,,, we may assume that y>O; write y as tuu for some u in GLn(R); then, we have tr(sy)=tr(ustu)~O. The following statement is an immediate consequence of this fact: Suppose that we have a Fourier series f(.)=

then, for every Yo> 0,

L a(s) e(tr(s-r)); s~o

L la(s)1 exp( -

2 n tr(s Yo))

s~o

is a dominant series for the above Fourier series provided that r is restricted to Im(r)~yo. We shall show that tjn is contained in the subset of 6 n defined by Im(.)~yo for some Yo >0. By Corollary 2 of Theorem 5 and Lemma 15, if ~ is in Rn and y=Im(r) for. in 6 n , we have n

~ y t~>-< LYi ~l >-~ t~ i=1

on Rn x tjn. We can restate the fact that ~ Im(.) t~>-~ t~ on Rnx tjn by saying that there exists a positive constant c such that Im(.)~c In for every. in tjn; hence we can take Yo =c In. As a consequence, we see that every modular form f of degree n (and of any weight) is bounded on tjn. Consider an arbitrary sequence S in tjn; if we write a general term r of S as x+iy, x is bounded in Xn; in fact, we have IXijl~! for all i,j. As for y, there exists a non-negative integer n' such that Yi' for 1 ~ i' ~ n' are bounded in R while Yi" for n' < i" ~ n are unbounded. In general, if a sequence (Y)y in Xn has the property that, for any Yo>O, we have y~Yo for almost all y in the sequence, we shall write y --+ 00. Let A.(y) denote the smallest eigenvalue of y; then we have y --+ 00 in Xn if and only if A.(y)--+ 00 in R. We shall show that if y--+ 00 and s~O in Xn (depending on y) is such that tr(s y)~const., then s --+ O. We have y~A.(y) In for every y in Xn; hence tr(s y)~A.(y) tr(s). On the other hand, s~O implies that (Si)2 ~Si sj~tr(sf, hence ISijl~tr(s), for every i,j. Therefore, if y--+oo, tr(sy)~const., we get tr(s) --+ 0; hence s --+ O. If we write

*)

y=Im(r)= ( Y'* y"

with y' of degree n', hence y" of degree n" = n - n', for the above sequence S in tjn' then y' and * are bounded in Xn' and Mn',n,,(R) respectively. We shall show that y" --+ 00 in Xn" for a §uitable subsequence of S. Let y = tu a u

§ 5. Modular Forms

201

denote the Jacobi decomposition of y; and put a' 0 ) a= ( 0 a" ,

U' U= ( 0

*)

u" ,

in which a", u" are of degree n"; then, we have y,,;;;;tu" a" u". By Corollary 1 of Theorem 5, u" varies in a compact subset of SLn" (R) while a" -+ 00 for a suitable subsequence; hence we have y" -+00.

Theorem 6. Consider a sequence S of points r = x + i y in 6 n such that x is bounded in x" and y M -reduced; assume that x', y' in

are convergent in x"" ~n' and y" -+ 00 in Xn" for n = n' + n" ;let the limit of r' = x' + i y' in 6 n,. Then, for any Fourier series f(r)= we have

r~

denote

L a(s) e(tr(s r»),

s;;;;o

S'

0)

( lim f(r) = LaO 0 e(tr(s' r~»), reS

S/~O

in which the summation is taken over the set of s';;;; 0 in Xn , such that the matrix s composed of s', 0, 0, 0 is in L. Proof We shall prove the existence of a dominant series for the given Fourier series restricted to S; the main part is the existence of a positive constant c such that

le(tr(s r»)1 ~exp( - 2n c tr(s») for every s;;;;O in Xn and almost all r in S. Let C denote an arbitrary compact subset of 6 n , =Xn , + i ~n' and C its projection to the ~n,-space; then, there exists a positive constant Co such that tr(s' y');;;; Co tr(s') for every s';;;;O in Xn , and y' in C. In fact, we have only to take as Co the minimum of A, (y') for y' in C. We observe that r' varies in such a compact subset C. We shall examine tr(s y) for y=Im(r), r in S. For the sake of simplicity, we shall use i' and i" to denote indices from 1 to n' and from n' + 1 to n' +n" =n; also we put c=t co. We shall first show that tr(sy» c· tr(s) for every s;;;;O and for almost all r in S. If we put S'

*)

s= ( * s" ,

we have

tr(sy)=tr(s' y')+tr(s" y")+2 L i~

i"

Si'i"

Yi'i";

202

V. Graded Rings of Theta Constants

we may assume (for our purpose) that s belongs to the compact subset of ~n defmed by s~O, tr(s) = 1. If the statement to be proved is false, there exists a subsequence of S such that tr(s y)~c for all y=Im(r), -r in the subsequence, and s depending on y; we may assume that S itself is this subsequence. Since s ~ 0, we have (Sij)2 ~ Si Sj for every i, j; since y is M-reduced, we have IYijl~tYi for every i
2.h"l sI'i" Yi'i,,1 ~ .h,,(Si')"~ (SI,,)1- Y6 hence

I,l

I,'

tr(s" y")=tr(sy)-tr(s' y')-2 L Si'i" Yi'i" i',i"

We observe that the right side is bounded; hence tr(s" y") is bounded. Since y" -+ 00, by our previous remark, we get S" -+ O. Therefore, the right side converges to c-co= -tco; since tr(s" y")~O, we get -tco~O; but this is a contradiction. By our agreement, the Fourier series converges at every point of 6 n, hence at tic In. Therefore, there exists a positive constant M such that la(s)1 exp( -1tctr(s»)~M for every s~O in L. Consequently, by what we have shown, we will have

L a(s) e(tr(s-r»)~M . L exp( -1t c tr(s») .~o

s~O

on S provided that the right side is convergent; we shall show that this is indeed the case. By our other agreement, there exists a positive integer d such that dL is contained in Mn(Z). We observe that, for any non-negative integer t, the number of s~O in L satisfying d· tr(s)=t is at most equal to (t+ l)n-1(2t+ 1)1- n (n-1) ~(2t + I)N-1

for N =tn(n+ 1); hence

L exp( -1tctr(s»)~ L (2t+l)N-1 exp( -1tcld· t);
t=O

.~o

this is a power series in exp( -1t cld), and it is clearly convergent. Since the given Fourier series restricted to S has a dominant series, we have lim f(-r) = L a(s) lim e(tr(s -r»). teS

.~o

teS

On the other hand, in the expression for tr(s y), both tr(s' y') and L Si'i" Yi'i" are bounded; hence y"-+OO implies that lim tr(sy)=oo, i.e., i',i"

_

S

§ 5. Modular Forms

203

lime(tr(sr))=O, unless s"=O. If s"=O, we have s;,;,,=O for all il,i"; s

hence s is composed of e(tr(sl o~)). q.e.d.

Sl,

0, 0, 0; moreover, we have lim e(tr(s-r))= s

The above proof shows that if we have a set of sequences such as S in which 0 remains constant on each Sand y" is independent of S, then 1

0)

Sl

( limf(o)= LaO 0 e(tr(slo/)) reS

s'!i;O

is uniform in 0 provided that it is restricted to a compact subset C of 6 n, . On the other hand, for any given 0 1 in C, we can take 1

.=

(0

0)

1

iA In" '

0

in which A-+ 00, as our sequence S. We shall show that if f is a modular form of degree n and of weight k, SI

0)

f'(o/)=s'~oa ( 0 0 e(tr(slo/))

is a modular form of degree nl and of weight k. We have seen that I' is a holomorphic function on 6 n ,. If (11 is an arbitrary element of SP2n' (Z), for (1=(11$ 12n", we have

0)

(1.

1 (0 0 i Ai n "

=

0)

((11 . 01 . 0 i A In" .

Therefore, if (yl b /) is the bottom row of (11, we have 1'«(11. 0/)= lim f ((11 . 01 . 0 ) )._00 0 lA In"

= limf ( )._00

(1'

0))

( 01 0 iA In"

=det(yl 01+ b/)k 1'(0 /).

If n = 1, I' clearly satisfies the additional condition to be a modular form. In this way, we get the following corollary: l

Corollary. If f is a modular form ofdegree n and of weight k, for every 0 in 6

n _1>

1'(0 /) =limf (0 )._00

1

0

.0 \ lAl

1

V. Graded Rings of Theta Constants

204

is well defined; and f' is a modular form of degree n -1 and of weight k. Moreover, if f(7:)= L a(s) e(tr(s 7:») sil!;O

is the Fourier expansion of f,

f' (7:')= s'~oa (~

~) e(tr(s' 7:'»)

is the Fourier expansion of f'. We observe that the correspondence f --. f' is C-linear; after Siegel, we shall denote this mapping by ~=~n.k; the kernel of ~".k is called a cusp form of degree n and of weight k. If f is a cusp form, we have

f(7:)=

L a(s)e(tr(s7:»).

s>o

Otherwise, we have a(s)=I=O for some s~O not satisfying s>O. Then, by an observation made after Lemma 12, there exists ~ =1=0 in Q" satisfying ~ st~ =0; we may assume that ~ is in zn and its coefficients relatively prime. Let u denote an element of SLn(Z) with ~ as its n-th row; then we have S' a(s)=a(ustu)=a ( 0 0 .

0)

Since f is a cusp form, the right side is 0; but this is a contradiction. Lemma 20. If f is a cusp form of degree n, there exist positive constants M, c, in which C is independent of f, such that

1f(7:) 1~M . exp( -2n C tr(y»)

for every 7:=x+i y in Proof. If 7: = we have

X

~n'

+ i Y is in ~n' by Corollary 2 of Theorem 5 and Lemma 15, y;';c·

C' ". J

;';co' ·1,

for some positive constants c, c'. Suppose that 7: satisfies just this condition. Then, since a(s) =1= 0 implies that s > 0, hence Sj ~ 1 for 1 ~ i ~ n, we have la(s) e(tr(s 7:»)1 = la(s)1 exp( - 2n tr(s y»)

~la(s)1 exp ( -2ncc'tr(S)-2nCjt/j(yj-C'») ~la(s)1 exp( -2ncc'tr(s») exp( -2nc(tr(y)-nc')).

§ 5. Modular Forms

205

Since the Fourier series is absolutely convergent at icc' 1n, we have exp(2nncc') L la(s)1 exp( -2ncc'tr(s»)~M 8>0

for some M>O; hence Ifft) I~M· exp( -2nctr(y»). q.e.d. If u is an element of SP2n(R) with (y 15) as its bottom row, we have

Im(u· -r) = t(y -r +15)-1 Im(-r)(y:r+I5)-1. Therefore, if f is a function on 6 n satisfying f(u·-r)=det(y-r+l5tf(-r) for every u in a subgroup r of SP2n(R), the function cP on 6 n defined by cP(-r) = det(Im(-r»)"/2 I f(-r) I is r-invariant, i.e., cP(u· -r)= cP(-r) for every u in r.

Lemma 21. If f is a cusp form of degree n and of weight k, the function cP on 6 n defined by

cP (-r) = det (1m (-r»)kI2 If(-r) I attains its maximum at some point of ~n. Proof Since cP is r-invariant and 6 n=r· ~n for r=SP2n(Z), we may restrict -r=x+iy to ~n. Let M denote the supremum of cP(-r); we may have M=oo; choose a sequence S of points in ~n such that cP(-r)-+M on S; we shall exclude the trivial case where M =0. We shall show that det(y) is bounded on S. We recall that det(Y)>
on ~n' hence on S. If det(y) is not bounded on S, by passing to a subsequence, we may assume that det(y)-+oo, hence Yn-+oo, on S. By Lemma 20, we have

n(V}/2 exp( - 2n c Yj») n

cP(-r)-< on

~n;

j=1

we also have

lim A,k12 exp( -2n d)=O.

)._a:>

Therefore we get cP (-r) -+ 0 on S, hence M = 0; but this is not the case. We have thus shown that det(y)~c' on S for some c'>O; then S is contained in the subset ~n(c') of 6 n, which is compact by Lemma 16. Hence, if -ro is the limit of any convergent subsequence of S, we have M = cP(-ro) < 00. q.e.d. We shall prove the following remarkable theorem discovered by Siegel:

V. Graded Rings of Theta Constants

206

Theorem 7. Let f denote a modular form of degree n and of weight k, and f(7:) = L a(s) e(tr(s 7:») 0;:;0

its Fourier expansion; then the condition that a(s)=O for all s satisfying in which we can take kj4n as
tr(s)~Kn
Proof. We recall that Kn has been defined in Lemma 18 as the supremum oftr(y-l) for 7:=x+iy in tjn. We shall show that if f has the additional property of being a cusp form, we get f =0. By Lemma 21, the function 4> on 6 n dermed by 4>(7:) = det (1m (7:»)",2 1f(7:) 1 attains its maximum M at some point, say 7:0' of tjn. We introduce two complex numbers z, q related by

q=e(z),

z=I/2ni·logq;

we put 7: = 7:0 + z . In and consider the function F of q defined by

F(q) = f(7:) e( -A tr(7:») for some A in C. If flo is the smallest eigenvalue of 1m (7:0), f(7:) becomes a holomorphic function of q in O
la(s) e(tr(s 7:»)1 = la(s) e(tr(s 7:0»)1 exp(2n fl tr(s») ~ la(s) e(tr(s 7:0»)1 exp(2n fl([Kn a] + 1»). If we take the summation with respect to s, we get 1f(7:) 1~Ml . exp(2n fl([Kn a] + 1»),

in which Ml is a constant. This implies that

IF(q)1 ~M2' exp(2n fl([Kna] -nA.+ 1»), in which M2 is a constant. Therefore F is bounded in Iql ~ 1 if (*)

nA~[Kna]+l.

§ 5. Modular Forms

207

We shall assume that A. satisfies this additional condition; then F is holomorphic in Iql
!q!;:;;PI

We choose Zl satisfying e(zl)=q1 and put r1 =rO+z1 . 1n; we have 1m (Zl) = -"1. Since PI > 1, we have IF(q1) I~ 1F(1)1; this can be rewritten as Hence we get

If(r1)1~lf(ro)1

exp(2n A. n"l)'

M~det(Im(r1»)k/2If(r1)1

in which

~det(yo -"1 1n)k/2If(ro)1 exp(2n A. n "1)= M . (}("1),

We observe that () can be expanded into its Taylor series as

(}(,,) = 1+(2n A. n- k/2· tr(Yo1»)" + .... By the definition of K n , the coefficient of " is certainly positive if (**)

In this case, we have (}(,,) > 1 provided that ,,>0 is sufficiently small. We may assume that the above "1 >0 has been taken small enough to satisfy (}("1) > 1. Then M~M· (}("1), M~O imply M =0; hence 4>=0, f=O. We observe that if we take rx=k/4n and determine A. as nA.=[Kn rx]+1, the two conditions (*), (**) are both satisfied. What remains to be shown is that if f satisfies the condition of the theorem, it is necessarily a cusp form. Since this is clearly true for n = 1, by what we have shown the theorem is true for n= 1. We shall assume that n ~ 2 and apply an induction on n. Considerf' = tP f; by the corollary of Theorem 6, f' is a modular form of degree n -1 and of weight k with f'(r')= Id(s')e(tr(s'r'»), 8';1;0

S'

0)

a'(s')=a ( 0 0

as its Fourier expansion. Since a(s)=O for all s satisfying tr(s)~Knrx and Kn_1~Kn by Lemma 18, we get a'(s')=O for all s' satisfying tr(s')~Kn_1 rx. Therefore, by the induction assumption, we get f' =0; hence f is a cusp form. q.e.d.

208

v. Graded Rings of Theta Constants

Put r=SP2n(Z) and consider the mappingf-(a(s»)s, in whichfis in A(r)k and s runs over the set of half-integer matrices satisfying s~O, tr(s)~1Cnk/41t. This mapping is clearly C-linear; by Theorem 7, it is injective. Therefore dim(A(r)k) is at most equal to the number of all such s. In particular, we get dim(A(r)o)~1, hence A(r)o=C. In general, for any non-negative integer t, the number of half-integer matrices s satisfying s~O, tr(s)~t is at most equal to

(t+ 1f(4t+ 1)"tn (n-l) ~(4t+ 1)N for N =tn(n+ 1). In this way, we get the following corollary: Corollary 1. We have dim (A (SP2n (Z»k)-
for N=tn(n+1) as k-oo.

By putting this corollary and Chap. III, Proposition 1 together, we see that any N + 2 modular forms of degree n are algebraically dependent over C. We shall show that this theorem is valid for every modular group r contained in SP2n(R). We recall that r contains r,.(l) for some positive integer 1 as a subgroup of finite index; and A(r) is a graded subring of A(r,.(l)); also A (SP2n(Z») = A (r,.(1») is a graded subring of A(r,.(l)). We shall show that A(r,.(l») is integral over A(r,.(1»). Since r,.(l) is a normal subgroup of r,.(1), as we have seen in Chap. III, § 5, r,.(1) acts on A(r,.(l)) as a group of degree-preserving automorphisms; and the action of r,.(l) is trivial. In this way, the finite group r,.(1)/r,.(l) acts on A(r,.(l)); and A(r,.(1») is precisely the invariant subring of A (r,.(l)) under r,.(1)/r,.(l). In general, if we have a commutative ring A (with the identity) on which a finite group G is acting as a group of automorphisms, A is integral over the subring AG of G-invariant elements of A. The proof is quite simple: If a is an arbitrary element of A and t a letter, the product of t - (1 • a for all (1 in G is a monic polynomial in t with coefficients in A G ; and it has a among its roots. As a consequence, the field of quotients of A(r,.(l)) is algebraic over the field of quotients of A (r,. (1»). By Corollary 1 and Chap. III, Proposition 1 (or the remark after its proof), the transcendence degree of the second field over C is at most equal to N + 1. Therefore, the transcendence degree of the first field, hence also that of the field of quotients of A (r), over C is at most equal to N+1. We observe that A(F)o is integral over A(r,.(1»)o=C. Since A(r)o is an integral domain containing C, we get A(r)o=C. We have thus obtained the following corollary: Corollary 2. If r is a modular group contained in SP2n(R), any tn(n+ 1)+2 modular forms of arbitrary weights relative to r are algebraically dependent over A(r)o::;;=C.

§ 6. The Group of Characteristics

209

Let fo,ft, ... ,fN denote elements of A(r)k which do not vanish simultaneously at any point of 6 n ; then, they give rise to a continuous mapping of r\ 6 n to &(C). Let X denote the image and Y the smallest Zariski closed set in &(C) which contains X. Then Y is a projective variety, i.e., an irreducible projective variety; and Corollary 2 shows that dim (Y)i£tn(n+ 1). We shall later need the following more general case: fo, f1' ... , fN are holomorphic functions on 6 n and, for every (1 in r with y, ~ as its bottom row, we can choose the sign of det(yt+~)t so that we have fi«(1· -r)=det(y t +~)tkfi(t)

for Oi£ii£N. (We have to make the usual restriction in the case where n = 1). Then fo, ft, ... ,fN are integral over A (r), hence they are algebraic over the field of quotients of A (r). Therefore, the same result as above holds for such modular forms of "half-integer weights". Remark. We observe that, in the above application, Theorem 7 has been used only in a very weak form: It is enough to know that "a(s)=O for tr(s)i£f3k implies that f=O", in which f is in A(~(1))k and 13 is independent of k. If we use Theorem 7 in full, i.e., if we use the fact that we can take KJ4n as 13, we will get, e.g., dimA(Ii(1))ki£1 for ki£10 and dim A (12 (1))k i£1 for ki£8. We leave the verification as an exercise. Actually, the structure of the graded ring A(~(1)) is known for n= 1, 2; in particular, we have dim A (Ii (1))12 = dim A (12 (l))10 =2.

§ 6. The Group of Characteristics We shall give an axiomatic treatment of the classical theory of "theta characteristics". We shall prove only those properties which are closely related to an application in § 7. Let P denote a rmite dimensional vector space over F2 = Z/2Z, Ji.2 the cyclic subgroup of C x of order 2, and e(r, s) a non-degenerate bilinear mapping of P x P to 112; this means that e(r1 + r2, s)=e(rl> s) e(r2' s), e(r, S1 + s2)=e(r, S1) e(r, S2) and, for any fixed r =l= 0 or s =l= 0, the corresponding homomorphism of P to Ji.2 is not trivial, hence surjective. We shall assume that e(r, s) is alternating, i.e., e(r, r)= 1 for every r in P; this implies that e(r, s) e(s, r) = 1. We shall show that there exists a mapping c: P -+ Ji.2 satisfying c(r) c(s)=e(r, s) c(r+s)

for every r, s in P. Suppose that we have such a mapping. Then, for every r1 , ••• , r1J in P, we have c

ct/

j)

= ]]i e(rj, r)'n c(rj)

V. Graded Rings of Theta Constants

210

with the usual understanding (that the product over an empty set is 1). After this observation, which will be used only as a motivation, we take an Frbase r1 , .•. , rm of P, assign ± 1 to each ri , and define c(ri1 + ... +ri ) by using the above formula; then we get a mapping c of P to J1.2 with the required property. Since the verification is purely combinatorial, we leave it as an exercise. Let T denote the set of all mappings such as c. If oc is an element of T and t an element of P, oc + t defined by (oc + t)(r) = e(t, r) oc(r) is also an element of T. Since every character of P can be written as r --+ e(t, /'), every element of T is of the form oc + t. Therefore, if we put (oc+r)+s=oc+(r+s), (oc+r)+(oc+s)=r+s, the disjoint union, say X, of P and T forms a commutative group which is isomorphic to Fi+ 1 • We observe that the structure of X does not depend on oc. Moreover, by the orthogonality of characters of P, we get

(I oc(r))2 = I oc(r) oc(s)= Ie(r, s)oc(r+s) r. s

r

t

r, s

r

I oc(r) = I (oc + t)(r)=(I e(t, r)) oc(r) = 2m(r =0), 1%

O(r =1= 0) .

t

The first identity shows that m is even, say m = 2 n; moreover, if we put

e(oc) gives a mapping of Tto J1.2; and we have e(oc) e(f3) = 1/2 2n • I oc (r) f3(s) = 1/2 2n • I oc(r) oc(s) e(oc + f3, s) r. s

r. s

= 1/2 2n • I oc(t) e(oc + f3 + t, s) =oc(oc + f3) = f3(oc + f3). s. t

We call X the group of characteristics of degree n; the elements of Pare called period characteristics and those of T theta characteristics. An element oc of Tis called even or odd according as e(oc) = 1 or e(oc)= -1. Let T+ and T_ denote the sets of even and odd characteristics; then we have card(T+)+card(L)=2 2n. The sum of all e(oc) can be calculated in two ways: card(T+)-card(L)= 1/2n • Ioc(r); 1%,

r

by the second identity, this is 2n. In this way, we get

§ 6. The Group of Characteristics

211

We shall consider the symplectic group of P with respect to e(r, s) and denote it by S(X). An element (I of S(X) is, therefore, an automorphism of P satisfying e«(I' r, (I. s)=e(r, s) for every r, s in P.

Proposition 1. Every (I in S(X) can be extended uniquely to an automorphism of X keeping T± invariant; in this way, X becomes an S(X)module. Proof. We choose an element (j of T and write T as (j+P. We shall prove the uniqueness of the extension. Let (1* denote an automorphism of X satisfying (1* . r = (I. r and e«(I* . a) = e(a) for every r in P and a in T; put (1*' (j=(j+r«(I). Then, by using the identity e(a)e(p)={J(a+{J), we get e((I* . «(j + r») e«(j + r) = (j(r«(I») (j«(I. r) (j(r) e(r«(I), (I. r) for every r in P. Since this has to be 1, by replacing r by 0 and (I-I . r, we get (j(r«(I»)= 1 and e(r«(I), r)=(j«(I-I. r)(j(r) for every r in P. This defines r«(I) uniquely; hence (1* is unique. We shall prove the extendability of (I. We observe that r--+(j«(I-l· r)(j(r) defmes a character of P; hence, it can be written as e(r«(I), r) for some r«(I) in P. Put (I*'r=(I'r and (I*·«(j+r)=(j+r«(I)+(I·r for every r in P; then (1* is an extension of (I to an automorphism of X. Moreover, by the calculation as above, we get e«(I* . a) =(j (r«(I») e(a) for every a in T. If we take the summation over T, we get 2n=(j(r«(I»)2n; hence (j(r«(I») = 1. Finally, since the extension (1--+ (1* is unique, we have «(11(12)* = (It (I! for every (11) (12 in S(X). Therefore, if we put (I. x=(I*' x for every (I in S(X) and x in X, X becomes an S(X)-module. q.e.d. Let 19 denote the unique isomorphism of 112 to F2 and t an arbitrary element of P; for every r in P, we put

We can easily verify that ((It)2. r= r, (It' (r+s)= (It' r+ (It' S, e«(lt' r, (It' s)= e(r, s) for every r, s in P; hence (It is an element of S(X). We shall show that the extension of (It to X, according to Proposition 1, is given by

(It' a =a +(1 + 19(a(t») t. Choose an element (j of T satisfying (j(t) = 1. Then, rewriting (j«(lt' r) by using (j(r) (j(s)=e(r, s) (j(r+ s) and (j(t) = 1, we get (j «(It' r) (j (r) =

e(r, 19(e(t, r» t) = e(t, r);

V. Graded Rings of Theta Constants

212

hence, in the notation of the proof of Proposition 1, we have r(O"t) = t. Therefore we get

O"t' (<5 +r)=<5+ t+O"t' r=(<5+ r)+(1 +lg(e(t, r))) t. Since e(t, r) = (<5 + r)(t), if we put a=<5+r, this becomes the aboveformula. Lemma 22. Let r1, ... , rk and Sl' ... , Sk denote two sequences of elements of P. Then there exists an element 0" of S(X) mapping the r-sequence to the s-sequence, i. e., satisfying 0"' rj = Sj for 1 ~ i ~ k, if and only if, under the mapping 'i -+ Sj, linearly dependent subsequences correspond to each other and e(1j, r)=e(sj, s)for 1~i,j~k.

Proof We have only to prove the if-part. Since it is trivially true for k = 0, we shall assume that k ~ 1 and apply an induction on k. There exists an element 0" of S(X) satisfying 0"' rj = Sj for 1 ~ i < k; hence we may assume that 1j=Sj for 1~i
t = 0, we see, by assumption, that neither r nor s is contained in H; also we have e(t, h)= 1 for every hinH. Therefore,in the case where e(r, s)= -1, O"t keeps H elementwise invariant and permutes rand s. In the case where e(r, s)= 1, we proceed as follows: Consider the system oflinear equations 19(e(h, u))=O for every h in Hand 19(e(r, u)) = 19(e(s, u))= 1 in the unknown element u of the vector space P. Since r, s are not in H, this system is compatible; and it has a solution u. By usinge(r, s)= 1, we can easily verify that O"t+u O"u keeps H elementwise invariant and permutes rand s. q.e.d.

We shall prove a similar statement for Tinstead of P. We observe that arbitrary elements r, s of P can be written as a + p, a + y for a fixed a and some p, y in T; moreover, we have

e(r, s)= e(a) e(p) e(y) e(a + p +y). We shall denote this expression by e(a, p, y); clearly, e(a, p, y) depends symmetrically on a, p, y; and it is an S(X)-invariant, i.e.,

e(O" 'a, 0"' p, 0"' y)=e(a, p, y) for every

0"

in S(X).

Proposition 2. Let al> ... , ak+1 and Pl> ... , Pk+1 denote two sequences of elements ofT. Then there exists an element 0" ofS(X) mapping the a-sequence to the p-sequence if and only if, under the mapping aj -+ Pj, linearly dependent subsequences correspond to each other and the functions e(a), e(a, p, y) take same values at corresponding elements and triples.

Proof Again, we have only to prove the if-part. Put a1=a, aj+1 =a+rj and P1 = p, Pj+ 1 = P + Sj for 1 ~ i ~ k. Then the two sequences r1, ... , rk and Sl, ... , Sk satisfy the conditions in Lemma 22; hence there exists an element

§ 6. The Group of Characteristics

213

a of SeX) satisfying a ri = Si for 1;£ i;£ k. Put a· a = fJ + t; then we have a·ai+1 =fJi+1 +t for l;£i;£k. Moreover, since we have e(fJ)=e(a)= e(aOaj)=e(fJj+t), we get fJj(t) = 1, hence atofJj=fJj+t for 1;£j;£k+1. Therefore ata transforms the a-sequence to the fJ-sequence. q.e.d. 0

Corollary. SeX) is doubly transitive on T±. We shall defme "quotients" of the group of characteristics. For any non-empty subset N of P, we shall denote by N.l the set of elements s of P satisfying e(r, s)= 1 for every r in N. Clearly N i is a subspace of P; and, if N is a subspace of P, we have dim(N) + dim (N i) = dim (P) = 2 n; moreover, since N is contained in (Ni}i and dim(N)=dim(Ni}.l), we have N = (Ni}i. We say that a subspace N of P is totally isotropic if N is contained in N i ; in this case, we have dim(N};£n. By Lemma 22, any two totally isotropic subspaces of the same dimension are conjugate under the action of SeX). Let N denote a totally isotropic subspace of P of dimension k. We shall construct a group of characteristics of degree no = n - k as a subgroup of X/No Put Po =N.l/N; then Po is a subspace of PIN of dimension(2n-k}k=2noo If N,.=N +r, N,=N +s are elements of Po, e(r, s} depends only on N,., N,; we put e(N,., N,)=e(r, s). Then e(N,., N,} gives a bilinear alternating mapping of Po x Po to 112; since N = (N i )\ e(N" Ns} is non-degenerate. Therefore, we can embed Po into a group of characteristics of degree no, say Xo. We shall show that Xo can be considered as a subgroup of X/No We observe that every a in T gives rise to a character of N; and a, fJ give the same character ofN if and only if a + fJ is in N i. Therefore, elements of T give rise to 2k distinct characters, hence all characters, of N. In particular, the set of all a's which give the trivial character of N forms a coset in X / N i; let To denote the set of all co sets Na=N +a contained in this coset; then we have card (To) = 22no If Na is in To and Nr in Po, a(r} depends only on Na and Nr; we put Na(Nr}=a(r}. The 22no mappings of Po to 112 so obtained are distinct; and if c is one of them, we have c(Nr} c(Ns} = e(N" Nr ) c(Nr+ Ns}. Therefore, we can identify To with the set of theta characteristics in Xo. For an obvious reason, we call Xo the quotient of X by N. We shall change our notation and use g instead of n. A sequence of 2 g elements u~, u~, u~, ... , u~ of P is called a canonical base of P if o

0

••

,

e(ui, ui'} = e(ui', ui)= -1 for 1;£ i ~ g and e (other pair) = 1. The existence of such a base can be proved by modifying the proof of Chap.,II, Lemma 5: We take an

214

V. Graded Rings of Theta Constants

element u~ =1=0 of P; then, the homomorphism of P to J.l2 defmed by v -+ e(u~, v) is surjective. Let Ii; denote an element of P satisfying e(~, Ii{)= -1; consider the subgroup, say P: of P whose elements v satisfy e(u~, v) = e(u1', v)= 1. Then P decomposes into the direct sum of F2 u~ +F2 u1' and P'; and P' is isomorphic to FI g- 2; moreover, the restriction of e(r, s) to P' x P' is a non-degenerate bilinear alternating mapping of P' x P' to J.l2' Therefore, by induction, we get a canonical base U2, ... , u~, u~, ... , Ii; of P'; then u~, ... , u~, u1', ... , u~ form a canonical base ofP. We observe that every element of S(X) transforms a canonical base into a canonical base. In this way, S(X) acts on the set of canonical bases; the action is transitive by Lemma 22. We map P isomorphically to FIg with respect to a fixed canonical base of P. Since the multiplication by t gives an isomorphism of Z/2Z to t Z/Z, we can map P isomorphically to the subgroup
This implies, in particular, that X depends only on g. By expressing T as ~ + P for a fixed b in T, we can introduce coordinates also in T; the coordinate of b + r is the coordinate of r in the above sense. Lemma 23. A canonical base of P determines a unique element b of T+ such that if m mod 1 is the coordinate of an arbitrary element 0( of T with respect to b, we have e(0()=e(2m"m"). Proof. Let m mod 1 denote the coordinate of an element r of P and put c(r)=e(2m' 'm"). Then, for every r, s in P, we have c(r+ s)=e(r, s) c(r) c(s);

hence c is an element of T. Also we have Lc(r)=( r

L

(-1)liq)g=2g;

p,q=O,l

hence e(c)= 1; and e(c+r)=e(c) e(c+r)=c(r) for every r in P. Therefore, we can take c as b. Conversely, suppose that ~ has the required property. Then we have b(r)=e(~)e(b+r)=e(~+r)=c(r) for every r in P; hence ~=c. q.e.d. Let u' and utI denote colunm vectors with ui and ui' as their i-th entries for l~i~g. Then every element (1 of S(X) gives rise to an element of Sp2g(Z/2Z) composed of a, b, c, d as

§ 6. The Group of Characteristics

215

If we associate this element of SP2g (Zj2Z) to a, we get an isomorphism of S(X) to Sp2g(Zj2Z).

Proposition 3. Let b denote the unique element of T+ determined by the canonical base of p .. for every r in P and a in S(X), let m mod 1, m" mod 1 denote the coordinates of b + r, a· (b + r). Then we have m""=:m

e~rl

+H(ctd)o(a'b)o) mod 1.

Proof. By definition, we have r=2m (::,) ,

a'(b+r)=b+2m" (::,).

(::,)

As in the proof of Proposition 1, we put a' b =b +r(a); then we have 2m" (::,) =2m (;

~rl

+r(a).

Therefore we have only to show that the coordinate of r( a) is

H(c 'd)o (a 'b)o) mod 1. We recall that r(a) is characterized by e(r(a),r)=b(a-l·r)b(r); we can rewrite the right side as e(J +a- l . r) e(J +r). If (m' m") mod 1 is the coordinate of r, (m' a + mil c m' b + mil d) is the coordinate of a- l . r. Therefore e(r(a), r) is equal to

e(2(m' a+m" c)l(m' b+m" d)+2m'lm")=e(c td)o 1m" -(a'b)o 1m'); hence H(c 'd)o (a 'b)o) mod 1 is the coordinate of r(a).

q.e.d.

Finally, we shall show that, for every 1~2, .fg(l)j.fg(1) is isomorphic to Sp2g(ZjIZ). We need the following lemma:

Lemma 24. Let I denote a power of a prime number and x, y elements ofzn satisfying xly=: 1 mod lfor some n~2. Then there exist x*, y* in zn satisfying x* I y* = 1, x* =: x, y* =: y mod I. Proof. Put x=(x l ... x n), y=(Yl ... Yn), xly= 1 +kl. Since I is a power of a prime number, for some i, Xi and I are relatively prime; by changing indices, we may assume that i = 1. Let Pb ... ,PI denote distinct prime factors of Xl; then no Pi divides I. Hence we can find ~i in Z satisfying X2 + 1~i =: 1 mod Pi; choose an element ~ of Z satisfying ~ =: ~i mod Pi for 1 ~ i ~ t. Then Xl and X2 + 1~ are relatively prime; hence we can find 1], , in Z satisfying Xl1] +(X2 +I~)' = - Y2 ~ -k. Put x!=x2+1~,

yf=YI+I1],

Y!=Y2+ 1C

xf = Xi for i =!= 2, yf = Yi for i =!= 1, 2, and x* = (xf ... x:), y* = (yf ... y:); then we have x* I y* = 1. q.e.d.

216

v. Graded Rings of Theta Constants

Lemma 25. The homomorphism of SP2g(Z) to Sp2g(Z/IZ) defined by CT-+CTmodl is surjective, hence I'g(1)/I'g(l) is isomorphic to Sp2g(Z/IZ),Jor every 1~2. Proof. If a ring A is a direct sum of two rings A' and A", Sp2g(A) decomposes into the direct product of Sp2g(A') and SP2g(A"). Since Z/I Z is the direct sum of Z/q1 Z, Z/Q2 Z, ... , in which Ql> Q2, ... are powers of distinct prime factors of I, we have only to consider the case where I itself is a power of a prime number. We represent a given element of Sp2g(Z/IZ) as CT mod I for some CT in M 2g (Z). Let =(e' e"), 11 =(rl' 1'(') denote the first and the (g + 1)-th row vectors of CT; then we have e' 111" e" 111' == 1 mod I. By Lemma 24, we may assume that f 111" - e" 111' = 1. As the proof of Chap. II, Lemma 5 shows, there exists an element CT1 of SP2g(Z) with e,11 as its first and the (g+ l)-th row vectors. Then CT 0"1-1 mod I is the (B-sum of 12 modI and an element of SP2g_2(Z/IZ). Therefore we have only to use an induction on g to complete the proof. q.e.d.

e

§ 7. Modular Varieties We recall that our objective has been to examine the image of Gz(e, 2e)\ 6 g under the injective holomorphic mapping 9 defined in Theorem 4. We shall formulate the key steps in three lemmas; the first of these is as follows: Lemma 26. Let U denote a connected complex manifold and Z a subset of U different from U such that,Jor every point a of U, there exists an open neighborhood V of a and elements!t, f2' ... of l!J(V) with Z n Vas the set of common zeros. Then U -Z is a connected dense open subset of u. Proof. We recall that l!J(V) is the ring of hoi omorphic functions on V; also the assumption, stated more precisely, is that a point z of V is contained in Z n V if and only if !t (z) = f2 (z) = ... = o. It follows from this that Z is a closed subset of U. We shall show that U -Z is dense in U. Let ZO denote the set of interior points of Z. Since ZO is the complement of the closure of U -Z, U -Z is dense in U if and only if ZO is empty. Suppose that ZO is not empty; let a denote a point of its closure and choose Vas in the lemma; we may assume that V is connected. Then we have f1=f2=···=0 on ZOnVH~, hence f1=f2=···=0 on V. This implies that a is in ZO; hence ZO is a closed subset of U. Since U is connected by assumption, we get ZO = U; but this is not the case. We shall show that U -Z is connected. If U -Z is not connected, it is the disjoint union of two non-empty open subsets, say U1 and U2 ; let ~ and il2 denote their closures. Since U - Z is dense in U, we have U = ~ U il2. Since U is connected, ill n il2 contains at least one point,

§ 7. Modular Varieties

217

say a; and every open neighborhood V of a intersects V1 and V2 • This shows that V - Z n V is not connected. In other words, if V - Z is not connected, we can realize the same situation locally. Let V denote a convex open subset ofC', Z a subset of V different from V, andfi,f2' ... elements of (!)(U) with Z as the set of common zeros. We have only to show that V - Z is connected. Let a, b denote arbitrary points of V - Z; then we have/;(a)/;(b)=1=0 for some i,j. If we have/; (b) =1=0 or/; (a) =1=0, we put f = /; or f = /;; otherwise, we put f = /; +/;. Let Vf denote the open subset of V defined by f(z) =1=0; then Vf contains a, b and is contained in V - Z. Consider the affine linear mapping of C to C' defined by t ~ a + (b -a) t; then the inverse image of V is a convex open subset, say T, of C. Define a functionf* on Tbyf*(t)=f(a+(b-a)t); thenf* is an element of (!)(T) satisfyingf*(O)f*(l)=I=O. Let Tf denote the open subset of T defined by f*(t) =1=0; then T-Tf is a discrete subset of T, hence Tf is connected. Therefore the image is also connected; and it is contained in Vf , hence in V - Z. Since Tf contains 0 and 1, the image of Tf contains a and b; hence V -Z is connected. q.e.d. We shall apply this lemma in the special case where V is the set of simple points of a projective variety Y in PN(C) and Z the intersection with V of a Zariski closed set in &(C) which has a dimension smaller than that of Y. The second lemma is concerned with "Gapel's systems" of theta characteristics. We consider (t Z/Z)2 g equipped with e(m, n)=e(2(m' t n" -nltm"»)

as a group of period characteristics of degree g and, at the same time, as the set of coordinates of theta characteristics with respect to -'(m)= e(2mlt m"). We shall write m instead of m mod 1; also we shall identify a theta characteristic with its coordinate. A set L of 2g even characteristics such that all their triple sums are even is called a Gopel system of degree g; e. g., the set of (m' 0) for all m' in
hence a +L spans a totally isotropic subspace. Since a +L already contains 2g elements by assumption, it is a maximal totally isotropic subspace; in particular, we have L + L + L= L' Conversely, if N is a maximal totally isotropic subspace, the quotient by N is a group of characteristics of degree 0; hence there exists a theta characteristic a unique modN such that L=a+N forms a set of even characteristics; then L is clearly a Gapel system. As we have seen in § 5, Sp2g(Z/2Z) acts transitively on the set of all maximal totally isotropic subspaces;

218

V. Graded Rings of Theta Constants

therefore Sp2g(Z/2Z) acts transitively on the set of all Gopel systems of degree g. Lemma 27. For every

m=(m~ ... m~m~ ... m;),

put

let 4> denote the "operator" under which m is replaced by mo or simply deleted according as m~=O or m~=t. Then, if we disregard the multiplicities of elements in the "image" under 4>, it "maps" a Gopel system to a Gopel system. Proof. Let L denote a Gopel system of degree g; define Li as the subset of L consisting of those m in which m~ = i/2 for i = 0, 1. If Li is not empty, we have Li+ Li+ Li= Li' hence (ai+ L;)+(ai+ Li)= Li+ Li= ai+ Li for any ai in Li' This implies that Li=a+N;, in which N; is a subgroup of (tZ/Z)2 g • Since N; is contained in the (maximal) totally isotropic subspace ai+ L, it is totally isotropic; hence card (Ii) is a power of 2. For a similar reason, if Li is not empty, the set of distinct mo for m in Li is a coset by a totally isotropic subspace of (tz/zf g - 2. Therefore, if we denote this set by Lt, card(Lt) is a power of 2, which is at most equal to 2g - 1 • Moreover, if Lo is not empty, we have L~=4>Lo=4>L; and this is a set of even characteristics of degree g - 1. We observe that if mo is in Lt, it determines m in Ll uniquely. In fact, since m is even, we have m;=O or m;=t according as mo is even or odd. Hence card(Ll)= card (Lt);;£2 g - 1 • This implies that Lo is never empty. Moreover, the inverse image in LO of every element of 4> LO consists of at most two elements. Therefore, if Ll is empty, 4>Lo is a Gopel system (of degree g-l). IfLl is not empty, since 2g =2g - 1 + 2g - 1 is the only way to express 2g as a sum of two powers of 2, we have card (II) = card (Lo) = 2g - 1 ; and 4> is injective on LO' Otherwise, there exist two distinct elements a, b of LO satisfying 4>a=4>b; this implies that a+b=(O ... Ot). Then a+b+ Ll is a subset ofL consisting of odd characteristics; but this is not the case. Therefore 4>Lo is a Gopel system. q.e.d.

Remark. We can formulate and prove Lemma 27 in the language of § 6: Let X denote a group of characteristics of degree g and P the subgroup of period characteristics. Let N denote a maximal totally isotropic subspace of P and 0( the unique theta characteristic mod N such that N + 0( consists of even characteristics. On the other hand, let r =1= 0 denote an element of P. Then the statement is that N +0( is "mapped" to a similar set by taking the quotient of X by {O, r}. In the proof, we have to separate two cases according as N does or does not contain r; these cases correspond to Ll = ~ and LI =I=~.

§ 7. Modular Varieties

219

Lemma 28. Let S denote a sequence of points. = x + i y in 6 g such that x is bounded in;Eg and y Minkowski reduced; assume that x', y' in x= (x*'

**),

*)

y= (Y' * y"

are convergent in ;Eg" ~g' and y" ....+(1) in ;Eg" for g = g' + gil; let .~ denote the limit of .' = x' + i y' in 6 g ,; and finally, for every m = (m' mil) in Q2 g, let mo = (m~ m~) denote the element of Q2 g' obtained from m by deleting mi, mi' for g' < i ~ g. Then we have

if mi==O mod 1 for g'
L

.;;;0

in which s=!t(p+m')(p+m'), p in zg, and a(s)=e(p+m') Im") or e(p+m') tm") +e( -(p+m') Im") depending on the case; and we have s

= (S'

0)

0 0

with s' in ;Eg' if and only if mi == 0 mod 1, Pi = - mi for g' < i ~ g. Therefore we get or 0 according as mi == 0 mod 1 for g' < i ~ g or otherwise. We shall prove the second part. By what we have shown and by Lemma 27, it is enough to show that, for any .~ in 6 g" there exists an element mo mod 1 of a given Gapel system of degree g' such that Omo (.~)::j:: O. We may assume that g' =g; we shall use instead of .~. We have observed that SP2 g(Zj2 Z) acts transitively on the set of all Gapel systems of degree g. Also, by Lemma 25, every element of Sp2g(Zj2Z) can be written as (J mod 2 for some (J in SP2 g(Z). Therefore there exists an element (J of SP2g(Z) which maps the Gapel system of (m' 0) mod 1 for all m' in ! zg to the given Gapel system; write. * = (J ••• By the corollary of Theorem 2 (and by Chap. I, § 10, (0.2)), if we put (J. (m' O)==n mod 1, we have

.*

220

V. Graded Rings of Theta Constants

t=l=O depends on (1,., and m'. On the other hand, by Chap. IV, Lemma 11, we have Om' 0(.)=1=0 for at least one m'; hence On('*) =1=0 for at least one n mod 1 in the given Gapel system. q. e. d.

Theorem 8. Let X denote th! image of Gz(e, 2e)\ 6, under the ini.!!ctive holomorphic mapping 8 and X the closure of X in &(C); then X is a projective variety of dimension t g(g+ 1); moreover X -X is contained in a Zariski closed set of dimension at most t g(g-1). Proof. Let Y denote the smallest Zariski closed set in &(C) which contains X; then Y contains X. Moreover, by Corollary 2 of Theorem 7, Y is a projective variety of dimension at most equal to t g(g+ 1). Let y. denote the set of simple points of Y; then, in the same way as in Chap. III, § 7, we see that X n y. is a non-empty open subset of y.; also we get g (g + 1). For the sake of completeness, we shall recall that dim (Y) argument: Since Yis the smallest Zariski closed set containing X, we get X n Y. =I=~. Let x denote an arbitrary point of X n y. and take an open neighborhood Vof x in &(C) satisfying Y n V= y'n V; then we have X n V c y'n v. We recall that y. is a submanifold of &(C) of complex dimension at most equal to t g(g + 1). On the other hand, by Theorem 4, there exists a submanifold of &(C) of complex dimension t g(g + 1) which contains x and is contained in X n V. Therefore we have dim(Y)= t g(g+ 1); also, if we make V smaller, we get X n V= y'n V; hence X n y. is open in y.. We shall examine the points of Bd(X)=X -X. We recall that

=t

(1 0)-1 Gz (e,2e) (10' 0)e

r= 0' e

is a subgroup of F,(1) containing F,(2e;); and we have Gz(e, 2e)\ 6 g =r\6,. Let ~ denote an arbitrary point of Bd(X); then there exists a sequence (x(v»)v in X which converges to ~. We choose a sequence (.*(V»)v in 6 g so that 8 maps r·.*(v) to xlv) for every v. We choose a sequence «(1(v»)v in F,(1) such that «(1(v»)-l .• *(v)=.(v) is in ij, for every v. As we have shown before Theorem 6, by passing to a subsequence if necessary, we may assume that the sequence (.(V»)v has properties similar to the sequence S in Lemma 28. We recall that .*(v) can be replaced by any other point of r·. * (V); hence q
§ 7. Modular Varieties

221

of '&(C), which is x(v), for every v. By Lemma 28, the limit f,,(-ro) = lim 8n (-r(v)) v_ao

exists for every n;f,,(ro) = 8no (-ro) if n; == 0 mod 1 for g' < i ~ g and 0 otherwise. Since e == 0 mod 2, in fact e == 0 mod 4, the set of n mod 1 contains a Gapel system of degree g; hence, by Lemma 28, f,,(-ro)=1=0 for at least one n. Therefore (e (q,n(CT)f" (-ro))n represents a point of '&(C); this is nothing else than ~. We have not really used the assumption that ~ is in Bd(X)=X -x. We shall show that, because ofthis assumption, we have g'
222

V. Graded Rings of Theta Constants

of the proof of Theorem 8 shows that Yis a projective variety of dimension

t g(g+ 1) and that X n Y, is a non-empty open subset of Y,. Let (T(V»)v

denote a sequence in 6 g which has properties similar to the sequence" S" in Lemma 28; for every v let T'(v) denote the g'-th principal monor of T(v) and TO the limit of T'(v) in 6 g, as v -+ 00; finally, let u denote an arbitrary element of I;(1). Since r is a modular group, it contains I;(l) for some I (as a subgroup of finite index); then A(r)k is contained in A(I;(l))k; and we have u- 1. A (r)k C u- 1. A (I; (l))k = A (I;(l))k' Consequently, in view of Theorem 6, for every i we get lim (u- 1 • Ii) (T(V») = Ii" (TO)'

v_

00

in which Ii" = CPg" (u- 1.Ii) is an element of A(I;,(l))k depending only on (rnI;(l)) u and k If, jor every u and S, there exists at least one i for which K(ToH=0, then X -X is contained in a Zariski closed set in 11(c) ofdimension at most g (g - 1). For this we refer to the proof of Theorem 8. Therefore, by its last part we get X =X. In other words, under the above assumption, "Theorem 8 holds" for the injective, locally biholomorphic mapping given by fo, fl' ... , fN' We shall prove a ring-theoretic version of Theorem 8. As before, let r denote the modular group defined by

t

r=

(1 0) (og1 0)-1 e Gz (e,2e) og e

for e=Omod 2; then we have r\ 6 g = Gz(e, 2e)\ 6 g • We define A( Gz(e, 2 e)) as A(r); an element of A(Gz(e, 2e))k is a holomorphic function f on 6 g satisfying f(u . T) =det(e- 1 (y T + (je))k f(T) for every u in Gz (e,2e) with (y (j) as its bottom row (plus a certain restriction in the case where g= 1). We have seen (after the proof of Lemma 8) that for every m', n' in zg e- 1/Z g , the product Om'oOn'O is an element of A(Gz(e, 2e))1 if e=O mod 4. The following theorem shows that such products generate nearly the entire ring A(Gz(e, 2e)) over C: Theorem 9. If e=O mod 4, the graded ring A(Gz(e, 2e)) is thenormalization of the ring C[Om'O On'oJm', n' which is generated over C by the products Om'O On'O for all m', n' in zg e- 1/Z g ; in particular, A(Gz(e, 2 e)) is finitely generated over C.

Proof Put R=C[Om'O On'O]'n',n' and S=A(Gz(e, 2e))=A(r); then R is a graded subring of S. We shall first show that, in the notation of Chap. III, § 1, we have F(R)=F(S). Let f denote an arbitrary homo-

§ 7. Modular Varieties

223

geneous element of S, say of degree k; we have only to show that f is in F(R). Consider all monomials of degree 2k in Om' 0 for m' in zg e-1/Z' and denote them by fo, f1> ... ,fNk; then, they give rise to a holomorphic mapping 9(k) of r\6 g to &)C); also fo, ... ,fNr fgive rise to a holomorphic mapping 9* of r\ 6, to &k+ 1(C); let X( ), X* denote the images of r\ 6, under 9(k),9* respectively. Then, by the remark after the proof of Theorem 8, the closures y(k), y* of X(k), X* are projective varieties of dimension !g(g+l) such that y(k)_X(k), Y*-X* are contained in Zariski closed sets of dimensions at most! g (g - 1). In fact, the verification of the conditions is straightforward. The rest of the proof is quite similar to the proof of Chap. III, Theorem 5: Let K denote a countable subfield of C which contains the coefficients of finite ideal bases for y(k), y* and for the Zariski closed subsets containing y(k) - X(k), y* - X*; let YJk), Yo* denote the Zariski open subsets of y(k), y* defined by fo9=O. Then YJk), Yo* are affine varieties defined over K. Let Y*=(Y1 ... YNk+ 1) denote a generic point of Yo* over K; then Y=(Yl ... YNk) is a generic point of YJ") over K. We shall show that K (y*) = K (y). Consider the conjugate field of K (y*) in Cover K (y) and denote the image of y* by y**; then y** is also a generic point of Yo* over K. Hence y*, y** are contained in X*. Let T*, T** denote points of 6, which are mapped to y*, y**. Since T*, T** have the same image y in X(k), they are r-equivalent; hence y*=y**. This implies that K(y*) = K(y). Put R*=R[f]; then, by what we have shown, we get Fo(R)=Fo(R*). Since R l 9=O, this implies that F(R)=F(R*); hence f is in F(R). We shall show that S is integral over R. As we have said, the proof is similar to that of Chap. III, Theorem 5. Let f denote an element of Sk; we have only to show that f is integral over R. By changing the notation, letfo,h, ... ,fN denote Om' 0 for all m' in Z' e- 1/Z g• We have only to show that f If?k is integral over R; = C [fol fi, ... ,jNIfi] for every i. Let l'i denote the Zariski open subset of Y defined by fi9=O; take any point of l'i and a sequence (x(Y»y in X which converges to We have only to show that (flfiz")(x(v)(l~v
e

e.

Y

r· T#(Y) to x(v) for every v; by passing to a subsequence, we may assume

that we have the same situation as in the proof of Theorem 8, e.g., T#(Y) = U • T(Y) for every v, but including the case where g' = g. Then we have (flfiZk) (x(Y»= f(T#(V»/fi(T#(V»Zk =(u- 1 ·f)(T(V»/(u- 1 ·fiz") (T(Y»

224

v. Graded Rings of Theta Constants

for every v; and, as above, lim (0-- 1 ·f)(r(v») = f"(r v

o), lim (o--l ..t;2k)(r(V»)= v

U;2k)" (ro). We know that (f/k)" (ro)=l=O for some j. Since ~ is in 1';, therefore, we get (.t;2k)" (ro) =1= 0; hence lim (f/.t;2k)(X(v»)=I= 00. We thus have a v contradiction. q.e.d.

Since S=A(G z (e,2e)) is finitely generated over C, by Chap.III, Lemma 3, we have S(d)=C[Sd] for some d. Since Sd contains R d , by the remark after the proof of Theorem 8, any C-base of Sd gives rise to an injective, locally biholomorphic mapping of Gz(e, 2e)\ 6 g to PM(C)' (M =dim(Sd)-1) such that "Theorem 8 holds" for the image. The closure of the image is called the standard compacti,[ication of Gz(e, 2e)\ 6 g ; it differs very little from the projective variety X in Theorem 8; this may be loosely called a modular variety. It is an extremely interesting variety which deserves closer examination. In the simplest case where g = 1 and e = 4, the results in Chap. III, § 6 imply that the modular variety is a plane curve defined by X4 + y4 = Z4; it is a smooth curve of genus 3.

Remark. It is well known that the standard compactification of Gz(e, 2e)\ 6 g contains its image as a Zariski open subset. If we use a general theorem of Zariski to the effect that normal varieties are everywhere" analytically irreducible," the above statement becomes an exercise of what we have shown. On the other hand, it is a rather easy corollary of Theorem 9 that for any modular group r the graded ring A(r) is finitely generated over C. In fact, we have only to recall a simple lemma which states that if A is a noetherian ring and G a finite group, the ring of G-invariants in A is also noetherian provided that card (G) ·1 is a unit of A. In the case where Gz(e, 2e)=Fg(r Z, 21'2), i.e., if e is the scalar matrix r21g, for some even positive integer 1', in view of the relation between 8m (r, I' z) and 8[m' 1'-10](1'2 T, 1'2 z), Theorem 8 and Theorem 9 will give results on the theta constants 8m (r) for I' m=O mod 1. We shall state this special case of Theorem 9 as its corollary: Corollary. Ifr is an even positive integer, the graded ring A(I;(r2, 2r2)) is the normalization of the ring C[Om On]m,n which is generated over C by the products 8m On for m, n running over a complete set of representatives ofZ 2g r- 1 /Z 2g• We observe that the action of I; (2) on C[8m On]m,n is known by the corollary of Theorem 2 and Theorem 3. Therefore, by the process of taking invariant subrings, we can derive similar results from the above corollary.

Sources (These are the references we actually used in writing this book.) 1. Baily, W. L.: Satake's compactification of v". Amer. J. Math. SO, 348-364 (1958). 2. Bargmann, V.: On a Hilbert space of analytic functions and an associated integral transform. I. Comm. Pure AppL Math. 14, 187-214 (1961). 3. Cartan, H. Seminaire: Fonctions automorphes. Paris: E. N. S. Secretariat math. 1957-58. 4. Cartier, P.: Quantum mechanical commutation relations and theta functions. Proc. Symp. pure Math. IX. Amer. Math. Soc. 361-383 (1966). 5. Igusa, J.: On the graded ring of theta-constants. Amer. J. Math. 86, 219-246 (1964); II, 88, 221-236 (1966). 6. - Modular forms and projective invariants. Amer. J. Math. 89, 817-855 (1967). 7. - Harmonic analysis and theta-functions. Acta Math. 120, 187-222 (1968). 8. Krazer, A.: Lehrbuch der Thetafunktionen. Leipzig: B. G. Teubner 1903. 9. - Wirtinger, W.: Abelsche Funktionen und allgemeine Thetafunktionen. EnzykL Math. Wiss. II B7, 604-873. 10. Maass, H.: Ober die Darstellung der Modulformen n-ten Grades durch Poincaresche Reihen. Math. Ann. 123, 125-151 (1951). 11. Mackey, G. W.: A theorem of Stone and von Neumann. Duke Math. J. 16, 313-326 (1949). 12. Mumford, D.: On the equations defining abelian varieties. I-III. Invent. Math. 1, 287-354 (1966); 3, 75-135, 215-244 (1967). 13. Pontrjagin, L.S.: Topologische Gruppen. I, II. Leipzig: B.G. Teubner 1957-58. 14. Pyatetski-Shapiro, 1.1.: Geometry of the classical domains and the theory of automorphic functions [in Russian]. Moscow: Fizmatgiz 1961. 15. Segal, I. E.: Transforms for operators and symplectic automorphisms over a locally compact abelian group. Math. Scand. 13, 31-43 (1963). 16. Siegel, C. L.: Gesammelte Abhandlungen. I-III. Berlin-Heidelberg-New York: Springer 1966; especially: Einftihrung in die Theorie der Modulfunctionen n-ten Grades, II, 97-137 (1939); Einheiten quadratischer Formen, II, 138-168 (1940); Moduln Abelscher Funktionen, III, 373-435 (1964). 17. - Vorlesungen tiber ausgewiihlte Kapitel der Funktionentheorie (Lect. Notes), Gottingen 1966. 18. v.d. Waerden, B.L.: Einftihrung in die algebraische Geometrie, New York: Dover 1945. 19. Weil, A.: L'integration dans les groupes topologiques et ses applications. Paris: Hermann 1953. 20. - Introduction a l'etude des varietes kiihleriennes. Paris: Hermann 1958. 21. - Sur certains groupes d'operateurs unitaires. Acta Math. 111, 143-211 (1964). 22. Zariski, 0.: Some results in the arithmetic theory of algebraic varieties. Amer. J. Math. 61, 249-294 (1939). 23. - Pencils on an algebraic variety and a new proof of a theorem of Bertini. Trans. Amer. Math. Soc. 50, 48-70 (1941).

Further References and Comments We shall start by providing further references to the Foreword. Mumford called to our attention to a paper by W. L. Baily entitled, "On the moduli of Abelian varieties with multiplications," J. Math. Soc. Japan 15, 367-386 (1963). On pp. 382-383, Baily gave (in a general case) a sketchy proof to the fact that a certain set of theta constants gives rise to a holomorphic mapping of the standard compactification to a projective variety with finite sets as fibers. Although this is only a part of the "fundamental lemma," it should have been mentioned in the Foreword. Tate never published his theory of p-adic theta functions; the reader can find a part of it in P. Roquette, " Analytic theory of elliptic functions over local fields," Hamburger Math. Einzelschriften (N.F.), 1 (1970). The theory by H. Morikawa appeared in the following two papers: Theta functions and Abelian varieties over valuation fields of rank one I, Nagoya Math. J. 20, 1-27 (1962); II, ibid. 21, 231-250 (1962). Also the following dissertation by J. P. McCabe can be mentioned: P-adic theta functions, Thesis at Harvard (1968). The work by A. Andreotti and A. L. Mayer appeared as the joint paper entitled, "On period relations for abelian integrals on algebraic curves," Ann. Scuola Norm. Sup. Pisa 21, 189-238 (1967). The recent results of H. E. Rauch and H. M. Farkas are contained in their joint paper, "Period relations of Schottky type on Riemann surfaces," Ann. Math. 92, 434--461 (1970); the reader can find further results of Rauch and Farkas in the references at the end of that paper. Also the following dissertation by J. D. Fay has to be mentioned: Special moduli and theta relations, Thesis at Harvard (1970). Chapter I. We inserted § 5 as a sort of digression to have a better perspective on the whole situation; for a more systematic treatment, the reader is refered to Cartier [4]. About Mackey's theorem, the reader will find a beautifully written introduction to his theory of induced representations in G. W. Mackey, "Induced representations of locally compact groups and applications," Functional Analysis and Related Fields, Springer-Verlag, 132-166 (1970). In § 14 of that paper, Mackey gave a group representational proof to the Frobenius theorem on the dimension of the vector space oftheta functions (cf. Chap. II, Theorem 4); compared to Cartier's proof, his proof is elementary and readily accessible to the

228

Further References and Comments

reader of this book. Also the following paper by L Satake should be mentioned: Fock representations and theta functions, Ann. Math. Studies 66, Princeton Univ. Press, 393-405 (1971). We discussed the conjugacy of maximal compact subgroups of some groups (cf. Theorems 5,8). Actually, the conjugacy property holds for any connected Lie group; for this we refer the reader to K. Iwasawa, "On some types of topological groups," Ann. Math. 50, 507-558 (1949). The bicontinuous isomorphism B(X)jA(X) ~ Sp (X) (Theorem 6) is basic in the theory of theta functions; it is worth while to know that, although the proof in this book breaks down, the above isomorphism remains valid for any locally compact commutative group X (cf. 7, p. 203). Chapter II. Large parts of this and the next chapters were taken from Weil [20]. The proof of the main existence theorem (Theorem 2) was applied to a more general case by Kodaira in G. de Rham and K. Kodaira, "Harmonic integrals," Lect. Notes, LA.S. (1950); see also P. Dolbeault, "Fonctions theta associees a un diviseur donne," Seminaire H. Cartan (1951). We did not say anything about the divisor group of a complex manifold. For this we refer the reader to a memorable paper by K. Kodaira and D. C. Spencer entitled, "Groups of complex line bundles over compact Kahler varieties," Proc. Nat. Acad. Sci. 39, 868-872 (1953). The proof of Theorem 6, especially the use of the scalar product in the vector space of theta functions, was taken from Siegel (3rd in 16). Chapter I II. In writing certain parts of this and the next chapters, we found our self-imposed restriction to be elementary almost untenable. It is highly recommended that the reader familiarizes himself with the material in J.-P. Serre, "Geometrie algebrique et geometrie analytique," Ann. Institut Fourier 6, 1-42 (1955-56), and substitutes elementary (but artificial) proofs in the book by natural proofs; in doing so, he will find out that certain results can be improved. In this connection, the reader is refered to Siegel's proof for the fact that the field of meromorphic functions on any connected compact complex manifold M is algebraic if the transcendence degree of the field over C is equal to dimdM); the proof is accessible to the reader of this book, and it is in Siegel [16], III, 216-222 under the title, "Meromorphe Funktionen auf kompakten analytischen Mannigfaltigkeiten". Chapter I V. The main theorem in this chapter states that an abelian variety embedded in a projective space is the intersection of quadrics. This was proved under the following assumption: If D is any "hyperplane section" of the abelian variety, the imaginary part A of the Riemann form of D, which does not depend on the choice of D, satisfies A = 0 mod 4, i.e., the smallest elementary divisor e 1 of A satisfies e 1 =0 mod 4. The

Further References and Comments

229

quadrics can be so chosen that their coefficients are quadratic polynomials in the theta constants with coefficients in Z. The proof was taken from Mumford [12]; actually, Mumford proved this important theorem under the sole restriction that the characteristic is different from 2. He later succeeded in making the following improvement: The above theorem holds if el ~4 instead of el =0 mod 4; there is no restriction on the characteristic; and, at every point of the abelian variety, the tangent space of the abelian variety is the intersection of the tangent spaces of the quadrics. For these we refer the reader to "Theorem 10" in his paper entitled, "Varieties defmed by quadratic equations," Questioni sulle varieta algebraiche, Corsi dal CL.M.E., Edizioni Cremonese, Roma (1969). It seems appropriate to mention here the following two remarkable papers: I. Barsotti, "Considerationi sulle funzioni theta," Instituto Nazionale di Alta Mathematica Symposia Mathematica 3 (1970); H. Morikawa, "On some results on theta constants I," Nagoya Math. J.37, 145-181 (1970). Chapter V. As we mentioned in the Foreword, we used Siegel's classical method to prove the algebraic dep~ndence of modular forms. If we "forget" about the difficult details, it amounts to finding an estimate for the dimension of the vector space of modular forms of a given weight so that a simple lemma (Chap. III, Proposition 1) can be applied. Inspired by this and by other works of Siegel (16, III, 216-222; 350-365), Andreotti and Grauert discovered the concept of " pseudoconcave manifolds". Their main result states that if M is a (connected) pseudoconcave manifold, for any invertible sheaf L on M, one has

diIDc r(M, L")-< 1(ii mc(M) as k --+ 00; this implies that the field of meromorphic functions on M is algebraic (if the transcendence degree of the field over C is equal to diIDc(M)). For the details and for the relevance to our problem, we refer the reader to A. Andreotti and H. Grauert, "Algebraische Korper von automorphen Funktionen," Gottingen Nachr. (1961), 3, 39-48; A.Andreotti, "Theoremes de dependence algebrique sur les espaces complexes pseudoconcave," Bull. Soc. Math. France 91, 1-38 (1963). We presented the theory of "characteristics" in the style of[6].0n this matter, we refer the reader also to D. Mumford, "Theta characteristics of an algebraic curve," Ann. Ecole Norm. Sup. 4, 181-192 (1971). The main theorem in this chapter (Theorem 8) is not definitive. Actually, one can show that the image X under 8 is Zariski open in its closure X. For this we refer the reader to Mumford [12] and another paper of his: The structure of the moduli spaces of curves and abelian varieties, Actes, Congres intern. math. 1, 1-9 (1970). The reader is adviced not to confuse this precise

230

Further References and Comments

result with the weaker result stating that X, more precisely its image, is Zariski open in the normalization of X. As we remarked, this is a consequence of Theorem 8 and a theorem on normalizations. Theorem 9, or rather its corollary, the corollary of Theorem 2, and Theorem 3 constitute the "fundamental lemma". We did not give any of its applications; for this we refer the reader to J. Igusa, "Geometric and analytic methods in the theory of theta functions," Proc. Bombay Colloq. on Algebraic Geometry, 241-253 (1968). Another application is the following: In "Automorphic forms with integral Fourier coefficients," Lect. Notes in Math. 155, 1-8, Springer-Verlag (1970), W. L. Baily proved the existence of a finite set of modular forms with integral Fourier coefficients which generates over Q a ring containing all such modular forms; this theorem depends, in effect, on the assumption that a set of modular forms satisfying certain conditions exists. That this assumption is indeed valid can be proved easily by the fundamental lemma. We did not mention the number theoretic aspect of theta functions; the reason is that this belongs to an area too rich in mysterious results to make any brief yet meaningful comment. We conclude this appendix by mentioning the following three recent publications which are related to this book: M. Eichler, "Projective varieties and modular forms," Lect. Notes in Math. 210, Springer-Verlag (1971); H. Maass, "Siegel's modular forms and Dirichlet series," Lect. Notes in Math. 216, SpringerVerlag (1971); D. Mumford, "Abelian varieties," Oxford Univ. Press (1970).

Index of Definitions abelian function 133 - variety, polarized 118 addition formula 139 affine variety, dimension 108 algebraic dependence of modular forms 208 automorphic form 115 automorphy factor 60, 112 canonical base 72, 213 characteristic: even or odd, group, period 210 -, theta 49,210 cusp form 204 divisor group 59 dominant series 199 dual of a group, of a homomorphism 8 - of a measure 9 elliptic function 133 embedding 115 -, projective 116 field of definition 108 Fock representation 35 Fourier transformation 8 Gaussian density 40 general linear group 5 generating function 87 generic point 108 Gapel system 217 graded module, graded ring 86 group of characteristics 210 - of characteristics, quotient 213 - ring 11 Hadamard's theorem 191 Heisenberg group 11 Hermite's theorem 191 hermitian form 64

Hilbert-Schmidt operator 13 Hodge decomposition theorem 54 induced representation

15

Jacobi decomposition

190

Koecher's theorem

198

localization 97 Mackey's theorem 19 maximal compact subgroup 23 minimal element 88, 118 Minkowski reduced 191 Minkowski's theorem 192 modular form, degree, weight 197 - group 114 - variety 224 modulus of an automorphism 3 - of a theta function 49 normal, normalization of A in B 91 orthogonal group 5 Pf 119 Pfaffian, reduced 72 Plancherel theorem 7 Poisson formula 44 polarized abelian variety 118 - abelian varieties, isomorphism - abelian variety, type of 119 Pontrjagin duality theory 8 positive divisor 58 - divisor, irreducible 106 - divisor, non-degenerate 128 - divisor, reduced 107 - divisor, support 105 principal congruence group, level 114 projective variety, dimension 116 properly discontinuous 117

232

Index of Definitions

quasi-hermitian form 64 - form, Her, hermitian part, Sym, symmetric part 65 relations, ideal 152, 163 -, quadratic 167 Riemann form 66 Riemann's theta formula 137 ring of theta constants 222 - of theta functions 119 SchrOdinger representation 11 Schwartz function, space 40 second degree character, associated with A 66 - degree character, strongly associated withA 67 Siegel reduced 194 Siegel's theorem 206 - upper-half space 24 simple point 108, 116

standard compactification symplectic group 5, 20

224

theta constant 173 - function 49, 60 - function, classical, local expression 49 - function, normalized 69 - function of type (Q, I, 1/1) 70 - function, trivial 61 torus (complex & real) 51 totally isotropic 213 unitary group 2, 5 - representation 2 Weierstrass polynomial, preparation theorem 103 Zariski closed set 107, 116 - topology 108

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