Andrew Pressley-Instructor's Solutions Manual to Elementary Differential Geometry-Springer (2012).pdf

Solutions to the Exercises in Elementary Differential Geometry Chapter 1 1.1.1 1.1.1 It is a parame parametriz trizatio ation n of the part of the parabola parabola with x 0. γ (t) = (sec t, tan t) with π/2 π/ 2  < t < π/2 π/ 2 and π 1.1.2 (i) γ (i) γγ  and  π /2  < t <  3π/  3 π/2. 2. Note that γ that  γγ  is defined defined on the union of two two disjoint disjoint interv intervals: als: this corresponds corresponds to the fact that 2 2 x = 1 is in two pieces, where y 1 and where y the hyperbola y 1. (ii) γ (ii) γγ (t) = (2 cos cos t, 3sin t). 1.1.3 (i) x + y = 1. (ii) y (ii)  y  = (ln x)2 . ˙  (t) = sin2t 1.1.4 (i) γγ  sin2t( 1, 1). ˙  (t) = (et , 2t). (ii) γγ  ˙  (t) = 3sin t cos t( cos t, sin t) vanishes where sin t   = 0 or cos t  = 0, i.e. t = 1.1.5 γγ  nπ/2 nπ/ 2 where n  is any any intege integer. r. These These points points corres correspond pond to the four cusps cusps of the astroid (see Exercise 1.3.3).

≥

 −

−

 ≥

 ≤ −

−

−

1.1.6 (i) The squares of the distances from p  to the foci are ( p cos t

± ǫp) ǫp)2 + q 2 sin2 t  = ( p2 − q 2 )cos2 t ± 2ǫp2 cos t + p2 =  p 2 (1 ± ǫ cos t)2 ,

so the sum of the distances is 2 p 2 p.. ˙   = (  p sin t, q cos ˙    = 0. Hence the q cos t) so if n = (q cos q cos t, p sin t) then (ii) γγ  th en n.γγ  the distances from the foci to the tangent line at γ at  γγ (t) are

−

( p cos t

ǫp,q sin t).n pq (1 ∓ ǫp,q sin (1 ∓ ǫ cos t) = 2 2 n ( p sin t + q 2 cos2 t)1/2 1

2

and their product is

p2 q 2 (1 ǫ2 cos2 t)  q 2 . ( p2 sin2 t+q 2 cos2 t) = (p ff 1  ).n

−

− −f  = (p f2  ).n . Compu tation on shows shows that p−ff 1   p−ff 2   Computati

(iii) It is enough to prove that both sides are equal to q .

1.1.7 When the circle has rotated rotated through through an angle t, its centre has moved to (at,a (at,a), ), so the point on the circle initially at the origin is now at the point (a(t

− sin t), a(1 − cos t))

(see the diagram above). 1.1.8 Suppose that a point (x,y,z (x,y,z)) lies on the cylinder if  x2 + y 2 = 1/4 and on the z sphere if (x (x  + 12 )2 +  y 2 + z2 = 1. From rom the second second equati equation on,, 1 1 2 1 3 so let z   = sin t. Subtra Subtractin ctingg the two two equatio equations ns gives gives x  + 4  + sin t = 4 , so 2 1 1 2 x  = 2 sin t  = cos t 2 . From either equation we then get y  = sin t cos t (or y  = sin t cos t, but the two solutions are interchanged by t π t). ˙   = ( 2sin t  + 2 sin 2t, 2cos t 2cos2t π/ 4. So th 1.1.9 γγ  2cos2t) = 2( 2 1, 1) at t = π/4. the 1 tangent line is y ( √ 2 1) = (x (x 2)/ 2)/( 2 1) and the normal line is

− ≤ ≤

 − − −

−

→ − √  √  − √  √  − −  − − − √  √  1 √  − − − − y ( 1) = (x 2)( 2 − 1). 1). 2

γ  (t) = (sec t, 1.1.10 1.1.10 (i) Putting Putting x   = sec t  gives y = sec t tan t so γ parametrizations of the two pieces of this curve (x (x 1 and x 2 3t 3t (ii) Putting y Putting  y  = tx  =  tx  gives x  = 1+t 1+t3 , y = 1+t 1+t3 .

 ±

≥

± sec t tan t) gives ≤ −1).

1.1.11 (i) From From x = 1 + cos t, y  = sin t(1 + cos t) we get y  = x  =  x sin t so y 2 =  x 2 (1

− (x − 1)2) = x3 (2 − x).

(ii) y (ii)  y  = tx  =  tx so x4 =  t 2 x3 + t3 x3 = y 2 x + y 3 =  y 2 (x + y ). ˙  (t) = ( sin t, cos t + cos ˙  (t) = 0  if and only if  t = nπ (n 1.1.12 1.1.12 (i) γγ  cos 2t) so γγ  ( 1)n + 1 = 0, so n  must be odd.

−

−

 ∈ Z) and

3

˙  (t) = (2t t(2+3t (ii) γγ  (2t + 3t2 , 3t2 + 4t3 ). This vanishes (2+3t) = 0 and t and t 2 (3+4t (3+4t) = 0, t  = 0. i.e. OP  make an angle θ  with the positive x-ax 1.1.13 (i) Let OP  make -axis. is. Then Then R  has coordinates γγ  (θ) = (2a (2a cot θ, a(1 cos2θ cos2θ)). y/ 2a, cos2 θ = (ii) From x = 2a cot θ, y = a(1  cos2θ  cos2θ ), we get sin2 θ = y/2 y/ 8a3 , so the Cartesian equation is y/2 y/ 2a + x2 y/8 y/ 8a3 = 1. cot2 θ sin2 θ  = x  =  x 2 y/8  a,, and the moving circle radius b 1.1.14 Let the fixed circle have have radius radius a radius b (so  (so that b that b < a in the case of the hypocycloid), and let the point P  point  P  of  of the moving circle be initially in contact with the fixed circle at (a, ( a, 0). When When the mo movin ving g circle has rotated rotated through an angle ϕ, the line joining the origin to the point of contact of the circles makes an angle θ  with the positive x-axis, where aθ = bϕ. The point point P  bϕ. The is then at the point

⇐⇒

 ⇐⇒  ⇐ ⇒

−

 −

γγ  (θ) = ((a ((a + b)cos θ =



− b cos(θ cos(θ + ϕ), (a + b)sin θ − b sin(θ sin(θ + ϕ)) a+b a+b θ , (a + b)sin θ (a + b)cos θ − b cos b b

 

 

in the case of the epicycloid,

and γγ  (θ) = ((a ((a =

b)cos θ + b cos(ϕ cos(ϕ

 −− (a

b)cos θ

− b cos

θ ), (a

− b sin(ϕ sin(ϕ − θ )) a−b , (a − b)sin θ

−− − a

b

b

θ

b)sin θ

  b

in the case of the hypocycloid. ˙  (t) = (et (cos t sin t), et (sin t + cos t)) so the angle θ   between γγ  (t) and γγ  ˙  (t) is 1.1.15 γγ  given by

−

cos θ  =

˙  γγ  .γγ  ˙  γγ   γγ 

       

e2t (cos2 t sin t cos t + sin2 t + sin t cos t) 1 , = = 2 e2t ((cos t sin t)2 + (sin t + cos t)2 )

−

−

4

so θ = π/3. ˙  (t) = (t cos t, t sin t) so a unit tangent vector is t  = (cos t, sin t). The distance 1.1.16 γγ  of the normal line at γγ (t) from the origin is

|γγ  (t)..t| = | cos2 t + t sin t cos t + sin2 t − t sin t cos t| = 1.   γγ ˙     = cosh t and the arc-length is s  = 0t cosh u du = sinh t. ˙    2 = 14 (1 + t) + 14 (1 − t) + 12 = 1. 1.2.2 (i)  γγ  2 2 2 9 2 2 γ  ˙    2 = 16 (ii)  γ 25  sin t + cos t + 25  sin t = cos t + sin t = 1. ˙   = (r˙ cos θ − r sin θ, r˙ sin θ + r cos θ) so  γγ  ˙    2 = r˙ 2 + r 2 . 1.2.3 Denoting d/dθ by a dot, γγ  Hence, γγ   is regular unless r = r = ˙ 0 for some value of  θ. It is unit-speed if and 2 2 only if r˙ = 1 − r , which gives r = ± sin(θ + α) for some constant α. To see

 

˙  (t) = (1, sinh t) so 1.2.1 γγ 

that this is the equation of a circle of radius 1/2, see the diagram in the proof  of Theorem 3.2.2. γ  ˙ .  u  = γ ˙    cos θ, where θ is the angle between γγ  ˙   and 1.2.4 Since u  is a unit vector, γγ  b b ˙ .  u ˙   . Then, (qq p)..u = (γγ  ˙ .  u dt ˙   dt. γγ  γ (b) γγ  (a))...u = a γγ  γγ  u, so γγ  a q p)/ q p  gives the result. Taking u  = (q γ  ˙  (t) = (6t, 1  9t2 ) so γ ˙  (t) = 36t2 + 1 18t2 + 81t4 = 1 + 9t2 . So the 1.2.5 γγ  arc-length is

 | |    −  −     √ 

 ≤   −  −

t

s =

 

 

−

−

 

 ≤   

(1 + 9u2 )du = t + 3t3 .

0

˙  (t) = ( 2sin t + 2 sin 2t, 2cos t 1.2.6 We have γγ 

 γγ ˙  (t)  =

−

− 2cos2t) so

 

 

8(1

− sin t sin2t − cos t cos2t =

So the arc-length is

x

s =

  0

t 4sin dt = 8 1 2

1.2.7 The cycloid is parametrized by γγ  (t) = a(t through which the circle has rotated. So ˙   = a(1 γγ 

 x 2

−  cos

8(1

− cos t) = 4sin 2t .

= 16 sin2

x . 4

− sin t, 1 − cos t), where t is the angle

− cos t, sin t),  γγ ˙    2 = a2(2 − 2cos t) = 4a2 sin2 2t ,

and the arc-length is 2π

  0

t 2a sin  dt = 2

 −

t 4a cos 2

 

t=2π

= 8a. t=0

5

1.2.8 The curve intersects the x-axis where cosh t = 3, say at t = ˙  (t) 2 = 2 cosh2 t 2 cosh t so the arc-length is have γγ 

 



s =

−

a

   

2

2 cosh t

−a

a

− 2 cosh t dt = 2

   

2 cosh2 t

0

±a where a > 0. We − 2 cosh tdt.

To evaluate the integral put u = cosh t. Then, 3

s = 2

   

u du = u+1

1

3 √  √   2 + √  √  √  √ 3 . u(u + 1) − ln( u + u + 1) = 2 3− 2−ln

 



1+

1

2

... b  + t 2 c, where a, b, c   are constant vectors. This implies 1.2.9 γγ   = 0 = γγ   = a  + tb that γγ   is contained in the plane passing through the point with position vector a  and parallel to the vectors b  and c   (i.e. perpendicular to b c). (If b  and c are parallel there are infinitely-many such planes.)

⇒

×

γ  γ   is not ˙   = sin2t( 1, 1) vanishes when t  is an integer multiple of  π/2, so γ 1.3.1 (i) γ regular. ˙   = 0  for 0 < t < π/2. (ii) γγ   is regular since γγ  ˙   = (1, sinh t) is obviously never zero, so γγ   is regular. (iii) γγ  1.3.2 x = r cos θ   = sin2 θ, y = r sin θ   = sin2 θ tan θ, so the parametrization in terms of  θ is θ   (sin2 θ, sin2 θ tan θ). Since θ sin θ   is a bijective smooth map ( π/2, π/2) ( 1, 1), with smooth inverse t sin−1 t, t   = sin θ is a reparametrization map. Since sin2 θ = t2 , sin2 θ tan θ = t3 / 1 t2 , so the reparametrized curve is as stated. ˙   = 0 at t = 0 m and n are both  2. If m > 3 the first components of  1.3.3 (i) γγ   ... γ   are both 0 at t = 0 so γγ  ¨  and γ ¨  and γγ  ¨  are linearly dependent at t = 0; similarly γγ  ... γ   ¨  or γ if  n > 3. So there are four cases: if (m, n) = (2, 2) or (3, 3) then either γγ  is zero at t  = 0, so the only possibilities for an ordinary cusp are (m, n) = (2, 3)  ... ¨  and γγ   are easily seen to be linearly independent at t  = 0. and (3, 2) and then γγ  3 ˙   = 0, γγ  ¨  = (2, 0), (ii) Using the parametrization γγ  (t) = t2 , √ 1t−t2 , we get γγ  ... γγ   = (0, 6) at t = 0 so the origin is an ordinary cusp. γ  ˜ (t˜) be a reparametrization of  γγ  (t), and suppose γγ    has an ordinary (iii) Let γ γ  γ  ˜ /dt˜2 = γγ  /dt˜ = (dγ γ /dt)(dt/dt˜) = 0 , d2 γ cusp at t = t0 . Then, at t = t0 , d˜ ˜ /dt˜3 = (d3γγ  /dt3 )(dt/dt˜)3 + 3(d2 γγ  /dt2 )(dt/d˜t)(d2 t/dt˜2 ). (d2γγ  /dt2 )(dt/dt˜)2 , d3 γγ  γ  ˜ /d˜t2 and d3 γ ˜ /dt˜3 are Using the fact that  dt/dt˜ = 0, it is easy to see that d2γγ  linearly independent when t = t 0 . ˜ (t) = γγ  γ (ϕ(t)), let ψ be the inverse of the reparametrization map ϕ. Then 1.3.4 (i) If γγ  ˜ (ψ(t)) = γγ  γγ  γ (ϕ(ψ(t))) = γγ  γ (t). ˜ (t) = γγ (ϕ(t)) and γγ  ˆ (t) = γγ  ˜ (ψ(t)), where ϕ and ψ  are reparametrization (ii) If  γγ  ˆ (t) = γγ  γ ((ϕ ψ)(t)) and ϕ ψ is a reparametrization map because it maps, then γγ  d  ˙ ˙ ψ(t) is smooth and dt (ϕ(ψ(t)) = ϕ(ψ(t)) ˙ = 0 as ϕ˙ and ψ are both = 0.

−

 

−

→  → −

→ 

 ⇐⇒

→ 

√  −

 ≥





 

◦

◦



 

6

˙  (t) = (2t, 3t3 ) which vanishes t = 0. So γγ   is not regular. 1.3.5 (i) γγ  ˙  (t) = ( sin t sin2t, cos t + cos 2t) so γγ  ˙  (t) = 2 cos 2t . Hence, γγ  ˙  (t) = (ii) γγ  t   is an odd multiple of  π. Thus, when t   is retricted to the interval 0 π < t < π, γγ   is regular.

−

−

⇐⇒

˙  (t) = 1.3.6 γγ 



 ⇐⇒

−

 

 



4t γ   is regular. Let 2, (1+t  is obviously never zero, so γ 2 )2 dϕ 1 dt = 1+sin t   is never zero so ϕ   is a reparametrization

−

Then, u = ϕ(t) we get γγ  (u) =

−



  2cos t 2 , cos2 t 1 + sin t 1 + (1+sin t)2

  =

  2cos t  2(1 + sin t)2 , 1 + sin t 2 + 2 sin t

ϕ(t) =

cos t 1+sin t .

map. Setting



γγ  (t), = ˜

˜ . so γγ   is a reparametrization of  γγ  ˙  (t) = (aω cos ωt,b cos t) = 0 ωt = (2k +1) π2 and t = 1.3.7 γγ  cos ωt = cos t = 0 (2l + 1) π2 , where k, l Z. This means that ω = 2k+1 2l+1  is the ratio of two odd integers.

⇐⇒

 ∈

t

  

⇐⇒

t˜

  

γ  ˜ /d˜ dγ γ /du du, s˜ = t˜0 dγγ  u d˜ u. By the chain rule, 1.3.8 We have s = t0 t γ  ˜ /d˜ ˜ /d˜ dγ γ /du = (dγγ  u)(d˜ u/du), so s = dγγ  u (d˜ u/du) du = s˜, the sign t0 being that of  d u ˜/du. γ  is unit-speed and that the tangent lines all pass 1.3.9 We can assume that the curve γγ  through the origin (by applying a translation to γγ ). Then, there is a scalar λ(t) ˙ γ  γ  γ  ˙  (t) = λ(t)γγ  ¨  = λγ ˙   = (λ˙ + λ2 )γγ  γ (t) for all t. Then, γ γ + λγ γ . This implies such that γγ  ¨  is parallel to γγ  ˙  . But these vectors are perpendicular since γγ   is unit-speed. that γγ  ¨  = 0, so γγ  b  where a , b  are constant vectors, i.e. γγ   is a straight γ (t) = taa +b Hence, γγ  line. If all the normal lines are parallel, so are all the tangent lines. If  γγ   is assumed to ˙   is constant, say equal to aa. Then γγ (t) = taa + b be unit-speed, this means that γγ  as before.



   ± 



 

 ±

γ  (t) for all t. Suppose that γγ  (t) = γγ  (u). 1.4.1 It is closed because γγ  (t + 2π) = γ Then cos3 t(cos3t, sin3t) = cos3 u(cos3u, sin3u). Taking lengths gives cos3 t = cos3 u so cos t = cos u, so u = t, π t, π + t or 2π t (up to adding multiples of 2π). The second possibility forces t = nπ/3 for some integer n and the third possibility is true for all t. Hence, the period is π and for the self-intersections we need only consider t = π/3, 2π/3, giving u = 2π/3, π/3, respectively. Hence, γ  (π/3) = ( 1/8, 0). there is a unique self-intersection at γ ˜ (t) = (cos(t3 + t), sin(t3 + t)) is a reparametrization of the circle 1.4.2 The curve γγ  γγ  (t) = (cos t, sin t) but it is not closed. 1.4.3 If  γγ   is T -periodic then it is kT -periodic for all k = 0 (this can be proved by induction on k if  k > 0, or on k if  k < 0). If  γγ  is T 1 -periodic and T 2 -periodic

±

±

−

−

−

 −



7

then it is k1 T 1 - and k2 T 2 -periodic for all non-zero integers k1 , k2 , so γγ  (t + k1 T 1  + k2 T 2 ) = γγ  γ (t + k1 T 1 ) γ (t) as γγ  is k1 T 1 -periodic. as γγ  is k2 T 2 -periodic, which = γγ  1.4.4 If  γγ   is T -periodic write T  = kT 0 + T 1  where k  is an integer and 0 T 1  < T 0 . By Exercise 1.4.3 γγ  is T 1 -periodic; if  T 1 > 0 this contradicts the definition of  T 0 .

≤

γ  is T 1 -periodic; then T 1  is not the smallest positive 1.4.5 (i) Choose T 1 > 0 such that γγ  number with this property, so there is a T 2 >   0 such that γγ   is T 2 -periodic. Iterating this argument gives the desired sequence. (ii) The sequence T r r≥1  in (i) is decreasing and bounded below, so must converge to some T ∞   0. Then γγ   is T ∞ -periodic because (using continuity of  γγ  ) γγ  (t + T ∞ ) = limr→∞ γγ  (t + T r ) = limr→∞ γγ  (t) = γγ  (t). By Exercise 1.4.3, γγ   is (T r T ∞ )-periodic for all r 1, and this sequence of positive numbers converges to 0. (iii) If  T r  is as in (i) and T r  0 as r , then by the mean value theorem ˙ + λT r ) for some 0 < λ < 1. Letting r 0 = (f (t + T r ) f (t))/T r = f (t  gives ˙ = 0 for all t, so f  is constant. f (t)

 { }  ≥

−

 ≥

 { }

 →

−

 → ∞

 → ∞

1.4.6 Following the hint, since T 0 = (ki /k)T i  is an integer multiple of  T i , each γ i is T 0 periodic. Let  be the union of the finite sets of real numbers T i , 2T i , . . . , ki T i γγ   is T ′ -periodic . over all i  such that γ i  is not constant, and let = T ′ Then   is finite (because  is) and non-empty (because T  ). The smallest ′ ′ γ  is T 0 -periodic (since element of    is the smallest positive number T 0  such that γγ  ′ ′ ′ if  γγ   is T  -periodic either T  > T  or T  ). By Exercise 1.4.4, T 0 = k′ T 0′ for some integer k′  and then there are integers ki′  such that T 0′ = k i′ T i  for all i such that γ i  is not constant. Then, ki T i /k = k′ ki T i so kk ′  divides each ki . As k is the largest such divisor, k′  = 1, so T 0  =  T 0′ .

T 

 P 

 {  P   {  ∈ T |  ∈ P 

 T 

 P 

} }

 ∈ P 

2

2

u −3 3 2 2 1.4.7 If  γγ  (t) = γγ  (u) then tt2− +1 = u2 +1 . This implies t = u . This shows that γγ    is not periodic and that for a self-intersection we must have t = u. The 2 −3) = 0, so t = 0 or γ ( t) implies t(tt2 +1 equation γγ  (t) = γγ  3. Hence, the unique γ ( self-intersection is at γγ ( 3) = γγ  3) = 0.

− √ 

−√ 

 −

√   ±

γ  (t) 2 = γγ  (u) 2 , which gives γ 1.4.8 If  γγ  (t) = γγ  (u) then sin t   = sin u. Also, 5 + 4 cos t = 5 + 4 cos u, hence cos t = cos u. So t = u + 2nπ for some integer n. We must have cos ω(t + 2nπ) = cos ωt so ω(t + 2nπ) = ωt + 2mπ for some integer m. Hence, ω = m/n   is rational. The period T  of  γγ    is the smallest positive number such that γγ  (t + T ) = γγ  (t) for all t. From the first part we know that T  = 2Nπ  for some integer N . Then, N  is the smallest positive integer N   such that ω(t+2Nπ) = ωt+2M π for some integer M , i.e. the smallest positive integer such that mN/n is an integer. If  m and n have no common factor, N  = n.



 



8

 x 1.5.1 x(1 x2 )  0 1 or 0  x  1 so the curve is in (at least) two pieces. The parametrization is defined for t 1 and 0  t  1 and it covers the part of the curve with y 0. γ (t) = (x(t), y(t), z(t)) is a curve in the surface f (x,y,z) = 0, differentiating 1.5.2 If γγ  ˙   is perpenf (x(t), y(t), z(t)) = 0 with respect to t gives xf  ˙ x + yf  ˙ y + zf  ˙ z  = 0, so γγ  dicular to f  = (f x , f y , f z ). Since this holds for every curve in the surface, f  is perpendicular to the surface. The surfaces  f  = 0 and g = 0 should intersect in a curve if the vectors f  and g are not parallel at any point of the intersection. 1.5.3 Let γγ  (t) = (u(t), v(t), w(t)) be a regular curve in R3 . At least one of u, ˙ ˙ v, w˙ is non-zero at each value of  t. Suppose that u(t ˙ 0 ) = 0 and x0 = u(t0 ). As in the ‘proof’ of Theorem 1.5.2, there is a smooth function h(x) defined for x near x0   such that t = h(x) is the unique solution of  x = u(t) for each t   near t0 . γ (t) is contained in the level curve f (x,y,z) = g(x,y,z) = 0, Then, for t  near t 0 , γγ  where f (x,y,z) = y v(h(x)) and g(x,y,z) = z w(h(x)). The functions f  and g   satisfy the conditions in the previous exercise, since f  = ( vh ˙ ′ , 1, 0), g = ( wh  ˙ ′ , 0, 1), a dash denoting d/dx. 1.5.4 The equation cannot be satisfied if  x = 1, so the line x = 1 separates the curve into at least two connected pieces. The curve γγ   is a parametrization because

−  ≥  ⇐⇒  ≤ −  ≥

 ≤  ≤  ≤ −

 ≤  ≤

 ∇∇

 ∇∇

∇

∇

 

 −

∇

 −

 ∇∇

−

(x

−

− 1)2(x2 + y2 ) = cos2 t((1 + cos t)2 + (sin t + tan t)2)

= cos2 t(1 + cos t)2 (1 + tan2 t) = (1 + cos t)2 = x 2 .

γ  must be defined on one But since tan t is undefined when t = π/2, 3π/2,..., γγ  of the intervals ( π/2, π/2), (π/2, 3π/2), etc. The restriction of  γγ   to the first of  these intervals parametrizes the part of the curve with x > 1, the second gives the part with x < 1. This shows that the two parts of the curve with  x < 1 and with x > 1 are connected. 1.5.5 Letting f (x,y,z) = x2 + y 2 = 1/4, g(x,y,z) = x2 + y 2 + z 2 + x 3/4, we find ∂f  ∂f  ∂g ∂g ∂g that ( ∂f  ∂x , ∂y , ∂z ) = (2x, 2y, 0), ( ∂x , ∂y , ∂z ) = (2x + 1, 2y, 2z). If these vectors are parallel, z = 0; if  y = 0 the vectors must then be equal and this is impossible; so y = 0 and then we must have x = 1/2 to satisfy both equations. Hence, the vectors are parallel only at the point (1/2, 0, 0). ˙  (t) = ( sin2t, cos2t, cos t). This For the parametrization γγ   in Exercise 1.1.8, γγ  2 2 ˙  (t) = 1 + cos t, so γγ   is regular. is never zero as γγ  1.5.6 Define Θ(t) = tan πθ(t)/2, where θ   is the function defined in Exercise 9.4.3. Then Θ is smooth, Θ(t) = 0 if  t 0, and Θ : (0, ) (0, ) is a bijection. The curve (Θ(t), Θ(t)) if t  0, γγ  (t) = ( Θ( t), Θ( t)) if  t 0,

±

−

±

−

 

 

−



≤



− − is a smooth parametrization of  y = |x|.

∞→ ∞ ≥ − ≤

9

There is no regular parametrization of  y = x . For if there were, there would ˜ (t), say, and we can assume that γγ  ˜ (0) = (0, 0). be a unit-speed parametrization γγ  1 1 ˜˙   would have to be either √  The unit tangent vector γγ  (1, 1) or √  (1, 1) when 2 2 1 ˜˙  (0) = √  x > 0, so by continuity we would have γγ  (1, 1). But, by considering the

 | | ±

 −

˜˙  (0) = part x < 0 in the same way, we see that γγ  are contradictory.

2

 ± √ 12 (1, −1). These statements

Chapter 2 2.1.1 (i) γγ   is unit-speed (Exercise 1.2.2(i)) so 1  1 ¨  = ( (1 + t)−1/2 , (1 κ = γγ  4 4

   

− t)−1/2, 0)  =

1 . 8(1 t2 )

 

−

¨  = ( 45  cos t, sin t, 35  cos t) (ii) γγ   is unit-speed (Exercise 1.2.2(ii)) so κ = γγ  = 1. t,0)×(0,cosh t,0) 2 cosh t (iii) κ = (1,sinh =  cosh 3 t  = sech t using Proposition 2.1.2. (1,sinh t,0)3 (iv) ( 3cos2 t sin t, 3sin2 t cos t, 0) ( 3cos3 t+6 cos t sin2 t, 6sin t cos2 t 3sin3 t, 0) (0,0,−9sin2 t cos2 t) 1 = (0, 0, 9sin2 t cos2 t), so κ = (−3 cos 2 t sin t,3sin 2 t cos t,0)3 = 3| sin t cos t| . This becomes infinite when t   is an integer multiple of  π/2, i.e. at the four cusps ( 1, 0) and (0, 1) of the astroid.

    −

−

×−

−

±

 

−

±

2.1.2 The proof of Proposition 1.3.5 shows that, if v(t) is a smooth (vector) function of  t, then v(t)  is a smooth (scalar) function of  t  provided v(t) is non-zero for all t. The result now follows from the formula in Proposition 2.1.2. The curvature of the regular curve γγ  (t) = (t, t3 ) is κ(t) = 6 t /(1 + 9t4 )3/2 , which is not differentiable at t  = 0. ˙  (t) = (1 cosh 2t, 2 sinht, 0) = 2sinh t( sinh t, 1, 0), γγ  ¨ (t) = 2.1.3 Working in R3 , γγ  ˙ ¨   γ  ×γ γ  γ  γ  ˙  γ ¨  = (0, 0, 4 sinh2 t cosh t) and κ = γγ ( 2sinh2t, 2cosh t, 0) which gives γ ˙  3 = γ 

 

 

||

−

−

×

−

1/2 sinh t cosh2 t. This is non-zero if  t > 0, but

 → 0 as t → ∞.

10

2.1.4 We have ˙  (t) = (sec t tan t, sec t tan2 t + sec3 t, 0), γγ  ¨ (t) = (2 sec3 t γγ 

− sec t, 6sec3 t tan t − sec t tan t),

˙   γγ  ¨  = sec4 t(0, 0, 2sec2 t 3). Hence, the curvature vanishes where which gives γγ  sec2 t = 3/2. If  π/2   < t < π/2, cos t >  0 so the curvature vanishes at the two values of  t  between π/2 and π/2 at which cos t = 2/3, i.e. at the point (3/2, 5/2).

×

√ 

 −

−

 −

ns  = 0 and use t˙  = κsns . 2.2.1 Differentiate t.n ˙   is smooth and hence so are t˙  and ns  (since ns  is obtained γ  is smooth, t  = γγ  2.2.2 If γγ  ns  is smooth. by applying a rotation to t). So κs = t˙ .n 2.2.3 For the first part, from the results in Appendix 1 it suffices to show that κ ˜s = κs if  M  is the reflection in a straight line l. But this is clear: if we take the fixed angle ϕ0  in Proposition 2.2.1 to be the angle between l and the positive x-axis, γ  ˜  have the same then (in the obvious notation) ϕ˜ = ϕ. Conversely, if  γγ   and γ non-zero curvature, their signed curvatures are either the same or differ in sign. In the first case the curves differ by a direct isometry by Theorem 2.2.5; in the latter case, applying a reflection to one curve gives two curves with the same signed curvature, and these curves then differ by a direct isometry, so the original curves differ by an opposite isometry.

−

 −

2.2.4 The first part is obvious as the effect of the dilation is to multiply s by a and leave ϕ unchanged. For the second part, consider the small piece of the chain between the points with arc-length s  and s + δs. The net horizontal force on this piece is (in the obvious notation) δ (T  cos ϕ), and as this must vanish T  cos ϕ must be a constant, say λ. The net vertical force is δ (T  sin ϕ), and this must balance the weight of the piece of chain, which is a constant multiple of  δs. This shows that T  sin ϕ = µs + ν  for some constants µ, ν , and ν  must be zero because ϕ = s  = 0 at the lowest point of  . From T  cos ϕ =  λ, T  sin ϕ = µs, we get tan ϕ = s/a  where a = λ/µ. Hence, sec2 ϕ dϕ = 1/a, so the signed curvature is ds

 C

κs = dϕ/ds = 1/a sec2 ϕ = 1/a(1 + tan2 ϕ) =

a . s2 + a2

Using the first part and Example 2.2.4 gives the result. γ  γ  ns /dt = (1  λκ s )ds/dt t, so the arc-length sλ γ λ /dt = dγ γ /dt + λdn 2.2.5 We have dγ γ λ is tλ = of  γγ  λ satisfies dsλ /dt = 1 λκs ds/dt. The unit tangent vector of γγ  γ  ns . Then, γ λ /dt)/(dsλ/dt) = ǫtt, hence the signed unit normal of  γγ λ is nλs = ǫn (dγ

 | −

|

−

11

the signed curvature κλs of  γγ λ is given by κλsnλs

dttλ dttλ /dt ǫ = λ = = 1 λκs (ds/dt) 1 λκs ds κs ǫκs κs ns  = nλs = = 1 λκs 1 λκs 1 λκs

| −

−

| −

dtt ds

| − | nλs . | − |

2.2.6 ǫ(s0 ) lies on the perpendicular bisector of the line joining γγ  (s0 ) and γγ  (s0  + δs). γ (s0 ) +γγ  γ (s0 + δs)))...(γγ  γ (s0 + δs) γγ  (s0 )) = 0. Using Taylor’s This gives (ǫǫ(s0 ) 12 (γγ  theorem, and discarding terms involving powers of δs higher than the second, this ˙  δs +  12 (ǫǫ.γγ  ¨   1 γγ .  γγ  ¨  )(δs)2 = 0. gives (with all quantities evaluated at s 0 ) (ǫǫ γγ  ).γγ  This must also hold when δs  is replaced by δs; adding and subtracting the ˙   = 0 and (ǫǫ γγ  ).γ¨γ   = 1. The first equation gives two equations gives (ǫǫ γγ  ).γγ  ns  for some scalar λ, and since γγ  ¨  = κ sns  the second gives λ = 1/κs . ǫ  = γγ  γ + λn 2.2.7 The tangent vector of  ǫ   is t  + κ1s ( κst) κκ˙ s2 ns = κκ˙ s2 ns   so its arc-length is s s ˙s κ 1 u= ǫ˙ ds = κ2 ds = u0 κs , where u0   is a constant. Hence, the unit s ns /du = tangent vector of  ǫ is ns   and its signed unit normal is t. Since dn 3 κst/(du/ds) = κκ˙ ss t, the signed curvature of  ǫ is κ3s /κ˙ s . ns (s) for some λ. Hence, Any point on the normal line to γγ   at γγ  (s) is γγ  (s) + λn the normal line intersects ǫ  at the point ǫ (s), where λ = 1/κs (s), and since the tangent vector of  ǫǫ  there is parallel to ns (s) by the first part, the normal line is tangent to ǫǫ at ǫǫ(s). Denoting   d/dt   by a dash, γγ  ′ = a(1 cos t, sin t) so the arc-length s of  γγ   is γ  γ /ds   = (sin(t/2), cos(t/2)). So n s = given by   ds/dt = 2a sin(t/2) and t = dγ 1 ( cos(t/2), sin(t/2)) and ˙t  = (dtt/dt)/(ds/dt) = 4a sin(t/2) (cos(t/2), sin(t/2)) = ns , so the signed curvature of  γγ  is 1/4a sin(t/2) and its evolute 1/4a sin(t/2)n

−

−

−  − −

−

   

 

 −

 −

−

− −

 −

 −

 −

 −

− −

−

 − ǫ(t) = a(t − sin t, 1 − cos t) − 4a sin(t/2)(− cos(t/2), sin(t/2)) = a(t + sin t, −1 + cos t). Reparametrizing ǫǫ by t˜ = π + t, we get a(t˜ − sin t˜, 1 − cos t˜) + a(−π, −2), so ǫǫ is obtained from a reparametrization of  γγ   by translating by the vector a(−π, −2). γ  at γγ  γ (s) and has length ℓ − s, hence the 2.2.8 The free part of the string is tangent to γγ  ˙   − γγ  ˙  + (ℓ − s)γγ  ¨  = κs (ℓ − s)n ns stated formula for ιι(s). The tangent vector of  ι  is γγ  (a dot denotes   d/ds). The arc-length v of  ι  is given by dv/ds = κs (ℓ − s) so ns /dv = its unit tangent vector is ns  and its signed unit normal is  −t. Now dn − 1 1  ˙ κ (ℓ−s) n s = ℓ−s t , so the signed curvature of  ιι is 1/(ℓ − s). √  ˜ (s) = (sinh−1 s, 1 + s2 ). 2.2.9 The arc-length parametrization of the catenary is γγ  s

The involute



s ˜˙  (s) = sinh−1 s − √  ι(s) =  ˜ γγ  (s) − sγγ 

1 + s2

,

1 1 + s2

√ 



= (u

− tanhu, sechu)

12

if  u = sinh−1 s. Thus, if (x, y) is a point on the involute ιι, u = cosh−1 (1/y) and x = cosh−1 (1/y) 1 y2. 2.2.10 The rotation ρ−θ(s)  takes the tangent line of  γγ   at γγ (s) to l and the line joining q  and γγ (s) to a line parallel to that joining Γ(s) to p + saa. Hence,

  − −

Γ(s)

− (pp + saa) = ρ−θ(s)(qq − γγ  (s)),

which gives the stated equation. Now, Γ˙ (s) = a +





d q ρ (q ds −θ(s)

− γγ  (s)) − ρ−θ(s)γγ ˙  (s).

The last term is clearly parallel to aa and as they are both unit vectors they are equal. So we want to prove that





d q ρ (q ds −θ(s)

− γγ  (s))...ρ−θ(s)(qq − γγ  (s)) = 0.

v)t dA v If  A = ρ −θ(s) , v  = q γγ  (s), we have to show (in matrix notation) (Av ds v  = 0, v i.e. vt At dA ds v   = 0. Since A is orthogonal, this follows from parts (i) and (ii) of  the hint. To prove (i), use components: vt S v  = i,j vi vj S ij = i,j v j vi S ji = i,j  vi vj S ij . t For (ii), differentiate A A = I . 2.2.11 If two unit-speed curves have the same non-zero curvature, their signed curvatures are either the same or differ in sign. In the first case the curves differ by a direct isometry by Theorem 2.2.6; in the latter case, applying a reflection to one curve gives two curves with the same signed curvature by Exercise 2.2.3, and these curves then differ by a direct isometry. 2.2.12 By applying a translation we can assume that all the normal lines pass through ns (t) for all t. the origin. Then, there is a scalar λ(t) such that γγ  (t) = λ(t)n ˙ ns λκs t  using Exercise 2.2.1. It follows that λ = ˙ Differentiating gives t  = λn 0, so λ is constant, and that λκs = 1, so κs = 1/λ  is constant. Hence, γγ   is a circle by Example 2.2.7. 2.2.13 The arc-length

−



−

s = =

 −

dt =

ekt (k cos t

k2 + 1ekt dt =

−



 −

        √      ˙  γγ 



− sin t, k sin t + cos t)  dt

k2 + 1 kt e + c, k

where c is a constant. Taking c = 0 makes s

→ 0 as t → ∓∞ if  ±k > 0.

13

˙  / γγ  ˙   = √ k12 +1 (k cos t sin t, k sin t + cos t), the signed unit normal Since t  = γγ  is ns = √ k12 +1 ( k sin t cos t, k cos t sin t). So

    −

−

−

dtt dtt/dt e−kt = = 2 ( k sin t ds ds/dt k +1

−

−

e−kt ns , k2 + 1

− cos t, k cos t − sin t) = √ 

hence κs = 1/ks. By Theorem 2.2.6, any other curve with the same signed curvature is obtained from the logarithmic spiral by applying a direct isometry. 2.2.14 Let δδ (t) be the foot of the perpendicular from p  to the tangent line of  γγ  at γγ (t). ns   for some scalar λ = (δδ  δ  p)..ns . But γγ   δδ   is parallel to the Then δδ   = p  + λn γ  δδ )  .ns  = 0. Hence, δδ .  ns = γγ .ns and tangent line, so is perpendicular to ns : (γγ  γ  γ  p)..ns . we get λ = (γ ˙   = κs (((γγ  ˙ = 0 γ  ns +((γ γ  p)..t)n γ  p)..ns )tt). So δδ  Using Exercise 2.1.1, we find that δδ  ns + ((γγ  γ  p)..t)n γ  p)..ns )tt) = 0. Since ns  and t if and only if either κs  = 0 or (((γγ  γ  p)..t  = (γγ  γ  p)..ns = 0, are perpendicular, the last equation can hold only if (γγ  which implies that γγ   = p. Hence, δδ    is regular if and only if  γγ    has nowhere vanishing curvature and does not pass through p. For the circle γγ (t) = (cos t, sin t), ns = γγ   and so

 −  −

−

−

−

 −

−

−

−

−

−

−

δδ  (t) = ( 2, 0) + (2 + cos t, sin t)..(cos t, sin t)(cos t, sin t)

−

= ((1 + 2 cost)cos t

− 2, (1 + 2 cost)sin t),

which is obtained by translating the lima¸con in Example 1.1.7 by ( 2, 0). ns . Since t  and ns  are perpendicular 2.2.15 (i) Differentiating γγ  = rtt  gives t  = rt ˙ t + κs rn vectors, it follows that κs  = 0 and γγ   is part of a straight line. γ   = rn ns  gives t = rn (ii) Differentiating γ ˙ ns  + r n˙ s = rn ˙ ns κs rtt  (Exercise 2.1.1). γ  is part Hence, r = ˙ 0, so r  is constant, and κ s = 1/r, hence κ s  is constant. So γγ  of a circle. (iii) Write γγ  = r(tt cos θ + ns sin θ). Differentiating and equating coefficients of t and ns   gives r˙ cos θ κs r sin θ = 1, r˙ sin θ + κs r cos θ = 0, from which r = ˙ cos θ and κs r = sin θ. From the first equation, r = s cos θ   (we can assume the arbitrary constant is zero by adding a suitable constant to s) so κ s = 1/s cot θ. γ  is obtained by applying a direct isometry to the logarithmic By Exercise 2.2.13, γγ  spiral defined there with k = cot θ. 1 1 √ 1+4t 2.2.16 For the parabola γγ  (t) = (t, t2 ), t = √ 1+4t 2t, 1), so 2 (1, 2t), n s = 2(

−

 −

−

−

−

−

−

γ λ

γ   (t) =

− t

2λ , t2 1+4t2

√ 

λ 1+4t2

+ √ 



λ

˙  = . This gives γγ 

− 1

−

2λ (1+4t2 )3/2



(1, 2t).

˙  λ (t) = 0 Hence, γγ  2λ  = (1 + 4t2 )3/2 . If  λ < 1/2 this has no solution for t, so γγ  λ is regular. If  λ = 1/2, the only solution is t = 0 so (0, 1/2) is the γ λ has unique singular point of  γγ  . If  λ > 1/2 there are two solutions for t, hence γγ  two singular points.

⇐⇒

14

˙ = 2γγ  ˙ 0) = 0 ˙ .  (γγ  ˙  (t0 ) is γ  c). So F (t0 ) = F (t γγ  (t0 ) c = R and γγ  2.2.17 F  perpendicular to γγ  (t0 ) c. This means that γγ  (t0 ) is on the circle   and that ˙  (t0 ) = t(t0 ) is parallel to the tangent to  at that point (since this tangent is γγ  perpendicular to the radius γγ (t0 ) c). ¨ = 2t˙. (γγ  ¨ 0) = 0 γ  c)+2 (since γγ   is unit-speed), so F (t γ (t0 ) c) = Now F  t˙ (t0 ).(γγ  ns (t0 ) is perpendicular to t(t0 ) (since γγ   is unit-speed), 1. Now t˙ (t0 ) = κs (t0 )n ˙ 0 ) = 0. So the condition F (t ¨ 0) = hence parallel to γγ  (t0 ) c  from the condition F (t ns (t0 ) for some scalar λ  and we have κ s (t0 )n ns (t0 )..λn ns (t0 ) = 0 gives γγ (t0 ) c  = λn ˙ 0 ) = F (t ¨ 0 ) = 0 are 1 which gives λ = 1/κs (t0 ). So the conditions F (t0 ) = F (t 1 γ (t0 ) + κs (t satisfied if and only if the centre of   is c  = γγ  ns (t0 ) and its radius is 0) γγ  (t0 ) c = λ = 1/ κs (t0 ) . Comparing with Exercise 2.2.6 shows that is γ  (t0 ). the osculating circle of  γγ  at γ

 −

−

−

−

−

 − 

 C

⇐⇒

−

−

−

− 

⇐⇒   C

−

−    | |

|

|

 C

 C

1 1 √ 1+4t 2.2.18 From the solution of Exercise 2.2.16, tt = √ 1+4t 2t, 1), 2 (1, 2t), ns = 2(

− √  ˙     = 1 + 4t2 (a dot denoting d/dt). and if  s is the arc-length of  γγ  ,  ds/dt =  γγ  Hence,

dtt = ds

1 t˙  = 2 1 + 4t

√ 

−

4t 2 , (1 + 4t2 )2 (1 + 4t2 )2



,

hence κs  = 2(1 + 4t2 )−3/2 . The evolute is therefore 1 1 ǫ  = γγ  γ + ns = (t, t2 ) + (1 + 4t2 )( 2t, 1) = κs 2

−

−

1 4t3 , 3t2 + 2



.

 

γ  ,   ds/dt = a2 sin2 t + b2 cos2 t = 1/λ, say. Then, 2.2.19 If  s   is the arc-length of  γ ˙   = λ( a sin t, b cos t), ns = λ( b cos t, a sin t), denoting d/ds by a dot, t = γγ 

− − − ns t˙  = λ[λ(−a cos t, −b sin t) − λ3 (a2 − b2 )sin t cos t(−a sin t, b cos t)] = λ 3 abn

after some simplification. Hence, κs = λ 3 ab = λ ǫ  = (a cos t, b sin t) + 3 ( b cos t, a sin t) = λ ab

−

−

ab   and (a2 sin2 t+b2 cos2 t)3/2



a2

so

− b2 cos3 t, b2 − a2 sin3 t a



b

.

ns . Then, γγ  ˙  λ = γγ  ˙  +λn˙ s = (1 λκs )n ns . Assuming that λκs = 1 (so γ λ = γγ  γ +λn 2.2.20 Let γγ  λ λκs , tλ = t, that γγ  λ is regular), we have (in an obvious notation) ds ds = 1 nλs = ns , where  is the sign of 1 λκs . Hence,

−

±

 ±

−

 | −

±t˙ = ± κsns = κsnλs . | − λκs| |1 − λκs | |1 − λκs|

dttλ = 1 dsλ

|

 

 ±

15

s So κλs = |1−κλκ and s|

ns + ǫλ = γγ  γ + λn

 |1 − λκs | (±ns) = γγ γ + λnns + 1 − λκs ns  = ǫǫ. κs

κs

2.2.21 (i) With the notation in Exercise 2.2.7, the involute of  ǫǫ is ι (u) = ǫǫ + (ℓ

dǫǫ 1 ns  = γγ  ns , − u) du γ + ns − (ℓ − u)n γ  − (ℓ − u0 )n = γγ  κs

since u = u 0 κ1s , so ιι  is the parallel curve γγ −(ℓ−u0 ) . (ii) Using the results of Exercise 2.2.8, the evolute of  ιι is

−

ι + (ℓ

− s)(−t) = γγ γ + (ℓ − s)tt − (ℓ − s)tt = γγ γ .

2.2.22 (i) If  γγ (θ) = (x(θ), y(θ)), we have x cos θ + y sin θ = p. Denoting d/dθ by a dot, we also have y˙ dy = = x˙ dx

θ − cos . sin θ

Differentiating the first equation and using the second gives

−x sin θ + y cos θ =  p.˙ Solving the first and third equations for x and y gives the stated formula for γγ . ˙   = ( p +  p)( ˙   =  p +  p¨ . Hence, γγ   is regular (ii) From (i), γγ  ¨ sin θ, cos θ), so γγ  if and only if  p +  p is ¨ never zero. In that case p +  p must ¨ be either always > 0 or always <   0. But the latter case is impossible since it implies that  p¨ < 0, 2π contradicting the fact that 0  p¨ dθ = 0 since p is smooth and 2π-periodic. d 1 θ + π2 = dθ (iii) If  s is the arc-length of  γγ  , the signed curvature is ds ds = ds/dθ = 1 1  p . γγ ˙   =  p+¨

−

     |

 

(iv) Since

ds dθ

|

 

= p + p, ¨ the length of  γγ  is 2π

  0

2π

( p + p)dθ = ¨

 

 p dθ.

0

(v) The first part is obvious. The foot of the perpendicular from the origin to the tangent line at γγ (θ) is p(θ)(cos θ, sin θ), so w(θ) is the distance between this

16

point and p(θ + π)(cos(θ + π), sin(θ + π)) = p(θ + π)( cos θ, sin θ) which is  p(θ) + p(θ + π). (vi) The function q (θ) = w(θ) w(θ + π) is smooth and q (0) = q (π/2) because w(0) = w(π). Hence, for some angle θ0  with 0 θ0 π/2, we have q (θ0 ) = 0, i.e. w(θ0 ) = w(θ0 +π/2). This means that the tangent lines at θ0 , θ0 +π/2, θ0 +π and θ0  + 3π/2 form a square. (vii) From (iv) and (v), the length is

−

−

 ≤  ≤

π

 

− −

π

( p(θ) + p(θ + π))dθ =

0

 

w(θ) dθ,

0

which is equal to πD if  w(θ) is equal to the constant D. kπ 2 kθ (viii) If  p(θ) is as stated, w(θ) = a cos2 kθ 2  + cos 2 + 2 (ix) If  k = 1 we find that

 | |

γγ  (θ) =

     1 a, 0 + 2

+ 2b = a + 2b.

1 a + b (cos θ, sin θ), 2

which is the circle with centre ( 12 a, 0) and radius 1 1 2 ka sin kθ,  p = k a cos kθ, so ¨ 2 2

−



1 2a

+ b. In general,  p˙ =

−

κs =

a cos2

which is constant

kθ 2

−

1 = 1 2 k a kθ + b cos 2

1 , 1 1 2 )a cos kθ a b k + + (1 2 2

−

 ⇐⇒ k = 0 or ± 1 (and k = 0 does not give a regular curve).

2.2.23 Parametrize the parabola by x = sinh t, y = 12  sinh2 t. The arc-length (starting at the origin) is t

s =

 

(cosh2 u + cosh2 u sinh2 u)1/2 du =

0

1 (t + sinh t cosh t). 2

In the notation of Exercise 2.2.10, θ  is the angle between (cosh t, cosh t sinh t) and the x-axis, so cos θ = secht, sin θ = tanht, p  = (0, 0), a  = (1, 0) and we get Γ  = (s, 0) +



cos θ sin θ

−

sin θ cos θ



− 1sinh t2 1 − 2  sinh t



.

Using the formulas for s and θ from above gives the stated formula for Γ  after simplification. γ (t) gives t(t+a) = t(a); rotating anti-clockwise by π/2 2.2.24 Differentiating γγ  (t+a) = γγ  ns (t + a) = κ s (t)n ns (t), gives ns (t + a) = ns (t); differentiating again gives κs (t + a)n so κs (t + a) = κ s (t).

17

1 γ   is unit-speed; 2.3.1 (i) tt = ( 21 (1 + t)1/2 , 12 (1 t)1/2 , √  ) is a unit vector so γ 2 t˙  = ( 14 (1 + t)−1/2 , 14 (1 t)−1/2 , 0), so κ = t˙ = 1/ 8(1 t2 ); 1 1 n  = 1 t˙  = √  ((1 t)1/2 , (1 + t)1/2 , 0); b  = t n  = ( 1 (1 + t)1/2 , 1 (1 t)1/2 , √  );

−

κ

−

2

−

−

 

  × −2

−

  −− 2

2

b˙ = ( 14 (1 + t)−1/2 , 14 (1  t)−1/2 , 0) so the torsion τ  = 1/ 8(1 t2 ). The equation n˙  = κtt + τ n  is easily checked. (ii) t  = ( 45  sin t, cos t, 35  sin t) is a unit vector so γγ   is unit-speed; t˙  = ( 45  cos t, sin t, 35  cos t), so κ = t˙ = 1; n  = κ1 t˙  = ( 45  cos t, sin t, 35  cos t); b  = t n  = ( 35 , 0, 54 ), so b˙  = 0 and τ   = 0. By the proof of Proposition 2.3.5, γγ   is a circle of radius 1/κ = 1 with centre γγ  + κ1 n  = (0, 1, 0) in the plane passing through (0, 1, 0) perpendicular to b  = ( 35 , 0, 54 ), i.e. the plane 3x + 4z = 0.

−

− ×

−

−

−

−

−

−

 

−

−

−

−

2.3.2 Let a = κ/(κ2 + τ 2 ), b = τ /(κ2 + τ 2 ). By Examples 2.1.3 and 2.3.2, the circular a b helix with parameters a and b has curvature a2 +b 2 = κ and torsion a2 +b2 = τ . By Theorem 2.3.6, every curve with curvature κ  and torsion τ   is obtained by applying a direct isometry to this helix. 2.3.3 Differentiating t.aa   (= constant) gives n.aa   = 0; since t, n, b  are an orthonormal b for some scalar µ; since a  is a unit vector, µ = sin θ; basis of  R3 , aa = t cos θ+µb differentiating a  = t cos θ b sin θ gives τ  = κ cot θ. Conversely, if  τ  = λκ, there b sin θ is a constant exists θ with λ = cot θ; differentiating shows that a  = t cos θ+b vector and t.aa  = cos θ so θ  is the angle between t  and a . For the circular helix in γ /dθ = ( a sin θ, a cos θ, b) Example 2.1.3, the angle between the tangent vector dγγ  and the z-axis is the constant cos−1 (b/ a2 + b2 ).

±

±

−

√ 

γ  n.(γ γ  a).(γγ  γ  a) = r 2 repeatedly gives t.(γγ  γ  a) = 0; t.tt +κn γ  a) = 2.3.4 Differentiating (γγ  2 2 γ  b).(γγ  0, so n.(γ ˙ , and so b.(γγ  ˙ ; γ  a) = 1/κ; n.tt +( κtt +τ b γ  a) = κ/κ γ  a) = κ/τκ 2 2 γ  a ) = (κ/τκ and finally b.tt τ n.(γγ  ˙ )˙, and so τ /κ = (κ/τκ ˙ )˙. Conversely, if Eq. 2 2 (2.22) holds, then ρ = σ(ρσ)˙, ˙ so (ρ + ( ρσ) ˙ )˙ = 2ρρ + ˙ 2(ρσ)( ˙ ρσ)˙ ˙ = 0, hence 2 2 2 n  + ρσb γ   + ρn ρ + (ρσ) ˙ is a constant, say r (where r >   0). Let a = γ ˙ b; then b) + (ρσ)˙b τ n) = 0  using Eq. (2.22); so aa is a a˙  = t  + ρn ˙ n + ρ( κtt + τ b ˙ b + ( ρσ)( ˙ 2 2 2 constant vector and γγ   a = ρ + ( ρσ) ˙ = r 2 , hence γγ   is contained in the sphere with centre a  and radius r. γ  ˙   so T = P tt and Γ˙ 2 = (P  γγ  ˙  ).(P  γγ  ˙  ) = γγ  ˙ .  γ ˙   since P   is orthogonal. Then, 2.3.5 Γ˙  =  P  γγ  ¨  = P γγ  ¨ , taking lengths shows that γγ  γ  and Γ  have the same curvature κ, and then Γ n. Then B = P tt P n n. If  P   corresponds to a direct dividing by κ gives N = P n b, but if  P   corresponds isometry (i.e. a rotation), this is equal to P (tt n) = P b n  = P b b  (Proposition A.1.6). to an opposite isometry, P tt P n

−

−

− −

−

−

−  −

−

−

−

−

−

−

   −      

×

×

−

×

vj . The vectors v1 , v2 , v3  are orthonormal if and only if  λij = δ ij 2.3.6 Let λij = vi .v (= 1 if  i = j   and = 0 if  i = j). So it is enough to prove that λij = δ ij vj   gives for all values of  s   given that it holds for s = s0 . Differentiating v i .v 3 λ˙ ij = k=1 (aik λkj + a jk λik ). Now λij = δ ij   is a solution of this system of  differential equations because aij +aji  = 0. But the theory of ordinary differential





19

µ, say. The principal normal of  δδ    is N = µ−1 ( t  + κτ  b) and its binormal is b + κτ  t). The torsion T  of  δδ   is given by dB B/ds = T N N, i.e. B  = T N  == µ −1 (b  ˙ = T N N. Computing the derivatives and equating coefficients of b  gives κ−1B 2 T  = (κτ ˙ τ  κ)/κ(κ ˙ + τ 2 ). 2.3.13 The curve is γγ  (t) = (eλt cos t, eλt sin t, eλt ). The angle between the tangent ˙   = eλt (λ cos t  sin t, λ sin t + cos t, λ) and the vector k = (0, 0, 1) is vector γγ  ˙  .k γγ  −1 √  λ , which is constant. cos−1 γγ  ˙   = cos 2λ2 +1 2.3.14 We find that

−

×  −  −

 −

 −

 

2(9b2 c2 t4 + 9a2 c2 t2 + a2 b2 )1/2 3abc κ = , τ  . = 9b2 c2 t4 + 9a2 c2 t2 + a2 b2 (a2 + 4b2 t2 + 9c2 t4 )3/2 γ  is a generalized helix if and only if there is a constant λ such that τ  = λκ, Now γγ  i.e. 3abc(a2 + 4b2 t2 + 9c2 t4 )3/2 = 2λ(9b2 c2 t4 + 9a2 c2 t2 + a2 b2 )3/2 , which is equivalent to (*)

9b2 c2 t4 + 9a2 c2 t2 + a2 b2 = µ(a2 + 4b2 t2 + 9c2 t4 )

where µ = (3abc/2λ)2/3. The last equation holds if and only if  9b2 c2 = 9µc2 , 9a2 c2 = 4µb2 , a2 b2 = µa 2 . The first and third equations give µ = b2 and the second gives µ = 9a2 c2 /4b2 . γ    is a generalized helix 9a2 c2 /4b2 = b2 , which is equivalent to the Hence, if  γ stated condition. Conversely, if this condition holds, taking µ = b2 (and hence γ  λ = 3ac/2b2 ) satisfies (*) so τ  = 3ac 2b2 κ and γ   is a generalized helix. ˜ (s˜) = γγ  (s) + a(s)n n(s), where a   is a scalar possibly 2.3.15 (i) From the definition, γγ  depending on s. Then, γ  n d˜ s ˜ dγ γ  da dn t  = + n + a = (1 ds ds ds ds

b. − κa)tt + da n  + τ ab ds

But we are given that n˜ = n, so ˜t   is perpendicular to n. This implies that da/ds = 0 and hence a is constant. (ii) We have d ˜ d˜ s n.˜t + κ ˜  = 0, (tt.t) = κn ˜t.n ds ds

 ±

since ˜t is perpendicular to  ˜n and hence to n, and similarly t is perpendicular to  ˜n. Hence, t.˜t is a constant, say cos α (it must lie between 1 and 1 as t and ˜t are unit

−

20

b, vectors). Then, since t  is perpendicular to n, we must have  ˜t  = cos αtt sin αb and we can assume the sign is  by changing α to α  if necessary. Then,

 −

 − ˜  =  ˜t × n ˜  = ±(cos αtt − sin αb b) × n  = ±(sin αtt + cos αb b). b

±

(iii) From (i) and (ii), d˜ s (cos αtt ds

− sin αbb) = (1 − κa)tt + τ abb.

Equating coefficients of tt and b  gives the first pair of equations. The second pair ˜ , which changes α to α and a to follows by interchanging the roles of  γγ   and γγ  a. (iv) This follows from the first pair of equations in (iii). (v) The first equation follows from the two equations in (iii) involving the torsion, s ds noting that d˜ = 1. The first equation follows similarly from the other two ds d˜ s equations in (iii). n for any constant a = 0. If  γγ   = γγ  γ +an 2.3.16 Let γγ  (s) be a unit-speed plane curve, and let ˜ s˜ ˜ , the unit tangent vector ˜t of  γ ˜γ   is given by d˜ t s˜ is the arc-length of  γγ  ds t  = (1 κa)t. ˜  = n. Hence ˜t = t  and so (since γγ   is planar) n 2.3.17 The ‘only if’ part was proved in Exercise 2.3.15(iv). Conversely, if aκ + bτ  = 1, γ  ˜  = γγ  n. Then, γ + an let γ

 −

−

±

±

d˜ s˜ t  = (1 ds Hence,  ˜t  =

 −

− κa)tt + τ abb = τ (btt + abb).

± √ a 1+b (btt + abb). Differentiating, 2

2

˜  = κ ˜n

b dtt db κb − τ a ds ± √ a21+ b2 ds (b + a ) = ± √  n. d˜ s ds ds s a2 + b2 d˜

˜  = n. It follows that n γ  ˜  = γγ  n  is a Bertrand 2.3.18 The solutions of Exercises 2.3.15 and 2.3.17 show that γ γ + an mate of  γγ   (a being a constant) if and only if  aκ + bτ   = 1 for some constant b, and ˜  are all the Bertrand mates of  γγ . Thus, γγ   has more than one that such curves γγ  Bertrand mate if and only if the equation aκ + bτ  = 1 is satisfied by more than one value of  a (and some corresponding values of  b). In that case κ and τ  must be γ  ˜  is then a circular γ  is a circular helix (Exercise 2.3.2). To see that γ constant, so γγ  helix with the same axis and pitch as  γγ , take γγ   to have axis the z-axis (this can be achieved by applying an isometry of  R3 ). Then, γγ (t) = (c cos λt,c sin λt, dλt), where c   is the radius of  γγ  , 2πd   is its pitch and λ = (c2 + d2 )−1/2 . Hence, n  = ( cos λt, sin λt, 0) so the most general Bertrand mate of  γγ  is

±

−

−

˜  = γγ  γγ  γ a + n  = ((c

− a)cos λt, (c − a)sin λt, dλt) ,

21

which is a circular helix with axis the z-axis, pitch 2πd and radius c a . 2.3.19 If κ is constant, Eq. (2.22) in Exercise 2.3.4 gives τ   = 0, so the curve is a circle (constant curvature and zero torsion). 2.3.20 Let γγ (t) be a unit-speed parametrization of  . If all the normal planes of   pass through a point p, there are scalars λ, µ (possibly depending on t) such that

 | − |

 C

 C

n + µb b  = p γγ  + λn for all t. Differentiating and using the Frenet-Serret equations, ˙ n + λ( κtt + τ b) + µb t + λn ˙b

−

Hence, 1

− λκ = 0, λ˙ − µτ  = 0,

− µτ n =

0.

λτ  + µ = ˙ 0.

Using the first and third equations, d 2 (λ + µ2 ) = 2λλ˙ + 2µµ = ˙ 2λ(µτ ) + 2µ( λτ ) = 0, ds

−

 

so γγ   p = λ2 + µ2 is constant. Hence, γγ   lies on a sphere with centre p. (An alternative is to make use of Exercise 2.3.4.) 2.3.21 (i) is obvious. ˙ 0 ) = 0 if and only if the tangent vector of  γγ   at γγ  ˙ .  N  so F (t γ (t0 ) is parallel (ii) F ˙ = γγ  to Π. Using (i) this gives (ii). ¨ = γγ  ¨ 0 ) = 0 if and only if nn(t0 ) is parallel to Π. Thus, ¨.  N = κn n.N, so F (t (iii) F  ˙ 0 ) = F (t ¨ 0 ) = 0 is that passing through the unique plane Π such that F (t0 ) = F (t γγ  (t0 ) and parallel to t(t0 ) and n(t0 ), i.e. perpendicular to b(t0 ). ˙ .  N ′ = γγ  ¨.  N′   = 0 so by (iii) (iv) If  γγ   is contained in the plane v.N′ = d′ , then γγ  the osculating plane Π of  γγ   at any point γγ  (t0 ) is perpendicular to N′ , and so is parallel to Π′ . But both Π and Π′  pass through γγ (t0 ), so they must coincide. γ (t) γγ  (t0 ))...b(t0 ). Then, (v) The (signed) distance from γγ  (t) to Π is d(t) = (γγ  d(t0 ) = 0 and

   −  

−

˙ 0 ) = t(t0 )..b(t0 ) = 0, d(t ¨ 0 ) = κ(t0 )n n(t0 )..b(t0 ) = 0, d(t ... ˙ n + κ( κtt + τ b))(t0 )..b(t0 ) = κ(t0 )τ (t0 ). d (t0 ) = (κn

−

If  τ (t0 ) = 0, Taylor’s theorem gives



d(t) =

1 κ(t0 )τ (t0 )(t 6

− t0 )3 + higher order terms.

22

As t passes from values < t0  to values > t0 , the leading term changes sign, so γγ  crosses Π. γ (t) = (a cos t, a sin t,bt), b  = √ a21+b2 (b sin t, b cos t, a), so 2.3.22 For the circular helix γγ  γ (t) γγ  (t0 ))...b(t0 ) = 0, i.e. by Exercise 2.3.21 the osculating plane at γγ (t0 ) is (γγ 

−

xb sin t0

−

− yb cos t0 + az = abt0.

2.3.23 Proceeding as in the preceding exercise, the osculating plane at γγ (t) is found to be 3t2 x 3ty + z = t 3 .

−

If  P 1 , P 2 , P 3  correspond to t = t1 , t2 , t3 , the three osculating planes intersect at a single point if and only if the determinant

 

  

3t21 3t22 3t23

−3t1 1 −3t2 1 = 0. −3t3 1 But the determinant equals 9(t1 − t2 )(t2 − t3 )(t3 − t1 ), and this is indeed non-zero as t1 , t2 , t3  are distinct. If  Q  is the point (X , Y , Z )  , then t1 , t2 , t3  are the roots of  t3

− 3t2 X  + 3tY  − Z  = 0.

On the other hand, if the plane Π passing through the points γγ  (t1 ), γγ  (t2 ) and γγ  (t3 ) has equation ax + by + cz + d = 0, then t1 , t2 , t3  are also the roots of  at + bt2 + ct3 + d = 0. The two cubic equations must be the same, so Π must be the plane 3Y x

− 3Xy + z =  Z,

and the point (X , Y , Z )  obviously lies on this plane. 2.3.24 We can assume that γγ   is unit-speed. If all the osculating planes pass through a point p, there are scalars λ, µ such that n  = p. γγ  + λtt + µn Proceeding as in Exercise 2.3.20 we find that µτ  = 0, λ˙

− µκ = −1,

λκ + µ = ˙ 0.

23

If  µ = 0 the third equation gives λ = 0, which contradicts the second equation. So µ = 0 and the first equation gives  τ  = 0, i.e. γγ   is planar. 2.3.25 Let γγ   be a unit-speed curve. Its orthogonal projection onto the normal plane at γγ  (t0 ) is γ  (γγ  γ .t(t0 ))tt(t0 ). Γ  = γγ 



−

Then,

Γ˙  = t

− (tt.t(t0))tt(t0 ) = 0

if  t = t 0 .

Hence, Γ  has a singular point at t = t 0 . Similarly, we find that ... ¨ (t0 ) = κ(t0 )n n(t0 ), Γ (t0 ) = κ(t n(t0 ) + κ(t0 )τ (t0 )tt(t0 ). Γ ˙ 0 )n These vectors are linearly independent if  κ(t0 ) and τ (t0 ) are non-zero, so in that case Γ  has an ordinary cusp at t = t 0 . The orthogonal projection of  γγ   onto its osculating plane at γγ (t0 ) is ˜  = γγ  Γ γ  so

˜˙ (t0 ) = t(t0 ) Γ

− (γγ γ.  b(t0 ))bb(t0),

− (tt(t0 )..b(t0 ))bb(t0) = t(t0 ) =

0,

˜. so γγ (t0 ) is a regular point of  Γ 2.3.26 (i) is obvious. ˙ ˙ 0 ) = 0 if and only if t(t0 ) is perpendicular to γγ  γ  γ  c) so F (t γ (t0 ) (ii) F  = 4tt.(γ i.e. γγ   is tangent to at γγ (t0 ). Next,

−

 S  ¨ = 2κn n. (γγ  F  γ  − c) + 2,

... γ  ˙ n.(γγ  F  = 2κn

− c,

− c) + 2κ(−κtt + τ b)..(γγ γ  − c).

... ˙ 0 ) = F (t ¨ 0 ) = F  Hence, F (t0 ) = F (t (t0 ) = 0 if and only if  γγ   is tangent to at γγ  (t0 ) and κ˙ n. (γγ  κn γ  c) = 1, κτ b.(γγ  γ  c) = κ (all quantities evaluated at t = t 0 ). These equations and the fact that γγ (t0 ) c is perpendicular to t(t0 ) are equivalent to

 S 

−

−

−

−

γγ  

− c  = − κ1 n + κκ2˙ τ b,

so the centre c of    is uniquely determined and its radius is

 S 

R = γγ  

 −c =

 

κ˙ 2 1 + 4 2 κ2 κ τ 

24

(again, all quantities evaluated at t = t0 ). The centre cc  is independent of  t0 if  and only if  d κ˙ 1 γγ  + n b =0 dt κ κ2 τ 



−



when t = t 0 . Using Frenet-Serret, the derivative is found to be

  −   τ  κ

d dt

κ˙ κ2 τ 

b.

By Exercise 2.3.4, c  is independent of  t0  if and only if  γγ   is spherical, in which case the solution of Exercise 2.3.4 shows that γγ   lies on the sphere with centre c and radius R, i.e. the osculating sphere. 2.3.27 Since b(t0 ) is perpendicular to the osculating plane at γγ  (t0 ), the centre of the b(t0 ) for some scalar λ. As this osculating circle will be of the form c′ = c(t0 ) + λb point lies on the osculating plane, it must be of the form γγ (t0 ) + µtt(t0 ) + ν n(t0 ) for some µ, ν . Hence, γγ  +

1 n κ

− κκ2˙ τ b + λbb = γγ γ + µtt + ν n, ˙ κ κ2 τ    and

all quantities being evaluated at t = t0 . So λ = osculating circle is b(t0 ) = γγ  γ (t0 ) + c(t0 ) + λb

the centre of the

1 n(t0 ). κ(t0 )

If  R is the radius of the osculating sphere, the radius r of the osculating circle is given by κ˙ 2 1 1 2 2 2 ′ r = R c c = 2 + 4 2 λ2 = 2 , κ κ τ  κ

− − 

 −

so r = 1/κ(t0 ).

Chapter 3 ˙   = ( sin t a sin2t, cos t + a cos2t), γγ  ˙   2 = 1 + a2 + 2a cos t 1 + a2 2 a = 3.1.1 γγ  ˙   = 2(1 + a cos t) so a )2 so γγ    is regular if  a = 1. If  a   = 1 then γγ  (1 the origin is a singular point of  γγ  . If  a = 0 then γγ    is a circle. If 0 < a < 1, then γγ  (t1 ) = γγ (t2 ) = 1 + a cos t1 = 1 + a cos t2 = cos t1  = cos t2 =  t 2 = t 1 or 2π t1 . In the latter case, γγ  (t2 ) = ((1 + a cos t1 )cos t1 , (1 + a cos t1 )sin t1 ) so γγ  (t1 ) = γγ  (t2 ) = sin t1 = 0 = t1   = 0 or π. In all cases, t2  t 1 is a multiple of 2π, so γγ   is a closed curve with period 2π  without self-intersections. If  a > 1, γγ   passes through the origin when cos t = 1/a, which has two roots

− −| |

−

 | |

−

     | |   | |

⇒ ⇒

  

⇒

⇒

 −

−

≥

−||  | | ⇒  −

25

with 0 t < 2π, say t1 < t2 , so the origin is a self-intersection. The picture is qualitatively similar to that in Example 1.1.7 (which is the case a = 2), so the γ  is the union of two bounded regions enclosed by complement of the image of γγ  the part of the curve with t1 t t2 , and an unbounded region. ˙   = ( sin t 2a sin t cos t, cos t + a(cos2 t sin2 t)) so 3.1.2 γγ 

 ≤

−

 ≤ ≤

−

−

tan ϕ =

t + a cos2t − cos  , sin t + a sin2t

dϕ 1 + 3a cos t + 2a2 ∴ , sec t = dt (sin t + a sin2t)2 dϕ a(cos t + a) 1 + 3a cos t + 2a2 . = = 1 + dt 1 + 2a cos t + a2 1 + 2a cos t + a2 2

∴

If  a < 1, γγ   is a simple closed curve of period 2π   (Exercise 3.1.1), so by the Umlaufsatz, 2π dϕ  dt = 2π, dt 0

 | |

 

giving

2π

  0

a(cos t + a) dt = 0. 1 + 2a cos t + a2

If  a > 1, it is clear geometrically that the tangent vector of  γγ   rotates by 4π on γ   (see the diagram in Example 1.1.7 for the case a = 2), so going once around γ

 | |

2π

  0

giving

2π

  0

dϕ  dt = 4π, dt

a(cos t + a) dt = 2π. 1 + 2a cos t + a2

v) = P vv +b b, where P  is a 3.2.1 By Appendix 1, any isometry M  of  R2 is of the form M (v γ  ˜γ   = M (γγ  ˜˙   = P  γγ  ˙  , so γ ), then γ 3 3 orthogonal matrix and b  is a constant vector. If  γ γ  ˜˙   = γγ  ˙   , which implies that γγ  and γγ  ˜  have the same length. If we think of  γγ  γ T  ˙   γγ  ¨ ).k dt, as a curve in the xy-plane in R3 , Eq. (3.2) can be written (γγ  γ ) = 0 (γγ  ˜  have the γ  and γγ  where k  = (0, 0, 1). It now follows from Proposition A.1.6 that γγ  same area (note that if  M   is opposite, the area appears to change sign, but it does ˜  is negatively-oriented when γγ   is positively-oriented). not because in that case γγ  3.2.2 Parametrizing the ellipse by γγ (t) = ( p cos t, q sin t), with 0 t 2π, its area is 2π  pq sin2 t + pq cos2 t dt = 2π  pq . By the isoperimetric inequality, the length 0 ℓ   of the ellipse satisfies ℓ 4π πpq  = 2π  pq , with equality if and only

×      

   

A

√  ≥ √  ×

≤ ≤

√ 

 

×

26

if the ellipse is a circle, i.e. p = q . 2π  p2 sin2 t + q 2 cos2 t dt. 0

   

But its length is

2π 0

 



˙  γγ 



dt =

γ (t)  t ′ t is a multiple of 2π, so 3.2.3 γγ  (t′ ) = γγ   cos t′  = cos t and sin t′  = sin t γγ   is simple closed with period 2π. Taking x = p cos t, y = q sin t in Eq. 3.2 gives 2π the area as 12 0   pq dt = πpq .

⇐⇒

⇐⇒ −

 

2

2

x y 3.3.1 Let (x1 , y1 ) and (x2 , y2 ) be points in the interior of the ellipse, so that  p2i  + qi2 < 1 for i = 1, 2. A point of the line segment joining the two points is

(tx1  + (1

− t)x2, ty1 + (1 − t)y2)

for some 0

≤ t ≤ 1. This is in the interior of the ellipse because (tx1  + (1 − t)x2 )2 (ty1  + (1 − t)y2 )2 + = t 2

 p2 x21 y12 + 2  p2 q 



< t2 + (1

−

q 2 x22 y22 2 + (1 t) + 2 + 2t(1  p2 q  x21 y12 x22 y22 2 t) + t(1 t) + 2 + 2 + 2  p2 q   p q 



−

−







− t)

≤



x1 x2 y1 y2 +  p2 q 2

t2 + (1



− t)2 + 2t(1 − t) = 1.

    √ 

˙   = ( sin t 2sin2t, cos t + 2 cos 2t) and γγ  ˙   = 5 + 4 cos t, so the angle ϕ 3.3.2 γγ  sin t−2sin2t t+2cos2t √ 5+4 γ  ˙   and the x-axis is given by cos ϕ = − √  between γ   , sin ϕ = cos . 5+4 cos t cos t

−

−

2 t+42 cos t+9) Differentiating the second equation gives ϕ˙ cos ϕ = − sin t(24cos , so (5+4 cos t)3/2 2

t(24cos t+42 cos t+9)   9+6 cos t ϕ˙ = sin γγ   κ (5+4 cos t)(sin t+2sin2t) = 5+4 cos t . Hence, if s  is the arc-length of  , s = dϕ/ds = (dϕ/dt)/(ds/dt) = (9 + 6 cost)/(5 + 4 cos t)3/2 , so

κ˙ s =

  12sin t(2 + cos t) . (5 + 4 cos t)5/2

This vanishes at only two points of the curve, where t = 0 and t = π. 3.3.3 From ǫ(s) = γγ (s) + κ1s ns  we get ǫ˙  = κ˙ s  = 0, i.e. where γγ   has a vertex.

 −κ˙ sns/κ2s , so ǫ  has a singular point where

3.3.4 Parametrize by γγ (x) = (x, f (x)) and denote d/dx  by a dot. The arc-length s is ds 1 ˙ and we given by dx = 1 + f ˙2 , so the unit tangent vector is tt = (1, f )  ˙ 2 find that

 

√ 1+f 

dtt 1 ˙ 1  ˙ f, ¨ f ¨). t = = ( f  2 2 ˙ ds ds/dx (1 + f  )

−

27

The signed unit normal ns = ¨ f  .  ˙ (1+f 2 )3/2

√ 1+1 f  ˙ (−f,˙ 1), so the signed curvature is κs = 2

Then,

...  ˙ f ¨2 f (1 + f ˙2 ) 3f  κ˙ s = . (1 + f ˙2 )5/2 ... dκs s f (1 + f ˙2 ) = 3f ˙f ¨2 . Hence, dκ = 0 = 0 ds dx ˙   = (a b cos t, b sin t), so if  ϕ is a turning angle for γγ , 3.3.5 γγ 

−

⇐⇒

−

⇐⇒

b sin t , a b cos t b(a cos t b) ∴ ϕ˙ sec2 ϕ = , (a b cos t)2 b(a cos t b) (a b cos t)2 . = a2 2ab cos t + b2 b cos t)2 + b2 sin2 t tan ϕ =

∴

ϕ = ˙

b(a cos t b) (a b cos t)2 (a

−

−

−

−

−

−

−

−

−

If  s  is the arc-length of  γγ , the signed curvature is κs = Hence,

dϕ  dϕ/dt = = ds ds/dt

ϕ˙ ˙  γγ 

  

=

(a2

b(a cos t b) . 2ab cos t + b2 )3/2

−

−

dκs ab sin t(2b2 a2 ab sin t cos t) . = dt (a2 2ab cos t + b2 )5/2

−

− −

Thus, γγ (t) is a vertex if and only if sin t = 0 or cos t =

2b2 a2 ab .

−

Hence, there are

vertices when t is an integer multiple of  π , and no other vertices if  On the other hand, if 

  ≤ 2b2 a2 ab

−

b   1, i.e. if  a− b

many values of  t such that cos t = points on the curve.

2b2 a2 ab ,

−

  2b2 a2 ab

−

> 1.

≤ 2ba ≤ a+b   , there are infinitelyb

and these all correspond to different

Chapter 4 4.1.1 Let U  be an open disc in R2 and = (x,y,z) R3 (x, y) U, z = 0 . If  W  = (x,y,z) R3 (x, y)  U  , then W  is an open subset of  R3 , and W  is homeomorphic to U  by (x,y, 0) (x, y). So   is a surface. σx±  is the intersection of the sphere with the open set x > 0 in 4.1.2 The image of σ R3 , and its inverse is the projection (x,y,z)  (y, z). Similarly for σσ y± and σ z± . A point of the sphere not in the image of any of the six patches would have to have x, y and z all zero, which is impossible.

 {

∈

|

 S   { ∈ } →

 ∈  S 

→

|

 ∈

}  S ∩

 ±

28

4.1.3 Multiplying the two equations gives (x2 z 2 )sin θ cos θ = (1 y 2 )sin θ cos θ, so x2 + y 2 z2 = 1 unless cos θ = 0 or sin θ  = 0; if cos θ = 0, then x = z and y = 1 and if sin θ = 0 then x = z and y = 1, and both of these lines are also contained in the surface. The given line Lθ  passes through (sin2θ, cos2θ, 0) and is parallel to the vector (cos 2θ, sin2θ, 1); it follows that we get all of the lines by taking 0 θ < π. Let (x,y,z) be a point of the surface; if  x = z, let θ   be such that cot θ = (1 y)/(x z); then (x,y,z) is on Lθ ; similarly if  x = z. The only remaining cases are the points (0, 0, 1), which lie on the lines Lπ/2 and L0 . To get a R3 by surface patch covering , define σ  : U 

−

−

−

 −

≤

−

−



−

 S 

±

 −

−

 →

σ (u, v) = (sin 2θ,

− cos2θ, 0) + t(cos2θ, sin2θ, 1).

By the preceding paragraph, this patch covers the whole surface. Let M ϕ  be the line (x z)cos ϕ = (1+y)sin ϕ, (x+z)sin ϕ = (1 y)cos ϕ. By the same argument as above, M ϕ  is contained in the surface and every point of the surface lies on some M ϕ  with 0 ϕ < π. If  θ + ϕ is not a multiple of  π, the lines −ϕ)  sin(θ−ϕ)  cos(θ+ϕ) Lθ and M ϕ  intersect in the point ( cos(θ sin(θ+ϕ) , sin(θ+ϕ) , sin(θ+ϕ) ); for each θ   with 0 θ < π, there is exactly one ϕ  with 0 ϕ < π such that θ + ϕ is a multiple of  π, and the lines L θ and M ϕ  do not intersect. If (x,y,z) lies on both L θ and Lϕ , with θ = ϕ, then (1 y)tan θ = (1 y)tan ϕ and (1 + y)cot θ = (1 y)cot ϕ, which gives both y = 1 and y = 1 (the case in which θ = 0 and ϕ = π/2, or vice versa, has to be treated separately, but the conclusion is the same). This shows that Lθ and Lϕ  do not intersect; similarly, M θ and M ϕ  do not intersect.

−

−

≤

≤

≤

 

−

 −

−

−

√ 

4.1.4 For the first part, let U  = (u, v) R2 0 < u2 + v 2 < π 2 , let r = u2 + v 2 , and define σ  : U  R3 by σσ(u, v) = ( ur , vr , tan(r π2 )). R3 , then S 2 would If  S 2 could be covered by a single surface patch σ : U  be homeomorphic to the open subset U  of  R2 . As S 2 is a closed and bounded subset of  R3 , it is compact. Hence, U  would be compact, and hence closed. But, since R2 is connected, the only non-empty subset of  R2 that is both open and closed is R2 itself, and this is not compact as it is not bounded. 4.1.5 If  σ α : U α R3  is an atlas for a surface , and if W  is an open subset of  R3 , 1 W  (one should discard then the restrictions σ U α ∩σ −  form an atlas of  α (W ) 1 the restrictions for which U α σ − α (W ) is empty). 4.1.6 Parametrize the curve by γγ (t) = (cos θ, sin θ, z) where θ and z are smooth func˙   = ( θ˙ sin θ, θ˙ cos θ, z) tions of  t, and assume that γγ   is unit-speed. Then, γγ  ˙ so γγ    interescts the rulings of the cylinder at a constant angle if and only if z˙ is constant, i.e. z = at + b for some constant a, b. Since γγ  γ  is unit-speed, θ˙2 + z˙ 2 = 1 ˙ a constant c such that a2 + c2 = 1 and θ =  ct + d where d is a constant. so θ is γ  (t) = (cos(ct + d), sin(ct + d), at + b). This is a circle if  a = 0, a straight Then, γ line if c = 0 and a circular helix in all other cases.

 {

 →

 {

 → }

 { |

 ∈

∩

}

|

}

−

 S 

→

 S ∩

−

29

4.1.7 Noting that (x,y,z) lies on the ellipsoid if and only if ( px , yq , zq ) S 2 , we can use the latitude-longitude parametrization of S 2 to get the parametrization

∈

σ (θ, ϕ) = ( p cos θ cos ϕ, q cos θ sin ϕ, r sin θ) of the ellipsoid. σ  is continuous and is a homeomorphism onto its image if (θ, ϕ) is restricted in the same way as for S 2 (Example 4.1.4). 4.1.8 If  x = sin u, y = sin v and z = sin(u + v), then x2

− y2 + z2 = sin2 u − sin2 v + sin2(u + v) = sin2 u − sin2 v + sin2 u cos2 v + cos2 u sin2 v + 2 sin u sin v cos u cos v = sin2 u + sin2 u cos2 v − sin2 v(1 − cos2 u) + 2 sin u sin v cos u cos v = (1 − sin2 v)sin2 u + sin2 u cos2 v + 2 sin u sin v cos u cos v = 2 sin2 u cos2 v + 2 sin u sin v cos u cos v = 2 sin u cos v(sin u cos v + cos u sin v) = 2 sin u cos v sin(u + v).

σ Hence, (x2 y 2 + z 2 )2 = 4 sin2 u(1 sin2 v)sin2 (u + v) = 4x2 (1 y 2 )z 2 . Now, σ 1 1 − − is clearly continuous, as is its inverse map (x,y,z) (sin x, sin y), so σσ is a homeomorphism.

−

−

4.2.1 σ   is obviously smooth and σ u regular.

 × σ v

−

→

= ( f u , f v , 1) is nowhere zero, so σ is

− −

±√  − −

√  − −

4.2.2 σ z±  is a special case of Exercise 4.2.1, with f  = 1 u2 v 2 ( 1 u2 v 2 is smooth because 1 u2 v 2 > 0 if (u, v) U ); similarly for the other patches. u, v˜) = (u, v), where The transition map from σ x+ to σ y+ , for example, is Φ(˜ y σ + (˜ u, v˜) = σ x+ (u, v); so u = 1 u ˜2 v˜2 , v = v˜, and this is smooth since u, v˜) U . 1 u ˜2 v˜2 > 0 if (˜ 4.2.3 (i) is clearly injective and is regular because σ  is smooth and σ u σ v = ( v, u, 1) is never zero. (ii) is injective but is not regular since σu σv = (0, 3v 2 , 2v) vanishes when v = 0. σ( u 1, v) and is also not regular since (iii) is not injective because σσ (u, v) = (σ σ u σ v = (0, 2v(1 + 2u), 1 + 2u) vanishes when u = 1/2. σ (θ, ϕ) = 4.2.4 This is similar to Example 4.1.4, but using the ‘latitude-longitude’ patchσ ( p cos θ cos ϕ, q cos θ sin ϕ, r sin θ).

− −

− −

 ∈ √  − −

∈

×

 × − −

×

− −

−

−

4.2.5 A typical point on the circle  has coordinates (a+b cos θ, 0, b sin θ); rotating this σ (θ, ϕ); the torus is covered about the z-axis through an angle ϕ gives the point σ

C

30

by the four patches obtained by taking (θ, ϕ) to lie in one of the following open sets: (i) 0 < θ < 2π, 0 < ϕ < 2π; (ii) 0 < θ < 2π, π < ϕ < π; (iii) π < θ < π, 0 < ϕ < 2π; (iv) π < θ < π, π < ϕ < π. Each patch is regular because

−

 −  −

−

σθ

 × σ ϕ = −b(a + b cos θ)(cos θ cos ϕ, cos θ sin ϕ, sin θ)

is never zero (since

  σ u × σ v  = b(a + b cos θ) ≥ b(a − b) > 0).

4.2.6 Suppose the centre of the propeller is initially at the origin. At time t, the centre is at (0, 0, αt) where α  is the speed of the aeroplane. If the propeller is initially along the x-axis, the point initially at (v, 0, 0) is therefore at the point (v cos ωt,v sin ωt,αt) at time t, where ω is the angular velocity of the propeller. Let u = ωt, λ = α/ω. Next, σ u = ( v sin u, v cos u, λ), σ v  = (cos u, sin u, 0), so the standard unit normal is N  = (λ2 + v 2 )−1/2 ( λ sin u, λ cos u, v). If θ is the angle between N  and the z-axis, cos θ = v/(λ2 +v 2 )1/2 and hence cot θ = v/λ, while the distance from the z-axis is v.

−

−

−

−

±

4.2.7 σ   is the tube swept out by a circle of radius a   in a plane perpendicular to γ  . σ s = (1  κa cos θ)tt τ a sin θn n + τ a cos θb b, γγ    as its centre moves along γ n + a cos θb b, giving σ s σ θ = a(1 κa cos θ)(cos θn n  + sin θb b); σ θ = a sin θn this is never zero since κa < 1 implies that 1 κa cos θ > 0 for all θ. The first fundamental form is ((1 κa cos θ)2 + τ 2 a2 ) ds2 + 2τ a2 dsdθ + a2 dθ 2 , so the area 2π s is s01 0 a(1 κa cos θ)dsdθ = 2πa(s1 s0 ).

 −

   

−

−

 −  −  ×  −  − − −

σ + a, where a  is a constant vector, then σ˜  is smooth if  σ  is smooth, and 4.2.8 If  σ˜  = σ σ˜ u = σ u , σ˜ v = σ v , so σ˜   is regular if  σ   is regular. If  A  is an invertible 3 3 σ , then σ˜  is smooth if  σ  is smooth and σ˜ u = Aσ σu , σ σv , ˜ v = Aσ matrix and σ˜ = Aσ so since A  is invertible σ˜ u and σ˜ v   are linearly independent if  σ u and σ v are linearly independent.

×

4.2.9 See Exercise 4.1.5. The restriction of a smooth map U  open subset of  R2 , to an open subset of  U  is smooth.

→ R2, where U   is an

4.2.10 The map σσ  is a diffeomorphism, since it is smooth, obviously bijective, and has smooth inverse (x,y,z) (x, y).

→ 4.2.11 We find that σ u × σ v = (− cos v cos(u + v), − cos u cos(u + v), cos u cos v). This is non-zero when −π/2 < u, v < π/2 since cos u  and cos v  are non-zero in this range.

4.2.12 The torus is covered by the patches σ : U n

 → R3 , n = 1, 2, 3, where σ  is as in

31

the question and U 1 = (θ, ϕ) 0 < θ < 2π, 0 < ϕ < 2π ,

{

U 2 = U 3 =

 

| } 2π 4π 4π 2π <θ< ,− <ϕ< (θ, ϕ) | − 3 3 3 3 (θ, ϕ)

| − 4π3 < θ < 2π3 , − 2π3 < ϕ < 4π3

 

, .

σ  : V  Suppose for a contradiction that the torus is covered by two patches  σ R3 R3 , where σ   is as before and V, V ′  are open subsets of  R2 . We and σ : V ′ might as well assume that

→

 →

V  = (θ, ϕ) α < θ < α + 2π, β < ϕ < β  + 2π ,

{ | } V  ′ = {(θ, ϕ) | α′  < θ < α ′  + 2π, β ′  < ϕ < β ′  + 2π }, for some α, α′ , β , β ′  ∈ R. But neither of the patches σ  : V  → R3 or σσ  : V  ′ → R3 ′

contains the point σσ (α , β ) (for example). 4.2.13 Taking f  = 3xy z(z + 4) + c, we have f  = (3y, 3x, 2z 4) which is zero only at the point (0, 0, 2). But f (0, 0, 2) = 0 only if  c = 4. So if  c = 4, f   = 0 is a smooth surface. 4.2.14 Taking f  = x 3 + 3(y 2 + z2 )2 2, we have f  = (3x2 , 12y(y 2 + z2 ), 12z(y 2 + z2 )) which is zero only at the origin. But f (0, 0, 0) = 0, so f  = 0 is a smooth surface.

 −

 ∇∇ −

−

−

∇  ∇

− −  −

   −



4.2.15 Take f  = x2/3 + y 2/3 + z 2/3 = 1. Not that f   is smooth provided x, y and z are never zero, i.e. if we exclude the points on the coordinate planes. Then f  = ( 23 x−1/3 , 23 y −1/3 , 23 z−1/3 ) which is non-zero. 4.2.16 If xyz  = 1 then x, y and z are all non-zero. By the intermediate value theorem, a smooth (or even just continuous) curve on the surface cannot connect a point with x >   0 (for example) to a point with x <   0 (as it would have to pass through a point with x = 0). Thus, two points in the same connected piece of  the surface must have x-coordinates of the same sign (and similarly for their yand z-coordinates). Note that the sign of any two of the coordinates determines that of the third, since xyz = 1. Consider the piece with x, y and z all > 0. This can be parametrized using the σ (u, v) = (u,v, 1/uv), defined on the open set (u, v) u > 0, v > 0 . single patch σ It is clear that σσ  is smooth and injective. 4.2.17 Consider the circular helix γγ  (θ) = (a cos θ, a sin θ,bθ). The mid-point of the chord joining γγ (θ) and γγ  (ϕ) is the point

∇

{



1  1  1 a(cos θ + cos ϕ), a(sin θ + sin ϕ), b(θ + ϕ) 2 2 2



|

= (v cos u, v sin u,bu),

}

32

where u = 12 (θ + ϕ), v = a cos 12 (θ ϕ). Hence, the set of mid-points of the chords of the helix is the subset of the helicoid in Exercise 4.2.6 with λ = b given a. by v

 −

 | | ≤ | | 4.3.1 If  S   is covered by a single surface patch σσ : U  → R3 , then f  : S → R  is smooth ˜  → R3 is if and only if  f  ◦ σ : U  → R  is smooth. We must check that, if  σ˜ : U  another patch covering S , then f  ◦ σ˜  is smooth if and only if  f  ◦ σ   is smooth. σ to This is true because f  ◦ σ˜ = (f  ◦ σ ) ◦ Φ, where Φ is the transition map from σ σ˜ , and both Φ and Φ −1 are smooth. The last part is true because if  σ : U  → R3 is a smooth map, where U  is an open subset of  R2 , then each component of  σ (which is a map U  → R) is smooth (this is because a vector function such as σ

is differentiated ‘componentwise’). 4.3.2 f   is not a diffeomorphism as it is not injective: f (0, y , z) = f (0, y , z + 2π). Take an atlas for the cone consisting of the patches σσ(u, v) = (u cos v, u sin v, u), defined on the open sets U 1 = (u, v) u > 0, 0 < v < 2π and U 2 = (u, v) u > 0, π < v < π

{

|

}

{

|

−

}

(call these σ 1 and σ 2 ), and parametrize the half-plane by π (u, v) = ( 0, u , v) with u >   0. If (0, a , b) is any point in the plane, assume first that b   is not an even multiple of  π, say 2nπ < b < 2(n + 1)π   for some integer n. Then, π(u, v)) = σ 1 (u, v 2nπ) if 2nπ < v < 2(n + 1)π. So f  is a diffeomorphism f (π from the open subset (0, y , z) 2nπ < z < 2(n + 1)π  of the half-plane to the cone with the half-line y = 0, x = z > 0 removed. Similarly if  b is not an odd multiple of  π. This proves that f  is a local diffeomorphism.

−  {

|

}

4.4.1 (i) At (1, 1, 0), σ u = (1, 0, 2), σ v = (0, 1, 2) so σ u σ v = ( 2, 2, 1) and the tangent plane is 2x + 2y + z = 0. (ii) At (1, 0, 1), where r = 1, θ = 0, σ r = (1, 0, 2), σ θ = (0, 1, 2) so σ r σ θ = ( 2, 2, 1) and the equation of the tangent plane is 2x 2y + z = 0. ˜ (˜ u, v˜) be a reparametrization of   σ . Then, 4.4.2 Let σ

−

 −

− −

 ×

−

 ×

 − −

σu =

∂ ˜ u ∂ v˜ ∂ ˜ u ∂ v˜ σ σ σ σ ˜ u˜  + ˜ v˜ , σ v = ˜ u˜  + ˜ v˜ , ∂u ∂u ∂v ∂v

˜ u˜  and ˜ σ v˜ . Hence, any linear combination so σ u  and σ v  are linear combinations of  σ of  σ u and σ v  is a linear combination of  σ˜ u˜ and σ˜ v˜ . The converse is also true since σσ  is a reparametrization of  σ˜ . d ˙  . Since F (γγ  γ (t)) = F x ˙x + F y ˙y + F z ˙z = F ..γγ  4.4.3 If  γγ  (t) = (x(t), y(t), z(t)) then dt ˙  (t0 ) for every curve F  is perpendicular to T p , it is perpendicular to γγ  S F  ˙   = F ..γγ  ˙   at p. If  γγ   on   passing through p  when t =  t 0 . It follows that S F ..γγ  the restriction of  F  to   has a local maximum or a local minimum at p , so does

∇  −∇∇  S 

S 

 S 

 ∇∇

 ∇∇  ∇∇

33

d γ  F (γγ  γ (t)) for all curves γγ  γ  on   passing through p, hence dt F (γ γ (t)) = 0 at p, which ˙  , and hence perpendicular to the tangent implies that F  is perpendicular to γγ  plane of    at p. This means that S F  = 0.

 S 

S 

 ∇∇

 ∇∇

γ  )/dt) = Dγγ  (t) f (γγ  ˙  (t)) is non-zero because γγ  ˙   is non-zero (γγ  γ  is regular) and 4.4.4 d(f  γ Dγγ  (t) f  is invertible (Proposition 4.4.6).

 ◦

− − √ 

√ 

4.4.5 We find that, at θ = ϕ = π/4, σσ θ = 12 b( 1, 1, 2), σσ ϕ = 12 (a 2 + b)( 1, 1, 0), so the unit normal of the torus at this point is parallel to

−

√ 

√  √  × (−1, 1, 0) = − 2(1, 1, 2). √  Hence, the tangent plane is x + y + 2z = 0. ( 1, 1, 2)

− −

θ) is defined on the union of the rectangles given 4.5.1 The transition map Φ(t, θ) = (t˜, ˜ by 0 < θ < π and π < θ < 2π (and 1/2 < t < 1/2). Obviously Φ(t, θ) = (t, θ) ˜ if 0 < θ < π. If   π < θ < 2π, we must have θ˜ = θ 2π. Since sin θ2 = sin θ2 , ˜ θ) forces t˜ = t. So Φ(t, θ) = (t, θ) if 0 < θ < π, cos θ2 = cos θ2 , σ (t, θ) = σ˜ (˜t, ˜ and = ( t, θ 2π) if   π < θ < 2π. The Jacobian determinant is +1 on the first rectangle, 1 on the second.

 −

−

 −

 −  − − −  − 4.5.2 Let {σ α : U α → R3 }  be an atlas for S   such that the transition map Φαβ   between σ α and σσ β   satisfies det(J (Φαβ )) >  0 for all α, β   (Definition 4.5.1). By Proposi˜, and the transition maps for this atlas are tion 4.3.1, {f  ◦ σ α }  is an atlas for S  1 the same as those for the atlas of  S , because (f  ◦ σ β )−1 ◦ (f  ◦ σ α ) = σ − β ◦ σα ˜  the structure of  (where this composite is defined). So the atlas {f  ◦ σ α }  gives S  an oriented surface.

σ (θ, ϕ) = (cos θ cos ϕ, cos θ sin ϕ, sin θ), we find that σσ θ σ ϕ = cos θσ σ (θ, ϕ). 4.5.3 If σ Hence, the standard unit normal points inwards if cos θ > 0 (if  π/2 < θ < π/2, for example) and outwards of cos θ < 0. Note that the points at which cos θ = 0 (i.e. the north and south pole) must be excluded for σσ  to be regular.

 ×

−

−

σ x± , for example, we find that (σ σ x± )u (σ σx± )v = √  −u12 −v2 σ x± , so For the patches σ 1 the standard unit normal of  σ x+  points outwards and that of  σ x−  points inwards.

×

˙  .v, g = γγ  ˙  .v ˜. 4.5.4 Apply the proof of Proposition 2.2.1 with f  = γγ  4.5.5 From σσ ∗ = σσ.σ , we get σ ∗u =

σ.σ )σ σu (σ

− 2(σσu.σ )σσ ,  σ 4

±

34

with a similar formula for σσ ∗v . This gives σ ∗u

× σ ∗v =  σ1 6  σ 2 (σσu × σ v ) − 2σσ × ((σσu.σ )σσv − (σσv.σ )σσu ) = = =

  ×       ×  × −          ×    − 1 σ 1 σ σu

σ

6

6

2

σu (σ

σ. (σ σu 2(σ

σv

σ σ v ) + 2σ

σ σ v ))σ

σ.N )σ σ 2(σ σ 2

 σ 4

2(σ .N )σ

σ 2

2

× (σσu × σ v ))

σu (σ



× σv)

N .

  

Hence, it suffices to show that length is σ.N )2 4(σ σ 2

σ

σ (σ

− N   is a unit vector.

σ.N)2 4(σ + σ 2

   −   





But its squared

 N 2= 1.

Chapter 5 5.1.1 (i) f x = 2x, f y = 2y and f z = 4z 3 vanish simultaneously only when x = y = z = 0, but this does not satisfy x2 + y 2 + z4 = 1. So by Theorem 5.1.1 this is a smooth surface. (ii) Let f (x,y,z) be the left-hand side minus the right-hand side; then, f x = 4x(x2 + y 2 + z 2

− a2 − b2), f y = 4y(x2 + y 2 + z2 − a2 − b2 ), f z = 4z(x2 + y 2 + z2 + a2 − b2 ); if  f z  = 0 then z = 0 since x2 + y 2 + z 2 + a2 − b2 > 0 everywhere on the torus;

if  f x = f y   = 0 too, then since the origin is not on the torus, we must have x2 + y 2 = a2 + b 2 , but then substituting into the equation of the torus gives (2a2 )2 = 4a2 (a2 + b2 ), a contradiction. For the last part, let σσ (θ, ϕ) = (x,y,z) be the parametrization in Exercise 4.2.5. Then, x2 + y 2 + z 2 + a2 b2 = 2a(a + b cos θ), so

−

(x2 + y 2 + z 2 + a2

− b2)2 = 4a2 (a + b cos θ)2 = 4a2(x2 + y2).

 

Conversely, if (x,y,z) satisfies the given equation, let r = x2 + y 2 . A little b so z = b sin θ algebra gives (r2 + z 2 a2 b2 )2 = 4a2 (b2 z2 ). Hence, z 2 2 2 2 for some θ R. Then we find r = a + b cos θ 2ab cos θ, so (since r 0) σ (θ, ϕ) for some ϕ R; with the r = a b cos θ. With the plus sign (x,y,z) = σ

±

 ∈

− −

−

±

 | | ≤

 ∈

 ≥

35

minus sign, (x,y,z) = σσ(π θ, ϕ) for some ϕ. Thus, the image of   σ : coincides with the set of solutions to the given equation.

−

R2

→ R3

5.1.2 See the solution of Exercise 4.4.3 for the first part. Since   has a (smooth) f   at each point, it is orientable. The solution of  choice of unit normal f  Exercise 4.4.3 also shows that if the restriction of  F  to   has a local maximum or a local minimum at p, then F  is perpendicular to the tangent plane of  at p. But f   is also perpendicular to the tangent plane. Hence, if the restriction of  F  to   has a local maximum or a local minimum at p, then F  is parallel to f  at p, i.e. F  = λ f  for some scalar λ.

∇

 ∇∇  S 

 ∇∇   ∇   ∇

S 

 S 

 S 

 ∇∇

 ∇∇

∇ 5.1.3 Let f (x,y,z) = xyz  − 1, F (x,y,z) = x2 + y 2 + z 2 . Then, f   = 0 is a smooth surface S   by Theorem 5.1.1 and F  defines a smooth function on S . To see that F  has a smallest value on S , let B  be the closed ball given by x2 + y 2 + z2 ≤ 3. Then, B ∩ S   is compact as it is closed and bounded and it is non-empty because it contains the point (1, 1, 1). Hence, the continuous positive function F   must attain its lower bound, say ℓ, on B ∩S , and ℓ ≤ 3 since F (1, 1, 1) = 3. Obviously F (x,y,z) > 3 if (x,y,z) ∈ / B, so ℓ is the smallest value of  F  on S . By Exercise 5.1.2, the local maxima or minima of  F  on S   occur where (2x, 2y, 2z) = λ(yz,xz,xy) for some λ. Since xyz = 1 on S   this gives x 2 = y 2 = z 2 = λ/2, so x,y,z are equal up to sign. Since their product is 1, there are four possibilities: (x,y,z) = (1, 1, 1), (1, −1, −1), (−1, 1, −1) or (−1, −1, 1). The value of  F  is 3 at each of these points, which is the smallest value of  F  on S √    from above. The 3 distance between any two of these points of  R is the same (2 2), so they form the vertices of a regular tetrahedron. 5.2.1 (i) ( p cos u cos v, q cos u sin v, r sin u) (cf. Exercise 4.2.4); (ii) see Exercise 4.1.3; (iii) (u,v,

  ±

2

1 +  pu2 +

v2 q 2 );

(iv), (v), (vi) see Exercise 4.2.1;

(vii) ( p cos u, q cos u, v); (viii) (  p cosh u, q sinh u, v); (ix) (u, u2 /p2 , v); (x) (0, u , v); (xi) (  p, u, v).

±

±

 − −        

  −

1 1/3 0 5.2.2 In the notation of Theorem 5.2.2, A = 1/3 1 0 . The eigenvalues 0 0 2 are 2/3, 4/3, 2 and the corresponding unit eigenvectors are the columns of  P  = x′ x 1/ 2 1/ 2 0 ′ = P  y , then z′ = z   and the quadric 1/ 2 1/ 2 0 . If  y z′ z 0 0 1 becomes 23 x′ 2 + 43 y ′ 2 2z′ 2 + 4z′ = c, i.e. 23 x′ 2 + 43 y ′ 2 2(z′ 1)2 = c 2. Comparing with the standard forms in Theorem 5.2.2 gives the stated results when c > 2 and c <  2. If  c = 2 we have a cone with axis the z-axis (which is the same as the z′ -axis), vertex at x′ = y ′ = 0, z ′  = 1, i.e. x = y = 0, z = 1, and 2 2 cross-section perpendicular to the z-axis an ellipse 23 x′ + 43 y ′ = constant.

 

√  − −√  √  √ 

−

−

 −

−

36

b  into the equation of the 5.2.3 Substituting the components (x,y,z) of  γγ  (t) = a  + tb quadric gives a quadratic equation for t; if the quadric contains three points on the line, this quadratic equation has three roots, hence is identically zero, so the quadric contains the whole line. For the second part, take three points on each of the given lines; substituting the coordinates of these nine points into the equation of the quadric gives a system of nine homogeneous linear equations for the ten coefficients a 1 , . . . , c of  the quadric; such a system always has a non-trivial solution. By the first part, the resulting quadric contains all three lines. 5.2.4 Let L1 , L2 , L3   be three lines from the first family; by the preceding exercise, there is a quadric  containing all three lines; all but finitely-many lines of the second family intersect each of the three lines; if  L′   is such a line,  contains three points of  L′ , and hence the whole of  L′   by the preceding exercise; so contains all but finitely-many lines of the second family; since any quadric is a closed subset of  R3 ,  must contain all the lines of the second family, and hence must contain .

 Q

 Q

 Q

 Q

  −  S  

y y 1 1 x q  p + q = uv, x = 2 p(u+v), y = 2 q (v u), so a parametrisation is σ u σ v = ( 12 q (u+v), 12 p(v u), pq ) is never σ (u, v) = ( 12 p(u+v), 12 q (v u), uv); σ σ  is regular. For a fixed value of  u, σ σ (u, v) = ( 12 pu, 12 qu, 0)+v( 21 p, 12 q, u) zero so σ

5.2.5 z =

x  p

−

×

−

−

−

−

is a straight line; similarly for a fixed value of  v; hence the hyperbolic paraboloid is the union of each of two families of straight lines. a c 5.2.6 Let A = . Since A is a real symmetric matrix, there is an orthogonal c b a′ 0 ′ ′ t matrix P   such that P AP  = A  is a diagonal matrix, say A =  (see 0 b′ Appendix 0). We can assume that det(P ) = 1 by changing the sign of one of the rows of  P  if necessary. Then the equation of the conic becomes

 

 

2

2

a′ x′ + b′ y ′ + d′ x′  + e′ y ′  + f  = 0, x′ x d′ d where = P  , = P  . In other words, we might as well ′ ′ y y e e assume that c = 0 from the beginning.

    

Assume then that c = 0. If  a = 0, we can assume that d = 0 by replacing x by x + ad  (a translation). Similarly, if  b = 0 we can assume that e = 0.

 



If  a and b are both = 0, we can therefore assume that d =  e  = 0 and we are in case (i) or (iii).

 

If  a = 0 and b = 0, the equation takes the form by 2 + dx + f   = 0. If  d = 0 replacing x by x + f d  we can assume that f  = 0 and then we are in case (ii). If  d = 0 we are in case (v). The case a = 0, b = 0 is similar.

 

 



37

If  a = b = 0 we have a straight line. 5.2.7 (i) From the classification in Theorem 5.2.2, the connected quadric surfaces are those of types (i), (ii), (iv), (v), (vii), (ix) and (x) (remember that for type (vi) to be a smooth surface, we have to remove the vertex which results in a disconnected surface). Type (i) is diffeomorphic to S 2 by (x,y,z) ( px , yq , zr ); type (ii) to the unit cylinder by (x,y,z)

   →    x p

2 x2  + yq2 p2

,

y q

2 x2  + yq2 p2

 → 

, z ; types (iv)

and (v) to a plane by (x,y,z) (x,y, 0); type (vii) to the unit cylinder by x y (x,y,z)  (  p , q , z); type (ix) to a plane by (x,y,z)  (x, 0, z); and type (x) is a plane. (ii) For types (iii) and (vi) the connected pieces are the parts of the quadric with z > 0 and with z < 0. Each part of type (iii) is diffeomorphic to the plane by (x,y,z)  (x,y, 0). Each part of type (vi) is diffeomorphic to the unit cylinder

→ 

→ →

by (x,y,z)

    →    x p

y2 x2  + 2 p q2

,

y q

y2 x2  + 2 p q2

 →

 | |

, ln z

. The two connected pieces of type

(viii) are given by x > 0 and x < 0; each piece is diffeomorphic to the plane by (x,y,z) (0, y , z).

→

5.3.1 From Example 5.3.2, the surface can be parametrized by σ (u, v) = (cosh u cos v, cosh u sin v, u),

∈ R and −π < v < π or 0 < v < 2π. 5.3.2  σ (u, v) 2 = sech2 u(cos2 v + sin2 v) + tanh2 u   = sech2 u + tanh2 u   = 1, so σ   parametrizes an open subset of  S 2 ; σ   is clearly smooth; and σ u ×  σσ v = −sech2 u σ (u, v) is never zero, so σ  is regular. Meridians correspond to the pawith u

rameter curves v = constant, and parallels to the curves u = constant.

γ  .a  = 0 so γ γ  ˜  is contained in the plane perpendicular to aa and passing through γ 5.3.3 (i)  ˜ the origin; (ii) simple algebra; (iii) v˜ is clearly a smooth function of (u, v) and the Jacobian matrix of the map 1 0 (u, v) (u, v˜) is , where a dot denotes d/du; this matrix is invertible ˙  .a 1 γγ  ˜  is a reparametrization of  γγ . so γγ  ˙  , σ v = δδ  ˙  (u) is perpendicular to the surface at ˙  +vδδ  δ  (a dot denotes d/du) so δδ  5.3.4 σ u = γγ  ˙ .  (γγ  ˙  ) = 0,  ˙δδ .  δδ   = 0. The second equation follows from δδ   = 1 ˙  + vδδ  σ (u, v) δδ  ˙  )/ δδ  ˙   2 . Hence, Γ(u) = γ  ˙  .δδ  v = (γ so the two conditions are satisfied ˙  )δδ  / δδ  ˙   2 . Using δδ  ˙ .  δδ   = 0 again, Γ˙ .δδ  ˙   = γγ  ˙   (γ ˙  )δδ  ˙ .  δδ  ˙  / δδ  ˙   2 = 0. γ  ˙  .δδ  ˙ .  δδ  ˙  .δδ  γγ   (γγ 



→

 ⇐⇒

−

  



 ⇐⇒

−

−

  

   

  

38

σ u  is tangent to the meridians, so a unit-speed curve γγ  γ  is a loxodrome 5.3.5 The vector σ ˙  .σ u /  σ ˙ = σ u = cos α, which gives u = if  γγ  ˙ cosh u cos α; since γγ   is unit-speed, γγ  ( u˙ sech u tanh u cos v v˙ sech u sin v, u˙ sech u tanh u sin v+v˙ sech u cos v, u˙ sech2 u) is a unit vector; this gives u˙ 2 + v˙ 2 = cosh2 u, so v˙ = cosh u sin α. The correponding curve in the uv-plane is given by  dv/du = v/ ˙ u˙ = tan α, and so is a straight line v = u tan α + c, where c is a constant. 5.3.6 The ruling that makes an angle θ   with the positive x-axis is given by y = x tan θ, z = f (θ), so a point at a distance u  from the z-axis has coordinates df  df  x = u cos θ, y = u sin θ, z = f (θ). σu σθ = ( dθ sin θ, dθ  cos θ, u) and this is non-zero if  u = 0 (i.e. if we exclude the line  itself from the surface). 5.3.7 (i) The assumption means that N   is constant. Hence, σ .N   is constant, since σ.N)u = σ σ u.N  = 0, etc. So σσ  is contained in a plane v.N  = constant. (σ N  for some scalar λ. (ii) If all the normal lines pass through a point p, σσ  = p + λN σ u.N  = λ u  since Nu.N  = 0 as N  is a unit vector. Nu  so 0 = σ Then, σ u = λ uN  + λN Similarly λ v  = 0. So λ is a constant and σσ  is part of the sphere v p = λ . (iii) By applying an isometry we can assume that the given straight line is the z-axis. Reparametrize the surface using polar coordinates in the xy-plane: N   is the poσ (r, θ) = (r cos θ, r sin θ, f (r, θ)), say. For some scalar µ, σ + µN sition vector of a point on the z-axis. If N = (a,b,c), setting the x- and N   equal to zero gives µa = r cos θ, µb = r sin θ, y-components of  σ + µN and hence either µ   = 0 (i.e., r   = 0) or a sin θ = b cos θ. But σ r  σσ θ = (f θ sin θ rf r cos θ, f θ cos θ rf r sin θ, r), so if  µ = 0 then

−

   

−

−

 ±

±

 ×



 L

 ±

−

  −  | |

−

−

−

−  sin θ(rf r cos θ − f θ sin θ) = cos θ(rf r sin θ + f θ cos θ),

−  ×

i.e. f θ   = 0. We have now proved that f θ  = 0 except possibly when r = 0; but then f θ  = 0 for all (r, θ) as f  is smooth (and hence continuous). Thus, f  depends σ  is the surface of revolution obtained by rotating the curve only on r, and hence σ r (r, 0, f (r)) in the xz-plane around the z-axis. 5.3.8 Use the parametrization γγ (u) = (cos u, sin u, 0) of the ‘waist’ of the hyperboloid. From Exercise 4.1.3, there is a straight line on the hyperboloid passing through γγ  (u), and it takes the form

 →

(x

− z)cos θ = (1 − y)sin θ,

(x + z)sin θ = (1 + y)cos θ.

Substituting x = cos u, y  = sin u, z  = 0 gives θ = u2 + π4 . The vector δδ  (u) = (cos2θ, sin2θ, 1) = ( sin u, cos u, 1) is parallel to this line, so a parametrization of the hyperboloid as a ruled surface is

−

σ (u, v) = (cos u, sin u, 0) + v( sin u, cos u, 1) = (cos u

−

− v sin u, sin u + v cos u, v).

˙   = 0, the line of striction is γγ   (see the solution to Exercise 5.3.4). ˙ .  δδ  Since γγ 

39

5.3.9 We use the notation in Theorem 5.2.2. (a) Types (vii) through (xi). (b) Types (vi) and (x). (c) Types (ii), (v) and (vi) through (xi). (d) Type (i) if  p2 , q 2 , r2 are not distinct; types (ii), (iii), (iv), (vi) and (vii) if   p2 = q 2 ; and types (x) and (xi). γ   is unit-speed and δ (u), where γ 5.3.10 Take the ruled surface to be σ (u, v) = γγ (u) + vδδ  δδ    is a unit vector. Consider the ruling corresponding to u = u0 , say. Since ˙  ) δδ   (a dot denoting d/du, evaluated at u = u 0 ), the union of  ˙  + vδδ  σ u σ v = (γγ  the normal lines along the ruling u = u 0  is the surface

×

×

˙  ) σ (u0 , v) + w(γγ  ˙  + vδδ  Σ(v, w) = σ

× δδ  = a + vbb + wcc + vwdd, ˙   × δδ   are constant vectors. ˙   × δδ   and d  = δδ  where a  = γγ  γ (u0 ), b  = δδ  δ (u0 ), c  = γγ 

By applying a translation we can assume that a   = 0. Then, by applying a ˙   is parallel to (0, 1, 0) (note rotation we can assume that b  = (1, 0, 0) and that δδ  ˙   is perpendicular to δδ  δ ). Then, c  is parallel to the yz-plane and d  is parallel that δδ  to the z-axis. ˙  (u0 ) = 0, Σ  is a plane. If d = (0, 0, λ) = 0  and c = (0, µ , ν )  , If d = 0, i.e. if  δδ  then Σ(v, w) = (v,µw,νw + λvw). If  µ = 0 we have the xz-plane. Otherwise, x λν   followed by a rotation by π/2 around the z-axis takes the translation x Σ  to a parametrization of the hyperbolic paraboloid z = λµ (x2 y 2 ).

 

→ −

−

5.4.1 Both surfaces are closed subsets of  R3 , as they are of the form f (x,y,z) = 0, R   is a continuous function (equal to x2  y 2 + z 4  1 and where f  : R3 x2 + y 2 + z 4 1 in the two cases). The surface in (i) is not bounded, and hence not compact, since it contains the point (1, a2 , a) for all real numbers a; that in (ii) is bounded, and hence compact, since  x 2 + y 2 + z4 = 1 = 1 x,y,z 1. The surface in (ii) is obtained by rotating the curve x 2 + z4 = 1 in the xz-plane around the z-axis:

−

→

−

⇒− ≤

−

 ≤

5.4.2 A closed curve γγ   with period T  can be identified with the unit circle by  γγ (t)

→

40

(cos(2πt/T ), sin(2πt/T )). This gives rise to a diffeomorphism from the tube around γγ   to a tube around the circle, i.e. a torus. We have to make the tube have a sufficiently small radius as otherwise it might intersect itself and then it would not be a surface. 5.5.1 Both parts are geometrically obvious. 5.5.2 Let (a,b,c) R3 with a and b non-zero. Then F t (a,b,c) as t  and as 2 2 t approaches p and q  from the left; and F t (a,b,c) as t  and as t 2 2 approaches p and q  from the right. From this and the fact that F t (a,b,c) = 0 is equivalent to a cubic equation for t, it follows that there exist unique numbers u,v,w   with u < p2 , p2 < v < q 2  and q 2 < w   such that F t (a,b,c) = 0 when t = u, v or w. The surfaces F u (x,y,z) = 0 and F w (x,y,z) = 0 are elliptic paraboloids and F v (x,y,z) = 0 is a hyperbolic paraboloid, and we have shown that there is one surface of each type passing through each point (a,b,c).

∈

→∞ →∞ → −∞ → −∞

 ∇∇



2x

2y

 −

To show that the system is orthogonal, note that F t =  p2 −t , q 2−t , 2 so if (a,b,c) is a point at the intersection of the surfaces F u (x,y,z) = 1 a n d F v (x,y,z) =, we have 4a2 4b2 F u. F v = 2 + u)( p2 v) (q 2 u)(q 2 ( p 4(F u (a,b,c) F v (a,b,c)) 4 = = u v u

∇ ∇

−

− − −

−

− v) + 4 − 4 = 0. −v

Thus, the first two families intersect orthogonally. Similarly for the other two pairs of families. To parametrize these surfaces, write F t (x,y,z) = 0 as the cubic equation x2 (q 2

− t) + y2( p2 − t) − 2z( p2 − t)(q 2 − t) + t( p2 − t)(q 2 − t) = 0, and note that the left-hand side must be equal to (t − u)(t − v)(t − w); putting t = −v)( p −w) ,  p2 , q 2 and then equating coefficients of  t 2 (say) gives x = ± ( p −u)( p q − p −v)(q −w) , z = 1 (u + v + w − p2 − q 2 ). y = ± (q −u)(q 2  p −q 5.5.3 (i) Let U  = xy − uz 2 , V  = x2 + y 2 + z 2 − v, W  = x2 + y 2 + z 2 − w(x2 − y 2 ). ∇U  = (y,x, −2uz), ∇∇V  = (2x, 2y, 2z), ∇W  = (2(1 − w)x, 2(1 + w)y, 2z). Then, ∇ Hence, at a point of intersection of the surfaces U  = 0, V  = 0, W  = 0, we have ∇U .∇V  = 4xy − 4uz2 = 0 by the equation U  = 0; ∇V ..∇W  = 4((1 − w)x2 + (1 + w)y2 + z2 ) = 4(x2 + y2 + z2 − w(x2 + y2)) = 0

 

2

2

2

2

2

 

2

2

2

by the equation V  = 0; and

∇W ..∇U   = 2(1 − w)xy + 2(1 + w)xy − 4uz2 = 4(xy − uz2 ) = 0

2

2

41

by the equation U  = 0. (ii) Let

  

 

 

 

− ux, V  = x2 + y2 + x2 + z2 − v, W  = x2 + y2 − x2 + z2 − w. ∇U  = (−u,z,y), ∇∇V  = √ xx+y + √ x x+z , √ x y+y , √ x z+z , ∇∇W  = Then,  ∇ √ xx+y − √ x x+z , √ xy+y , − √ x z+z . Hence, at a point of intersection of the ∇U .∇V  = (yz − ux) √ x 1+y + √ x 1+z = surfaces U  = 0, V  = 0 and W  = 0, ∇ 0 by the equation U  = 0; ∇V ..∇W  = x x+y − x x+z + x y+y − x z+z = 0; ∇W ..∇U  = (yz − ux) √ x 1+y − √ x 1+z = 0 by the equation U  = 0. U  = yz



2

2

2

2

2

2

2

2

2

2

2

2

2

2



2

2

2

2



2



2

2

2

2

2

2

2

2



2

2

2

2

2

2

2

2



2



5.5.4 Two families of plane curves U (x, y) = u and V (x, y) = v   form an orthogonal system if  U . V  = 0 at each point of intersection of a curve of the first family ∂U  with a curve of the second family. Here, U  = ∂U  ∂x , ∂y , etc. (i) U (x, y) = x, V (x, y) = y, i.e. the two families of straight lines parallel to the y- and x-axis, respectively. The orthogonality is obvious. (ii) U (x, y) = y cos u x sin u, V (x, y) = x 2 + y 2 v 2 , i.e. the family of straight lines passing through the origin and the family of circles with centre the origin. The orthogonality is obvious. 5.5.5 The function F t (a, b) is a continuous function of  t (for fixed non-zero real numbers a, b) in each of the open intervals ( , p2 ), ( p2 , q 2 ) and (q 2 , ). Also, F t (a, b) 0 as t , and F t (a, b)  as t approaches p 2 or q 2 from the left (+ sign) or right (  sign). It follows that there is at least one value of  t in each of the , p2 ) and ( p2 , q 2 ) such that F t (a, b) = 1. On the other hand, open intervals ( F t (a, b) = 1 is equivalent to a quadratic equation for t which has at most two real , p2 ) and v ( p2 , q 2 ) roots. It follows that there are unique numbers u ( such that F u (a, b) = F v (a, b) = 1. The conics F u (x, y) = 1 and F v (x, y) = 1 are ellipses and hyperbolas, respectively, and we have shown that there is one of each passing through each point on the plane that does not lie on one of the coordinate axes. To show that this is an orthogonal system of curves, we calculate F u. F v at a point (a, b) that lies on the conics F u   = 1 and F v   = 1. Since F u = 2y 2x  p2 −u , q 2 −u , we get

 ∇ ∇

 ∇∇

−

→ ±∞ −





−

→ ±∞

−∞

∞

−∞

 ∈ −∞





→

 ∈

 ∇∇ ∇  ∇∇

4a2 4b2 F u. F v = 2 + u)( p2 v) (q 2 u)(q 2 v) ( p 4(F u (a, b) F v (a, b) 4 4 = = = 0. u v u v

∇ ∇

−

− − −

−

− −

−

42 2

 t . Argu For the last part, let F t (x, y ) =  p2x−t  2y  2 y  +  t. Arguin ingg as in Exerc Exercis isee 5.5 5.5.2, .2, for any point (a, (a, b) R2 not on one of the coordinate axes, there are unique , p2 ) and v  ( p numbers u  (  ( p2 , ) such that F u (a, b) =  F v (a, b) = 1. Both systems of curves F u (x, y ) = 1 and F v (x, y ) = 1 are parabolas. At (a, (a, b) we find that

 ∈ −∞

 ∈

∇F u.∇F v = ( p2 −

 ∈

4a2 u)( p )( p2

− v)

 −

∞

+4 =

4(F  4(F u (a, b) F v (a, b)) 4 = u v u

− −

− 4 = 0, −v

showing that the parabolas form an orthogonal system. 5.5.6 Let U ( U (x, y ) = u and V ( V (x, y ) = v   be an orthogonal system of plane curves (see  u ,  V (  V  (x, y ) = Exercise 5.5.4). We will show that the system of surfaces U  surfaces  U ((x, y ) = u, v , z = w  is orthogonal. orthogonal. The first two families of surfaces surfaces consist consist of generalized generalized cylinders, cylinders, the third consists consists of parallel planes. planes. Suppose Suppose that (a,b,c (a,b,c)) is a point of intersection of a surface from each family. We have to prove that the vectors ∂U  ∂U  ∂V  ∂V  (0, 0, 1) are mutually perpendicular at (a,b,c (a,b,c). ). ∂x , ∂y  , 0 , ∂x , ∂y  , 0   and (0, But it is obvious that the third vector is perpendicular to the other two, and U (x, y ) = u and V  (x, the first two are perpendicular because the curves U (  (x, y ) = v intersect orthogonally at (a, (a, b).







V  be the smooth bijective  F −1 :  V  W  5.6.1 Let F  Let F  :  W  bijective map with smooth inverse inverse F  1 γ  − constr construct ucted ed in the proof proof of Proposi Proposition tion 4.2.6. 4.2.6. Then, Then, (u(t), v(t)) =  F  (γ  (t)) is smooth.

 →  →

→

F (x, y ) = (x, f ( f (x, y )); then F  5.6.2 Suppose, Suppose, for example, example, that f  that f y = 0 at (x (x0 , y0 ). Let F ( then  F  1 f x is smooth and its Jacobian matrix  is invertible at (x (x0 , y0 ). By the in0 f y verse function theorem, F  theorem,  F  has  has a smooth inverse G inverse  G defined on an open subset of  R2 F (x0 , y0 ) = (x0 , 0), and G  must be of the form G(x, z) = (x, g (x, z )) containing F ( for some smooth function g . Then γ Then  γγ (t) = (t, g (t, 0)) is a parametrization of the f (x, y ) = 0 containing (x level curve f ( (x0 , y0 ).

 

The matrix



f x gx

f y gy

f z gz



 

 has rank 2 everywhere; suppose that, at some point





f y f z (x0 , y0 , z0 ) on the the lev level curv curve, the the 2 2 submatrix submatrix   is invertib invertible. le. g y gz F (x,y,z) x,y,z ) = (x, f ( f (x,y,z) x,y,z), g (x,y,z)) x,y,z)) is smooth and its JaThen, Then, the functio function n F ( 1 0 0 f x f y f z   is inv x,u,v) = cobian matrix invertib ertible le at (x0 , y0 , z0 ). Let G(x,u,v) gx gy gz x,u,v ), ψ(x,u,v)) x,u,v)) be the smooth inverse of  F  defined  F  defined near (x (x, ϕ(x,u,v) ( x0 , 0, 0). Then γγ  (t) = (t, ϕ(t, 0, 0), f (x,y,z) x,y,z) = 0), ψ (t, 0, 0)) is a parametrization of the level curve f ( g (x,y,z) x,y,z) = 0 containing (x (x0 , y0 , z0 ).

 ×

 

 

43

f (u, v ), g (u, v), h(u, v )). The 5.6.3 Let σ(u, v ) = (f ( The condition condition σ u σ v = 0 at (u0 , v0 ) f u f v means that the matrix gu gv  has rank 2 at (u (u0 , v0 ), so at least one 2 2 hu hv f u f v F (u, v ) = (f ( f (u, v ), g (u, v)), then as submatrix is invertible, say . If  F ( gu gv in the proof of Proposition 4.2.6 there is an open subset V  of  R2 containing F ( F (u0 , v0 ) and an open subset W  of  U   U   containing (u V  (u0 , v0 ) such that F  :  W   σ to W   W   is injective. is bijective, in particular injective. Then the restriction of  σ f (u, v ), g (u, v), h(u, v )). The conditi 5.6.4 Let σ (u, v) = (f ( cond ition on that tha t N N(u0 , v0 ) is not parf u f v xy -plane means that the matrix allel to the xy-plane  is invertible at (u (u0 , v0 ). gu gv F (u, v) = (f ( f (u, v), g (u, v )), then as in the proof of Proposition 4.2.6 there If  F ( F (u0 , v0 ) and an open subset W  of  U  is an open subset V  of  R2 containing F ( V  is bijectiv containing (u (u0 , v0 ) such that F  : W  bijectivee with smooth inve inverse rse.. If  − 1 F  (u, v) = (α(u, v), β (u, v)), then near (x (x0 , y0 , z0 ) the surface coincides with xy -plane, then the graph z = h(α(x, y ), β (x, y )). If N(u0 , v0 ) is parallel to the xy-plane,  σ (u, v) at least one of the other two 2 2 submatrices of the Jacobian matrix of  σ is invertible, and then the surface coincides near (x ( x0 , y0 , z0 ) with a graph of the form x  = ϕ  =  ϕ((y, z ) or y  = ϕ  =  ϕ((x, z ). γ  ˙  (t0 ) = 0, so at least one compo5.6.5 Let γ Let  γγ (t) = (γ 1 (t), . . . , γn  (t)). As γ As  γγ  is   is regular, γ ˙  (t0 ) is non-zero nent of  γγ  non-zero,, say γ  ˙ 1 (t0 ) (the proof is the same in the other cases). By the inverse function theorem for real-valued function of one variable, there is a smooth function α, say, defined on an open interval containing γ 1 (t0 ), and  t  for all t some ǫ some  ǫ > 0 such that α that  α((γ 1 (t)) =  t for all t (t0 ǫ, t0 + ǫ). This implies that the restriction restriction of γ  of  γ 1  to the interval (t (t0 ǫ, t0 + ǫ) is injective. Hence, the restriction  γγ  to of  γ   to the same interval is injective.

 

 



 ×



×



 →  →





→

×

 

−

∈ −

Chapter 6 6.1.1 (i) Quadric Quadric cone x2 + z 2 =  y 2 ; we have σ u  = (cosh u sinh v, cosh u cosh v, cosh u), σ v  = (sinh u cosh v, sinh u sinh v, 0), 0),

 σ u 2 = 2 cosh cosh2 u cosh2 v,

σv  = 2 sinh σ u .σ sinh u cosh u sinh v cosh v,

 σ v 2 = sinh2 u cosh2v, cosh2v, and the first fundamental form is dudv + 2 cosh cosh2 u cosh2 v du2 + sinh sinh 2u sinh2v sinh2v dudv  + sinh2 u cosh2v cosh2v dv2 .

44

dudv + (ii) Paraboloid of revolution; (2 + 4u 4u2 ) du2 + 8uv 8uv dudv  + (2 + 4v 4v 2 ) dv 2 . (iii) Hyperbolic cylinder; (cosh2 u + sinh2 u) du2 + dv 2 . dudv + (iv) Paraboloid of revolution; (1 + 4u 4u2 ) du2 + 8uv 8uv dudv  + (1 + 4v 4v 2 ) dv 2 . 6.1.2 6.1.2 Applyin Applyingg a transla translation tion to a surfac surfacee patch patch σ  does not change σ u or σ v . If  P  is P (σ u ), (P σ )v = P ( P (σ v ), and P   P   preserves a 3 3 orthogonal matrix, (P (P σ )u = P ( 3 p).P (q q) = p.q  for all vectors .P (q dot products (P  (P (p (p vect ors p, q R ). Applying the dilation σ  by a x,y,z) a(x,y,z), x,y,z), where a (x,y,z) where  a is  is a non-zero constant, multiplies σ multiplies σ  by  a and  and hence 2 the first fundamental form by a .

×

 ∈

→

6.1.3 Since both sides define bilinear forms forms on the tangent tangent plane, it suffices suffices to prove prove that the two two sides agree when v, w  belong to the basis σ u , σ v . This This is easi easily ly du(σ u ) =  dv(  dv (σ v ) = 1, du( du(σ v ) = dv(  dv (σ u ) = 0. checked using du(

 {

}

∂u ∂v σ ˜ σ ∂  σ ∂v ˜ u˜ = σ u ∂ ˜ ˜ ˜ = σ u ∂u 6.1.4 By the chain rule, σ ∂ u u ∂ ˜ ∂ v ∂ v ˜ +  σ v ∂ ˜ ˜, v ˜ +  σ v ∂ ˜ ˜ , which gives E  = ∂u 2 ∂u ∂v ∂v 2 ˜ and G ˜  can be ˜ u˜ =  E  ∂ ˜ σ˜ u˜ .σ F  G F  + 2F  2 + ∂ u ∂ ˜ ∂ u ∂ u ∂ ˜ ∂ u ˜ ˜ ∂ ˜ ˜ ˜ . Similar expressions for found; multiplying out the matrices shows that these formulas are equivalent to the matrix equation in the question.





Ed u2 + 2Fdudv Following the procedure given, Edu 2Fdudv +  + Gdv2 = E 



∂u ˜ ∂ ˜ ∂ u ˜ du

2 2 ∂u ∂u ∂u ∂v ∂v ∂v ∂v d v F  d u d v d u d v G d u d v ˜ +2F  +2 ˜ + ˜ ˜ + ˜ + ˜ + ˜ . The ∂ ˜ ∂ v ∂ ˜ ∂ u ∂ ˜ ∂ v ∂ ˜ ∂ u ∂ ˜ ∂ v ∂ ˜ ∂ u ∂ ˜ ∂ v ˜ ˜ ˜ ˜ ˜ ˜ ˜ ∂u 2 u ∂v ∂v 2  d u G ∂ ˜ of  d ˜2 is E  is  E  ∂ ˜ + 2F  ∂∂ ˜ ∂ u ∂ u ∂ u ∂ u ˜ ˜ ∂ ˜ ˜  + ˜ , which agrees with the expres-

+

  



 



coefficient coefficient ˜  found ˜ and G ˜. sion for E    found above. Similarly for F 



 σσu .σ uv =  σσ v .σ uv = 0  σσuv  is 6.1.5 (i)  (ii): E v = Gu = 0  i s paral par alle lell to N.  u 0 , u = u Consider the quadrilateral bounded by the parameter curves u curves  u =  = u  =  u 1 , v  = v1 v0 , v = v1 . The The length length of the side side given given by u = u0 is v0 (σ v (u0 , v) dv = v1 G(u0 , v )dv. dv . v0 (i) =   (iii) (iii):: If  Gu = 0, G  depends only on v  so the last integral is unchanged when u0  is replaced by u1 . So the two two sides sides u = u0 and u = u1  have the same length; and similarly for the other two sides. v dv  is independent of  u; (iii) =  (i): If the lengths lengths are equal equal then then v01 G(u, v)dv is v1 Gu dv  = 0 for all v0 , v1 , so Gu = 0; differentiating with respect to u  gives v0 2√ G dv = and similarly similarly E v = 0.

 ⇐⇒  ⇐ ⇒

 ⇐⇒  ⇐ ⇒

 ⇐⇒  ⇐ ⇒

  

   

 

⇒

     

⇒

  

  

E (u) du, du, v˜ = G(v ) dv. dv. Assuming conditions (i) - (iii) are satisfied, set u˜ = u Then, (u, (u, v) (u, ˜, v˜) is a reparametrization map because its Jacobian matrix E  0 EG.. The first  has non-zero determinant EG first fundame fundament ntal al form of  G 0

 √ 

 →  √ 

√ 

 d˜  du2

2F  dud˜ u ˜dv˜ + dv˜2 . EG

σ (u, u  σ (u, v) is the reparametrization ˜ reparametrization  ˜ ˜, v˜) of  σ i s ˜ + √ 

 

F  <  1 Now √ 2EG

G F 2 >  0, so there is a smooth function θ u, u, v˜) with 0 < π  such that since E since EG function θ(˜ (˜ 0  < θ < π such F  2 2 θd ud˜ u cos θ  = √ EG . This gives the first fundamental form as du ˜ + 2 cos cos θd˜ ˜dv˜ + dv˜2 .

−

45

u + vˆ), v˜ = 12 (ˆ u Since u ˜  = 12 (ˆ

− vˆ), the first fundamental form becomes

1 1 1 u + dˆ v )2 +  cos θ(dˆ u2 dˆ v 2 ) + (dˆ u dˆ v )2 (dˆ 4 2 4 θ 2 θ 2 1 1 u2 + (1 cos θ)dˆ v 2 = cos2 dˆ u + sin2 dˆ v . = (1 + cos θ)dˆ 2 2 2 2

−

−

−

6.1.6 (i) σ u   = (cos v, sin v, 1/u), σ v = ( u sin v, u cos v, 0) so E  = 1 + u −2 , F  = 0, G = u2 , The surface is obtained by rotating the curve z = ln x in the xz-plane around the z-axis. σ v = ( u sin v, u cos v, 1) so E  = 1, F  = 0, G = 1 + u2 . (ii) σσ u  = (cos v, sin v, 0), σ This is a helicoid (Exercise 4.2.6). σ u  = (sinh u cos v, sinh u sin v, 1), σ σ v = ( cosh u sin v, cosh u cos v, 0) so E  = (iii) σ 2 2 1 + sinh u = cosh u, F  = 0, G = cosh2 u. This is a catenoid (Exercise 5.3.1). 6.1.7 The first fundamental form is 2du2 + u2 dv 2 . Hence, the length is

−

−

−

π

 

(2(λeλt )2 + (eλt )2 )1/2 dt =

0

√ 2λ2 + 1 λ

(eλπ

− 1).

A ruling can be parametrized by u = t, v = constant, so the angle is cos−1



λt

2(λe

+

u2

(0))

 

√ 

(2λ2 + 1)e2λt 2



= cos−1

 

2λ2 , 2λ2 + 1

which is constant and independent of the ruling. ˙  + vb˙  = t τ vn n, σσ v = b , so since t, n  and b are 6.1.8 Denoting d/du by a dot, σσ u = γγ  2 2 perpendicular unit vectors, E  = 1 + τ  v , F  = 0, G = 1. σ u.σ uu = 2σ σu.σ uu . Hence, σ u.σ uu = 6.1.9 From E  = σ u.σ u  we get E u = σ uu.σ u + σ 1 2 E u . Similarly, differentiating E   with respect to v and G   with respect to u and v   gives σ u .σ uv = 12 Gu , σ u .σ uv = 12 E v , σ v .σ vv = 12 Gv . Differentiating σ u.σ v  with respect to u gives F u = σ σuu.σ v  + σ u.σ uv  and using the formula F  = σ 1 1 σ u .σuv = 2 E v   gives σ uu .σ v = F u σ .σ 2 E v . The formula for u vv   is proved similarly.

−

 −

 √  → 

√ 2sin √ v , 0



v ,u 2

u 2cos √  6.2.1 The map is σ (u, v) = σ˜ (u, v), say. The im2 age of this map is the sector of the xy-plane whose polar coordinates (r, θ) satisfy 0   < θ < π 2. The first fundamental form of  σ is 2 du2 + u2 dv 2 ; v v v v 2 √  √  √  ˜ ˜ σ˜ u = , , σ u , u , σ 2cos √  2sin 0 , = sin cos 0 , so = v u 2 2 2 2

√  √  √  −   σv = 0,   σ v 2 = u 2 and the first fundamental form of  σ˜ is 2 du2 + u2 dv 2 . 2, σ u .σ √  6.2.2 No: the part of the ruling (t, 0, t) with 1  ≤ t ≤  2 (say) has length 2 and is mapped to the straight line segment (t, 0, 0) with 1 ≤ t ≤ 2, which has length 1.









46

6.2.3 A straightforward calculation shows that the first fundamental form of  σt is σ (u, v) σ t (u, v) cosh2 u(du2 +dv 2 ); in particular, it is independent of  t. Hence, σ is an isometry for all t. Taking t = π/2 gives the isometry from the catenoid to the helicoid; under this map, the parallels u = constant on the catenoid go to circular helices on the helicoid, and the meridians v  = constant go to the rulings of the helicoid.

→

˙  )δδ  / δδ  ˙   2 (Exercise 5.3.4), where in this ˙  .δδ  6.2.4 The line of striction is given by v = (γγ  ˙  . Since γγ   is unit-speed, γγ  ˙  .γ¨γ   = 0 so v  = 0 and we get the curve γγ   itself. δ  = γγ  case δδ  For the second part, we can assume that u 0  = 0 and by applying an isometry of  γ  (0) = 0, t(0) = iii, n(0) =  j, R3 that γ j b(0) = k  (in the usual notation). Then, using ... γγ  (0) = κ(0)j jj, γγ  (0) = ( κ(0)2 , ˙κ(0), κ(0)τ (0)) so, neglecting higher Frenet-Serret,  ¨ ... ˙  ( 0)+  12 γγ  ¨ (0)u2 +  16 γγ  (0)u3 +. . . = γ (u) = γγ  γ ( 0)+uγγ  powers of  u in each component, γγ  (u, 12 κ(0)u2 , 16 κ(0)τ (0)u3 ). The intersection of the surface with the plane perpendicular to t(0) = iii  is given by setting the x-component of  σ (u, v) equal to zero. This gives v = u+ higher terms, so neglecting such terms, u = v. Then the σ ( v, v) = γγ  ˙  ( v) = (0, 21 κ(0)v 2 , 13 κ(0)τ (0)v 3 ). γ ( v) + vγγ  intersection is Γ(v) = σ

−

  

−

 −

−

−

−

−

 −

σ(u, v) = γγ  γ (u) +vaa, where we can 6.2.5 A generalized cylinder can be parametrized by σ ˙  .a  = 0, assume (see Exercise 5.3.3) that γγ   is unit-speed, a  is a unit vector, and γγ  ˙  , σ v = a   so the first fundamental where the dot denotes  d/du. Then, σ u = γγ  2 2 form is du + dv . This is the same as the first fundamental form of the plane (u,v, 0), so by Corollary 6.2.3 the map σσ (u, v) (u,v, 0) is a local isometry.

→

By applying an isometry of  R3 (a translation), we can assume that the vertex of the cone is the origin. It can then be parametrized by σσ (u, v) = vγγ (u), where ˙  , σ v = γγ  so the first fundamental γγ   is a unit-speed curve on S 2 . Then, σ u = v γγ  form is v 2 du2 + dv2 . This is the same as the first fundamental form of the plane in polar coordinates (v cos u, v sin u, 0). 6.2.6 Assume that f   satisfies the given condition and denote d/du  by a dot. Define 1 f ˙2 du; then g   is smooth since the term inside the square root g(u) = is always >   0. The curve γγ  (u) = (f (u), 0, g(u)) in the xz-plane is unit-speed, and by Example 6.1.3 if we rotate γγ   around the z-axis, the resulting surface of  revolution has first fundamental form du2 + f (u)2 dv 2 . By Corollary 6.2.3, the given surface is locally isometric to this surface of revolution.

   −

  

U (u) du. The given condition on U  implies that U (u) > 0 for all 6.2.7 Define u˜  = u/du is never zero. It follows values of  u, so u ˜  is a smooth function of  u and d˜ u, v) is a reparametrization map. The first fundamental form in that (u, v) (˜ ˜ u)dv2 , where U (˜ ˜ u) = U (u). By Exercise 6.2.6, this u, v) is d˜ u2 + U (˜ terms of (˜ df  surface patch is locally isometric to a surface of revolution if  d˜ u <  1 for all

 →

 

47

values of u ˜, where f  =

   

˜ But, U .

˜ df  1 dU  1 dU/du 1 dU   . = = = d˜ u d˜ u d˜ u /du du 2U  ˜ U  2 2 U 

√ 

6.3.1 If the first fundamental forms of two surfaces are equal, they are certainly proportional, so any isometry is a conformal map. Stereographic projection is a conformal map from S 2 to the plane, but it is not an isometry since λ = 1 (see Example 6.3.5). 6.3.2 The first fundamental form of the given surface patch is (1 +u2 +v 2 )2 (du2 +dv 2 ); this is a multiple of  du2 + dv2 so the patch is conformal.

 

6.3.3 The first fundamental form of  σ˜ (u, v) is

  dψ du

2

du2 + cos2 ψ(u)dv 2 . So σ˜ is

 

 dψ/du = cos ψ. Taking the plus sign, we get u = sec ψdψ = conformal ψ −u =   1+sin ψ + cos ψ = u u ln(sec ψ +tan ψ), so   1+sin cos ψ = e . Then 2 cosh u = e + e cos ψ 1+sin ψ 2sec ψ. Hence, cos ψ = sechu, sin ψ = tanhu and σ˜ (u, v) is the patch in Exercise 5.3.2. 6.3.4 Φ is conformal if and only if  f u2 + g u2 = f v2 + g v2 and f u f v + g u gv   = 0. Let z = f u + ig u , w = f v + ig v ; then Φ is conformal if and only if  z¯ z = w ¯ w and z ¯ w + z¯w = 0, where the bar denotes complex conjugate; if  z = 0, then w = 0 and all four equations are certainly satisfied; if  z = 0, the equations give z 2 = w 2 , so z = iw; these are easily seen to be equivalent to the first pair of equations in the statement of the exercise if the sign is +, and to the second pair if the f  gu sign is . We have det(J (Φ)) = u = (f u2 + f v2 ), with a plus sign if the f v gv first pair of equations hold and a minus sign if the second pair of equations hold. 6.3.5 Let   be an orientable surface. Fix a smooth choice of unit normal at each point of  , and let   be the atlas for   consisting of all the surface patches for whose standard unit normal agrees with the chosen normal. On the other hand, by Theorem 6.3.6   has an atlas consisting of conformal parametrizations; let ˜  be the maximal such atlas (i.e. the set of all conformal parametrizations of  ˜  is an atlas for . Indeed, if p ) . Then, , let σ   be any conformal parametrization of    containing p. If  σ   has the wrong orientation (so that ˜ (u, v) = σ ( u, v) is a conformal parametrization containing p σ / ˜), then σ that has the correct orientation. Thus, in any case there is a surface patch of    containing p   that is both conformal and correctly oriented. Let Φ be the ˜. Then, Φ is a transition map between two of the patches in the atlas conformal diffeomorphism between open subsets of  R2 . By Exercise 6.3.4, Φ is either holomorphic or anti-holomorphic, and in the latter case det(J (Φ)) < 0, contradicting the fact that Φ is the transition map between two correctly oriented surface patches. Hence, Φ must be holomorphic.

 ⇐⇒

 ±

 

 ±

 

 −

S   S 

±

 S 

 S 

 S 

A S 

 A ∩ A

∈A

S 

 A

 

 −

S 

 S 

 ∈ S 

−

 A ∩ A

48

y x ˜ , z+1 , 0). Identifying (u, v) 6.3.6 Following Example 6.3.5, we find Π(x,y,z) = ( z+1 2

1−|w| ˜ 1 (w) = ( |w2w C, we find σ with w = u +  iv |2+1 , 1+|w|2 ). Then σ 1 (w) = σ˜ 1 (1/w), ¯ so the transition map is w 1/w. ¯ This is not holomorphic, so the atlas 2 ˜ 1  does not give S  the structure of a Riemann surface. If  σˆ 1 (w) = σ˜ 1 (w), σ1, σ ¯ the transition map between σ 1 and σˆ 1 is w  1/w. This is holomorphic (when ˆ 1 gives w = 0, which holds on the overlap of the two patches), so the atlas σ 1 , σ S 2 the structure of a Riemann surface. Any circle on S 2 is the intersection of  S 2 with a plane, and so (see Appendix 2) has equation of the form aw + a ¯w + ¯ bz = c, where a C, b, c R  are constants |ξ|2 −1   gives , z (and a and b are not both zero). Substituting w = |ξ|2ξ = 2 +1 |ξ|2 +1 2 ¯ = b + c, which is the equation of a Circle in C∞  (a line if  aξ  (b c) ξ  + 2aξ  + 2¯ b = c, a circle otherwise). The converse is proved similarly. ˜ 1 of  S 2 in ExThe expression of the map Π−1 M  Π in terms of the atlas σ 1 , σ M (w), w ercise 6.3.6, which consists of conformal patches, is of the form w −1 −1 M (1/w), M (w) , or w  M (1/w) ¯ w ¯ , i.e. a M¨ obius or conjugate-M¨ obius transformation. Since such transformations are conformal (Appendix 2), the result follows. The first fundamental form is (1 + f ˙2 )du2 + u2 dv 2 , where a dot denotes  d/du. u2 1, i.e. if and only if  f (u) = So σ   is conformal if and only if  f ˙ = ( 12 u u2 1 12  cosh−1 u) + c, where c is a constant. ˙  + v 2 δδ  ˙  .δδ  ˙  )du2 + 2γ γ  ˙  .δδ  ˙  .δδ  dudv + dv2 . So σ is The first fundamental form is (1 + 2vγγ  ˙  + v 2δδ  ˙  .δδ  ˙   = 1 and γγ  ˙  .δδ  ˙  .δδ   = 0 for all u, v; the first conformal if and only if 1 + 2vγγ  ˙   = 0, so δδ   is constant, and the second condition then says that condition gives δδ  γγ  .δδ   is constant, say equal to d. Thus, σσ  is conformal if and only if  δδ   is constant δ  = d. In this case, σσ  is a generalized cylinder. and γγ   is contained in a plane v.δδ  From σσ ∗u = σ1.σ σ u 2((σσ..σσ)u2) σ , we get (in the obvious notation)

R2

{

∈

 →

}

 →

 

6.3.7

∈

{

∈

}

∈

− ||

6.3.8

◦ ◦

 →

6.3.9

6.3.10

6.3.11

{

}  →

σ .σ u )2 σ .σ u )2 4(σ 4(σ + = σ.σ )3 σ.σ )3 (σ (σ

E  . σ 4

 →

→

±√  −

± √  − −

−

σ ∗u.σ ∗u = E ∗  = σ

E  σ.σ )2 (σ

 − = F/  σ 4 , G∗

  

Similarly, we find that F ∗ = G/ σ 4 . Hence, the first fundamental form of  σσ∗  is a multiple of that of  σ . 6.3.12 By Example 6.1.3, any surface of revolution has an atlas consisting of surface σ (u, v) whose first fundamental form is  du 2 + f (u)2 dv 2 for some positive patches σ du smooth function f (u). Define u˜ = f (u) . Then, the first fundamental form of  ˜ (˜ u, v) of  σ (u, v) is f 2 (d˜ u2 + dv 2 ). This is a conformal the reparametrization σ surface patch.

 

 

6.4.1 Parametrize the paraboloid by σ (u, v) = (u,v,u2 + v 2 ); its first fundamental form is (1 + 4u2 )du2 + 8uv dudv + (1 + 4v 2 )dv2 . Hence, the required area is

49

  

6.4.2

the integral 1 + 4(u2 + v 2 ) dudv, taken over the disc u2 + v 2 < 1. Let 1 u = r sin θ, v = r cos θ; then the area is 2π 0 1 + 4r2 r dr = π6 (53/2 1). This is less than the area 2π of the hemisphere. If    is a sphere with centre the origin and radius R, the map S 2   given 2 p   multiplies the first fundamental form by R , and so is conformal Rp by p but multiplies areas by R2 . It follows from Theorem 6.4.7 that the sum of the angles of a spherical triangle of area A on is π + A/R2 . In this case, R is the radius of the earth and A is  the area of Australia, so the sum of the angles is 30 π + (7 500 000)/(6 500)2 = π + 169   radians. Hence, at least one angle of the 10 triangle must be at least one third of this, i.e. π + 169   radians. Take a point pp inside the polygon and join it to each vertex of the polygon by an arc of a great circle. This gives n triangles whose sides are arcs of great circles. The sum of their angles is the sum of their areas (i.e. the area of the polygon) minus nπ   by Theorem 6.4.7, and is also the sum of the angles of the polygon plus 2π (the angle around p). The sum of the angles around any vertex is 2π, so the sum of the angles of all the polygons is 2πV . By the preceding exercise, the sum of the angles of a polygon with n sides is (n 2)π  plus its area. Summing over all polygons gives 2πV  = 4π + polygons (n 2)π, since the sum of the areas of all the polygons is the area 4π of the sphere. Since two polygons meet along each edge, polygons n = 2E , and since there are F  polygons altogether we get 2πV  = 2πE  2πF  + 4π, which is equivalent to V  E  + F  = 2. (i) is obvious as a local isometry preserves E , F, G and hence EG F 2 . ˜ = λE , F  ˜ = λF  and G = λG, ˜ ˜  ˜ ˜ 2 = EG F 2 , then λ2 = 1 and G F  If  E  and if  E  ˜ are > 0). This proves (ii). so λ = 1 (as E, E 

 S 

6.4.4

→ S 

 S 

 ≥

− −



6.4.5

−

 →

≥

6.4.3

  √ 



 −

−

−

− √  −

The map from S 2 to the unit cylinder in the proof of Theorem 6.4.6 is an equiareal map that is not a local isometry. 6.4.6 Let σσ  : U  R3 ; f  is equiareal

 →

 

(E 1 G1

R

 ⇐⇒

− F 12)1/2 dudv =

 

(E 2 G2

R

− F 22)1/2 dudv

for all regions R

⊆ U .

E 1 G1 F 12 = This holds  the two integrands are equal everywhere, i.e. E 2 G2 F 22 . 6.4.7 Since N N is perpendicular to the tangent plane, N σ u  is parallel to the tangent σ u + βσ σ v  for some α, β . Now (N σ u  = 0, (N N σ u ).σ N σ u )..σ v = plane, and so = ασ σu  σσv )..N = σ u  σσ v (σ N.N = EG F 2 by Proposition 6.4.2. This gives the two equations αE  + βF  = 0, αF  + βG = EG F 2 , which imply α = F/ EG F 2 , β  = E/ EG F 2 . The formula for N σ v   is proved similarly.

−

 ⇐⇒

 ×   × √  − −

 ⇐⇒



√  −

√  −

× ×

√  −

−

×

 ×

50

6.4.8 By Exercise 6.1.6(ii), the first fundamental form is du2 + (1 + u 2 )dv 2 . So the area is 2π

1

     

1+

0

√ 

u2 dudv = π(

2 + ln(1 +

√ 

2)).

0

(The integral can be evaluated by putting u = tan θ.) 6.4.9 If we count the m edges that meet at each vertex, and then sum over all vertices, we will have counted each edge twice, once from each end. Hence, mV  = 2E . Similarly, counting the n edges of each polygon, and then summing over all the polygons, gives twice the total number of edges as each edge is an edge of two polygons. This gives nF  = 2E .

−1

1 1 From V  E  + F  = 2 and V  = 2E/m, F  = 2E/n we get E  = m + n1 , 2 and hence values for V  and F . The desired inequality follows from the fact that V, E  and F   are all >  0.



 −

 −



The inequality can be written (m 2)(n 2) <  4. Thus, m 2 and n 2 are two positive integers whose product is less than 4, giving the 5 possibilities 1 1, 1 2, 2 1, 3 1, 1 3.

−

×

×

×

−

−

−

×

×

6.4.10 We can assume that the sphere is the unit sphere S 2 . If such curves exist they would give a triangulation of  S 2 with 5 vertices and 5 4/2 = 10 edges, hence 2 + 10 5 = 7 polygons. Since each edge is an edge of two polygons and each polygon has at least 3 edges, 3F   2E ; but 3 7 >  2 10. If curves satisfying the same conditions exist in the plane, applying the inverse of the stereographic projection map (Example 6.3.5) would give curves satisfying the conditions on S 2 , which we have shown is impossible.

−

 ≤

×

× ×

6.4.11 Such a collection of curves would give a triangulation of the sphere with V  = 6, E   = 9, and hence F   = 5. The total number of edges of all the polygons in the triangulation is 2E   = 18. Since exactly 3 edges meet at each vertex, going around each polygon once counts each edge 3 times, so there should be 18/3 = 6 polygons, not 5. 6.4.12 Parametrize the surface by σ (u, v) = (ρ(u)cos v, ρ(u)sin v, σ(u)), where γγ  (u) = (ρ(u), 0, σ(u)). By Example 6.1.3, the first fundamental form is du 2 + ρ(u)2 dv 2 , ρ(u) dudv = 2π ρ(u) du. so the area is

 

 

π/2

 

(i) Take ρ(u) = cos u, σ(u) = sin u, with π/2 u π/2; so 2π −π/2 cos u du = 4π is the area. (ii) For the torus, the profile curve is γγ  (θ) = (a + b cos θ, 0, b sin θ), but this is γ  ˜ (u) = a + b cos ub , 0, b sin  ub not unit-speed; a unit-speed reparametrisation is γ 2πb a + b cos ub du = 4π 2 ab is the area. with 0 u 2πb. So 2π 0

≤ ≤

  

−



≤ ≤





6.4.13 Using the parametrization in Exercise 4.2.7, the first fundamental form is found

51

to be ((1

− κa cos θ)2 + τ 2 a2) ds2 + 2τ a2 dsdθ + a2 dθ2, so the area is 2π

s1

    s0

a(1

0

− κa cos θ)dsdθ = 2πa(s1 − s0).

6.4.14 If E 1  = λE 2 , F 1 = λF 2 , G1  = λG 2 , and if  E 1 G1 F 12 = E 2 G2 F 22 , then λ2 = 1 so λ = 1 (since λ > 0). R2 is obviously smooth so M   is a diffeomorphism if  6.4.15 (i) Any linear map R2 and only if  M  is bijective. This holds if and only if  M  takes a basis of  R2 , such as i, j jj , to another basis. Hence, M  is a diffeomorphism if and only if  u, v is a basis of  R2 , i.e. if and only if u  and v  are linearly independent. u + vvv. The first fundamental form of  σ (ii) M   takes (u, v) R2 to σ (u, v) = uu 2 2 2 2 u.v)dudv+ v du + 2(u dv . This is equal to du2 + dv 2 if and only if  is u u = v = 1 and u.v  = 0, i.e. if and only if u  and v  are perpendicular unit vectors. u.v)dudv+ v 2 dv 2 is a multiple of  du 2 + dv2 if and only if  (iii) u 2 du2 + 2(u u = v  and u.v  = 0, i.e. if and only if u  and v  are perpendicular vectors of  equal length. (iv) Since E  = u 2 , F  = u.v, G = v 2 , M   is equiareal if and only if  u.v)2 = 1. From the proof of Proposition EG F 2 = 1, i.e. u 2 v 2 (u 6.4.2, the left-hand side is equal to  uu v 2 . So M  is equiareal if and only if  u v = 1. 6.4.16 From the solution to Exercise 6.3.9, the first fundamental form of  σ  is (denoting d/du by a dot) (1+ f ˙2 )du2 +u2 dv 2 . So σ  is equiareal if and only if  u2 (1+ f ˙2 ) = 1, √  df  1−u2 i.e. du = . Thus, u

−

−

→

 { }

 { }

               

 ∈

 

 

          −   ×  

−  × 

±

f (u) =

√  −   ± 1

u

u2

du =

   ± − 1

u2

− cosh−1

  1 u

(up to adding an arbitrary constant). This is a parametrization of the   tractrix , so σσ  is equiareal  σσ  is a reparametrization of the   pseudosphere  (see 8.3). 6.4.17 The first fundamental form of  σ 3   is cos2 θdθ 2 + 2f ˙ cos θdθdϕ + (1 + f ˙2 )dϕ2 , where the dot denotes   d/dθ. With E   = cos2 θ, F  = f ˙ cos θ, G = 1 + f ˙2 , EG F 2 = cos2 θ. As this does not depend on f , the fact that the map σ 1 (θ, ϕ) σ 3 (θ, ϕ) is equiareal follows from the fact, proved in Theorem 6.4.6, that the map σ 1 (θ, ϕ) σ 2 (θ, ϕ) is equiareal.

 ⇐⇒

 §

−

→

→

6.5.1 If the internal angles are equal to α, Theorem 6.4.7 gives 3α π = 4π/4, so α = 2π/3. Corollary 6.5.6 then gives the length of a side as A = cos−1 ( 1/3). 6.5.2 Using the notation following Proposition 6.5.8, there is a rotation R 1 of  S 2 that takes a′   to a; then a further rotation R2  around the diameter through a  that

−

−

52

b′ ) coincide with the side through a  and b. By makes the side through a  and R 1 (b Corollary 6.5.6 the two triangles have sides of the same length, so we must have b′ ). If c  and c ′  are on the same side of the plane containing the side b  =  R 2 R1 (b through a  and b, we shall then have c  = R 2 R1 (cc′ ) and the isometry R2 R1  takes the triangle with vertices a′ , b′ , c′  to the triangle with vertices a, b, c; if they are on opposite sides the isometry R3 R2 R1  does this, where R3  is reflection in the plane containing the side through a  and b. 6.5.3 By applying an isometry of  R3 , which leaves lengths and areas unchanged, we can assume that p  is the north pole (0, 0, 1). The spherical circle of radius R and centre p  is then the circle of latitude ϕ = π/2 R (Example 4.1.4), which is a circle of radius sin R. The area inside it is, by Example 6.1.3 and Proposition 2π π/2 6.4.2, 0 π/2−R cos θdθdϕ = 2π(1 cos R). The maximum value of  R is π; if  π/2 R π, one replaces R by π R in (i) and (ii).

−

   

≤ ≤

′

− −

′

w+b 6.5.4 (i) If  M ′ (w) = ac′ w+d obius transformation, (M ′ M )(w) = ′  is another unitary M¨ Aw+B ′ ′ ′ ′ ′ ′ ′ ′ Cw+D , where A = a a + b c, B = a b + b d, C  = c a + d c, D = c b + d d. Thus, ¯ a¯′ a ¯ Inverses are dealt A = ¯ + b¯′ c¯ = d ′ d + ( c′ )( b) = D  and similarly C  = B. with similarly. C, the plane through (ii) Denoting (x,y,z) R3 by (ξ, z) with ξ  = x + iy  w¯a  + 2bz   = 0, and reflection in it the origin perpendicular to (a, b) i s wa + ¯ | w2 −1 wa+w¯ a+2bz ¯ 2w F (ξ, z) z) b). ξ  z is = (ξ, 2 |a|2 +b2 (a, Taking = |w|2+1 , = |w|2 +1 , we

◦

− −  ∈

 −  ∈

 −

a+b( w 2 1)) 2( a 2 +b2 )w 2a(wa+w¯ ¯ , z ( w 2 +1)( a 2 +b2 ) 2 ′ +b2 w a2 w+ab ¯ giving w = 1 ξ z′ = b2abww2+b¯ aw+baw+ ¯ a2

find that F (ξ, z) = (ξ ′ , z ′ ), where ξ ′ = a+b( w ( a 2 +b2 )( w 2 1) 2b(wa+w¯ ¯ ( w 2 +1)( a 2 +b2 ) ( aw+b)(bw+a) ¯ aw+b ¯ (bw+¯ ¯ a)(bw+a) = bw+¯ ¯ a .

|| −

| |2 −1)) ,

| |− − | | || −

|| ′

− | | −

|| − | | | |

| |− −

′ =

|| =

(iii) By Proposition 6.5.7, if F  is any isometry of  S 2 , F ∞ = (M 1 J ) (M 2 J )

◦ ◦ · · · ◦ (M k ◦ J )

◦ ◦

for some k.

Since J  M  J   is easily seen to be a unitary M¨ obius transformation if  M  is one, part (i) implies that F ∞  is a unitary M¨obius transformation if  k is even, and of  the form M  J   with M  unitary M¨obius if  k  is odd. C, b R, call the unitary M¨ (iv) If  a obius transformation M (w) = −aw+b bw+¯ a special unitary . Then M  = F ∞ J  where F  is as in (ii). Since J  = R ∞  where R is reflection in the yz-plane, M  = (F  R)∞  corresponds to the isometry F  R of  S 2 . It therefore suffices to prove that every unitary M¨ obius transformation is a composite of finitely-many special unitary M¨obius transformations. If  M ′ (w) = Aw+B obius tranformation, where A, B C, let B = be iθ with ¯  ¯  is any unitary M¨ −Bw+ A b, θ R. Then M ′ = ρ M  ρ−1 , where ρ(w) = e iθ and M (w) = −Aw+b  ¯  are both bw+A iθ/2 , b = 0). special unitary M¨obius transformations (ρ(w) = −aw+b bw+¯ a   with a = e

◦ ◦  ◦  ∈  ∈

 ∈

◦

◦

◦

 ∈

◦ ◦

R−1 v, followed by a suitable translation, takes the sphere of  6.5.5 The dilation v radius R to S 2 . It takes a triangle on   with sides of length A, B,C   to a triangle

→

S 

 S 

53

on S 2 with sides of length A/R, B/R, C/R and the same angles. Applying the A B A B cosine rule to the triangle on S 2 gives cos C  R   = cos R  cos R  + cos γ sin R  sin R , where γ   is the angle opposite the side of length C   (and two similar formulas). 2

x x x If  R is large, cos R  is approximately equal to 1 2R 2  and sin R  is approximately x equal to R . Substituting into the cosine rule gives

−

1

−

C 2 = 2R2

A2 2R2

B2 2R2

 −  −  1

1

2

2

+

AB cos γ, R2

which gives C 2 = A 2 + B 2 2AB cos γ  A2RB2   . Letting R  gives the usual cosine rule for a plane triangle with sides of length A, B,C   and with γ  the angle opposite the side of length C . This happens because when R and A, B,C  are fixed, the triangle becomes more and more nearly planar. (When standing at a point on a sphere of very large radius, one seems to be standing on a plane - this is why the Ancients thought the Earth is flat!) 6.5.6 Let n, a, b  be the North Pole, Athens and Bombay, respectively, and let R = 6500km be the radius of the Earth. The angle at n  of the spherical triangle with vertices n, a, b  is 73◦ 24◦ = 49◦ , and the lengths of the sides opposite b  and a are obtained by multiplying by R the radian equivalents of 90◦ 19◦ = 71◦ and 90◦ 38◦ = 52◦ , respectively. If  d is the spherical distance between Athens and Bombay, applying the cosine rule in the preceding exercise thus gives

−

−

→∞  → ∞

−

−

cos

−

d  = cos52◦ cos71◦ + sin 52◦ sin 71◦ cos49◦ = 0.6892. R

Hence, d/R = 0.8104 (radians) and d = 0.8104R = 5267km. 6.5.7 Drawing the diagonal of the square gives two spherical triangles with angles α/2, α/2 and α, the sides opposite these angles being A, A and D (the length of  the diagonal). Applying the cosine rule to one of these triangles gives  1 cos A = cos A cos D + sin A sin D cos α, 2 2 2 cos D = cos A + sin A cos α. Using the sine rule (Proposition 6.5.3(ii)) gives sin D   = sin A sin α/ sin 12 α = 2sin A cos 12 α. Substituting the formulas for cos D and sin D into the first equation above gives 1 cos A = cos3 A + cos A sin2 A cos α + 2 sin2 A cos2 α. 2 Hence, cos A sin2 A(1 mula for cos A.

− cos α) = 2sin2 A cos2 12 α, and this gives the desired for-

54

6.5.8 (i) This is obvious by the sine rule. (ii) This follows immediately from Corollary 6.5.6. Similarly one proves that cos β  + cos α cos γ  = λ sin γ sin A cos B. (iii) By (ii), (cos α + cos β )(1 + cos γ ) = cos α + cos β cos γ  + cos β  + cos α cos γ  = λ sin γ sin B cos A + λ sin γ sin A cos B by (ii), which = λ sin γ (sin A cos B + cos A sin B) = λ sin γ sin(A + B). (iv) Writing the formula in (iii) in terms of half-angles gives  1  1 cos (α + β ) cos (α 2 2

− β ) = λ tan 12 γ sin 12 (A + B)cos 12 (A + B).

Similarly, the first of the two equations in (i) can be written  1  1 sin (α + β ) cos (α 2 2

− β ) = λ sin 12 (A + B)cos 12 (A − B).

Dividing this equation by the preceding one gives the first formula in (iv). The second follows similarly from (iii) and the second equation in (i). (v) Applying the first formula in (iv) to the polar triangle, we get  1 tan (π 2 i.e.,

−A+π−

cos 12 (π B) = cos 12 (π

− α − π + β ) cot  1 (π − C ), − α + π − β ) 2

cos 12 (α β )  1  1 C. tan (A + B) =  tan 2 2 cos 12 (α + β )

−

Similarly,  1 tan (A 2

−

sin 12 (α β )  1 B) = C.  tan 2 sin 12 (α + β )

−

a, a′ ) around the line through the 6.5.9 If a′ = a  a rotation through an angle 12 dS 2 (a origin parallel to a a′  takes a′  to a. If a′ = a  the same thing is achieved by applying the isometry v v. Thus, in all cases there is an isometry of  S 2 that takes a′  to a. Since composites of isometries are isometries, it is sufficient to prove the result when a′ = aa. If  θ is the angle between the great circles passing through a  and b  and through a and b′ , a rotation by θ about the line through the origin and a  fixes a  and takes b′  to a point b′′  on the great circle passing through a  and b. Since b  and b′ are equidistant from a , so are b  and b′′ . It follows that either b′′ = b  or b′′  is obtained

   ±

×

 → −

 −

55

by reflecting b  in the plane passing through the origin and a  perpendicular to the great circle passing through a  and b. It is therefore sufficient to prove the theorem when aa = a′  and b  = b′ . The great circle passing through a  and c  makes the same angle with that passing through a  and b  as does the great circle passing through a  and c ′ . Hence, either these great circles coincide or one is obtained from the other by reflecting in the plane containing a, b   and the origin. It therefore suffices to prove the result in the case where the two great circles coincide. Since cc  and c ′   are the same distance from a  on the same great circle, either they coincide or cc′  is obtained from c  by reflecting in the plane passing through the origin and a  perpendicular to the great circle passing through a  and c .

Chapter 7 7.1.1 σ u = (1, 0, 2u), σ v = (0, 1, 2v), so N = λ( 2u, 2v, 1), where λ = √ 1+4u12 +4v2 ; σ uu = (0, 0, 2), σ uv = 0, σ vv = (0, 0, 2), so L = 2λ, M  = 0, N  = 2λ, and the second fundamental form is 2λ(du2 + dv2 ). σu  = 0; similarly, Nu .σ σv = Nv .σ σu = Nu = σ uu .N N  (since σ u .N N  = 0), so Nu .σ 7.1.2 σ u .N σv   = 0; hence, Nu   and Nv   are perpendicular to both σ u and σ v , and so Nv .σ are parallel to N. On the other hand, Nu  and Nv  are perpendicular to N  since N   is a unit vector. Thus, Nu = Nv = 0, and hence N   is constant. Then, σ.N σ .N N)u = σσu .N N  = 0, and similarly (σ N)v  = 0, so σ .N N  is constant, say equal to (σ N  = d. d, and then σσ  is an open subset of the plane vv.N ∂v ˜ = N, the sign being that of det(J ). From σ˜ u˜ = σ u ∂u + σ σ v ∂ ˜ 7.1.3 From 4.5, N ∂ ˜ u u, ∂v σ u ∂u σ˜ v˜ = σ ∂ ˜ v + σ v ∂ ˜ v , we get

− −

 −

 §

 ±

σ˜ u˜u˜ ˜ = So L

∂ 2 u ∂ 2 v σ u 2 + σ v 2 + σ uu = σ ∂ ˜ u ∂ ˜ u ∂u 2 ∂ ˜ u

2

    ∂u ∂ ˜ u

∂u ∂v σ uv + 2σ + σ vv ∂ ˜ u ∂ ˜ u

  ∂v ∂ ˜ u

2

2

  ±  

.

∂v ∂v N = σ v .N N   = 0. This, to+ 2M ∂u , since σ u .N ∂ ˜ u ∂ ˜ u + N  ∂ ˜ u ˜ and N , ˜ are equivalent to the matrix equation gether with similar formulas for M  in the question. 7.1.4 Let σσ  be a surface patch, P  a 3 3 orthogonal matrix, a R3 a constant vector, σ  = P σ +aa. Then, σ˜ u = P σ u , ˜ σ v = P σ v , so ˜ σ u σ˜ v = σ u σ v   (Proposition and ˜ P vv+aa is direct and if it is opposite. A.1.6), the sign being + if the isometry v ˜ = L, M  ˜ = M, N  ˜ = N . The It follows that (in the obvious notation), L v, where a   is a non-zero constant, multiplies σ by a and hence av dilation v multiplies each of   L, M,N  by a. 7.1.5 σ u  = (cos v, sin v, 0), σ v = ( u sin v, u cos v, 1), so σ u σ v   = (sin v, cos v, u) 1 v, cos v, u). Then, σ uu = 0, σσ uv = ( sin v, cos v, 0), σσ vv = and N  = √ 1+u 2 (sin

L

×

→

 →

−

−

×

 ±

∈ ± × −  ±

 ×

−

 ±

−

56

1 − √ 1+u ,

( u cos v, u sin v, 0). This gives L = σ uu.N = 0, M  = σ uv .N = √ 2dudv σ vv .N  = 0. So the second fundamental form is − N  = σ  . 1+u2

−

−

2

7.1.6 The tangent developable is parametrized by σ (u, v) = γγ  (u) + vtt(u), where t = γ  dγ γ /du. Using the standard notation and the Frenet-Serret equations, we find n, σσ v = tt, σ u σ v = κvb b, N  = |vv| b, σ u = t  + κvn

×

n + σ uu = κn

−

−

dκ n + κv( κtt + τ b), σ uv = κn n, σ vv = 0. vn du

−

2

Hence, L = κτ  |vv| , M  = N  = 0. So the second fundamental form is κτ  v du2 . This vanishes at all points of the surface if and only if  τ   = 0 at all points γ  , which holds if and only if  γγ    is planar (Proposition 2.3.3). In that case, of  γ the tangent developable is part of a plane. Thus, for tangent developables, the second fundamental form vanishes everywhere if and only if the surface is part of a plane. This is a special case of Exercise 7.1.2.

−

− ||

 − x2 − y2 and

7.2.1 The paraboloid is the level surface f   = 0 where f (x,y,z) = z x ,f y ,f z ) N  = (f  (f x ,f y ,f z )  is the corresponding unit normal. So (x,y,z) =

 G

( 2x, 2y,z) . (4x2 +4y 2 +1)1/2

− −

7.2.2 This is obvious since changing the orientation changes the Gauss map

 G to −G.

7.2.3 The hyperboloid of one sheet x2 + y 2  z 2 = 1 is obtained by rotating the hyperbola x2 z2 = 1 in the xz-plane around the z-axis. For this hyperbola, dz = xz 1 as x . The unit normal to the hyperbola therefore makes dx an angle θ   with the x-axis, where the angle θ   takes all values in the range π < θ < π4 . Rotating around the z-axis, it follows that the image of the Gauss 4 map of the hyperboloid is the region of  S 2 for which the latitude θ   satisfies π < θ < π4 . This is an annulus on S 2 . 4

→±

−

−

 → ±∞

− −

The hyperboloid of two sheets x 2 y 2 z2 = 1 is obtained by rotating the same hyperbola as above around the x-axis. The image of   is the union of two ‘caps’ on S 2 , namely the parts of the sphere x2 + y 2 + z2 = 1 with x > 1/ 2.

− −

 G

√ 

 | |

7.3.1 Let t be the parameter for γγ  , let s be arc-length along γγ  , and denote d/dt by a dot ds 2 γ ′′  d 2 s γ ′ ′ , γγ  γ  ˙   = ds ¨ γ     γ   + dt2 γ   . By Proposition 7.3.5, and d/ds by a dash. Then, γγ  = dt dt 1 1 γ  ˙  ,  ˙γγ   /(ds/dt)2 = γγ  ˙  , γγ  ˙   / γγ  ˙  , γγ  ˙   . For the κn = γγ  ′ , γγ  ′ = (ds/dt)  ˙γγ  , (ds/dt)  ˙γγ   = γ





 

  

    3

γ  ′′ .(N N γγ  ′ ) = 0, we have γγ  ¨ .(N N γγ  ˙  ) = ds N γ second part, since γγ  ′. (N dt ˙  ,  ˙γγ   3/2 κg . γγ  7.3.2 Let γγ    be a unit-speed curve on the sphere of centre a   and radius r. γ  γ  ˙  .(γ ¨ .(γγ  γ  a).(γ γ  a ) = r 2 ; differentiating gives γγ  γ  a ) = 0, so γγ  γ  a) = (γγ  γ  γ (t) 1. At the point γγ  (t), the unit normal of the sphere is N = 1r (γ γ  .N  = 1r γγ  ¨ .(γγ  κn =  ¨ γ γ  a) = 1r .

×

  −

−

−

×

−

±

−

∓



−  ±

× γγ  ′) = Then, ˙  .γγ  ˙ = γγ  a), so

 − −

57

7.3.3 If the sphere has radius R, the parallel with latitude θ has radius r = R cos θ; if  p  is a point of this circle, its principal normal at p  is parallel to the line through p  perpendicular to the z-axis, while the unit normal to the sphere is parallel to the line through p   and the centre of the sphere. The angle ψ   in Eq. 7.10 is therefore equal to θ or π θ so κ g = 1r sin θ = R1  tan θ. Note that this is zero if and only if the parallel is a great circle. ˙ σ v −F σ u )+v(F  √  ˙ σ v −Gσ u) , ˙   = uσ ˙   = u(E  7.3.4 We have γγ  ˙ σ + vσ ˙ σ , so by Exercise 6.4.7, N γγ 

−

u

±

±

×

v

2σ

2σ

EG F 2

−

¨  = u γγ  ¨σ u  + v¨σ v  + u˙ σ uu + 2u˙ vσ ˙ σ uv  + v˙ σ vv . Hence, ¨ .(N N κg = γγ 

× γγ ˙  ) = (u¨ ˙ v − v¨ ˙ u)

where

 

EG

− F 2 + Au˙ 3 + B u˙ 2 v + ˙ C  u˙ v˙ 2 + Dv˙ 3 ,

 − F σ u) = E ((σσu.σσv )u − σ u.σσuv ) − 21 F (σσu.σσu)u 1 1 = E (F u − E v ) − F E u , 2 2

σ uu .(Eσ σv A = σ

with similar expressions for B , C, D. If  F   = 0, we find by this method that A = 12 E v E/G, B = Gu 1 E  G/E , C  = 12 Gv E/G E v G/E , D = 12 Gu G/E . 2 u

 

 

N1 .n n, κ2 = κN N2 .n n, so 7.3.5 κ1 = κN κ1 N2

−

 

 −

   

 

E/G

−

− κ2N1 = κ((NN1 .nn)NN2 − (NN2 .nn)NN1) = κ(NN1 × N2 ) × n.

Taking the squared length of each side, we get κ21  + κ22

− 2κ1κ2N1 .NN2 = κ2  (NN1 × N2 ) × n 2 . N2  = cos α; γγ  ˙   is perpendicular to N1  and N2 , so N1 × N2  is parallel to Now, N1 .N ˙  , hence perpendicular to n; hence,  (N N1 × N 2 ) × n  =  N 1 × N 2  n  = sin α. γγ 

7.3.6 A straight line has a unit-speed parametrization γγ  (t) = p  + t q  (with q  a unit ¨  = 0   and hence κn = γγ  ¨ .N   = 0. In general, κn = 0 ¨  is vector), so γγ  γγ  perpendicular to N N  is perpendicular to n N  is parallel to b  (since N is perpendicular to t).

⇐⇒

⇐⇒

⇐⇒ √  2 2 2 7.3.7 The second fundamental form is ( −du + u dv )/u 1 + u2 , so a curve γγ  (t) = ˙ u˙ = σ (u(t), v(t)) is asymptotic if and only if  −u˙ 2 + u 2 v˙ 2 = 0, i.e.   dv/du = v/ ±1/u, so ln u = ±(v + c), where c is a constant.

7.3.8 By Proposition 7.3.2, if  κn = κ g  = 0 then κ = 0. But a curve with zero curvature everywhere is part of a straight line.

58

7.3.9 We find that, at the point where u = v = 1, the first and second fundamental √ 5 , respectively. Taking u = v = t, so forms are 32 du2 + 2dudv + 32 dv 2 and 2dudv that u = ˙ v = ˙ 1, Exercise 7.3.1 gives the normal curvature as 2u˙ v˙ 5

√ 

3 2 u 2  ˙

+ 2 u˙ v + ˙ 32  ˙v 2

=

2 √  . 5 5

7.3.10 From Examples 6.1.3 and 7.1.2, the first and second fundamental forms are 2 ˙g f ¨ g)du du2 + f (u)2 dv 2 and (f ¨ ˙ + f  gdv ˙ 2 , respectively (a dot denoting  d/du. We use the formulas in Exercise 7.3.1. (i) For a meridian we use u   as the parameter, so the curve is γγ  (u) = σ (u, v) (with v   constant). Note that γγ   is unit-speed. The second formula in Exercise ˙g f ¨ g.  ˙ cos v, f ˙ sin v, g), ˙   = (f  ¨  = ( ¨ f  cos v, f ¨ sin v, g¨), 7.3.1 gives κn = f ¨ ˙ Now, γγ  ˙ γγ  ˙ (from Example 7.1.2 again), giving κg = γγ  ¨.  (N N γγ  ˙  ) = 0. N  = ( g˙ cos v, g˙ sin v, f ) (ii) For a parallel we use v   as the parameter, so the curve is Γ(v) = σ (u, v) Γ/dv = (with u   constant). Note that Γ   is not unit-speed in general. Now, dΓ ¨ . (N ˙ N Γ˙ ) = f 2 f . ( f  sin v, f  cos v, 0), d2Γ/dv 2 = ( f  cos v, f  sin v, 0), giving Γ 2 ˙ ˙ Hence, κg = f f 3f  = f  f  .

 −

−

−

−

×

−

−

−

×

7.3.11 The paraboloid z = x 2 + y 2 is obtained by rotating the parabola z = x2 in the xz-plane around the z-axis. We could therefore use the preceding exercise but we would need a unit-speed parametrization of the parabola, which is complicated. σ (u, v) = (u cos v, u sin v, u2 ). Proceeding directly, parametrize the paraboloid by σ 2 +2u2 dv2 The first and second fundamental forms are (1+4u2 )du2 +u2 dv 2 and 2du√ 1+4u , 2 respectively. A curve u   = constant can be parametrized by γγ  (v) = σ (u, v). Hence, κn =

2u2 √ 1+4u 2

u2

.

˙   = ( u sin v, u cos v, 0), γγ  ¨  = Now, denoting   d/dv   by a dot, we find that γγ  1 ( u cos v, u sin v, 0), and N  = √ 1+4u2 ( 2u cos v, 2u sin v, 1), which gives

−

¨ .(N N γγ 

−

−

2

2

u u × γγ ˙  ) = √ 1+4u . Hence, κg = u1  × √ 1+4u 3

2

2

−

−

1 = u√ 1+4u 2.

˙  ,  σ σ v = δδ  σ u = tt, σσ v = δδ  δ , so along γγ  γ  (where v  = 0) we have σ δ  and hence 7.3.12 σ u = t + vδδ  ×δδ  . Hence, N  = tsin θ δδ   cos θtt (tt δδ  ) t , N t  = = sin θ sin θ so t˙ .δδ   N t) = κg = t˙ .(N , sin θ

×

× ×

−

×

since t˙ .t   = 0. Differentiating t.δδ    = cos θ   gives t˙ .δδ   = stated formula for κg .

 − sin θθ˙ − t.δδ ˙  , hence the

59

7.3.13 A curve v = constant is parametrized by γγ  (u) = σ (u, v), so (denoting d/du by ˙   = σ u , γγ  ¨  = σ uu and κ′g = σ uu.(N N σ u ). Now, σ u.σ u = σ v .σ v = 1, a dot) γγ  ×σ v and σ u.σ v  = cos θ, so N  = σ usin θ 1 1 N σu = σ u σ v ) σu = σ v cos θσ σu ). (σ (σ sin θ sin θ Hence, σuu.σ u σ uu.σ v cos θσ κ′g = . sin θ By Exercise 6.1.9, σ uu.σ u = 0, σ uu.σ v = θu sin θ. Inserting these formulas in the preceding equation for κ′g   gives κ′g = θu . The formula for κ′′g   is proved similarly. 7.3.14 From the solution to Exercise 7.3.4, X  κg = EG(u¨ , ˙v u ¨v) ˙ + EG where 1 1 1 1 X  = EE v ˙u3 + EGu GE u u˙ 2 v + EGv GE v u˙ v˙ 2 + GGu ˙v 3 ˙ 2 2 2 2 1 1 E v (1 Gv˙ 2 )u˙ + EGu GE u u˙ 2 v˙ = 2 2 1 1 EG v GE v u˙ v˙ 2 + Gu (1 E  u˙ 2 )v˙ + 2 2 1 1 1 G(E u ˙u + E v ˙v)u˙ v + E (Gu ˙u + Gv ˙v)u˙ v. = (Gu v˙ E v ˙u) ˙ ˙ 2 2 2 Substituting into the formula for κg  gives the desired result. σ (u, v), but this γ (u) = σ 7.3.15 A parameter curve v = constant can be parametrized by γγ  curve will not be unit-speed in general. Let s be the arc-length of  γγ   and denote d/ds by a dot. Then,  ds/du = σ u = E , so u = ˙ 1/ E . Since v˙ = v¨ = 0, the formula in the preceding exercise gives E v ˙u E v κ′g = . = 2 EG 2E  G The formula for κ′′g  is proved similarly.

×

×

×

×

 −

 −

 −  −

√ 

√ 

−



  −   − −   −  −

− −

−

−

   √  − √ 

√ 

=

√ 

 √ 

EG(u¨ ˙v

√ 

− √ 

√ 

We have cos θ = u˙ E , sin θ = v˙ G, so ˙ cos θ d (sin θ) sin θ d (cos θ) θ = ds ds v˙ = u˙ E  v¨ G + (Gu ˙u + Gv ˙v) 2 G

√ 

 −



−

√ 

− u¨v)˙ + 21 ˙uv˙

 − √   √     

u˙ v˙ G u ¨ E  + (E u ˙u + E v ˙v) 2 E 

E  (Gu ˙u + Gv ˙v) G

−

√ 



G (E u ˙u + E v ˙v) . E 



60

Using the preceding exercise and the first part of this exercise, we get ˙ θ = κ g

√  √  − 2√ 1EG (Gu v˙ − E v ˙u) = κg + κ′g ˙u E + κ′′g ˙v G = κg − κ′g cos θ − κ′′g sin θ.

γ (t) be a unit-speed parametrization of  , and let p  = γγ  γ (t0 ). Let N  be a unit 7.3.16 Let γγ  ˜  = γγ   (γγ  N. Denoting γ .N)N normal of    at p. Then, a parametrization of  ˜ is γγ  γ  γ  γ  .N)N ˜˙   = γ ˙   (γ ˙ .  N)N N, γγ  ˜¨  = γγ  ¨  (¨ N, so the curvature of  γγ  ˜  d/dt by a dot as usual, γγ  γ at p is

C

 S 

 −

C  −

 −

˙  (t0 ) κ ˜  = γγ 

× (γγ ¨ (t0) − (γγ ¨ (t0)..N)NN)  =  γγ ˙  (t0 ) × (NN × (γγ ¨ (t0 ) × N)) , γ  ˜˙  (t0 ) = γγ  ˙  (t0 ) is a unit vector. Now, γγ  ¨ (t0 ) = κ g N × γγ  ˙  (t0 ) + κn N, where κg since γ ¨ (t0 ) × N = and κn   are the geodesic and normal curvatures of  γγ   at p. Hence, γγ  N × γγ  ˙  (t0 )) × N  = κ g ˙γγ  (t0 ), so κg (N ˙  (t0 ) × (κg N × γγ  ˙  (t0 ))  = |κg |  γγ  ˙  (t0 )  N × γγ  ˙  (t0 )  = |κg | κ ˜  =  γγ  

˙  (t0 ) are perpendicular unit vectors. as N and γγ  2

2

−dv , so a curve γγ  (t) = σ (u(t), v(t)) is 7.3.17 The second fundamental form is √ du 1+u2 +v 2 asymptotic if and only if u˙ 2  ˙v 2 = 0, i.e. u + v or u v is constant. If  u + v = c, say, where c   is a constant, then σ (u, v) = (u, c u, 12 c(2u  c)), which is a parametrization of a straight line. Similarly if  u v is constant.

−

−  − −

 −

Geometrically, the surface is a hyperbolic paraboloid and is doubly ruled (Exercise 5.2.5). Both of the straight lines on the surface passing through a given point are asymptotic curves (Exercise 7.3.6). 7.3.18 In the usual notation, σσ u = t + v( κtt + τ b) = t  along γγ ; σ v = n. Hence, along n + v τ ˙ b  = κn n γγ  ,  σσu σ v = b. Next, we find that σσuu = v κt ˙ t + (κ (κ2 + τ 2 )v)n along γγ  . The parameter curve v  = 0 is asymptotic if and only if  L  = 0. But, σuu .N  = κn n.b  = 0. along γγ , L = σ 7.3.19 Parametrize the surface by σ (u, v) = (u2 cos v, u2 sin v, 2u). The second funda2 +2u2 dv2 mental form is −2du√ 1+u , so the asymptotic curves are given by 2

−

×

−

−

−2u˙ 2 + 2u2 v˙ 2 = 0. This gives v = ln u+ c, where c is an arbitrary constant. The projection of this asymptotic curve on the xy-plane is given by x = u2 cos v, y = u2 sin v. Since u2 = e±2(v−c) , this is a logarithmic spiral. γ   is asymptotic γ  ¨ .N = γ 7.3.20 Let γγ   be a unit-speed parametrization of  . Then, γ 0 n.N  = 0, where N  is a unit normal to   and n  is the principal normal of 

±

 ⇐⇒

 C

 S 

 ⇐⇒

61

γ  ˙   = tt is perpendicular to N, γγ   is asymptotic γγ  . Since γ N  is perpendicular to both t  and n. But the osculating plane is spanned by t  and n. 7.3.21 If every curve on a surface is asymptotic, L u˙ 2 + 2M  u˙ v + ˙ N  v˙ 2 = 0 for all curves γγ  (t) = σ (u(t), v(t)) on the surface. This is possible only if  L = M  = N  = 0 everywhere. By Exercise 7.1.2 this implies the surface is an open subset of a plane. 7.3.22 N.n   = cos ψ, N.t   = 0, so N.b   = sin ψ; hence, N = n cos ψ + b sin ψ   and B = t (n n cos ψ + b sin ψ) = b cos ψ n sin ψ. Hence,

 ⇐⇒

×

−

˙ n sin ψ + b cos ψ) N˙  = n˙ cos ψ + b˙ sin ψ + ψ( ˙ n sin ψ + b cos ψ) = ( κtt + τ b)cos ψ nτ  sin ψ + ψ(

−

− − − ˙ b cos ψ − n sin ψ) = −κ cos ψtt + (τ  + ψ)(b = −κn t  + τ g B.

The formula for B˙  is proved similarly. Since t, N , B is a right-handed orthonormal basis of  R3 , Exercise 2.3.6 shows that the matrix expressing t˙ , N˙ , B˙  in terms of t, N, B  is skew-symmetric, hence the formula for t˙ . 7.3.23 By Exercise 7.3.6, b  is parallel to N, so b  = N ; then, B  = t N  = n. Hence, b) = κtt τ N; comparing with the formula for B˙  in the B˙  = n˙ = ( κtt + τ b preceding exercise shows that τ g = τ   (and κn = κ).

{

 ∓  ∓ −

 ± −

}

±

7.4.1 v˜  is a smooth function of  t and lies in T γγ  (ϕ(t))

×

∓

±

S = T γγ ˜ (t)S , so ˜v  is a tangent vector

˜ field along ˜γγ  . The formula follows from Eq. (7.11) and the fact that ddtv = ddvt˜ dϕ  , dt ˜  = 0 where t˜ = ϕ(t). The last part follows since ϕ˙ = 0 so γγ  ˜  v γγ  v  = 0. 7.4.2 If p  and q  correspond to the parameter values t =  a and t =  b, respectively, let qp Γ(t) = γγ  (a + b t) (thus, Γ is γγ   ‘traversed backwards’). We show that Π is Γ pq the inverse of Πγγ    . Let w T p   and let v  be the tangent vector field parallel pq w along γγ   such that v(a) = w. Then, Πγ γ   (w) = v(b). By Exercise 7.4.1, V(t) = qp v qp V v(a + b t) is parallel along Γ so Π (v (b)) = Π (V (a)) = V(b) = v(a) = w. Γ Γ qp pq This proves that Π Πγγ   is the identity map on T p . One proves similarly Γ pq qp (or by interchanging the roles of  γγ   and Γ) that Πγγ   Π is the identity map Γ on T q . 7.4.3 Let α, β,γ   be the internal angles of the triangle at p, q, r, respectively. Since the arc through p   and q  is part of a great circle, the tangent vector of the arc is parallel along the arc (Example 7.4.7). So the result of parallel transporting vv0 to q  along the arc pq  through p  and q  is a vector v1  tangent to pq  at q. Now v1 makes an angle π β  with the arc qr  at q, so parallel transporting v1  along qr



−

 ∇

⇐⇒ ∇

 ∈ S 

−

◦

S 

−

◦

S 

62

to r  gives a vector v2  which makes an angle (π β ) + (π γ ) with the arc rp  at r. Parallel transporting v2  along rp  to p  then gives a vector v3  which makes an angle (π β ) + (π γ ) + (π α) with v0 . Since v0 , v1 , v2 , v3  all have the same length (Proposition 7.4.9(ii)), the result follows from Theorem 6.4.7. 7.4.4 Putting E  = G  = 1, F  = cos θ in the formulas in Proposition 7.4.4 gives Γ111 = θu cot θ, Γ211 = θu cosec θ, Γ122 = θv cosec θ, Γ222 = θ v cot θ, Γ112 = Γ212 = 0. 7.4.5 In the notation of Proposition 7.4.5, the tangent vector σ v   to the parameter curves u   = constant are given by α = 0, β   = 1, and a curve v   = constant σ v  is parallel along the curves corresponds to v = ˙ 0 (and u˙ = 0). From Eq. 7.13, σ 1 2 v   = constant if and only if Γ12 = Γ12   = 0; from the formulas in Proposition 7.4.4, this holds if and only if 

−

−

−

−

−

−

−



GE v = F Gu and EG u = F E v .

(*)

If the parameter curves form a Chebyshev net,  E v = Gu  = 0 (Exercise 6.1.5), so (*) holds. Conversely, if (*) holds than (EG

− F 2 )Gu = G(EGu) − F (F Gu) = G(F E v ) − F (GE v ) = 0

and so Gu  = 0; then GE v  = 0 by (*) again, so E v  = 0. Hence, if (*) holds the parameter curves form a Chebyshev net. This proves that (i) is equivalent to (ii). That (i) is equivalent to (iii) is proved similarly. .σ v = F/ EG, so A = EG EG cos2 θ = EG sin θ. 7.4.6 We have cosθ = σσuu σv  Differentiating the first equation with respect to u and using the second equation gives 2EGF u F GE u EF Gu EG F  θu = . = A 2AEG EG u

√ 

√ 

−

√ 

√  −

 √  

−

−

On the other hand, the formulas in Proposition 7.4.4 give Γ211 Γ112 1 2EF u EE v F E u GE v F Gu + = + E  G E  G 2A2 2EGF u F GE u EF Gu . = 2A2 EG



−

−

−

 −

−



7.4.7 From A2 = E G F 2 we get 2AAu = EGu  + GE u 2F F u . On the other hand, u −2F F u the formulas in Proposition 7.4.4 give Γ111 + Γ212 = GE u +EG . By the 2A2 2 preceding formula, this is equal to 2AAu /2A = Au /A. The other formula is proved similarly.

−

−

63

Chapter 8 8.1.1 Parametrize the surface by σ (x, y) = (x,y,f (x, y)). Then, σ x = (1, 0, f x), σ y = σxx = (0, 0, f xx), σ σ xy = (0, 0, f xy ), (0, 1, f y ), N  = (1 + f x2 + f y2 )−1/2 ( f x , f y , 1), σ σ yy = (0, 0, f yy ). This gives E  = 1 + f x2 , F  = f x f y , G = 1 + f y2 and L = (1 + f x2 + f y2 )−1/2 f xx , M   = (1 + f x2 + f y2 )−1/2 f xy , N   = (1 + f x2 + f y2 )−1/2 f yy . By

− −

Corollary 8.1.3, K  =

2 f xx f yy f xy (1+f x2 +f y2 )2 , H 

−

=

(1+f y2 )f xx 2f x f y f xy +(1+f x2 )f yy . 2(1+f x2 +f y2 )3/2

−

8.1.2 For the helicoid σσ (u, v) = (v cos u, v sin u,λu), σ u = ( v sin u, v cos u, λ), σ v  = (cos u, sin u, 0), 1 N  = ( λ sin u, λ cos u, v), λ2 + v 2 σ uu = ( v cos u, v sin u, 0), σ uv = ( sin u, cos u, 0), σ vv = 0.

−

√  − −

− −

−

√ 

This gives E  = λ 2 + v 2 , F  = 0, G = 1 and L = N  = 0, M  = λ/ λ2 + v 2 . Hence, K  = (LN  M 2 )/(EG F 2 ) = λ2 /(λ2 + v 2 )2 . For the catenoid σσ (u, v) = (cosh u cos v, cosh u sin v, u),

 −

−

−

σ u  = (sinh u cos v, sinh u sin v, 1), σ v = ( cosh u sin v, cosh u cos v, 0), N  = sech u( cos v, sin v, sinh u),

−

−

−

σ uu  = (cosh u cos v, cosh u sin v, 0), σ uv = ( sinh u sin v, sinh u cos v, 0),

− σ vv = (− cosh u cos v, − cosh u sin v, 0).

This gives E  = G   = cosh2 u, F   = 0 and L = 1, M  = 0, N   = 1. Hence, K  = (LN  M 2 )/(EG F 2 ) = sech4 u. σ  is smooth and σ σ u σ v  is never zero, N  = σ σ u σ v / σ u σ v is smooth. 8.1.3 Since σ Hence,   E,F,G,L,M  and N   are smooth. Since EG F 2 >  0 (by the remark following Proposition 6.4.2), the formulas in Corollary 8.1.3 show that H  and K  are smooth. ˙  .N  = 0 ˙  .((tt +vδδ  ˙  ) δδ  ) = 0 ˙  .(tt δδ  ) = δδ  δδ  δδ  8.1.4 From Example 8.1.5, K  = 0 ˙   = κtt + τ b b, t δδ   = b, so K  = 0 δ  = n, δδ  τ  = 0 γγ   is planar (by 0. If δδ  ˙   = τ n, t δδ  = n, so again K  = 0 τ  = 0. Proposition 2.3.3). If  δδ  = b , δδ 

 −

−

 −

×

−

−

⇐⇒ ×

×  ×   −

⇐⇒

−

×

 ⇐⇒ −

×

 ⇐⇒

⇐⇒

⇐⇒

×

64

8.1.5 The dilation (x,y,z) (ax,ay,az), where a is a non-zero constant, multiplies 2 E , F , G by a and L, M,N  by a, hence H  by a −1 and K  by a −2 (using Corollary 8.1.3).

 →

8.1.6 This follows immediately from Definition 8.1.1 and the hint. 8.1.7 Suppose that the cone is the union of the straight lines joining points of a curve  to a vertex v. It is clear that the Gauss map  is constant along the rulings of  the cone, so the image of the cone under  is the same as the image of    under , which is a curve. 8.1.8 By Eq. 8.2, the area of  σ (R) is

C G

 G

   R

Nu

  |

× Nv  dudv =

 G

 C

|  σ u × σ v  dudv =

K 

R

  | R

K  d σ .

|A

8.1.9 Using the parametrization σ  in Exercise 4.2.5, we find that E  = b 2 , F  = 0, G = cos θ (a+b cos θ)2 and L = b, M  = 0, N  = (a+b cos θ)cos θ. This gives K  = b(a+b cos θ) . It follows that + and −  are the annular regions on the torus given by

 S 

 S 

−π/2 ≤ u ≤ π/2 and π/2 ≤ u ≤ 3π/2, respectively.

S +

S −

It is clear that as a point p  moves over + (resp. − ), the unit normal at p covers the whole of  S 2 . Hence, S + K  d = S − K  d = 4π by the preceding exercise; since K  = K  on ± , this gives the result. wu .N )N N so 8.1.10 uw  = wu (w

∇

 | | ±

  |

 S 

 S  |A

  | S | A

− ∇v (∇uw) = wuv − (wwuv.N)NN − (wwu.Nv )NN − (wwu.N)NNv − (wwuv.N)NN = w uv

 −

wuv .N)N N + (w wu.N v )N N + (w wu.N)(N Nv .N)N N + (w wuv .N)N N (w wu.N)N Nv . (w

−

65

Interchanging u and v and subtracting gives the first formula. Replacing w by w  in this formula gives λw wv .N)N Nu (w wu .N)N Nv +λv (w w.N)N Nu λu (w w.N)N Nv = λ (w wv .N)N Nu (w wu .N)N Nv λ (w

{

−

}

−

{

−

}

since w.N  = 0. It is also obvious that

∇v (∇u(ww1 + w2)) − ∇u (∇v (ww1 + w2 )) = (∇v (∇uw1 ) − ∇u(∇vw1 )) + (∇v (∇u w2 ) − ∇u (∇v w 2 )) for any two tangent vector fields w1 , w2 . σ uv .N)N σ uu.N )N Nu (σ Nv = M N Nu LN Nv  (in the Now v ( u σ u ) u ( v σ u ) = (σ σ u  + bσ σ v ) L(cσ σ u  + dσ σv ) usual notation). Using Proposition 8.1.2, this = M (aσ and using the explicit expressions for a, b, c, d in Proposition 8.1.2 this becomes σv F σ u ). Similarly, v ( uσ v ) σ u  + F σ v ). K (Eσ u ( v σ v ) = K ( Gσ

 ∇ ∇

−∇ ∇

 −

If 

−

 ∇ ∇

 − −

−∇ ∇

−

∇v (∇uw) − ∇u (∇v w) =

(*)

0

for all w, then taking w = σ u   gives K  = 0 since E  = 0, σ v = 0. Conversely, if  K  = 0 then (*) holds for w = σ u and σ v , and hence by the first part of the σu + βσ σ v  for all smooth functions α, β  of (u, v). But every exercise it holds for ασ tangent vector field w  is of this form.

 

 

σ vv = 0, σ uv = (0, 0, 1). When u = v = 1, 8.1.11 σ u = (1, 1, v), σσv = (1, 1, u), σσuu = σ we find from this that E  = 3, F  = 1, G = 3 and L = N  = 0, M  = 1/ 2. Hence, K  = (LN  M 2 )/(EG F 2 ) = 1/16, H  = (LG 2M F  + N G)/2(EG F 2 ) = 1/8 2.

√ 

− −

−

−

− √ 

−

−

8.1.12 We write the equation of the quadric in the form z = f (x, y), where f (x, y) =

  − c 1

x2 a

−

y2 b



. The result of the calculation will give the value of the Gauss-

ian curvature K  on the part of the quadric where z > 0. Similar calculations (with the same result) give K  on the parts of the surface where z < 0, x > 0, x < 0, y > 0 and y < 0. Together these regions cover the whole surface. Taking f (x, y) =

  − c 1

x2 a

−

y2 b



, and applying the first formula in Exercise

8.1.1, we get, after straightforward calculation, 1 K  = abc

x2 a



x2 a2

+ +

y2 b y2 b2

+ +

z2 c z2 c2



2.

67

8.1.14 If the circle in the xz-plane has radius a   and touches the z-axis at the point (0, 0, c), it can be parametrized by γγ (u) = (a(1 + cos u), 0, a sin u + c). Rotating γγ  (u) around the z-axis through an angle v gives the point (a(1 + cos u)cos v, a(1 + cos u)sin v, a sin u + c). Since the circle rotates at constant angular velocity and moves parallel to the z-axis at constant speed, the distance it moves parallel to the z-axis in the time taken to rotate through an angle v   is a constant multiple of  v, say bv. Hence, after the circle has rotated through an angle v  the point initially at γγ  (u) will have moved to the point σ (u, v) in the statement of the exercise. σ(u, v) = a((1+cos u)cos v, (1+cos u)sin v, sin u + v). If  a = b and c = 0 we have σ We find that the first fundamental form is a2 (du2 + 2 cos ududv + ((1 + cos u)2 + 1)dv 2 ).

 √ EG − F 2 , so that A2 = 2a4(1 + cos u).

Let A = σ u σ v = fundamental form is

  ×

a3  ((1 + cos u)du2 A

The second

− 2sin2 ududv + (1 + cos u)2 cos udv2 ).

The first formula in Corollary 8.1.3 now gives a6 ((1 + cos u)3 cos u K  = A4

− sin4 u) .

Writing c = cos u we get c(1 + c)3 (1 c2 )2 c(1 + c) (1 K  = = 4a2 (1 + c)2 4a2

− − c)2 =   3cos u − 1 .

− −

4a2

Since d = a(1 + cos u), this agrees with the formula in the statement of the exercise. ˙   δδ  . ˙   δδ  + vδδ  8.1.15 Using the parametrization in Example 8.1.5, we have σσ u σ v = γγ  Recalling that σ u σ v = EG F 2 , the formula for K   in that example becomes ˙ .  (γγ  ˙   δδ  ))2 (δδ  K  = . ˙   δδ   4 ˙   δδ  + vδδ  γγ 

× × × √     ×  − − ×  × ×   ˙  × δδ    =  , K  → 0 as v → ∞ (with u fixed); thus, if  K  is constant we must have If  δδ  ˙   × δδ   = , δδ  ˙   is parallel to δδ  , hence δδ  ˙  .(γγ  ˙    × δδ  ) = 0 K   = 0. On the other hand, if  δδ  0

0

and again K  = 0.

68

8.1.16 We use the parametrization σσ (s, θ) in Exercise 4.2.7, a dot to denote d/ds, and the standard notation relating to the Frenet-Serret data for γγ  . Then, σ s = n + τ a cos θb b, σσθ = a sin θn n + a cos θb b. Hence, (1 κa cos θ)tt τ a sin θn

−

−

−

σs

× σ θ = −a(1 − κa cos θ)(cos θnn + sin θbb), n  + sin θb b), E  = (1 − κa cos θ)2 + τ 2 a2 , F  = τ a2 , G = a2 and so N = −(cos θn EG − F 2 = a 2 (1 − κa cos θ)2 . Next, n σ ss = (κτ a sin θ − κa ˙ cos θ)tt + (κ(1 − κa cos θ) − τ˙ a sin θ − τ 2 a cos θ)n b, + (τ˙ a cos θ − τ 2 a sin θ)b n − τ a sin θb b, σ sθ = κa sin θtt − τ a cos θn n − a sin θb b. σ θθ = −a cos θn Hence, L = aτ 2 − κ cos θ(1 − κa cos θ), M  = τ a, N  = a, Thus,  −κa cos θ(1 − κa cos θ) = −κ cos θ . K  = a2 (1 − κa cos θ)2 a(1 − κa cos θ) (i) We have dA =  σ s × σ θ    dsdθ = a(1 − κa cos θ)dsdθ, so 2π

ℓ

    0

(ii) Since

2π 0

  |

2π

ℓ

Kd

A=

0

    0

0

( κ cos θ)dsdθ = 0.

−

π/2

     

cos θ dθ = 2 −π/2 cos θdθ = 4, we get

|

2π

ℓ

0

ℓ

|K |dA = 4

0

 

κ(s)ds.

0

8.1.17 It follows from Exercises 6.1.2 and 7.1.4 that applying a direct isometry of  R3 leaves both H  and K   unchanged, while applying an opposite isometry leaves K  unchanged but changes the sign of  H . γ  is unit-speed its normal curvature is κn = 8.1.18 From the proof of Proposition 7.3.3, if γγ  γ  ˙   = (γγ  ˙  ).γγ  ˙  . Hence, γγ   is asymptotic if and only if  (γ ˙  ) is perpendicular to N˙ .γγ  γ  ˙   at each point of  γγ  . Since ˙  ) is the tangent vector of the image of  γγ   under γγ  (γ the Gauss map of  , the result follows. ˙  ), and that   is self-adjoint (Corollary 7.2.4), we have 8.1.19 Recalling that N˙  = (γγ  to prove that 2 ˙  ) 2H  (γγ  ˙  ) + K γγ  ˙  , γγ  ˙   = 0. (γγ 

−

W 

 W 

 −W 

 S  −W  W 

W  − W 

This follows from the preceding exercise.



69

γ  We can assume that γγ   is unit-speed. As in the solution of Exercise 8.1.18, if γγ  ˙ ˙  ˙ ˙   = 0 and the first part gives N.N  = K . But by Exercise is asymptotic then N.γγ  γ  is asymptotic, N  = b so N˙  = τ n  (where n  is the principal normal 7.3.6, if γγ   ˙  = τ 2 . of  γγ   and b  is its binormal). Hence, N˙ .N

 ±

−

 ∓

8.2.1 For the helicoid σσ (u, v) = (v cos u, v sin u,λu), the first and second fundamental forms are (λ2 + v 2 )du2 + dv 2 and 2λdudv/ λ2 + v 2 , respectively. Hence, the κ(λ2 + v 2 ) √ λ2λ+v2 principal curvatures are the roots of  = 0, i.e. κ = √ λ2λ+v 2 κ λ/(λ2 + v 2 ).

√ 

 − 

±

−

 

For the catenoid σσ (u, v) = (cosh u cos v, cosh u sin v, u), the first and second fundamental forms are cosh2 u(du2 +dv 2 ) and du2 +dv2 , so the principal curvatures 1 κ cosh2 u 0 are the roots of  = 0, i.e. κ = sech2 u. 2 0 1 κ cosh u

 − −   W     ⇐⇒ 

8.2.2 This is obvious, since

˙  ) = (γγ 

−

−N˙ .

−

 

±

˙   = uσ γγ  8.2.3 γγ   is a line of curvature ˙ σ u + vσ ˙ σ n  is a principal vector for all t L M  E F  = κ for some scalar κ. Writing this matrix equation as two L N  F G scalar equations and then eliminating κ gives the stated equation.



 ⇐⇒

For the second part, if the second fundamental form is a multiple of the first, the Weingarten map is a scalar multiple of the identity map, so every tangent vector is principal and every curve on the surface is a line of curvature. If F  = M  = 0 the matrices I  and II    are diagonal, hence so is the matrix I −1 II    of the Weingarten map with respect to the basis σ u , σ v . This means that σ u and σ v   are principal vectors, i.e. that the parameter curves v = constant and u = constant are lines of curvature.

 F 

 F 

 {

}

 F  F 

Conversely, if every parameter curve is a line of curvature, the stated equation must hold if u = ˙ 0 and if v = ˙ 0. This gives EM  = F L and F N  = GM , which imply that (EN  GL)F  = E GM  EGM   = 0 and so either EN  = GL or F  = 0. If  F   = 0 then GM   = 0 so M   = 0 and (ii) holds. If EN  = GL   the equation in the exercise implies that every curve is a line of curvature, so every tangent vector is principal, so (i) holds.

−

−

Condition (i) implies that the two principal curvatures are equal everywhere, i.e. every point is an umbilic, so σ  is an open subset of a plane or a sphere by Proposition 8.2.9. From Examples 6.1.3 and 7.1.2, the first and second fundamental forms of a ˙g f ¨ g˙ )du2 + f  gdv surface of revolution are du 2 + f (u)2 dv2 and (f ¨ ˙ 2 , respectively. Since the terms dudv are absent, the vectors σσ u and σσ v  are principal; but these are tangent to the meridians and parallels, respectively.

−

70

8.2.4 Let N1  be a unit normal of  . Then, K  = 0 N˙ 1.(tt N 1 ) = 0. Since N˙ 1 is ˙1 N perpendicular to N1  and N1  is perpendicular to t, this condition holds ˙   for some scalar λ. Now use Exercise 8.2.2. is parallel to t, i.e. N˙ 1 = λγγ 

 S   ⇐⇒ −

 ⇐⇒

×

 ⇐⇒

γ is a unit-speed parametriza8.2.5 Let N1  and N2  be unit normals of the two surfaces; if γγ  ˙ γ  ˙   for some scalar λ1   by Exercise 8.2.2. If   is a line tion of  , then N1 = λ1 γ ˙   for some scalar λ2 , and then (N N1 .N N2 )˙ = of curvature of  2 , then N˙ 2 = λ2 γγ  ˙  .N2 λ2 γγ  ˙  .N1  = 0, so N1 .N N2   is constant along γγ  , showing that the angle λ1γγ   ˙ 2 = 0 N2  is constant, then N1 .N between 1 and 2  is constant. Conversely, if N1 .N N2 = λ1 γγ  ˙  .N 2  = 0; thus, N˙ 2  is perpendicular to N1 , and is also persince N˙ 1 .N ˙   is also perpendicular to N1 and pendicular to N2  as N2  is a unit vector; but γγ  ˙ N2 ; hence, N2   must be parallel to γγ  ˙  , so there is a scalar λ2   (say) such that ˙ 2 = λ2 γ γ  N ˙  .

−

 C  −  S   −  S   S   −

 C

 −

−

8.2.6 (i) Differentiate the three equations in (8.5) with respect to w, u and v, respectively; this gives σ v  + σ u .σ σ vw = 0, σ uv .σ σ w  + σ v .σ σ uw = 0, σ vw .σ σu  + σ w .σ σuv = 0. σ uw .σ σ vw = 0, Subtracting the second equation from the sum of the other two gives σ u .σ σuw = σ σ w .σ σ uv = 0. and similarly σσ v .σ σw   = 0, it follows that the matrix I   for the u = u0   surface is (ii) Since σ v .σ diagonal (and similarly for the others). Let N  be the unit normal of the u = u 0 σ v σ w  by definition, and hence to σ σ u  since σσu , σσ v and surface; N  is parallel to σ σu   = 0, hence σ vw .N N  = 0, proving that the σ w   are perpendicular; by (i), σ vw .σ matrix II  for the u = u 0  surface is diagonal. (iii) By part (ii), the parameter curves of each surface u = u0  are lines of curvature. But the parameter curve v = v0 , say, on this surface is the curve of  intersection of the u = u 0  surface with the v = v 0  surface.

 F 

×

 F 

8.2.7 On the open subset of the ellipsoid with z = 0, we can use the parametrization 2 y2 σ (x, y) = (x,y,z), where z = r 1  px2 q 2 . By Proposition 8.1.2 and the remarks following Proposition 8.2.1, the condition for an umbilic is that II  = κ I    for some scalar κ. The formulas in the solution of Exercise 8.1.1 lead to the equations zxx = λ(1 + zx2 ), zxy = λzx zy , zyy = λ(1 + zy2 ), where

   ± −

   −

 F 

F 

λ = κ

 

1 + zx2  + zy2 . If  x and y  are both non-zero, the middle equation gives λ =

−1/z, and substituting into the first equation gives the contradiction p2 = r2. Hence, either x = 0 or y = 0. If x = 0, the equations have the four solutions

x = 0, y =

  −

±q 

q 2 q 2

−

 p2 , z= r2

±r

  − r2 r2

−

 p2 . q 2

71

Similarly, one finds the following eight other candidates for umbilics:

  − ± −   −

x =  p x =  p

±

 p2

q 2

 p2

r2

 p2

r2

 p2

, y = 0, z =

− q 2 ,

  −

q 2 q  2 q 

y =

±

±r −

  − r2 r2

−

q 2 ,  p2

r2 , z = 0.  p2

Of these 12 points, exactly 4 are real, depending on the relative sizes of  p2 , q 2 and r 2 . If  p =  q  =  r, the only umbilics are the two points (0, 0, r). If  p =  q  = r   every point of the ellipsoid (now a sphere) is an umbilic. 8.2.8 By Proposition 8.2.3, the principal curvatures are the roots of  κ2 2Hκ+K  = 0, H 2 K . If there are no umbilics, we must have H 2 > K , and then i.e. H  the principal curvatures are smooth because H  and K   are (Exercise 8.1.3). The second part follows from Exercises 6.1.4 and 7.1.3 and Proposition 8.2.6. 8.2.9 Setting x = r cos θ, y  = r sin θ gives z  = aθ  so a parametrization of the surface is σ (r, θ) = (r cos θ, r sin θ,aθ). Then σ r  = (cos θ, sin θ, 0), σ θ = ( r sin θ, r cos θ, a), so E  = 1, F  = 0, G = r2 + a2 . Next, σ r  σσ θ = (a sin θ, a cos θ, r) so N = G−1/2 (a sin θ, a cos θ, r). From σ rr = 0, σ rθ = ( sin θ, cos θ, 0), σ θθ = ( r cos θ, r sin θ, 0) we get L = 0, M  = aG−1/2 , N   = 0. The principal curvatures are the roots of 

 

±

−

 ± √  −

−

 ×

−

−

−

 − −

κ aG−1/2

i.e. κ =

±a/G = r ±+aa 2

2

=

−aG−1/2 −κG

a x2 +y 2 +a2 .

±

8.2.10 In the notation of Proposition 8.2.3, H 2

 

−

−

−

= 0,

− K  = 14 (κ1 + κ2 )2 − κ1 κ2 = 14 (κ1 − κ2 )2.

8.2.11 The condition H 2 = K  gives x 2 + y 2 + z2 = 3. We showed in Exercise 5.1.3 that the points (x,y,z) which satisfy this condition as well as xyz  = 1 are precisely the four points (1, 1, 1), ( 1, 1, 1), ( 1, 1, 1) and (1, 1, 1), and we found in Exercise 8.1.13 that these are the points where the Gaussian curvature attains its maximum value. 8.2.12 If K > 0, then since K  = κ 1 κ2 , either κ1 > 0 and κ2 > 0, or κ 1 < 0 and κ 2 < 0. It follows from Euler’s theorem 8.2.4 that the normal curvature κn  of any curve γγ   on the surface is > 0 in the first case and < 0 in the second (since sin θ and cos θ cannot be zero simultaneously). By Eq. 7.9, the curvature of  γγ  is > 0. 8.2.13 Changing the sign of N   changes the sign of the Weingarten map . With the notation in Proposition 8.2.1, ( )(tt1 ) = ( κ1 )tt1 , ( )(tt2 ) = ( κ2 )tt2 , which shows that κ1 and κ2  are the principal curvatures and t1  and t2 are corresponding principal vectors.

− −

 −

−

 −

−

− −

−W 

−

−W 

W  −

72

σ (u, v) be a surface patch, let T (v v) = P vv + a  be an isometry of  R3 , where P  8.2.14 Let σ σ ). Then, σ˜ u = P (σ σ u ), etc. is an orthogonal matrix and a R3 , and let σ˜ = T (σ ˜ = P (σ σu )..P (σ σ u ) = σ σ u.σ u  (since P  is orthogonal) so so in the obvious notation E  ˜ = εN ˜ = E . Similarly, F  ˜ = F, G ˜ = G. By Proposition A.1.6, N N, the sign ε E  ˜ = εL, M  ˜ = εM , N  ˜ = εN . being + if  T  is direct and if  T  is opposite, so L The result now follows from Proposition 8.2.6. 8.2.15 Let κ 1 and κ 2  be the principal curvatures of a surface patch of    at p  and θ the oriented angle between a principal vector corresponding to κ1  and the tangent vector of one of the m curves at p. By Euler’s Theorem 8.2.4, the sum of the normal curvatures of the m curves at p is

 ∈

 −

 S 

m 1

−



κ1 cos

r=0

2



rπ rπ θ + + κ2 sin2 θ + m m



Using cos θ = 12 (eiθ + e−iθ ), sin θ = 1 2 But,



1 iθ 2i (e

m 1

−

1 (κ1  + κ2 ) + (κ1 4 r=0



m 1 r=0



− κ2 )

−



2i(θ+rπ/m)

e



+ e−2i(θ+rπ/m) .

r=0

1=e±2iπ  = 1 e±2iπ/m

−

.

− e−iθ ), etc., the sum becomes

m 1

− e±2i(θ+rπ/m) = e ±2iθ 1 2



0, so the desired sum is simply

m 1

−

1 (κ1  + κ2 ) = m(κ1  + κ2 ) = mH, 2 r=0



where H  is the mean curvature of    at p. 8.2.16 Using Proposition 6.2.5 and Exercise 7.1.6 (and the notation there), E  = 1+v 2 κ2 , F  = G = 1, L = κτ v, M  = N  = 0 (we work in the part of the surface in which v >   0; the region v <   0 is treated similarly). By Exercise 8.2.3 the lines of  curvature are given by κτ v u˙ 2 + κτ v u˙ v˙   = 0, so u˙   = 0 or u + ˙ v˙   = 0, i.e. u = constant or u + v = constant; the first of these corresponds to the tangent lines to the given curve. N1 (u) in the notation of Exercise γ (u)+vN 8.2.17 The surface is parametrized by σ (u, v) = γγ  ˙  , where κ is the principal 8.2.4. Since γγ   is a line of curvature, we have N˙ 1 = κγγ  ˙  . curvature corresponding to the principal vector γγ  (i) If the surface is a generalized cone, the rulings all pass through some fixed point, so there is a function v(u) such that σ (u, v(u)) is independent of  u. Dif˙  + vN ˙   = 0. Since γγ  ˙   is perpendicular ferentiation with respect to u  gives γγ  ˙ N1 κvγγ  to N1 , this gives 1  κv   = 0 and v˙   = 0, so κ = 1/v  is a non-zero constant. Conversely, if  κ is constant, σ (u, 1/κ) is constant so the surface is a generalized cone.

 S 

−

−

 −

−

73

(ii) If the surface is a generalized cylinder, the rulings are all parallel, so N N1 is ˙ ˙ ˙   we have κ  = 0. Conversely, constant along γγ  . So N1 = 0, and since N1 = κγγ  if  κ = 0 then N˙ 1  = 0 so N1  is constant along γγ   and the rulings are parallel.

 −

8.2.18 The image of  γγ   under the Gauss map is the curve N N, so the condition is that ˙   is parallel to N˙ , i.e. N˙ = λγγ  ˙   for some scalar λ. The result now follows from γγ  Exercise 8.2.2. The second part follows from the last sentence of Proposition 8.2.1. 8.2.19 If every curve on  is a line of curvature, every tangent vector of    at p is a principal vector. By Proposition 8.2.6, it is enough to prove that every point of    is an umbilic. Suppose for a contradiction that p   is not an umbilic, and let κ1 = κ2  be the principal curvatures at p  with corresponding non-zero principal vectors t1   and t 2 . If    is the Weingarten map of    at p, we have (t1 + tt2 ) = κ1 t1 + κ 2 t2 . Since every tangent vector is principal, there is a scalar λ such that (t1  + t2 ) = λ(tt1  + t2 ); then (κ1 λ)tt1  + (κ2 λ)tt2 = 0. But t 1  and t 2  are linearly independent, so this equation forces κ1 = λ = κ2 , a contradiction. 8.2.20 In the notation of Exercise 7.3.22, τ g = N˙ .(tt N). By Exercise 8.2.2, if  γγ   is a line of curvature N˙   is parallel to t, so τ g   = 0. Conversely, if  τ g   = 0 then  ˙ t) = 0; but N˙  and t  are both perpendicular to N, so N˙ t  is parallel to N.(N γ  is a line of curvature by Exercise N; hence, N˙ t  = 0, so N˙  is parallel to t  and γγ  8.2.2 again.

 S 

 S 

 S 

 ∈ S 

 

 W 

W 

 S 

 W 

 −

 −

×

×

×

×

γ  and let N  be the unit normal of  . 8.2.21 We use the usual Frenet-Serret notation for γγ   ˙ . If  κ is the principal curvature n.N  = b.N We are given that b.N  is constant, so τ n of    corresponding to γγ  , we have N˙ = κtt  so b.N  = 0. If  τ  = 0 at some point γγ  (t0 ) of  γγ  , say, then for some ǫ > 0 we have τ (t) = 0 if t0 ǫ < t < t 0  + ǫ. For values of  t in this interval, n  is perpendicular to N, so b  is parallel to N, hence b = N. But then b˙ = N˙ so τ n = κtt; this forces τ  = κ = 0, contradicting γ   is a plane curve. assumption. It follows that τ  = 0 at all points of  γγ , so γ

S 

 S 

 −

 ±

 ±

 

−

 

 ±

The converse is a special case of Exercise 8.2.5, since every curve in a plane is a line of curvature of the plane. 8.2.22 In the notation of the proof of Euler’s Theorem 8.2.4, t  = cos θtt1  + sin θtt2 , so N˙  = (t) = κ1 cos θtt1 κ2 sin θtt2 . Since t 2 N  = t1  and t 1 N  = t2 , we have t N  = cos θtt2 +sin θtt1  so since t1  and t2  are perpendicular unit vectors, τ g = N˙ .(tt N ) = κ1 sin θ cos θ + κ2 sin θ cos θ = (κ2 κ1 )sin θ cos θ.

 −W   − × − × −

−

×

×

 −

−

8.2.23 By Euler’s Theorem 8.2.4, for an asymptotic curve we have κ1 cos2 θ +κ2 sin2 θ = 0, so κ1 κ2 = κ22 tan2 θ   0 (if  θ   is a multiple of  π/2 Euler’s Theorem gives κ1  = 0 or κ2  = 0, and the condition again holds). If    is ruled, we must have κ1 κ2 0 because the rulings are asymptotic curves (Exercise 7.3.6).

 −

 ≤

≤

If  κ 1 κ2 < 0, Euler’s theorem gives tan2 θ =

 S 

−κ1 /κ2 , so there are two asymptotic

74

curves each of which makes an angle tan−1

8.2.24 8.2.25 8.2.26 8.2.27

 −

κ1 /κ2  with a principal vector corresponding to κ1 , so the angle between them is as stated. If  κ1 = κ2  = 0 everywhere, every curve is asymptotic. If  κ1   = 0 but κ2 = 0, say, Euler’s Theorem gives sin θ  = 0 so there is one asymptotic curve through each point, namely the line of curvature corresponding to κ1   (it is proved in Proposition 8.4.2 that these curves are, in fact, straight lines). Part (i) follows from the fact that ,  is symmetric; part (ii) follows from the fact that it is linear in its second argument (since it is actually bilinear). This is an immediate consequence of Proposition 7.3.3. t1 , t 2 = (t1 ), t2 = κ1 t1 , t2 . But if κ1 =  κ2  then t1  is perpendicular to tt2 (Proposition 8.2.1). By Proposition 8.2.1, we can assume that t1  is perpendicular to t2  (whether or θtt1  + sin ˜ θtt2 , so not κ1 and κ2  are distinct). Then t  = cos θtt1  + sin θtt2 ,  ˜t = cos ˜

 

  



 W 







 

t, ˜t = cos θ cos ˜θt1, t1  + (cos θ sin ˜θ + cos ˜θ sin θ)t1 , t2 + sin θ sin ˜θt2 , t2. Now ti , tj  = W (ti ), tj  = κi ti , tj  = κi if  i = j  and = 0 otherwise. Hence, t, ˜t = κ1 cos θ cos ˜θ + κ2 sin θ sin ˜θ, from which the result follows.

8.2.28 Using the notation and result of the preceding exercise,

1 1 1 1 + = + κn κ ˜n κ1 cos2 θ + κ2 sin2 θ κ1 cos2  ˜ θ + κ2 sin2  ˜ θ θ 1 + tan2 θ 1 + tan2  ˜ = + κ1  + κ2 tan2 θ κ1  + κ2 tan2  ˜ θ

− 

1 κκ21 tan2 θ 1 κκ12 1 1 = + + + κ2 κ1 κ2 κ1  + κ2 tan2 θ κ1  + κ12 cot2 θ = 8.2.29

−

1 1 + . κ1 κ2

v, ˙γγ   = v, W (γγ ˙  ) = −v, N˙  = −v.N ˙  = v˙ .N  (since v.N  = 0).

8.2.30 The formulas for E ∗ , F ∗ and G∗   were proved in Exercise 6.3.11, where it was also shown that σ u.σ )σ σ σu 2(σ σ ∗u = . σ 2 σ 4

    −   

Hence,

σu.σ )σ σu σ uu 4(σ  − σ ∗uu = 2 4

 σ 

 σ  −

σuu.σ  + σ u.σ u )σ σ 8(σ σ u.σ )2σ 2(σ . + σ 4 σ 6

  

  

75

Using the formula for N∗  in Exercise 4.5.5, this gives L∗ =



σ uu σ 2

σu .σ )σ σu 4(σ σ 4

    −    −

σ uu.σ  + σ u.σ u )σ σ 8(σ σ u . σ )2 σ 2(σ + σ 4 σ 6

  

  

σ.N )σ σ 2(σ . σ 2



   − N

which simplifies to give L∗ =

σu .σ u ) σ N)σ L 2E (σ .N − (σσ uuσ .N2 ) − 2(σσ.NN )(σ −  − =  σ 2 4 E. σ  4

Similar calculations give the stated formulas for M ∗ and N ∗ . In the usual notation, the preceding formulas can be written p.N) F I ∗ =  p1 4 F I , F II ∗ = −  p1 2 F II  −  2(p  p 4 F I . Hence, for any κ

∈ R,

F II ∗ + ( p 2 κ + 2(pp.N))F I ∗ = −  p1 2 (F II  − κF I ). Parts (i) and (ii) follow immediately from this equation, and part (iii) is an immediate consequence of (ii). σu + bσ σv , 8.2.31 In the notation of the proof of Proposition 8.1.2, we have N Nu = aσ σ u  + dσ σv , so Nv = cσ

F II I  = =



Nu Nu .N Nv N u .N

Nv Nu .N Nv Nv .N



Ea 2 + 2F ab + Gb2 Eac + F (ad + bc) + Gbd Eac + F (ad + bc) + Gbd Ec2 + 2F cd + Gd2

     a b c d

E F

F  G

a b

c d



−F I −1 F II )tF I (−F I −1F II ) = F II F I −1 F I F I −1 F II  = F II F I −1 F II . =

=(

σ   is unit-speed. From Examples 6.1.3 8.2.32 We can assume that the profile curve of σ ˙g  = f¨/g (if g˙ = 0) and g/f . and 7.1.2, the principal curvatures are f ¨ g + ˙ f ¨ ˙ ˙ (i) If g(u ˙ 0 ) = 0 the first principal curvature must be non-zero when u = u0 , so g¨(u0 ) = 0. Since ¨g   is a continuous function of  u, there is an ǫ >   0 such that g¨(u) = 0 if 0 < u u0 < ǫ. Then, g(u) ˙ = 0 if 0 < u u0 < ǫ by the intermediate value theorem. (ii) Assume that g˙   is never zero. Then the first principal curvature must be ¨ = 0, i.e. f (u) = au + b  for some constants a, b. Then, zero everywhere, so f  

 −

  

 | − |

−

 

 

 | − |



,

76

√  −

g˙ 2 = 1 f ˙2 = 1 a2 so (up to a sign) g(u) = 1 a2 u + c for some constant c. Since g˙ = 0, a2 < 1. If  a = 0 we have a circular cylinder; if  a = 0 then making the reparametrization u˜ = au + b and a suitable translation parallel to the z-axis u, v) = u˜(cos v, sin v, 1 a2 ). gives the circular cone σ˜ (˜ 8.2.33 The first and second fundamental forms are

−  

(1 +

−

 

√  −

zx2 )dx2

+ 2zx zy dxdy + (1 +

At an umbilic we have equivalent to

zxx dx2 + 2zxy dxdy + zyy dy 2

zy2 )dy 2 ,

 

1+

zx2  +

zy2

 F II  = κF I , where κ  is the principal curvature.

.

This is

zxx = λ(1 + zx2 ), zxy = λzx zy , zyy = λ(1 + zy2 ), where λ = κ

 

1 + zx2  + zy2 .

8.2.34 By Exercises 6.1.2 and 7.1.4, applying an isometry of R3 does not change the first fundamental form, and either leaves the second fundamental form unchanged (if  the isometry is direct) or changes its sign (if the isometry is opposite). It follows that the principal curvatures stay the same (if the isometry is direct) or both change sign (if the isometry is opposite). All the assertions in the exercise follow immediately from this. 8.2.35 (i) By applying an isometry of  R3 , we can assume (Theorem 5.2.2) that the hyperboloid is of the form x2 y 2 z2 + 2 = 2  p2 q  r

 − 1,

where p, q, r   are positive constants, and by the preceding exercise it suffices to consider this case. Differentiating implicitly we find

zxx

r2 = 2  p z

r2 x r2 y zx = 2 , zy = 2 ,  p z q  z r 2 x2 r4 xy r2 , zxy = , zyy = 2  p2 z 2  p2 q 2 z 3 q  z

−  1

−

By Exercise 8.2.33 the conditions for an umbilic are r2  p2 z

r 2 x2 r 4 x2 1 = λ 1 + 4 2  p2 z 2  p z r4 xy λr4 xy , =  p2 q 2 z 3  p2 q 2 z2 r2 y 2 r4 y 2 1 = λ 1 + 4 2 q 2 z 2 q  z

−  



,



.

−

r2 q 2 z

−  

r2 y 2 q 2 z2

−  1

.

77

The second equation gives x = 0, y  = 0 or λ = 1/z; but substituting λ = 1/z in the first equation leads to r 2 /p2 = 1, which is absurd; so x = 0 or y = 0.

−

−

−

If  x = 0, the first equation gives λ = r 2 /p2 z and then the last equation gives r2 y2 q 2 z 2

r4 y2 1+ 4 2 q  z

−  

 p2 1 2

2

2

Using y /q  = z /r

2

= q 2

− 1, this leads to y  =

  ± q 



.

 p2 q 2 q 2 +r2 , z  =

−

  ±

  ± r

 p2 +r 2 q 2 +r2  (with

q 2  p2  p2 +r2 , z

combination of signs). Similarly, y  = 0 leads to x =  p

−

=

any

  ± r

q 2 +r2  p2 +r2 .

If  p2 =  q 2 , this gives four distinct (real) points of the surface; if  p2 = q 2 we get the two points (0, 0, r). (ii) We can take the surface to be



±

x2 y2 z = 2 + 2.  p q  Proceeding as in (i), we find that the conditions for an umbilic are 2 4x2 = λ 1 + 4  p2  p





4λxy 2 4y 2 , 0 = 2 2 , 2 = λ 1 + 4  p q  q  q 





.

The second equation gives x = 0, y = 0 or λ = 0, and λ = 0 is impossible by the other two equations. If  x = 0 we get y = 12 q   p2 q 2 , z = 14 ( p2 q 2 ). If  y = 0 we get x = 12 p q 2  p2 , z = 14 (q 2  p2 ). If  p2 = q 2 this gives two (real) points of the surface; if p2 = q 2 we get the origin as the only umbilic. (iii) The calculation is simplified by noting that these surfaces are ruled, so that, if  κ1 , κ2   are the principal curvatures, κ1 κ2   0 everywhere (Exercise 8.2.23). Hence, if  κ1 = κ2  we must have κ1 = κ2  = 0, and hence II    = 0, so we are looking for the points at which the second fundamental form is zero.

 ±

  −

 ±  ± −

  −

≤



 ±

−

 F 

For the hyperboloid of one sheet x2 y 2 z2 + 2 = 2 + 1,  p2 q  r we can use x, y as parameters on the open subset of the surface where z = 0. The formulas for zxx , zxy and zyy  are the same as in part (i); the condition zxy = 0 forces x   = 0 or y   = 0, but neither of these is consistent with the conditions zxx = zyy  = 0. Hence, there are no umbilics for which z = 0.

 

 

To deal with the ‘waist’ z  = 0 of the hyperboloid, we can use x, z as parameters if  x = 0 and y, z if  y = 0. In the first case, for example, the condition for an

 

 

78

umbilic is yxx = yxz = yzz  = 0; proceeding as above, we find that there are no umbilics. Similarly in the last case. Finally, for the hyperbolic paraboloid x2 z= 2  p

 −

y2 , q 2

zxx = 2/p2 is never zero, so there are no umbilics. 8.2.36 We have to show that the umbilics on the surface xyz  = 1 are the four points (1, 1, 1), (1, 1, 1), ( 1, 1, 1) and ( 1, 1, 1). Using x, y   as parameters, we find as in the preceding exercise that the conditions for an umbilic are

− −

−

−

− −





,





,

2 1 = λ 1 + x3 y x4 y 2 λ 1 , = x2 y 2 x3 y 3 2 1 = λ 1 + xy 3 x2 y 4

for some λ. The second equation gives λ = xy  and then the other equations give x4 y 2 = x 2 y 4 = 1; these equations force x2 = y 2 = 1, so x = 1, y = 1 (with any combination of signs). Then z = 1/xy = 1 and we obtain the four points required. 8.2.37 On the open subset of    on which z = 0 we can use x, y  as parameters. Differentiating implicitly we find that

 ±

 ±

 S 

zxx =

 ±

 

(y 2 + z 2 )(2x2 z2 y 2 z2 z3 (x2 + y 2 )2

−

− x2 y 2 ) ,

so z xx  = 0 when x 2 = y 2 = z 2 = 1, and in particular at p. Similarly, zxy = z yy = 0 at p. It follows that the second fundamental form of    is zero at p  (Exercise 8.2.33), i.e., p  is a planar point. The argument actually proves that the eight points ( 1, 1, 1) are all planar, but this can also be deduced from Exercise 8.2.34 and the fact that   is preserved by the isometries (x,y,z) ( x, y, z) 3 of  R (with any combination of signs). (i)   is a level surface of  f (x,y,z) = x 2 y 2 + y 2 z 2 + z2 x2 , so

 S 

± ± ±

S 

→ ± ± ±

 S 

∇f  = (2x(y2 + z2 ), 2y(x2 + z2 ), 2z(x2 + y2)) n  is normal is normal to at (x,y,z). Taking x = y = 1 we find that (4, 4, 4) = 4n to   at p, and hence so is n. Since t1  and t2  are perpendicular to n, they are tangent to   at p.

 S 

S   S 

79

 X,, Y , Z  as (ii) Since the vectors tt1 , t2 , n  are linearly independent, scalars  X   as in the statement of the exercise exist, and are small if (x,y,z (x,y,z)) is near p. When When we  Z  in the equation substitute x  = 1 + X  +  X  + Z   +  Z , y = 1 X  + Y   +  Y  + Z , z = 1 Y  + Z  in of  , we know that all quadratic and lower order terms will cancel, and the quartic terms can be discarded as they will be small compared to the cubic terms. terms. Retaining Retaining only the cubic terms, we find after some algebra the equation equation 2 2 6Z  3X  Y  + 3X 3X Y  = 0, hence the stated equation.  ˜  + ˜ , Y  = X  ˜  + ˜ , Z  ˜ =  Z , Y  Y  (iii) Making the linear transformation X  transformation  X  = X   + 3Y ,  + 3Y , ˜ = X  ˜ (X   ˜ 2 3Y  ˜ 2 ), which is a monkey saddle. we get Z 

−

 S   S 

−

 −  −

√ 

−

−

√ 

v˜, v = u 8.3.1 (i) Setting u˜ = v, v˜ = w = e−u , we have u = ln v, ˜   so, in the nota1 0 ˜ . Since v J   is invertible, (u, tion of Exercise 6.1.4, J  = ince J   (u, v) (v, w) 1 0 is a repa repara rame metr triza izatio tion n ma map. p. The The first first fund fundam amen ental tal form form in terms terms of  v, w is 1 ˜ ˜ E F  0 1 1 0 0 E  F  t v ˜ given by = 1 2 ˜ G ˜ = J  F G J  = f ( f  u 0 0 ( ) 1 0 F  v ˜ 1 0 w2 , so the first fundamental form is (dv (dv 2 + dw2 )/w2 . 1 0 w2 (ii) We find that the matrix

−  −       −    





˜= J 

∂v ∂V  ∂w ∂V 

∂v ∂W  ∂w ∂W 

=

−

v (w + 1) 1 2 (w + 1) 2 ) 2 (v

−

 →



1 2 v 2(

− (w + 1)2 )

v (w + 1)

˜t so the the first first fund fundam amen ental tal form form in term termss of  V  and W  is J  (v 2 +(w +(w+1)2 )2 I  = (1 U 24 V 2 )2 I , 4w 2



1 w2

0

−





, 0

1 w2



˜ = J 

after some tedious algebra. In (i), u <   0 and π < v < π   corresponds to π < v < π and w > 1, a vw -plane. semi-infinite rectangle in the upper half of the vw-plane.

− −

 −

 −

To find the corresponding region in (ii), it is convenient to introduce the complex  v  + iw, iw, Z  =  U  + iV  . numbers z =  v +   . Then, the equations in (ii) are equivalent to

80

i Z  = zz− +i , z =

+1 Z +1 i(Z  1) .

−

vw -plane corresponds to z + z¯ = 2π The line v  = π  =  π  in the vw-plane

(the bar denoting complex conjugate), i.e.

+1 Z +1 i(Z  1)

−

+1 − i(Z Z  ¯¯+1 −1) = 2π, which simplifies

V -plane with to Z  (1 πi ) 2 = π12 ; so v  = π  =  π  corresponds to the circle in the U V -plane centre 1 πi   and radius π1 . Similar Similarly ly,, v = π  corresponds to the circle with centre 1 + πi   and radius π1 . Finall Finally y, w   = 1 corresponds to z z¯ = 2i, i.e. ¯ +1 +1 Z +1 Z +1 This simplifie simplifiess to Z  12 2 = 14 ; so w  = 1 corresponds to  ¯ −1) = 2i. This i(Z −1) + i(Z  V -plane. The required region the circle with centre 1/ 1/2 and radius 1/ 1/2 in the U V -plane. in the U the  U V  -plane   -plane is that bounded by these three circles:

 | − − | −

 −

 −

 |  −  − |

For (iii) (iii) we follo follow w the the hin hint and and ma mak ke use use of polar polar coo coord rdin inate atess on the the disc disc,, r  ¯  ¯ ¯ ¯ V  = r cos θ, W  = r sin θ , V  = r¯ cos θ, W  = r¯ sin θ. We find find that that r¯ = r22+1 , θ¯ = θ. Suppo Suppose se that that the the first first fund fundam amen ental tal form form in term termss of these these para paramm2 −r ) r¯dθ¯ + Gdθ¯2 . Since ddrr¯ = 2(1 eters eters is E dr¯2 + 2F drd first funda fundamen mental tal (1+r (1+r 2 )2 , the first 4(1 r 2 )2 4(1 r 2 ) 2 + E dr Fdrdθ  + Gdθ  Gdθ 2 . (1+r (1+r 2 )4 (1+r (1+r2 )2 Fdrdθ + 4(dr 4(dr2 +r2 dθ2 ) (1+r (1+r2 )4 1 E  F  , w e get = 2 2 (1 r ) (1 r2 )4 = (1 r¯2 )2 ,

form is

−

−

Equa Equatin ting g this to 2

4(dV  4(dV 2 +dW 2 ) (1 V  2 W 2 )2 = 2 = 1 r¯ r¯2 . ConCon¯  ¯ 2 ¯

− − −

4r = 0, G = (1− r2 )2 ¯ ), we have rd¯ ¯ + W dW , r¯ dθ = W  r¯dr¯ = V¯ dV  verting back to the parameters ( V¯ , W ),

−

−

−

81

 ¯ V¯ dW 

¯ , so the first fundamental form becomes   −− W¯ dV , V  ¯ +W ¯ dW )  ¯ )2 + (1 − r¯2 )(V¯ dW   ¯  − ¯ )2 W  V   − W¯ dV ) (V¯ dV  r¯2 (1 − r¯2 )2 ¯ 2 + (1 − r¯2 )W   ¯ 2 )dV  ¯ 2 + 2¯ ¯W ¯ dV¯ dW   ¯ + (W   ¯ 2 + (1 − r¯2 )V  ¯ 2 )dW   ¯ 2 (V  2r¯2 V  = r¯2 (1 − r¯2 )2 ¯ 2 )dV  ¯ 2 + 2 V  ¯W ¯ dV¯ dW   ¯ + (1 − V  ¯ 2 )dW   ¯ 2 (1 − W  . = ¯ 2 − W  ¯ 2 )2 (1 − V 

f (u)cos v, f ( f (u)sin v, g (u)), f  8.3.2 The parametris parametrisation ation is σ (u, v) = (f ( )),  f ((u) =  e u , g (u) = < u <  0. 1 e2u cosh−1 (e−u ), f (u) =  eu , so has length 2πe (i) A parallel u  = constant is a circle of radius f ( 2πe u .  f (u)dudv and dudv  and the (ii) From Example 7.1.2, E  = 1, F  = 0, G  = f   =  f ((u)2 , so d σ =  f ( 2π 0 u dudv  = 2π . area is 0 −∞ e dudv = f ˙g¨ f ¨ ˙ g˙  = (iii) From Examples 7.1.2 and 8.1.4, the principal curvatures are κ are  κ 1 = f ¨ f¨/g˙  = (e−2u 1)−1/2 , κ2 =  f  ˙ g/f  ˙ 2 = g/f  ˙ = (e−2u 1)1/2 .  <  0, κ2 >  0. (iv) κ1  < 0, γ (u) = (f ( f (u), 0, g (u)) and denote d/du f ¨+ K f  = 8.3.3 Let γ Let γγ  denote  d/du by  by a dot; by Example 8.1.4, f +

√  −

−

−

    −

 −∞

A

−

−

−

√ −Ku √ −Ku − 0. If  K <   0, the general solution is f  = ae +  be where a, b are

f (π/2) π/ 2) =  f (  f ( π/2) π/ 2) = 0 forces a = b constants constants;; the condition condition f (  =  b =  = 0, so γ so  γγ  coincides with the z -axis, contradictin contradictingg the assumptions. assumptions. If K  If  K  = 0, f  =  a + bu and bu  and again a  = b  =  b =  = 0 is forced. So we must have K >  0 and f  =  a cos K u+b sin K u. This π/ 2) = f (  f ( π/2) π/ 2) = 0 and a, time, f  time,  f ((π/2) and a, b not both zero implies that the determinant

−

−



√ 

√  √ 

π/ 2 cos K π/2 π/ 2 cos K π/2

√ 

√  − √ 



√ 

π/ 2 sin K π/2 = 0. π/ 2 sin K π/2

This gives sin K π  = 0, so K  = n2 for some integer n = 0. If  n = 2k  is even, f  = b sin2ku f (0) = 0, contradic sin2ku,, but then f (0) contradicting ting the assumptions. assumptions. If n If  n  = 2k + 1 is odd, f  = a cos(2k cos(2k + 1)u 1)u and f ( which h contr contrad adic icts ts the the f (π/2(2 π/ 2(2k k   + 1)) = 0, whic 2 assumptions unless k unless  k =  = 0 or 1, i.e. unless K  unless K  = (2k (2k +1) = 1. Thus, f  Thus,  f  =  a cos u, γ  ˙   = (f˙, 0,  ˙g) is perpendicular to the zg˙ = 1 f ˙2 = 1 a2 sin2 u. No Now, w, γ

  −   −

 

−

√  −

g˙   = 0. So the assum axis assumpt ption ionss give give 1 a2 = 0, i.e. a = 1. Then Then,, γγ  (u) = ( cos u, 0, sin u) (up to a translation along the z -axis) and   is the unit sphere.

 ⇐⇒  ⇐ ⇒ ±

±

 ±

 S 

u 8.4.1 Let σ˜ (u, ˜, v˜) be a patch of    cont contai aini ning ng p = σ˜ (u ˜0 , v˜0 ). The Gaussian Gaussian curvature curvature K  of  is <   0 at p; sinc K   is a smooth function of (˜ u, u, v˜) (Exercise 8.1.3), sincee K   ˜  containing K (u, u  <  0 for (˜ u, u, v˜) in some open set U  u0 , v˜0 ); then every point of  ˜, v˜)  < 0   containing (˜ ˜ ) is hyperbolic. Let κ1 , κ2  be the principal curvatures of  σ˜ , let 0 < σ˜ (U ) U  π/ 2 0  < θ < π/2

S 

 S   S 

82

 −

κ1 /κ2 , and let e1  an be such that tan θ  =  and e 2  be the unit tangent vectors of  σ˜  making angles θ angles  θ and θ , respectively, with the principal vector corresponding to κ1  (see Theorem 8.2.4). Applying Applying Proposition 8.4.3 gives gives the result. For the last part, part, put put v˙  = 0 in the formula for κn   in Proposition Proposition 7.3.5: this shows shows that L  = 0 if the parameter curves v  = constant N  = 0 if  constant are asymptotic. asymptotic. Similarly Similarly N  = the parameter curves u  = constant are asymptotic. 8.4.2 The first first and second fundamen fundamental tal forms are (2 + v 2 )du2 + 2uvdudv +(2+ u2 )dv 2 and 4dudv/ 4 + 2u 2 u2 + 2v 2 v 2 . By Exercise 8.2.3, the lines of curvature are given by (2 + v 2 )u˙ 2 + (2 + u2 )v˙ 2 = 0,

 −

√ 

 −

−

so

 

and hence

dv = 2 + v2

  ±

du , 2 + u2

√ v2 ± sinh−1 √ u2 = constant. constant. v u v u  − We can take s   = sinh−1 √   + sinh−1 √  , t   = sinh−1 √  sinh−1 √  , so u = 2 2 2 2 √ 2sinh 1 (s − t), v = √ 2 sinh sinh 12 (s + t). 2 sinh−1

γ (u) + vδδ  δ (u), 8.4.3 8.4.3 Assum Assumee that that γγ   is is unit-speed. unit-speed. Then, Then, the ruled surface surface is σ is  σσ (u, v ) = γγ  ˙   (a where δδ   = = t N, t  = γγ  (a dot denoting d/du denoting d/du)), and N  is the unit normal of (a patch ˙ .  N  = 0, i.e., (tt N˙ )..N = 0. of) of ) . By Example Example 8.1.5, the flatnes flatnesss conditi condition on is δδ  ˙  is tangent to , so t N ˙  is Now, N  i s para pa rall llel el to N. Hence, ˜ is   is flat if and only if  ˙ ˙ ˙  . By Exercise 8.2.2, this is precisely t N  = 0, i.e. if and only if  N  is parallel to γγ  the condition for γ for  γγ  to   to be a line of curvature of  .

 S  ×

×

 S 

×

S 

×

 S   S 

8.5.1 8.5.1 By Corollary Corollary 8.1.3 8.1.3 and the fact that σ that σσ  is conformal, the mean curvature of  σ is L+N  H  = 2E  , so σ  is minimal

(*)

⇐⇒ L + N  = 0 ⇐⇒ (σ uu + σ vv )..N  = 0.

σuu +σ vv = 0. For the converse, we have to Obviously, then, σ  is minimal if ∆σ ∆σ  = σ  =  σ show that ∆σ ∆σ = 0  if (*) holds. It is enough to prove that ∆σ ∆σ .σ u = ∆σ .σ v = 0, 3 σ vv .σ u = since σ u , σ v , N   is a basis of  R . We compu compute te ∆σ ∆σ.σ u = σ uu.σ u + σ 1 1 (σ v .σ uv ) = 2 (σ u.σ u σ v .σ v )u + (σ ( σ v .σ u )v . But, But, sinc sincee 2 (σ u.σ u )u  + (σ v .σ u )v σ  is conformal, σ u .σ u = σ v .σ v and σ u .σ v   = 0. He Henc nce, e, ∆σ ∆σ .σ u   = 0. Similar Similarly ly,, ∆σ .σ v = 0. The first fundamental form of the given surface patch is

 {

}

 −

 −

(1 + u2 + v 2 )2 (du2 + dv 2 ), so it is conformal, and σ and  σσ uu + σ vv = ( 2u, 2v, 2) + (2 (2u, 2v, 2) = 0.

−

− −

83

f (x, y ) = ln cos 8.5.2 Using the formula formula in Exercise 8.1.1 with f ( cos y H  =

− lncos x  gives

sec2 x(1 + tan2 y ) sec2 y (1 + tan2 x) = 0. 2(1 + tan2 x + tan2 y )3/2

−

σ u + wN u , Σv = σ σv  + w 8.5.3 Σu = σ  +  wN Nv , Σw = N N. Σu .Σw  = 0 since σ u .N  = Nu .N  = 0, and an d simi si mila larl rly y Σv .Σw  = 0. Finally, σu .σ σv  + w (σ u .Nv  + σ v .Nu ) + w2 Nu .Nv Σ u .Σv = σ

 − 2wM  + w2Nu.Nv = w 2Nu .Nv .  − L σ u , Nv = − N  By the proof of Proposition Proposition 8.1.2, Nu = − E  G σv, =  F 

so Nu .Nv = LN  F   = 0. Ever Every y surf surface ace u = u0   (a constant) is ruled as it is the union of  EG F   the straight lines given by v  = constant; by Exercise 8.2.4, this surface is flat provided the curve γγ  (v ) = σ (u0 , v ) is a line of curvature of  , i.e i.e. if  σ v is a principal vector; but this is true since the matrices I I  and II  II    are diagonal. Similarly for the surfaces v  = constant. δ (u), where γγ   is unit-speed 8.5.4 8.5.4 We take take the ruled ruled surfa surface ce to be σ (u, v) = γγ  (u) + vδδ  ˙   by  d/du  by a dot and γγ  and δδ   is a unit unit vector. vector. Denotin Denotingg  d/du by by t, the coefficients of  ˙   2 , F  = t.δδ  , G = 1, the first and second fundamental forms are E  = t  + v  +  v δδ  ˙ δ   )×δδ   . Then (t+vδ ¨ ˙ ˙ L = (t  + v N   = 0, wher  +  v δδ  ). ).N, M  = δδ .  N, N   wh eree N = Then,, the mean mean ˙ (t+vδδ  )×δδ   curvature ˙   + LG 2M F  + N E  (t˙ 2(t.δδ  )δδ  + vδ¨δ  ). ).N H  = . = 2(E 2(E G F 2 ) 2(EG 2(EG F 2 )

 S   S   F 

 F 



−

−

  

−

− ˙  (u0 ) =  H (u0 , v) → 0 as v   for some value of  u u0 ,  H ( If  δδ  as  v  → ±∞, since the numerator H   is independen in the expression for H   independentt of  v   while the denominator → ∞ as v → ±∞. As H  is H  is constant, H  = H  = 0 everywhere, contradicting the assumption. 0

˙   = 0  everywhere, δ δ  is Hence, δδ   everywhere,  δδ  is constant, and the surface is a generalized cylinder. γ   lies in a plane perpendicular to δδ   (Exercise 5.3.3). We can now assume that γ Then Th en,, t δδ   is is a unit vector and the formula for the mean curvature simplifies to

×

H  =

t˙ .(tt δδ  ) . 2 t δδ   2

×  ×   

Now ˙t  = κ  =  κn n  where κ  where  κ is  is the curvature of γ of γγ   an and n  is its it s principal normal. Also, t δδ   is perpendicular to t  and parallel to the plane containing γγ  ; hence, henc e, t δδ   = n. κ/2, so if H  It follows that H  = κ/2, if  H  is  is a non-zero constant then κ  is constant and the plane curve γ curve  γγ  is   is a circle. Thus, the surface is a circular cylinder. 8.5.5 8.5.5 We use use the notation notation and result resultss of Propositio Proposition n 8.5.2. If γ If γγ   is a curve on σ , the  λN. No ‘corresponding’ curve γγ  λ on the parallel surface σ λ is γγ  λ = γγ   +  λN Now w, γγ  λ

±

×

×  ±

84

λ ˙  λ for some scalar κ (Exercise 8.2.2) is a line of curvature on σ λ N˙ = κγγ   ˙  = κ(γγ   ˙ ). But this last equation holds if and only if  N˙  is a scalar ˙  + λN  ǫN ˙  , i.e. if and only if γγ  γ   is a line of curvature on σ . In this case, multiple of  γγ  λ  ˙  = (1 κλ)γ γ  ˙   = γγ  ˙   + λN ˙   is parallel to γγ  ˙  . γγ 

⇐⇒

⇐⇒

 −

−

 −

σ (u, v)+λ(u, v)N N(u, v) is σ (u, v) = σ 8.5.6 Let σ (u, v) be a patch of  . We are given that ˜ σ  at σ (u, v), a patch of  ˜, for some function λ, where N(u, v) is the unit normal of σ σ u + λN Nu + λuN; since and that N(u, v) is also normal to ˜ at σ˜ (u, v). Now σ˜ u = σ σ u , σ˜ u   and N u  are all perpendicular to N, this implies that λu   = 0; similarly λv  = 0. Hence, λ is constant and ˜ = λ is a parallel surface of  .

S 

S 

S  S  S 

 S 

Chapter 9 9.1.1 By Exercise 4.1.3, there are two straight lines on the hyperboloid passing through (1, 0, 0); by Proposition 9.1.4, they are geodesics. The circle z = 0, x2 + y 2 = 1 and the hyperbola y = 0, x2 z2 = 1 are both normal sections, hence geodesics by Proposition 9.1.6.

−

¨ .(N N γ ) = γγ  9.1.2 Let κ(γγ  Γ) = then κ(Γ

 × dϕ dt

˙  ). Note that if Γ(t) = γγ  (ϕ(t)) is a reparametrization of  γγ  , γγ 

3

γ  κ(γ γ ). In particular, κ(γγ  γ ) = 0

⇐⇒ κ(ΓΓ) = 0.

For (i), let γγ   be a pre-geodesic and let Γ  be a geodesic reparametrization of  γγ  . ˜ (t) = Γ(t/v) is By Proposition 9.1.2 Γ   has constant speed, say v, and then Γ ˜ ) = 0, hence κ(Γ Γ) = 0, hence a unit-speed geodesic. By Proposition 9.1.3, κ(Γ γ  κ(γγ  γ ) = 0. Conversely, if  κ(γ γ ) = 0 and if Γ  is a unit-speed reparametrization of  Γ) = 0 so Γ  is a geodesic by by Proposition 9.1.3. γγ  , then κ(Γ Part (ii) is obvious. For (iii), let γγ   be a constant speed pre-geodesic, say with speed v. Then Γ(t) = γγ  (t/v) is a unit-speed pre-geodesic, hence a geodesic by (i) and Proposition 9.1.3. ¨ , γγ  ¨  =  v 2Γ ¨  is perpendicular to the surface, so γγ   is a geodesic. Since γγ  Finally, (iv) follows from (iii) and Proposition 9.1.2. 9.1.3 Let Πs   be the plane through γγ  (s) perpendicular to t(s); the parameter curve s  = constant is the intersection of the surface with Π s . From the solution to Exercise 4.2.7, the standard unit normal of  σ  is N = (cos θ n  + sin θ b). Since this is perpendicular to t, the circles in question are normal sections.

 −

2

2

2

9.1.4 Take the ellipsoid to be  px2 +  yq2 + zr2  = 1; the vector ( px2 , qy2 , rz2 ) is normal to the ellipsoid by Exercise 5.1.2. If  γγ (t) = (f (t), g(t), h(t)) is a curve on the ellipsoid, 2 2 2 2 ˙2 ˙2 ¨  R = ( pf 2 + gq˙ 2 + hr2 )−1/2 , S  = ( pf 4 + gq4 + hr4 )−1/2 . Now, γγ   is a geodesic γγ  f, g¨, ¨h) = λ( f 2 , g2 , h2 ) for some scalar λ(t). From is parallel to the normal ( ¨ f 2 g2 h2  +  + 2 2  p q r 2  =

1 we get

 ⇐⇒

˙ f f ˙ g g˙ hh  p2  + q 2 + r2  =

 ⇐⇒

 p

0, hence

q r ˙2 ¨ f ˙2 g˙ 2 f  ¨ f  g¨ g h hh  +  +  + 2 2 2  p q r  p2  + q 2 + r2  =

0, i.e.

85

f ˙2  p2

+

g˙ 2 q2

+

˙2 h r2

+λ



f 2  p4

+

g2 q4

  || h2 r4

+

f 2  p4

 ¨2 + g¨2 + ¨h2 )1/2 = λ γ   = ( f   ¨γ

  

1 d 2 dt

   1 R2 S 2

=

+

g2 q4

f f ˙ g g˙ hh˙ + 4 + 4  p4 q  r +

=

= 0, which gives λ =

1 R2





2

+



h2 r4



1/2

−S 2 /R2. The curvature

λ| = |S  =

f ˙2 g˙ 2 h˙ 2 + 2+ 2  p2 q  r

2

2

f  g h + +  p4 q 4 r4

f f ˙ g g˙ hh˙ + 4 + 4  p4 q  r



Finally,



¨ f ˙f ¨ g¨ ˙ g h˙ h + 2 + 2  p2 q  r

  +

S  R2 .

λ S 2



f f ˙ g g˙ hh˙ + 4 + 4  p4 q  r



= 0,

since λ = S 2 /R2 . Hence, RS  is constant. γ  lies in the plane vv.a  = b, where a  and b are constants. 9.1.5 Suppose that a geodesic γγ  ˙  .a = γγ  ¨.  a   = 0. Since γγ  ¨  is parallel to N   (the unit normal of the surface), Then γγ  ˙   and N˙  are all parallel to the plane and the last N.a  = 0, so N˙ .a  = 0. Since N, γγ  two vectors are perpendicular to the first, they are parallel. Hence γγ   is a line of curvature by Exercise 8.2.2. Conversely, if  γγ  is both a geodesic and a line of  γ  has unit-speed (for the unit-speed reparametrization curvature, we may assume γγ  ˙  . Then of  γγ    would still be a geodesic and a line of curvature). Let a = N γγ  ˙  +N N γγ  ¨  = 0 since the first term vanishes because γγ  γ  is a line of curvature a˙  = N˙ γγ  ˙ .  a  = 0 so γγ  .a is and the second because γγ   is a geodesic. So a  is constant. And γγ  γ   lies in the plane v.a  = b. a constant, say b. Hence γ γγ   is a non-zero vector parallel to both N1  and N2 , so N1  and N2 9.1.6 For (i) note that ¨ must be parallel. For an example, take 1 and 2  to be the sphere and cylinder in Theorem 6.4.6. ˙  (ii) Now suppose that 1 and 2   intersect perpendicularly. Then, N1 , N2 and γγ   ˙ 2  = 0. If γ γ  ˙ .  N2  = 0 we get γγ  ¨.  N2  + γγ  ˙ .  N γ  is are perpendicular unit vectors. From γγ  ˙ γ  ¨.  N2  = 0 since γγ  ¨  is parallel to N1 , so N2   is perpendicular to a geodesic on 1 , γ ˙  . Since N˙ 2  is also perpendicular to N2 , it must be parallel to N1 . Conversely, if  γγ   ˙ 2  = 0 so γγ  ˙ .  N ¨  is perpendicular to N2 . Since γγ  ¨  is also N˙ 2  is parallel to N1 , then γγ  ˙  , it must be parallel to N1 . Finally, if  γγ   is a geodesic on both perpendicular to γγ  ˙ ˙ 1 and 2 , then N1   is parallel to N2 and N2   is parallel to N1 . It follows that N1 N2 )˙ = N˙ 1 N 2 + N1 N˙ 2 = 0  so N1 N2  is a constant vector. Since γγ  ˙   is (N ˙   is constant, so  is part of a straight line. a unit vector parallel to N1 N2 , γγ  9.1.7 The ellipses formed by the intersections of the ellipsoid with the three coordinate planes are normal sections. Indeed, a normal to the ellipsoid is  px2 , qy2 , rz2 . The plane z = 0 (say) has normal (0, 0, 1), so the intersection with the plane z = 0 is a normal section if and only if   px2 , qy2 , rz2 .(0, 0, 1) = z/r 2 = 0 when z = 0, which is true.

−

×

×

×

 S 

 S 

 S 

 S 

 S 

S 

×

 S 

×

× ×

×





 C





86

9.1.8 The intersections of the surface with each of the planes x = y, y = z and z = x are normal sections. Indeed, by the solution to Exercise 8.2.37, a normal to the surface is (x(y 2 + z 2 ), y(x2 + z 2 ), z(x2 + y 2 )). The planes x = y   have normals (1, 1, 0) so the intersection of the surface with x = y  is a normal section if and only if 

 ±

±

∓

 ±

x(y 2 + z 2 )

∓ y(x2 + z2 ) = 0

when x =

 ±  ±

±y,

which is evidently true. Similarly for the other four planes. 9.1.9 Assume that γγ   is unit-speed with tangent vector t. Then (with the usual notan.a  = 0. Since κ = 0, n  is perpendicular tion), t.a  is constant, so t˙ .a  = 0, i.e. κn to a. As γγ    is a geodesic, t˙   is parallel to N, so n   is parallel to N. Hence, a is perpendicular to N , hence tangent to .

 

 S 

9.1.10 From the solution to Exercise 7.3.3., a circle of radius r on a sphere of radius R r=R has geodesic curvature κg = R2 r 2 /rR. Hence, κg = 0 the circle is a great circle. 9.1.11 The osculating plane is perpendicular to the binormal b of  γγ  , and so is perpendicular to   at p  if and only if b  is perpendicular to N  at p. Since t  and N are γγ   is a geodesic. perpendicular, b  is perpendicular to N n  is parallel to N ¨  is perpendicular to N; γγ   is a geodesic ¨  is parallel γγ  γγ  9.1.12 γγ   is asymptotic γ  ¨  = 0, so γ  is both asymptotic and a geodesic, we must have γ to N. Hence, if γγ  γ   is a straight line. γγ  (t) = taa + b  for some constant vectors a, b, i.e. γ ˙  ) δδ   when v = 0, i.e. σ  ¨ ˙  + vδδ  γγ   is parallel to σσ u σ v = (γγ  9.1.13 γγ   is a geodesic on σ ¨  (γγ  ˙   δδ  ) = 0. Since γγ  ¨  is perpendicular to γγ  ˙  ,  γγ   ¨ γ .  δδ   = 0 γγ  γ  is a geodesic γ δδ   ¨  = κn n. is perpendicular to γγ  9.1.14 Denoting by t, n  and b  the unit tangent vector, principal normal and binormal of Γ, and by κ its curvature, we have n  = N  since Γ  is a geodesic so

√  −

 S 

⇐⇒

 ⇐⇒

× ×

 ⇐⇒

 ⇐⇒

 ⇐⇒

⇐⇒  ⇐⇒

×

×  ⇐⇒

⇐⇒

± N × N˙ ) = t.(n n × n˙ ) = t.(n n × (−κtt + τ b b)) = t.(κb b + τ t) = τ. Γ˙ . (N

We can assume that γγ   is unit-speed; denote by a dash differentiation with respect to the arc-length of  γγ . By Exercise 7.3.22, the geodesic torsion of  γγ  is

× N)..N′ = γγ γ ′.(NN × N′). Since γγ   and Γ  touch at p, γγ  ′ = ±Γ˙  at p, and we can assume that the sign is + ′ ′ ′ τ g = (γγ  γ ′

by changing the sign of the parameter of  γγ   if necessary. Since γγ  = u σ u  + v σ v , N′ = u′N u  + v ′Nv , etc., it follows that u′ = u˙ and v ′ = v at ˙ p, and hence that ′ ˙ ˙ ˙ N = N  at p. Hence, τ g = Γ.(N N N) and the result now follows from the first part of the exercise.

×

87

Since Γ  touches itself at each of its points (!), its torsion is equal to its geodesic torsion. 9.1.15 If  γγ    is an asymptotic curve, then (with the usual notation) b = N so τ  = b˙ .n  = N˙ .n. Since n  = b t  = N t, we get

−

∓

±

×

± × N × t) = t. (N N × N˙ ). τ  = −N˙ . (N

The last part follows from the preceding exercise. 9.1.16 We use the notation of Exercise 9.1.14 and assume that γγ   is a line of curvature. ˙  γ ′  and N′ = N˙  at p. By Rodrigues’ formula, N˙  = κγγ  We can assume that Γ˙  = γγ  γ  , so where κ   is the principal curvature corresponding the line of curvature γ ′ ′ γ  γ  . Hence, the torsion of Γ is N = κγ

−

−

N τ  = Γ˙ . (N

× N˙ ) = −κγγ γ ′.(NN × γγ  ′) = 0.

9.1.17 (i) By Exercise 9.1.14, the torsion τ  of a geodesic γγ   on a surface   is equal to its geodesic torsion which, by Exercise 8.2.22, is equal to (κ2 κ1 )sin θ cos θ in the notation there. At an umbilic κ1 = κ2 so τ  = 0. (ii) By (i), the torsion is given by τ  = (κ2 κ1 )sin θ cos θ. If a second geodesic ˜ ˜  intersects γγ   at right angles at p, then in the obvious notation θ = θ γγ  π/2 and hence τ˜  = τ . γ  is a geodesic, its curvature κ is equal, up to sign, to its normal curvature, (iii) If γγ  so κ = (κ1 cos2 θ + κ2 sin2 θ) by Euler’s Theorem 8.2.4. Taking the + sign,

−

 S 

−

±

−

±

κ

− κ1 = (κ2 − κ1 )sin2 θ, κ − κ2 = (κ1 − κ2 )cos2 θ,

so (κ

− κ1 )(κ − κ2 ) = −(κ1 − κ2 )2 sin2 θ cos2 θ = −τ 2 .

Taking the  sign gives the other expression for τ 2 . (iv) If the surface is flat, either κ1   = 0 o r κ2   = 0 at p. If  κ2   = 0, then κ = κ1 cos2 θ, τ  = κ1 sin θ cos θ, so τ  = κ tan θ. The other formula results if  κ1  = 0 at p. 9.1.18 We can assume that γγ   is unit-speed as conditions (i) - (iii) are unchanged by reparametrization. Let δδ (u) be a non-zero vector parallel to the ruling through γγ  (u). Then (i) says that γγ    is a geodesic, and by Exercise 9.1.13 this holds ˙   = 0. And (iii) obviously holds t˙.δδ   = 0. By Exercise 5.3.4, (ii) holds t.δδ  t.δδ   is a constant. Everything now follows immediately. 9.1.19 Suppose that every geodesic on a surface   is a plane curve. By Proposition 8.2.9, it suffices to show that every point of    is an umbilic. Suppose for a contradiction that p   is not an umbilic, let κ1 , κ2   be the distinct principal

 −

 ±

 −

⇐⇒ ⇐⇒

 ∓

 ⇐⇒  S 

 ∈ S 

 S 

88

curvatures of    at p, and let t  be a non-zero principal vector corresponding to the principal curvature κ1  at p. Choose an angle θ such that (i) θ is not an integer multiple of  π/2, and (ii) κ 1 cos2 θ + κ2 sin2 θ = 0.

 S 

 

Let γγ   be the unit-speed geodesic on   passing through p  and making an angle θ with t. By condition (ii) and Euler’s Theorem 8.4.2, γγ   has non-zero curvature at p, and γγ   is a plane curve by the hypothesis. By Exercise 9.1.5, γγ   is a line of  curvature, but this contradicts property (i).

 S 

9.1.20 Assume that γγ    is unit-speed, and let r   be the radius of the sphere. Then, γ  ˙  .(γ ¨.  (γγ  ˙  .γγ  ˙   = 1. But, as γγ   is γ  c)..(γγ  γ  c) = r2 so γγ  γ  c) = 0, γγ  γ  c) = γγ  (γγ  ¨  = κN N, where κ is the curvature of  γγ  , so the last equation gives a geodesic, γγ  N.(γγ  γ  c) = 1/κ, i.e. 1 κ = , γ  c) N.(γγ 

 −

 −  ± − ±

 −

 −

|

 −

 −

− |

as required. 9.2.1 If p and q  lie on the same parallel of the cylinder, there are exactly two geodesics  joining them, namely the two circular arcs of the parallel of which p  and q are the endpoints. If p  and q  are not on the same parallel, there are infinitely-many circular helices joining p  and q  (see Example 9.2.8). 9.2.2 Take the cone to be σσ (u, v) = (u cos v, u sin v, u). By Exercise 6.2.1, σσ  is locally isometric to an open subset of the xy-plane by σ (u, v)

 √   →

√  √ 



v v u 2cos , u 2sin ,0 . 2 2

√ 

By Corollary 9.2.7, the geodesics on the cone correspond to the straight lines in the xy-plane. Any such line, other than the axes x = 0 and y = 0, has equation ax + by = 1, where a, b are constants; this line corresponds to the curve



cos v sin v 1 √  √   → √ 2(a cos √  , , v v v v v v + b sin √  ) 2(a cos √  + b sin √  ) 2(a cos √  + b sin √  )

v

2

2

2

2

− and y-axes correspond to straight lines on the cone.

2

2

the x

9.2.3 Parametrize the cylinder by σ (u, v) = (cos u, sin u, v). Then, E  = G = 1, F  = 0, so the geodesic equations are u ¨ = v¨  = 0. Hence, u = a + bt, v = c + dt, where a,b,c,d are constants. If b = 0 this is a straight line on the cylinder; otherwise, it is a circular helix.



;

89

9.2.4 For the first part, (E  u˙ 2 + 2F  u˙ v + ˙ Gv˙ 2 )˙ = (E u ˙u + E v ˙v) u˙ 2 + 2(F u ˙u + F v ˙v)u˙ v + ˙ (Gu ˙u + Gv ˙v)v˙ 2 + 2E  u¨ ˙ u + 2F (u¨ ˙ v + u¨v) ˙ + 2Gv¨ ˙v = u(E  ˙ u ˙u2 + 2F u ˙uv + ˙ Gu ˙v 2 ) + v(E  ˙ v ˙u2 + 2F v ˙uv + ˙ Gv ˙v 2 ) + 2E  u¨ ˙ u + 2F (u¨ ˙ v + u ¨v) ˙ + 2Gv¨ ˙v u = 2(E u ˙u + F  v˙ )˙u˙ + 2(F  u˙ + Gv)˙ ˙ v + ˙ 2(E ˙u + F  v˙ )¨ + 2(F  u˙ + Gv)¨ ˙ v by the geodesic equations = 2[(E ˙u + F  v) ˙ u]˙+ ˙ 2[(F  u˙ + Gv) ˙ v]˙ ˙ = 2(E ˙u2 + 2F  u˙ v + ˙ Gv˙ 2 )˙. ˙   2 = E  u˙ 2 + 2F u˙ v˙ + Gv˙ 2 is constant. Hence, (E ˙u2 + 2F  u˙ v˙ + Gv˙ 2 )˙= 0 and so γγ  Suppose now that (i) and (ii) hold. Differentiating E  u˙ 2 +2F u˙ v˙ +Gv˙ 2 = constant gives

   

(E u ˙u + E v ˙v)u˙ 2 + 2(F u ˙u + F v ˙v))u˙ v + ˙ (Gu ˙u + Gv ˙v)v˙ 2 =

−2(E ˙u + F  v˙ )¨u − 2(F  u˙ + Gv)¨ ˙ v,

i.e., (E u ˙u2 + 2F u ˙uv + ˙ Gu v˙ 2 )u˙ + (E v ˙u2 + 2F v ˙uv + ˙ Gv ˙v 2 )v˙ =

−2(E ˙u + F  v˙ )¨u − 2(F  u˙ + Gv)¨ ˙ v.

Using (i) we get 2u˙

 d v  = (E  u˙ + F  v) ˙ + 2(E ˙u + F  v˙ )¨ dt

−2(F u + ˙ Gv)¨ ˙ v − (E v ˙u2 + 2F v ˙uv + ˙ Gv ˙v 2 )v. ˙

The left-hand side of this equation equals 2

 d (u(E  ˙  u˙ + F  v˙ )) = dt

d −2 dtd (F  u˙ v + ˙ Gv˙ 2 ) = −2(F u˙ + Gv)¨ ˙ v − 2v˙ (F  u˙ + Gv). ˙ dt

Combining the last two equations gives (ii) provided v˙ = 0. γ  is unit-speed u˙ 2 +(1+u2 )v˙ 2 = 1. The second 9.2.5 E  = 1, F  = 0, G = 1+u2 , so γγ  d a equation in (9.2) gives dt ((1+u2 )v) ˙ = 0, i.e. v˙ = 1+u 2 , where a is a constant. So

⇐⇒

 

2

a dv a − 1+u  and, along the geodesic, du = uv˙˙ = ± √  . If  a = 0, (1−a +u )(1+u ) √  then v = constant and we have a ruling. If a = 1, then dv/du =  ±1/u 1 + u2 , which can be integrated to give v =  v 0 ∓ sinh−1 u1 , where v0  is a constant.

u˙ 2 = 1

2

2

2

2

90 2

2

2

2

For the last part, note that du = (u +1)(ua2 +1−a ) so dv a2 1 and this is the minimum distance (i) if  a 2 > 1 then du/dv = 0 for u = of the geodesic from the z-axis; (ii) if a2 < 1 then du/dv > a−2 1 so u will decrease to zero and the geodesic will cross the z-axis; (iii) if  a 2 = 1 then du/dv = (u2 +1) so u = tan(v + c) where c is a constant.



 |

|

±√  −

−

±

±

˙  .σ u = u˙   (since The information given implies that, when u = D, cos α = γγ  E  = 1, F   = 0) so a2 = (1 + D2 )sin2 α. Then, a2 is > 1,  < 1 or = 1 according as D is >, < or = cot α. 9.2.6 This is straightforward algebra. 9.2.7 By Exercise 5.3.3 we can parametrize the generalized cylinder by σ (u, v) = γγ  (u) + vδδ  δ , where γγ   is a unit-speed curve, δδ   is a constant unit vector parallel to γ  ˙   = dγ γ /du is perpendicular to δδ   for all values of  u. Then the the rulings, and γγ  first fundamental form of  σ is du 2 + dv 2 , so the map σσ(u, v) (u,v, 0) is a local isometry from   to the xy-plane. The curves on   that make a constant angle with δδ   correspond to the curves in the plane that make a constant angle with the axes. These are, of course, exactly the straight lines in the plane, i.e. the geodesics in the plane. It follows from Corollary 9.2.7 that the geodesics on are exactly the curves on   making a constant angle with δδ .

 S 

 S 

 S 

→

 S 

 S 

9.2.8 Let γγ    be a parameter curve v   = constant, and assume that γγ    is unit-speed. Then, in the usual notation, E  u˙ 2 = 1 so u = ˙ 1/ E . By Theorem 9.2.1, γγ  is a geodesic if and only if 

± √ 

(*)

d d 1 1 (E ˙u) = E u ˙u2 and (F  u) ˙ = E v ˙u2 . dt dt 2 2

The second equation in (*) holds

 ⇐⇒

± √   F  E 

i.e.

u = ˙ u

  √  √  1 E 

F  E 

u

=

E v , 2E  E v . 2E 

This is equivalent to the stated condition. One checks in the same way that the first equation in (*) holds identically. 9.2.9 Let γγ   be the unit-speed curve on the surface corresponding to u + v = c, where c is a constant. Then, u + ˙ v = ˙ 0 so the unit-speed condition gives (1 + u2 + 2uv + 1 + v 2 )u˙ 2 = 1,

91

i.e., (2 + c2 )u˙ 2 = 1, so u (and ˙ hence also v) ˙ is constant. The geodesic equations are d ((1 + u2 )u˙ dt

− uvv)˙ = u(u ˙ u˙ − v v), ˙ d (−uv u˙ + (1 + v 2 )v) ˙ = v(v ˙ v˙ − uu). ˙ dt Using v = ˙

−u the ˙ first equation becomes d ((1 + cu)u) ˙ = c u˙ 2 , dt

u   = 0. Similarly, the second geodesic equation is equivalent to i.e. (1 + cu)¨ v are (1 + cv)¨v  = 0. Both equations are obviously satisfied when u and ˙ ˙ constant. 9.2.10 The geodesic equations u ¨  =

d 1 1 Gu v˙ 2 , (Gv) ˙ = Gv ˙v 2 dt 2 2

must be satisfied if  v = cu and c is any constant. It is easy to see that these equations are satisfied if  c = 0 (since in that case u˙ = 1 from the unit-speed condition), so we assume that c = 0 from now on. Then the second equation is d ˙ = 12 cGv ˙u2 , i.e. dt (Gu) 1 G¨ u + (Gu  + cGv )u˙ 2 = 0. 2 Using the first geodesic equation and writing c = v/u we get the stated equation for G. Substituting G = f (u)/v 2 leads to the ordinary differential equation

 ±

 

(f  + 2u2 )

df  = 2uf, du

which can be solved for the function f (u). (The solution is 2u3 = 3f 3 ln af , where a is an arbitrary constant.) 9.2.11 Assume that the geodesic is unit-speed. The second geodesic equation in Theov = rem 9.2.1 gives F  u+G ˙ ˙ Ω, a constant. Using this and the unit-speed condition E ˙u2 + 2F  u˙ v + ˙ Gv˙ 2 = 1 gives u = ˙

 

±

G Ω2 . EG F 2

− −

Hence, either u = constant or, along the geodesic, dv v˙ Ω F u˙ = = = du u˙ Gu˙

−

F  G

−  ±

Ω G

 

EG F 2 . G Ω2

− −

92

b(u), where b  is the binormal γ (u) + vb 9.2.12 The surface is parametrized by σσ (u, v) = γγ  n, σ v = b  and the first of  γγ   (we assume that γγ   is unit-speed). Then σ u = tt τ vn fundamental form is (1 + τ v 2 )du2 + dv 2 .

−

Let Γ(t) be a unit-speed geodesic on , and denote   d/dt   by a dot. The first geodesic equation in Theorem 9.2.1 gives

 S 

(1 + τ 2 v 2 )u = c, ˙ where c   is a constant. To determine c, note that, at the intersection point p (say), Γ˙ .t  = cos α, which gives (u(t ˙ t

− τ vnn) + vb ˙ b)..t  = cos α,

i.e. u = ˙ cos α at p. Since v = 0 at p, we have c = cos α and so u = ˙

cos α . 1 + τ 2 v 2

The unit-speed condition (1 + τ 2 v 2 )u˙ 2 + v˙ 2 = 1 now gives sin2 α + τ 2 v 2 v˙ = , 1 + τ 2 v 2 2

and hence du = dv

±

cos α

 

2

(sin α +

.

τ 2 v 2 )(1 + τ 2 v 2 )

If  τ  = 0, du/dv  decreases as v  increases and, for large v , du/dv  is approximately cos α/τ 2 v 2 . It follows that u  approaches limits u ±  (say) as v  and that u is always between these limits.

   |

|

 | |

 | |  |

|  → ±∞

If  τ  is always zero, then du/dv = cot α. If cot α = 0 then u as v and then γγ   is not contained between two rulings. If cot α = 0 then u  is constant and γγ   is a ruling.

±



→ ±∞

√  u E 

 → ±∞

√  v G

9.2.13 Writing E  = (U  + V )P , G = (U  + V )Q, we have cos θ = ˙ , sin θ = ˙ 2 2 V E  u˙ is constant along γγ . Now we have to prove that UGv˙

so

−

2

 d (UGv˙ 2 dt

− V E ˙u2) = 2U  v˙ dtd (Gv)˙ − 2V u˙  ddt (E ˙u) + 2Gv˙ dtd (U  v)˙ − 2E  u˙  ddt (V u). ˙

93

Using the geodesic equations in Theorem 9.2.1, this becomes d  d − uV (E  u2 + Gu ˙v 2 ) + 2Gv˙ (U  v) ˙ ˙ − 2E ˙u (V u) ˙ u ˙ dt dt = u˙ 2 v(U ˙ E v − 2EV v ) + u˙ v˙ 2 (2GU u − V Gu ) + 2GU  v¨ ˙ v − 2EV u¨ ˙u + Gv U  v˙ 3 − E u V u˙ 3 = u˙ 2 v(U ˙ E v − 2EV v ) + u˙ v˙ 2 (2GU u − V Gu ) d d + U  (Gv˙ 2 ) − U Gu ˙uv˙ 2 − V  (E ˙u2 ) + V E v ˙u2 v˙ dt dt = u˙ 2 v((U  ˙  + V )E v − EV v ) + u˙ v˙ 2 (GU u − (U  + V )Gu ) d d + U  (Gv˙ 2 ) + U u Gu˙ v˙ 2 − V  (E  u˙ 2 )V v E  u˙ 2 v˙ dt dt d = (U Gv˙ 2 − V E ˙u2 ) + u˙ 2 v((U  ˙ + V )E v − EV v ) + u˙ v˙ 2 (GU u − (U  + V )Gu ). dt

vU (E  u2 + Gv ˙v 2 ) ˙ v ˙

Thus, d (U Gv˙ 2 dt

− V E ˙u2) = u˙ 2 v((U  ˙  + V )E v − EV v ) + u˙ v˙ 2 (GU u − (U  + V )Gu ). But, since U  and P   depend only on u, E v = V v P   and so (U  + V )E v − EV v = (U  + V )(V v P  − V v P ) = 0. Similarly, GU u − (U  + V )Gu = 0.

9.2.14 The first part can be verified directly. Alternatively, this parametrization can be deduced from that in (5.12) by putting a =  p 2 u, b = q 2 u, c = r2 u and then reparametrizing by u ˜  = u v, v˜ = u w. Straightforward algebra gives the first fundamental form as

−

(u

− v)



udu2 (a + u)(b + u)(c + u)

−

 −

−

−

−

vdv 2 (a + v)(b + v)(c + v)

Thus, the quadric is a Liouville surface with U  = u, V  = follows from the preceding exercise.

 −v.



.

The result now

9.3.1 They are normal sections. 2

2

9.3.2 (i) Let the spheroid be obtained by rotating the ellipse  px2 + zq2   = 1 around the z-axis, where p, q >  0. Then, p is the maximum distance of a point of the spheroid from the z-axis, so the angular momentum Ω of a geodesic must be  p (we can assume that Ω 0). If Ω = 0, the geodesic is a meridian. If 0 < Ω < a, the geodesic is confined to the annular region on the spheroid contained between Ω2 the circles z = q  1  p 2  , and the discussion in Example 9.3.3 shows that the geodesic ‘bounces’ between these two circles (see diagram below).

≥

  ± −

 ≤

94

If Ω = p, Eq. 9.10 shows that the geodesic must be the parallel z = 0. (ii) Let the torus be as in Exercise 4.2.5. If Ω = 0, the geodesic is a meridian (a circle). If 0 < Ω < a b, the geodesic spirals around the torus. If Ω = a b, the geodesic is either the parallel of radius a  b   or spirals around the torus approaching this parallel asymptotically (but never crossing it):

−

0 < Ω < a

−

 −

−b

Ω = a

−b

If  a b < Ω < a + b, the geodesic is confined to the annular region consisting of the part of the torus a distance Ω from the axis, and bounces between the two parallels which bound this region:

−

 ≥

95

If Ω = a + b, the geodesic must be the parallel of radius a + b.

   ±   −  − 1 Ω2

9.3.3 The two solutions of Eq. 9.14 are v = v0

self-intersection is that, for some w > 1, 2 k > 0. This holds

  ⇐⇒  − 2

1 Ω2

1 Ω2

w2 , so the condition for a

w 2 = 2kπ   for some integer

1 > 2π, i.e. Ω < (1 + π 2 )−1/2 . In this case, there

are k self-intersections, where k is the largest integer such that 2kπ < 2

   − 1 Ω2

1.

−i , where z = v + iw. 9.3.4 From the solution to Exercise 8.3.1, Z  = U  + iV  = zz+i This is a M¨obius transformation, so it takes lines and circles to lines and circles and preserves angles (Appendix 2). Since the geodesics on the pseudosphere correspond to straight lines and circles in the vw-plane perpendicular to the vaxis, they correspond in the V W -plane to straight lines and circles perpendicular z −i to the image of the V -axis under the transformation z z+i , i.e. the unit circle 2 2 V  + W  = 1. ¯ + bW   ¯ = c   in the V  ¯ W -plane ¯ A straight line aV  (where   a,b,c   are constants) corresponds to the curve 2aV  + 2bW  = c(V 2 + W 2 + 1) in the V W -plane. If  c = 0 this is a straight line through the origin, which corresponds to a geodesic on the pseudosphere by the first part. If  c = 0 it is the equation of a circle with centre (a/c, b/c) and squared radius (a2 + b2 c2 )/c2 . This circle intersects the boundary circle V 2 + W 2 = 1 orthogonally because the square of the distance between the centres of the two circles is equal to the sum of the squares of their radii. Hence this circle also corresponds to a geodesic on the pseudosphere. This ¯ W -plane ¯ proves that every straight line in the V  corresponds to a geodesic on the pseudosphere. That every geodesic on the pseudosphere arises from a straight ¯ W -plane ¯ line in the V  in this way can be proved by similar arguments, or by ¯ W -plane ¯ noting that there is a straight line in the V  passing through any point  2 2 ¯ + W  ¯ < 1 in any direction and using Proposition 9.2.4. of the disc V 

 →



−

9.3.5 By Proposition 9.3.1(ii), a surface of revolution for which every parallel is a geodesic would have   df/du   = 0 for all values of  u, and hence f   = constant.

96

Then the surface would be a circular cylinder. 9.3.6 By Exercise 6.1.6(iii), the first fundamental form is cosh2 u(du2 + dv 2 ), so the result is a special case of Exercise 9.2.11. 9.3.7 By Clairaut’s Theorem 9.3.2, ρ is constant along every geodesic. It follows that ρ is actually a constant (and hence the surface is a circular cylinder). For if not, then  dρ/dz = 0 for some value of  z, say z = z0 . But then a geodesic passing through a point on the parallel z = z0   and not tangent to that parallel would have non-constant ρ, contrary to hypothesis.

 

9.3.8 By Clairaut’s Theorem, a geodesic on the unit cylinder (for which ρ is a constant) must have ψ  = constant, i.e. the geodesic must intersect the meridians (i.e. the rulings) of the cylinder at a constant angle. By Exercise 4.1.6, the curves on the cylinder that have this property are the straight lines (rulings), circles (parallels) and helices. 9.3.9 The first fundamental form is du2 + k2 cos2 udv 2 . By Clairaut’s Theorem, (k2 cos2 u)v = ˙ Ω, where Ω is a constant. As in Exercise 9.2.11, we find that, along any geodesic that is not a parallel, 2

  dv du

Ω2 = 2 k cos2 u(k2 cos2 u

− Ω2 ) .

γ   makes an angle α with the parallel u = 0 at the point σσ (0, 0) = The geodesic γ (k, 0, 0). Hence, γ  ˙ .  σ v γ = cos α, k which gives v = ˙ cos α/k at (k, 0, 0). On the other hand, v = ˙ Ω/k2 cos2 u = Ω/k 2 at (k, 0, 0). Hence, Ω = k cos α and we get 2

  dv du

cos2 α = 2 k cos2 u(cos2 u

− cos2 α) .

This gives dv k = du Integrating,

sec2 u . 2 2 tan α tan u

± √ 

−

 

tan u kv = ± sin−1 , tan α

97

and hence tan u = tan α sin kv. Since sin kv  1, the maximum value of  u along the geodesic is α, so the maximum height above the xy-plane is

 ±

 |

α

    − 1

|≤

k2 sin2 θdθ.

0

9.3.10 (i) We note that all geodesics γγ  (t) on the pseudosphere, except meridians, are defined only for t in some finite interval α t β , say, whereas meridians are γ  γ (t)) is a geodesic defined on the defined on a semi-infinite interval. Since f (γ γ  (t), it follows that f   takes meridians to meridians, i.e. if  v is same   interval as γ constant, so is v˜. Hence, v˜ does not depend on w. (ii) f  preserves angles and takes meridians to meridians, so must take parallels to parallels. Hence, w does ˜ not depend on v. (iii) The parallel w  = constant has length 2π/w  by Exercise 8.3.2(i) (w = e−u ). As f  preserves lengths, part (ii) implies that 2π/w = 2π/ w, ˜ so w = w. ˜ σ(u, v)) = σ σ(F (v), w) for some smooth function F (v). (iv) We now know that f (σ 2 σ (F (v), w) is w−2 dF  dv 2 + dw2 ; since f   is an The first fundamental form of σ dv

 ≤  ≤

 



isometry, this is equal to w −2 (dv2 + dw2 ), hence dF/dv = 1, so F (v) = v + α, where α is a constant. If the sign is +, f  is rotation by α around the z-axis; if  the sign is , f  is reflection in the plane containing the z-axis making an angle α/2 with the xz-plane.

±

±

 −

9.4.1 From Exercise 6.2.1, the cone is isometric to the ‘sector’ vertex at the origin and angle π 2:

√ 

 S   of the plane with

Geodesics on the cone correspond to possibly broken line segments in : if a line segment meets the boundary of    at a point A, say, it may continue from the point B  on the other boundary line at the same distance as A from the origin and with the indicated angles being equal:

 S 

S 

98

(i) TRUE: if two points P  and Q can be joined by a line segment in   there is no problem; otherwise, P  and Q can be joined by a broken line segment satisfying the conditions above:

S 

To see that this is always possible, let p1 , p 2 , q 1  and q 2  be the indicated distances, and let R and S   be the points on the boundary of the sector at a distance ( p2 q 1  + p1 q 2 )/( p2  + q 2 ) from the origin. Then, the broken line segment joining P  and R followed by that joining S  and Q is the desired geodesic. (ii) FALSE:

(iii) FALSE: many meet in two points, such as the two geodesics joining P  and Q in the diagram in (ii). (iv) TRUE: the meridians do not intersect (remember that the vertex of the cone has been removed), and parallel straight lines that are entirely contained in

 S 

99

do not intersect. (v) TRUE: since (broken) line segments in   can clearly be continued indefinitely in both directions. (vi) TRUE: a situation of the form

S 

in which the indicated angles are equal is clearly impossible. But the answer to this part of the question depends on the angle of the cone: if the angle is α, instead of  π/4, lines can self-intersect if   α < π/6, for then the corresponding sector in the plane has angle < π:

9.4.2 We consider the intersection of S 2 with the plane passing through p  and q and making an angle θ with the xy-plane, where π/2 < θ < π/2. This intersection is a circle θ  but it is not a great circle unless θ = 0. Hence, if θ = 0, the short segment of  θ   joining p  and q   is not a geodesic and so has length > π/2 (the length of the shortest geodesic joining p  and q). Since the length of  θ is 2π, the length of the long segment of  θ  joining p  and q  has length < 3π/2 if  θ = 0, i.e. strictly less than the length of the long segment of the geodesic 0   joining p  and q. So the long geodesic segment is not a local minimum of the length of  curves joining p  and q.

 −

 C  C

 

 C  ≤    C

 C

2

2

9.4.3 (i) This is obvious if  n 0 since e−1/t 0 as t 0. We prove that t−n e−1/t 0 as t  0 by induction on n   0. We know the result if  n = 0, and if  n > 0

 →

≥

 ≥

→

→

→

100

we can apply L’Hopital’s rule: limt→0

t−n  = e1/t2

limt→0

nt−n−1 2 1/t2  = 3e t

limt→0

n t−(n−2) 2 e1/t2 ,

which vanishes by the induction hypothesis. (ii) We prove by induction on n that θ is n-times differentiable with dn θ = dtn



P n (t) 1/t2 e 3n t

−

0

if  t = 0, if  t = 0,



where P n  is a polynomial in t. For n   = 0, the assertion holds with P 0 = 1. n+1 2 P n′ 3nP n n e−1/t if  t = Assuming the result for some n 0, ddtn+1θ = −t3n+1 + t3n + t2P  3n+3



≥

0, so we take P n+1 = (2 3nt2 )P n +t3 P n′ . If  t = 0, P n (0) limt→0

−1/t2

e t3n+1  =

−

dn+1 θ dtn+1  =

limt→0



P n (t) 1/t2 e 3n+1 t

−



=

0 by part (i). Parts (iii) and (iv) are obvious.

9.4.4 For the circular cylinder, the answers are: (i) true; (ii) false; (iii) false; (iv) true; (v) true; (vi) false; (vii) true. For the sphere: (i) true; (ii) false; (iii) false; (iv) false; (v) true; (vi) false; (vii) true. R

σ r   = 1, so 0 σ r .σ σ r dr = R. Differentiating with 9.5.1 Since γγ  θ is unit-speed, σ r .σ R σ rθ  dr = 0, and then integrating by parts gives respect to θ gives 0 σ r .σ

 

σr r=R σ θ .σ r=0

|

    − R

σ rr  dr = 0. σ θ .σ

0

Now σ (0, θ) = p   for all θ, so σ θ = 0  when r   = 0. So we must show that the ¨ θ , the dot denoting the integral in the last equation vanishes. But, σ rr = γγ  derivative with respect to the parameter r of the geodesic γγ  θ , so σ rr  is parallel σ rr  = 0. The first N  = 0, it follows that σσ θ .σ to the unit normal N of  σ ; since σ θ .N σr  = 1 and σ σ r .σ σ θ = 0. fundamental form is as indicated since σσr .σ

    1 0

    ≥

1 γ  between p  and q  is f ˙2 + Gg˙ 2 dt f ˙2 dt = 9.5.2 (i) The length of the part of γγ  0 f (1) f (0) = R. (ii) Use the hint, noting that the length of the part of  γγ   between p  and q′  is R. (iii) If the part of  γγ    between p  and q   has length R, then it must stay inside the geodesic circle with centre p  and radius R  by (ii), and then we must have 1 1  ˙ 2 dt. Then Gg = f ˙2 + Gg˙ 2 dt = 0 f  ˙ 0 for all t (0, 1), so g = ˙ 0 (as G > 0) 0 and so g is a constant which must be α as γγ   passes through q . This means that γγ   is a parametrization of the radial line θ = α.

−

   

   

≥

∈

9.5.3 The first fundamental form is dr2 + Gdθ 2 and a geodesic circle is a parameter curve r = constant. By Exercise 7.3.15, its geodesic curvature is G r /2G. We are

101

given that this is a function of  r only, say A(r). Then, ∂  (ln G) = A(r) ∂r ln G =

∴

∴

where f (r) = e

 

A(r)dr

 

A(r)dr + B(θ) (say)

G = f (r)g(θ),

and g (θ) = eB(θ) .

Chapter 10

10.1.1 The matrix of the Weingarten map with respect to the basis σ u , σ v is I −1 II  = −1 cos2 v 0 cos2 v 0 = = I , so Nu = σ u , Nv = σ v . Thus, N = 0 1 0 1 σ a, where a  is a constant vector. Hence, σ a = 1, showing that the surface is an open subset of the sphere   of radius 1 and centre a . The standard latitudelongitude parametrization σσ (u, v) of  S 2 has first and second fundamental forms both given by du2 + cos2 udv 2 , so the parametrization σ (v, u) + a of    has the given first and second fundamental forms (the second fundamental form changes sign because σ v σ u  = σ u σ v ).



 −

−

− S 



{

} F  F 

 −

 − 

 S 

 ×

− ×

10.1.2 Γ122  = sin u cos u and the other Christoffel symbols are zero; the second Codazzi– Mainardi equation is not satisfied. 10.1.3 The Christoffel symbols are Γ111  = 0, Γ211  = 1/w, Γ112 = 1/w, Γ212  = 0, Γ122  = 0, Γ222 = 1/w. Using the first equation in Proposition 10.1.2 we get K  = 1. The Codazzi-Mainardi equations are Lw = (L + N )/w, N v   = 0. Hence, N  depends only on w, and since 1 = K  =   LN/EG, we have LN  = 1/w4 so L also depends only on w; the first Codazzi-Mainardi equation gives dL/dw = L/w + 1/Lw5 , which is the stated differential equation. Putting P  = Lw 2 we get dP/dw = (1 + P 2 )/wP  which integrates to give 1 + P 2 = Cw2 , where C > 0 Cw2 1/w. Hence, the second fundamental form is is a constant, i.e. L = C −1/2 . only defined for w C −1/2 or w

−

 −

 −

 −

−

 ≥

√   ±

 −

 −

−  ≤ −

The first fundamental form is this exercise is the same as that of a suitable parametrization of the pseudosphere (see Exercise 8.3.1(i)). We saw that the pseudosphere corresponds to (part of) the region w > 1. 10.1.4 The Christoffel symbols are Γ111 = E u /2E , Γ211 = E v /2G, Γ112 = E v /2E , Γ212 = Gu /2G, Γ122 = Gu /2E , Γ222 = Gv /2G. The first Codazzi–Mainardi equation is LE v E v L N  1 Lv = N  , = E v + E  G 2E  2G 2

−

−

−

− 





102

and similarly for the other equation. Finally,



E v L N  (κ1 )v = + 2E  E  G

−



LE v E v N  = E 2 2E  G

 −



L E v = (κ2 E  2E 

− κ1 ),

and similarly for (κ2 )u . 10.1.5 If   E, F, G, L, M,N   are constant, the Christoffel symbols are all zero so Eqs. 10.1 are obviously satisfied, and the Gauss equations (Proposition 10.1.2) are satisfied K  = 0, i.e. LN  = M 2 . Of course, we must also have E > 0, G > 0 and EG F 2 > 0. a b E F  L M  Let C  be the constant matrix , and let I  = , II  = . c d F G M N  u, v˜) leads to the By Exercises 6.1.4 and 7.1.3, the reparametrization (u, v) (˜ stated first and second fundamental forms if and only if 

⇐⇒ −

 

t

C 

F I C  = I ,

  F  

F 

 →

t

C 

F II C  =

  κ 0 0 0



.

To prove the existence of an invertible matrix C   satisfying these conditions, note first that since I    is symmetric, there is an orthogonal matrix P   such λ 0 that P t I P  = , where λ, µ  are the eigenvalues of  I   (see Appendix 0 µ 0). Since the trace E  + G  and the determinant EG F 2 of  I   are both > 0, it follows that the sum and product of  λ, µ are both > 0, and hence that λ, µ λ−1/2 0 t t are both >  0. Letting A = − 1/2 , we have A P  I P A = I . The µ 0 t t matrix A P  II P A is symmetric, so there is an orthogonal matrix Q such that κ1 0 Qt (At P t II P A)Q = , for some (constants) κ1 , κ2 . The matrix C  = 0 κ2 κ1 0 P AQ has the property that C t I C  = I , C t II C  = . 0 κ2 By Proposition 8.2.6, the principal curvatures are the roots of 

 F 

F 

F 

F 

−







F 



F 

 F   F 



F 



F II  − κF I ) = 0.

det(

Since C   is invertible, these are the same as the roots of  det(C t

F II C  − κC tF I C ) = 0,

i.e. κ1 , κ2 . Since K  = κ 1 κ2 , one of  κ1 , κ2  is zero, and we might as well assume 0 1 that κ2  = 0 (if  κ1  = 0, we can replace C  by C  ). This gives the stated 1 0 result, with κ1 = κ.

 

103

If  κ = 0, the surface has the same first and second fundamental forms as a plane, so by Theorem 10.1.3 the surface can be obtained from a plane by applying a direct isometry of  R3 . It follows that the surface is a plane. κ −1/2 v we can assume that κ = 1. If κ = 0, then by the reparametrization v By Examples 6.1.4, 7.1.2 and 10.1.4, in both cases there is a parametrization of  the unit cylinder with these first and second fundamental forms. By Theorem 10.1.3 again, the surface is obtained by applying to this cylinder a direct isometry of  R3 , and so is a circular cylinder. E v E v Gu 2 1 2 u 10.1.6 (i) The Christoffel symbols are Γ111 = E  2E , Γ11 = 2G , Γ12 = 2E , Γ12 = 2G , Gv 2 u L = N  = 0, M  = 1) now state that Γ122 = G 2E  , Γ22 = 2G . Eqs. 10.1 (with (E/G)u  = 0 and (E/G)v  = 0. Hence, E/G  is constant. (ii) If  E/G = λ4 , the reparametrization u˜ = λu, v˜ = λ−1 v   leads to the first ˜ (d˜ u2 + d˜ v 2 ) and 2d˜ ud˜ v, respectively, where and second fundamental forms E  2 − ˜ = λ E . E  (iii) If  E  = G, the first of the Gauss equations in Proposition 10.1.2 gives



 → | |

±

 −

 −

− i.e.

1 = E 

2

2

2

            − − − −     E v 2E 

v

E u 2E 

+

u

E u 2E 

E u 2E 

+

u

E v 2E 

E v 2E 

=

v

+

E v 2E 

E u 2E 

2

,

2 , E 

which is the stated equation. 10.1.7 The assumption is that L = N  = 0. The Codazzi-Mainardi equations are

−M u = M (Γ212 − Γ111 ),

M v = M (Γ222

− Γ112). 10.1.8 (i) By Proposition 8.4.1, there is a patch σ (U, V ) of  S   containing p  such that

F  = M   = 0. If κ1 = L/E  is a constant κ, Exercise 10.1.4 implies that E V  = 0 (since the principal curvatures are distinct by assumption). So E  is a function E dU , v = V . Then the first and second fundamental of  U   only. Define u = forms are of the form du2 + Gdv2 and κdu2 + N dv 2 , respectively. (ii) The Christoffel symbols are Γ 111 = Γ211 = Γ112   = 0, Γ212 = Gu /2G, Γ122 = Gu /2G, Γ222 = G v /2G. (iii) The parameter curve v = constant is parametrized by Γ(u) = σσ (u, v). De¨ = σ uu = κN noting   d/du  by a dot, we have Γ˙ = σ u , Γ ... N, by Proposition 7.4.4 Nu = κ2 σ u . Since and part (ii). By the proof of Proposition 8.1.2, Γ = κN σ u , σ v / G, N  is a right-handed orthonormal basis of  R3 ,

  √ 

− {

 −

√  }

 Γ˙ × Γ¨  =  √ κ σ v  = |κ|. G  Γ˙ 3

104

The torsion of Γ is ... ¨) σ v / G) Γ . (Γ˙ Γ ( κ2 σ u )..( κσ = = 0, ¨ 2 κ2 Γ˙ Γ

×

−

 × 

√ 

−

since σσ u.σ v  = 0. It follows that Γ  is a circle of radius r = 1/ κ . σ uuu = κN σ uu + r12 u is a function of  v only. Hence, σ σ (u, v) = Nu = κ2 σ u , so σ (iv) σ u u γγ  (v) + C(v)cos r + D(v)sin r   for some smooth functions C , D, γγ   of  v. We may assume that the curve γγ   is unit-speed, since replacing v  by a smooth function of  v   does not change the coefficients   E,F,L,M  (as F  = M   = 0). By (iii), σ (u, v) 2 = r2 for all u, v. This implies that C C   and D   are perpendicular vectors of length r. Let c  = r −1C , d  = r −1D. (v) Using the fact that c(v) and d(v) are perpendicular unit vectors for all values of  v, we get γ  dγ γ   u  u dcc σ u .σ v = . c sin + d cos + r .d dv r r dv γ   γ   dγ dγ for all u, v. Hence, .c  = .d  = 0 (= dc .d). It follows that, for some angle α

||

−





−

dv



dv

dv

(possibly depending on v), we have c  = n cos α + b sin α,

±d = −n sin α + b cos α,

where n   and b   denote the principal normal and binormal of  γγ  , respectively. γ   dγ (The condition dv .d  = 0 implies that dα/dv = τ , where τ  is the torsion of  γγ .) Writing s = v, θ = α κu shows that σ  is a reparametrization of the tube of  γ   (Exercise 4.2.7). radius r around γ 10.1.9 Since p−1 Σu , q −1 Σv and r−1 Σw  are perpendicular unit vectors,

−

±

Σu  + bΣ Σv  + cΣ Σw , Σuu = aΣ for some scalars a, b, c. We have  p2 a = Σuu.Σu  =

1 1 Σu .Σu )u = ( p2 )u = ppu , (Σ 2 2

so a = p u /p. Next, Σu.Σv )u q 2 b = Σuu.Σv  = (Σ

− Σu.Σuv = − 12 (ΣΣu.Σu)v = − ppv ,

so b =  ppv /q 2 . Similarly, c =  ppw /r2 , hence the stated formula for Σuu . The other formulas for the second derivatives of Σ  are proved similarly. Σuu )v = We now use these formulas to compute both sides of the equation (Σ Σuv )u ; both sides are then linear combinations of Σ u , Σ v  and Σw . We find that (Σ

−

−

106

(iii) Using the formula for K   in Corollary 10.2.3(ii) and the expression for the first fundamental form of  σ  in Exercise 9.5.1, we get

 

T 

α

Kd σ =

A

f (θ)

0

0

α

=

0

∂  G ∂r

∂ 2

√ G √ 

1 G ∂r 2

    − √  √      −

Gdrdθ

r=f (θ)

α

dθ =

ψ ′  +

√  ∂  G

0

r=0

√ 

  

√  −

∂r

  

dθ.

r=0

By Exercise 10.2.2, G = r +o(r) so ∂  G/∂r = 1 at r = 0, where o(r)/r 0 as r 0. Hence, ABC  K d σ = ψ(α) ψ(0)+α = γ  (π β ) +α = α +β +γ  π.

 

 →

A

 →

− −

−

10.2.5 With the notation of Example 4.5.3 we have, on the median circle t = 0, σt =

−

 θ sin  cos θ, 2

−

 θ  θ sin  sin θ, cos 2 2

hence E  = 1, F  = 0, G = 1 and N  = σ tt = 0, σ tθ =

−

−

cos



, σθ = ( sin θ, cos θ, 0),

θ 2  cos

1  θ  θ  cos  cos θ + sin  sin θ, 2 2 2

−

−

θ,

− sin

θ 2  sin θ,

1  θ  cos  sin θ 2 2

−

− sin

θ 2



;

 θ sin  cos θ, 2

−

1  θ  sin 2 2

giving L = 0, M  = 12 . Hence, K  = (LN  M 2 )/(EG F 2 ) = 1/4. Since K  = 0, the Theorema Egregium implies that the M¨ obius band is not locally isometric to a plane.

−

−

−

 

10.2.6 The catenoid has first fundamental form cosh2 u(du2 + dv 2 ) and its Gaussian curvature is K  = sech4 u (Exercise 8.1.2). If  f  is an isometry of the catenoid, let σ(u, v)) = σ σ (˜ f (σ u, v˜). By the Theorema Egregium, sech4 u = sech4 u˜, so u ˜  = u; reflecting in the plane z = 0 changes u  to u, so assume that the sign is +. Let v˜ = f (u, v); the first fundamental form of   σ (u, f (u, v)) is

−

 −

(cosh2 u + f u2 )du2 + 2f u f v dudv + f v2 cosh2 udv 2 .

 ±



,

107

Hence, cosh2 u = cosh2 u + f u2 , f u f v  = 0 and f v2 cosh2 u = cosh2 u. So f u = 0, f v = 1 and f  = v + α, where α  is a constant. If the sign is + we have a rotation by α about the z-axis; if the sign is  we have a reflection in the plane containing the z-axis and making an angle α/2 with the xz-plane. 10.2.7 Using Corollary 10.2.3(i), we find that

 ±

 ±

 −

1 K  = 4



m(2 m) n(2 n) + un v 2 u2 v m

−

−



.

Hence, the surface is flat if and only if (m, n) = (0, 0), (2, 0), (0, 2) or (2, 2). If  m = n = 0 the first fundamental form is du2 + dv 2 so the surface is locally isometric to a plane (Corollary 6.2.3). If (m, n) = (2, 0), we get v 2 du2 + dv 2 , which is the first fundamental form pf a cone (Example 6.1.5), and this is locally isometric to a plane by Exercise 6.2.5. Similarly if (m, n) = (0, 2). The case in which m = n  = 2 is less obvious, but the reparametrization U  = 2−1/2 uv, V  = ln uv  transforms the first fundamental form v 2 du2 +u2 dv 2 into dU 2 +U 2 dV 2 , which is the first fundamental form of a cone. 10.2.8 Using the Christoffel symbols calculated in Exercise 7.4.4, the second of the Gauss equations (Proposition 10.1.2) gives K  cos θ = 0

− (θu cot θ)v + 0 − (−θucosec θ)(−θv cosec θ) = −θuv cot θ,

hence the result. 10.2.9 By Exercise 6.2.5 and the Theorema Egregium, K  = 0 for a generalized cylinder and a generalized cone. But K  = 0 for a sphere so by the Theorema Egregium again, there can be no local isometry between a sphere and a generalized cylinder or cone. ˜ are 10.2.10 The first fundamental forms of   σ and σ

 

  1+

1 u2

du2 + u2 dv 2 and du2 + (u2 + 1)dv 2 ,

σ˜ (u, v) is not an respectively. Since these are different, the map σ (u, v) isometry. Nevertheless, by Corollary 10.2.3(i), the Gaussian curvatures are equal:

→ 

˜ = K  = K 

∂  1 2 u2 + 1 ∂u

− √ 

 √   2u 1 + u2

=

− (1 +1u2)2 .

1 σ (u, v) σ˜ (˜ u, v˜) is an isometry, the Theorema Egregium tells us that (1+u If σ 2 )2 1 = ˜ = u; l e t v˜ = f (u, v). The first fundamental form of  u2 )2 , s o u (1+˜ σ˜ ( u, f (u, v)) is (1 + (1 + u2 )f u2 )du2 + 2(1 + u2 )f u f v dudv + (1 + u2 )f v2 dv 2 ; this

− ±

→

±

−

108

is equal to the first fundamental form of   σ (u, v) 1 + (1 + u2 )f u2  = 1 + 1/u2 , f u f v  = 0 and (1 + u2 )f v2 = u 2 . The middle equation gives f u  = 0 or f v  = 0, but these are both impossible by the other two equations. Hence, the isometry does not exist. 10.2.11 From the solution of Exercise 8.1.9, the Gaussian curvature of the torus (in the parametrization of Exercise 4.2.5) is

⇐⇒

K  =

cos θ 1 = 2 b(a + b cos θ) b

 − b2 (a +ab cos θ) .

˜ ϕ) σ (θ, If  σ (θ, ϕ) ˜ is an isometry of the torus, the Theorema Egregium implies θ, and hence θ˜ = θ   (up to adding integer that we must have cos θ   = cos ˜ multiples of 2π). Since the first fundamental form is b2 dθ 2 + (a + b cos θ)2 dϕ2 , ϕ/∂θ = 0, ∂ ˜ ϕ/∂ϕ = 1. Hence, we must therefore have dϕ˜2 = dϕ 2 , and hence ∂ ˜ ϕ = ϕ + constant. ˜ 10.2.12 Since all parameter curves are pre-geodesics we have, by Exercise 9.2.8,

→

±

±

±

EE v  + F E u = 2EF u , GGu  + F Gv = 2GF v . If the parameter curves intersect orthogonally, F   = 0 so these equations give E v = Gu  = 0, so the first fundamental form is E (u)du2 +G(v)dv2 . ReparametrizE (u)du, V  = G(v)dv, the first fundamental form becomes ing by U  = 2 2 dU  + dV  . Hence, the surface is locally isometric to the plane, and so is flat by the Theorema Egregium. The result is not true without the assumption of orthogonality. Example 10.4.3 gives a parametrization σ (u, v) of  S 2 (actually of a hemisphere) with the property that the pre-geodesics on S 2 correspond exactly to the straight lines in the uvplane. In particular, all the parameter curves are pre-geodesics.

  

  

10.3.1 Arguing as in the proof of Theorem 10.3.4, we suppose that J  attains its maxiσ  of  . We can assume mum value > 0 at some point p   contained in a patch σ that the principal curvatures κ1 and κ2 of  σ   satisfy κ1 > κ2 >   0 everywhere. Since H  = 12 (κ1  + κ2 ), κ1 > H  and J  = 4(κ1 H )2 . Thus, J   increases with κ1 when κ1 > H , so κ1  must have a maximum at p, and then κ2 = 2H  κ1 has a minimum there. By Lemma 10.3.5, K  0 at p, contradicting the assumption that K > 0 everywhere. 10.3.2 We start with the parametrization σ (U, V ) = (f (U )cos V, f (U )sin V, g(U )), where f (U ) = e U , g(U ) = 1 e2U dU . The first and second fundamental forms are U  dU 2 + e2U dV 2 and −e 2U  dU 2 + eU  1 e2U dV  2 , respectively. In the notation

∈ S 

S 

−

 ≤

  √  −

√ 1−e

√  −

√  − √  −

 −

√  −

of the proof of Proposition 10.3.2, κ1 = 1/eU  1 e2U , κ 2 = e−U  1 e2U . So we are in case (ii) of the proof and tan ω = e−2U  1. We find that e(U ) =

−

109

dU  ˜ = V  and U  ˜ = E/ sin2 ω = 1−1e2U  , g(V ) = G sec2 ω   = 1. So V  = 1−e2U  ˜ + c)). Hence, θ = cosh−1 (e−U ) c for some constant c. So U  = ln(cosh(U  ˜ + c) 1 = 2tan−1 sinh(U  ˜ + c). 2ω   = 2tan−1 e−2U  1 = 2tan−1 cosh2 (U 

−

√ −

−

˜ + V ), ˜ v = 1 (V  ˜ Finally, u = 12 (U  2

 −

 

 √ 

−

˜ so U  ˜ = u − v.  − U ),

√  −

10.3.3 We have E  = G = 1, F   = cos θ and θuv   = sin θ. Hence, EG F 2 = θuv . Consider the quadrilateral bounded by the parameter curves u = u0 , u = u1 , v = v 0 and v = v 1 , and let α1  be the angle between the parameter curves α2  be the angle between the parameter curves α3  be the angle between the parameter curves α4  be the angle between the parameter curves

u = u 0 u = u 0 u = u 1 u = u 1

and v = v 0 , and v = v 1 , and v = v 0 , and v = v 1 .

The area of the quadrilateral is v1

u1

    v0

v1

θuv dudv =

u0

 

(θv (u1 , v)

− θv (u0, v))dv v = θ(u1 , v1 ) − θ(u1 , v0 ) − θ(u0 , v1 ) + θ(u0 , v0 ) = α 4 − (π − α3 ) − (π − α2 ) + α1 = α 1  + α2  + α3  + α4 − 2π. 0

˜ be a local diffeomorphism that takes unit-speed geodesics to unit10.4.1 Let f  : γ (t) speed geodesics. Let p   and let 0 = v T p . There is a unique geodesic γγ  ˙  (0) = v/ v . Then γγ   is unit-speed so γγ  ˜  = f  γγ   on   such that γγ (0) = p and γγ  ˜˙   = Dp f (v v/  vv ) is a unit vector, i.e. is a unit-speed geodesic. In particular, γγ  v) = v . This means that Dp f  : T p  Dp f (v  T f (p)  ˜  is an isometry, so f  is a local isometry.

S → S 

 S 



∈ S 

 ∈ S   

 ◦

    S→ S 

    

10.4.2 Local isometries take geodesics to geodesics by Corollary 9.2.7. If we apply a v, where a = 0 is a constant, to a surface, the first fundamental av dilation v form is multiplied by a2 so the Christoffel symbols are unchanged (see Proposition 7.4.4). By Proposition 9.2.3, the geodesic equations are unchanged. It follows that dilations take geodesics to geodesics. Hence, any composite of local isometries and dilations also takes geodesics to geodesics.

 →

 

The converse is false: the map from the xy-plane to itself given by (x, y) (x, 2y) takes geodesics to geodesics (as it takes straight lines to straight lines) but is not the composite of a dilation and a local isometry.

→

10.4.3 (i) This is true because F   is conformal.

110

σ (u, v0 ) is a geodesic on σ   for any fixed v0 by (ii) The parameter curve u σ . Since F  is a geodesic local diffeomorphism, construction of the geodesic patch σ σ(u, v0 )) is a pre-geodesic on σ˜ . Hence, for some smooth function u(t), u F (σ σ(u(t), v0 )) is a geodesic on σ˜ . The second geodesic equation in Theorem t F (σ 9.2.1 gives λv ˙u2 = 0, so λv  = 0 and λ is independent of  v. σ (u(t), v(t)); we can assume that γ γ   is unit-speed. Since the first (iii) Let γγ (t) = σ fundamental form of  σ is du2 + Gdv 2 , the parameter curves v  = constant and σ u and u = constant intersect orthogonally and unit vectors parallel to them are σ ˙   and the curve v = constant σ v / G, respectively. If the oriented angle between γγ  − 1/2 σu  + G σ v = uσ ˙   = cos θσ is θ, we have γγ  sin θσ ˙ σ u  + vσ ˙ σ v . Hence, u = ˙ cos θ and sin 1 θ v˙ = √ G . The first geodesic equation in Theorem 9.2.1 gives u¨ = 2 Gu ˙v 2 , i.e. Gu ˙v dθ θ˙ θ˙ sin θ = 12 Gu v˙ 2 . Substituting for v gives ˙ = dv v˙ = 2 sin θ = G u /2 G. (iv) Apply (iii) to F  γγ   and use the fact that F   is conformal. (v) Parts (iii) and (iv) imply (λG)u = λGu , hence λu G = 0, hence λu  = 0, i.e. λ is independent of  u. By (ii), λ is constant. (vi) If  Dλ−1/2  is the dilation by a factor λ−1/2 , the composite Dλ−1/2 F   preserves the first fundamental form and so is a local isometry, say . Then, F  = Dλ1/2 .

 →

 → →

√ 

√ 

 ◦

G

◦

◦G

Chapter 11 11.1.1 Let l meet the real axis at b and suppose that Re(a) > b (the case Re(a) < b is similar). The semicircle with centre c on the real axis and radius a d  passes d b , i.e. provided through a   and does not meet l  provided that a d 2 2 d (a b )/(Re(a) b).

≥||−

 | − | ≤ | − |

−

 | − |

11.1.2 Suppose that a  and b  lie on a half-line geodesic, say a = r + is,b = r + it, where t d r,s,t R and t > s. Then, dH (a, b) = s dw w   = ln(t/s) = d, say, so t/s = e .

 

 ∈

On the other hand, the formula in Proposition 11.1.4 gives 2tanh−1 2tanh−1

  ed 1 ed +1

−

= 2tanh−1 (tanh d2 ) = d.

  t s t+s

−

=

11.1.3 The required hyperbolic line cannot be a half-line, so must be a semicircle with centre the origin, and it must have radius a .

 | |

11.1.4 By Proposition 11.1.4, z

 

⇐⇒   2tanh−1 zz−−aa¯ = R. If  λ   = tanh(R/2),

∈ Ca,R a − λ2 a)z − (a − λ2 a z  + (1 − λ2 )|a|2 = 0. this is equivalent to (1 − λ2 )|z |2 − (¯ ¯)¯ According to Proposition A.2.3, this is the equation of a circle provided that λ2 < 1, which is obvious, and |a ¯ − λ2 a|2 > (1 − λ2 )2 |a|2 . This condition reduces to 2|a|2 > a2 + ¯a2 . Writing a = | a|eiθ this becomes cos 2θ < 1, which is true because a ∈ H  implies 0 < θ < π. Cic,R   will be a circle with centre on the imaginary axis, say at ib. Then Ca,R intersects the imaginary axis at the points i(b ± r), so these two points must

111

be a hyperbolic distance 2R   apart, i.e. 2R   = 2tanh−1 2ir 2ib . This gives r = b+r 1 b tanhR, which is equivalent to R = 2  ln b−r . Next, the points i(b ±  r) are −ic i(b−r)−ic the same hyperbolic distance from ic, so i(b+r) i(b+r)+ic = i(b−r)+ic . This gives

    

    

(b + r c)(b r + c) = (c + r b)(c + r + b), which simplifies to c2 = b 2 r2 . θ, and denoting d/dθ by a dot, the Parametrizing ic,R  by v = r cos √  θ, w = b+r sin 2π v˙ 2 +w˙ 2 2π dθ = 0 b+rrsin θ dθ = √ b2πr circumference of  ic,R is 0 = 2π sinh R. 2 −r2 w2

−

−

−

C

 C The area inside Cic,R is =

2π r sin θ dθ = 0 b+r sin θ

 

−

   

= Cic,R dv w   by Green’s theorem, which 2π = 2π(cosh R 1).

int(

√ b2πb 2 −r 2  −

Cic,R )

dvdw w

−

−

The circumference is 2π sinh R = 2π(R+ 61 R3 +o(R3 )) and the area is 2π(cosh R 1 R4 + o(R4 ))). Since K  = 1 these formulas are consitent with 1) = 2π( 21 R2 + 24 those in Exercise 10.2.7. (See Exercise 10.2.4 for the o( ) notation.) Suppose that l is the imaginary axis for simplicity. If  a = reiθ , with 0 < θ < π/2, the distance of  a from l is the distance of  a  from the point ir on l. From the proof of Proposition 11.1.4, this distance is lntan θ2 . Hence, the points at a fixed distance from l are those with a constant value of  θ, i.e. the straight halflines passing through the origin. Of course, there are two such half-lines for any given distance, each being the (Euclidean) reflection of the other in l. These lines are not geodesics (unless the distance is zero, in which case they coincide with l), as they are not perpendicular to the real axis. The region (u, v) u < 0, π < v < π   in the parametrization of the pseuπ < v < π, w > 1 in , a dosphere corresponds to the region (v, w) semi-infinite rectangle in the upper half of the vw-plane. By Proposition 11.2.3, there is an isometry that takes a to i and b to ir, say, where r >  0. Since isometries leave distances unchanged, we need only prove the result for a = i, b = ir. Assume that r > 1 (the case r < 1 is similar). Then, dH (a, b) = ln r. On the other hand, if γγ  γ (t) = v(t) + iw(t) is any curve in γ  (t0 ) =  i, γ γ  (t1 ) = ir, say, the length of the part of  γγ   between a and b is with√ γ t1 v˙ 2 +w˙ 2 t1 w˙ r dw dt dt = r. w 1 w = ln t0 t0 w By applying an isometry we can assume that l  is a half-line, and then the result was proved in Exercise 11.1.1. By applying an isometry, we can assume that l   is the imaginary axis. Then, m  must be the semicircle with centre the origin and radius a . Let a = ρeiθ , where ρ > 0, π < θ < π, and let c = it, t >   0. Since tanh−1 x   is a strictly −it > a−iρ if  t = ρ. The increasing function of  x, we have to show that aa+it a+iρ

−

−

11.1.5

 −

11.1.6

11.2.1

 {

|

−

{

}

|−

}

H

 H

11.2.2 11.2.3

 

≥

−

 

 

 | |

    2

 

−sin θ , and the difference is 2(ρ−t)2 sin θ , which is second expression equals 11+sin |a+it| 1+sin θ θ > 0 if  t =  ρ. 11.2.4 (i) If  a , let l′ and m′   be the unique hyperbolic lines passing through a and perpendicular to l and m, respectively (Exercise 11.2.3). Let b and c be

 ∈H

112

the intersections of  l′ and m′   with l and m, respectively. We are given that F (b) = b and F (c) = c, so F (l′ ) = l ′ and F (m′ ) = m ′ as l ′ and m ′  are the unique hyperbolic lines passing through b and c and perpendicular to l and m. Since a is the unique point of intersection of  l′ and m′ , we must have F (a) = a. (ii) Either F  or 0,1 F   fixes l, m  and the interior of the semicircle m. Next, either F , 0,1 F , 0 F  or 0 0,1 F   fixes l, m, the interior of the semicircle Re(z) > 0  to the right of  l. Let G be this m  and the region >0 = z isometry. Then, G fixes each point of  m  because there is a unique point of  m at any given distance d > 0 from i in the region >0 . Similarly, G  fixes each point of  l. Hence, G is the identity by (i). (iii) Let F  be any isometry of  . By the proof of Proposition 11.2.3, there is an isometry G  that is a composite of elementary isometries and which takes F (i) to i and F (l) to l. Then, G F  is an isometry that fixes l and i. As m is the unique hyperbolic line intersecting l  perpendicularly at i, G F   fixes m. By (iii), G F  is one of four composites of elementary isometries. It follows that F  is a composite of elementary isometries. (iv) By (iii) it suffices to prove that every elementary isometry is a composite of reflections and inversions in lines and circles perpendicular to the real axis. For reflections and inversions there is nothing to prove, so we need only consider translations and dilations. But, if  a R, a = 0 1 , where 1  is the reflection in the line Re(z) = a/2; and if  a > 0, a = 0 , where   is inversion in the circle with centre the origin and radius a. 11.2.5 (i) This is obvious from the proof of Proposition A.1.2(ii). ad−bc (ii) If   a,b,c,d R, a calculation shows that Im(M (z)) = |cz+d|2 Im(z). If   ad  bc >   0 this is >   0 whenever Im(z) >   0. Conversely, suppose that M  takes   to itself. Assume that c = 0 and d = 0 (the cases in which c = 0 or d   = 0 are similar but easier). Then, M   must take the real axis to itself  (as it must take the lower half-plane   to itself), i.e. az+b R if  z R. cz+d R, say. Letting z R, Taking z   = 0 gives b/d = λ   gives a/c = µ λ−µ say. Then, az+b R, cz+d = µ + dc z+1 . This is real whenever z  is real, so c/d = ν  say. Hence,   a,b,c,d are, up to an overall multiple, equal to the real numbers µν,λ,ν, 1, respectively. The condition ad bc > 0 now follows from the previous calculation. (iii) Following the proof (and the notation) of Proposition A.2.2, we have M  = b/d a/d if  c = 0 (and then d = 0), while if c = 0,

 I   ◦  I  ◦  R ◦  R ◦I  ◦  H  {  ∈ H |

}  H

 H ◦

◦

◦

∈ T  R ◦R  D √   I  ◦ I 

 −

R  I 

∈

 H



∈



 −H

→∞

∈

∈  ∈  ∈

−

T  ◦ D

  M  = T a/c ◦ D(ad−bc)/c  ◦ (−K ) ◦ T d/c . Hence, it suffices to show that −K   is a composite of elementary isometries. But −K  = R0 ◦ I 0,1  in the notation of Exercise 11.2.4. (iv) J   is reflection in the imaginary axis and hence an isometry of  H , so the 2

result follows from (i).

113

(v) This follows from the fact that, if  M  is a real M¨ obius transformation, so is J  M  J . For example, if M 1 , M 2  are real M¨ obius transformations, then

 ◦  ◦

(M 1 J ) (M 2 J ) = M 1 (J  M 2 J )

◦ ◦

◦

◦  ◦

◦

is a composite of real M¨ obius transformations, hence a real M¨obius tranformation by (i). (vi) By (v) and Exercise 11.2.4(iii), it suffices to prove that every elementary isometry of  is a M¨ obius isometry. If  a R, a  is real M¨ obius and a = 2a J . If  a > 0 then a  is real M¨ obius. Finally, if  a R, r > 0 then

H  D

∈ T  ∈  I a,r = T a ◦ Dr  ◦ I 0,1 ◦ T −a

R

T  ◦

2

(see the proof of Proposition 11.2.1), so it suffices to prove that 0,1  is a M¨obius isometry; but 0,1  = ( K ) J , where K (z) = 1/z is real M¨ obius.

 I 

− ◦

 −

 I 

−

11.2.6 By Proposition 11.1.3(i), there is a unique hyperbolic line l, say, passing through a and b. By Proposition 11.2.3, by applying a suitable isometry of    we can assume that l is the imaginary axis. Let and ′  be the hyperbolic circles with centres a and b, respectively, which pass through c (Exercise 11.1.4). These are Euclidean circles with centres on the imaginary axis.

 C

 H

 C

Suppose for a contradiction that the stated inequality is false, and let a = iw, b = iw′ , where w′ > w   (interchanging the roles of  a, b if necessary). Then,  intersects the imaginary axis at a point ir, say, with r > w; we must have r < w′   since r w′  implies that d(a, b) d(a,ir) = d(a, c), which contradicts our assumption. Similarly, ′  intersects the imaginary axis at a point is, say, between a and b. Since and ′   intersect, we must have s r; in fact, s < r for if  s =  r  then and ′  touch, contradicting the fact that a, b, c do not lie on a geodesic. Then,

C

 ≥

 C

 C  C

 C

 ≤

 C

 ≤

d(a, c) + d(c, b) = d(a,ir) + d(is,b) > d(a,ir) + d(ir,b) = d(a, b), a contradiction. 11.2.7 Let F   be an isometry of    that takes l   to the imaginary axis. By Exercise 11.1.5, the equidistant curves are then a pair of Euclidean half-lines l 1 , l 2  passing through the origin. The isometry F −1 of  , being a composite of reflections and inversions (Exercise 11.2.4), takes Euclidean lines and circles to Euclidean lines and circles (Appendix 2). Since the real axis is the only Euclidean straight line passing through a and b, F −1 must take l1 , l2  to a pair of circular arcs passing through a and b. We observed in Exercise 11.1.5 that these are not geodesics.

 H

H

11.2.8 By applying a suitable isometry, we can assume that a and d are on the imaginary axis. The centre of the semicircle geodesic passing through a and b is (R, 0),

114

where a = iR cot 12 α, and its radius is Rcosec 12 α. Let d = it. We must have Rcosec 12 α < R + t. Then, d(a, d) = ln



 

R  1 cot α t 2

< ln

cot 12 α cosec 12 α 1

−



= f (α).

11.2.9 As in the proof of Theorem 11.2.4, we can assume that a = a′ , that a, c and c′   are all on the imaginary axis, that b and b′   are on the same side of the imaginary axis, and that c and c′  are either both ‘above’ or both ‘below’ a on the imaginary axis. Then, clearly, c = c′ . Since the angles of the two triangles at a are equal, the geodesic through a and b is the same as that through a and b′ . Since d(a, b) = d(a, b′ ) and since b and b′  are on the same side of the imaginary axis, we must have b = b ′ . 11.2.10 The proof is the same as in Euclidean geometry. Let a, b , let d   be the mid-point of the geodesic l  passing through a and b, and let m be the geodesic passing through d   perpendicular to l. If  c   is any point on m   then, by the preceding exercise, the triangles with vertices a, d, c and b, d, c are congruent. In particular, d(a, c) = d(b, c). Hence, m  is the set of points equidistant from a  and b.

 ∈ H



−

−

−

−



1 1 11.3.1 The distance we want is 2tanh−1 P  1 (b)−P  1 (a) . Now use  P −1 (z) = P  (b)−P  (a)

The algebra is straightforward.

z+1 i(z 1) .

−

11.3.2 By Proposition 11.2.3, there is an isometry F   that takes l to the real axis and the point of intersection of  l and m to the origin. Then F   must take m to the imaginary axis as this is the unique hyperbolic line through the origin perpendicular to the real axis. The number of such isometries is the number of isometries that take the real axis to itself and the imaginary axis to itself. If  G is such an isometry, then either G, 0 G, 1 G or 1 0 G fix the real and imaginary axes and also each quadrant into which the disc P  is divided by the axes. If  H  is this isometry, the argument used in the solution of Exercise 11.2.4(ii) shows that H  must be the identity map. Hence, G = 0 , 1 , 0 1  or the identity map.

 R ◦  R ◦

 R ◦R ◦  D R  R  R ◦ R

11.3.3 Let us call a M¨ obius transformation of the type in the statement of the exercise a   hyperbolic  M¨ obius transformation. Since   is a M¨obius transformation, the M¨ obius transformations that take P    to itself are those of the form M  −1 , where M   is a M¨ obius transformation that takes  to itself, i.e. a real M¨ obius az+b R and transformation (Exercise 11.2.5). If  M (z) = cz+d , where   a,b,c,d

 D

ad

 P 

 H

 P  P  ∈

−c))z+a−d−i(b+c) . Since − bc > 0, we find that P M P −1 (z) = (a+d+i(b − −i(b−c) d+i(b+c))z+a+d (a

|a + d + i(b − c)|2 − |a − d − i(b + c)|2 = 4(ad − bc) > 0,

115

P M P −1 is hyperbolic.

Conversely, we have to show that if  M  is a hyperbolic M¨ obius transformation, then −1 M    is a real M¨ obius transformation. The calculation is similar to that already given.

P 

P 

11.3.4 By Exercise 11.2.5(iii), the isometries of   are of the form M  or M  J   where M  is real M¨obius and J (z) = z¯. Hence, the isometries of  P  are M  −1 and J  −1 . But M  −1 is hyperbolic and J  −1 (z) = (M  J ) −1 = M  −1 z+1 z¯+1 −1 (¯z )) = z¯. J  i(z −1) = i(¯ z −1) = (

 H P  P 

 ◦  P  P  P  P 

−  D P  ◦ P  P  P  ◦P  P  P  P  P  P  11.3.5 We know that every isometry of  H  is the composite of reflections Ra  and inversions I a,r  with a ∈ R, r > 0 (Exercise 11.2.4(iv)). Now, I a,r = T a ◦Dr ◦I 0,1 ◦T −a and any translation T a  (a ∈ R) is the composite of reflections R0 ◦Ra/2 . It therefore suffices to show that, if  F  is any isometry of  H  of the form Ra (a ∈ R), Da (a > 0) or I 0,1 , then P ◦ F  ◦P −1 is a composite of isometries of  DP  of the types in Proposition 11.3.3. We find that (a) if a   = 0,  P ◦ Ra ◦ P −1 = I b,r , where −1 is reflection in the real axis; (c) P◦I 0,1 ◦P −1 b = 1+ia ia , r = 1/|a|; (b) P◦R0 ◦P  is reflection in the imaginary axis; (d) if  a > 0,  P ◦ Da ◦ P −1 is the composite of  two inversions of the type in Proposition 11.3.3(i), namely I c,√ c −1 ◦ I b,√ b −1 ,

   

2

2

2

where b, c are real numbers such that b2 > 1, c2 > 1 and a = f (c)/f (b) where f (x) = xx+1 −1 . (We can take b to be any real number > 1 and distinct from 1/f (a), then choose c =  f (af (b)); then, f (c) = f (f (af (b))) = af (b) using the property f (f (x)) = x for all x = 1.)



11.3.6 Suppose first that γ  = π/2. The cosine rule gives cosh C  = cosh A cosh B and cosh A = cosh B cosh C  sinh B sinh C  cos α. Eliminating cosh C  gives cos α = cosh A sinh B/ sinh C . Hence,

 −

sin2 α sinh2 C   = sinh2 C 

 − cosh2 A sinh2 B = cosh2 A cosh2 B − 1 − cosh2 A sinh2 B = cosh2 A − 1 = sinh2 A,

so

sin α sinh A

=

1 sinh C  .

Interchanging the roles of  A and B  gives

sin β sinh B

=

1 sinh C  .

In the general case suppose that the hyperbolic line through the vertex of the triangle with angle α intersects the opposite side at a point which divides that side into segments of lengths A′ and A′′ , so that β   is the angle between the sides of lengths C  and A′ . Suppose also that this hyperbolic line segment has length D. Then the original triangle is divided into two triangles, one with angles π/2, β , α′  and sides of lengths C, D, A′  and the other with angles π/2, γ , α′′ and sides of lengths B , D, A′′ . Applying the first part to each of these triangles gives sin β/ sinh D = 1/ sinh C   and sin γ/ sinh D = 1/ sinh B. Hence, sin β/ sinh B = sin γ/ sinh C . The other equation is proved by interchanging the roles of  A and B (for example). The case in which the hyperbolic line through a vertex meets

116

the hyperbolic line through the other two vertices in a point outside the triangle is similar. 11.3.7 (i) This was established in the solution of the preceding exercise. B B cosh C −cosh A α =   coshsinh (ii) Using the sine and cosine rules, sin β  =   sinh . sinh C  and cos B sinh C  2 α   cosh B cosh C −cosh A cosh A−cosh A Hence, cos =   cosh B sinh = cosh A. 2B sin β = sinh2 B (iii) Using sin β  = sinh B/ sinh C  and the cosine rule for cos β , we get cot β  = cosh A cosh C −cosh B   sinh2 A cosh B   sinh A = sinh A sinh B sinh A sinh B = tanhB . 11.3.8 If  γ  = π/2 the formula we want is that in Exercise 11.3.7(ii). In the general case, we use the method (and notation) of the solution of Exercise 11.3.6. In the case where the hyperbolic line through a vertex perpendicular to the opposite side meets that side at a point inside the triangle, applying Exercise 11.3.7(ii) to the two right-angled triangles gives cosh A′  = cos α′ / sin β , cosh A′′  = cos α′′ / sin γ , so cosh A = cosh(A′  + A′′ ) = cosh A′ cosh A′′ + sinh A′ sinh A′′ cos α′ cos α′′   tanh2 D = + sin β sin γ  tan β tan γ  using Exercise 11.3.7(iii). By Exercise 11.3.7(ii), cosh D =

cos β sin α′

=

cos γ  sin α′′

so

cosh A sin β sin γ   = cos α + sin α′ sin α′′  + tanh2 D cos β cos γ  = cos α + sech2 D cos β cos γ   + tanh2 D cos β cos γ  = cos α + cos β cos γ. The case in which the perpendicular meets the opposite side at a point outside the triangle is similar. γ  γ (t) = (x(t), y(t), z(t)) be a curve on S 2 . Then, Π(γ γ (t)) = (u(t), v(t)), where 11.3.9 Let γγ  (1−z)x+x ˙ z˙ y x u = 1−z , v = 1−z . Denoting d/dt by a dot, u = ˙ with a similar formula (1−z)2 for v, ˙ which give u˙ 2 + v˙ 2 1 4 = (1 1 + u2 + v 2 (1 z)2



− z)2 (x˙ 2 + y˙ 2) + (x2 + y2)z˙2 + 2(xx˙ + yy)˙ z(1 ˙ − z)



. − y = Using x 2 +y 2 = 1 − z2 , which implies x x+y ˙ ˙ −zz, ˙ this expression simplifies to 2 2 2 x˙ +y˙ +z˙ . Hence, √  the length of Π◦γγ  calculated using the given first fundamental 2

2

+v˙ form on R2 is 21+uu˙ 2 +v 2 dt = Π is an isometry.

 

  

x˙ 2 + y˙ 2 + z˙ 2 dt, which is the length of γγ  γ . Hence,

−1 (l1 ) and −1 (l2 ) are geodesics in . By 11.3.10 Since : P  is an isometry, Proposition 11.2.3, there is an isometry F  of   that takes −1 (l1 ) to −1 (l2 )

 P   H → D

 P 

 P   H

 P 

 H  P 

117

and −1 (z1 ) to and z1 to z2 .

 P 

 P −1 (z1 ). Then, P F P −1 is an isometry of  DP   that takes l 1 to l 2

11.3.11 Draw the radii of  P  given by arg(z) = 2πk/n with k = 0, 1, . . . , n 1, and let a0 , a1 , . . . , an−1   be the points at a Euclidean distance r (with 0 < r < 1) from the origin on these radii. For each value of  r, we then have a regular hyperbolic n-gon with vertices a0 , a1 , . . . , an−1 . It is clear that the interior angle α of this n-gon is an increasing function of  r   and depends continuously on r. When r tends to zero, the area of the n-gon also tends to zero, so α must approach its Euclidean value (Theorem 11.1.5), namely (n 2)π/n. On the other hand, as r tends to 1, α tend to zero. By the intermediate value theorem, when 0 < r < 1, α take every value in the interval 0 < α < (n 2)π/n. By Exercise 11.3.8,

 D

−

− −

2 cos 2π n  + cos cosh A = sin2 α2 ∴

α 2

  2cos2 πn 1 + cosh A = sin2 α2 cos nπ  A ∴   cosh . = 2 sin α2

11.3.12 Let D be the length of the diagonal of the quadrilateral, and let β   be the angle between the diagonal and the side of length B. By Corollary 11.3.7, cosh D = cosh A cosh B, and by the cosine rule (Theorem 11.3.6), cosh C   = cosh A cosh D

− sinh A sinh D sin β.

By the sine rule (Exercise 11.3.6), sinh A = sinh D sin β  so cosh C   = cosh A cosh D

− sinh2 A = cosh2 A cosh B − sinh2 A.

By the sine rule, the other two (equal) angles α of the quadrilateral are given by sin α =

 sinh D cos β  . sinh C 

By the cosine rule, cosh A = cosh B cosh D

− sinh B sinh D cos β.

118

Hence, sin α =

  cosh B cosh D cosh A  cosh A sinh B . = sinh B sinh C  sinh C 

−

11.3.13 Let n be one of the geodesics bisecting the angle between l and m, and let a be a point on n. We will show that a is equidistant from l and m. By applying an isometry we can assume that n is the real axis in P . Let j and k be the geodesics through a meeting l and m  at right angles, and let b and  c  be the points in which j meets l and k  meets m, respectively. Reflection R in the real axis is an isometry of  P  that interchanges l and m  and fixes each point of  n. If  R(c) =  b, the hyperbolic triangle with vertices a, b and R(c) would have two angles equal to π/2, contradicting Theorem 11.1.5. It follows that R interchanges b and c, and hence that d(a, b) = d(a, c). Consider a hyperbolic triangle with vertices a, b, c and let l, m, n be the geodesics bisecting the angles at a, b, c, respectively. Let d  be the point of intersection of  l and m. Then, d is equidistant from all three sides of the triangle, and hence lies on n. 11.3.14 By applying an isometry of  P , we can assume that l is the imaginary axis, m is the real axis, a = 0 and c = 1. By Proposition 11.1.4, the hyperbolic circle with centre b > 0 and radius R is

 D

 D



 D

|z − b|  = tanh R . |1 − bz| 2 If this circle passes through the origin, b = tanh R 2  so the hyperbolic circle is

|z − b| = b|1 − bz|, which can be written as

|z|2 − 1 +b b2 (z + z¯) = 0. As b

→ 1 this tends to the circle |z|2 = 12 (z + z¯), i.e.

 − z

1 2



2

=

1 , 4

which is a circle Γ touching  at 1. ˜ that intersects  perpendicularly at 1 also intersects Γ perpendicularly A circle Γ ˜ is also at right angles. at 1, and hence the other intersection of Γ and Γ

C

 C

11.4.1 Let l and m be two distinct hyperbolic lines in  that do not intersect at any point of  . If  l and m are both half-lines they are parallel as they do not have

 H

 H

119

a common perpendicular. If at least one of  l and m is a semicircle, then l and m   are parallel if they intersect at a point of the real axis, and ultra-parallel otherwise. 11.4.2 We work in  and assume that l is the imaginary axis (by applying a suitable isometry). If  a = v + iw, the semicircle geodesic through a   intersects l at −v−iw i v 2 + w2 = ir, say. The distance of  a  from l  is 2tanh−1 ir ir−v+iw . Setting this

 H

√ 





equal to a constant, say D, gives (after some algebra) v 2 /w2 = 2sinh2 (D/2). This is the equation of a pair of lines passing through the origin. As they are not perpendicular to the real axis (unless D = 0), they are not hyperbolic lines. 11.4.3 We work in P . By applying an isometry we can assume that a is the origin and b > 0. Suppose that the hyperbolic triangle with vertices a, b, c has internal angles α, β,γ , and assume that Re(c) > 0. The hyperbolic line through b and c is part of a circle Γ with centre d and radius r, say. The line through b and d  makes an angle π/2 β   with the real axis, so d = b + r sin β  + ir cos β . Similarly, d = c + r sin(α + γ ) ir cos(α + γ ) = c + r sin(A + β ) + ir cos(A + β ) using A = π α β  γ . Writing c = v + iw we get v = b r sin A cos β + r(1 cos A)sin β ,  w = r(1 cos A)cos β + r sin A sin β . Then, v(1 cos A) + w sin A = (b + 2r sin β )(1 cos A). But, since γ   intersects the boundary of  P  perpendicularly, r 2 + 1 = d 2 = (b + r sin β )2 + r2 cos2 β  = A r 2 + b2 + 2br sin β , so 2br sin β  = 1 b2 . Hnece, v(1 cos A) + w sin A = 1−cos . b This is the equation of a straight line that interescts the real axis at 1/b and makes an angle A/2 with the (negative) real axis. The set of points c for which the triangle with vertices a, b, c has area A is the union of this line together with its reflection in the real axis. These lines are not hyperbolic lines as they do not pass through the origin. 11.4.4 In the first two parts of this exercise it is slightly easier to work in (by applying −1 ). (i) By applying a suitable isometry of  , we can assume that one of the sides of the triangle is the imaginary axis. Then the other sides must be a semicircle R, together with the half-line passing through the origin and some point a Re(z) = a. By further applying a suitable translation parallel to the real axis and a positive dilation (which are isometries of  ), we can assume that the three sides are the half-lines Re(z) = 1 and the semicircle z 2 = 1. Then, the area of the triangle is

 D

−

− −

− − − − − || −

−

C D

−

−

H

P 

 H

 ∈ H

 ±

1

 | |

∞ dw 1 dv √  dv √  = = π. 2 1−v 2 w −1 −1 1 − v 2

   

 

(ii) Let a be the vertex of the biasymptotic triangle in , and let l be the geodesic through a  perpendicular to the opposite side, meeting it at  a ′ , say. By applying an isometry of  , we can assume that l is the imaginary axis and that a′ = i.

H

 H

120

Then, the side opposite a is the semicircle z 2 = 1. The side joining 1 and a is an arc of a circle with centre b < 0, say, where

 | |

cos

 α b . = 2 1 b

− −

The equation of the circle is (v b)2 + w2 = (1 b)2 , so the area of the triangle is 1 1 1 dv. 2 2 v 1 v)(1 v (1 + 2b) 0

−

   √ 

−

−  −

  −



−

The second integral can be evaluated using the substitution u = v gives the area as 1 b

π

−

du (1 b)2

    − − 2

−b

− u2

 − b.

du = 2 sin−1 = π

−b 1−b

− 2cos−1

−  b

1

−b

= π

This

− α.

That a bisymptotic triangle with angle α   exists for any 0   < α < π  is proved using the method of Exercise 11.3.11. (iii) We now switch back to P . Assume that the vertex at which the angle is α is 0, let b be the other vertex of the triangle in P . Let the side of the triangle passing through 0 and b meet at c, and let the other side passing through 0 meet at d (these sides are radii of  ). The area of the asymptotic triangle with vertices 0, b , d  is the difference between the areas of the biasymptotic triangles with vertices 0, c , d and b, c, d, which by part (ii) is

 D

 C

 C

(π

 D

 C

− α) − (π − (π − β )) = π − α − β.

Let a be a point on the radius joining 0 and d   and consider the triangle with vertices 0, a , b. Let the angle of this triangle at b be β ′  and the angle at a  be α′ . Thus, α ′  tends to zero and β ′  tends to β  as a tends to d along the radius (‘tends to’ in the Euclidean sense). By Exercise 11.3.8, the length A of the side joining 0 and b  is given by cos α′  + cos α cos β ′ . cosh A = sin α sin β ′ Letting a tend to c, we obtain cosh A =

 1 + cos α cos β  . sin α sin β 

121

11.4.5 (i) Let a, b P    be the vertices of one of the triangles, let l   be the geodesic passing through a, b and let m be the other side of the triangle passing through a; define a ′ , b′ , l′ , m′  similarly for the other triangle. By applying an isometry to one of the triangles, we can assume that a = a′ , m = m′  and that l and l′ are on the same side of  m. As the triangles have the same angle at a, l = l ′  and by Exercise 11.4.4(iii) the lengths of the finite sides of the two triangles are equal. It follows that b = b ′ , and since the two triangles have the same angle at b  their third sides must also coincide. (ii) As in (i), by applying an isometry to one of the triangles we can assume that their vertices in P    coincide, as do the two sides passing through this vertex. But then the two triangles coincide. (iii) This was established in the solution of Exercise 11.4.4(i).

 ∈ D

 D

11.4.6 We consider the biasymptotic triangle with vertices 0, 1 and e3iπ/4 , where the two sides meeting at the origin are radii of  P  and the side joining 1 and e 3iπ/4 is an arc of a circle meeting  at right angles at 1 and e3iπ/4 . The ‘altitude’ of  this triangle passing through 1 is a circular arc that meets at i, and that through e3iπ/4 is a circular arc meeting at e5iπ/4 . These two geodesics are clearly ultraparallel (one is confined to the region v > 0, the other to the region v < 0). If we now replace the vertices 1 and e 3iπ/4 with nearby points a, b P , the altitudes of the hyperbolic triangle with vertices 1, a , b passing through a and b will still be ultraparallel.

 D

 C

 C  −

 C

∈D

11.5.1 From Example 6.3.5, Π−1 (v, w) = 2(v,w) v 2 +w 2 +1 .



v 2 +w2 1 2v 2w , , 2 2 2 2 v +w +1 v +w +1 v 2 +w 2 +1



− , so K(v, w) =

11.5.2 From Appendix 2, every M¨ obius transformation is a composite of transformations z + λ, z λz (where λ = 0 in the last case). Hence, it of the form z 1/z, z suffices to establish Eq. 11.11 when M  is of this form. For the last two types this (a−1 −c−1 )(b−1 −d−1 ) is obvious; for the first, (a−1 , b−1 ; c−1 , d−1 ) = (a −1 −d−1 )(b−1 −d−1 ) = (a, b; c, d) on multiplying numerator and denominator by   abcd. This argument is only valid provided none of   a,b,c,d is 0 or , but a similar argument works in the other cases. For example, if  a = but   b,c,d = 0, we have to show that 1 −1 −1 − (0, b ; c , d ) = ( , b; c, d). This is proved by multiplying numerator and c−1 (b−1 −d−1 ) denominator of (0, b−1 ; c−1 , d−1 ) = − −d−1 (b−1 −c−1) by bcd.

 →

 →

∞

 →



 ∞ ∞



C∞   is a bijection satisfying Eq. 11.11, let  b,c,d C∞ be If  M  : C∞ such that M (b) = 1, M (c) = 0, M (d) = . Then, M (z) = (M (z), 1; 0, ) = (M (z), M (b); M (c), M (d)) = (z, b; c, d), so M  is a M¨obius transformation.

→

∈

 ∞

∞

, b′ = 0, c′   = 1. For if  M  11.5.3 It is enough to prove the existence when a′ = and M ′   are M¨obius transformations taking (a,b,c) and (a′ , b′ , c′ ) to ( , 0, 1), −1  M   is a M¨obius transformation taking (a,b,c) to (a′ , b′, c′ ). But then M ′ M (z) = (a, b; c, z) is a M¨ obius transformation that takes (a,b,c) to ( , 0, 1). For

∞

◦

∞

∞

122

the uniqueness, note that if  M  is a M¨ obius transformation that takes (a,b,c) to ( , 0, 1), then M (z) = ( , 0; 1, M (z)) = (a, b; c, z).

∞ a; b, −1/¯b) = 11.5.4 (a, −1/¯

∞

− − a1¯ + 1¯b ) − a¯1 −b) =

(a b)( (a+ 1¯b )(

(a b)(¯ ¯) b a (1+¯ ab)(1+a¯ b)

 K R

K

−

−

=

−   a b ab 1+¯

−

2

=

− tan2 12 d by

Proposition 6.5.2. 11.5.5 The reflection in the line through the origin making an angle θ with the real axis 2iθ z¯ 2iθ is (z) = e2iθ z¯. Then, ( (z)) = |2e (z) = ( (z)). z |2 +1 = e

 R

RK

11.5.6 This follows from Exercises 11.3.5 and 11.5.5 and Proposition 11.5.4. 11.5.7 Since = pr Π−1 , and since Π is conformal (Example 6.3.5), it suffices 2 to show that pr : −   preserves the angles between curves intersecting at pr(0, 0, 1) = (0, 0, 0). But the first fundamental form of the patch 2 pr−1 (x,y, 0) = (x,y, 1 x2 y 2 ) of  − is 1 ((1 y 2 )dx2 + 2xydxdy + (1 x2 )dy 2 ), 2 2 y 1 x which reduces to dx2 + dy 2 when x = y = 0. 11.5.8 Suppose first that a, b, c, d lie on a Circle . Let M  be the M¨obius transformation such that M (a) = , M (b) = 0, M (c) = 1 (Exercise 11.5.3), and let M (d) = z. Now M ( ) is a Circle ˜, say (Proposition A.2.4); since ˜ passes through it must be a straight line, and as this line passes through 0 and 1 it must be the real axis; hence z R. But (a, b; c, d) = ( , 0; 1, z) = z  (Exercise 11.5.2). Conversely, assume that (a, b; c, d) R, and let M  be as above. Then, M (d) R and 0, 1, , M (d) lie on a Circle , namely the real axis. But then a, b, c, d lie on the Circle M −1 ( ). 11.5.9 Let M  be the M¨obius transformation taking , 0, 1 to a, b, c, respectively. Then, M (d) = λ. By Exercise 11.5.2, if we subject  a,b,c,d  to any permutation, the resulting cross-ratio will be the cross-ratio of the points obtained by subjecting , 0, 1, λ to the   same   permutation. It therefore suffices to check that the six stated values are the values of the cross-ratios of the points , 0, 1, λ taken in any order. This is straightforward.

K

 ◦ S  → D − − − − − − −

  C

C

 ∈

∞

−

C

∞

C

 S 

∈  C

C

 ∞

∞

∈

∞

∞

 ∞

Chapter 12 κ21 = 0 12.1.1 κ1  + κ2 = 0 = κ2 = κ1 = K  = κ1 κ2 = κ21 0. K  = 0 κ1 = κ2 = 0  the surface is an open subset of a plane (by Proposition 8.2.9). σ u σ v , where κ1 , κ2   are the 12.1.2 From Eq. 8.15, σ λu σ λv = (1  λκ 1 )(1  λκ 2 )σ principal curvatures of  σ . Since σσ  is minimal, κ2 = κ1 so

⇒ ⇐⇒

Aσ

 −

⇒

 ×

λ

 −  ≤

 −

(U ) =

 

 ⇐⇒

 −

 × − (1 − λ2 κ21 )  σ u × σ v    dudv

U 

=

Aσ (U ) − λ2

 

U 

κ21

  σ u × σ v    dudv.

 ⇐⇒

123

Since the integrand in the last integral is  0 everywhere, the stated inequality follows. Equality holds  the last integral vanishes, which happens the κ1  = 0 everywhere. In that case κ2 = integrand vanishes everywhere, i.e. κ1  = 0 also, and σ  is an open subset of the plane by Proposition 8.2.9. 12.1.3 By Proposition 8.6.1, a compact minimal surface would have K > 0 at some point, contradicting Exercise 12.1.1.

 ≥

 ⇐⇒

 ⇐⇒

⇐⇒

−

12.1.4 The first part follows from Exercises 6.1.2 and 7.1.4. The map which wraps the plane onto the unit cylinder (Example 6.2.4) is a local isometry, but the plane is a minimal surface and the cylinder is not. 12.1.5 If  κ1 , κ2  are the principal curvatures at the point, we have κ1  + κ2  = 0 as the surface is minimal, and κ1  =  κ 2  as the point is an umbilic. Hence, κ1 = κ2 = 0. 12.1.6 Let Σ0 be the minimal surface and let = Σλ in the notation of Definition 8.5.1. Let k1 , k2  be the principal curvatures of Σ. By Proposition 8.5.2(ii), we have

S 

1 1 1 λk1 1 λk2 + = + = κ1 κ2 ǫk1 ǫk2

−

since

1 k1

+

1 k2

−

− 2λǫ ,

= 0 (because k1  + k2  = 0).

12.1.7 The condition for a minimal surface is LG 2M F  + N E   = 0 (Corollary 8.1.3). From the solution of Exercise 8.1.16, E  = (1 κa cos θ)2 +τ 2 a2 , F  = τ a2 , G = a 2 , L = aτ 2 κ cos θ(1 κa cos θ), M  = τ a, N  = a, and we find that the condition for the tube to be a minimal surface is

−

− −

−

a(1

− κa cos θ)(1 − 2a cos θ) = 0,

which obviously cannot hold at all points of the surface. 12.1.8 Let t1 , t2  be orthonormal principal vectors of    at p   corresponding to principal curvatures κ1 , κ2   (Corollary 8.2.2). Then, t = a1 t1  + a 2t2  for some a1 , a2 R, so (t) = κ1 a1 t1 + κ2 a2 t2  (where   is the Weungarten map of    at p). Hence,

 S 

 W 

 W   S  t, t = W (t), W (t) = κ21a21 + κ22 a22 . t, t t, t a21  + a22 But, as S   is a minimal surface, κ2 = −κ1 , so t, t = κ2 = −κ1 κ2 = −K. t, t 1

 ∈

Alternatively, this exercise can be deduced from Exercise 8.1.6. 12.2.1 By the solution of Exercise 8.1.2, the helicoid σσ (u, v) = (v cos u, v sin u,λu) has

124

E  = λ2 + v 2 , F  = 0, G = 1, L = 0, M  = λ/(λ2 + v 2 )1/2 , N   = 0, so H  = LG−2MF +NG = 0. 2(EG −F 2 )

12.2.2 A straightforward calculation shows that the first and second fundamental forms of   σ t are cosh2 u(du2 + dv 2 ) and cos t du2 2sin tdudv +cos t dv 2 , respectively, 2 u+cos t cosh2 u so H  = − cos t cosh = 0. 2 cosh4 u γ  (u) + vaa, 12.2.3 From Example 5.3.1, the cylinder can be parametrized by σ (u, v) = γ γ  is unit-speed, a = 1 and γγ  γ  is contained in a plane Π perpendicular to where γγ  ˙   = t  (a dot denoting d/du), σσ v = a, so E  = 1, F  = 0, G = 1; a. We have σ u = γγ  σ vv = 0, so L = κn n, σ uv = σ n.(tt a), M  = N   = 0. Now N  = t a, σ uu = t˙  =  κn t a  is a unit vector parallel to Π and perpendicular to t, hence parallel to n; κ = 0 γγ   is part of a straight line so L = κ and H  = κ/2. So H  = 0  the cylinder is an open subset of a plane. 12.2.4 The first fundamental form is (cosh v + 1)(cosh v  cos u)(du2 + dv 2 ), so σ is conformal. By Exercise 8.5.1, to show that σ   is minimal we must show that σ vv = 0; but this is so, since σ uu  = (sin u cosh v, cos u cosh v, sin  u2  sinh v2 ) = σ uu +σ σ vv . σ (0, v) = (0, 1 cosh v, 0), which is the y-axis. Any straight line is a geodesic. (i) σ (ii) σ (π, v) = (π, 1 + cosh v, 4 sinh v2 ), which is a curve in the plane x = π such that z2 = 16sinh2 v2   = 8(cosh v   1) = 8(y 2), i.e. a parabola. The d d geodesic equations are dt (E  u) ˙ = 12 E u (u˙ 2 + v˙ 2 ), dt (E ˙v) = 12 E v (u˙ 2 + v˙ 2 ), where a dot denotes the derivative with respect to the parameter t   of the geodesic and E   = (cosh v + 1)(cosh v cos u). When u = π, the unit-speed condition is E ˙v 2 = 1, so v˙ = 1/(cosh v + 1). Hence, the first geodesic equation is 0 = 12 E u v˙ 2 , which holds because E u  = sin u(cosh v +1) = 0 when u = π; the second geodesic d equation is dt (cosh v + 1) = (cosh v + 1) sinh v v˙ 2 = sinh v v, ˙ which obviously holds. σ (u, 0) = (u sin u, 1 cos u, 0), which is the cycloid of Exercise 1.1.7 (in the (iii) σ xy-plane, with a = 1 and with t replaced by u). The second geodesic equation is satisfied because E v   = sinh v(2 cosh v + 1  cos u) = 0 when v   = 0. The unit-speed condition is 2(1 cos u)u˙ 2 = 1, so u˙ = 1/2sin u2 . The first geodesic d d equation is dt (4sin2 u2 u) ˙ = sin uu˙ 2 , i.e. dt (2 sin  u2 ) = cos u2  ˙u, which obviously holds. 12.2.5 Using Exercise 8.1.1, the surface is minimal

−

−

  

× × ± ⇐⇒ −

×

±

⇐⇒

⇐⇒  −

−

−

 −

 −

−

−

−

−

 −

 ⇐⇒

2  ¨ (1 + g ′ )f   + (1 + f ˙2 )g ′′ = 0,

where a dot denotes d/dx and a dash denotes d/dy; hence the stated equation. Since the left-hand side of the stated equation depends only on x and the righthand side only on y, we must have f ¨ = a, 1 + f ˙2

g ′′ = 1 + g ′2

−a,

125

˙ then f ¨ =   rdr/df  and for some constant a. Suppose that a = 0. Let r = f ; the first equation is rdr/df  = a(1 + r 2 ), which can be integrated to give af  = 1 2 2  ln(1+r ), up to adding an arbitary constant (which corresponds to translating e2af  1, which integrates to the surface parallel to the z-axis). So df /dx = give f  = a1  ln cos a(x + b), where b   is a constant; we can assume that b = 0 by translating the surface parallel to the x-axis. Similarly, g = a1  ln cos ay, after translating the surface parallel to the y -axis. So, up to a translation, we have

 

 ±√  −

 −

az = ln

cos ay , cos ax

 

which is obtained from Scherk’s surface by the dilation (x,y,z)  a(x,y,z). If  a = 0, then f ¨ = g ′′  = 0 so f  = b + cx, g = d + ey, for some constants b, c, d, e, and we have the plane z = b + d + cx + ey. 12.2.6 Using Exercise 8.1.1, taking f (x, y) = sin−1 (sinh x sinh y), and writing λ = 1/ 1 sinh2 x sinh2 y, we find that f x = λ cosh x sinh y, f y = λ cosh y sinh x, f xx = λ 3 sinh x sinh y cosh2 y, f xy = λ 3 cosh x cosh y, f yy = λ 3 sinh x sinh y cosh2 x. It is easy to check that (1 + f y2 )f xx 2f x f y f xy  + (1 + f x2 )f yy = 0.

  −

 →

−

2 12.3.1 (i) From the proof of Theorem 12.3.2, the Gauss map is conformal = λ..id, where λ is a smooth function on . If  H  = 0 at p, then H  = 0 on an open 2 +K  subset of    containing p. By Exercise 8.1.6, = λ 2H  .id, so every point of   is an umbilic and  is an open subset of a plane or a sphere by Proposition 8.2.9. Since H  = 0 the planar case is impossible. Part (ii) is now obvious. For (iii), assume that   is not minimal. Then, there is a point p   at which H  = 0, say H  = µ. The argument in (i) and (ii) shows that the set µ  of points of    at which H  = µ is a (non-empty) open subset of  ; it is also a closed subset because H  is a continuous function on . Since   is connected, µ  = . Hence, every point of    is an umbilic, and so   is an open subset of a sphere (the planar case is impossible as µ = 0). 12.3.2 (i) From Example 12.1.4, N = ( sech u cos v, sech u sin v, tanh u). Hence, if  N(u, v) = N(u′ , v ′ ), then u = u′  since u  tanh u  is injective, so cos v = cos v ′ and sin v   = sin v ′ , hence v = v ′ ; thus, N   is injective. If N = (x,y,z), then x2 +y 2 = sech2 u = 0, so the image of N  does not contain the poles. Given a point (x,y,z) S 2 other than the poles, let u = sech−1 x2 + y 2 , the sign being that of  z, and let v be such that cos v = x/ x2 + y 2 , sin v = y/ x2 + y 2 ; then, N(u, v) = (x,y,z). (ii) By the solution of Exercise 8.1.2, N  = (λ2 + v 2 )−1/2 ( λ sin u, λ cos u, v). Since N(u, v) = N(u + 2kπ,v) for all integers k, the infinitely-many points

 S 

 O  S   O  

O

   W 

 

 S 

  S 

S 

∈



 S  S 

−

 S 

 →



 −

 ⇐⇒ W 

 ∈ S   S   S  S 

S 

−

    ±

σ(u, v) + (0, 0, 2kπ) σ (u + 2kπ,v) = σ

 − −

 

−

126

of the helicoid all have the same image under the Gauss map. (Of course, this is geometrically obvious because the helicoid itself is left unchanged by the translation by 2π parallel to the z-axis.) If N  = (x,y,z), then x2 +y 2 = λ 2 /(λ2 +v 2 ) = 0, so the image of N  does not contain the poles. If (x,y,z)  S 2 and x2 + y 2 = 0, let v = λz/ x2 + y 2 and let u  be such that sin u = x/ x2 + y 2 , cos u = y/ x2 + y 2 ; then N(u, v) = (x,y,z).

−

∈  −

   −  

 

 

S 2 is a conformal 12.4.1 By Proposition 12.3.2, if K  = 0 then the Gauss map : p) local diffeomorphism. Let R be a rotation of  R3 about the origin that takes (p 2 to the south pole of  S  (or any point other than the north pole). There is an open subset  of    containing p  such that ( ) does not contain the north pole. By Example 6.3.5, Π R  is a conformal diffeomorphism from  to an open 2 subset U  of  R . The inverse of this diffeomorphism is the desired surface patch σ. 12.4.2 Writing x = r cos θ, y = r sin θ  gives z = aθ, so we have the parametrization σ  is dr 2 +(r2 +a2 )dθ2 , σ (r, θ) = (r cos θ, r sin θ,aθ). The first fundamental form of σ so σσ  is not a conformal parametrization. However, noting that

 O S 

 

 G  S →

G

GO

◦ ◦G

 O

dr r  = sinh−1 , a r 2 + a2

  √ 

we let ϕ = sinh−1 ar  so that r = a sinh ϕ. The reparametrization σ (a sinh ϕ, θ) = (a sinh ϕ cos θ, a sinh ϕ sin θ,aθ) ˜ (θ, ϕ) = σ σ has first fundamental form (r 2 + a2 )(dϕ2 + dθ 2 ) = a 2 cosh2 ϕ(dθ 2 + dϕ2 ), so σ˜  is a conformal parametrization of the surface. 12.5.1 We have σu ϕ  = σ

− iσσv = (1 − u2 + v 2 − 2iuv, 2uv − i(1 − v 2 + u2 ), 2u + 2iv) = (1 − ζ 2 , −i(1 + ζ 2 ), 2ζ ).

So the conjugate surface is, up to a translation, σ˜ (u, v) = Re

  −    −  (i(1

ζ 2 ), 1 + ζ 2 , 2iζ ) dζ 

ζ 3 2 ζ 3 , ζ  + , iζ  = Re i ζ  3 3 v3 u3 2 v + u v ,u+ uv 2 , 2uv . = 3 3

−

−



−

−



127

− v)/√ 2, V  = (u + v)/√ 2, σ˜(U, V ) = σσ(u, v); then, 1 1 1 σ˜ (U, V ) = √  U  − V  + U V 2 − U 2 V  + V 3 − U 3 , 3 3 2

Let U  = (u

 

 √  1 2

2

U  + V  + U V  + U  V 

Applying the π/4 rotation (x,y,z) 1 3 U  3

2

1 3 V  3

+ U V 2 , V 

  →

σ = Re = Re

ζ 2 ), 2i (1

ζ −4 )(1 1 2

ζ 

1 (ζ  6

ζ 3 3

−

 

1 3 U  , U 2 3

 2

− V 



.

√ 12 (x + y), √ 12 (y − x), z to  ˜σ (U, V ) then

+ U 2 V, U 2

  −  −  − − −    − −   − −

gives U  again. 12.5.2 ϕ  = 12 (1

 −

1 3 V  3



− V 2

ζ −4 )(1 + ζ 2 ), ζ (1



, which is Enneper’s surface

− ζ −4 ) , so



ζ −3 i ζ 3 ζ −3  ζ 2 ζ −2 − − 1 1 ζ  + , ζ  + , + ζ  + + 3

2 3 i  1 ζ −1 )3 , (ζ  + ζ −1 )3 , (ζ  + ζ −1 )2 , 6 2



3



2

2



˜ ˜ v . Then, σσ(u, v) = σ˜ (˜ u, v˜), where up to a translation. Put ζ  = eζ , ζ  =u ˜ + i˜

−



4 ˜  4i  cosh3 ζ, ˜ 2 cosh2 ζ  ˜ ˜ (˜ σ u, v˜) = Re  sinh3 ζ, 3 3 1 v (cosh2 u = 4 sinh u ˜ cos˜ ˜ sin2 v˜  sinh2 u˜ cos2 v˜) , 3 1 4 sinh u˜ sin v˜(  sinh2 u ˜ sin2 v˜ cosh2 u ˜ cos2 v˜), 3



−

− 2(cosh2 u ˜ cos2 v˜ − sinh2 u ˜ sin2 v˜)



.

12.5.3 The first part is obvious. σ qv = a(σ σ u iσ σ v ) implies that σ σ a = aσ σ + a, where (i) If  a R, the identity σσau iσ a σ  by applying the dilation Da a  is a constant vector. Hence, σσ is obtained from σ followed by the translation T a  (Appendix 1). σ  (Proposition (ii) If f  and g are the functions in the Weierstrass representation of σ a 12.5.4), those in the Weierstrass representation of  σ are af  and g   (see Eq. 12.23). By Eq. 12.25, replacing f  by af   leaves the first fundamental form σ a (u, v) is an isometry, and by Eq. 12.26 N unchanged, so the map σ (u, v) σ a at corresponding points does not depend on f , so the tangent planes of  σ and σ are parallel. it it σ eu σ ev = eit (σ σu iσ σ v ). Since σ σ (u, v) = cosh u cos v, cosh u sin v, u), iσ 12.5.4 We have σ we get

∈

−

−

 →

−

it

σ eu   = (cos t sinh u cos v it

σ ev

−

− sin t cosh u sin v, sin t cosh u cos v + cos t sinh u sin v, cos t), = (− cos t cosh u sin v − sin t sinh u cos v, cos t cosh u cos v − sin t sinh u sin v, − sin t).

128

Integrating gives it

σ e (u, v) = cos t(cosh u cos v, cosh u sin v, u) + sin t( sinh u sin v, sinh u cos v, v) σ (u, v) + sin tσ ˆ (u, v), = cos tσ

−

−

say (up to a translation). In the notation of Exercise 6.2.3, σ˜ (sinh u, π2 + v) = ( sinh u sin v, sinh u cos v, π2 +v). Reflecting in the xy-plane and then translating ˜ (sinh u, π2 + v) to σˆ (u, v). by π/2 along the z-axis takes σ 12.5.5 (i) ϕ  is never zero since we have arranged that F ′ and G′  are never both zero. Condition (ii) in Theorem 12.5.2 is obvious. ∂  d  ˙ g, ˙   (a (ii) When v = 0, F ′ (z) = ∂u ˙ 0) = γγ  F (u, 0) = F u , etc, so du σ (u, 0) = (f, dot denoting d/du). This proves (ii).  ˙ ˙ f,0) ˙ g, i f   ˙ 2 + g˙ 2 ). Using Eq. 12.26, N (iii) When v = 0, ϕ  = (f, ˙ N = (−g, . Then,

−

 

  −

√ f ˙ +g˙ 2

2

 ˙ 2 + g˙ 2 ) and finally γγ  ˙   = (0, 0, ¨.  (N N γγ  ˙  ) = 0. It follows that γγ   is a f  N γγ  pre-geodesic on σσ  (Exercise 9.1.2). σv = ϕ = (iv) If  γγ   is the cycloid, F (z) = z sin z, G(z) = 1 cos z, so σ u iσ (1 cos z, sin z, 2i sin z). This gives

 ×

 ×

 −

−

−

 −

− cos u cosh v, sin u cosh v, −2cos u2  sinh v2 ),  u  v σ v = (− sin u sinh v, − cos u sinh v, −2sin  cosh ). 2 2

σ u = (1

σ (u, v) = (u sin u cosh v, cos u cosh v, 4sin u2  sinh v2 ), up to Integrating gives σ a translation. Translating by (0, 1, 0) gives Catalan’s surface. ϕ  = ( 21 if (1 g 2 ), 2i if (1 + g 2 ),ifg), which 12.5.6 ϕ  = ( 21 f (1 g 2 ), 2i f (1 + g 2 ), f g) = iϕ corresponds to the pair if  and g. 12.5.7 From Example 12.5.3, for the catenoid we have ϕ1   = sinh ζ , ϕ2 = i cosh ζ , ϕ3  = 1, so f  = ϕ1 iϕ2  = sinh ζ  cosh ζ  = e−ζ , g = ϕ3 /(ϕ1 iϕ2 ) = eζ . Since the helicoid is the conjugate surface of the catenoid (Example 12.5.3 again), the preceding exercise gives f  = ie−ζ , g = eζ  for the helicoid. 12.5.8 By Example 6.3.5, Π(x,y,z) = (x + iy)/(1 z). From Eq. 12.26, (ζ ) = 1 2 1), so |g|2+1 (g + g¯, i(g g¯), g

−

−

−

⇒

 −

−

 − −

 − −  −

− − ||−

π( (ζ )) =

G

−

 −

− −||−

12.5.9 By Eq. 12.22, ϕ  =

−

1 2 (ζ  2



− ζ −2 ), i (ζ 2 + ζ −2 ), 1 , 2

 −

G

g + g¯ + g g¯ = g(ζ ). g 2 + 1 ( g 2 1)

||

 −

129

so the parametrization is, up to a translation, σ (u, v) = Re = Re =

  −

−

ϕ(ζ )dζ  (u + iv)3 6

1 3 (u 6

u iv  i(u + iv)3 i(u iv) , , u + iv 2(u2 + v 2 ) 6 2(u2 + v 2 ) u v  1 2 3 , v + v ,u . ( 3u ) 2(u2 + v 2 ) 6 2(u2 + v 2 )

−

−

− 3uv2) −

−

−

−

−

By Proposition 12.5.5, the Gaussian curvature is

− 64|ζ |6 K  = , (1 + |ζ |4 )4 √  and this tends to zero as |ζ | = u2 + v 2 → ∞.

12.5.10 Using the conformal parametrization in Exercise 12.4.2, we get σθ ϕ  = σ

 − iσσϕ = a(− sinh ϕ sin θ − i cosh ϕ cos θ, sinh ϕ cos θ − i cosh ϕ sin θ, 1) = a(−i cos ζ, −i sin ζ, 1),

where ζ  = θ + iϕ. Then, f  = ϕ 1 g =

− iϕ2 = −iae−iζ ,

ϕ3 = ie iζ . ϕ1 iϕ2

−

By Proposition 12.5.5, K  = = =

iζ  2 − a2|e−iζ 16|2(1|e +| |eiζ |2 )4

−

16e−2ϕ a2 e2ϕ (1 + e−2ϕ )4 16 a2 (eϕ + e−ϕ )4

− = −a−2 sech4 ϕ.

 

130

Chapter 13

 

γ  is a   simple   closed geodesic, Theorem 13.1.2 gives int(γγ  ) Kd = 2π; since 13.1.1 If γγ  K  0, this is impossible. The parallels of a cylinder are not the images under a σ  : U  R3 of a  simple   closed curve π π  in the plane such that int(π π) surface patch σ is contained in U . Note that the whole cylinder can actually be covered by a single patch (see Exercise 4.1.4) in which U  is an annulus, but that the parallels correspond to circles going ‘around the hole’ in the annulus. 13.1.2 Since the unit normal N of  S 2 is equal to n, the geodesic curvature κg   of n n n′ ). Let t  be the arc-length of  γγ   and denote   d/dt by is, up to a sign, n′′. (n b = κ2 + τ 2 = R, say, where t = γγ  ˙  . κtt + τ b a dot. Then, ds/dt = n˙ = −κt+τ b = d b)/R, n n′ = (κb b + τ tt)/R, and n′′ = R1 dt Then, n ′ = ( κtt + τ b R

 ≤

 →

 ±   √ 

A

×     − − × −R−1(κ/R)˙tt + R−1 (τ /R)˙bb − R−2 (κ2 + τ 2 )nn. These formulas give 3 n × n′ ) = −R−2 τ (κ/R)˙ + R−2 κ(τ /R)˙= (κτ ˙ − τ  κ)/R . n′′. (n ˙ −τ κ = ± d  tan −1 τ  . Applying Theorem 13.1.2 Since κ˙ = Rκ′ , etc, κg = ± κτ  κ +τ  ds κ ′





′

2

2

ℓ(n ) κg dt = 0 to the curve n on S 2 , and noting that K   = 1 for S 2 and that 0 n)-periodic function (where ℓ(n n) is the length because κ g  is the derivative of an ℓ(n of the closed curve n), we get that the area inside n is 2π. 13.1.3 Before smoothing K   = 0 so the total curvature is zero. To compute the total curvature after smoothing, we apply Theorem 13.1.2 to the smoothed cone with γγ    being the circle on the cone given by z = a   for some constant a >  0. The radius of this circle is a tan α so its curvature is κ = 1/a tan α. Since the angle of the cone is α, the geodesic curvature of  γγ   is κg = κ sin α = a−1 cos α. The part of the cone with z > a still has K   = 0 so does not contribute to the total curvature. Integrating K  over the part of the cone with z < a, Gauss-Bonnet gives the total curvature as

 

 

Kd

A = 2π

  −

γγ  

κg ds = 2π

− (a−1 cos α)(2πa tan α) = 2π(1 − sin α).

13.2.1 The parallel u = u1   is the circle γγ  1 (v) = (f (u1 )cos v, f (u1 )sin v, g(u1 )); if  s γ  1 ,   ds/dv = f (u1 ). Denote  d/ds   by a dot and   d/du by is the arc-length of  γ γ  ˙   = ( sin v, cos v, 0), γγ  ¨  = f (u1 1 ) (cos v, sin v, 0), and the unit a dash. Then, γ normal of the surface is N = ( g ′ cos v, g ′ sin v, f ′ ). This gives the geodesic ′ γ  1 ) ℓ(γ (u1 ) γ  N  ˙γγ  ) = f f (u γ   κg ds = curvature of  γγ   as κ g =  ¨γγ  .(N  . Since ℓ(γ ) = 2πf (u ), 1 1 ) 0 1 2πf ′ (u1 ). Similarly for γγ  2 . By Example 8.1.4, K  = f ′′ /f , so R Kd σ = 2π u2 f ′′ ′ f ′ (u2 )). Hence, f  fdudv = 2π(f  (u1 ) 0 u1 γ  1 ) γ  2 ) ℓ(γ ℓ(γ κg ds κg ds = Kd σ .

−

×

    −

−

−

 −

 −

−

  0

  

  − 0

  R

A

A

131

This equation is the result of applying Theorem 13.2.2 to the curvilinear polygon shown below.

13.2.2 Applying Theorem 12.2.2 to the curvilinear n-gon, and noting the get γ  γ )) (n 2)π αi . (int(γ

 −K  ≥  1, we

A

≤ −

 − i

If  n 2 the right-hand side is < 0, which is absurd. If  n = 3 the right hand-side is < π. 13.2.3 Consider a small curvilinear quadrilateral on   bounded by parameter curves u = a, u = a + h, v = b, v = b + k, say, where h and k  are small. The internal angles are α, α, π α, π α, where α is the angle between the parameter curves u = a and v = b. Let  be the (small) area of the quadrilateral. Since the sides of the quadrilateral are geodesics, Theorem 13.2.2 gives

≤

 S 

−

 A

−

K (a, b)

A = (4 − 2)π − (α + α + π − α + π − α) = 0.

13.3.1 This can be proved by expressing a, b, c  in component form and computing both sides. Alternatively, one may observe that both sides of the equation are linear in each of a, b, c   (separately), and change sign when any two of a, b, c are interchanged. This means that it is enough to prove the formula when a  = i, b  =  j j, j c  = k, when both sides are obviously equal to 1. 13.3.2 Define ϕk  as in the hint. The stated properties are easily checked.

 

13.4.1 By Corollary 13.4.8, S  K d = 4π(1 g), and g = 1 since   is diffeomorphic to T 1 . By Proposition 8.6.1, K > 0 at some point of  . 13.4.2 K > 0 = S  Kd > 0 = g < 1 by Corollary 13.4.8; since g is a non-negative integer, g   = 0 so   is diffeomorphic to a sphere. The converse is false: for

  ⇒

A S 

A

⇒

−

 S 

 S 

132

example, a ‘cigar tube’ is diffeomorphic to a sphere but K  = 0 on the cylindrical part.

13.4.3 The ellipsoid is diffeomorphic to S 2 by the map (x,y,x) (x/a, y/a, z/b), so the genus of the ellipsoid is zero. Hence, Corollary 13.4.8 gives S  K d = 4π(1 0) = 4π. Parametrising the ellipsoid by σ (θ, ϕ) = (a cos θ cos ϕ, a cos θ sin ϕ, b sin θ) (cf. the latitude-longitude parametrization of S 2 ), the first and second fundamental forms of  σ are

 →

 

−

(a2 sin2 θ+b2 cos2 θ)dθ 2 +a2 cos2 θdϕ2 and respectively. This gives K  =

A

Hence,

S 

     

a2 sin θ + b2 cos2 θ

(dθ 2 +cos2 θdϕ2 ),

a2 sin2 θ + b2 cos2 θdθdϕ. π/2

2π

Kd σ =

A

2

b2 , (a2 sin2 θ + b2 cos2 θ)2

d σ = a cos θ

 

ab

 

A

0

−π/2

ab2 cos θ dθdϕ . (a2 sin2 θ + b2 cos2 θ)3/2

13.4.4 The surface in Exercise 5.4.1(ii) is diffeomorphic to S 2 , hence its Euler number is 2. 13.5.1 Assuming that every country has  6 neighbours, the argument in the proof of  E  + E  Theorem 13.5.1 gives E  3F  and 2E  3V , so V  E  + F  2E  3 3 = 0, contradicting V  E  + F  = 2.

 ≥

 −

 ≥

 ≥

 −

 ≤

−

13.5.2 3F  = 2E   because each face has three edges and each edge is an edge of two faces. From χ = V   E  + F , we get χ = V   E  + 23 E , so E  = 3(V   χ).

−

−

−

133

Since each edge has two vertices and two edges cannot intersect in more than one vertex, E  12 V (V  1); hence, 3(V  χ) 12 V (V  1), which is equivalent to V 2  7V  + 6χ   0. The roots of the quadratic are 12 7 49 7χ , so 1 1 V  49 7χ or V  49 7χ . Since χ = 2, 0, 2, . . . , the 2 7 2 7+ first condition gives V    3, which would allow only one triangle; hence, the second condition must hold. ˜  takes an n-coloured map on to 13.5.3 This is obvious since a diffeomorphism an n-coloured map on ˜. 13.5.4 Since three edges meet at each vertex and each edge has two vertices, we have 3V  = 2E . As each edge is an edge of two countries,

 ≤  − − √  ≥ ≤ − − ≥ ≤





 −  ≤ √  −



 −

 ± √  − 



−

 S → S 

S 

2E  =

∞



 S 

ncn .

n=2

Hence, E  =

1 2



ncn , V  =

∞

χ =



n=2

1 3



1 n 3

 −

ncn , and since F  =

−



∞

cn  we get

1 1 n + 1 cn = (6 2 6 n=2

n)cn .

13.5.5 (i) In the notation of the preceding exercise, for a soccer ball we have cn = 0 unless n = 5 or 6 so the formula in the preceding exercise gives c5  = 12 (since χ = 2 for a sphere). (ii) This time we are given that cn  = 0 unless n = 4 or 6, so the formula gives 2c4  = 12. 13.6.1 The circle θ = θ0   is a circle in the plane z = b sin θ0   with centre on the zaxis, so its principal normal is a unit vector perpendicular to the circle and in σ this plane, hence equal (up to a sign) to (cos ϕ, sin ϕ, 0). The unit normal of σ is N = ( cos θ cos ϕ, cos θ sin ϕ, sin θ), so the angle between N  and n at a point of the circle θ = θ0 is θ0 . The radius of the circle is a + b cos θ0 , so its sin θ0 κg ds = 2π sin θ0   and the holonomy is geodesic curvature is a+b cos θ0 . Hence, 2π 2π sin θ0 . The circles ϕ = constant are geodesics (as they are meridians on a surface of revolution) so κg  = 0 and the holonomy is 2π 0 = 2π. 13.6.2 For the circle v   = 1 on the cone, the radius is 1 and the angle between the principal normal n   of the circle and the unit normal N   of the cone is π/4, so the holonomy is 2π 2π/ 2 = (2 2)π. The cone is flat so if the converse of  Proposition 13.6.5 were true the holonomy around any closed curve on the cone would be zero. γ  ) dϕ ℓ(γ 13.6.3 For a general closed curve, 0 ds ds   is an integer multiple of 2π, say 2nπ. Then, (13.28) is replaced by γ  ) ℓ(γ κg ds. 2nπ

−

−

−

 

−

−

√ 

− √ 

 

  − 0

−

134

13.6.4 As in the proof of Proposition 13.6.1, vv = cos ϕtt + sin θtt1 , where t1  = N v˙  = ϕ( ˙ sin ϕtt + cos ϕtt1 ) + cos ϕt˙  + sin ϕt˙ 1 .

× t, so

−

By Exercise 7.3.22, t˙  = κ n N + κg t1 , t˙ 1 =

−κg t + τ g N, so

N  = (κn cos ϕ + τ g sin ϕ)N N, v˙  = (ϕ˙ + κg )( sin ϕtt +cos ϕtt1 ) + (κn cos ϕ + τ g sin ϕ)N

−

by Proposition 13.6.1, and hence

 v˙ 2 = (κn cos ϕ + τ g sin ϕ)2. 13.7.1 Take the reference tangent vector field to be ξξ   = (1, 0), and take the simple closed curve γγ (s) = (cos s, sin s). At γγ (s), we have V  = (α, β ), where α + iβ  =



 (cos s + i sin s)k (cos s i sin s)−k

−

if  k > 0, if  k < 0.

By de Moivre’s theorem, α = cos ks, β  = sin ks in both cases. Hence, the angle ψ ξ  is equal to ks, and Definition 13.7.2 shows that the multiplicity between V  and ξξ  is k. u, v˜), where (˜ u, v˜) 13.7.2 If  σ (u, v) = σ˜ (˜ (u, v) is a reparametrization map, then u ∂ ˜ v ˜ ∂ ˜ u ∂ ˜ v ˜σ˜ v˜ = α σ u  + βσσ v = α V = ασ ˜σ˜ u˜  + β  ˜ = α ∂ ˜ ˜ ∂u + β ∂u , β  = α ∂v + β ∂v . Hence, α ˜ and β  are smooth if  α and β   are smooth. Since the components of the vectors σ u and σσ v  are smooth, if V  is smooth so are its components. If the components σv + βσ σ v  are smooth, then V.σ σu  and V.σ σ v  are smooth functions, hence of V  = ασ u )−F (V.σ v ) v )−F (V .σ u ) α = G(V.σEG , β  = E (V.σEG are smooth functions, so V is −F 2 −F 2 smooth. ˜ the angle between V 13.7.3 If  ψ is V and  ˜ξξ  , we have ψ˜ ψ = θ  (up to multiples of 2π); so γ  ) ˙ ℓ(γ θ ds = 0 (a dot denotes d/ds). This is not obvious since we must show that 0 θ   is not a well defined smooth function of  s   (although dθ/ds   is well defined). ξ .ξ˜ξ  /  ξξ  ξ   ˜ξξ   . Now, However, ρ = cos θ is well defined and smooth, since ρ = ξξ  γ  ) ρ˙ ℓ(γ ρ = θ˙ sin θ, so we must prove that 0 ds = 0. Using Green’s theorem, ˙ 2

→ 

⇒

−

 

−

        √         −     − − 1 ρ

this integral is equal to ρu du + ρv dv

    − 1

π

ρ2

=

−

int(π )

∂  ∂u

ρv

1

∂  ∂v

ρ2

ρu

1

ρ2

dudv,

σ (π π (s)); and this line integral vanishes where π  is the curve in U   such that γγ  (s) = σ because ∂  ∂u



ρv

  − 1

ρ2

  ∂  = ∂v

ρu

  − 1

ρ2



ρuv (1 ρ2 ) + ρρu ρv = (1 ρ2 )3/2

− −



.

135

˜ be R  be a smooth function on a surface , let p 13.8.1 Let F  : , let σ and σ u0 , v˜0 ) = p, and let f  = F  σ and patches of    containing p, say σσ (u0 , v0 ) = σ˜ (˜ ∂u ∂v ∂v f ˜ = F   ˜ σ . Then, f ˜u˜ = f u ∂ ˜u + f v ∂ ˜u , f ˜v˜ = f u ∂u ∂ ˜ v + f v ∂ ˜ v , so if  f u = f v  = 0 at u0 , v˜0 ). (u0 , v0 ), then f ˜u˜  = f ˜v˜  = 0 at (˜

 S →  S   ◦

 S 

 ∈ S 

 ◦

2 ∂v ∂v 2 Since f u = f v  = 0 at p, we have f ˜u˜u˜ = f uu ∂u + 2f uv ∂u ∂ ˜ u ∂ ˜ u ∂ ˜ u + f vv ∂ ˜ u , with similar expressions for f ˜u˜v˜ and f ˜v˜v˜ . This gives, in an obvious notation, ˜ =

J t J , where J  =

H



∂u ∂ ˜ u ∂v ∂ ˜ u

∂u ∂ ˜ v ∂v ∂ ˜ v







H

 is the Jacobian matrix of the reparametrization

→ (u, v). Since J  is invertible, H˜  is invertible if  H  is invertible. Since the matrix H is real and symmetric, it has eigenvectors v1 , v2 , with eigen j. 2 Then, 2if  values λ1 , λ2 , say, such that vtivj   = 1 if  i = j   and = 0 if  i = t v = α1v1  + α2v2  is any vector, where α1 , α2  are scalars, v Hv = λ1 α1  + λ2 α2 ;  ⇐⇒ λ1 and λ2 are both > 0 (resp. both hence, vt Hv  > 0 (resp. < 0) for all v = < 0)  ⇐⇒ p  is a local minimum (resp. local maximum); and hence p  is a saddle point ⇐⇒ vt Hv  can be both > 0 and < 0, depending on the choice of v. Since ˜ v˜  = v˜t J t HJ v˜  = vt Hv. J  is invertible, a vector v˜  =  ⇐⇒ v  =  J v˜  = ; and v˜t H u, v˜) map (˜

0

0

0

The assertions in the last sentence of the exercise follow from this.

13.8.2 (i) f x = 2x

− 2y, f y = −2x + 8y, so f x = f y  = 0 at the origin. f xx = 2, f xy = −2, f yy  = 8, so H = −22 −82 . H  is invertible so the origin is non-degenerate; √  and the eigenvalues 5 ± 13 of  H  are both > 0, so it is a local minimum. 2 4 (ii) f x = f y   = 0 and  H =   at the origin; detH = −16 <   0, so the 4 0 eigenvalues of  H  are of opposite sign and the origin is a saddle point. (iii) f x = f y  = 0 and H = 0 at the origin, which is therefore a degenerate critical





 

point.

σ  in Exercise 4.2.5 (with a = 2, b = 1) gives f (θ, ϕ) = 13.8.3 Using the parametrization σ σ(θ, ϕ)) = (2+cos θ)cos ϕ+3. Then, f θ = sin θ cos ϕ, f ϕ = (2+cos θ)sin ϕ; F (σ since 2 + cos θ > 0, f ϕ = 0 = ϕ = 0 or π, and then f θ = 0 = θ = 0 or π; so there are four critical points, p = (3, 0, 0), q = (1, 0, 0), r = ( 1, 0, 0) and cos θ cos ϕ sin θ sin ϕ 1 0 s = ( 3, 0, 0). Next, = = sin θ sin ϕ (2 + cos θ)cos ϕ 0 3 1 0 1 0 1 0 at p, =  at q, =  at r, and = at s; hence, p is a 0 1 0 1 0 3 local maximum, q  and r  are saddle points, and s is a local minimum (all of which is geometrically obvious).

−

 H



−



 

⇒ − −



−

−

−

 − 

 

⇒ −

−

13.8.4 Since the torus is compact, any smooth function f   on the torus must achieve its maximum and minimum values, and these are of course critical points of  multiplicity 1. Since the Euler number of the torus is zero, Theorem 13.8.6 implies that f   must have at least two saddle points (more if  f  has other local

136

maxima or minima). 13.8.5 The position of the two rods can be specified by giving the angles θ and ϕ  made by the rods with the vertical. The bijection takes this configuration of the rods σ(θ, ϕ) of the torus, in the notation of the solution of Exercise 4.2.5. to the point σ The potential energy is E  = 12 aα cos θ + (a + 12 b)β cos ϕ + constant, where a, b are the lengths of the two rods and α, β   are positive constants; write this as E  = λ cos θ + µ cos ϕ + ν   for simplicity (where λ, µ > 0 and ν  are constants). For a critical point we must have ∂E/∂θ = ∂E/∂ϕ = 0, which gives sin θ = sin ϕ = 0. Thus, there are four critical points, namely (θ, ϕ) = (0, 0), (π, 0), (0, π), (π, π). In the notation of Definition 13.8.3,

H=

−

λ cos θ 0

0 λ cos ϕ

−



,

which is obviously invertible if  θ, ϕ = 0 or π, so all the critical points are nondegenerate. At (0, 0) the matrix  has two negative eigenvalues, at (π, 0) and (0, π) it has one, and at (π, π) it has none. So (0, 0) is a local maximum, (π, 0) and (0, π) are saddle points and (π, π) is a local minimum. The left-hand side of the equation in the statement of Theorem 13.8.6 is 1 2 + 1 = 0, which is indeed the Euler number of the torus (Example 13.8.7).

 H

−