Amitabha Lahiri_Lecture Notes on Differential Geometry for Physicists 2011.pdf

1 1 0 2 s t s i c i s y h P r o f y r t e m o e G l a i t n e r e f f i D n o s e t o N e r u t c e L : i r i h a L a h b a t i m A c 

Chapter 1

Topology We start by defining a topological space. • A topological space is a set S together with a collection O of subsets called open sets such that the following are true: i) the empty set ∅ and S are open, ∅ , S ∈ O ii) the intersection of a finite numb number er of open open sets sets is open; open; if U 1, U 2 ∈ O , then U 1 ∩ U 2 ∈ O iii) the union of any number of open sets is open, if U i ∈  U i ∈ O irrespective of the range of i.

O ,

then 

i

It is the pair {S , O } which is, precisely speaking, a topological space, or a space with topology. But it is common to refer to S as a topological space which has been given a topology by specifying O . Example: S = R, the real line, with the open sets being open intervals ]a, b[ , i.e. i.e. the the sets sets {x ∈ R | a < x < b } and their unions, plus ∅ and R itself. Then (i) above is true by definition. For two such open sets U 1 = ]a1 , b1 [ and U 2 = ]a2 , b2 [ , we can b 1  a 2 , the intersection U 1 ∩ U 2 = ∅ ∈ O . suppose a 1 < a2 . Then if b Otherwise U 1 ∩ U 2 = ]a2 , b1 [ which is an open interval and thus U 1 ∩ U 2 ∈ O . So (ii) is true. And (iii) is also true by definition.  n Similarly R can be given a topology via open rectangles, i.e. via the sets {(x1 , · · · , xn ) ∈ Rn | ai < xi < bi }. This This is call called ed the the n standard or usual topology of R . • The trivial topology on S consists of O = {∅ , S} .  1

2 1 1 0 2 s t s i c i s y h P r o f y r t e m o e G l a i t n e r e f f i D n o s e t o N e r u t c e L : i r i h a L a h b a t i m A c 

Chapte Chapterr 1. Topology opology

• The discrete topology on a set S is is defined by O = {A | A ⊂ S }, S . i.e., O consists of all subsets of S  • A set A is closed if its complement in S , also written S\ A or  as A , is open.  n Closed rectangles in R are closed sets as are closed balls and single point sets. A set can be neither open nor closed, or both open and closed. In a discrete topology, every set A ⊂ S is both open and closed, whereas in a trivial topology, any set A  = ∅ or S is neither open nor closed. The collection C of closed sets in a topological space S satisfy the following: i) the empty set ∅ and S are open, ∅ , S ∈ C ii) the union A 1 , A2 ∈ C , union of a finite number number of open sets is open; if A then A1 ∪ A2 ∈ C iii) the intersect A i ∈ C , inte ion of any number number of open sets is open, if A  rsection then Ai ∈ C irrespective of the range of i. i

Closed sets can also be used to define a topology. Given a set S with a collection C of subsets satisfying the above three properties of closed sets, we can always define a topology, since the complements of closed sets are open. (Exercise!) • An open neighbourhood of a point P in a topological space S is an open set containing P . A neighbourhood of P is a set containing an open neighbourhood of P . Ne Neigh ighbou bourh rhoods oods can can be defined for sets as well in a similar fashion.  Examples: For a point x ∈ R, and for any  > 0 , x, ]x − , x + [ is an open neighbourhood of x, x, [x − , x + [ is a neighbourhood of x, {x −   y < ∞} is a neighbourhood of x, [x, x + [ is not a neighbourhood of x.  • A topolog topological ical space space is is Hausdorff if two distinct points have disjoint neighbourhoods.  Topology is useful to us in defining continuity of maps. • A map f : S 1 → S 2 is continuous if given any open set U ⊂ S 2 its inverse image (or pre-image, what it is an image of) f −1(U ) ⊂ S 1 is open.  m n When this definition is applied to functions from R to R , it is


the same as the usual  − δ definition definition of continuity, which says that m n • f : R → R is continuous at x0 if given  > 0 , we can always find a δ > 0 such that | f (x) − f (x0 )| <  whenever | x − x0 | < δ .  For the case of functions from a topological space to Rn , this definition says that • f : S → Rn is continuous at s0 ∈ S if given  > 0, we can always find an open neighbourhoood U of s s0 such that |f (s)−f (s0 )| <  whenever s ∈ U.  • If a ma map p f : S 1 → S 2 is one-to-o one-to-one ne and onto, onto, i.e. i.e. a bijection, − 1 and both f and f are continuous, f is called a homeomorphism and we say that S 1 and cs2 are homeomorphic.  Proposition: The composition of two continuous maps is a continuous map. f : S 1 → S 2 and g : S 2 → S 3 are continuous maps, and Proof: If f U is some open set in S 3 , then its pre-image g −1 (U ) is open in S 2 . So f −1 (g −1 (U )), which is the pre-image of that, is open in S 1 . Thus (g ◦ f )−1 (U ) = f −1(g −1(U )) is open in S 1 . Thus g ◦ f is continuous. 


Chapter 2

Manifolds Now that we have the notions of open sets and continuity, we are ready to define the fundamental object that will hold our attention during during this course. course. • A manifold is a topological space which is locally like Rn .  That is, every point of a manifold has an open neighbourhood with a one-to-one map onto some open set of Rn . More re prec precis isel ely y, a topol topologi ogica call spac spacee M is a smooth n• Mo dimensional manifold if the following are true: i) We can cover the space with open sets U α , i.e. every point of M lies within some U α . one-to-on -onee and onto ii) ∃ a map ϕα : U α → Rn , where ϕα is one-to n −1 some open set of R . ϕα is continuous, ϕ α is continuous, i.e. ϕα → V α ∈ Rn is a homeomorphism for V α . (U α , ϕα ) is called a chart (U α is called the domain of the chart). The collection of charts is called an atlas. iii) In any intersection U α ∩ U β , the maps ϕα ◦ ϕ β −1 , which are

called transition functions and take open sets of Rn to open sets of Rn , i.e. ϕα ◦ ϕβ −1 : ϕβ (U α ∩ U β ) → ϕα (U α ∩ U β ), are smooth maps.  • n is called the dimension of M.  We have defined smooth manifolds. A more general definition is that of a C k manifold, in which the transition functions are C k , i.e. k times differentiable. Smooth means k is large enough for the purpose at hand. hand. In practi practice, ce, k is taken to be as large as necessary, up to ∞ C . We get real analytic manifolds when the transition functions 4


are real analytic, i.e. have have a Taylor aylor expansion expansion at each each point, which conver converges. ges. Smoothness Smoothness of a manifold is useful useful because then we can say unambiguously if a function on the manifold is smooth as we will see below. • A complex analytic manifold is defined similarly by replacing Rn with Cn and assuming the transitions functions ϕ α ◦ ϕβ −1 to be holomorphic holomorphic (complex analytic). analytic).  • Given a chart (U α , ϕα) for a neighbourhood neighbourhood of some point P, the n image (x1 , · · · , xn ) ∈ R of P is called the coordinates of P in the chart (U α , ϕα ). A chart is also called a local coordinate system . In this language, a manifold is a space on which a local coordinate system can be defined, and coordinate transformations between different local coordinate systems are smooth. Often we will suppress U and write only ϕ for a chart around some point in a manifold. We will always mean a smooth manifold when we mention a manifold. Examples: Rn (with the usual topology) is a manifold.  The typical typical example example of a manifo manifold ld is the sphere. sphere. Consid Consider er the n n +1 sphere S as a subset of R : (x1 )2 + · · · + ( xn+1 )2 = 1

(2.1)

It is not possible to cover the sphere by a single chart, but it is possible to do so by two charts.1 For the two charts, we will construct what is called the stereographic projection. It is most convenient to draw this for a circle in the plane, i.e. S 1 in R2 , for which the equatorial ‘plane’ is simply an infinit infinitee straig straight ht line. Of course course the construc construction tion works works for any any 1 n S . consider the ‘equatorial plane’ defined as the x = 0, i.e. i.e. the the n+1 n 2 set { (0, x , · · · , x )}, which is simply R when we ignore the first zero. We will find homeomorphisms homeomorphisms from open sets on S n to open sets on this Rn . Let us start start with the north pole N , defined as the point (1, 0, · · · , 0). We draw a straight line from N to any any point on the spher sphere. e. If 1 that point is in the upper hemisphere (x > 0) the line is extended till it hits the equatorial plane. The point where it hits the plane is the 1

The reason reason that that it is not possible possible to cove coverr the sphere sphere with with a single single chart chart is that the sph sphere ere is a compac compactt space, space, and the image image of a compac compactt space space under a continuous continuous map is compact. Since R is non-compact, there cannot be a homeomorphism between S and R . n

n

n


Chapte Chapterr 2. Manifo Manifolds lds

image of the point on the sphere which the line has passed through. For points on the lower hemisphere, the line first passes through the equatorial plane (image point) before reaching the sphere (source point). point). Then Then using using similarit similarity y of triangles triangles we find (Exerci (Exercise! se!)) that that n the coordinates on the equatorial plane R of the image of a point on S n \{N} is given by ϕN

 :

1

n+1

2

x ,x ,··· ,x

 → 

x2 xn+1 , , · · · 1 − x1 1 − x1



. (2.2)

Similarly, the stereographic projection from the south pole is ϕS : S n \{S} → Rn ,



x1 , x2 , · · · , xn+1

 → 

x2 xn+1 , , · · · 1 + x1 1 + x1



. (2.3)

If we write z =



x2 xn+1 ,··· , 1 − x1 1 − x1



,

(2.4)

we find that 2

|z | ≡



x2 1 − x1

2



+ · · · +



xn+1 1 − x1



2

=

1 − (x1 )2 1 + x1 = (2.5) (1 − x1 )2 1 − x1

The overlap between the two sets is the sphere without the poles. Then the transition function between the two projections is ϕS ◦ ϕN : Rn \{0} → Rn \{0},

z →

z . |z |2

(2.6)

These are differentiable functions of z in Rn \{0}. This shows that the sphere is an n -dimensional differentiable manifold.  • A Lie group is a group G which is also a smooth (real analytic for the cases we will consider) manifold such that group composition written as a map (x, y) → xy −1 is smooth.  Anot Anothe herr way of defin definin ingg a Lie Lie grou group p is to star startt with with an nparameter parameter contin continuous uous group group G which is a group that can be parametrized by n (and only n ) real continuous variables. n is called the dimension of the group, n = dim G. (This (This is a differen differentt definition nition of the dimensi dimension. on. The paramet parameters ers are global, global, but do not in general form a global coordinate system.)


Then any element of the group can be written as g (a) where a = (a1 , · · · , an ) . Since the composition of two elements of G must be another element of G, we can write g (a)g (b) = g (φ(a, b)) where φ = (φ1 , · · · , φn ) are n functions of a and b. Then for a Lie group, the functions φ are smooth (real analytic) functions of a and b . These definitions of a Lie group are equivalent, i.e. define the same objects, if we are talking about finite dimensional Lie groups. Further, it is sufficient to define them as smooth manifolds if we are interested only in finite dimensions, because all such groups are also real analytic manifolds. Apparently there is another definition of a Lie group as a topological group (like n-parameter continuous group, but without an a priori restriction on n , in which the composition map (x, y) → xy −1 is continuous) in which it is always possible to find an open neighbourhood of the identity which does not contain a subgroup. Any of these definitions makes a Lie group a smooth manifold, an n-dimensional Lie group is an n-dimensional manifold.  The phase space of N particles is a 6N -dimensional manifold, 3N coordinates and 3N momenta.  The M¨ obius strip is a 2-dimensional manifold.  The space of functions with some specified properties is often a manifold. For example, linear combinations of solutions of Schrödinger equation which vanish outside some region form a manifold.  Finite dimensional vector spaces are manifolds.  Infinite dimensional vector spaces with finite norm (e.g. Hilbert spaces) are manifolds.  • A connected manifold cannot be written as the disjoint union of open sets. Alternatively, the only subsets of a connected manifold which are both open and closed are ∅ and the manifold itself.  SO(3), the group of rotations in three dimensions, is a 3dimensional connected manifold. O(3), the group of rotations plus reflections in three dimensions, is also a 3-dimensional manifold, but is not connected since it can be written as the disjoint union SO(3)∪PSO(3) where P is reflection.  ↑ L+ , the group of proper (no space reflection) orthochronous (no time reflection) Lorentz transformations, is a 6-dimensional connected manifold. The full Lorentz group is a 6-dimensional manifold,


Chapter 2. Manifolds

not connected.  Rotations in three dimensions can be represented by 3 × 3 real orthogonal matrices R satisfying RT R = I. Reflection is represented by the matrix P = − I. The space of 3 × 3 real orthogonal matrices is a connected manifold.  The space of all n × real non-singular matrices is called GL(n, R). This is an n2 -dimensional Lie group and connected manifold. 


Chapter 3

Tangent vectors Vectors on a manifold are to be thought in terms tangents to the manifold, which is a generalization of tangents to curves and surfaces, and will be defined shortly. But a tangent to a curve is like the velocity of a particle at that point, which of course comes from motion along the curve, which is its trajectory. And motion means comparing things at nearby points along the trajectory. And comparing functions at nearby points leads to differentiation. So in order to get to vectors, let us first start with the definitions of these things. • A function f : M → R is differentiable at a point P ∈ M if in a chart ϕ at P, the function f ◦ ϕ−1 : Rn → R is differentiable at ϕ(P ).  −1 This definition does not depend on the chart. If f ◦ ϕ α is −1 differentiable at ϕα (P ) in a chart (U α , ϕα ) at P , the f ◦ ϕ β is differentiable at ϕβ (P ) for any chart (U β , ϕβ ) because f ◦ ϕβ −1 = (f ◦ ϕα −1 ) ◦ (ϕα ◦ ϕβ −1 )

(3.1)

and the transition functions (ϕα ◦ ϕβ −1 ) are differentiable. This should be thought of as a special case of functions from one manifold to another. Consider two manifolds M and N of dimension m and n , and a mapping f : M → N , P → Q. Consider local charts (U, ϕ) around P and (W, ψ) around Q. Then ψ ◦ f ◦ ϕ −1 is a map from Rm → Rn and represents f in these local charts. • f is differentiable at P if ψ ◦ f ◦ ϕ−1 is differentiable at ϕ (P ). In other words, f is differentiable at P if the coordinates y i = f i (xµ ) of Q are differentiable functions of the coordinates xµ of P .  −1 • If f is a bijection (i.e. one-to-one and onto) and f and f are 9


Chapter 3. Tangent vectors

both differentiable, we say that f is a diffeomorphism and that M and N are diffeomorphic.  k In all of these definitions, differentiable can be replaced by C or smooth. • Two Lie groups are isomorphic if there is a diffeomorphism between them which is also a group homomorphism.  • A curve in a manifold M is a map γ of a closed interval R to M. (This definition can be given also when M is a topological space.)  We will take this interval to be I = [0, 1] ⊂ R. Then a curve is a map γ : I → M. If γ (0) = P and γ (1) = P  , for some γ, we say that γ joins P and P  . • A manifold M is connected (actually arcwise connected)1 if any two points in it can be joined by a continuous curve in M .  As for any map, a curve γ is called smooth iff its image in a chart is smooth in Rn , i.e., iff ϕ ◦ γ : I → Rn is smooth in Rn . Note that the definition of a curve implies that it is parametrized. So the same collection of points in M can stand for two different curves if they have different parametrizations. We are now ready to define tangent vectors and the tangent space to a manifold. There are different ways of defining tangent vectors. i) Coordinate approach: Vectors are defined to be objects satisfying certain transformation rules under a change of chart, i.e. coordinate transformation, (U α , ϕα ) → ( U β , ϕβ ). ii) Derivation approach: A vector is defined as a derivation of func-

tions on the manifold. This is thinking of a vector as defining a “directional derivative”. iii) Curves approach: A vector tangent to a manifold is tangent to

a curve on the manifold. The approaches are equivalent in the sense that they end up defining the same objects and the same space. We will follow the third approach, or perhaps a mix of the second and the third approaches. Later we will briefly look at the derivation approach more carefully and compare it with the way we have defined tangent vectors. Consider a smooth function f : M → R. Given a curve γ : I → M, the map f ◦ γ : I → R is well-defined, with a well-defined 1

It can be shown that an arcwise connected space is connected.


df

derivative. The rate of change of f along γ is written as . dt Suppose another curve another curve µ(s) meets γ (t) at some point P , where s = s 0 and t = t 0 , such that d d   (f ◦ γ ) = (f ◦ µ) dt ds P P 



∀f ∈ C ∞ (M)

(3.2)

That is, we are considering a situation where two curves are tangent to each other in geometric and parametric sense. Let us introduce a convenient notation. In any chart ϕ containing the point P, let us write ϕ(P ) = (x1 , · · · , xn ). Let us write f ◦ γ = (f ◦ ϕ−1 ) ◦ (ϕ ◦ γ ), so that the maps are f ◦ ϕ−1 :

n R

→ R, ϕ ◦ γ : I → Rn ,

→ f (x) or f (xi ) t → { xi (γ (t))}.

(3.3)

x

(3.4)

The last are the coordinates of the curve in Rn . Using the chain rule for differentiation, we find d d ∂f dxi (γ (t)) (f ◦ γ ) = f (x(γ (t))) = . dt dt ∂x i dt Similarly, for the curve µ we find

(3.5)

d d ∂f dxi (µ(s)) (f ◦ µ) = f (x(µ(s))) = (3.6) . ds ds ∂x i ds Since f is arbitrary, we can say that two curves γ, µ have the same tangent vector at the point P ∈ M (where t = t 0 and s = s 0 ) iff dxi (γ (t))  dxi (µ(s))  . =   dt ds t=t0 s=s0 



(3.7)

We can say that these numbers completely determine the rate of change of any function along the curve γ or µ at P. So we can define the tangent to the curve. • The tangent vector to a curve γ at a point P on it is defined as the map ˙ P : C ∞ (M) → γ

R,

˙ P (f ) ≡ f → γ

d (f ◦ γ )|P . (3.8) dt

As we have already seen, in a chart with coordinates { xi } we can write using chain rule dxi (γ (t)) ∂f  ˙ P (f ) = γ dt ∂x i ϕ(P ) 

(3.9)


Chapter 3. Tangent vectors

dxi (γ (t))  The numbers are thus the components of γ ˙ P . We will  dt ϕ(P ) often write a tangent vector at P as v P without referring to the curve 

it is tangent to. We note here that there is another description of tangent vectors based on curves. Let us write γ ∼ µ if γ and µ are tangent to each other at the point P . It is easy to see, using Eq. (3.7) for example, that this relation ∼ is transitive, reflexive, and symmetric. In other words, ∼ is an equivalence relation, for which the equivalence class [γ ] contains all curves tangent to γ (as well as to one another) at P . • A tangent vector at P ∈ M is an equivalence class of curves under the above equivalence relation.  The earlier definition is related to this by saying that if a vector v is tangent to some curve γ at P , i.e. if v = γ ˙ , we can write v = [γ ]. P P

P

P


Chapter 4

Tangent Space • The set of all tangent vectors (to all curves) at some point P ∈ M is the tangent space T P M at P.  Proposition: T P M is a vector space with the same dimensionality n as the manifold M. Proof: We need to show that T P M is a vector space, i.e. X P + Y P ∈ T P M ,

(4.1)

aX P ∈ T P M ,

(4.2)

∀X P , Y P ∈ T P M, a ∈

R.

That is, given curves γ, µ passing through P such that X P = ˙ , Y P = µ˙ , we need a curve λ passing through P such that γ λ˙ (f ) = X P (f ) + Y P (f )∀f ∈ C (M). Define λ : I → Rn in some chart ϕ around P by λ = ϕ ◦ γ + ϕ ◦ µ − ϕ(P ). Then λ is a curve in Rn , and P

P

∞

P

λ = ϕ ◦ λ : I → M

(4.3)

is a curve with the desired property.  Note: we cannot define λ = γ + µ − P because addition does not make sense on the right hand side. The proof of the other part works similarly. (Exercise!) To see that T P M has n basis vectors, we consider a chart ϕ with coordinates xi . Then take n curves λk such that





ϕ ◦ λk (t) = x1 (P ), · · · , xk (P ) + t, · · · , xn (P ) ,

(4.4)

i.e., only the k-th coordinate varies along t. So λk is like the axis of the k -th coordinate (but only in some open neighbourhood of P ). 13


Chapter 4. Tangent Space

Now denote the tangent vector to λk

  ∂ ∂x k

f = λ˙ k (f )

P



=

P

  at by  ( ◦ ) ∂ ∂x k

P

d f λk dt

, i.e.,

P

(4.5)

.

P

This notation makes sense when we remember Eq. (3.9). Using it we can write

 =   ∀ ∈ (M) (4.6)   Note that is notation. We should understand this as       = ◦ (4.7)  ≡    λ˙ k (f ) ∂ ∂x k

∂ ∂x k

∂f ∂x k

P

∞

f C

.

P

P

f

P

∂ f ϕ ∂x k

in a chart around P . The

∂ ∂x k

1

−

ϕ(P )

∂f ∂x k

ϕ(P )

are defined only when this chart

P

is given, but these are vectors on the manifold at P , not on Rn . Let us now show that the tangent space at P has λ˙ k | as a basis. Take any vector v ∈ T P M , which is the tangent vector to some curve γ at P . (We may sometimes refer to P as γ ( 0) or as t = 0.) Then P

P

 

d vP (f ) = ( f ◦ γ ) dt

d = ((f ◦ ϕ dt

(4.8) t=0

1

−

 )

 )) 

◦ (ϕ ◦ γ ϕ(P )

.

(4.9)

t=0

Note that ϕ ◦ γ : I → Rn , t → ( x1 (γ (t)), · · · , xn (γ (t))) are the coordinates of the curve γ , so we can use the chain rule of differentiation to write ∂ (f ◦ ϕ vP (f ) = ∂x i

=

∂ (f ◦ ϕ ∂x i

1

−

 )  )

ϕ(P )

1

−

d i (x ◦ γ ) dt vP (xi ) .

 

(4.10)

t=0

(4.11)

ϕ(P )

The first factor is exactly as shown in Eq. (4.7), so we can write vP (f ) =

  ∂ ∂x k

P

f vP (xi )

∀f ∈ C (M) ∞

(4.12)


i.e., we can write i

vP = v P

  ∂ ∂x k

∀v ∈ T P M

P

(4.13)

P

where v i = v (xi ). Thus the vectors ∂x∂ P span T P M . These are to be thought of as tangents to the coordinate curves in ϕ. These can be shown to be linearly independent as well, so ∂x∂ P form a basis of T P M and v i are the components of v in that basis. The ∂x∂ P are called coordinate basis vectors and the set

  k

P

P

  k

P

P

     is called the k

∂ ∂x k P

coordinate basis.

It can be shown quite easily that for any smooth (actually C 1 ) function f a vector v defines a derivation f →  v (f ) , i.e., satisfies linearity and Leibniz rule, P

P

vP (f + αg ) = vP (f ) + αvP (g )

(4.14)

vP (f g ) = vP (f )g (P ) + f (P )vP (g )

∀f, g ∈ C 1 (M)and α ∈

R

(4.15) (4.16)

1 1 0 2 s t s i c i s y h P r o f y r t e m o e G l a i t n e r e f f i D n o s e t o N e r u t c e L : i r i h a L a h b a t i m A

Chapter 5

Dual space ∗

•

The dual space T P M of T P M is the space of linear mappings ω : T P M → R.  We will write the action of ω on vP ∈ T P M as ω (vP ) or sometimes as  ω | v P  . Linearity of the mapping ω means ω (uP + avP ) = ω (uP ) + aω (vP ) ,

∀uP , vP ∈ T P M and a ∈

(5.1) R.

The dual space is a vector space under the operations of vector addition and scalar multiplication defined by a1 ω1 + a2 ω2 : v P →  a 1 ω1 (vP ) + a2 ω2 (vP ) .

(5.2)

∗

•

The elements of T P M are called dual vectors, covectors, cotangent vectors etc.  A dual space can be defined for any vector space V as the space of linear mappings V → R (or V → C if V is a complex vector space). Example:

Vector column vectors kets |ψ  functions

Dual vector row vector bras  φ| linear functionals, etc.

c 

•

Given a function on a manifold f : M → R , every vector at P produces a number, v (f ) ∈ R ∀v ∈ T P M . Thus f defines a P

P

16


covector df , given by df (v ) = v (f ) called the differential or gradient of f .  Since v is linear, so is df , P

P

P

df (v + aw ) = (v + aw )(f ) P

P

P

P

= v (f ) + aw (f ) ∀v , w ∈ T P M, a ∈ P

(5.3)

P

P

P

R.

∗

Thus df ∈ T P M .

∗ Proposition: T P M is also n-dimensional. Proof: Consider a chart ϕ with coordinate functions x i . Each x i

is a smooth function xi : M → R . then the differentials dxi satisfy dx

i

    ∂ ∂x j

=

P

∂ ∂x j

∂ (x ) = xi ◦ ϕ−1 j ∂x P i





ϕ(P )

= δ ji . (5.4)

The differentials dxi are covectors, as we already know. So we have constructed n covectors in T P M . Next consider a linear combination of these covectors, ω = ω i dxi . If this vanishes, it must vanish on every one of the basis vectors. In other words, ∗

ω = 0 ⇒ ω

  =0   ∂ ∂x j

⇒ ωi dxi

P

∂ ∂x j

=0

P

⇒ ωi δ ji = 0 i.e. ω j = 0 .

(5.5)

So the dxi are linearly independent. Finally, given any covector ω , consider the covector λ = ω − ∂ d xi . Then letting this act on a coordinate basis vector, we ω ∂x P get

  i

λ

  ∂ ∂x j

P

=ω =ω

    ∂ ∂x j ∂ ∂x j

−ω

P

P

−ω

    d   = 0∀  

So λ vanishes on all vectors, since the

∂ ∂x i ∂ ∂x i

∂ ∂x j

xi

P

δ ji

∗

j

P

(5.6)

P

form a basis. Thus

P

the dxi span T P M , so T P M is n-dimensional. ∗

∂ ∂x j


Chapter 5. Dual space ∗

as

Also, as we have just seen, any covector ω ∈ T P M can be written ω = ω i dx

i

where ωi = ω

  ∂ ∂x i

,

(5.7)

P

so in particular for ω = df , we get ωi ≡ (d f )i = df

    ∂ ∂x i

∂f ∂x i

=

P

(5.8)

ϕ(P )

This justifies the name gradient. It is straightforward to calculate the effect of switching to another overlapping chart, i.e. a coordinate transformation. In a new chart ϕ where the coordinates are y i (and the transition functions are thus y i (x)) we can use Eq. (5.8) to write the gradient of y i as 

i

dy =

∂y i ∂x j

 

dx j

(5.9)

P

This is the result of coordinate transformations on a basis of covectors. ∂ Since is the dual basis in T P M to {dxi }, in order i

      for to be the dual basis to {d } we must have      ∂x

∂ ∂y i

P

yi

P

∂ ∂y i

=

P

∂x j ∂y i

∂ ∂x j

P

(5.10)

P

These formulae can be generalized to arbitrary bases. Given a vector v , it is not meaningful to talk about its dual, but given a basis { ea}, we can define its dual basis { ω a } by ω a(eb ) = δ ba . We can make a change of bases by a linear transformation, ea → e a = (A−1 )ba eb , (5.11)

ω a → ω a = A ab ω b ,

with A a non-singular matrix, so that ω a (eb ) = δ ba . Given a 1-form λ we can write it in both bases, 



λ = λ a ω a = λ a ω a = λ a Aab ω a , 

from which it follows that λa = (A

1 )b λ

−

a b

.

(5.12)


Similarly, if v is a vector, we can write v = v a ea = v a ea = v a (A−1 )ba eb ,

(5.13)

and it follows that va = A abv b . Quantities which transform like λa are called covariant, while • those transforming like v a are called contravariant. 


Chapter 6

Vector fields •

Consider the (disjoint) union of tangent spaces at all points,

 T M .

T M =

P

(6.1)

P ∈M

This is called the tangent bundle of M.  • A vector field v chooses an element of T P M for every P , i.e.  v(P ) ≡ v P ∈ T P M. v : P →  We will often write v(f )|P = v P (f ). Given a chart, v has components v i in the chart, vP = v

i

 ∂  ∂x

i

(v i )P = v P (xi ) .

,

(6.2)

P

The vector field v is smooth if the functions v i = v(xi ) are • smooth for any chart (and thus for all charts).  ∗ • A rule that selects a covector from T P M for each P is called a one−form (often written as a 1−form ).  1 Given a smooth vector field v (actually C is sufficient) we can • define an integral curve of v, which is a curve γ in M such that γ ˙ (t)|P = vP at every P ∈ γ. (One curve need not pass through all P ∈ M.)  Suppose γ is an integral curve of a given vector field v, with γ (0) = P. Then in a chart containing P , we can write γ ˙ (t) = v ⇒

d i x (γ (t)) = v i (x(t)) , dt

(6.3)

with initial condition xi (0) = x i |P . This is a set of ordinary first order differential equations. If vi are smooth, the theory of differential 20


equations guarantees, at least for small t (i.e. locally), the existence of exactly one solution. The uniqueness of these solutions implies that the integral curves of a vector field do not cross. One use of integral curves is that they can be thought of as co 0, it ordinate lines. Given a smooth vector field v such that v |P = i is possible to define a coordinate system {x } in a neighbourhood ∂ around P such that v = . ∂x i • A vector field v is said to be complete if at every point P ∈ M the integral curve the integral curve γ (t) of v passing through P can be extended to all t ∈ R .  The tangent bundle T M is a product manifold, i.e., a point in T M is an ordered pair (P, v) where P ∈ M and v ∈ T P M. The topological structure and differential structure are given appropriately. • The map π : T M → M, (P, v) → P (where v ∈ T P M) is called the canonical projection (or simply projection).  −1 For each P ∈ M, the pre-image π (P ) is T P M. It is called the • fiber over P . Then a vector field can be thought of as a section of the tangent bundle.  Given a smooth vector field v, we can define an integral curve γ through any point P by γ ˙ (t) = v , i.e., d i x (γ (t)) = v i (γ (t)) ≡ v(xi (γ (t))) , dt γ (0) = P .

(6.4) (6.5)

We could also choose γ (t0 ) = P. Then in any neighbourhood U of P we also have γ Q , the integral curve through Q. So we can define a map φ : I × U → M given by φ(t, Q) = γ Q (t) where γ Q (t) satisfies d i x (γ Q (t)) = v(xi γ Q (t)) , dt γ Q (0) = Q .





(6.6) (6.7)

• This φ defines a map φt : U → M at each t by φt (Q) = φ(t, Q) = γ Q (t) , i.e. for given t, φt takes a point by a parameter distance t along the curve γ Q (t). This φt is called the local flow of v.  The local flow has the following properties: i) φ0 is the identity map of U ;


Chapter 6. Vector fields

ii) φs ◦ φt = φ s+t for all s, t, s + t ∈ U ; iii) each flow is a diffeomorphism with φt −1 = φ −t . The first property is obvious, while the second property follows from the uniqueness of integral curves, i.e. of solutions to first order differential equations. Then the integral curve passing through the point γ Q (s) is the same as the integral curve passing through Q, so that moving a parameter distance t from γ Q (s) finds the same point on M as by moving a parameter distance s + t from γ Q (0) ≡ Q . A vector field can also be thought of as a map from the space of differentiable functions to itself v : C ∞ (M) → C ∞ (M), f →  v(f ) , with v(f ) : M → R, P →  v P (f ) . Often v(f ) is called the Lie derivative of f along v and denoted £v f . The map v : f →  v(f ) has the following properties: v(f + αg) = v(f ) + αv(g)

(6.8)

v(fg) = fv(g) + v(f )g

∀f, g ∈ C ∞ (M),

(6.9) α ∈

R

The set of all (real) vector fields V (M) on a manifold M has the structure of a (real) vector space under vector addition defined by (u + αv)(f ) = u(f ) + αv(f ),

u, v ∈ V (M),

α ∈

R.

(6.10) It is possible to replace α by some function on C ∞ (M). If u, v are vector fields on M and α is now a smooth function on M , define u + αv by (u + αv)P (f ) = u P (f ) + α(P )vP (f )

∀f ∈ C ∞ (M),

P ∈ M . (6.11)

This looks like a vector space but actually it is what is called a module. A ring R is a set or space with addition and multiplication • defined on it, satisfying (xy)z = x(yz) , x(y + z) = xy + xz , (x + y)z = xz + yz , and two special elements 0 and 1, the additive and multiplicative identity elements, 0 + x = x + 0 = x , 1x = x1 = x . A module X is an Abelian group under addition, with scalar multiplication by elements of a ring defined on it.


•

A module becomes a vector space when this ring is a commu tative division ring, i.e. when the ring multiplication is commutative, xy = yx, and an inverse exists for every element except 0. Given a smooth function α, in general α−1 ∈ / C ∞ (M), so the space of vector fields on M is in general a module, not a vector space. Given a vector field v, in an open neighbourhood of some P ∈ M and in a chart, and for any f ∈ C ∞ (M) , we have

 v(f )

i

P

= v P (f ) = v P

 ∂f  ∂x

i

, where vP i = v P (xi ) . (6.12)

P

Thus we can write v = v i

∂ with v i = v(xi ) , i ∂x

(6.13)

as an obvious generalization of vector space expansion to the module V (M). The v i are now the components of the vector field v, and ∂x∂ i are now vector fields, which we will call the coordinate vector fields. Note that this is correct only in some open neighbourhood on which a chart can be defined. In particular, it may not be possible in general to define the coordinate vector fields globally, i.e. everywhere on M , and thus the components v i may not be defined globally either.


Chapter 7

Pull back and push forward Two important concepts are those of pull back (or pull-back or pullback) and push forward (or push-forward or pushforward) of maps between manifolds. • Given manifolds M1 , M2 , M3 and maps f : M1 → M2 , g : M2 → M3 , the pullback of g under f is the map f g : M1 → M3 defined by ∗

f ∗ g = g ◦ f .

(7.1)

So in particular, if M 1 and M 2 are two manifolds with a map f : M 1 → M2 and g : M 2 → R is a function on M2 , the pullback of g under f is a function on M 1 , 

f ∗ g = g ◦ f .

(7.2)

While this looks utterly trivial at this point, this concept will become increasingly useful later on. • Given two manifolds M1 and M2 with a smooth map f : M1 → M2 , P → Q the pushforward of a vector v ∈ T P M1 is a vector f v ∈ T Q M2 defined by ∗

f ∗ v (g ) = v (g ◦ f )

for all smooth functions g : M2 → Thus we can write

(7.3)

R.

f ∗ v (g ) = v (f ∗ g ) .

24



(7.4)


The pushforward is linear, f ∗ (v1 + v2 ) = f ∗ v1 + f ∗ v2

f ∗ (λv ) = λf ∗ v .

(7.5)

(7.6)

And if M1 , M2 , M3 are manifolds with maps f : M1 → M2 , g : M2 → M3 , it follows that (g ◦ f ) = g f , (g ◦ f ) v = g f v ∗

∗

∗

∗

i.e.

∗

∗

∀v ∈ T P M1 .

(7.7)

Remember that we can think of a vector v as an equivalence class of curves [γ ]. The pushforward of an equivalence class of curves is f ∗ v = f ∗ [γ ] = [f ◦ γ ]

(7.8)

Note that for this pushforward to be defined, we do not need the original maps to be 1-1 or onto. In particular, the two manifolds may have different dimensions. Suppose M 1 and M 2 are two manifolds with dimension m and n respectively. So in the respective tangent spaces T P M1 and T Q M2 are also of dimension m and n respectively. So for a map f : M1 → M2 , P → Q , the pushforward f will not have an inverse if m  = n . Let us find the components of the pushforward f v in terms of the components of v for any vector v . Let us in fact consider, given charts ϕ : P → ( x1 , · · · , xm ) , ψ : Q → ( y 1 , · · · , yn ) the pushforward of the basis vectors. For the basis vector ∂x∂ P , we want the pushforward f ∂x∂ P , ∗

∗

 

    M so we can expand it in the basis which is a vector in       ∗

i

T Q

f ∗

∂ ∂x i

2 ,

=

i

∂ ∂y i

f ∗

P

∂ ∂x i

µ

∂ ∂y µ

P

Q

,

(7.9)

Q

In any coordinate basis, the components of a vector are given by the action of the vector on the coordinates as in Chap. 4, µ vP = v P (y µ )

(7.10)

Thus we can write µ

   f ∗

∂ ∂x i

P

= f

∗

  ∂ ∂x i

P

(y µ )

(7.11)


Chapter 7. Pull back and push forward

But f ∗ v(g ) = v (g ◦ f ) ,

(7.12)

so f ∗

  ∂ ∂x i

  ∂ ∂x i

µ

(y ) =

P

(y µ ◦ f ) .

(7.13)

P

But y µ ◦ f are the coordinate functions of the map f , i.e., coordinates around the point f (P ) = Q . So we can write y µ ◦ f as y µ (x) , which is what we understand by this. Thus µ

∂y µ (x) y f ( ◦ )= ∂x i P

     f ∗

∂ ∂x i

∂ ∂x i

=

P

µ

 

.

(7.14)

P

Because we are talking about derivatives of coordinates, these are actually done in charts around P and Q = f (P ) , so the chart maps are hidden in this equation. • The right hand side is called the Jacobian matrix (of y µ (x) = y µ ◦ f with respect to x i ). Note that since m and n may be unequal, this matrix need not be invertible and a determinant may not be defined for it.  For the basis vectors, we can then write f ∗

  ∂ ∂x i

∂y µ (x) = ∂x i P

    P

∂ ∂y µ

(7.15)

f (P )

Since f is linear, we can use this to find the components of (f v )Q for any vector v , ∗

∗

P

        ( )   ( )  

i f ∗ vP = f ∗ vP

= v i f P

=v

i

P

(f v )µ = v i

⇒

∗

P

P

∗

∂ ∂x i P ∂ ∂x i P

∂y µ x ∂x i ∂y µ x ∂x i

∂ ∂y µ

P

.

(7.16)

f (P )

(7.17)

P

Note that since f is linear, we know that the components of f v should be linear combinations of the components of v , so we can ∗

∗


already guess that (f v )µ = A µi v i for some matrix A µi . The matrix is made of first derivatives because vectors are first derivatives. Another example of the pushforward map is the following. Remember that tangent vectors are derivatives along curves. Suppose v ∈ T P M is the derivative along γ . Since γ : I → M is a map, we can consider pushforwards under γ , of derivatives on I . Thus for γ : I → M , t → γ (t) = P , and for some g : M → R , ∗

P

P

P

γ ∗

  d dt

d (g ◦ γ )|t=0 dt = γ ˙ P (g )|t=0 = v P (g ) ,

g =

t=0

so γ ∗

  d dt

= v

(7.18)

P

(7.19)

t=0

• We can use this to give another definition of integral curves. Suppose we have a vector field v on M . Then the integral curve of v passing through P ∈ M is a curve γ : t → γ (t) such that γ (0) = P and γ ∗

  d dt

t

= v |γ (t)

(7.20)

for all t in some interval containing P .  Even though in order to define the pushforward of a vector v under a map f , we do not need f to be invertible, the pushforward of a vector field can be defined only if f is both one-to-one and onto. If f is not one-to-one, different points P and P may have the same image, f (P ) = Q = f (P ) . Then for the same vector field v we must have 



f ∗ v |Q = f ∗ (vP ) = f ∗ (vP ) ,



(7.21)

which may not be true. And if f : M → N is not onto, f v will be meaningless outside some region f (M) , so f v will not be a vector field on N . If f is one-to-one and onto, it is a diffeomorphism, in which case vector fields can be pushed forward, by the rule ∗

∗

(f v )f (P ) = f (v ) . ∗

∗

P

(7.22)


Chapter 8

Lie brackets A vector field v is a linear map C (M) → C (M) since it is basically a derivation at each point, v : f →  v (f ) . In other words, given a smooth function f , v (f ) is a smooth function on M . Suppose we consider two vector fields u , v . Then u(v (f )) is also a smooth function, linear in f . But is uv ≡ u ◦ v a vector field? To find out, we consider ∞

∞

u(v (f g )) = u(f v (g ) + v (f )g )

= u(f )v (g ) + f u(v (g )) + u(v (f ))g + v (f )u(g) . (8.1) We reorder the terms to write this as uv(f g ) = f uv (g ) + uv (f )g + u(f )v(g ) + v (f )u(g ) ,

(8.2)

so Leibniz rule is not satisfied by uv . But if we also consider the combination v u , we get vu(f g ) = f (vu(g ) + vu (f )g + v (f )u(g ) + u(f )v(g ) .

(8.3)

Thus (uv − vu)(f g ) = f (uv − vu)(g) + ( uv − vu)(f )g , (8.4) which means that the combination [u , v ] := uv − vu

(8.5)

is a vector field on M , with the product uv signifying successive operation on any smooth function on M . 28


This combination is called the commutator or Lie bracket of • the vector fields u and v .  In any chart around the point P ∈ M , we can write a vector field in local coordinates v (f ) = v i

∂ f , ∂x i

(8.6)

so that





∂ i ∂ f u(v (f )) = u v ∂x j ∂x i ∂v i ∂f ∂ 2 f = u j j i + u j v i j i , ∂x ∂x ∂x x i 2 j ∂u ∂f j i ∂ f v (u(f )) = v + u v j i . ∂x j ∂x i ∂x x j

(8.7)

Subtracting, we get u(v (f )) − v (u(f )) = u j

i ∂v i ∂f j ∂u ∂f − v , ∂x j ∂x i ∂x j ∂x i

(8.8)

from which we can read off the components of the commutator, i

[u , v ]

i j ∂v = u ∂x j

−

i j ∂u v ∂x j

(8.9)

The commutator is antisymmetric, [u , v ] = − [v , u] , and satisfies • the Jacobi identity [[u , v ] , w] + [[v , w] , u] + [[w , u] , v] = 0 .

(8.10)

The commutator is useful for the following reason: Once we have a chart, we can use

  ∂ ∂x i

as a basis for vector fields in a neighbour-

hood. Any set of n linearly independent vector fields may be chosen as a basis, but they need not form a coordinate system. In a coordinate system,



∂ ∂ , ∂x i ∂x j



=0,

(8.11)

because partial derivatives commute. So n vector fields will form a coordinate system only if they commute, i.e., have vanishing commutators with one another. Then the coordinate lines are the integral


Chapter 8. Lie brackets

curves of the vector fields. For analytic manifolds, this condition is sufficient as well. A simple example is the polar coordinate system in R2 . The unit vectors are er = ex cos θ + ey sin θ eθ = −ex sin θ + ey cos θ ,

with ex =

(8.12)

∂ ∂ and ey = being the Cartesian coordinate basis ∂x ∂y

vectors, and

cos θ =

x , r

y r

sin θ = ,

r =



x2 + y 2

(8.13)

 0 , so Using these expressions, it is easy to show that [er , eθ ] = {er , eθ } do not form a coordinate basis.


Chapter 9

Lie algebra An (real) algebra is a (real) vector space equipped with a bi• linear operation (product) under which the algebra is closed, i.e., for an algebra A i) x y ∈ A

∀x, y ∈

•

A

ii) (λx + µy ) z = λx z + µy z x (λy + µz ) = λx y + µx z •

•

•

•

•

•

∀x , y , z ∈

A ,

λ, µ ∈

R.

If λ, µ are complex numbers and A is a complex vector space, we get a complex algebra.  A Lie algebra is an algebra in which the operation is • i) antisymmetric, x y = − y x , and •

•

ii) satisfies the Jacobi identity ,

(x y) z + ( y z ) x + ( z x) y = 0 . •

•

•

•

•

•

(9.1) 

The Jacobi identity is not really an identity — it does not hold for an arbitrary algebra — but it must be satisfied by an algebra for it to be called a Lie algebra. Example:

i) The space M n = {all n × n matrices} under matrix multiplication, A B = AB . This is an associative algebra since matrix multiplication is associative, (AB )C = A (BC ) . •

ii) The same space M n of all n × n matrices as above, but now

31


Chapter 9. Lie algebra

with matrix commutator as the product, A B = [A , B ] = AB − BA .

•

(9.2)

This product is antisymmetric and satisfies Jacobi identity, so M n with this product is a Lie algebra. iii) The angular momentum algebra in quantum mechanics. If Li are the angular momentum operators with [Li , L j ] = iijk Lk , we can write the elements of this algebra as L

      =

ζ i Li |ζ i ∈

a =

C

(9.3)

i

If a =



ai Li and b =

a b ≡ [ a , b] = •

bi Li , their product is

ai b j [Li , L j ] = i

ijk ai b j Lk . (9.4)

This is a Lie algebra because it [a , a] = 0 and the Jacobi identity is satisfied. iv) The Poisson bracket algebra of a classical dynamical sys-

tem consists of functions on the phase space, with the product defined by the Poisson bracket, f g = [f , g ]P.B. . •

This is a Lie algebra. dimensional.

(9.5)

As a vector space it is infinite-

v ) Vector fields on a manifold form a real Lie algebra under the

commutator bracket, since the Jacobi identity is a genuine identity, i.e. automatically satisfied, as we have seen in the previous chapter. This algebra is infinite-dimensional. (It can be thought of as the Lie algebra of the group of diffeomorphisms, Diff (M)) .

1 1 0 2 s t s i c i s y h P r o f y r t e m o e G l a i t n e r e f f i D n o s e t o N e r u t c e L : i r i h a L a h b a t i m A

Chapter 10

Local flows We met local flows and integral curves in Chapter 6. Given a vector field v , write its local flow as φt . • The collection φt for t <  (for some  > 0, or alternatively for t < 1) is a one −parameter group of local diffeomorphisms .  Consider the vector field in a neighbourhood U of a point Q ∈ M . Since φt : U → M, Q →  γ Q (t) is local diffeomorphism , i.e. diffeomorphism for sufficiently small values of t , we can use φt to push forward vector fields. At some point P we have the curve φ t (P ) . We push forward a vector field at t =  to t = 0 and compare with the vector field at t = 0 . We recall that for a map ϕ : M1 → M2 the pullback of a function f ∈ C (M2 ) is defined as ∞

ϕ f = f ◦ ϕ : M1 → ∗

R,

(10.1)

and ϕ f ∈ C (M1 ) if ϕ is C . The pushforward of a vector vP is defined by ∗

∞

∞

ϕ vP (f ) = v P (f ◦ ϕ) = vP (ϕ f )

(10.2)

∗

∗

vP ∈ T P M1 ,

ϕ vP ∈ T ϕ(P ) M2 .

∗

(10.3)

If ϕ is a diffeomorphism, we can define the pushforward of a vector field v by

c 

ϕ v (f )|ϕ(P ) = v ( f ◦ ϕ)|P ∗

i.e.

ϕ v (f )|Q = v ( f ◦ ϕ)|ϕ−

1

∗

= v ( ϕ f )|ϕ−

Q

∗

1

33

Q

.

(10.4)


Chapter 10. Local flows

We can rewrite this definition in several different ways, (ϕ v )(f ) = v (f ◦ ϕ) ◦ ϕ

1

−

∗

= (ϕ = (ϕ

1

−

∗

−

∗

) (v (f ◦ ϕ)) 1 ) (v (ϕ f )) .

∗

(10.5)

• If ϕ : M1 → M2 is not invertible, ϕ v is not a vector field on M2 . If ϕ 1 exists but is not differentiable, ϕ v is not differentiable. But there are some ϕ and some v such that ϕ v is a differentiable vector field, even if ϕ is not invertible or ϕ 1 is not differentiable. Then v and ϕ v are said to be ϕ−related.  Proposition: Given a diffeomorphism ϕ : M1 → M2 (say both C manifolds) the pushforward ϕ is an isomorphism on the Lie algebra of vector fields, i.e. ∗

−

∗

∗

−

∗

∞

∗

ϕ [u , v] = [ϕ u , ϕ v] . ∗

∗

∗

(10.6)

Proof:

ϕ [u , v](f ) = [u , v] (f ◦ ϕ) ◦ ϕ

−

1

∗

= u (v (f ◦ ϕ)) ◦ ϕ 1 − u ↔ v , [ϕ u , ϕ v ](f ) = ϕ u (ϕ v (f )) − u ↔ v = u (ϕ v (f ) ◦ ϕ) ◦ ϕ 1 − u ↔ v −

while

∗

∗

∗

(10.7)

∗

−

∗

 ( ◦

= u v f ϕ) ◦ ϕ = u (v (f ◦ ϕ)) ◦ ϕ

1

−

1

−

◦ ◦ ϕ

ϕ

1

−

−u ↔v.

−u ↔v

(10.8) 

•

A vector field v is said to be invariant under a diffeomorphism ϕ : M → M if ϕ v = v , i.e. if ϕ (vP ) = v ϕ(P ) for all P ∈ M .  We can write for any f ∈ C (M) ∗

∗

∞

(ϕ v ) (f ) = ϕ 1 (v (ϕ f )) ϕ ((ϕ v ) (f )) = v (ϕ f ) ,

  −

∗

⇒ ⇒

∗

∗

∗

∗

∗

∗

∗

ϕ ◦ϕ v = v◦ϕ .

∗

(10.9)

So if v is an invariant vector field, we can write ϕ ◦ v = v ◦ ϕ . ∗

∗

(10.10)


This expresses invariance under ϕ , and is satisfied by all differential operators invariant under ϕ . Consider a vector field u , and the local flow (or one-parameter diffeomorphism group) φt corresponding to u , φt (Q) = γ Q (t) ,

˙ Q (t) = u (γ Q (t)) . γ

(10.11)

But for any f ∈ C (M) , ∞

d f ◦ γ Q (t) dt d = ( f ◦ φt (Q)) dt d = ( φt (f )) = u γ Q (t) (f ) ≡ u(f ) dt

γ ˙ Q (f ) =







∗

At t = 0 we get the equation d ( φ (f )) dt t ∗

We can also write



t=0

  ( )

= u f

(10.13)

Q

d ( φ f ) (Q) = u (f ) (φt (Q)) = φ t u(f )(Q) . dt t ∗

(10.12)

γ Q (t)

∗

(10.14)

This formula can be used to solve linear partial differential equations of the form n ∂ f ( x, t) = ∂t



v i (x)

i=1

∂ f (x, t) ∂x i

(10.15)

with initial condition f (x, 0) = g (x) and everything smooth. This is an equation on Rn+1 , so it can be on a chart for a manifold as well. We can treat v i (x) as components of a vector field v . Then a solution to this equation is f (x, t) = φt g (x) ∗

≡ g (φt (x)) ≡ g ◦ φt (x) ,

(10.16)

where φt is the flow of v . Proof:

∂ d ∂ f f (x, t) = ( φt g ) = v (f ) ≡ v i i , ∂t dt ∂x ∗

(10.17)


Chapter 10. Local flows

using Eq. (10.13) .  Thus the partial differential equation can be solved by finding the integral curves of v (the flow of v ) and then by pushing (also called dragging ) g along those curves. It can be shown, using well-known theorems about the uniqueness of solutions to first order partial differential equations, that this solution is also unique. Example: Consider the equation in 2+1 dimensions ∂ f (x, t) = (x − y ) ∂t



∂f ∂f − ∂x ∂y



(10.18)

with initial condition f (x, 0) = x2 + y 2 . The corresponding vector field is v (x) = (x − y, −x + y ) . The integral curve passing through the point P = (x0 , y0 ) is given by the coordinates γ (t) = (vx (P )t + x0 , vy (P )t + y0 ) ,

(10.19)

so the integral curve passing through (x, y) in our example is given by γ (t) = ((x − y )t + x, (−x + y )t + y )

(10.20)

= Φt (x, y) , the flow of v . So the solution is f (x, t) = Φt f (x, 0) = f (x, 0) ◦ Φt (x, y ) ∗

= [(x − y)t + x]2 + [(−x + y )t + y ]2 = (x − y)2 t2 + x2 + 2(x − y)xt + ( x − y )2 t2 + y 2 − 2(x − y)yt = 2(x − y)2 t2 + (x2 + y 2 )(1 + 2t) − 4xyt. (10.21)


Chapter 11

Lie derivative Given some diffeomorphism ϕ, we have Eq. (10.5) for pushforwards and pullbacks, ∗ (ϕ∗ v )f = ϕ−1 v (ϕ∗ (f )) . (11.1)

  We will apply this to the flow of a vector field     ( ) = ( ) φt

d φ∗ f dt t

t=0

u f

u , defined by

.

(11.2)

(11.3)

Q

Applying this at −t , we get

1 φ−t∗ v (f ) = φ− −t

∗

   ( )  ( ) = v φ∗−t f

φ∗t v φ∗−t f

,

1 where we have used the relation φ − = φ −t . Let us differentiate this t equation with t ,

d d (φ−t∗ v )(f ) = φ∗t v φ∗−t (f ) (11.4) dt dt t=0 t=0 On the right hand side, φ ∗t acts linearly on vectors and v acts linearly on functions, so we can imagine A t = φ ∗t v as a kind of linear operator acting on the function f t = φ∗−t f . Then the right hand side is of



the form d At f t dt

 

t=0



 

      = +        = +   − ( ( ))  = ( ( ))   =[ ]( )

d d At f t At f t dt dt t=0 t=0 d ∗ d ∗ At φ v f t φ−t (f ) dt t dt t=0

u v f

u , v f

v u f

t=0

t=0

.

37

t=0

 

t=0

(11.5)


Chapter 11. Lie derivative

The things in the numerator are numbers, so they can be compared at different points, unlike vectors which may be compared only on the same space. We can also write this as lim

φt∗ vφt (P ) − vP t

t→0

= [ u , v] .

(11.6)

This has the look of a derivative, and it can be shown to have • the properties of a derivation on the module of vector fields, appropriately defined. So the Lie bracket is also called the Lie derivative, and written as £u v = [u , v ] . (11.7) The derivation on functions by a vector field u : C ∞ (M) → C ∞(M) , f →  u (f ) , can be defined similarly as φ∗t f − f u(f ) = lim . t→0 t

(11.8)

• So this can also be called the Lie derivative of f with respect to u , and written as £u f .  Then it is easy to see that £u (f g )

and So

£u is

£u (f + ag )

= (£u f ) g + f ( £u g ) , =

£u f + a£u g .

(11.9)

a derivation on the space C ∞(M) . Also, £u (v + aw )

and

£u (f v )

=

£u v + a£u w ,

= (£u f ) v + f £u v

). ∀f ∈ C ∞(M(11.10)

So £u is a derivation on the module of vector fields. Also, using Jacobi identity, we see that £u ( v • w )

= (£u v) w + v (£u w ) , •

•

(11.11)

where v w = [v , w] , so £u is a derivation on the Lie algebra of vector fields. Lie derivatives are useful in physics because they describe invariances. For functions, £u f = 0 means φ∗t f = f , so the function does not change along the flow of u . So the flow of u preserves f , or leaves f invariant. •


If there are two vector fields u and v which leave f invariant, £u f = 0 = £v f . But we know from the Eq. (11.8), which defines the Lie derivative of a function that £u+av f = £u f + a£v

and

=0

[£u , £v ]f = £[u ,v] f = 0 .

∀a ∈

R

(11.12)

So the vector fields which preserve f form a Lie algebra. Similarly, a vector field is invariant under a diffeomorphism ϕ if ϕ∗ v = v , as mentioned earlier. Using the flow of u , we find that a vector field v is invariant under the flow of u if φ−t v = v

⇒

£u v

= v.

(11.13)

So if a vector field w is invariant under the flows of u and v , i.e. if £u w = 0 = £v w , we find that 0 = £u £v w − £v £u w =

£[u ,v ] w .

(11.14)

Thus again the vector fields leaving w invariant form a Lie algebra. • Let us also define the corresponding operations for 1-forms. As we mentioned in Chap. 6, a 1−form is a section of the cotangent bundle

T ∗ M =



∗ T P M.

(11.15)

P

Alternatively, a 1-form is a smooth linear map from the space of vector fields on M to the space of smooth functions on M , ω : v → ω (v ) ∈ C ∞(M),

ω (u + av ) = ω (U ) + aω (v) . (11.16)

A 1-form is a rule that (smoothly) selects a cotangent vector at each point.  Given a smooth map ϕM1 → M2 (say a diffeomorphism, for • convenience), the pullback ϕ∗ ω is defined by (ϕ∗ ω ) (v ) = ω (ϕ∗ ω ) .

(11.17)

We have already seen the gradient 1-form for a function f : • M → R , which is a linear map from the space of vector fields to functions, df (u + av ) = u (f ) + av (f ) , (11.18)



and which can be written as ∂f i dx ∂x i

df =

(11.19)

in some chart.  For an arbitrary 1-form ω , we can write in a chart and for any vector field v , ω = ω i dxi ,

v = v i

∂ , ∂x i

ω (v ) = ω i v i .

(11.20)

All the components ωi , v i are smooth functions, so is ωI v i . The space of 1-forms is a module. Since the function ω (v ) is chartindependent, we can find the components ωi of ω in a new chart by noting that (11.21) ω (v ) = ω i v i = ω i v i . 



Note that the notation is somewhat ambiguous here – i also runs from 1 to n , and the prime actually distinguished the chart, or the coordinate system, rather than the index i . If the components of v in the new chart are related to those in the old one by v i = A ji v j , it follows that 





ωi A ji v j = ω j v j



ωi A ji = ω j

⇒





(11.22)

Since coordinate transformations are invertible, we can multiply both sides of the last equation by A−1 and write ωi = A−1

j ω . i j

 



(11.23) 

For coordinate transformations from a chart { xi } to a chart { xi } , 

∂x i , = ∂x j ∂x i j vi = v , ∂x j  A ji





so

A−1

j i

  ωi = 

=

∂x j (11.24) ∂x i 

∂x j ω j . (11.25) ∂x i 

We can define the Lie derivative of a 1-form very conveniently by going to a chart, and treating the components of 1-forms and vector fields as functions, £u ω (v )

∂ ωi v i j ∂x ∂ω ∂v i i i j j u v u ω . = + i ∂x j ∂x j

= £u ωi v i = u j

 

 

(11.26)


But we want to define things such that £u ω (v )

= (£u ω ) (v ) + ω (£u v) .

(11.27)

We already know the left hand side of this equation from Eq. (11.26), and the right hand side can be calculated in a chart as (£u ω ) (v ) + ω (£u v) = (£u ω )i v i + ωi (£u v)i = (£u ω )i v i + ωi [u , v ]i i

= (£u ω )i v + ωi

i j ∂ v u ∂x j



i j ∂u − v ∂x j



. (11.28)

Equating the right hand side of this with the right hand side of Eq. (11.26), we can write ∂ωi (£u ω )i = u j j ∂x

∂u j + ω j i . ∂x

(11.29)

These are the components of £u ω in a given chart { xi } . For the sake of convenience, let us write down the Lie derivatives of the coordinate basis vector fields and basis 1-forms. The coordinate basis vector corresponding to the i-th coordinate is v =

∂ ∂x i

⇒

j

v j = δ i .

(11.30)

Putting this into the formula for Lie derivatives, we get £u

∂ j ∂ = [ ] u , v ∂x i ∂x j j j k ∂v k ∂u = u − v ∂x k ∂x k ∂u j ∂ = 0 − δ ik k ∂x ∂x j ∂u j ∂ . =− ∂x i ∂x j

 



∂ ∂x j



 

(11.31)

Similarly, the 1-form corresponding to the i-th basis coordinate is dxi = δ ji dx j ,

i.e.

dxi

 = j

δ ji .

(11.32)

Using this in the formula Eq. (11.29) we get £u dx

i

= δ ki

∂u k j ∂u i j dx = j dx . ∂x j ∂x

(11.33)



There is also a geometric description of the Lie derivative of 1forms, £u ω |

P

= lim

t→0

=

1 t

  

d ∗ φ ω dt t

∗

φt ω |φ

t (P )

.

− ωP



(11.34)

P

We will not discuss this in detail, but only mention that it leads to the same Leibniz rule as in Eq. (11.27), and the same description in terms of components as in Eq. (11.29).


Chapter 12

Tensors So far, we have defined tangent vectors, cotangent vectors, and also vector fields and 1-forms. We will now define tensors. We will do this by starting with the example of a specific type of tensor. • A (1, 2) tensor AP at P ∈ M is a map AP : T P M × T P M × T P ∗ M → R

(12.1)

which is linear in every argument. So given two vectors uP , vP and a covector ωP , AP : (uP , vP , ωP ) → A P (uP , vP ; ωP ) ∈



R.

(12.2)

Suppose {ea }, {λa } are bases for T P M, T P M . Write ∗

Acab = A P (ea , eb ; λc ) .

(12.3)

Then for arbitrary vectors u P = u a ea , vP = v a ea , and covector ω P = ωa λa we get using linearity of the tensor map,



AP (uP , vP ; ωP ) = AP ua ea , v b eb ; ωc λc

= ua vb ωc Acab .



(12.4)

It is a matter of convention whether A as written above should be called a (1, 2) tensor or a (2, 1) tensor, and the convention varies between books. So it is best to specify the tensor by writing indices as there is no confusion about Acab . A tensor of type ( p, q ) can be defined in the same way, ∗ ∗ A p,q : T P M × · · · × T P M × T P M × · · · × T P M P





q times

 



p times

43



→

R

(12.5)


Chapter 12. Tensors

in such a way that the map is linear in every argument. • Alternatively, AP is an element of the tensor product space ∗ ∗ AP ∈ T P M ⊗ · · · ⊗ T P M ⊗ T P M ⊗ · · · ⊗ T P M





 

p times





q times

(12.6)

We can define the components of this tensor in the same way that we did for the (1, 2) tensor. Then a ( p, q ) tensor has components which can be written as a1 ···ap 1 ···bq

Ab

.

Some special types of ( p, q ) tensors have special names. A (1, 0) • tensor is a linear map AP : T P M → R , so it is a tangent vector. A (0, 1) tensor is a cotangent vector. A ( p, 0) tensor has components with p upper indices. It is called a contravariant p −tensor. A (0, q ) tensor has components with q lower indices. It is called a covariant q −tensor.  It is possible to add tensors of the same type, but not of different types, a a a a a a Ab bqp + Bb bqp = (A + B )b bqp . (12.7) ∗

1 ···

1 ···

1 ···

1 ···

1 ···

1 ···

•

A tensor field is a rule giving a tensor at each point.  We can now define the Lie derivative of a tensor field by using Leibniz rule in a chart. Let us first consider the components of a tensor field in a chart. For a (1, 2) tensor field A , the components in a chart are ∂ ∂ (12.8) Akij = A ( i , j ; dxk ) . ∂x

∂x

The components are functions of x in a chart. Thus we can write this tensor field as A = A kij dxi ⊗ dx j ⊗

∂ , ∂x k

(12.9)

where the × indicates a ‘product’, in the sense that its action on two vectors and a 1-form is a product of the respective components,



∂ dx ⊗ dx ⊗ ∂x k i

j



(u, v; ω) = u i v j ωk .

(12.10)

Thus we find, in agreement with the earlier definition, A(u, v ; ω ) = A kij ui v j ωk .

(12.11)




Under a change of charts, i.e. coordinate system xi → x i , the components of the tensor field change according to 

A = A kij dxi ⊗ dx j ⊗

Since

∂ ∂ = Aki j dxi ⊗ dx j ⊗ k k ∂x ∂x 









dx

∂x i dxi , = i ∂x

i





∂ ∂x i ∂ = ∂x i ∂x i ∂x i 

(12.12)





(12.13)



(i and i are not equal in general), we get 

Akij



∂ ∂x i ∂x  j ∂x k ∂ k i j ⊗ ⊗ dx ⊗ dx ⊗ A dx dx . = i j ∂x k ∂x i ∂x j ∂x k ∂x k i



j







(12.14) Equating components, we can write 

Akij



∂x i ∂x  j ∂x k ∂x i ∂x j ∂x k ∂x i ∂x j ∂x k . ∂x i ∂x  j ∂x k

=

 Aki j 

=

Akij



(12.15)

(12.16)



 Aki j 





∂

∂f

From now on, we will use the notation ∂ i for and ∂ i f for ∂x i ∂x i unless there is a possibility of confusion. This will save some space and make the formulae more readable. We can calculate the Lie derivative of a tensor field (with respect to a vector field u, say) by using the fact that £u is a derivative on the modules of vector fields and 1-forms, and by assuming Leibniz rule for tensor products. Consider a tensor field T = T am······b n ∂ m ⊗ · · · ⊗ ∂ n ⊗ dxa ⊗ · · · ⊗ dxb .

(12.17)

Then £u T

= (£u T am b n ) ∂ m ⊗ · · · ⊗ ∂ n ⊗ dxa ⊗ · · · ⊗ dxb +T am b n (£u ∂ m ) ⊗ · · · ⊗ ∂ n ⊗ dxa ⊗ · · · ⊗ dxb + · · · +T am b n ∂ m ⊗ · · · ⊗ ∂ n ⊗ (£u dxa ) ⊗ · · · ⊗ dxb , + · · · ···

···

···

···

···

···

(12.18) where the dots stand for the terms involving all the remaining upper and lower indices. Since the components of a tensor field are functions on the manifold, we have m···n

£u T a···b

= u i ∂ i T am b n , ···

···

(12.19)


Chapter 12. Tensors

and we also know that ∂u i £u ∂ m = − ∂ i , ∂x m

∂u a i £u dx = dx . ∂x i a

(12.20)

Putting these into the expression for the Lie derivative for T and relabeling the dummy indices, we find the components of the Lie derivative, (£u T )m a

n b

···

···

= ui ∂ i T am b n ···

···

− T ai nb ∂ i um − · · · − T am b i ∂ i un + T im b n ∂ a ui + · · · + T am i n ∂ b ui . (12.21) ··· ···

···

···

···

···

···

···


Chapter 13

Differential forms There is a special class of tensor fields, which is so useful as to have a separate treatment. There are called differential p −forms or p−forms for short. • A p −form is a (0, p) tensor which is completely antisymmetric, i.e., given vector fields v1 , · · · , v p , ω (v1 , · · · , vi , · · · , v j , · · · , v p ) = −ω (v1 , · · · , v j , · · · , vi , · · · , v p ) (13.1) for any pair i, j .  A 0-form is defined to be a function, i.e. an element of C (M) , and a 1-form is as defined earlier. The antisymmetry of any p-form implies that it will give a nonzero result only when the p vectors are linearly independent. On the other hand, no more than n vectors can be linearly independent in an n-dimensional manifold. So p  n . Consider a 2-form A . Given any two vector fields v 1 , v2 , we have A(v1 , v2 ) = − A(v2 , v1 ) . Then the components of A in a chart are ∞

Aij = A (∂ i , ∂ j ) = − A ji .

(13.2)

Similarly, for a p-form ω , the components are ωi1 ip , and components are multiplied by (−1) whenever any two indices are interchanged. n It follows that a p-form has independent components in n p dimensions. Any 1-form produces a function when acting on a vector field. So given a pair of 1-forms A,B, it is possible to construct a 2-form ω ···

 47


Chapter 13. Differential forms

by defining ω(u, v) = A(u)B(v) − B(u)A(v),

∀u , v .

(13.3)

This is usually written as ω = A ⊗ B − B ⊗ A , where ⊗ is called • the outer product.  • Then the above construction defines a product written as ω = A ∧ B = − B ∧ A ,

(13.4)

and called the wedge product . Clearly, ω is a 2-form.  Let us work in a coordinate basis, but the results we find can be generalized to any basis. The coordinate bases for the vector fields, {∂ i } , and 1-forms, {dxi } , satisfy dxi (∂ j ) = δ ji . A 1-form A can be written as A = A i dxi , and a vector field v can be written as v = v i ∂ i , so that A(v) = A i vi . Then for the ω defined above and for any pair of vector fields u,v, ω(u, v) = A(u)B(v) − B(u)A(v) = Ai ui B j v j − Bi ui A j v j = (Ai B j − Bi A j ) ui v j .

(13.5)

The components of ω are ωij = ω(∂ i , ∂ j ) , so that ω(u, v) = ω(ui ∂ i , v j ∂ j ) = ω ij ui v j .

(13.6)

Then ω ij = A i B j − Bi A j for the 2-form defined above. We can now construct a basis for 2-forms, which we write as dx i ∧ dx j , dxi ∧ dx j = dx i ⊗ dx j − dx j ⊗ dxi .

(13.7)

Then a 2-form can be expanded in this basis as ω =

1 ωij dxi ∧ dx j , 2!

(13.8)

because then 1 ωij dxi ⊗ dx j − dx j ⊗ dxi (u, v) 2! 1 = ωij ui v j − u j v i = ω ij ui v j . 2!

ω(u, v) =

 





(13.9)


Similarly, a basis for p −forms is dxi1 ∧ · · · ∧ dxip = dx [i1 ⊗ · · · ⊗ dxip ] ,

(13.10)

where the square brackets stand for total antisymmetrization: all even permutations of the indices are added and all the odd permutations are subtracted. (Caution: some books define the ‘square brackets’ as antisymmetrization with a factor 1/p! .) For example, for a 3-form, a basis is dxi ∧ dx j ∧ dxk = dxi ⊗ dx j ⊗ dxk − dx j ⊗ dxi ⊗ dxk +dx j ⊗ dxk ⊗ dxi − dxk ⊗ dx j ⊗ dxi +dxk ⊗ dxi ⊗ dx j − dxi ⊗ dxk ⊗ dx j . (13.11) Then an arbitrary 3-form Ω can be written as Ω=

1 Ωijk dxi ∧ dx j ∧ dxk . 3!

(13.12)

Note that there is a sum over indices, so that the factorial goes away if we write each basis 3-form up to permutations, i.e. treating different permutations as equivalent. Thus a p−form α can be written in terms of its components as α =

1 αi p! 1

ip dx

···

i1

∧ · · · ∧ dxip .

(13.13)

Examples: A 2-form in two dimensions can be written as

1 ωij dxi ∧ dx j 2! 1 = ω12 dx1 ∧ dx2 + ω21 dx2 ∧ dx1 2! 1 = (ω12 − ω21 ) dx1 ∧ dx2 2! = ω12 dx1 ∧ dx2 .

ω=



 (13.14) 

A 2-form in three dimensions can be written as 1 ωij dxi ∧ dx j 2! = ω12 dx1 ∧ dx2 + ω23 dx2 ∧ dx3 + ω31 dx3 ∧ dx1 (13.15)

ω=


Chapter 13. Differential forms 

i

i

In three dimensions, consider two 1-forms α = α i dx , β = β i dx . Then 1 α ∧ β = (αi β j − α j β i ) dx i ∧ dx j 2! i j = αi β j dx ∧ dx (α1 β 2 − α2 β 1 ) dx1 ∧ dx2

=

+ (α2 β 3 − α3 β 2 ) dx2 ∧ dx3 + (α3 β 1 − α1 β 3 ) dx3 ∧ dx1 .

(13.16)

The components are like the cross product of vectors in three dimensions. So we can think of the wedge product as a generalization of the cross product. • We can also define the wedge product of a p−form α and a q −form β as a ( p + q )−form satisfying, for any p + q vector fields v1 , · · · , v p+q , 1 α ∧ β (v1 , · · · , v p+q ) = (−1)deg P α ⊗ β (P (v1 , · · · , v p+q )) . p!q ! P (13.17) Here P stands for a permutation of the vector fields, and deg P is 0 or 1 for even and odd permutations, respectively. In the outer product on the right hand side, α acts on the first p vector fields in a given permutation P , and β acts on the remaining q vector fields.  The wedge product above can also be defined in terms of the components of α and β in a chart as follows. 1 α = αi1 ip dxi1 ∧ · · · ∧ dxip p! 1 β = β j1 jq dx j1 ∧ · · · ∧ dx jq q ! 1 α ∧ β = αi i β j j dxi1 ∧ · · · ∧ dxip ∧ dx j1 ∧ · · · ∧ dx jq . p!q ! 1 p 1 q (13.18)



···

···

···

···



 



Note that α ∧ β = 0 if p + q > n, and that a term in which some i is equal to some j must vanish because of the antisymmetry of the wedge product. It can be shown by explicit calculation that wedge products are associative, α ∧ (β ∧ γ ) = (α ∧ β ) ∧ γ . (13.19)


Cross-products are not associative, so there is a distinction between cross-products and wedge products. In fact, for 1-forms in three dimensions, the above equation is analogous to the identity for the triple product of vectors, a × (b × c )

= (a × b ) × c .

(13.20)

For a p-form α and q -form β , we find α ∧ β = (−1) pq β ∧ α .

(13.21)

Proof: Consider the wedge product written in terms of the com-

ponents. We can ignore the parentheses separating the basis forms since the wedge product is associative. Then we exchange the basis 1-forms. One exchange gives a factor of − 1 , dxip ∧ dx j1 = −dx j1 ∧ dxip .

(13.22)

Continuing this process, we get dxi1 ∧ · · · ∧ dxip ∧ dx j1 ∧ · · · ∧ dx jq = (−1) p dx j1 ∧ dxi1 ∧ · · · ∧ dxip ∧ dx j2 ∧ · · · ∧ dx jq = · · · = (−1) pq dx j1 ∧ · · · ∧ dx jq ∧ dxi1 ∧ · · · ∧ dxip . (13.23) Putting back the components, we find α ∧ β = (−1) pq β ∧ α

(13.24)

as wanted.  The wedge product defines an algebra on the space of differential • forms. It is called a graded commutative algebra .  Given a vector field v , we can define its contraction with a • p-form by ιv ω = ω(v, · · · ) (13.25) with p − 1 empty slots. This is a ( p − 1)-form. Note that the position of v only affects the sign of the contracted form.  Example: Consider a 2-form made of the wedge product of two 1-forms, ω = λ ∧ µ = λ ⊗ µ − µ ⊗ λ . Then contraction by v gives ιv ω = ω(v, ) = λ(v)µ − µ(v)λ = −ω( , v) . •

•

(13.26)


Chapter 13. Differential forms

If we have a p-form ω = p1! ωi1 ip dxi1 ∧ · · · ∧ dxip , its contraction with a vector field v = v i ∂ i is 1 ιv ω = ω ii2 ip v i dxi2 ∧ · · · ∧ dxip . (13.27) ( p − 1)! ···

···

Note the sum over indices. To see how the factor becomes ( p 11)! , we write the contraction as 1 ιv ω = ω i1 ip dxi1 ∧ · · · ∧ dxip v i ∂ i . (13.28) p! Since the contraction is done in the first slot, so we consider the action of each basis 1-form dxik on ∂ i by carrying dxik to the first position and then writing a δ iik . This gives a factor of (−1) for each exchange, but we get the same factor by rearranging the indices of ω , thus getting a +1 for each index. This leads to an overall factor of p . given a diffeomorphism ϕ : M1 → M2 , the pullback of a 1• form λ (on M 2 ) is ϕ λ , defined by −

 

···

∗

ϕ λ(v) = λ(ϕ v)

(13.29)

∗

∗

for any vector field v on M 1 .  i i Then we can consider the pullback ϕ dx of a basis 1-form dx . For a general 1-form λ = λ i dxi , we have ϕ λ = ϕ (λi dxi ) . But ∗

∗

∗

ϕ λ(v) = λ(ϕ v) = λ i dxi (ϕ v) . ∗

∗

∗

(13.30)

Now, dxi (ϕ v) = ϕ dxi (v) and the thing on the right hand side is a function on M 1 , so we can write this as ∗

∗

ϕ λ(v) = (ϕ λi )ϕ dxi (v) , ∗

∗

∗

(13.31)

where ϕ λi are now functions on M 1 , i.e. ∗

(ϕ λi )|P = λi |ϕ(P )

(13.32)

∗

So we can write ϕ λ = (ϕ λi ) ϕ dxi . For the wedge product of two 1-forms, ∗

∗

∗

ϕ (λ ∧ µ)(u, v) = (λ ∧ µ)(ϕ u , ϕ v) ∗

∗

∗

= λ ⊗ µ(ϕ u , ϕ v) − µ ⊗ λ(ϕ u , ϕ v) ∗

∗

∗

∗

= λ(ϕ u)µ(ϕ v) − µ(ϕ u)λ(ϕ v) = ϕ λ(u)ϕ µ(v) − ϕ µ(u)ϕ λ(v) ∗

∗

∗

∗

∗

∗

= (ϕ λ ∧ ϕ µ)(u , v) . ∗

∗

∗

∗

(13.33)


Since u, v are arbitrary vector fields it follows that ϕ (λ ∧ µ) = ϕ λ ∧ ϕ µ ϕ (dxi ∧ dx j ) = ϕ dxi ∧ ϕdx j . ∗

∗

∗

∗

∗

(13.34)

Since the wedge product is associative, we can write (by assuming an obvious generalization of the above formula)

      ϕ dx ∧ dx ∧ dx = ϕ dx ∧ dx ∧ dx   = ϕ dx ∧ dx ∧ ϕ dx ∗

i

j

k

i

∗

j

i

∗

j

k

∗

k

= ϕ dxi ∧ ϕ dx j ∧ ϕ dxk , (13.35) ∗

∗

∗

and we can continue this for any number of basis 1-forms. So for any p-form ω , let us define the pullback ϕ ω by ∗

ϕ ω(v1 , · · · , v p ) = ω (ϕ v1 , · · · , ϕ v p ) ,

∗

∗

∗

(13.36)

and in terms of components, by ϕ ω = ∗

1 ϕ ωi1 p!



∗

∗

ip

···

i1

 ϕ dx

∧ · · · ∧ dxip .

(13.37)

We assumed above that the pullback of the wedge product of a 2-form and a 1-form is the wedge product of the pullbacks of the respective forms, but it is not necessary to make that assumption – it can be shown explicitly by taking three vector fields and following the arguments used earlier for the wedge product of two 1-forms. Then for any p-form α and q -form β we can calculate from this that ϕ (α ∧ β ) = ϕ α ∧ ϕ β . (13.38) ∗

∗

∗

Thus pullbacks commute with (are distributive over) wedge products.


Chapter 14

Exterior derivative The exterior derivative is a generalization of the gradient of a function. It is a map from p-forms to ( p + 1)-forms. This should be a derivation, so it should be linear, d(α + ω ) = dα + dω

∀ p-forms α , ω .

(14.1)

This should also satisfy Leibniz rule, but the algebra of p-forms is not a commutative algebra but a graded commutator algebra, i.e., involves a factor of (−1) pq for exchanges. So we need d(α ∧ β ) = dα ∧ β + ( −1) pq dβ ∧ α ,

(14.2)

d(α ∧ β ) = dα ∧ β + ( −1) p α ∧ dβ .

(14.3)

or alternatively,

This will be the Leibniz rule for wedge products. Note that it gives the correct result when one or both of α, β are 0-forms, i.e., functions. The two formulas are identical by virtue of the fact that dβ is a (q + 1)-form, so that α ∧ dβ = (−1) p(q+1) dβ ∧ α .

(14.4)

We will try to define the exterior derivative in a way such that it has these properties. Let us define the exterior derivative of a p-form ω in a chart as dω =

1 ∂ i ωi p!

i

1 ··· p

dxi ∧ · · · ∧ dxi

54

1

p

(14.5)

55 1 1 0 2 s t s i c i s y h P r o f y r t e m o e G l a i t n e r e f f i D n o s e t o N e r u t c e L : i r i h a L a h b a t i m A

This clearly has the first property of linearity. To check the (graded) Leibniz rule, let us write α ∧ β in components. Then 1 ∂ i αi i β j j dxi ∧ dxi ∧ · · · ∧ dx j p!q ! 1 = ∂ i αi i β j j + αi i ∂ i β j j dxi ∧ dxi ∧ · · · ∧ dx j p!q ! 1 = ∂ i αi i β j j dx i ∧ dxi ∧ · · · ∧ dxi ∧ dx j ∧ · · · dx j p!q ! 1 + ( −1) p αi i ∂ i β j j dxi ∧ · · · ∧ dxi ∧ dxi ∧ dx j ∧ · · · dx j p!q ! = dα ∧ β + ( −1) p α ∧ dβ . (14.6)

  

d(α ∧ β ) =

1 ··· p

1 ··· q

 

1 ··· p

1 ··· p



1

1 ··· q

1 ··· p



1 ··· q



1

1 ··· q

1 ··· p

q



1 ··· q



1

1

p

1

q

1

p

A third property of the exterior derivative immediately follows from here, d2 = 0 . (14.7) To see this, we write 1 d ∂ i ωi p! 1 = ∂ j ∂ i ωi p!



d(dω ) =

i

1 ··· p

i

1 ··· p

dxi ∧ dxi ∧ · · · dxi 1

p



dx j ∧ dxi ∧ dxi ∧ · · · dxi . (14.8) 1

p

But the wedge product is antisymmetric, dx j ∧ dx i = −dxi ∧ dx j , and the indices are summed over, so the above object must be antisymmetric in ∂ j , ∂ i . But that vanishes. So d2 = 0 on all forms. Note that we can also write dω =

1 dωi p!



i

1 ··· p



i1

∧ dx

ip

∧ · · · dx

,

(14.9)

where the object in parentheses is a gradient 1-form corresponding to the gradient of the component. Consider a 1-form A = A µ dxµ where A µ are smooth functions on M . Then using this definition we can write dA = (dAν ) ∧ dxν

c  ⇒

(dA)µν

= ∂ µ Aν dxµ ∧ dxν 1 = ( ∂ µ Aν − ∂ ν Aµ ) dxµ ∧ dxν 2 = ∂ µ Aν − ∂ ν Aµ .

q

(14.10)

q


Chapter 14. Exterior derivative

We can generalize this result to write for a p-form, 1 αµ µ dxµ ∧ · · · ∧ dxµ (14.11) p! 1 dα = ∂ µ αµ µ dxµ ∧ · · · ∧ dxµ p! 1 ∂ [µ αµ µ ] dxµ ∧ dxµ ∧ · · · ∧ dxµ = ( p + 1)! (14.12) µ = ∂ [µ αµ µ ] α=

1 ···



1

1 ···

p



1 ···

⇒

(dα)µµ

1 ···

1 ···

p

p

p

1

p

1

p

p

p

Example: For p = 1 i.e. for a 1-form A we get from this formula

(dA)µν = ∂ µ Aν − ∂ ν Aµ , in agreement with our previous calculation. For p = 2 we have a 2-form, call it α. Then using this formula we get (dα)µνλ = ∂ [µ ανλ ] = ∂ µ ανλ − ∂ ν αµλ + ∂ ν αλµ − ∂ λ ανµ + ∂ λ αµν − ∂ µ αλν . (14.13) Note that d is not defined on arbitrary tensors, but only on forms. 

By definition, d 2 = 0 on any p -form. So if α = dβ , it follows that dα = 0 . But given a p-form α for which dα = 0 , can we say that there must be some ( p − 1)-form β such that α = dβ ? This is a good place to introduce some terminology. Any form • ω such that dω = 0 is called closed, whereas any form α such that α = dβ is called exact.  So every exact form is closed. Is every closed form exact? The answer is yes, in a sufficiently small neighbourhood. We say that every closed form is locally exact. Note that if a p-form α = dβ , we cannot uniquely specify the ( p − 1)-form β since for any ( p − 2)-form γ , we can always write α = dβ , where β = β + dγ . Thus a more precise statement is that given any p-form α such that dα = 0 in a neighbourhood of some point P , there is some neighbourhood of this point and some ( p−1)-form β such that α = dβ in that neighbourhood. But this may not be true globally. This statement is known as the Poincare´ lemma.  2 Example: In R remove the origin. Consider the 1-form 

α =



xdy − ydx . x2 + y 2

(14.14)


Then 1 2x2 1 dα = dx ∧ dy − − 2 2 2 2 x + y x + y 2 (x2 + y 2 ) 2 x2 + y 2 dx dx ∧ dy = 0 . = 2 ∧ dy − 2 x + y 2 (x2 + y 2 )2







−

2y 2 (x2 + y 2 )2



dy ∧ dx

(14.15)

Introduce polar coordinates r, θ with x = r cos θ , y = r sin θ . Then dx = dr cos θ − r sin θdθ

dy = dr sin θ + r cos θdθ

r cos θ (sin θdr + r cos θdθ ) r sin θ (cos θdr − r sin θdθ ) − r2 r2 r 2 cos2 θ + sin2 θ dθ = = dθ . (14.16) r2

α=





Thus α is exact, but θ is multivalued so there is no function f such that α = df everywhere. In other words, α = dθ is exact only in a neighbourhood small enough that θ remains single-valued.


Chapter 15

Volume form The space of p-forms in n dimensions is



n dimensional. So the p

space of n-forms in n dimensions is 1-dimensiona, i.e., there is only one independent component, and all n-forms are scalar multiples of one another.  0 at some point Choose an n-form field. Call it ω . Suppose ω = P . Then given any basis {eµ } of T P M , we have ω (e1 , · · · en ) =  0 since ω  = 0 . Thus all vector bases at P fall into two classes, one for which ω (e1 , · · · en ) > 0 and the other for which it is < 0 . Once we have identified these two classes, they are independent of ω . That is, if ω is another n -form which is non-zero at P , there must  0 such that ω = f ω . Two bases which gave be some function f = positive numbers under ω will give the same sign — both positive or both negative — under ω and therefore will be in the same class. So every basis (set of n linearly independent vectors) is a mem• ber of one of the two classes. These are called righthanded and  lefthanded. A manifold is called orientable if it is possible to define a con• tinuous n -form field ω which is non-zero everywhere on the manifold. Then it is possible to choose a basis with the same handedness everywhere on the manifold continuously.  Euclidean space is orientable, the Möbius band is not. An orientable manifold is called oriented once an orientation • has been chosen, i.e. once we have decided to choose basis vectors with the same handedness everywhere on the manifold.  • It is necessary to choose an oriented manifold when we discuss the integration of forms. On an n-dimensional manifold, a set of n 





58


linearly independent vectors define an n -dimensional parallelepiped. If we define an n-form ω  = 0 we can think of the value of these vectors as the volume of this parallelepiped. This ω is called a volume form. 

Once a volume form has been chosen, any set of n linearly independent vectors will define a positive or negative volume. The integral of a function f on Rn is the sum of the values of f , multiplied by infinitesimal volumes of coordinate elements. Similarly, we define the integral of a function f on an oriented manifold as the sum of the values of f , multiplied by infinitesimal volumes. The way to do that is the following. Given a function f , define an n-form in a chart by ω = f dx1 ∧···∧ dxn . To integrate over an open set U , divide it up into infinitesimal ‘cells’, spanned by vectors



∂ ∂ ∂ ∆x1 1 , ∆x2 2 , · · · , ∆xn n ∂x ∂x ∂x



,

where the ∆xi are small numbers. Then the integral of f over one such cell is approximately f ∆x1 ∆x2 · · · ∆xn = f dx1 ∧ · · · ∧ dxn ∆x1 ∂ 1 , · · · , ∆xn ∂ n

= ω (cell ) .





(15.1)

Adding up the contributions from all cells and taking the limit of cell size going to zero, we find

 U

ω =



f dn x .

(15.2)

ϕ(U )

The right hand side is the usual integration in calculus of n variables, and the left hand side is our notation which we are defining. The right hand side can be seen to be independent of the choice of coordinate system. If we choose a different coordinate system, we get a Jacobian, but also a redefinition of the region ϕ(U ) . Let us check that the left hand side is also invariant of the choice of the coordinates. We will do this in two dimensions with ω = f dx1 ∧ dx2 .


Chapter 15. Volume form

In another coordinate system (y1 , y 2 ) corresponding to ϕ (U ) 

∂x 1 1 ∂ x1 2 dx = dy + dy ∂y 1 ∂y 2 ∂x 2 1 ∂ x2 2 2 dx = dy + dy ∂y 1 ∂y 2 ∂x 1 ∂x 2 ∂x 1 ∂x 2 1 2 ∧ − dx dx = ∂y 1 ∂y 1 ∂y 2 ∂y 2 1

⇒



= J dy1 ∧ dy2 ,



dy 1 ∧ dy 2

(15.3)

and J is the Jacobian. So what we have here is

  = (  = ( 

f x1 , x2 )dx1 ∧ dx2

ω

U

U

f y 1 , y 2 )Jdy 1 ∧ dy 2

U

f (y 1 , y2 )Jd2 y ,

=

(15.4)

ϕ (U )

so we get the same result both ways. Given the same f , if we choose a basis with the opposite orientation, the integral of ω will have the opposite sign. This is why the choice of orientation has to be made before integration. Manifolds become even more interesting if we define a metric.


Chapter 16

Metric tensor A metric on a vector space V is a function g : V × V → • which is i) bilinear:

R

g (av1 + v2 , w) = ag (v1 , w) + g (v2 , w) g (v, w1 + aw2 ) = g (v, w1 ) + ag (v, w2 ) ,

(16.1)

i.e., g is a (0,2) tensor; ii) symmetric: g (v, w) = g (w, v );

(16.2)

iii) non-degenerate:

∀w

g (v, w) = 0

⇒ v = 0 .

(16.3) 

If for some v, w  = 0 , we find that g(v, w) = 0 , we say that v, w • are orthogonal.  • Given a metric g on V , we can always find an orthonormal = ν and ± 1 if µ = ν . basis {eµ } such that g (eµ , eν ) = 0 if µ   • If the number of (+1)’s is p and the number of (−1)’s is q , we say that the metric has signature ( p, q ) . We have defined a metric for a vector space. We can generalize this definition to a manifold M by the following. A metric g on a manifold M is a (0, 2) tensor field such that if • (v, w) are smooth vector fields, g(v, w) is a smooth function on M , and has the properties (16.1), (16.2) and (16.3) mentioned earlier.  61


Chapter 16. Metric tensor

It is possible to show that smoothness implies that the signature is constant on any connected component of M , and we will assume that it is constant on all of M . A vector space becomes related to its dual space by the metric. Given a vector space V with metric g , and vector v defines a linear map g(v, ·) : V → R , w →  g (v, w) ∈ R . Thus g(v, ·) ∈ V where V is the dual space of V . But g(v, ·) is itself linear in v , so the map V → V defined by g (v, ·) is linear. Since g is non-degenerate, this map is an isomorphism. It then follows that on a manifold we can use the metric to define a linear isomorphism between vectors and 1-forms. In a basis, the components of the metric are gµν = g (eµ , eν ) . This is an n × n matrix in an n -dimensional manifold. We can thus write g (v, w) = gµν v µ wν in terms of the components. Non-degeneracy implies that this matrix is invertible. Let gµν denote the inverse matrix. Then, by definition of an inverse matrix, we have ∗

∗

∗

gµν g νλ = δ µλ = g λν gµν .

(16.4)

Then the linear isomorphism takes the following form. i) If v = v µ eµ is a vector field in a chart, and {λµ } is the dual basis to { eµ } , (16.5) g (v, ·) = v µ λµ ,

where vµ = g µν vν . ii) If A = Aµ λµ is a 1-form written in a basis {λµ } , the corresponding vector field is Aµ eµ , where Aµ = g µν Aν .

This is the isomorphism between vector fields and 1-forms. (We could of course define a similar isomorphism between vectors and covectors without referring to a manifold.) A similar isomorphism holds for tensors, e.g. in terms of components, T µν ←→ T µν ←→ T µ ν ←→ T µν T µνρ

···

←→ T µν ρ

···

←→ T µν ρ ←→ T µνρ ···

···

(16.6)

←→ · · · (16.7)

These correspondences are not equalities — the components are not equal. What it means is that, if we know one set of components, say T µνρ , and the metric, we also know every other set of components. ···


Using the fact that a non-degenerate metric defines a 1-1 linear • map between vectors and 1-forms, we can define an inner product of 1−forms, by A | B  = g µν Aµ Bν (16.8) for 1-forms A,B . This result is independent of the choice of basis, i.e. independent of the coordinate system, just like the inner product of vector fields ,

v | w  = g (v, w) = g µν v µ wν .

(16.9) 

Given a manifold with metric, there is a canonical volume form dV (sometimes written as v ol) , which in a coordinate chart reads

 = | det

dV

gµν |dx1 ∧ · · · ∧ dxn .

(16.10)

Note that despite the notation, this is not a 1-form, nor the gradient of some function V . This is clearly a volume form because it is an n-form which is non-zero everywhere, as gµν is non-degenerate. We need to show that this definition is independent of the chart. Take an overlapping chart. Then in the new chart, the corresponding volume form is 

dV

 = | det

 gµν |dx1 ∧ · · · ∧ dxn .

(16.11)



We wish to show that dV = dV . In the overlap, 

dx

µ

∂x µ ν dx = A µν dxν (say) = ν ∂x

(16.12)

Then dx 1 ∧ · · · ∧ dx n = (det A)dx1 ∧ · · · ∧ dxn . On the other hand, if we look at the components of the metric tensor in the new chart, 



 = g(∂ µ , ∂ ν  ) gµν

∂x α ∂x β ∂ , ∂ β α ∂x µ ∂x ν

  =      =    = g

A−1

A−1

α µ

α ∂ , µ α

β

β

A−1 ν ∂ β

A−1 ν g αβ .

(16.13)


Chapter 16. Metric tensor

Taking determinants, we find −2



det gµν = (det A) Thus 

 | det

gµν | = | det A| 

(det gµν ) .

−1

 | det

gµν | ,

(16.14)

(16.15)

and so dV = dV . • This is called the metric volume form and written as dV =

 | |

g dx1 ∧ · · · ∧ dxn

(16.16)

in a chart.  When we write dV , sometimes we mean the n-form as defined above, and sometimes we mean |g |dn x , the measure for the usual integral. Another way of writing the volume form in a chart is in terms of its components,



dV =

 | | g

n!

µ

µ

1 ···

n

dxµ ∧ · · · ∧ dxµ 1

n

(16.17)

where  is the totally antisymmetric Levi-Civita symbol, with 12 n = +1 . Thus |g | µ µ are the components of the volume form. ···



1 ···

n


Chapter 17

Hodge duality We will next define the Hodge star operator. We will defineit in a chart rather than abstractly. The Hodge star operator, denoted  in an n-dimensional • manifold is a map from p-forms to (n − p)-forms given by (ω )µ1

µn−p

···

≡

 | | g

p!

µ1 ···µn g µn

p+1 ν 1

−

· · · g µn ν p ων 1

ν p , (17.1)

···

where ω is a p-form.  The  operator acts on forms, not on components. gµν = Example: Consider R3 with metric +++, i.e. µν diag(1, 1, 1) . Then |g | ≡ g = 1 , g diag(1, 1, 1) . Write the coordinate basis 1-forms as dx,dy, dz . Their components are clearly (dx)i = δ i1 , (dy )i = δ i2 , (dz )i = δ i3 ,

(17.2)

the δ ’s on the right hand sides are Kroenecker deltas. So (dx)ij = ijk g kl (dx)l =  ijk g kl δ l1 =  ijk g k1 1 1 dx = (dx)ij dxi ∧ dx j = ijk g k1 dxi ∧ dx j ⇒ 2! 2! k1 g = 1 for k = 1 , 0 otherwise 1 dx = dx2 ∧ dx3 − dx3 ∧ dx2 = dx 2 ∧ dx3 = dy ∧ dz . ⇒ 2! (17.3)



Similarly, dy = dz ∧ dx ,



dz = dx ∧ dy .

65




Chapter 17. Hodge duality Example: Consider p = 0 (scalar), i.e. a 0-form ω in n dimen-

sions. (ω )µ1

···

(1)µ1

···

⇒

⇒

µn µn

 | |  = | |  | | =

g µ1 ···µn ω g µ1 ···µn

(1) =

g

n! = dV

µ1 ···µn dxµ1 ∧ · · · ∧ dxµn

(17.4) 

Example: p = n . Then

(ω ) =

 | | g

n!

µ1 ···µn g µ1 ν 1 · · · g µn ν n ων 1 ···ν n .

(17.5)

For the volume form, dV

 | | = !  = | | g

µ1 ···µn dxµ1 ∧ · · · ∧ dxµn

n (dV )ν 1 ···ν n g ν 1 ···ν n |g | (dV ) = µ1 ···µn g µ1 ν 1 · · · g µn ν n ν 1 ···ν n n! |g | |g | n! n!(det g )−1 = = = sign(g ) = (−1)s , (17.6) n! n! g

where s is the number of (−1) in gµν . So we find that



(1) = dV = (−1)s ,

(17.7)

and (dV ) = (−1)s (1) = (−1)s dV ,

(17.8)

i.e., ()2 = (−1)s on 0-forms and n-forms. In general, on a p -form in an n -dimensional manifold with signature (s, n − s) , it can be shown in the same way that ()2 = (−1) p(n

p)+s

−

.

(17.9)

In particular, in four dimensional Minkowski space, s = 1, n = 4 , so ()2 = (−1) p(4

p)+1

−

.

(17.10)


It is useful to work out the Hodge dual of basis p -forms. Suppose we have a basis p-form dxI 1 ∧ · · · ∧ dxI p , where the indices are arranged in increasing order I p > · · · > I 1 . Then its components are I p!δ µI 11 · · · δ µpp . So  dxI 1 ∧ · · · ∧ dxI p



 | | = !  = | | g

p



ν 1 ···ν n−p

ν 1 ···ν n



p µ1 ···µp

−

g ν 1 ···ν n

p µ1 ···µp

−

I



g µ1 µ1 · · · g µp µp p! δ µI 1 · · · δ µp 



p

1

g µ1 I 1 · · · g µp I p .

(17.11)

We will use this to calculate ω ∧ ω . For a p-form ω , we have ω =

=

1 ωµ p! 1



µp dx

···

µ1

∧ · · · ∧ dxµp

ωI 1 ···I p dxI 1 ∧ · · · ∧ dxI p

(17.12)

I

where the sum over I means a sum over all possible index sets I = I 1 < · · · < I p , but there is no sum over the indices {I 1 , · · · , I p } themselves, in a given index set the I k are fixed. Using the dual of basis p−forms, and Eq. (13.13), we get

 (  | |  =

ωI 1 ···I p  dxI 1 ∧ · · · ∧ dxI p )

ω =

I

I

g ν ···ν (n − p)! 1 n

−

p µ1 ···µp

g µ1 I 1 · · · g µp I p ωI 1 ···I p dxν 1 ∧ · · · ∧ dxν n

−

p

.

(17.13) The sum over I is a sum over different index sets as before, and the Greek indices are summed over as usual. Thus we calculate ω ∧ ω =

 | | 

g (n − p)!

ν 1 ···ν n

p µ1 ···µp

−

g µ1 I 1 · · · g µp I p ωI 1 ···I p ×

I,J

dxν 1 ∧ · · · ∧ dxν n

p

−

=

 | | 

g (n − p)!

ν 1 ···ν n

p µ1 ···µp

−



∧ ωJ 1

J 1 J p dx

···

∧ · · · ∧ dxJ p

g µ1 I 1 · · · g µp I p ωI 1 ···I p ωJ 1 ···J p ×



I,J

dxν 1 ∧ · · · ∧ dxν n

p

−

∧ dxJ 1 ∧ · · · ∧ dxJ p

(17.14)


Chapter 17. Hodge duality

We see that the set { ν 1 , · · · , ν n p } cannot have any overlap with the set J = {J 1 , · · · , J p }, because of the wedge product. On the other hand, { ν 1 , · · · , ν n p } cannot have any overlap with { µ1 , · · · , µ p } because  is totally antisymmetric in its indices. So the set {µ1 , · · · , µ p } must have the same elements as the set J = { J 1 , · · · , J p } , but they may not be in the same order. Now consider the case where the basis is orthogonal, i.e. g µν is diagonal. Then g µk I k = g I k I k etc. and we can write −

−

ω ∧ ω =

 | | 

g (n − p)!

ν 1 ···ν n

−

p I 1 ···I p

g I 1 I 1 · · · g I p I p ωI 1 ···I p ωJ 1 ···J p ×

I,J

dxν 1 ∧ · · · ∧ dxν n

p

−

∧ dxJ 1 ∧ · · · ∧ dxJ p . (17.15)

We see that in each term of the sum, the indices {I 1 · · · I p } must be the same as {J 1 · · · J p } because both sets are totally antisymmetrized with the indices { ν 1 · · · ν n p }. Since both sets are ordered, it follows that we can replace J by I , −

ω ∧ ω =

 | | 

g (n − p)!

ν 1 ···ν n

−

p I 1 ···I p

g I 1 I 1 · · · g I p I p ωI 1 ···I p ωI 1 ···I p ×

I

dxν 1 ∧ · · · ∧ dxν n

p

−

=

 | | 

g (n − p)!

ν 1 ···ν n

−

p I 1 ···I p

∧ dxI 1 ∧ · · · ∧ dxI p

ω I 1 ···I p ωI 1 ···I p ×

I

dxν 1 ∧ · · · ∧ dxν n

p

−

∧ dxI 1 ∧ · · · ∧ dxI p . (17.16)

In each term of this sum, the indices {ν 1 · · · ν n p } are completely determined, so we can replace them by the corresponding ordered set K = K 1 < · · · < K n p , which is completely determined by the set I , so that −

−

ω ∧ ω

  = | | g

K 1 ···K n

−

p I 1 ···I p

ω I 1 ···I p ωI 1 ···I p ×

I

dxK 1 ∧ · · · ∧ dxK n

p

−

p . ∧ dxI 1 ∧ · · · ∧ dxI (17.17)

The indices on this  are a permutation of { 1, · · · , n} , so  is ± 1. But this sign is the same as that for the permutation to bring the basis to the order dx1 ∧ · · · ∧ dxn , so the overall sign to get both to


the standard order is positive. Thus we get

 | |   1 = | |

ω ∧ ω =

g

ω I 1 ···I p ωI 1 ···I p 1···n dx1 ∧ · · · ∧ dxn

I

g

=

ω µ1 ···µp ωµ1 ···µp dx 1 ∧ · · · ∧ dxn

p!

1 µ1 ω p!

µp

···

ωµ1 ···µp ( vol )

(17.18)

If we are in a basis where the metric is not diagonal, it is still symmetric. So we can diagonalize it locally by going to an appropriate basis, or set of coordinates, at each point. In this basis, the components of ω may be ωµ1 µp , so we can write 

···

ω ∧ ω =

1

p!

ω

µ1 ···µp

ωµ1 ···µp 







(vol )

(17.19)

But both factors are invariant under a change of basis. So we can now change back to our earlier basis, and find Eq. (17.18) even when the metric is not diagonal. Note that the metric may not be diagonalizable globally or even in an extended region.


Chapter 18

Maxwell equations We will now consider a particular example in physics where differential forms are useful. The Maxwell equations of electrodynamics are, with c = 1 , ∇

· E = ρ

× B − ∂ ∂tE = j ∇·B = 0 ∂ B = 0. ∇ × E + ∂t ∇

(18.1)

(18.2) (18.3)

(18.4)

The electric and magnetic fields are all vectors in three dimensions, but these equations are Lorentz-invariant. We will write these equations in terms of differential forms. Consider R4 with Minkowski metric gµν = diag(−1, 1, 1, 1) . For the magnetic field define a 2-form B = B x dy

∧ dz + B

y dz

∧ dx + B dx ∧ dy . z

(18.5)

For the electric field define a 1-form E = E x dx + E y dy + E z dz .

(18.6)

Combine these two into a 2-form F = B + E ∧ dt . Let us calculate dF = d (B + E ∧ dt) = dB + dE ∧ dt . As usual, We will write 1, 2, 3 70


for the component labels x,y, z . dB = d(B1 dy

∧ dz + B dz ∧ dx + B dx ∧ dy) = ∂ B dt ∧ dy ∧ dz + ∂ B dx ∧ dy ∧ dz +∂ B dt ∧ dz ∧ dx + ∂ B dy ∧ dz ∧ dx +∂ B dt ∧ dx ∧ dy + ∂ B dz ∧ dx ∧ dy . t

2

3

1

1

1

t

2

2

2

t

3

3

3

(18.7)

And d(E

∧ dt) = d(E dx ∧ dt + E dy ∧ dt + E dz ∧ dt) = ∂ E dy ∧ dx ∧ dt + ∂ E dz ∧ dx ∧ dt +∂ E dx ∧ dy ∧ dt + ∂ E dz ∧ dy ∧ dt +∂ E dx ∧ dz ∧ dt + ∂ E dy ∧ dz ∧ dt . (18.8) 1

2

2

3

1

3

1

1

2

3

2

1

3

2

3

Thus, remembering that the wedge product changes sign under each exchange, we can combine these two to get dF = (∂ t B1 + ∂ 2 E 3

− ∂ E ) dt ∧ dy ∧ dz + (∂ B + ∂ E − ∂ E ) dt ∧ dz ∧ dx + (∂ B + + ∂ E − ∂ E ) dt ∧ dx ∧ dy + (∂ B + ∂ B + ∂ B ) dx ∧ dy ∧ dz = (∂ B + ( ∇ × E ) ) dt ∧ dy ∧ dz + (∂ B + ( ∇ × E ) ) dt ∧ dz ∧ dx + (∂ B + ( ∇ × E ) ) dt ∧ dx ∧ dy + (∇ · B ) dx ∧ dy ∧ dz . t

t

2

t

3

1

1

3

1

3

1

2

1

2

3

2

1

2

2

3

1

3

1

t

2

2

t

3

3

(18.9)

Thus two of Maxwell’s equations are equivalent to dF = 0 . For the other two equations we need F . Using the formula (17.11) for dual basis forms, it is easy to calculate that (dx

∧ dy) = dt ∧ dz , (dx ∧ dt) = dy ∧ dz ,

(dy

∧ dz ) = dt ∧ dx , (dy ∧ dt) = dz ∧ dx ,

(dz

∧ dx) = dt ∧ dy , (dz ∧ dt) = dx ∧ dy .

(18.10)

We use these to calculate F = (B + E

∧ dt)

= B1 dt ∧ dx + B2 dt ∧ dy + B3 dt ∧ dz

+E 1 dy ∧ dz + E 2 dz ∧ dx + E 3 dx ∧ dy . (18.11)


Chapter 18. Maxwell equations

Then in the same way as for the previous calculation, we find dF = (∇ E ) dx

∧ dy ∧ dz + (∂ E − (∇ × B ) ) dt ∧ dy ∧ dz + (∂ E − (∇ × B ) ) dt ∧ dz ∧ dx + (∂ E + ( ∇ × B ) ) dt ∧ dx ∧ dy . ·

t

1

1

t

2

2

t

3

3

(18.12)

We need to relate this to the charge-current. Define the current four-vector as j µ ∂ µ = ρ∂ t + j 1 ∂ 1 + j 2 ∂ 2 + j 3 ∂ 3 .

(18.13)

Then there is a corresponding one-form jµ dxµ with jµ = g µν j ν . So in terms of components, jµ dxµ =

−ρdt + j dx

1

1

+ j2 dx2 + j3 dx3 .

(18.14)

Then using Eq. (17.11) it is easy to calculate that j =

−ρ dx ∧ dy ∧ dz + j dt ∧ dy ∧ dz + j dt ∧ dz ∧ dx + j dt ∧ dx ∧ dy . 1

2

3

(18.15)

Comparing this equation with Eq. (18.12) we find that the other two Maxwell equations can be written as dF =

−j .

(18.16)

Finally, using Eq. (17.18), we see that the action of electromagnetism can be written as

− 12



F

∧ F

(18.17)

This expression holds in both flat and curved spacetimes. For the latter, with local coordinates (t , x , y , z) we find

∧ F = (B − E )√ −g dt ∧ dx ∧ dy ∧ dz .

F

2

2

(18.18)


Chapter 19

Stokes’ theorem We will next discuss a very beautiful result called Stokes’ formula. This is actually a theorem, but we will not prove it, only state the result and discuss its applications. So for us it is only a formula, but still deep and beautiful. • A submanifold S is a subset of points in M such that any point in S has an open neighbourhoood in M for which there is some chart where (n − m) coordinates vanish. S is then m-dimensional.  • Suppose U is a region of an oriented manifold M . The bound ary ∂ U of U is a submanifold of dimension n − 1 which divides M in such a way that any curve joining a point in U with a point in U c must contain a point in ∂ U . Now suppose U has an oriented smooth boundary ∂ U . Then ∂ U is automatically an oriented manifold, by considering the restrictions of the charts on U to ∂ U . • Consider a smooth (n − 1) form in M . Stokes’ formula says that

  dω =

U

ω.

(19.1)

∂ U

If M is a compact manifold with boundary ∂ M , this formula can be applied to all of M . If ω vanishes outside some compact region we can again set U = M . Also, U can be a submanifold in another manifold, like a 2-surface in a 3-manifold.  Example: Let U = [0, 1] . Then a function f : M → R is a 0-form, and df = f  (x)dx is a 1-form. Take the orientation of M to be from 0 to 1. Then ∂ M consists of the points x = 0 and x = 1 , 73


Chapter 19. Stokes’ theorem

and Stokes’ formula says that

  df =

M

f

∂ M

1

i.e.



f  (x)dx = f (1) − f (0) .

(19.2)

0

Example: Consider a 2-d disk D in

2 , with

R

boundary ∂D .

Take a 1-form A . Then Stokes’ formula says

  A =

∂D

dA .

(19.3)

D

Let us seee this equation in a chart. We can write A = Ai dxi dA = ∂ i A j dx i ∧ dx j A evaluated on ∂D can be written as A

to ∂D . So we can write A



Ai dxi =

    d dt

= A i

(19.4)



d d where is tangent dt dt

dxi dt , and dt

∂ i A j dxi ∧ dx j

D

∂D

=

(∂ 1 A2 − ∂ 2 A1 ) dx1 ∧ dx2

D

=

(∂ 1 A2 − ∂ 2 A1 ) d2 x .

(19.5)

ϕ(D)

Similarly for higher forms on higher dimensional manifolds. • Gauss’ divergence theorem is a special case of Stokes’ theorem. Before getting to Gauss’ theorem, we need to make a new definition. Consider an n -form ω  = 0 on an n-dimensional manifold. We can write this in a chart as ω = f dx1 ∧ · · · ∧ dxn

=

1 f µ n!

µ

1 ···

n

dx µ ∧ · · · ∧ dxµ . 1

n

(19.6)


Given a vector field v , its contraction with ω is 1 ωµ µ ···µ v µ dxµ ∧ · · · ∧ dxµ (n − 1)! = f v 1 dx2 ∧ · · · ∧ dxn − f v 2 dx1 ∧ dx3 ∧ · · · ∧ dxn + · · · (19.7)

ιv ω = ω (v, · · · ) =

1

1

2

2

n

n

Then we can calculate d(ιv ω ) = dω (v, · · · ) = ∂ 1 (f v 1 ) dx1 ∧ dx2 ∧ · · · ∧ dxn

+∂ 2 (f v 2 ) dx1 ∧ dx2 ∧ · · · ∧ dxn + · · · + ∂ n (f vn ) dx1 ∧ dx2 ∧ · · · ∧ dxn = ∂ µ (f v µ ) dx1 ∧ dx2 ∧ · · · ∧ dxn 1 = ∂ µ (f v µ ) ω . f

(19.8)

In particular, if ω is the volume form, we can write

 

|g | µ ···µ dxµ ∧ · · · ∧ dxµ , n! 1 ∂ µ (vµ |g |)(vol ) . |g |

ω = d(ιv (vol )) =

•

1

1

n

n



(19.9)

This is called the divergence of the vector field v . There is another expression for the divergence. Remember that given a vector field v , we can define a one-form, also called v , with components defined with the help of the metric, 

vµ = g µµ v µ

(19.10)



Consider v , which has components (v)µ

1 ···µn−1

µ

µg

µµ

|g |µ

µ

µv

µ

µ

(−1)n−1 µ = (n − 1)!

µ

= ⇒

|g |µ

    

=

v =

1 ··· 1 ···

n−1

n−1

|g | µ (n − 1)!

1 ···

dv = ∂ µ

n

n−1

vµ



. µv

|g | µ (n − 1)!

1 ···

1 ···

n−1

µ

dxµ ∧ · · · ∧ dxµ 1

n−1

    µv

µ

µ

(19.11)

n−1

∂ µ

n

µ

dxµ ∧ dxµ ∧ · · · ∧ dxµ n

|g |v µ

1

n−1

dxµ ∧ · · · ∧ dxµ 1

n

(19.12)


Chapter 19. Stokes’ theorem

Both µ and µn must be different from ( µ1 , · · · , µn−1 ) , so µ = µn . Thus in each term of the sum, the choice of (µ1 , · · · , µn−1 ) automatically selects µn (= µ ) , so a sum over ( µ1 , · · · , µn ) overcounts n times. So we can write (−1)n−1 |g |v µ dxµ ∧ · · · ∧ dxµ µ ···µ ∂ µ (n − 1)! 1 ∂ µ ( |g |v µ )(vol ) . = (−1)n−1 (19.13) |g |

    

dv =

1

1

n

n

Since this is an n-form in n dimensions, we can calculate from here that dv =

(−1)n+s−1 ∂ µ ( |g |vµ ) , |g |





(19.14)

where as before s is the signature of the manifold, i.e. the number of negative entries in the metric in a locally diagonal form. Let us now go back to Stokes’ formula. Take a region U of M which is covered by a single chart and has an orientable boundary ∂ U as before. Then we find 1 ∂ µ ( |g |vµ )(vol ) = |g |

  

U

=

 

d(ιv (vol ))

U

ιv (vol ) .

(19.15)

∂ U

Now suppose b is a 1-form normal to ∂ U , i.e. b d



d = 0 for any dt

vector tangent to ∂ U , and α is an (n − 1)-form such that b ∧ α = dt (vol ) . Since all n-forms are proportional, α always exists. For the same reason, if b  = 0 on ∂ U , it is unique up to a factor. And b  = 0 on ∂ U because ∂ U is defined as the submanifold where one coordinate is d constant, usually set to zero, so that one component of vanishes dt at any point on ∂ U , and therefore the corresponding component of b can be chosen to be non-zero. So b is unique up to a rescaling b → b  = f b for some nonvanishing function f . But we can always scale α → α  = f α so that b  ∧ α = b ∧ α . Further, if we restrict α to ∂ U , i.e. if α acts only on tangent −1


vectors to ∂ U , we find that α is an (n − 1)-form on an (n − 1)dimensional manifold, so it is unique up to scaling. Therefore, α is unique once b is given. Finally, for any vector v , ιv (vol )



∂ U



= ι v (b ∧ α)

(19.16)

∂ U

is an (n − 1)-form on ∂ U which acts only on vectors tangent to ∂ U . Then



ιv (b ∧ α)

∂ U



= b (v)α

(19.17)

∂ U

because all terms of the form b ∧ ι v α gives zero for any choice of (n − 1) vectors on ∂ U . Then we have 1 ∂ µ ( |g |v µ )(vol ) = |g |

  

U

=

 

b(v )α

∂ U

(nµ v µ ) α .

(19.18)

∂ U

Usually b is taken to have norm 1. Then α is the volume form on ∂ U , and we can write 1 ∂ µ ( |g |v µ )(vol ) = |g |

  

U



∂ U

(nµ v µ )



|g(∂ U ) |dn−1 x . (19.19)


Chapter 20

Lie groups We start a brief discussion on Lie groups, mainly with an eye to their structure as manifolds and also application to the theory of fiber bundles. • A Lie group is a group which is also an analytic manifold.  We did not define a Lie group in this way in Chap. 2 , but said that a Lie group was a manifold in which the group product is analytic in the group parameters, or alternatively the group product and group inverse are both C ∞ . The definition above comes from a theorem that given a continuous group G in which the group product and group inverse are C ∞ functions of the group parameters, it is always possible to find a set of coordinate charts covering G such that the overlap functions are real analytic, i.e. are C ∞ and their Taylor series at any point converge to their respective values. A Lie subgroup of G is a subset H of G which is a subgroup of • G , a submanifold of G , and is a topological group, i.e., a topological space in which the group product and group inverse are continuous maps.  • Sometimes this expressed in terms of another definition. P is an immersed submanifold of M if the inclusion map j : P → M is smooth and at each point p ∈ P its differential dj p is one to one, with dj p being defined by dj p : T p P → T j ( p) M such that dj p (v)(g ) = v (g · j p ) .  We have mentioned some specific examples of Lie groups earlier. Let us mention some more examples. Example: n is a Lie group under addition. So is n . 78


Example: n

n

\{0} is a Lie group under multiplication. So is

\{0} . Example: The direct product of two Lie groups is itself a Lie

group, with multiplication (g1 , h1 )(g2 , h2 ) = (g1 g2 , h1 h2 ) .  Example: The set of all n × n real invertible matrices forms a group under matrix multiplication, called the General Linear group GL(n, ) . This is also the space of all invertible linear maps of n to itself. We can similarly define GL(n, ) .  The next few examples are Lie subgroups of GL(n, ) . Example: The Special Linear group SL (n, ) is the subset of GL(n, ) for which all the matrices have determinant +1 , i.e., SL (n, ) = { A ∈ GL (n, )| det A = 1} . One can define SL (n, ) in a similar manner.  Example: The Orthogonal group O(n) = {R ∈ GL(n, ) | T R R = } .  † Example: The Unitary group U (n) = { U ∈ GL (n, ) | U U = }.  Example: The Symplectic group Sp (n) , defined as the subgroup of U (2n) given by AT A = , where =



−

0 n×n

n×n

0

  T

Example: O( p, q ) = { R ∈ GL ( p + q, ) | R η p,q R = η p,q } , where

η p,q =



0

p× p

0

−

q×q

 

†

Example: U ( p, q ) = { U ∈ GL ( p + q, ) | U η p,q U = η p,q } . Example The Special Orthogonal group SO (n) is the subgroup of O(n) for which determinant is +1. Similarly, the Special unitary group SU (n) is the subgroup of U (n) with determinant

+1. Similarly for SO ( p.q ) and SU ( p, q ) . The group U (1) is the group of phases U (1) = { eiφ |φ ∈ } . As a manifold, this is isomorphic to a circle S 1 . The group SU (2) is isomorphic as a manifold to a three-sphere 3 S . These are the only two spheres (other than the point S 0 ) which admit a Lie group structure.


Chapter 20. Lie groups

An important property of a Lie group is that the tangent space of any point is isomorphic to the tangent space at the identity by an appropriate group operation. Of course, the tangent space at any pooint of a manifold is isomorphic to the tangent space at any other point. For Lie groups, the isomorphism between the tangent spaces is induced by group operations, so is in some sense natural. For any Lie group G , we can define diffeomorphisms of G labelled by elements g ∈ G , called g  → gg  ; • Left translation lg : G → G    • g → g g . Right translation r g : G → G  These can be defined for any group, but are diffeomorphisms for Lie groups. We see that lg− lg (g  ) = l g− (gg  ) = g −1 gg  = g  1







rg− rg (g ) = r g− (g g ) = g gg 1

1

−1

= g

(lg )−1 = l g−

⇒ ⇒

1



1

−1

(rg )

= r g− . (20.1) 1

It is easy to check that lg lg = l g 1

2

1

g2

rg rg = r g 1

2

2

g1

(20.2)

Further, lg− (g ) = e and rg− (g ) = e , so any element of G can be moved to the identity by a diffeomorphism. The tangent space at the identity forms a Lie algebra, as we shall see. The left and right translations lead to diffeomorphisms which relate the tangent space at any point to this Lie algebra, as we shall see now. 1

1


Chapter 21

Tangent space at the identity A point on the Lie group is a group element. So a vector field on the Lie group selects a vector at each g ∈ G . Since left and right translations are diffeomorphisms, we can consider the pushforwards due to them. • A left-invariant vector field X is invariant under left ttranslations, i.e.,

∀g ∈ G .

X = l g∗ (X )

(21.1)



In other words, the vector (field) at g is pushed forward by l g to the same vector (field) at l g (g ): 

lg∗ (X g ) = X gg 

•



∀g, g ∈ G .



(21.2)

Similarly, a right-invariant vector field X is defined by X = rg∗ (X )

i.e.

rg∗ (X g ) = X g g 



∀g ∈ G , ∀g, g ∈ G . 

(21.3)

A left or right invarian vector field has the important property that it is completely determined by its value at the identity element e of the Lie group, since

∀g ∈ G ,

lg∗ (X e ) = X g

(21.4)

and similarly for right-invariant vector fields. Write the set of all left-invariant vector fields on G as L (G) . Since the push-forward is linear, we get lg∗ (aX + Y ) = al g∗ X + lg∗ Y ,

81

(21.5)


Chapter Chapter 21. Tangent angent space at the identity identity

so that if both X and Y are left-invariant, + Y ) ) = aX + Y , lg∗ (aX +

(21.6)

so the set of left-invariant vector fields form a real vector space. We also know that push-forwards leave the Lie algebra invariant, i.e., for lg , ∗

[lg X , lg Y ] ] = l g [X , Y ] ] . ∗

∗

∗

(21.7)

Thus if X, Y ∈ L (G) , ] = [lg∗ X , lg∗ Y ] ] = [X , Y ] ] , lg∗ [X , Y ]

(21.8)

so [X , Y ] ] ∈ L (G) . Thus the set of all left-invariant vector fields on G forms a Lie algebra. • This L(G) is called the Lie algebra of G .  G because The dimension of this Lie algebra is the same as that of G of the Theorem: L(G) as a real vector space is isomorphic to the tangent space T e G to G at the identity of G . Proof: We will show that left translation leads to an isomorphism. For X ∈ T e G , define the vector field LX on G by LX g ≡ L X g := l g∗ X



∀g ∈ G

(21.9)

(21.10)



Then for all g, g ∈ G , X lg ∗(LX g ) = l g ∗ (lg ∗ X ) = l g g ∗ X = L g g . 





Note that for two diffeomorphisms ϕ 1 , ϕ2 , we can write (ϕ1 (ϕ2 v ))(f ) = (ϕ2 v )(f ◦ ϕ1 ) = v (f ◦ ◦ ϕ1 ◦ ϕ2 ) ∗

∗

∗

= ((ϕ1 ◦ ϕ2 ) v )(f ) ∗

⇒

ϕ1∗ (ϕ2∗ v ) = (ϕ1 ◦ ϕ2 )∗ v

(21.11)

Since left translation is a diffeomorphism, lg ∗ (lg∗ X ) = (lg ◦ lg )∗ X = (lg g∗ )X 





(21.12)


So it follows that LX is a left-invariant vector field, and we have a map T e G → L (G) . Since the pushforward is a linear map, so is the map X → L X . We need to prove that this map is 1-1 and onto. If LX = L Y , we have Y LX g = L g

∀g ∈ G ,

(21.13)

so lg

−1

∗

LX g = l g

−1

∗

LY g

⇒

X = Y

(∈ T e G) . (21.14)

So the map X → L X is 1-1. Now given LX , define X e ∈ T e G by X e = l g

−1

∗

LX g

for any any g ∈ G .

(21.15)

We can also write X e = L X e .

(21.16)

Then lg∗ X e = l g∗ lg

−1

∗

X LX g = L g .

(21.17)

So the map X →  L X is onto. Then we can define a Lie bracket on T e G by [u , v] = [ Lu , Lv ]e .



(21.18)

The Lie algebra of vectors in T e G based on this bracket is thus the Lie algebra of the group G . It follows that dim L(G) = dim T e G = dim G .

(21.19)

Note that since commutators are defined for vector fields and not vectors, the Lie bracket on T e G has to be defined using the commutator mutator of left-inv left-invarian ariantt vector vector fields on G and the isomorphism isomorphism T e G ↔ L (G) . If for an n-dimensional Lie group G , {t1 , · · · , tn } is a set of basis • vectors on T e G  L (G) , the Lie bracket of any pair of these vectors must be a linear combination of them, so [ti , t j ] =

 k

k C ij t k

(21.20)


Chapter Chapter 21. Tangent angent space at the identity identity

k for some set of real numbers C ij . These numbers are known as the  structure constants of the Lie group or algebra. Since L (G) is a Lie algebra, with the Lie bracket as the product, the Lie bracket is antisymmetric,

[t j , ti ] [ti ,k t j ] =  k

⇒

C ij t k =

k

⇒

k C ij

=

C ji tk

k k C ji ,

(21.21)

and the structure constants satisfy the Jacobi identity [ti , [t j , tk ]] + [t j , [tk , ti ]] + [tk , [ti , t j ]] = 0 l m l l m C ij C kl C ilm + C ki C jl ⇒ + C jk = 0.

(21.22)

A similar construction can be done using a set of right-invariant vector fields defined by RgX := r g∗ X

and its ‘inverse’ X e = r g

−1

∗

RX g .

for X ∈ T e G

(21.23)


Chapter 22

One parameter subgroups There is another characterization of T e G for a Lie group G as the set of its one parameter subgroups, which we will now define. This is also called the “infinitesimal” description of a Lie group, and what Lie called an infinitesimal group. • A one parameter subgroup of a Lie group G is a smooth homomorphism from the additive group of real numbers to G , γ : (R, +) → G . Then γ : R → G is a curve such that γ (s + t) = γ (s)γ (t) , γ (0) = e , and γ (−t) = γ (t)−1 .  Also, since this is a homomorphism, the one parameter subgroup is Abelian. Example: For G = (R\{0}, ×) the multiplicative group of nonzero real numbers, γ (t) = et is a 1-p subgroup. γ (t) = e it . Example: G = U (1) , cos t sin t γ (t) = . Example: G = SU (2) , − sin t cos t cos t sin t 0 γ (t) = − sin t cos t 0 . Example: G = GL(3, R) , 0 0 et The relation between 1-p subgroups and and T e G is given by the ˙ ( 0) = γ ˙ e defines a 1-1 correspon γ Theorem: The map γ → dence between 1-p subgroups of G, and T e G . Proof: For any X ∈ T e G define LX = l g∗ X as the corresponding left-invariant vector field. We need to find a smooth homomorophism from R to G using L X . This homomorphism is provided by the flow or integral curve of L X , but let us work this out in more detail.





 



85

 


Chapter 22. One parameter subgroups

Denote the integral curve of LX by γ X (t) , i.e., γ X (0) = L X e = X and

γ ˙ X (t) = L X γ (t) = l γ (t)∗ X

(22.1)

X Since LX is left-invariant, lg ∗ LX g = L g g . Consider the equation 



d γ (t) = L X γ (t) = l γ (t)∗ X ≡ γ ∗ dt

 d  . dt 

(22.2)

t

Given some τ , replace γ (t) by γ (τ + t) to get γ (τ + t) = l γ (τ +t)∗ X .

(22.3)

Remember that γ (t) is an element of the group for each t . Now replace γ (t) in Eq. (22.2) by γ (τ )γ (t) to get

γ (τ )γ (t)  d  = L ∗

dt

X γ (τ )γ (t)

(22.4)

We see that γ (t + τ ) and γ (τ )γ (t) are both integral curves of L X , i.e. both satisfy the equation of the integral curve of LX , and at t = 0 both curves are at the point γ (τ ) . Thus by uniqueness these two are the same curve, γ (τ + t) = γ (τ )γ (t) ,

(22.5)

and t → γ (t) is the homomorphism R → G that we are looking for. Thus for each X ∈ T e G we find a 1-p subgroup γ (t) given by the integral curve of LX , γ ˙ (t) = L X γ (t) = l γ (t)∗ X ,

(22.6)

where, as mentioned earlier, (γ ˙ (0) = X ) .  In a compact connected Lie group G , every element lies on some 1-p subgroup. This is not true in a non-compact G , i.e. there are elements in G which do not lie on a 1-p subgroup. However, an Abelian non-compact group will always have a 1-p subgroup, so this remark applies only to non-Abelian non-compact groups. For matrix groups, every 1-p subgroup is of the form

    γ (t) = e  M fixed, t ∈ . tM

R

(22.7)


Let us see why. Suppose {γ (t)} is a 1-p subgroup of the matrix group. Then γ (t) is a matrix for each t , and γ (s)γ (t) = γ (s + t) .

(22.8)

Differentiate with respect to s and set s = 0 . Then γ ˙ (0)γ (t) = γ ˙ (t) .

(22.9)

Write γ ˙ (0) = M . Since G is a matrix group, M is a matrix. Then the unique solution for γ is γ (t) = etM .

(22.10)

The properties of M are determined by the properties of the group and vice versa, not every matrix M will generate any group. The allowed matrices { M } for a given group G are the { γ ˙ (0)} for all the 1-p subgroups γ (t) , so these are in fact the tangent vectors at the identity. The allowed matrices {M } for a given matrix group G thus form a Lie algebra with the Lie bracket being given by the matrix commutator. This Lie algebra is isomorphic to the Lie algebra of the group G . (We will not a give a proof of this here.) We can find the Lie algebra of a matrix group by considering elements of the form γ (t) = e tM for small t , i.e., γ (t) = I + tM

(22.11)

for small t . Conversely, once we are given, or have found, a Lie algebra with basis {ti } , we can exponentiate the Lie algebra to find the set of 1-p subgroups i

γ (a) = exp a t  i

(22.12)

• This is the infinitesimal group, for compact connected groups this is identical to the Lie group itself. So in such cases, the entire group can be generated by exponentiating the Lie algebra. Noncompact groups cannot be written as the exponential of the Lie algebra in general.  Example: Consider SO(N ) , the group of N × N real orthogonal matrices R with RT R = I , det R = 1 . Write R = I + A , then AT = −A , i.e. the Lie algebra is spanned by N × N real antisymmetric matrices. Let us construct a basis for this algebra.


Chapter 22. One parameter subgroups

An N × N antisymmetric matrix has N (N − 1)/2 independent elements. So we define N (N − 1)/2 independent antisymmetric matrices, labelled by µ, ν = 1, · · · , N , M µν = −M νµ µ , ν are not matrix indices (M µν )ρσ = (M µν )σρ , ρ , σ are matrix indices . (22.13) A convenient choice for the basis is given by (M µν )ρσ = δ µρ δ νσ − δ µσ δ νρ .

(22.14)

Then the commutators are calculated to be [M µν , M αβ ] = δ να M µβ − δ µα M νβ + δ µβ M να − δ νβ M µα . (22.15) This defines the Lie algebra. Example: For SU (N ) , the group of N × N unitary matrices U with U † U = I , det U = 1 , the 1-p subgroups are given by γ (t) = etM with M † + M = 0 in the same way as above, and det(I + tM ) = 1 ⇒ Tr M = 0 . So the SU (N ) Lie algebra consists of traceless antihermitian matrices. Often the basis is multiplied by i to write γ (a) = exp(ia j t j ) , where t j are now Hermitian matrices, with [ti , t j ] = if abctc .

(22.16)


Chapter 23

Fiber bundles Consider a manifold M with the tangent bundle T M =

 P ∈M

T P M .

Let us look at this more closely. T M can be thought of as the original manifold M with a tangent space stuck at each point P ∈ M . Thus there is a projection map π : T M → M , T P M →  P , which associates the point P ∈ M with T P M . Then we can say that T M consists of pooints P ∈ M and vectors v ∈ T P M as an ordered pair (P, v ) . Then in the neighbourhood of any point P , we can think of T M as a product manifold, i.e. as the set of ordered pairs (P, v ) . This is generalized to the definition of a fiber bundle. Locally a fiber bundle is a product manifold E = B × F with the following properties. • B is a manifold called the base manifold, and F is another manifold called the typical fiber or the standard fiber. • There is a projection map π : E → B , and if P ∈ B , the preimage π−1 (P ) is homeomorphic, i.e. bicontinuously isomorphic, to the standard fiber.  E is called the total space, but usually it is also called the bundle, even though the bundle is actually the triple (E , π , B ) . • E is locally a product space. We express this in the following way. Given an open set U i of B , the pre-image π −1 (U i ) is homeomorphic to U i × F , or in other words there is a bicontinuous isomorphism ϕi : π −1 (U i ) → U i × F . The set { U i , ϕi } is called a local trivializa tion of the bundle.  If E can be written globally as a product space, i.e. E = B × F , • it is called a trivial bundle.  P

P

89


Chapter 23. Fiber bundles

This description includes a homeomorphism π −1 (P ) → F for • each P ∈ U i . Let us denote this map by h i (P ) . Then in some overlap U i ∩ U j the fiber on P , π−1 (P ) , has homeomorphisms hi (P ) and h j (P ) onto F . It follows that h j (P ) · h i (P )−1 is a homeomorphism F → F . These are called transition functions. The transition functions F → F form a group, called the structure group of F .  Let us consider an example. Suppose B = S 1 . Then the tanP , where gent bundle E = T S 1 has F = R and π(P, v ) →  1 1 1 P ∈ S , v ∈ T S . Consider a covering of S by open sets U i , and let the coordinates of U i ⊂ S 1 be denoted by λ i . Then any vector at d (no sum) for P ∈ U i . dλi  So we can define a homeomorphism hi (P ) : T P S 1 → R, v → ai (fixed i). If P ∈ U i ∩ U j there are two such homeomorphisms T S 1 → R , and since λi and λ j are independent, ai and a j are also T P S 1 can be written as v = a i

independent. Then hi (P ) · h j (P )−1 : F → F (or R → R) maps a j to ai . The homeomorphism, which in this case relates the component of the vector in two coordinate systems, is simply multiplication by the ai ∈ R\{0} . So the structure group is R\{0} with number rij = a j

multiplication. For an n-dimensional manifold M , the structure group of T M is GL(n, R) . • A fiber bundle where the standard fiber is a vector space is called a vector bundle .  A cylinder can be made by glueing two opposite edges of a flat strip of paper. This is then a Cartesian product of acircle S 1 with a line segment I . So B = S 1 , F = I and this is a trivial bundle, i.e. globally a product space. On the other hand, a M¨ obius strip is obtained by twisting the strip and then glueing. Locally for some open set U  S 1 we can still write a segment of the Möbius strip as U × I , but the total space is no longer a product space. As a bundle, the M¨ obius strip is non-trivial. Given two bundles (E 1 , π1 , B1 ) and (E 2 , π2 , B2 ) , the relevant • or useful maps between these are those which preserve the bundle structure locally, i.e. those which map fibers into fibers. They are called bundle morphisms.  A bundle morphism is a pair of maps ( F, f ) , F : E 1 → E 2 , f :


B1 → B2 , such that π2 ◦ F = f ◦ π1 . (This is of course better

understood in terms of a commutative diagram.) Not all systems of coordinates are appropriate for a bundle. But it is possible to define a set of fiber coordinates in the following way. Given a differentiable fiber bundle with n -dimensional base manifold B and p -dimensional fiber F , the coordinates of the bundle are given by bundle morphisms onto open sets of Rn × R p .  Given a manifold M the tangent space T P M , consider AP = • (e1 , · · · , en ) , a set of n linearly independent vectors at P . AP is a basis in T P M . The typical fiber in the frame bundle is the set of all bases, F = { AP } .  Given a particular basis AP = (e1 , · · · , en ) , any basis AP may be expressed as j

ei = a i e j .

(23.1)

The numbers a ji can be thought of as the components of a matrix, which must be invertible so that we can recover the original basis from the new one. Thus, starting from any one basis, any other basis can be reached by an n × n invertible matrix, and any n × n invertible matrix produces a new basis. So there is a bijection between the set of all frames in T P M and GL(n , R) . Clearly the structure group of the typical fiber of the frame bundle is also GL(n , R) . • A fiber bundle in which the typical fiber F is identical (or homeomorphic) to the structure group G, and G acts on F by left translation is called a principal fiber bundle.  1 Example: 1. Typical fiber = S , structure group U (1). 2. Typical fiber = S 3 , structure group SU (2). 3. Bundle of frames, for which the typical fiber is GL(n , R) , as is the structure group. • A section of a fiber bundle ( E , π , B ) is a mapping s : B → E , p → s( p) , where p ∈ B , s( p) ∈ π −1 ( p) . So we can also say π ◦ s = identity.  Example: A vector field is a section of the tangent bundle, v : P →  v . Example: A function on M is a section of the bundle which locally looks like M × R (or M × C if we are talking about complex functions). P



•

Starting from the tangent bundle we can define the cotangent bundle, in which the typical fiber is the dual space of the tangent space. This is written as T ∗ M . As we have seen before, a section of T ∗ M is a 1-form field on M .  π • Remember that a vector bundle F → E → B is a bundle in which the typical fiber F is a vector space.  A vector bundle (E, π , B , F , G) with typical fiber F and struc• ture group G is said to be associated to the principal bundle (P , π , B , G) by the representation {D (g )} of G on F if its transition functions are the images under D of the transition functions of P .  That is, suppose we have a covering { U i } of B , and local trivialization of P with respect to this covering is Φ i : π −1 (U i ) → U i × G , which is essentially the same as writing Φi,x : π −1 (x) → G , x ∈ U i . Then the transition functions of P are of the form



gij = Φi ◦ Φ j −1 : U i ∩ U j → G .

(23.2)

The transition functions of E corresponding to the same covering of B are given by φi : π −1 (U i ) → U i × F with φi ◦ φ j −1 = D (gij ) . That is, if vi and v j are images of the same vector vx ∈ F x under overlapping trivializations φi and φ j , we must have



vi = D (gij (x)) v j .

(23.3)

A more physical way of saying this is that if two observers look at the same vector at the same point, their observations are relatted by a group transformation ( p, v)  ( p, D(gij v ) . These relations are what are called gauge transformations in • physics, and G is called the gauge group. Usually G is a Lie group for reasons of continuity.  Fields appearing in various physical theories are sections of vector bundles, which in some trivialization look like U α × V where U α is some open neighborhood of the point we are interested in, and V is a vector space. V carries a representation of some group G , usually a Lie group, which characterizes the theory. To discuss this a little more concretely, let us consider an associated vector bundle (E, π , B , F , G) of a principal bundle (P , π , B , G). Then the transition functions are in some representation of the group G . Because the fiber carries a representation { D (g )} of G , there are




always linear transformations T x : E x → E x which are members of the representation {D (g )} . Let us write the space of all sections of this bundle as Γ(E ) . An element of Γ(E ) is a map from the base spacce to the bundle. Such a map assigns an element of V to each point of the base space. We say that a linear map T : Γ(E ) → Γ(E ) is a gauge trans • formation if at each point x of the base space, T x ∈ {D (g )} for some g , i.e. if T x : (x, v )α → ( x, D (g )v )α ,

(23.4)

for some g ∈ G and for (x, v )α ∈ U α × F . In other words, a gauge transformation is a representation-valued linear transformation of the sections at each point of the base space. The right hand side is often written as (x,gv)α .  This definition is independent of the choice of U α . To see this, consider a point x ∈ U α ∩ U β . Then (x, v )α = (x, gβα v )β .

(23.5)

In the other notation we have been using, v α and v β are images of the same vector vx ∈ V x , and vβ = D(gβα )vα . A gauge transformation T acts as T x : (x, v )α → ( x,gv)α .

(23.6)

(x,gv)α = (x, gβα gv )β

(23.7)

But we also have

using Eq. (23.5) . So it is also true that T x : (x, gβα v )β →  ( x, gβα gv )β .

(23.8)

Since F carries a representation of G , we can think of gv as a change of variables, i.e. define v  = gβα v . Then Eq. (23.8) can be written also as T x : (x, v  )β →  ( x, g  v  )β ,

(23.9)

where now g  = g βα gg βα −1 . So T is a gauge transformation in U β as well. The definition of a gauge transformation is independent of the



choice of U α , but T itself is not. The set of all gauge transformations G is a group, with (gh )(x) = g (x)h(x) , The groups G and • different people.

G arre

(g −1 )(x) = (g (x)) −1 .

(23.10)

both called the gauge group by 


Chapter 24

Connections There is no canonical way to differentiate sections of a fiber bundle. That is to say, no unique derivative arises from the definition of a bundle. Let us see why. The usual derivative of a function on R is of the form f (x + ) − f (x) . →0 

f  (x) = lim

(24.1)

But a section of a bundle assigns an element of the fiber at any point P ∈ M to the base point P . Each fiber is isomorphic to the standard fiber but the isomorphism is not canonical or unique. So there is no unique way of adding or subtracting points on different fibers. Thus there are many ways of differentiating sections of fiber bundles. Each way of taking a derivative, i.e. of comparing, is called a • connection. Let us consider a bundle π : E → B , where Γ(E ) is the space of all sections. Then a connection D on B assigns to every vector field v on B a map Dv : Γ(E ) → Γ(E ) satisfying Dv (s + αt) = Dv s + α Dv t Dv (f s) = v (f )s + f Dv s Dv+f w (s) = Dv s + f Dw s ,

(24.2)

where s, t are sections of the bundle, s, t ∈ Γ(E ) , v, w are vector fields on B , f ∈ C ∞(B ) and α is a real number (or complex, depending on what the manifold is).  Note that this is a connection, not some unique connection. In other words, we need to choose a connection before we can talk about it. In what follows, whenever we refer to a connection D , we 95


Chapter 24. Connections

mean that we have chose a connection D and that is what we are discussing. • We call Dv s the covariant derivative of s .  To be specific, let us consider the bundle to be a vector bundle on a manifold M , and try to understand the meaning of D by going to a chart. Consider coordinates x µ in an open set U ⊂ M , with ∂ µ the coordinate basis vector fields. Write Dµ = D∂ . Also choose a basis for sections, which is like a basis for the fiber (vector space) at each point of M , i.e. like a set of basis vector fields, but the vectors are not along M , but along the fiber at M . Call this basis {ei } , then {ei (x)} is a basis for the fiber at P ∈ M , with {x} being the set of coordinates at P . Any element of V  F x can be written uniquely as a linear combination of ei (x). But then Dµ e j can be expressed uniquely as a linear combination of the e i , µ

i Dµ e j = A µj ei .

(24.3)

• These Aiµj are called components of the vector potential or the connection one -form.  i i ∞ Given a section s = s ei with s ∈ C (M) , we can write Dv s = D v

µ

∂ µ s = v

µ

Dµ s .

(24.4)

Also, Dµ s = D µ (si ei ) = ∂ µ si ei + si Dµ ei

    = ∂ s e + s A   µ

i

i

i

j µi e j

= ∂ µ si + Aµji s j ei ,

(24.5)

so writing Dµ s = (Dµ s)i ei , we can say (Dµ s)i = ∂ µ si + Aµji s j .

(24.6)

We have considered connections on an associated vector bundle, which may be a principal fiber bundle such as a frame bundle. So we should be able to talk about gauge transformations. Remember that a gauge transformation is a linear map T : E →  (x,gv) for some g ∈ G and for all v ∈ V  F x . Let us E , (x, v ) → apply this idea to the section s . We claim that given a connection D , there is a connection D  on E such that D  (gφ ) = gDv φ ,

(24.7)


where v is a vector field on M and φ ∈ Γ(E ) (i.e. φ = s ). Let us first check if the definition makes sense. Since g (x) ∈ G for all x ∈ M , we know that g −1 (x) exists for all x. So Dv (φ) = Dv gg −1 φ



= gD v

 g φ , −1

(24.8)

and thus D  is defined on all φ for which D v φ is defined, i.e. D  exists because D does. We have of course assumed that g (x) is differentiable as a function of x. Let us now check that D  is a connection according to our definitions. D  is linear since Dv (φ1 + αφ2 ) = gD v g −1 (φ1 + αφ2 )

 g

   φ + αgD g

= gD v −1 1 = Dv φ1 + αDv φ2 .

v

−1

φ2



(24.9)

And it satisfies Leibniz rule because Dv (f φ) = gD v g −1 f φ

= gD v

  f g φ −1

= gv (f )g −1 φ + gf Dv g −1 φ



= v (f )φ + f gDv g −1 φ



= v (f )φ + f Dv (φ) .





(24.10)

Similarly, Dv +αw φ = gD v+αw g −1 φ





= g Dv g −1 φ + αDw g −1 φ

 



= gD v (g −1 φ) + αgD w  φ. = Dv φ + αDw

  g φ −1

(24.11)

So D  is a connection, i.e. there is a connection D  satisfying Dv (gφ ) = g (Dv φ) . Since φ is a section, i.e. φ ∈ Γ(E ), so is g −1 φ, and thus Dv g −1 φ ∈ Γ(E ) and also gDv g −1 φ ∈ Γ(E ) . Therefore, D v maps sections to sections, Dv : Γ(E ) → Γ( E ) . This completes the definition of the gauge transformation of the connection. We can now write









Dµ φ = ∂ µ φi + Aµ i j φ j ei .





(24.12)


Chapter 24. Connections

Using the dual space, let us write i

Aµ = Aµ j e i ⊗ θ j , Aµ = Aµ i j e i ⊗ θ j ,

(24.13)

where θi is the dual basis to {ei } . The gauge transformation is then given by

 

Dv φ = gD v g −1 φ −1



⇒ ⇒

    D φ = gD g φ  e = g ∂ g φ + A g φ  e , µ

µ

i

i µj

∂ φ + A µ



i

−1

µ

i

i µj

j

−1

i

(24.14)

where, as always, the g ’s are in some appropriate representation of G. Then we can write the right hand side as



     ∂ φ + g∂ g φ + gA g φ e     φ e . = ∂ φ + g∂ g + gA g µ

i

µ

µ

i

−1

i j

µ

j

−1

µ

µ

−1

−1

i j

i j

j

j

i

i

(24.15)

From this we can read off Aµ = gAµ g −1 + g∂ µ g −1 .

•

(24.16)

A connection which transforms like this is also called a Gconnection.  Example: Consider G = U (1) . Suppose E is a trivial complex line bundle over M , i.e. E = M× C , so that the fiber over any point p ∈ M is C . A connection D on E may be written as D µ = ∂ µ + Aµ . We can make E into a U (1) bundle by thinking of the fiber C as the fundamental representation space of U (1) . Then sections are complex functions, and a gauge transformation is multiplication by a phase. 


Chapter 25

Curvature We start with a connection D , two vector fields v, w on B , and a section s , all on some associated vector bundle of some principal G-bundle E . Then Dv , Dw are both maps Γ(E ) → Γ( E ) . We will define the curvature of this connection D as a rule F which, given two vector fields v , w , produces a linear map F (v, w) : Γ(E ) → Γ( E ) by F (v, w)s = D v Dw s − Dw Dv s − D[v,w] s .

(25.1)

Remember that Dv s = D v

= v

µ

µ

µ

∂ µ s = v



i

∂ µ s +

Dµ s

i Aµ j s j



ei

= v (si )ei + v µ Aµij s j ei .

(25.2)

Dµ s is again a section. So we can act with D on it and write Dv Dw

  s = D w (s )e + w A s e      = v w s e + v w A s e     v



j

j

j



j

ν

ν k

ν

j

k

j k ν k

j

j

+ w s j + wν Aν j k s k v µ Aµij s j ei .

(25.3)

Since the connection components Aν j k are functions, we can write



ν

j

v w Aν k s

k



= v (wν )Aν j k s k +wν v (Aν j k )sk +wν Aν j k v (sk ) . (25.4) 99


Chapter 25. Curvature

Inserting this into the previous equation and writing Dw Dv s similarly, we find i

Dv Dw s − Dw Dv s = [v , w](si )ei + [ v , w]µ Aµ j s j ei

+



    w v A s e −v w A   ν

i ν j

i µj

µ

j

i

+ v µ wν Aµij Aν j k − Aν ij Aµj k sk ei . (25.5) Also,

i

D[v ,w] s = [v , w](si )ei + [ v , w]µ Aµ j s j ei ,

(25.6)

so that µ

F (v, w)s = v w

ν



∂ µ Aν ij −

i ∂ ν Aµ j +

i k Aµ k Aν kj − Aν ik Aµ j



s j ei .

(25.7)

Thus we can define F µν by F (∂ µ , ∂ ν )s = F µν s = (F µν s)i ei = (F µν )i j s j ei ,

(25.8)

so that (F µν )i j = ∂ µ Aν ij − ∂ ν Aµij + Aµik Aν kj − Aν ik Aµkj .

(25.9)

Note that since coordinate basis vector fields commute, [∂ µ , ∂ ν ] = 0 , F µν s = F (∂ µ , ∂ ν )s = D µ Dν s − Dν Dµ s = [Dµ , Dν ]s . (25.10)

It is not very difficult to work out that the curvature acts linearly on the module of sections, •

F (u, v )(s1 + f s2 ) = F (u, v )s1 + f F (u, v )s2 ,

(25.11)

where f ∈ C ∞ (B ) . Also, F (u, v + f w) s = F (u, v )s + f F (u, w)s . •

(25.12)

For coordinate basis vector fields [∂ µ , ∂ ν ] = 0 , so F µν s = F (∂ µ , ∂ ν )s = D µ Dν s − Dν Dµ s = [Dµ , Dν ]s . (25.13)

Since F (∂ µ , ∂ ν )s is a section, so is Dλ (F µν s) = D λ [Dµ , Dν ]s .

(25.14)


Similarly, since Dλ s is a section, so is F µν Dλ s = [Dµ , Dν ]Dλ s .

(25.15)

Thus Dλ (F µν s) − F µν Dλ s = [Dλ , [Dµ , Dν ]]s .

(25.16)

Considering C ∞ sections, and noting that maps are associative under map composition, we find that [Dλ , [Dµ , Dν ]]s + cyclic = 0 .

(25.17)

On the other hand, i

F µν s = (F µν s) ei =



i F µν j s j



ei ,

(25.18)

where F µν ij and si are in C ∞ (B ) . So we can write Dλ (F µν s) = ∂ λ



i F µν j s j

  +

i F µν j s j



Dλ ei

  ∂ s  e + F s e + F = ∂ F    i λ µν j

j

i µν j

i

λ

j

i j k µν j s Aλ i ek

i

= ∂ λ F µν ij + F µν kj Aλik s j ei + F µν ij ∂ λ s j ei i

k

j

− F µν k Aλ j s

i



j

ei + F µν j Aλ k sk ei

= (Dλ F µν )i j s j ei + F µν ij (Dλ s) j ei ,

(25.19)

where we have defined (Dλ F µν ) by (Dλ F µν )i j in this. Then this is a Leibniz rule, Dλ (F µν s) = (Dλ F µν ) s + F µν ( Dλ s) .

(25.20)

Then we can write Dλ (F µν s) − F µν (Dλ s) + cyclic = 0 ⇒

(Dλ F µν ) s + cyclic = 0

⇒

Dλ F µν + cyclic = 0 .

∀s

(25.21)

This is known as the Bianchi identity . Given D and g such that g (x) ∈ G , we have D  given by Dv φ = gDv g −1 φ .







(25.22)

Amitabha Lahiri_Lecture Notes on Differential Geometry for Physicists 2011.pdf

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