A First Course in Differential Geometry Surfaces in Euclidean Space (Instructor Solution Manual, Sol

SOLUTION MANUAL

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A FIRST COURSE IN DIFFERENTIAL GEOMETRY Woodward and Bolton

Solutions to exercises Chapter 1 1.1. A sketch of the astroid is given in Figure 1(a). It is clear that all points in the image of α satisfy the equation of the astroid. Conversely, if x2/3 + y 2/3 = 1, then there exists u ∈ R such that (x1/3 , y 1/3 ) = (cos u, sin u).

Thus every point of the astroid is in the image of α.

Trigonometric identities may be used to show that α0 = (3/2) sin 2u(− cos u, sin u), which is zero only when u is an integer multiple of π/2. The corresponding points of the astroid are the cusps in Figure 1(a).

The required length is Z 3 π/2 3 sin 2u du = .

2 0 2

(a) Astroid

(b) An epicycloid

Figure 1

1.2. A sketch of the trace of an epicycloid is given in Figure 1(b). Trigonometric identities may be used to show that

α0 = 4r sin(u/2) sin(3u/2), cos(3u/2) .

R 2π So, for 0 ≤ u ≤ 2π, |α0 | = 4r sin(u/2), and required length is 4r 0 sin(u/2)du = 16r.

1.3. When r = 1, a calculation shows that α0 = tanh u sech u(sinh u, −1), so that, for u ≥ 0, t = sech u(sinh u, −1). It follows that α + t = (u, 0). A sketch of the trace of a tractrix is given in Figure 2(a).

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1.4. Here, |α0 | = (1 + g 0 )1/2 and t = (1 + g 0 )−1/2 (1, g 0 ). Hence n = 2 2 (1 + g 0 )−1/2 (−g 0 , 1). A calculation shows that t0 = g 00 (1 + g 0 )−3/2 (−g 0 , 1), so that dt 1 g 00 = 0 t0 = n. ds |α | (1 + g 0 2 )3/2

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SOLUTIONS TO EXERCISES

Figure 2: (a) shows a tractrix and (b) shows three catenaries

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Hence κ = g 00 (1 + g 0 )−3/2 .

Taking x(u) = u, y(u) = g(u) in the formula given in Exercise 1.8 gives the same formula for κ.

1.5. Use the method of Example 2 of §1.3. For u ≥ 0, |α0 | = tanh u and t = (tanh u, −sech u). It follows that dt/ds = (|α0 |)−1 t0 = n/ sinh u. Hence κ = cosech u.

1.6. EITHER: use Exercise 1.4 to show that the curvature of the catenary α(u) = (u, cosh u) is given by κ = sech2 u, OR: use the method of Example 2 of §1.3, and proceed as follows:α0 = (1, sinh u), so that |α0 | = cosh u and t = (sech u, tanh u). Hence n = (−tanh u, sech u), and 1 1 1 dt = 0 t0 = n.

2 (−tanh u, sech u) = ds |α | cosh u cosh2 u Hence κ = sech2 u. A sketch of the traces of three catenaries is given in Figure 2(b).

1.7. Differentiating with respect to u, we see that, using Serret-Frenet, α` 0 = α0 + `n0 = |α0 |(t − κ`t) = |α0 |(1 − κ`)t .

It follows that |α` 0 | = |α0 | |1 − κ`| and t` = t, where = (1 − κ`)/|1 − κ`|.

Hence n` = n, so, if s` denotes arc length along α` , we have 1

dt` = 0 t0 = 0 t0 . ds` |α | |1 − κ`| ` |α | |1 − κ`| Using Serret-Frenet, t0 = |α0 |κn = |α0 |κ n` , from which the result follows.

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1.8. Since α0 = (x0 , y 0 ), we have that |α0 | = (x0 + y 0 )1/2 . Hence t = 2 2 2 2 (x , y 0 )/(x0 + y 0 )1/2 and n = (−y 0 , x0 )/(x0 + y 0 )1/2 . Hence 0

t0 =

α00 (x0 2 + y 0 2 )1/2

−

α0 (x0 x00 + y 0 y 00 ) , (x0 2 + y 0 2 )3/2

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and a short calculation shows that

dt 1 0 00 0 0 00 0 0 00 00 0 y (x y − x y ), x (x y − x y ) = 02 2 ds (x + y 0 )2 x0 y 00 − x00 y 0 = 02 n, (x + y 0 2 )3/2 and the result follows.

1.9. (i) Let sα be arc length along α measured from u = 0. Since α0 = (1, sinh u) we see that dsα /du = |α0 | = cosh u. Hence sα (u) = sinh u and tα = (sech u, tanh u). The result follows from formula (1.9) for the involute. (ii) The evolute of α is given by β =α+

1 nα . κα

Here, we have (from Exercise 1.6) that κα = sech2 u and nα = (−tanh u, sech u).

A direct substitution gives the result.

A short calculation shows that β 0 = 0 if and only if u = 0, so this gives the only singular point of β (where the curve β has a cusp). A sketch of the traces of α and β is given in Figure 3.

Figure 3: A catenary and its evolute 1.10. Let sα denote arc length along α starting at u = u0 . Then, using (1.9) and the notation used there, we see that β 0 = α0 − sα 0 tα − sα tα 0 = −sα tα 0 .

It follows that β 0 = −sα |α0 |κα nα , so the only singular point of β is when sα = 0, that is at u = u0 .

1.11. For ease, assume that κα > 0, and restrict attention to u0 < u1 < u.

Then, from (1.12), we have that κ0 = 1/s0 and κ1 = 1/s1 .

Let ` be the length of α measured from α(u0 ) to α(u1 ). Then ` = s0 −s1 > 0, so the definition of involute gives that β 1 = β 0 + (s0 − s1 )tα = β 0 + `n0 . Hence β 1 is a parallel curve to β 0 , and κ0 1/s0 1 1 = = = = κ1 . |1 − κ0 `| |1 − `/s0 | |s0 − `| s1 1.12. A sketch of the trace of α is given in Figure 4.

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SOLUTIONS TO EXERCISES

Figure 4: The curve in Exercise 1.12 A short calculation shows that |α0 | =

√

3eu and

1 t = √ (cos u − sin u, sin u + cos u, 1) .

3 When z = λ0 we have that u = log λ0 , and when z = λ1 we have that u = log λ1 . So, required length is √ √ Z log λ1 u 3 e du = 3(λ1 − λ0 ) . log λ0

Also, using the method of Example 2 of §1.5, we find that dt 1 = u (− sin u − cos u, cos u − sin u, 0) , ds 3e so that κ =

√

2/(3eu ) and 1 n = √ (− sin u − cos u, cos u − sin u, 0) .

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It then follows that 1 b = t × n = √ (sin u − cos u, − cos u − sin u, 2) , 6 so that

1 db = √ (sin u + cos u, − cos u + sin u, 0) . ds 3 2eu

Hence τ = −1/(3eu ).

0

√ 1.13. Calculations similar to√those of Example2 2 of §1.5 show that |α | = 2 cosh u, t = (tanh u, 1, sech u)/ √ 2, κ = (1/2)sech u, n = (sech u, 0, −tanh u), and b = (−tanh u, 1, −sech u)/ 2. Differentiating one more time, we find that db/ds = (1/2)sech2 u(−sech u, 0, tanh u), so that τ = −(1/2)sech2 u.

1.14. Assume that α(s) is a smooth curve in R3 parametrised by arc length.

If α has zero curvature then dt/ds = 0, so that t is a constant unit vector t0 , say. Since dα/ds = t0 , it follows that α(s) = st0 + v 0 , for some constant vector v 0 . Hence α is a line. Conversely, if α is the line through v 0 in direction of unit vector t0 , say, then α may be parametrised as α(s) = st0 + v 0 , and it quickly follows that κ = 0.

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1.15. Using Serret-Frenet, we have α0 = |α0 |t ,

α00 = κ|α0 |2 n + |α0 |0 t ,

so that α0 × α00 = κ|α0 |3 b. Equating the lengths of both sides, we obtain the required formula for κ. Differentiating one more time, we find that α000 = −κτ |α0 |3 b + terms involving t and n , so that (α0 × α00 ).α000 = −κ2 τ |α0 |6 . Using the expression we have just found for κ now gives the required expression for τ .

1.16. (i) Assume first that α(u) = (a cos u, a sin u, bu). The tangent √ vector to α(u) makes angle φ with the generating lines where cos φ = b/ a2 + b2 .

Hence φ is a constant different from 0 and π/2. We saw in Example 2 of §1.5 that α(u) has non-zero torsion, so it follows that α(u) is a helix with value (a, 0, 0) when u = 0.

Conversely, assume that α(v) = (a cos θ(v), a sin θ(v), v + c) describes a helix with value (a, 0, 0) when v = 0. Then c = 0 and, denoting differentiation with respect to v by 0 , we have α0 = (−aθ0 sin θ, aθ0 cos θ, 1) , so the angle φ of the tangent vector to the z-axis is given by cos φ = (1 + 2 a2 θ0 )−1/2 .

Hence, our assumption that α(v) is a helix implies that θ0 is a non-zero constant, so that θ = c1 v + c2 , where c1 6= 0 and c2 are constants. The initial condition α(0) = (a, 0, 0) shows that c2 is an integer multiple of 2π, so if we let u = c1 v then α(u) = (a cos u, a sin u, u/c1 ), which is of the required form. (ii) Assume that α(s) has constant κ and τ . It follows immediately from SerretFrenet that the derivative of τ t − κb is zero, so that τ t − κb is constant, X 0 , say.

Let a = κ/(κ2 + τ 2 ). Then, using Serret-Frenet, d κ (α + an) = t + 2 (−κt − τ b) ds κ + τ2 τ = 2 (τ t − κb) . κ + τ2 Hence α + an has constant rate of change τ X 0 /(κ2 + τ 2 ), so that τ α + an = Y 0 + s 2 X0 , κ + τ2 for some constant vector Y 0 . Since n is perpendicular to t and to X 0 , it follows that α lies on the cylinder S of radius a whose axis of rotation is the line through Y 0 in direction X 0 . Finally, we note that t.X 0 = τ , which is constant, so that α is a helix on S.

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SOLUTIONS TO EXERCISES

1.17. Assume there is a unit vector X 0 such that t.X 0 = c, a constant.

Then n.X 0 = 0, so that X 0 = ct + c1 b for some constant c1 . Then 0 = X 0 0 = |α0 |(cκ + c1 τ )n, so that κ/τ = −c1 /c which is constant.

Conversely, if κ/τ = k, a constant, the Serret-Frenet formulae show that

(t − kb)0 = |α0 | κn − (κ/τ )τ n = 0 , so that t − kb is constant. The result follows since t.(t − kb) = 1.

1.18. The assumption on α implies the existence of a smooth function r(u) such that α + rn = p. If we differentiate this expression, use Serret-Frenet, and then equate the coefficients of t, n and b to zero, we find that r0 = 0 (so that r is a non-zero constant), κ = 1/r, and τ = 0. The result now follows from Lemma 1 of §1.5 and Example 8 of §1.3.

1.19. The given information implies that nα = ±nβ . (i) Differentiating tα .tβ (with respect to u), and using Serret-Frenet, we find that (tα .tβ )0 = |α0 |κα nα .tβ + |β 0 |κβ tα .nβ = 0 , so that tα .tβ is constant. (ii) The given information implies the existence of a smooth function r(u) such that β = α + rnα . Differentiating this, using Serret-Frenet, and taking inner product with nα gives that r0 = 0. Hence result. √ √ 1.20. Here, |α0 | = a2 + b2 and tα = (−a sin u, a cos u, b)/ a2 + b2 . Hence √ sα (u) = u a2 + b2 , from which it follows that β(u) = (a cos u + au sin u, a sin u − au cos u, 0) , so, in particular, the third component of β is zero. The circle of intersection of the plane z = 0 with the cylinder x2 + y 2 = a2 may be parametrised as γ(u) = (a cos u, a sin u, 0), and the exercise may now be completed using the formula for the involute by noting that γ 0 = (−a sin u, a cos u, 0), so that tγ = (− sin u, cos u, 0) and sγ (u) = au.

1.21. First assume that α(s) lies on a sphere with centre p and radius r, or, equivalently, that (α − p).(α − p) = r2 . We shall differentiate repeatedly to find an expression for α − p in terms of t, n and b.

So, differentiate once to find that (α − p).t = 0. Differentiating again and using Serret-Frenet, we obtain (α − p).n = −1/κ. Differentiating this and using Serret-Frenet gives (α − p).b = −κ0 /(τ κ2 ). It now follows that 1 κ0 α − p = − n − 2b . κ τκ The derivative of the left hand side, and hence of the right hand side, of the above equation is t. In particular, the coefficient of b of the derivative of the

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right hand side is zero, which gives the desired relation between κ and τ .

Conversely, assume that κ and τ for a regular curve α(s) are related as in the given formula, and let p(s) = α +

κ0 1 n + 2b . κ τκ

The given relation between κ and τ may be used to show that p0 = 0 so that p is constant. It now follows (again by differentiating) that (α − p).(α − p) is also constant. Since α is not constant we have that |α − p| = r for some positive constant r, so that α lies on the sphere with centre p and radius r.

Chapter 2 2.1. The line through (u, v, 0) and (0, 0, 1) may be parametrised by α(t) = 2 t(u, v, 0) + (1 − t)(0, 0, 1). This line intersects S 2 (1) when tu, tv, (1 − t) = 1, and a short calculation gives that t = 0 or t = 2/(u2 + v 2 + 1). Since t = 0 corresponds to (0, 0, 1), we quickly see that x(u, v) is as claimed.

The formula for F follows from consideration of similar triangles, OR we may use the following method which is similar to the one used in the earlier part of the solution to this exercise. The line through (x, y, z) and (0, 0, 1) may be parametrised by β(t) = t(x, y, z) + (1 − t)(0, 0, 1). For z 6= 1, this line cuts the xy-plane when t = (1 − z)−1 , which gives the point (1 − z)−1 (x, y, 0) on the line β. The formula for F now follows.

That F x(u, v) = (u, v) is a routine calculation (and also follows from the geometrical construction). That x is a local parametrisation as claimed is now immediate from conditions (S1) and (S2), taking U = R2 and W = R3 \ P .

2.2. Let V be an open subset of Rn such that X = V ∩ S is non-empty. If p ∈ X, let x : U → Rn be a local parametrisation of S whose image contains p. Then x−1 (V ) = {(u, v) ∈ U : x(u, v) ∈ V } is an open subset of R2 and the restriction of x to x−1 (V ) is a local parametrisation of X whose image contains p. To see this, if F : W → R2 is a map satisfying condition (S2) for x then consider the restriction of F to V ∩ W .

2.3(a). (i) There are many ways. The one which perhaps is closest to that given in Example 4 of §2.1 is to cover the cylinder by four local parametrisations.

2 + 3 Firstly, let U = {(u, v) √ ∈ R : −1 < u < 1, v ∈ R} and let x 3 : U → R be + 2 given by x (u, v) = ( 1 − u , u, v). If we let W = {(x, y, z) ∈ R : x > 0} and let F : W → R2 be given by F (x, y, z) = (y, z), then conditions (S1) and (S2) are satisfied. Hence x+ is a local parametrisation of the given cylinder S, and the corresponding coordinate neighbourhood is shown in the left hand picture of Figure 5. The whole of S may be covered by x+ and an additional three local parametrisations with domain U given by p x− (u, v) = (− 1 − u2 , u, v) ,

8

SOLUTIONS TO EXERCISES y + (u, v) = (u,

p

1 − u2 , v) ,

y − (u, v) = (u, −

p

1 − u2 , v) .

In fact, the whole of S may be covered by just one parametrisation. For instance, let U = R2 \ {(0, 0)} and define x̃ : U → R3 by

x̃(u, v) =

u v 1 2 2 , , log(u + v ) . (u2 + v 2 )1/2 (u2 + v 2 )1/2 2

Then, if we take W to be R3 , and let F : W → R2 be given by F (x, y, z) = (xez , yez ), it is routine to show that (S1) and (S2) both hold. (ii) If x2 + y 2 = 1 then (x cos u − y sin u)2 + (x sin u + y cos u)2 = 1 so

Figure 5: Coordinate neighbourhoods on a cylinder that the cylinder is setwise invariant under rotations about the z-axis. It is obtained by rotating the curve α(v) = (1, 0, v) about the z-axis. The coordinate neighbourhoods corresponding to the two local parametrisations x and y as discussed in §2.2 are shown in Figure 5. The coordinate neighbourhood for x is the cylinder with one line deleted (shown in broken bold face) in the centre picture of Figure 5, and for y is the cylinder with one line deleted (shown in bold face) in the picture on the right hand side of Figure 5. (iii) If f (x, y, z) = x2 + y 2 − 1 then grad f = (2x, 2y, 0), which is zero on and only on the z-axis. Since f maps the z-axis to −1, and since, for example, f (1, 0, 0) = 0, the equation f (x, y, z) = 0 defines a surface by Theorem 1 of §2.4. √ 2.3(b). (i) Use the local parametrisation x(u, v) = (u, v, u2 + v 2 + 1) (as given in Example 2 of §2.1) to cover the upper sheet, taking W = {(x, y, z) ∈ R3 : z > 0}, and F (x, y, z) = (x, y) for all (x, y, z) ∈√W .

Cover the lower sheet by using y(u, v) = (u, v, − u2 + v 2 + 1), taking W = {(x, y, z) ∈ R3 : z < 0} and F (x, y, z) = (x, y) for all (x, y, z) ∈ W . (ii) If x2 +y 2 = z 2 −1 then (x cos u−y sin u)2 +(x sin u+y cos u)2 = z 2 −1, so that this hyperboloid of two sheets is setwise invariant under rotations about the zaxis. It is obtained by rotating the union of the curves α(v) = (sinh v, 0, cosh v), v ≥ 0, and β(v) = (sinh v, 0, − cosh v), v ≥ 0, about the z-axis. (iii) If f (x, y, z) = x2 + y 2 − z 2 + 1, then the given hyperboloid of two sheets S has equation f (x, y, z) = 0. This is non-empty (eg, f (0, 0, 1) = 0) and grad f is zero at (0, 0, 0) only, which is not on S.

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2.4. Note that f isn’t smooth on the z-axis, so in order to apply Theorem 1 of §2.4 we must restrict the domain of definition of f to W = R3 \ z-axis. On W, ! p p x2 + y 2 − a x2 + y 2 − a ,y p ,z . grad f = 2 x p x2 + y 2 x2 + y 2 Hence, grad f = 0 on and only on the circle x2 + y 2 = a2 in the plane z = 0, and all these points are mapped by f to 0.

If 0 < b then, by Theorem 1 of §2.4, the set S of points of W satisfying the equation f (x, y, z) = b2 is a surface. If 0 < b < a then every point satisfying the equation f (x, y, z) = b2 is actually in W , so that S is equal to the set of points in R3 satisfying f (x, y, z) = b2 , so this latter set is indeed a surface. It is a torus of revolution since it is setwise invariant under rotations about the z-axis, and intersects the half-plane y = 0, x > 0 in the circle centre (a, 0, 0) radius b (and, in particular, x > 0 on this circle).

If 0 < a < b, there are points on the z-axis satisfying f (x, y, z) = b2 , so we cannot conclude that the set of points of R3 satisfying this equation form a surface. In fact, this set of points is obtained by rotating the circles of Figure 6 round the z-axis. You will see that “non-surface like” things happen on the z-axis.

Figure 6: f (x, y, z) = b2 when 0 < a < b

2.5(a). Since grad f = 2(x + y + z − 1)(1, 1, 1), it is clear that grad f = 0 exactly when x + y + z = 1. These are the points of R3 which are mapped to zero under f . We can conclude that the equation f (x, y, z) = k defines a surface for any k > 0 (namely, the union of two parallel planes), while, if k < 0 then f (x, y, z) = k is the empty set (and so not a surface). Note that, although grad f vanishes at EVERY point satisfying f (x, y, z) = 0, this set is still a surface, namely the plane x + y + z = 1.

This example shows that Theorem 1 of §2.4 is not an “if and only if” theorem.

2.5(b). If f (x, y, z) = xyz 2 , then grad f = 0 if and only if either z = 0 or x = y = 0. It follows that if k 6= 0 then f (x, y, z) = k defines a surface. The set xyz 2 = 0 is the union of the three coordinate planes, which is not a surface.

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SOLUTIONS TO EXERCISES 2.6. We have α0 = |α0 |t

and

α00 = |α0 |t0 + |α0 |0 t .

Hence,

xu = |α0 | + v|α0 |0 t + v|α0 |t0

and

xv = |α0 |t .

If κ(u) = 0 then t0 is zero, so that xu and xv are linearly dependent. If κ(u) 6= 0 then t0 = |α0 |κn, so that xu and xv are linearly dependent if and only if v = 0.

The next statement is immediate from Theorem 1 of §2.5.

2.7. The ellipsoid S is defined by f (x, y, z) = 1, where f (x, y, z) = (x2 /a2 )+ (y /b2 ) + (z 2 /c2 ). A routine check shows that grad f is never zero on S, so that S is a surface. (i) One suitable choice of F : W → R2 is obtained by taking 2

W = {(x, y, z) ∈ R3 : |z| < c, and if y = 0 then x > 0} , and

z y . , arcsin a b c (ii) We check conditions (1), (2) and (3) of that theorem. Firstly, f x(u, v) = cos2 v + sin2 v = 1, so (1) holds. Secondly, if x(u1 , v1 ) = x(u2 , v2 ) then comparing the third component gives that v1 = v2 . The first two components then show that u1 = u2 . Finally, it is clear that xu and xv are linearly independent unless cos v = 0, which never happens for the given range of values of v.

F (x, y, z) = Arg

x

+i

2.8. (i) Use Theorem 1 of §2.4 to show that S is a surface by checking that if f (x, y, z) = x2 − y 2 − z then grad f doesn’t vanish on the non-empty subset of R3 defined by equation f (x, y, z) = 0. We use Theorem 4 of §2.5 to show that x is a local parametrisation. We first note that f x(u, v) = 0 for all (u, v) ∈ R2 , so that the first condition holds. If x(u1 , v1 ) = x(u2 , v2 ) then the first two components show that cosh u1 − sinh u1 = cosh u2 − sinh u2 , so the third component shows that v1 = v2 from which it quickly follows that u1 = u2 .

Finally, since sinh u 6= cosh u, a comparison of the first two components of xu and xv shows that xu and xv are linearly independent at all points. (ii) This may be done using similar methods. We first note that the components of x(u, v) satisfy the given equation for all 0 < u < 2π, v ∈ R, so that the first condition holds. If x(u1 , v1 ) = x(u2 , v2 ) then the first and third components show that v1 = v2 , and then, using all three components, we may prove that sin u1 = sin u2 and cos u1 = cos u2 so that u1 = u2 . Finally, it may be checked that xu × xv is never zero so that xu and xv are linearly independent at all points.

2.9. Show that S is a surface by applying Theorem 1 of §2.4 to the function f (x, y, z) = x sin z − y cos z. It is clear that f x(u, v) = 0 for all (u, v) ∈ R2 , so that the image of x is a subset in S. To show that the image of x is the whole of S, we note that if (x, y, z) ∈ S then x(z, x cos z + y sin z) = (x, y, z).

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Part (i) is straightforward; for instance, xu ×xv = (− sin u, cos u, −v), which is never zero. For (ii), we note that our checking of the surjectivity of x shows that we can take F (x, y, z) = (z, x cos z + y sin z).

2.10. The statements in the given hint are straightforward to check, and it follows from these that f u and f v are linearly independent unless f u = 0.

However, if f u = 0 then the third component of f u shows that either v = 0 or u = (2n + 1)π. If v = 0 then f u = 2α0 6= 0, while if u = (2n + 1)π then the second component of f u is equal to ±v/2. It follows that f u is never zero, so that f u and f v are linearly independent for all values of u and v.

We now define two maps x and y by restricting the domain of definition of f . Specifically, we let x(u, v) = f (u, v) ,

−π < u < π , −1 < v < 1 ,

y(u, v) = f (u, v) ,

0 < u < 2π , −1 < v < 1 .

It is clear that S is the union of the images of x and y. We now show that x and y are both injective. There are several ways - here is a geometrical one. The line segment on S through 2α(u) projects down (under orthogonal projection)

to the line segment 2 + v cos(u/2) α(u) in the xy-plane and it is clear that, since −1 < v < 1, no two distinct segments intersect. Injectivity of x and y now follows.

The exercise may be completed by applying Theorem 4 of §2.5 to show that x(u, v) and y(u, v) are both local parametrisations of S. 2.11. First check that all points in the image of x satisfy the given equations.

Then check that x is injective (on its domain of definition). Finally check that xu and xv are linearly independent at all points. The image of x is the flat torus with two circles omitted. The whole of the flat torus may be covered by considering three further parametrisations which have the same formula as x but have domains (0, 2π) × (0, 2π), (0, 2π) × (−π, π), and (−π, π) × (0, 2π).

Chapter 3 3.1. Here, xu = (1, 0, 2u) and xv = (0, 1, 3v 2 ). Since these vectors span the tangent plane, we would need both (1, 0, 2u).(−1, 1, 0) and (0, 1, 3v 2 ).(−1, 1, 0) to be zero. Clearly, neither expression can be zero!

3.2. Clearly x(2, −1) = (2, −1, 3) ∈ S. At this point, xu = (1, 0, 4) and xv = (0, 1, 3). Also (1, −1, 1) = (1, 0, 4) − (0, 1, 3) and so is in T(2,−1,3) S. For the last part, we need scalars λ and µ such that λ(1, 0, 4)+µ(0, 1, 3) is orthogonal to (1, −1, 1). This gives 5λ + 2µ = 0, so we could take λ = 2 and µ = −5. Hence 2(1, 0, 4) − 5(0, 1, 3) = (2, −5, −7) is a vector in T(2,−1,3) S which is orthogonal to (1, −1, 1).

3.3. Let f (x, y, z) = 2x2 − xy + 4y 2 . Then grad f = (4x − y, −x + 8y, 0).

In particular, (grad f )(0, 1/2, 2) = (−1/2, 4, 0), so the unit normal there is

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SOLUTIONS TO EXERCISES

√ (−1, 8, 0)/ 65. A basis for the tangent plane of S at (0, 1/2, 2) is provided by any pair of linearly independent vectors orthogonal to the normal, for instance (0, 0, 1) and (8, 1, 0).

3.4. Let f (x, y, z) = (x2 /a2 ) + (y 2 /b2 ) + (z 2 /c2 ). Then the ellipsoid S has equation f (x, y, z) = 1. Also, grad f = 2(x/a2 , y/b2 , z/c2 ), which is never zero on S so that (grad f )(p) is a non-zero vector orthogonal to S at p. This is parallel to (1, 1, 1) if and only if (x/a2 ) = (y/b2 ) = (z/c2 ). Using this to eliminate x and y in the equation of the ellipsoid, we find that z 2 = c4 /(a2 + b2 + c2 ). So the tangent plane √ at a point (x, y, z) of the ellipsoid is orthogonal to (1, 1, 1) at ±(a2 , b2 , c2 )/ a2 + b2 + c2 .

3.5. Let f (x, y, z) = x2 + y 2 − z 2 . Then S has equation f (x, y, z) = 1. Also grad f = (2x, 2y, −2z), which is never zero on S so if p = (x, y, z) ∈ S then (grad f )(p) is a non-zero vector orthogonal to S at p. So, if p = (x, y, z) ∈ S, then T(x,y,z) S is orthogonal to (1, 0, −1) if and only if x2 + y 2 − z 2 = 1, x = z, y = 0, which is not possible. However T(x,y,z) S is orthogonal to (1, 1, 1) if and only if x2 + y 2 − z 2 = 1 and x = y = −z. The two possibilities are ±(1, 1, −1).

3.6. . Let f (x, y, z) = x2 + y 2 + z 2 − ax and g(x, y, z) = x2 + y 2 + z 2 − by.

Then grad f = (2x − a, 2y, 2z) and grad g = (2x, 2y − b, 2z). Grad f = 0 only at (a/2, 0, 0), and since f (a/2, 0, 0) = −a2 /4 6= 0 it follows that the equation f (x, y, z) = 0 defines a surface S1 in R3 . Similarly, g(x, y, z) = 0 defines a surface S2 . If (x0 , y0 , z0 ) ∈ S1 ∩ S2 then x0 2 + y0 2 + z0 2 = ax0 = by0 , from which it follows that, at (x0 , y0 , z0 ), grad f.grad g = 0. It follows that S1 and S2 intersect orthogonally at all points of intersection. The intersections of S1 and S2 with x2 + y 2 + z 2 = cz are dealt with in a similar manner.

3.7. We use the notation from the solution of Exercise 3.4. Since (grad f )(p) √ is orthogonal to√S at p, the equation of the tangent plane based at (a/2, b/2, c/ 2) is x/a + y/b + 2z/c = 2.

3.8. Here, xu = (cos v, sin v, 1) and xv = (−u sin v, u cos v, − tan v). Hence E = xu .xu = 2, and, similarly, F = − tan v, and G = u2 + tan2 v.

3.9. Let f1 (x1 , x2 , x3 , x4 ) = x1 2 + x2 2 and let f2 (x1 , x2 , x3 , x4 ) = x3 2 + x4 2 .

Then grad f1 = (2x1 , 2x2 , 0, 0) and grad f2 = (0, 0, 2x3 , 2x4 ), which are linearly independent unless either x1 = x2 = 0 or x3 = x4 = 0. Since f (r1 , 0, r2 , 0) = (r1 2 , r2 2 ), it follows that (r1 2 , r2 2 ) is a regular value of f for each pair of positive real numbers r1 , r2 .

We now check the conditions for the application of Theorem 4 of §2.5. Firstly, it is clear that f x(u, v) = (r1 2 , r2 2 ) so that (1) holds. That (2) holds is immediate. Finally, xu = (−r1 sin u, r1 cos u, 0, 0) and xv = (0, 0, −r2 sin v, r2 cos v), so that xu and xv are linearly independent for all (u, v).

The coefficients of the first fundamental form are E = r1 2 , F = 0, G = r2 2 .

3.10. Routine calculations show that the surface S has E = G = cosh2 v,

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F = 0. That xu = x̃v and xv = −x̃u is straightforward.

Also, Eθ = (cos θ xu + sin θ x̃u ).(cos θ xu + sin θ x̃u ) = (cos θ xu − sin θ xv ).(cos θ xu − sin θ xv ) = E cos2 θ − 2F cos θ sin θ + G sin2 θ = cosh2 v .

That Fθ = 0 and Gθ = cosh2 v are proved in a similar manner.

Finally, we note that the tangent plane to Sθ is spanned by cos θ xu −sin θ xv and cos θ xv + sin θ xu , which is parallel to the plane spanned by xu and xv .

Ru p 3.11. The length of αv from t = u0 to t = u1 is u01 E(t, v) dt, which is independent of v since E is independent of v.

3.12. (i) Here α0 = xu + xv so that |α0 |2 = E + 2F + G, where the right hand side of the equation is evaluated at (t, t). This gives that |α0 | = 2 cosh t, R1 so the required length is 2 0 cosh t dt = 2 sinh 1. (ii) The two given curves intersect when t = r = 0, which gives u = v = 0. The result follows since α0 (0).β 0 (0) = (xu + xv ).(xu − xv )(0, 0) = (E − G)(0, 0) = 0.

3.13. √The curves intersect when t = π/3 and √ r = 1/2, which gives u = 1/2 and v = 3/2.√ At this point, E = 2, F = 1/ 3, G = 4/3. √ We also have that α0 (π/3) = −( 3/2)xu + (1/2)xv , while β 0 (1/2) = xu + 3xv . A calculation using |α0 (π/3)||β 0 (1/2)| cos θ = α0 (π/3).β 0 (1/2) yields the given answer.

R π/2 p 3.14.Here, L(u) = −π/2 G(u, t) dt. Hence, differentiating under the in√ R π/2 tegral sign, 2L0 (u) = −π/2 Gu / G dt. A short calculation shows that Gu = (b2 − a2 ) sin2 v sin 2u, which is identically zero when u = −π/2, 0, π/2, and π.

Each curve αu forms one half of the intersection of S with the plane in R3 containing the z-axis and having equation xb sin u = ya cos u. This is an ellipse p with principal semi-axes of lengths a2 cos2 u + b2 sin2 u and c. The critical lengths |a| and |b| of the first of these semi-axes correspond to u = 0, π and u = ±π/2 respectively.

3.15. Here xu =

2 (−u2 + v 2 + 1, −2uv, 2u) , (u2 + v 2 + 1)2

so that E=

4 (u2 + v 2 + 1)4

(−u2 + v 2 + 1)2 + 4u2 v 2 + 4u2 =

4 . (u2 + v 2 + 1)2

That F = 0 and G = 4(u2 + v 2 + 1)−2 may be shown in a similar manner.

3.16. Let α(u)R = x(u, 0), u ∈ R. Then |α0 |2 = E(u, 0), so the required ∞ length is given by −∞ 2/(u2 + 1) du = 2[arctan u]∞ −∞ = 2π.

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SOLUTIONS TO EXERCISES

3.17. Let f (u + iv) = x(u, v) + iy(u, v), where x(u, v) and y(u, v) are the real and imaginary parts of f . Then x(u, v) = u, v, x(u, v), y(u, v) so that xu = (1, 0, xu , yu ) and xv = (0, 1, xv , yv ). That E = G, and F = 0, follows from the Cauchy-Riemann equations, xu = yv and xv = −yu .

3.18. Here, E = 1 + v 2 , F = 1 and G = 2. Also, F is given by φ(u, v) = constant, where φ(u, v) = u+v. It follows that the tangent vectors to the family F are scalar multiples of xu − xv .

π π π √ , , then u = π/6 A short calculation shows that if x(u, v) = 2 3 6 2 and v = π/3, at which point E = 1 + π 2 /9, F = 1 and G = 2. Then |xu ||xu − xv | cos θ = xu .(xu − xv ) , so that E(E − 2F + G) cos2 θ = (E − F )2 . Substituting the values of E, F , G at (π/6, π/3), we find that (1 + π 2 /9)2 cos2 θ = (π 2 /9)2 . It follows that cos θ is as claimed.

We also see that β(r) = x u(r), v(r) is an orthogonal trajectory of F if and only if (xu du/dr + xv dv/dr).(xu − xv ) = 0 which holds if and only if Edu/dr + F (dv/dr − du/dr) − Gdv/dr = 0 which quickly gives that 1 dv du = 2 . dr v dr Integrating, we see that the orthogonal trajectories of F are given by u + 1/v = constant. Since x(0, 1) = (1, 0, 1), we see that the orthogonal trajectory of F through (1, 0, 1) has equation u+1/v = 1. Since x(π/2, −π/2) = (0, −π/2, 0), we see that the orthogonal trajectory of F through (1, 0, 1) does not pass through (0, −π/2, 0).

3.19. Here, E = G = 1 and F = 0. The family F is given by φ(u, v) = constant, where φ(u, v) = v sin u/(1 − cos u), and it follows that if x u(t), v(t) is a member of F then (v 0 sin u + vu0 cos u)(1 − cos u) = u0 v sin2 u .

This simplifies to give u0 v = v 0 sin u, so that the tangent vectors to F are scalar multiples of sin u xu + vxv . It then follows that β(r) = x u(r), v(r) is an orthogonal trajectory of F if and only if du dv sin u = −v . dr dr Integrating, we see that the orthogonal trajectories of F are given by ψ(u, v) = constant where ψ(u, v) = 2 cos u − v 2 . Since |x(u, v)|2 = 2(1 − cos u) + v 2 , we see that |x(u, v)| is constant on each of these trajectories.

3.20. A non-zero vector X bisects the angle between the coordinate curves if and only if X.xu /|xu | = ±X.xv /|xv |. This quickly leads to the given condition.

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Let x(u, v) = (u, v, u2 − v 2 ). We note that x(1, 1) = (1, 1, 0), and, since xu (1, 1) = (1, 0, 2) and xv (1, 1) = (0, 1, −2), we see that E(1, 1) = G(1, 1) = 5 and F (1, 1) = −4. This leads to α = ±β, so that X is any non-zero scalar multiple of either (1, 1, 0) or (1, −1, 4).

3.21. Here, E = v 2 + 1, F = 0, G = 1. So, using the result of Exercise 3.20, it follows that a curve β(r) = x u(r), v(r) bisects the angles between the coordinate curves if and only if dv 1 du =± √ . dr dr v 2 + 1 Integrating, we see that u ± arcsinh v = constant.

3.22. If α(t) = x u(t), v(t) is a member of F, then α0 (t) = u0 xu + v 0 xv where u0 φu + v 0 φv = 0. Hence u0 /v 0 = −φv /φu , and it follows that tangent vectors to the family F at x(u, v) are the scalar multiples of φv xu − φu xv .

If x(u, v) = u, v, φ(u, v) then E = 1 + φu 2 , F = φu φv , G = 1 + φv 2 . The family F of level curves of φ (or, in other words, the contours on S) are given by φ(u, v) = constant, and the paths of steepest descent are the orthogonal

trajectories of F. So, if β(r) = x u(r), v(r) is a path of steepest descent then

dv du xu + xv . (φv xu − φu xv ) = 0 . dr dr Using the expressions for E, F and G given above, this equation simplifies to give du dv φv − φu = 0 . dr dr 2 dv If φ(u, v) = u3 + v 3 , then the orthogonal trajectories have v 2 du dr − u dr = 0, so that 1 dv 1 du − 2 =0.

2 u dr v dr Integrating, we see that the paths of steepest descent are given by ψ(u, v) = 1 1 constant, where ψ(u, v) = − . u v

3.23. The image of x is a cone having vertex at the origin with axis of rotation being the z-axis. That xu and xv are linearly dependent if and only if v = 1 is quick. The curve α is the intersection of the cone with the plane z = −1, and x(u, 1) = (0, 0, 0) for all u ∈ R. 3.24. The given line lies on S if and only if, for all λ ∈ R, (p1 + λv1 )(p2 + λv2 ) = (p3 + λv3 ) .

Taking into account that (p1 , p2 , p3 ) ∈ S, we see that the above holds if and only if v1 p2 + p1 v2 = v3 and v1 v2 = 0. This implies that (v1 , v2 , v3 ) is a scalar multiple of either (0, 1, p1 ) or (1, 0, p2 ). This shows that S is a doubly ruled

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SOLUTIONS TO EXERCISES

surface, and the rulings intersect orthogonally if and only if p1 p2 = 0. Hence result.

3.25. (i) The left hand side of the equation is equal to (xu × xv )(u0 , v0 ). (ii) Let γ(u) = α(u) + λ(u)β(u) for some smooth function λ(u). A short calculation shows that γ 0 .β 0 = 0 if and only if λ = −α0 .β 0 /|β 0 |2 . (iii) Since β 0 is orthogonal to both γ 0 and β, it follows that β 0 is parallel to γ 0 × β. If we put γ 0 × β = µβ 0 , then a short calculation shows that y u × y v = 0 if and only if µ(u) = v = 0 (and v = 0 corresponds to the striction curve).

3.26. Let S be a ruled surface of revolution, and assume first that the rulings of S are complete lines (rather than just line segments). Let ` be a line of the ruling, let a ≥ 0 be the perpendicular distance of ` from the axis of rotation, and let θ be the angle between ` and the axis of rotation.

By applying a rigid motion of R3 , we may assume that the axis of rotation is the z-axis, and that ` is the line parametrised by α(t) = (a, 0, 0) + t(0, sin θ, cos θ).

If a = 0 we have a cone or a plane, so we now assume a > 0. When the point α(t) is rotated about p the z-axis to be in the half-plane y = 0, x > 0, we obtain the point β(t) = ( a2 + t2 sin2 θ, 0, t cos θ), so that β is a parametrisation of the curve in the xz-plane with equation x2 − z 2 tan2 θ = a2 . If θ = 0, then S is the cylinder x2 + y 2 = a2 ; if θ = π/2, then S is a plane (with, arguably, the possibility that a disc centred on the origin may be removed). Otherwise, S is the hyperboloid of revolution x2 /a2 + y 2 /a2 − z 2 /c2 = 1, where c2 = a2 cot2 θ.

So, the surfaces of revolution ruled by complete lines are planes, cones, cylinders, and those hyperboloids of one sheet discussed in Example 2 of §3.6 for which a = b. Similar arguments show that surfaces of revolution ruled by line segments are rotationally invariant open subsets of the surfaces just described.

3.27. If we parametrise S 2 (1) by rotating (cos v, 0, sin v), −π/2 < v < π/2, about the z-axis, then E = cos2 v, F = 0 and G = 1. Hence the area of the southern hemisphere is (as expected!)

Z 0

Z π cos v du dv = 2π .

−π/2

−π

3.28. Here, EG − F 2 = (1 − u2 )−1 (1 − v 2 )−1 . Required area is π 2 /4.

3.29. We parametrise S 2 (1) as in the solution to Exercise 3.27, and let E, F and G be the coefficients of the first fundamental form.

If we define a local parametrisation of the cylinder by letting x̃(u, v) = f x(u, v) = (cos u, sin u, sin v), then routine calculations show that Ẽ G̃ − F̃ 2 = EG − F 2 , and the result follows from (3.20).

R 2π R 2π 3.30. Here E = r1 2 , F = 0 and G = r2 2 . Required area is 0 r1 r2 dudv = 0 4π 2 r1 r2 as claimed.

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3.31. (i) If the coordinate curves Rof xpform a Tchebycheff net, then, for u each fixed u0 < u1 , the length L(v) = u01 E(t, v) dt of the coordinate curve u 7→ x(u, v), u0 < u < u1 ,√is independent of v. By differentiating under the integral sign, we see that ( E)v = 0, and hence Ev = 0. Similar reasoning shows that Gu =√ 0. This argument may be reversed √ to give the converse. (ii) Here, ũu = E, ũv = 0, ṽu = 0 and ṽv = G. In particular, if u (resp. v) is constant then so is ũ (resp. ṽ). The statement concerning the families of coordinate curves now follows. It also follows that √ √ xu = x̃ũ E , xv = x̃ṽ G , p so that E = ẼE and G = G̃G. Hence Ẽ = G̃ = 1, and cos θ = F̃ / Ẽ G̃ = F̃ .

Chapter 4 4.1. Here, xu = (− sinh v sin u, sinh v cos u, 1) and xv = (cosh v cos u, cosh v sin u, 0).

It follows that xu × xv = cosh v(− sin u, cos u, − sinh v). Hence N = sech v(− sin u, cos u, − sinh v) .

This is not injective since it is periodic in u.

4.2. If we parametrise the surface of revolution as usual using a generating

curve f (v), 0, g(v) , f (v) > 0 for all v, then the formula in Example 2 of §4.1 2 shows p that the parallel of pS through v = v0 maps to the parallel of S (1) through 2 2 2 2 (g 0 / f 0 + g 0 , 0, −f 0 / f 0 + g 0 )(v0 ), while the meridian u = u0 maps to the meridian of S 2 (1) through (cos u0 , sin u0 , 0). p 2 2 4.3.(i) Grad √ f = (2x, 2y, −1), so that N = (2x, 2y, −1)/ 4x + 4y + 1 = (2x, 2y, −1)/ 4z + 1. Hence the image of N is contained in the lower hemisphere of S 2 (1). However, if X 2 + Y 2 + Z 2 = 1 with −1 ≤ Z < 0, then

X Y 1 2 N − ,− , (1 − Z ) = (X, Y, Z) , 2Z 2Z 4Z 2 so it follows that the image of N is the lower hemisphere (excluding the equator).

ALTERNATIVELY: Since S is a surface of revolution, the image of N will be obtained by rotating the image under N of the generating curve (v, 0, v 2 ), 2 v ≥ 0 of S. Since N (v, 0, v 2 ) is the unit p vector in direction (grad f )(v, 0, v ) 2 2 we find that N (v, 0, v ) = (2v, 0, −1)/ (1 + 4v ). If (X, Z) = (cos θ, sin θ) for −π/2 ≤ θ < 0, then, when v = (−1/2) cot θ ≥ 0, N (v, 0, v 2 ) = (X, 0, Z). Hence the image of N is as obtained above. (ii) This surface is obtained by rotating the curve (cosh v, 0, sinh v), v ∈ R, about the z-axis. Also, N (cosh v, 0, sinh v) = (cosh v, 0, − sinh v)(1 + 2 sinh2 v)−1/2 .

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SOLUTIONS TO EXERCISES

√ The third component takes all values between ±1/ 2. Hence result as given. (iii) This surface is obtained by rotating the union of the two curves (sinh v, 0, ± cosh v), v ≥ 0, about the z-axis. Also, N (sinh v, 0, ± cosh v) = (sinh v, 0, ∓ cosh v)(2 cosh2 v − 1)−1/2 . √ It follows that the image of the Gauss map is {(x, y, z) ∈ S 2 (1) : |z| > 1/ 2}.

4.4. We may use the method of the alternative solution to Exercise 4.3(i), since the catenoid is obtained by rotating (cosh v, 0, v), v ∈ R, about the z-axis.

Here N (cosh v, 0, v) = (1, 0, − sinh v)/ cosh v, and this equals (cos θ, 0, sin θ), −π < θ ≤ π if and only if −π/2 < θ < π/2 and sinh v = − tan θ. The result follows.

4.5. Here, h is the restriction to S of the map g : R3 → R given by the same formula. Since g is linear, it is smooth and is equal to its own derivative at each point. It follows that h : S → R is smooth and if X ∈ Tp S then dhp (X) = X.v.

This map is identically zero if and only if v is orthogonal to S at p.

4.6. This is similar to the above, but with g(p) = (q − p).(q − p). This is smooth on R3 and the derivative is dgp (X) = 2X.(p − q). Now argue as in the solution to Exercise 4.5.

4.7. The hypotheses imply the existence of a point q0 ∈ R3 and a function f : S → R, such that, if p ∈ S, then p + f (p)N (p) = q0 . The result of the previous exercise shows that f is smooth (except at q0 if this latter point is in S). If we differentiate the equation, we find that if X is a tangent vector to S then X + df (X)N + f dN (X) = 0. Taking the component of this orthogonal to S, we deduce that df (X) = 0, so that f is constant r, say. Since r is clearly non-zero, we see that q0 ∈ / S and S is an open subset of the sphere centre q0 and radius r.

4.8. Let (X, Y, Z) ∈ T(x,y,z) S 2 (1). Then xX+yY +zZ = 0, and df(x,y,z) (X, Y, Z) = (aX, bY, cZ). The vector (x/a, y/b, z/c) is orthogonal to S̃ at (ax, by, cz), and the inner product of (x/a, y/b, z/c) with (aX, bY, cZ) is zero.

4.9. One example is f (x, y, 0) = a cos(x/a), a sin(x/a), y . We check this is a local isometry using the conditions given in Proposition 2 of §4.3. First parametrise the plane using x(u, v) = (u, v, 0), so that f (u, v) = (a cos(u/a), a sin(u/a), v).

Then fu .fu = 1 = E, fu .fv = 0 = F , and fv .fv = 1 = G. This shows that f is a local isometry, while f is surjective because if x2 + y 2 = a2 then (x/a, y/a) is on the unit circle and so is equal to (cos θ, sin θ) for some −π < θ ≤ π. Hence f (aθ, z, 0) = (x, y, z).

4.10. We first note that f is well-defined since, if (r cos θ, r sin θ) = (r̃ cos φ, r̃ sin φ) then r = r̃ and φ and θ differ by an integer multiple of 2π, so that (r cos nθ, r sin nθ, br) = (r̃ cos nφ, r̃ sin nφ, br̃). It is clear that the image of f is on S̃, while f is surjective since every point of S̃ may be written in the form (r cos µ, r sin µ, br),

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r > 0. We parametrise the plane S (minus the non-positive real axis) by using polar coordinates; x(r, θ) = (r cos θ, r sin θ, 0), r > 0, −π < θ < π. Then f (r, θ) = (1/n)(r cos nθ, r sin nθ, br), so that fr = (1/n)(cos nθ, sin nθ, b) and √ fθ = (−r sin nθ, r cos nθ, 0). It follows that if b = n2 − 1 then fr .fr = (1 + b2 )/n2 = 1 = E, fr .fθ = 0 = F , and fθ .fθ = r2 = G. The result follows from Proposition 2 of §4.3.

To see the geometry of f , note that f (r̃eiθ̃ ) = f (reiθ ) if and only if r̃ = r and θ̃ = θ + (2k/n)π for some integer k. So, to model the map, cut a line from the edge of the paper to the centre, then make a cone with vertex at the centre of the paper by sliding the edges of the cut past each other until they line up again after n − 1 circuits.

4.11. It is clear that the image of the restriction of f to S lies on S̃. Surjectivity follows from the fact that f (r1 cos θ, r1 sin θ, r2 φ) = (r1 cos θ, r1 sin θ, r2 cos φ, r2 sin φ).

If we parametrise S by x(u, v) = (r1 cos u, r1 sin u, v), then E = r1 2 , F = 0 and G = 1. Also, fu = (−r1 sin u, r1 cos u, 0, 0) and fv = 0, 0, − sin(v/r2 ), cos(v/r2 ) , so that fu .fu = r1 2 , fu .fv = 0 and fv .fv = 1. The result follows from Proposition 2 of §4.3.

4.12. Let p and q be points of S̃, and let f be the map described in Example 4 of §4.3. By first applying a translation of the plane, we may assume that f (πr1 , πr2 , 0) = p. Then there exists 0 < u ≤ 2πr1 , 0 < v ≤ 2πr2 such that f (u, v, 0) √ = q. The line segment joining (πr1 , πr2 , 0) to (u, v, 0) has length at most π r1 2 + r2 2 , and the image of this line segment under f is a curve on S̃ joining p to q.

4.13. If we parametrise the helicoid as in Example 2 of §4.5, then E = G = cosh2 v, F = 0. A calculation gives that N u .N u = N v .N v = cosh−2 v and N u .N v = 0, so Proposition 3 of §4.4 shows that N is conformal with conformal factor λ = cosh−2 v.

4.14. The formula for f may be proved as in the solution to Exercise 2.1.

We note that a = 1/r1 > r2 /r1 = b, so that Ta,b is indeed a torus of revolution. A calculation shows that p if x1 2 + x2 2 = r1 2 and x3 2 + x4 2 = r2 2 , and if f (x1 , x2 , x3 , x4 ) = (x, y, z), then ( x2 + y 2 − a)2 + z 2 = b2 . Since r2 2 < 1, S ⊂ X, so that f maps S to Ta,b . If we parametrise S by taking

x(u, v) =

u v v u r1 cos , r1 sin , r2 cos , r2 sin r1 r1 r2 r2

,

then E = G = 1 and F = 0. −2 We now show conformality. Calculations show that fu .fu = 1−r2 sin(v/r2 ) and fu .fv = 0. A little more work using r1 2 + r2 2 = 1 shows that fv .fv = −2 1 − r2 sin(v/r2 ) , so the conformality of f follows from Proposition 3 of §4.4.

It also follows that fu and fv are linearly independent at each point, so the discussion at the beginning of §4.3 implies that if we show that f is bijective then it will follow that f is a diffeomorphism. To show that f is bijective, it is

20

SOLUTIONS TO EXERCISES

enough to show that the map

r1 , 0, r2 cos rv2 v v r1 , 0, r2 cos , r2 sin 7→ r2 r2 1 − r2 sin rv2 defines a bijective map from the circle x1 = r1 , x2 = 0, x3 2 + x4 2 = r2 2 in R4 onto the circle (x − a)2 + z 2 = b2 , y = 0, in R3 . We do this by showing (with some work) that the equation

r1 , 0, r2 cos rv2 r2 1 = (cos φ, 0, sin φ) + , 0, 0 1 − r2 sin rv2 r1 r1 has the unique solution

v (r1 sin φ, 0, r2 + cos φ) v . = cos , 0, sin r2 r2 1 + r2 cos φ 4.15. That the correspondence is an isometry is immediate from Proposition 1 of §4.5. Also N θ is a positive scalar multiple of (cos θxu + sin θx̃u ) × (cos θxv + sin θx̃v ) = (cos θxu − sin θxv ) × (cos θxv + sin θxu ) = xu × xv , which is independent of θ.

4.16. We use the notation of Example 1 of §4.6. The coordinate plane x = 0 in R3 intersects S 2 (1) in the meridian through (0, 1, 0), and the image of this under stereographic projection is the line u = 0 in the complex plane. This line is mapped by f to the line u = 1, and x maps this line to the intersection of S 2 (1) with the plane x + z = 1. The image of the equator of S 2 (1) is found in a similar manner. (Brief details: The image of the equator under f may be parametrised by (1+cos θ, sin θ) and the image of this under x consists of points of the form (2 + 2 cos θ, 2 sin θ, 1 + 2 cos θ) .

3 + 2 cos θ These points form the circle of intersection of S 2 (1) with the plane 2x − z = 1.)

Figure 7 shows the curves of intersection of S 2 (1) with (i) x = 0, (ii) z = 0,

Figure 7: For the solution to Exercise 4.16

A FIRST COURSE IN DIFFERENTIAL GEOMETRY

21

and also shows their images under f˜. The circle x2 + z 2 = 1, y = 0, is left setwise fixed.

4.17. (i) If p ∈ S then B(p, p) = −1, so, differentiating, we find that B(X, p) = 0. (ii) The line through (0, 0, −1) and (u, v, 0) may be parametrised as (0, 0, −1) + λ(u, v, 1). If u2 + v 2 < 1, this line intersects S at the unique point at which λ = 2/(1 − u2 − v 2 ). At this point (0, 0, −1) + λ(u, v, 1) is the given formula for f (u, v). (iii) The map f is also surjective since, if (x1 , x2 , x3 ) ∈ R3 , then the line through (0, 0, −1) and (x1 , x2 , x3 ) intersects the plane x3 = 0 at (x̃1 , x̃2 , 0) = (1 + x3 )−1 (x1 , x2 , 0), and a quick check shows that if (x1 , x2 , x3 ) ∈ S then (x̃1 )2 + (x̃2 )2 < 1. Finally, a calculation shows that fu =

2 (1 + u2 − v 2 , 2uv, 2u) , (1 − u2 − v 2 )2

and another calculation shows that hfu , fu i = B(fu , fu ) = 4/(1 − u2 − v 2 )2 = Ẽ.

Similar calculations show that hfu , fv i = 0 = F̃ , while hfv , fv i = G̃. The result now follows from Proposition 2 of §4.3. √ 4.18. First show that α0 (v) = (−v −2 , 0, v −2 v 2 − 1).

It is clear that the image of f is equal to S, and calculations similar to those used to compute the coefficients of the first fundamental form of a surface of revolution show that fu .fu = fv .fv = 1/v 2 , and fu .fv = 0. The result now follows from Proposition 3 of §4.4.

4.19. (i) Here, f˜ = xf x−1 so the conformal factor of f˜ at x(w) is the product of the conformal factors of x−1 at x(w), f at w and x at f (w). The result follows. (ii) This follows from (i) using the fact that f 0 (w) = (ad − bc)/(cw + d)2 . (iii) This follows since the limit as w → ∞ of the conformal factor of f˜ at x(w) is |ad − bc|/(|a|2 + |c|2 ), which is non-zero. (iv) Assuming that ad − bc = 1, the map f˜ is an isometry if and only if ¯ + (aw + b)(āw̄ + b̄) for all w. Comparing constants, 1 + ww̄ = (cw + d)(c̄w̄ + d) ¯ 0 = cd¯+ ab̄, and coefficients of w, and coefficients of ww̄, we obtain 1 = bb̄ + dd, 1 = aā + cc̄. The result follows.

4.20. We compute the conformal factor of the map f given in Example 4 of Appendix 2. We have that g(fu , fu ) =

(ad − bc)2 (cz + d)2 (cz̄ + d)2 (im f (z))2

.

A short calculation shows that if z = u + iv then the imaginary part of f (z) is equal to (ad − bc)v(cz + d)−1 (cz̄ + d)−1 . It then follows that g(fu , fu ) = 1/v 2 , so that the conformal factor of f is equal to 1.

22

SOLUTIONS TO EXERCISES

Chapter 5 √ 5.1. In this case, N = (−gu , −gv , 1)/ D, where D = 1 + gu 2 + gv 2 . Also, xuu = (0, 0, guu √ that L = √), xuv = (0, 0, guv ), xvv = √(0, 0, gvv ). It follows guv / D, and N = gvv / D. xuu .N = guu / D, and similarly M = √ √ For the given function g, we have that L = 2/ D, M = 0, and N = 2/ D, where D = 1 + 4u2 + 4v 2 . Hence (LN − M 2 ) = 4/D > 0, so that K = (LN − M 2 )/(EG − F 2 ) > 0.

5.2. Finding E, F and G is straightforward. Further calculations show that N = (−2u, 2v, −D + 2)/D, where D = 1 + u2 + v 2 . Since xuu = (−2u, 2v, 2), xuv = (2v, 2u, 0), and xvv = (2u, −2v, −2), more calculations show that L = xuu .N = 2, and similarly, M = 0 and N = −2. That H = 0 follows immediately from Lemma 2 of §5.4.

5.3. The coefficients of the first fundamental form are quickly found to be E = G = cosh2 v, F = 0. Also, N = (− sin u, cos u, − sinh v)/ cosh v, while xuu = (− sinh v cos u, − sinh v sin u, 0) , xuv = (− cosh v sin u, − cosh v cos u, 0) , xvv = (sinh v cos u, sinh v sin u, 0) .

Hence L = xuu .N = 0, and similarly M = 1 and N = 0. Hence K = (LN − M 2 )/(EG − F 2 ) = − cosh−4 v, and, using Lemma 2 of §5.4, we find that H = 0. It then follows from Lemma 2 of §5.5 that the principal curvatures are ± cosh−2 v.

5.4. We found in Example 1 of §3.2 that E = 1 + gu 2 ,

F = gu gv ,

G = 1 + gv 2 ,

while L, M and N were found in Exercise 5.1. It follows that K = (LN − M 2 )/(EG − F 2 ) = (guu gvv − guv 2 )/D2 , where D = 1 + gu 2 + gv 2 . Also, from (5.19), gvv (1 + gu 2 ) − 2gu gv guv + guu (1 + gv 2 ) .

H= 2D3/2 5.5. (a) Here, xu = α0 + vα00 , xv = α0 , so that N is the unit vector in direction α00 × α0 . This is independent of v. (b) Here, N is the unit vector in direction α0 × α. (c) Here, N is the unit vector in direction α0 × e.

5.6. Let x(u, v) = α(u) + vβ(u) sweep out a ruled surface as described in §3.6. Then xvv = 0, so that N v .xv = −N .xvv = −N = 0. Hence K = −M 2 /(EG − F 2 ) so that K = 0 if and only if M = 0. Also, N v .xu = −M , so that N v = 0 if and only if M = 0.

A FIRST COURSE IN DIFFERENTIAL GEOMETRY

23

5.7. Differentiating, we find that if X is a tangent vector to S, then df λ (X) = X + λdN (X). (i) Since dN maps vectors tangential to S to vectors tangential to S, we see that N λ f λ = N . (ii) It follows that, if X is tangential to S, then dN λ (df λ (X)) = dN (X). In particular, if dN (X) = −κX then df λ (X) = (1 − λκ)X, so that

−κX = dN (X) = dN λ df λ (X) = (1 − λκ)dN λ (X) , and the result follows. (iii) Let κ1 and κ2 be the principal curvatures of S. Then, since K λ is the product of the principal curvatures of S λ , we have that Kλ =

κ1 κ2 κ1 κ2 , = (1 − λκ1 )(1 − λκ2 ) 1 − λ(κ1 + κ2 ) + λ2 κ1 κ2

which leads to the desired expression for K λ . The expression for H λ is obtained in a similar manner. (iv) This is a direct substitution.

5.8. Corollary 5 of §5.5 says that K = −f 00 /f . (i) Here, f 00 + f = 0, so, replacing v by v + c for a suitable constant c, we can take f (v) = A cos v for a positive constant A and −π/2 < v < π/2. However, since |α0 | = 1 we need |f 0 (v)| < 1 (since K = 0 if g 0 (v) = 0). Hence the domain of the generating curve is as stated in the exercise.

Having determined f (v), then g(v) is given by the indefinite integral Z q 2 g(v) = 1 − f 0 (v) dv .

Figure 8 gives sketches of the requested generating curves. Note that the

Figure 8: For the solution to Exercise 5.8(i) case A = 1 gives the unit sphere. (ii) Here f 00 = 0, so that f (v) = Av + B for suitable constants A and B, and we need −1 ≤ A ≤ 1. If A = 0 then B > 0, g(v) = ±v+C, and we√obtain a cylinder.

If A 6= 0, we may assume that B = 0. In this case, g(v) = ±v 1 − A2 + C, and hence S is a plane if A = 1 and a cone otherwise.

24

SOLUTIONS TO EXERCISES

(iii) Here, f 00 − f = 0, so that f (v) = λev + µe−v .

If λ and µ are both non-zero and have the same sign (which must be positive), then, replacing v by v +c for a suitable constant c, we may write f (v) = A cosh v for some positive constant A. Then the domain of the generating curve is (−v0 , v0 ) where A sinh v0 = 1.

If one of λ and µ is zero, we may assume that f (v) = e−v , v > 0. The corresponding surface is the pseudosphere.

If λ and µ are both non-zero and have opposite signs, then, replacing v by ±v + c for a suitable constant c, we may write f (v) = B sinh v, v > 0, for some positive constant B. This time the domain of the generating curve is (0, v0 ) where B cosh v0 = 1. In particular, 0 < B < 1.

Figure 9 gives sketches of the requested generating curves.

Figure 9: For the solution to Exercise 5.8(iii)

5.9. We may find κ1 and κ2 by putting f (v) = v in (5.23), and the expression for K = κ1 κ2 is then immediate. In particular, we note that the sign of K is the same as the sign of g 0 g 00 . The given surface is the surface of revolution ob tained by rotating the curve v, 0, 1/(1 + v 2 ) , v > 0, about the z-axis (together with a pole at (0, 0, 1)) and, from working out the derivatives, it follows that √ the sign of K is equal to√the sign of 1 − 3v 2 . So, K > 0 when 0 < v < 1/ 3, and K < 0 when v > 1/ 3. At the pole, the curvature is obtained by taking the limit as v → 0, and this gives curvature 4 at the pole.

Figure 10 gives a sketch of the generating curve and its reflection in the

Figure 10: For the solution to Exercise 5.9 z-axis, and indicates those parts of the curve which give points of positive Gaus-

A FIRST COURSE IN DIFFERENTIAL GEOMETRY

25

sian curvature and points of negative Gaussian curvature on the corresponding surface of revolution.

5.10. This is a short calculation using the formula for H given in Proposition 4 of §5.6.

5.11. It follows from the formulae in Example 6 of §5.6 that the torus of revolution has points where H = 0 if and only if we can find a value of v such that 2b cos v = −a. Such a v exists if and only if a ≤ 2b.

5.12. Parametrise S by taking x(u, v) = (af (v) cos u, bf (v) sin u, v) .

Calculations show that N = (b cos u, a sin u, −abf 0 )/D, where D is the modulus of the numerator. Then L = xuu .N = −abf /D, M = 0, N = −abf 00 /D. Hence K is a positive scalar multiple of −a2 b2 f f 00 , and the result follows.

5.13. (i) If we parametrise the (upper half of the) cone as usual by taking x(u, v) = (v cos u, v sin u, v) ,

v>0,

√ then calculations give that E = v 2 , F = 0, G = 2, while L = −v/ 2, M = N = t 2t 0. At α(t) we have u = t and √ v = e , so that, at α(t), we have E = e , F = 0 and G = √ 2, while L = −et / 2, M = N = 0. Formula (5.33) now shows that κn = −(3 2et )−1 . (ii) Methods described in Chapter 1 show that if t is the unit tangent vector to α then dt 1 = t (− sin t − cos t, cos t − sin t, 0) . ds 3e √ Usual methods show that the unit normal to the cone at α(t) is (cos t, sin t, −1)/ 2, and a calculation using the formula for κn given in the exercise yields the same answer as that found in (i).

5.14. Let x(u, v) be a local parametrisation whose image is V . Then (x.N )u = xu .N = 0, and, similarly, (x.N )v = 0. Hence x.N is constant, and V is contained in a plane.

2 5.15. Here E = 1 + 4u2 , F = 4uv, √ and G = 1 + 4v . In the √ solution to Exercise 5.1 we found that L = 2/ D, M = 0, and N = 2/ D, where D = 1 + 4u2 + 4v 2 . Hence, using (5.41), we may show that the given curve is a line of curvature if and only if the given equation holds.

So, either uu0 + vv 0 = 0 or u0 v − uv 0 = 0. The first integrates up to give 2 u + v 2 = constant and the second gives u/v = constant.

In fact, S is a surface of revolution, and the families of curves we have found are the parallels and the meridians. We noted near the start of §5.9 that these are indeed the lines of curvature on a surface of revolution.

5.16. Let x(u, v) be a local parametrisation of S whose image contains no umbilics and whose coordinate curves are lines of curvature. Lemma 1 of §5.6

26

SOLUTIONS TO EXERCISES

shows that F = M = 0. It follows that K = LN/EG, so we may assume that L = 0 and N 6= 0. It then follows that N u = 0 and N v = −(N/G)xv . For each fixed v0 , consider the coordinate curve α(u) = x(u, v0 ). Then N is constant along α, and we now show that xuu is a scalar multiple of xu which will show that α is a straight line. To do this, we note that xuu .N = −L = 0, while xuu .xv is zero if and only if xuu .N v = 0. However, this latter follows because xuu .N v = (xu .N v )u − xu .N uv = 0.

5.17. We found E, F , G; L, M and N for Enneper’s surface in Exercise 5.2. (i) Since F = M = 0, the principal curvatures are L/E and N/G. (ii) This is immediate from Lemma 1 of §5.6. (iii) It follows from (5.43) that the asymptotic curves are given by u02 − v 02 = 0, that is, u ± v = constant.

Fix k1 and k2 and consider the curves u + v = k1 and u − v = k2 . These intersect at the point corresponding to u = (k1 + k2 )/2, v = (k1 − k2 )/2.

5.18. In the solution to Exercise 5.3, we found that L = N = 0 and M = 1. Hence the asymptotic curves satisfy u0 v 0 = 0, and so are the families of coordinate curves. These are the z-axis, the lines of the ruling of the helicoid, and the helices obtained by intersecting the helicoid with the cylinders whose axis is the z-axis.

5.19. Routine calculations show that E = sech2 v, F = 0 and G = tanh2 v.

We may use (5.22) to show that L = −N = −sech v tanh v, and M = 0. It now follows from (5.43) that the asymptotic curves are u ± v = constant. The tangent vectors to the asymptotic curves are xu ∓ xv , and the angle θ between these is given by cos θ =

sech2 v − tanh2 v (xu + xv ).(xu − xv ) = = 2 sech2 v − 1 . |xu + xv | |xu − xv | sech2 v + tanh2 v

5.20. A line has zero curvature, so the normal curvature of a line on a surface is also zero. It follows from Proposition 4 of §5.7 that the principal curvatures at a point of a ruled surface cannot both have the same sign. Hence K ≤ 0.

5.21. Let α be an asymptotic curve on a surface S, and assume that α is parametrised by arc length. Then α00 .N = 0, so that the principal normal n of α is also orthogonal to N . Hence, if b is the binormal of α then (by choosing the correct sign for N ), we have that b = N . Hence b0 = N 0 = τ n, so, if θ is the angle between α0 and the principal direction with principal curvature κ1 , then, by (5.35), τ 2 = |N 0 |2 = κ1 2 cos2 θ + κ2 2 sin2 θ. Since α is an asymptotic curve, 0 = II(α0 ) = κ1 cos2 θ + κ2 sin2 θ, so that τ 2 = −κ1 κ2 (sin2 θ + cos2 θ) = −K. 5.22. Let e1 and e2 be unit vectors in the two principal directions, and let κ1 and κ2 be the corresponding principal curvatures. Then cos θe1 + sin θe2 is in an asymptotic direction if and only if κ1 cos2 θ + κ2 sin2 θ = 0, which happens

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27

if and only if tan2 θ = −κ1 /κ2 . This implies that cos θe1 + sin θe2 is in an asymptotic direction if and only if cos(−θ)e1 + sin(−θ)e2 is in an asymptotic direction, so the result follows.

Since lines of curvature intersect orthogonally, the equivalence of the three conditions follows because θ = π/4 if and only if κ1 = −κ2 .

5.23. The vector α00 is a linear combination of α0 and n. Hence n.N = 0 if and only if α00 .N = 0. This holds if and only if II(α0 ) = 0.

5.24. Let α(t) be a parametrisation of the regular curve C. Then N 1 .N 2 = const ⇐⇒ N 1 0 .N 2 + N 1 .N 2 0 = 0 ⇐⇒ κα0 .N 2 + N 1 .N 2 0 = 0 .

However, α0 is orthogonal to N 2 , and N 1 .N 2 0 = 0 if and only if N 2 0 is orthogonal to both N 1 and N 2 . This happens if and only if N 2 0 is a scalar multiple of α0 .

5.25.√ Using the notation of the solution √ √ to Exercise 5.1, we have that L = 6u/ D, M = −6v/ D, and N = −6u/ D. Hence LN − M 2 < 0 except at u = v = 0, in which case LN − M 2 = 0.

5.26. Let α(t) = x(t), y(t), z(t) be a regular curve on S. It follows from (5.34) that the sign of the normal curvature κn of α is the same as that of α00 .grad f .

Differentiating x4 + y 4 + z 4 = 1 with respect to t we find that x3 x0 + y 3 y 0 + z 3 z 0 = 0 , and, differentiating again, 2

2

2

3(x2 x0 + y 2 y 0 + z 2 z 0 ) + x3 x00 + y 3 y 00 + z 3 z 00 = 0 .

Hence, 2

2

2

α00 .grad f = 4(x3 x00 + y 3 y 00 + z 3 z 00 ) = −12(x2 x0 + y 2 y 0 + z 2 z 0 ) .

Thus κn is always negative unless: (i) any two of x, y, z are zero, in which case all normal curvatures are zero; (ii) exactly one of x, y, z is zero. If x = 0, then the rate of change of N in direction (1, 0, 0) is zero, but, for instance, the rate of change of N in direction (0, −z 3 , y 3 ) is non-zero.

The result now follows.

5.27. For a general local parametrisation x(u, v) we have (u2 xuu + 2uvxuv + v 2 xvv ).N = u2 L + 2uvM + v 2 N = II(uxu + vxv ) .

The result follows by applying this to the given local parametrisation at (0, 0).

28

SOLUTIONS TO EXERCISES 5.28. Parametrise S by setting x(u, v) = (v cos u, v sin u, v 2 ) ,

v > 0 , −π < u < π .

Then E = v 2 , F = 0, G = 1 + 4v 2 . Also, using Exercise 5.9, K = 4(1 + 4v 2 )−2 .

Hence, ZZ Z Z ∞

π

4v(1 + 4v 2 )−3/2 du dv = 2π.

|K|dA = S

−π

0

This is equal to the area of the lower hemisphere of S 2 (1), which, as found in the solution to Exercise 4.3, is the image of the Gauss map of the paraboloid of revolution.

5.29. We use the argument and notation of the proof of Theorem 4 of §5.12. If α(t) is a curve on S with α(0) = p0 we have that α0 (0).q0 = 0 and α00 (0).q0 ≤ 0. It follows that N (p0 ) = q0 , and all the normal curvatures of S at p0 are non-positive. Hence p0 isn’t a hyperbolic point.

Chapter 6 6.1. No; for instance, there is a local isometry from the plane to the cylinder.

6.2. The expressions for the Christoffel symbols follow immediately from equations (6.4). The Gauss formula (6.9) now shows that GK = −Guu /2 + Gu 2 /4G, from which the result follows. The formula for the Gaussian curvature may also be obtained using equation (6.10) for K in orthogonal coordinates.

6.3. Equations (6.4) may be used to show that Γ122 = Γ222 = 0, and that 1 Γ12 = −Γ212 = v/(1 + v 2 ). The Gauss formula (6.9) may then be used to obtain the desired expression for K. Alternatively, since the Gauss formula is difficult to remember(!), we may mimic the method of proof of that formula to find the expression for K as follows.

LN = LN .N N = (xuu − Γ111 xu − Γ211 xv ) .(xvv − Γ122 xu − Γ222 xv ) = (xuu − Γ111 xu − Γ211 xv ) .xvv = xuu .xvv , while M 2 = M N .M N , which, after a little work gives M 2 = xuv .xuv −

v2 .

1 + v2

Hence LN − M 2 = xuu .xvv − xuv .xuv +

v2 .

1 + v2

Then, using (6.8), we find that LN −M 2 = −1/(1+v 2 ), so obtaining the desired expression for K.

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6.4. This is very quick using the formula for the Gaussian curvature of a Tchebycheff parametrisation given in Example 2 of §6.2.

6.5. Let F = cos θ. Then Fu = −θu sin θ, Fv = −θv sin θ, and Fuv = −θuv sin θ − θu θv cos θ. The required formula now follows from the formula for the Gaussian curvature of a Tchebycheff parametrisation given in Example 2 of §6.2.

6.6. The given parametrisation x of the helicoid has E = 1 + v 2 , F = 0, √ 2 G = 1, L = N = 0, M = 1/ 1 + v . Hence K = −(1 + v 2 )−2 . The formula given in Exercise 5.9 may be used to show that if K̃ is the Gaussian curvature of S̃ then K̃ = −(1 + v 2 )−2 also. However, the given map isn’t an isometry since Ẽ = v 2 + v −2 6= E. 6.7. If L = M = N = 0, then N is constant and S is an open subset of a plane. (i) Equations (6.4) show that all the Christoffel symbols are zero, so that xuu = LN , xuv = M N , and xvv = N N . So, from (6.7) and (6.8), we see that LN − M 2 = 0. Hence K = 0. (ii) If L = N = 0 then M = 0 also. Hence, assuming that S is not a plane, we may assume that L 6= 0.

From now on, we assume that L 6= 0. (iii) Then au = LM N −LM N = 0 and similarly av = 0. Since S is connected, it follows that a is a non-zero constant vector. (iv) That a.a = L(EN − 2F M + GL) is quick using LN − M 2 = 0. (v) We first show that bu .xu = 0. In fact xu .a a.xu + cN u .xu bu .xu = xu .xu − a.a (EM − F L)2 =E− − cL , L(EN − 2F M + GL) which, on substituting for c and simplifying, is seen to be zero.

Similar calculations show that bu .xv = bv .xu = bv .xv = bu .N = bv .N = 0, so that bu = bv = 0. It now follows that b is constant. (vi) Just show that a.b = 0. (vii) It now follows from part (v) that (b1 , b2 , 0) = (x1 , x2 , x3 ) − (0, 0, x3 ) + cN , so that |(b1 − x1 , b2 − x2 , 0)|2 = c2 , and the result follows.

6.8. It follows from Lemma 1 of §5.6 that the coordinate curves are lines of curvature if and only if F = M = 0. In this case κ1 = L/E and κ2 = N/G, and the first Codazzi-Mainardi equation (6.16) becomes 2Lv = Ev (κ1 + κ2 ).

However, Lv = (Eκ1 )v = Ev κ1 + E(κ1 )v , so we see that the first CodazziMainardi equation holds if and only if the first of the given equations holds.

The equivalence of the other pair of equations is proved similarly (or simply interchange u and v).

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SOLUTIONS TO EXERCISES

6.9. Using the equations from Exercise 6.8, we see that the first CodazziMainardi equation gives (L − N )v = {E(κ1 − κ2 )}v = Ev (κ1 − κ2 ) + E(κ1 − κ2 )v = −2E(κ1 )v + E(κ1 − κ2 )v = −E(κ1 + κ2 )v .

Hence, if κ1 + κ2 is constant then (L − N )v = 0. Interchanging u and v shows that (L − N )u = 0 also, so that L − N is constant.

6.10. We saw in Example 2 of §6.2 that if a surface S has a Tchebycheff parametrisation with E = G = 1 and F = cos θ, with θ satisfying the sineGordon equation, then S has constant Gaussian curvature K = −1. If we also want the coordinate curves of x to be asymptotic curves then, by (5.43), we need to have L = N = 0. In order to satisfy (EG − F 2 )K = LN − M 2 with K = −1 we also need to have M = ± sin θ, and, reflecting if necessary, we assume M = sin θ (using the unit normal in direction xu × xv ).

We now show we can apply Bonnet’s theorem for the above values of E, F , G, L, M , N . Example 2 of §6.2 showed that the right hand side of (6.9) is equal to −θuv sin θ, so (6.9) reduces to 1 − cos2 θ = θuv sin θ, which holds since θuv = sin θ.

We see from the first pair of equations of (6.5) that Γ111 sin θ = θu cos θ so the right hand side of (6.14) reduces to −θu cos θ, as does the left hand side. A similar argument (or just reverse the roles of u and v) shows that (6.15) also holds. Hence we can apply Bonnet’s theorem to prove the required result.

6.11. Let λ and µ be the principal curvatures of a compact connected surface S in R3 with constant mean curvature H and everywhere positive Gaussian curvature K. Since S is compact, the globally defined continuous function (λ − µ)2 = 4(H 2 − K) attains its maximum at some point p0 ∈ S. If p0 is an umbilic point, then every point of S is an umbilic point and S is a sphere. So, assume that, at p0 , λ > µ > 0. Then λ − µ has a maximum at p0 and, since λ + µ is constant, λ has a maximum at p0 and µ has a minimum. Hence, at p0 , λu = λv = 0, λuu ≤ 0, λvv ≤ 0, and µu = µv = 0, µuu ≥ 0, µvv ≥ 0. If we differentiate (6.19) at p0 , we find that, at p0 , Evv (λ − µ) = −2Eλvv ≥ 0 ,

and

Guu (λ − µ) = 2Gµuu ≥ 0 .

Continuing as in the proof of Liebmann’s theorem (Theorem 1 of §6.4), we find that K(p0 ) ≤ 0, which contradicts one of our initial assumptions.

6.12. If K > 0 then Theorem 5 of §6.4 shows that S is a sphere. Hilbert’s theorem shows that K can’t be a negative constant, so we now assume that K = 0. In this case, by Massey’s theorem, S is a generalised cylinder, and so has a parametrisation of the form y(u, v) = α(u) + ve , where α0 is never a scalar multiple of the constant non-zero vector e. Assuming without loss of generality that e = (0, 0, 1), we see that if α(u) =

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x(u), y(u), z(u) then the curve β(u) = (x(u), y(u), 0) is also regular, and, by replacing v with v + z(u), we may assume that y(u, v) = β(u) + v(0, 0, 1) .

Now assume that β(u) is parametrised by arc length, and let n(u) = (−y 0 (u), x0 (u), 0) be its principal normal. If κ is the curvature of β and N is the unit normal of S, then y u = β 0 , y v = (0, 0, 1), N = −n, y uu = β 00 = κn, y uv = y vv = 0.

Hence E = 1, F = 0, G=1, L = −κ, M = 0, N = 0. Since H is constant, so is κ. Hence the trace of β is a line or a circle, so that S is a plane or a round cylinder.

Chapter 7 7.1. We note that dα/ds = α0 /|α0 |, so that

α00 1 d 1 d2 α = 02+ 0 α0 . ds2 |α | |α | dt |α0 | It follows from (7.1) that κg = (d2 α/ds2 ).(N × dα/ds), and the desired formula follows if we substitute the expressions obtained above for dα/ds and d2 α/ds2 .

7.2. The Gauss map of S 2 (1) is the identity map, so if α is a regular curve on S 2 (1) then the first formula of §7.1 shows that α00 = κg α × α0 + κn α. Hence, if κg = c then v 0 = α × α00 + cα0 = −cα0 + cα0 = 0 .

Also, α.v = c, so that α lies on the plane√ perpendicular to v whose perpendicular distance from the origin is c/|v| = c/ 1 + c2 .

7.3. This may be done using a very similar method to that employed in Example 2 of §7.1. Specifically, α0 = r(− cos v0 sin t, cos v0 cos t, 0), so that |α0 | = r cos v0 and α00 = −r(cos v0 cos t, cos v0 sin t, 0). Also, N = (1/r)α so a short calculation using (7.2) shows that κg = (1/r) tan v0 as required.

7.4. Parametrise Cv as usual by setting

α(u) = f (v) cos u, f (v) sin u, g(v) .

Then find α0 , α00 and N (using, for instance, Example 2 of §4.1). Then use (7.2) to find κg .

7.5. (a) Note that (g 0 , 0, −f 0 ) is orthogonal to S at (f, 0, g), and the line through (f, 0, g) in direction (g 0 , 0, −f 0 ) is given by (f, 0, g) + λ(g 0 , 0, −f 0 ). This meets the z-axis when λ = −f /g 0 , at which point the z-coordinate is given by h = g + f f 0 /g 0 . (b) A short calculation shows that |(0, 0, h) − (f cos u, f sin u, g)|2 = r2 , so

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SOLUTIONS TO EXERCISES

that Cv lies on S̃. Also, using (a), the outward unit normal Ñ to S̃ at (f cos u, f sin u, g) is given by Ñ =

f 1 (f cos u, f sin u, g − h) = 0 (g 0 cos u, g 0 sin u, −f 0 ) . r gr

The unit normal N of S is given by N=

(g 0 cos u, g 0 sin u, −f 0 ) p , f 0 2 + g0 2

so that Ñ = N , where = g 0 /|g 0 | . which proves (b). (c) It follows from Exercise 7.3 that if −π/2 < µ < π/2 is such that tan µ = (g − h)/f then κ̃g = (1/r) tan µ .

However, 2

2

r2 = f 2 + (g − h)2 = f 2 + (f f 0 /g 0 )2 = (f 0 + g 0 )(f /g 0 )2 , while tan µ = (g − h)/f = −f 0 /g 0 , from which the given formula for κ̃g follows. (d) It follows from Exercise 7.4 that κ̃g = κg , which is in accordance with Proposition 3 of §7.1 since Ñ = N .

7.6. That α lies on the given cone is easy to check. Also, α0 = et (cos t − sin t, sin t + cos t, 1) , so that |α0 | =

√

3et ,

while α00 = et (−2 sin t, 2 cos t, 1) .

The unit normal N to the cone at (x, y, z) is in direction grad (x2 + y 2 − z 2 ), so that √

N α(t) = (cos t, sin t, −1)/ 2 .

A calculation using formula (7.2) for geodesic curvature now shows that √ κg = 1/( 6et ).

7.7. Let α(u) = (u, v0 , u2 − v0 2 , 2uv0 ). A quick check shows that x(u, v) is an orthogonal (in fact, isothermal) parametrisation, so that X = xv /|xv | is a unit vector tangential to S and orthogonal to α0 . Formula (7.3) for geodesic curvature quickly gives that κg = −4v0 (1 + 4u2 + 4v0 2 )−3/2 .

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7.8. (a) Let α(s) √ = x(u(s), v0 ) be a parametrisation of v = v0 by arc length.

Then du/ds = 1/ E, and, since xu and xv are orthogonal, N × dα/ds = xv /|xv |. Hence, using (7.1) (or (7.3) for a surface in Rn ),

xv 1 Ev 1 d 1 d2 α xv √ √ κg = . x = √ xuu .xv = − √ , = u .√ ds2 |xv | E du E G E G 2E G which leads to the given answer. (b) Substitute the expressions for E, F and G in terms of f and g. (c) Using (a), we √ see that the geodesic curvature of the coordinate curve u = constant is (1/2 E)(log G)u , and the result follows. (d) The standard parametrisation of a surface of revolution has F = 0 and Gu = 0.

7.9. Here, xv = b, and xs = α0 + vb0 . Hence α00 .xv = κn.b = 0, while, when v = 0, α00 .xs = α00 .α0 = 0. Hence α00 is orthogonal to S and the result follows.

7.10. (a) Let Ñ be the (constant) unit normal to P . Then Ñ and α0 are non-zero orthogonal vectors in Tα(t) S, and hence span Tα(t) S. However, 0 α00 .Ñ = α0 .Ñ = 0, while, since α is parametrised proportional to arc length, α00 .α0 = 0. It follows that α00 is orthogonal to S, and so α is a geodesic on S. (b) Let f (x, y, z) = x4 + y 6 + z 8 . The unit normal to S at (x, y, z) is in direction grad f = (4x3 , 6y 5 , 8z 7 ). If P is the yz-plane, then Ñ = (1, 0, 0) which is orthogonal to N (0, y, z). The result for P now follows from (a), and the other coordinate planes are handled similarly.

7.11. (a) Using the standard notation, the assumptions imply that N 0 = λα and α00 = µN for some scalar functions λ and µ. It follows that (N ×α0 )0 = 0 so that N × α0 is constant. It now follows that α.(N × α0 ) is also constant, so that α lies on a plane with normal N × α0 and the result follows. (b) This time the assumptions imply that, if Ñ is the (constant) unit normal to P , then α0 .Ñ = 0 and α00 = µN with µ never zero. Then 0 = α00 .Ñ = µN .Ñ , which implies that N .Ñ = 0, so that N 0 .Ñ = 0. Since Ñ is tangential to S and orthogonal to α0 , it follows that N 0 is a scalar multiple of α0 and the result follows.

0

7.12. First show that xu is orthogonal to xv .

(i) and (iii) Let γ(t) = x(π, t) = π, 1 + cosh t, 4 sinh(t/2) /2. This is the parabola x = π/2, y − 1 = z 2 /4.

Since xu and xv are orthogonal, to show that (when parametrised proportional to arc length) the coordinate curve u = π is a geodesic, it is enough to prove that γ 00 (t) is orthogonal to xu γ(t) . This is a routine calculation. (ii) and (iii) This time we take γ(t) = (t − sin t, 1 − cos t, 0)/2. This is a cycloid

in the xy-plane, and it is quick to show that γ 00 (t) is orthogonal to xv γ(t) .

7.13. Every planar point is an umbilic. If p ∈ S is not a planar point, then,

by Exercise 7.11, there are an infinite number of lines of curvature through p.

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SOLUTIONS TO EXERCISES

This implies that p is also an umbilic, so the result follows from Theorem 1 of §5.8.

7.14. Assume that α(t) is a closed geodesic on a helicoid, and let it attain its maximum height at α(t0 ), say. Then α is tangential to the straight line on the helicoid through α(t0 ), so, by uniqueness of geodesics, the trace of α(t) must be this line. Hence α(t) is not a closed curve.

7.15. (a) The segment of the unit circle which lies in the sector has length φ, and when we bend the sector to form the cone, this segment maps to the

Figure 11: For the solution to Exercise 7.15(a) 2 2 2 parallel on the cone p which is a circle of radius p r where r + β r = 1 (Figure 2 2 11). Hence r = 1/ 1 + β so that φ = 2π/ 1 + β , and the result follows. (b),(c) and (d) Cut the plane along a half-line to the origin, and then form a cone by sliding the edges of the cut past each other until the line segments forming an angle φ at the origin line up to give a generator of the cone. A line ` in the plane not intersecting the cut gives a maximal geodesic on the cone, and two points p1 and p2 on ` give the same point on the geodesic if and only if they are equidistant from the origin 0 and the angle ∠p1 0p2 is an integer multiple of φ.

If φ ≥ π, then no geodesic has self-intersections. If π > φ ≥ π/2 then every geodesic which is not a meridian has exactly one self-intersection. If π/n > φ ≥ π/(n + 1) for some integer n then every geodesic which is not a meridian has exactly n self-intersections. Figure 12 illustrates this for n = 2.

In this figure, the rays from the origin drawn in a solid line map to the same meridian of the cone, and those drawn in a dashed line map to the opposite meridian. The points p1 and p2 (resp. q1 and q2 ) give a point where the geodesic has a self-intersection.

The result now follows from (a).

Figure 12: The horizontal line gives a geodesic on the cone (Exercise 7.15) 7.16. (a) A short calculation using equation (7.11) shows that (u0 /v 2 )0 = 0 if and only if (7.11) holds. We also note that, in this case, u00 = 2u0 v 0 /v.

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(b) The curve u(t), v(t) in H is parametrised proportional to arc length if and only if 0 2 0 2 u v + = const .

(1) v v However, a calculation shows that if u00 = 2u0 v 0 /v then !

0 2 0 2 !

2 2 d u v v0 u0 00 0 + +v − . =v dt v v v v Hence (1) holds if and only if either (7.12) holds or v is constant. However, if v is constant, then (7.12) shows that u is also constant. (c) The curves (u0 , v0 ekt ) are geodesics. If u0 = cv 2 with c 6= 0, and if u02 +v 02 = kv 2 with k 6= 0, then √ k − c2 v 2 v0 = , u0 |c|v so that

Z

√

u − u0 = |c|

v dv , k − c2 v 2

which integrates up to give the desired answer.

7.17. Let α(s) = u(s), v(s) be a curve on H parametrised by arc length s, and let 0 denote differentiation with respect to s. Then u0 + v 0 = v 2 ,

2

2

(2)

κg = (2u0 /v) + φ0 .

(3)

while (7.14) gives that However, cos φ = u0 /v so a short calculation gives that φ0 =

u0 v 0 − u00 v . vv 0

So, if κg is equal to a constant, k, say, then (3) quickly gives that u00 − 2

u0 v 0 + kv 0 = 0 . v

(4)

We now solve (2) and (4). We begin by noting that (4) integrates to give u0 = v 2 c + (k/v) for some constant c, so, from (2), we deduce that 2

2

u0 =

v 0 (cv + k)2 .

1 − (cv + k)2

Taking square roots and integrating, we find that the curves in H of constant geodesic curvature are the intersection with the upper half-plane of (Euclidean) circles in R2 (together with the lines u = constant).

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SOLUTIONS TO EXERCISES

7.18. The map f (u, v) = (cos u, sin u, cos v, sin v) is a local isometry of the plane onto the flat torus, so the geodesics on the latter are the images under f of straight lines in the plane. The geodesics through (1, 0, 1, 0) come from lines through the origin, and so take the form t 7→ (cos kv, sin kv, cos v, sin v). This returns to (1, 0, 1, 0) (in which case the geodesic closes up) if and only if k is a rational number.

7.19. Use either Exercise 7.8 or (7.14) to see that νg = −

Ev √ , 2E G

µg =

Gu √ .

2G E

Hence, from (7.14), if α(s) = x u(s), v(s) , then √ du √ dv dφ κg = νg E + µg G + . ds ds ds However, cos φ = α0 . and, similarly, sin φ =

√

√ du xu du E √ = E = , |xu | ds E ds

Gdv/ds. The result now follows.

7.20. Here we are rotating (v, 0, βv) about the z-axis, so Clairaut’s relation gives v cos θ = (1/β) cos θ0 . In particular, z(t) = βv ≥ cos θ0 . Note that this result may also be shown using the ideas of Exercise 7.15 - the parallels of the cone come from circles with centre at the origin of the plane; now use the fact that if ` is a line in the plane cutting the unit circle at angle θ0 then the perpendicular distance of ` from the origin is cos θ0 .

7.21. Here we are rotating (cosh v, 0, v) about the z-axis. Let x(u, v) be the standard parametrisation of the surface, and let α(t) = x u(t), v(t) be a closed geodesic. Assume that cosh v(t) attains its maximum at t = t0 . Then |v(t)| also attains its maximum at t = t0 . Hence v 0 (t0 ) = 0 so that α is tangential to the parallel at α(t0 ). Hence from Corollary 3 of §7.6, cosh v(t) also attains its minimum at t0 , so that cosh v(t), and hence v(t), is constant and α is a parallel.

However, Lemma 1 of §7.6 shows that the parallel v = v0 is a geodesic if and only if sinh v0 = 0, so that the only closed geodesic is the parallel v = 0 (when parametrised proportional to arc length).

7.22. We may obtain this surface by rotating (cosh v, 0, sinh v) about the z-axis. The solution is now word-for-word the same as for Exercise 7.21.

7.23. Here we are rotating (v, 0, 1/v 2 ) about the z-axis. Arguing as in

the solution to Exercise 7.21, we see that if α(t) = x u(t), v(t) is a closed geodesic and if v(t) attains its maximum at t = t0 , then v 0 (t0 ) = 0 so that α is tangential to the parallel at α(t0 ). It follows from Corollary 3 of §7.6 that α must be a parallel, but, for this surface, Lemma 1 of §7.6 shows that no parallels are geodesics.

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7.24. As usual, we assume that S is obtained by rotating f (v), 0,

g(v) , f (v) > 0 ∀v, about the z-axis. If α(t) = x u(t), v(t) , then f v(t0 ) is the maximum value of f v(t) so that fv v(t0 ) v 0 (t0 ) = 0. If fv v(t0 ) = 0 then, by Lemma 1 of §7.6, the parallel through α(t0 ) is a geodesic (when parametrised proportional to arc length). Now assume that v 0 (t0 ) = 0. Then α is tangential to the parallel at α(t0 ), so Corollary 3 of §7.6 shows that α is a parallel (which is thus a geodesic).

The result does not hold if “maximum” is replaced by “minimum”. Take, for example, any non-equatorial great sphere on S 2 (1) or any geodesic on the cone which is not a meridian.

7.25. Clairaut’s relation says that, along the geodesic α(t) = x u(t), v(t) , we have (a + b cos v) cos θ = (a + b) cos θ0 .

If α also intersects the parallel v = π at an angle θ1 , say, then (a − b) cos θ1 = (a + b) cos θ0 .

For this to have a solution θ1 we need that cos θ0 ≤ (a − b)/(a + b). However, if we have equality, then cos θ1 = 1 so that the geodesic α is tangential to the parallel v = π. Since this parallel is itself a geodesic, the uniqueness theorem for geodesics would imply that α is also the parallel v = π, which gives a contradiction.

Conversely, assume that for some > 0 we have that cos θ0 = (a − b − )/(a + b) .

2

We also assume that α is parametrised by arc length, which implies that u0 E + √ 02 0 v G = 1 and that cos θ = u E. Since G = b2 , these equations imply that 2

v 0 b2 = 1 − cos2 θ .

(5)

Clairaut’s relation gives that (a + b cos v) cos θ = (a + b) cos θ0 = a − b − , so that cos θ =

a−b−

≤1− , a + b cos v a−b

so, from (5),

2 v 0 b2 ≥ 1 − 1 −

a−b

2

≥1− 1−

a−b

=

. a−b

It follows that v 0 is bounded away from zero, so that |v| tends to infinity as |t| tends to infinity. In particular, α must cross the parallel v = π.

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SOLUTIONS TO EXERCISES

7.26. The generating curve f (r), 0, g(r) is the geodesic from p with initial vector (f 0 (0), 0, g 0 (0)). Since the pole isn’t a singular point it follows that g 0 (0) = 0, so if we parametrise the generating curve by arc length, then the generating curve is α(1,0,0) (r), the geodesic with initial vector (1, 0, 0). For each θ, let αθ be the meridian αθ (r) = f (r) cos θ, f (r) sin θ, g(r) . Then α0θ (0) = (cos θ, sin θ, 0) so αθ (r) = α(cos θ,sin θ,0) (r). Hence x(r, θ) = x̃(r cos θ, r sin θ) = α(cos θ,sin θ,0) (r) = f (r) cos θ, f (r) sin θ, g(r) , as required.

7.27. (a) We have that xu = βv 0 . Since xv (0, 0) = α0 (0) we see that xu (0, 0) and xv (0, 0) are orthogonal unit vectors. It follows from Theorem 3 of §2.5 that x is a local parametrisation of S on a sufficiently small open neighbourhood U of (0, 0). We also have that E = 1, while Fu = xuu .xv + Ev /2 = β v 00 .xv + 0 = 0 .

However, F (0, v) = β v 0 (0).α0 (v) = 0, so it follows that F = 0.

Finally, G(0, v) = α0 .α0 = 1, and, since Gu = 2xuv .xv = −2xu .xvv , we see that Gu (0, v) = −2xu (0, v).xvv (0, v) = −2β v 0 (0).α00 (v) = 0 . (b) In this case we have that α(v) = (cos v, sin v, 0). Also, β 0 (u) = (cos u, 0, sin u) while β v (u) is obtained from β 0 (u) by rotation about the z-axis through angle v. Hence x(u, v) = (cos u cos v, cos u sin v, sin u) , which is the standard parametrisation of the unit sphere as surface of revolution, but with u and v interchanged. In particular, E = 1, F = 0 and G = cos2 u. (c) We prove Minding’s theorem by showing that E, F and G are uniquely determined when K is a given constant. Arguing as in the proof of Minding’s √ theorem given in §7.7, it is enough to show that if K is constant and if g = G then the solutions of guu + Kg = 0 are uniquely determined by the conditions on G determined in (a).

CASE 1: K = 0. The general solution of guu = 0 is g(u, v) = uh1 (v) + h2 (v).

The conditions on G then imply that g(u, v) = 1.

CASE 2: K = k 2 > 0. The general solution of guu + k 2 g = 0 is g(u, v) = h1 (v) √ sin ku + h2 (v) cos ku. The conditions on G then imply that g(u, v) = cos Ku. CASE 3: K = −k 2 < 0. The general solution of guu − k 2 g = 0 is g(u, v) = h1 (v)√sinh ku + h2 (v) cosh ku. The conditions on G then imply that g(u, v) = cosh −Ku. 7.28. This is clear since a local isometry maps a geodesic on S1 parametrised by arc length to a geodesic on S2 parametrised by arc length.

7.29. We use geodesic polar coordinates R π px(r, θ). (i) Equation (7.33) says that `(r) = −π G(r, θ) dθ. If K is constant, then the proof of Minding’s theorem shows that: if K = 0 then G = r2 so that `(r) = 2πr;

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if K = k 2 > 0 then G = (1/k)2 sin2 kr so that `(r) = (2π/k) sin kr; if K = −k 2 < 0 then G = (1/k)2 sinh2 kr so that `(r) = (2π/k) sinh kr. (ii) Using the given formula for κg we see that if K = 0 then κg = ±1/r; if K = k 2 > 0 then κg = ±k cot kr; while if K = −k 2 < 0 then κg = ±k coth kr. (iii) Integrating the results obtained in (i), we find that if K = 0 then A(r) = πr2 ; if K = k 2 > 0 then A(r) = 2π(1 − cos kr)/(k 2 ); while K = −k 2 < 0 then A(r) = 2π(cosh kr − 1)/(k 2 ).

The final statement is a routine check.

7.30. Let p = (0, 0, 1) ∈ S 2 (1). The circumference of the geodesic circle in S (1) radius r is 2π sin r. Hence, 2

lim

r→0

2πr − 2π sin r r − sin r 2π π 2πr − L(r) = lim = 2π lim = = . r→0 r→0 r3 r3 r3 3!

3

7.31. For rolling without slipping, we have 0

Ñ .(Ñ × t̃) = (A0 N + AN 0 ).A(N × t) = −AN .A0 (N × t) + N 0 .(N × t) . (6) If there is no twisting, then by Proposition 1 of §7.9, we have that κg = κ̃g .

Also, the equation displayed in the Exercise follows immediately from condition (7.36) for no twisting, and equation (6) above.

Assume now that κg = κ̃g and the equation displayed in the Exercise holds.

Then, using the proof of Proposition 1 of §7.9, we find that A0 (N × t).At = 0 .

Also, using (6) and the equation displayed in the Exercise, we have A0 (N × t).AN = 0 , while A0 (N × t).A(N × t) is also zero. It now follows that A0 (N × t) = 0, so the rolling takes place without twisting.

The surface S rolls without twisting (or slipping) on a straight line in a plane if and only if κg = 0 and N 0 .(N × t) = 0, which happens if and only if α is a geodesic and N 0 is a scalar multiple of t (so that α is a line of curvature). The final statement is now immediate from Exercise 7.11.

Chapter 8 8.1. The first set has non-compact interior. The boundary of the second set is not a disjoint union of piecewise regular simple closed curves. The third one is not the boundary of an open set.

8.2. The interior angles at v1 , v2 , v3 and v4 are π/2, 3π/2, π/2 and 3π/2 respectively. The exterior angles are π/2, −π/2, π/2, −π/2 respectively. Figure 13 shows the exterior angles at v1 , v2 and v4 for the indicated orientation. Note

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SOLUTIONS TO EXERCISES

Figure 13: For the solution to Exercise 8.2 that the exterior angles at v2 and v4 are negative, which reflects the fact that the angle of rotation from the incoming to outgoing vectors at these vertices is in the opposite direction to that given by the orientation.

8.3. A generalisation of the Theorem of Turning Tangents is obtained by replacing “Let T be a triangle contained in the image of a local parametrisation x : D → S of a surface S in Rn ” by “Let R be a regular region of a surface S in Rn which is the image of an n-gon in D under a local parametrisation x : D → S”, and then replacing the final equation by Z X dθ ds = αi − (n − 2)π , ∂R ds where the sum is taken over the interior angles at the vertices of ∂R. The proof is by induction on n. It may be obtained R by dividing the n-gon into an (n − 1)-gon and a triangle, and noting that dθ ds ds cancels over the common edge of the (n − 1)-gon and the triangle.

8.4. (i) Two edges of T are meridians, and hence have zero geodesic curvature. The third edge is a segment of a parallel, and may be parametrised by

α(u) = f (v1 ) cos u, f (v1 ) sin u, g(v1 ) , u0 ≤ u ≤ u1 .

Then, along α,

nin = − f 0 (v1 ) cos u, f 0 (v1 ) sin u, g 0 (v1 ) , since these are the unit vectors pointing back down the meridians. A routine calculation shows that d2 α 1 =− (cos u, sin u, 0) , 2 ds f (v1 ) so that, using (8.2), κg = f 0 (v1 )/f (v1 ). This is constant as expected, since rotations of S about the z-axis are isometries. The length of α is equal to (u1 − u0 )f (v1 ) so that Z κg ds = (u1 − u0 )f 0 (v1 ) . ∂T

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41

(ii) The Gaussian curvature K was found in Corollary 5 of §5.5 to be given by K = −f 00 /f , so ZZ Z v1 Z u1 00 p f KdA = − f 2 du dv f u0 0 T Z v1

= −(u1 − u0 ) f 00 dv = (u1 − u0 ) f 0 (0) − f 0 (v1 ) .

0

Since the pole of S is not a singular point, S intersects its axis of rotation orthogonally (Example 6 of §4.2), so that g 0 (0) = 0. Hence f 0 (0) = 1, so that ZZ

KdA = (u1 − u0 ) 1 − f 0 (v1 ) .

T

(iii) The interior angles of T are π/2, π/2 and u1 −u0 , so we see that (8.4) holds for T , which confirms the Gauss-Bonnet theorem for a triangle in this case.

8.5. Since S is compact, Theorem 4 of §5.10 says that S has an elliptic point p0 , say. Hence, if S had no hyperbolic points the Gauss-Bonnet theorem would imply that the Euler characteristic of S was positive. In this case, the Classification theorems in §8.4 would imply that S is diffeomorphic to a sphere.

So, let p1 be a hyperbolic point, and join p0 to p1 by a curve on S. Applying the Intermediate Value theorem to K along this curve shows there is a point on the curve with K = 0.

8.6. Parametrise the surface S as a surface of revolution, x(u, v) = (v cos u, v sin u, v 2 ) ,

−π < u < π , v > 0 .

Routine calculations show that K = (LN − M 2 )/(EG − F 2 ) = 4(1 + 4v 2 )−2 (or use (5.25)), so that

ZZ Z aZ π p 1 2 √ .

KdA = K EG − F du dv = 2π 1 − 1 + 4a2 R 0 −π The boundary ∂R is the parallel v = a, and nin = −xv /|xv |. Hence, using (8.3), xv Ev 1 1 √ = √ xuu . = . κg = − |xu |2 |xv | a 1 + 4a2 2E G √ R The length of the parallel is 2πa so that ∂R κg ds = 2π/ 1 + 4a2 . Since χ(R) = 1, we now see that the Gauss-Bonnet theorem holds for R. 8.7. We assume that the generating curve is parametrised by arc length.

Along the parallel v = v0 we have that nin = xv , so, using (8.3) (or Exercise 7.4), we see that along v = v0 we have κg = xuu .xv /(|xu |2 ) = −f 0 (v0 )/f (v0 ).

Similarly, along v = v1 we have κg = f 0 (v1 )/f (v1 ).

Hence, Z

κg ds = 2π f 0 (v1 ) − f 0 (v0 ) . ∂R

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SOLUTIONS TO EXERCISES

Also, from Corollary 5 of §5.5, we have that K = −f 00 /f . Hence ZZ Z v1

KdA = −2π f 00 dv = 2π f 0 (v0 ) − f 0 (v1 ) .

R

v0

This verifies the Gauss-Bonnet Theorem in this case, since R is an annulus and hence has Euler characteristic zero.

8.8. This surface is diffeomorphic to a cylinder. Moreover, using the result of Exercise 5.12 (or just do the calculations) we see that the sign of K is opposite to the sign of cosh00 z. Hence K < 0 everywhere on S. It now follows from Theorem 6 of §8.6 that S has at most one simple closed geodesic. The surface intersects the plane z = 0 orthogonally (with the curve of intersection being an ellipse) so it follows from Exercise 7.10 that this ellipse is a geodesic on S. 8.9. It follows from Exercise 5.26 that K ≥ 0, and K = 0 only on the intersection of S with the coordinate planes. Since S is compact, Theorem 1 of §8.6 shows that S is diffeomorphic to S 2 (1). If geodesics Γ1 and Γ2 don’t intersect then they bound a region R homeomorphic RR to an annulus. On this annulus, K ≥ 0, and K > 0 somewhere. Hence R KdA > 0, so the GaussBonnet theorem would imply that χ(annulus) > 0 which gives a contradiction.

Chapter 9 9.1. Let S be a CMC surface in R3 with constant mean curvature c 6= 0, and let f : R3 → R3 denote scalar multiplication by c. If S̃ is the image of S under f , it is quick to prove that S̃ is a surface in R3 (if x(u, v) is a local parametrisation of S then x̃(u, v) = cx(u, v) is a local parametrisation of S̃).

Since f is linear, its derivative at any point is equal to f itself, from which it follows that Tf (p) S̃ = Tp S for each point p ∈ S. It follows from this that Ñ f = N , so that c dÑ f (p) = dN p . Hence the mean curvature of S̃, being minus half the trace of dÑ , is equal to c/c = 1.

ALTERNATIVELY: in the standard notation, it is quick to show that Ẽ = (1/c2 )E, F̃ = (1/c2 )F , G̃ = (1/c2 )G, L̃ = (1/c)L, M̃ = (1/c)M , Ñ = (1/c)N , so it follows from (9.3) that S̃ has constant mean curvature 1.

9.2. This follows by taking h = H in equation (9.4) to see that A0 (0) < 0 for the given variation of any such open neighbourhood.

9.3. (i) This is immediate.

2 (ii) Integrating the given equation, we find that log(f 0 + 1) = 2(log f + log k) for some positive constant k, so that 2

f 0 + 1 = (kf )2 . (iii) It follows that f0 p = ±1 , (kf )2 − 1

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43

so, integrating, we find that arccosh (kf ) = ±k(v + c) for some constant c, so that f = (1/k) cosh k(v + c) . Hence the graph of f is a catenary.

9.4. (i) In Example 1 of §3.2 we found that, for the standard parametrisation of S as a graph, E = 1 + gu 2 ,

G = 1 + gv 2 ,

F = gu gv ,

while the solution of Exercise 5.1 gives that if D = 1 + fu 2 + fv 2 then √ M = fuv / D ,

√ L = fuu / D ,

√ N = fvv / D .

The result now follows quickly from equation (9.3). (ii) If f (x, y) is a function of x only, then the minimal graph equation becomes f 00 = 0, which integrates to give f (x) = ax + b for some √ constants a and b.

Hence S has equation z − ax = b so that N = (−a, 0, 1)/ 1 + a2 . Conversely, √ if N = (−a, 0, 1)/ 1 + a2 for some constant a then S has equation z − ax = b for some constant b, and S is the graph of f (x) = ax + b. (iii) This time, denoting differentiation with respect to the appropriate variable by 0 , we obtain 2

2

(1 + h0 )g 00 + (1 + g 0 )h00 = 0 .

Hence g 00 = 0 if and only if h00 = 0, and in this case the graph of f is a plane.

Otherwise, using the usual separation of variables argument, 2

2

(1 + h0 )/h00 = −(1 + g 0 )/g 00 = c for some non-zero constant c.

The first equation integrates to give

h(y) = −c log cos (y − k̃)/c + `˜ , where k̃ and `˜ are constants, while the second one gives

g(x) = c log cos (x − k)/c + ` , where k and ` are constants.

Hence f takes the form given in the exercise, and the result follows.

9.5. The conjugate minimal surface is given by the imaginary part of the C3 -valued function ψ given in Example 7 of §9.8, namely y(u, v) =

1 2

v − u2 v +

v3 u3 ,u + − uv 2 , 2uv 3 3

.

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SOLUTIONS TO EXERCISES

If x̃(u, v) = x̃1 (u, v), x̃2 (u, v), x̃3 (u, v) , then

1 u3 v3 u− + uv 2 + v − + u2 v , x̃1 = √ 3 3 2 2

3 u 1 v3 u− x̃2 = √ + uv 2 − v + − u2 v , 3 3 2 2 1 2 x̃3 = (u − v 2 ) .

2

A calculation now shows that if ũ and ṽ are as given, then x̃(ũ, ṽ) = y(u, v) as required.

9.6. If y is the harmonic conjugate of x then xθ = cos θ x + sin θ y. Hence (xθ )u = cos θ xu + sin θ y u = cos θ xu − sin θ xv , with a similar expression for (xθ )v . The unit normal N to Sθ is independent of θ so that Lθ = −(xθ )u .N u = L cos θ − M sin θ , and the expressions for Mθ and Nθ may be found in a similar manner.

9.7. (i) Complex bilinearity shows that φ.φ = (xu .xu − xv .xv ) + 2ixu .xv , which equals zero if and only if x is isothermal. Similarly, φ.φ̄ = (xu .xu + xv .xv ) , and the result follows. (ii) We note that φ is holomorphic if and only if both xuu = −xvv and xuv = xvu . The second equation always holds, and the first holds if and only if x is harmonic. The final statement follows immediately from equation (9.12).

9.8. With f and g as given, short calculations show that φ(z) =

1 1 − cos z, sin z, −2i sin(z/2) , 2

so that Z ψ(z) =

φ(z)dz =

1 z − sin z, − cos z, 4i cos(z/2) + (c1 , c2 , c3 ) , 2

where c1 , c2 , c3 are (complex) constants of integration. Hence, x(u, v) =

1 u − sin u cosh v, − cos u cosh v, 4 sin(u/2) sinh(v/2) + (k1 , k2 , k3 ) , 2

and we take k1 = k3 = 0 and k2 = 1/2 to obtain the given formula for x.

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9.9. Here, 1 2

φ(z) =

ψ(z) =

1 2

3

z 1 z3 1 + ,i − , 2z , 3 z 3 z

so that

z2 −

1 1 2 , i z + , 2 , z2 z2

which gives that 1 x(u, v) = 2

3 u u v3 v 2 2 , −u v + , 2u . − uv + 2 − 2 3 u + v2 3 u + v2

9.10. Arguing as in Example 7 of §9.8, we see after some calculation that x̃ is the real part of the function ψ̃ defined in the exercise. The final equation is now clear.

9.11. Routine calculations show that, for the parametrisation of Enneper’s surface given in Exercise 9.5, N = (2u, 2v, u2 + v 2 − 1)/(u2 + v 2 + 1) .

It follows that πN = u + iv, so that Theorem 2 of §9.9 is checked in this case.

9.12. The given parametrisation of Enneper’s surface arises from taking f (z) = 1, g(z) = z. It follows from Proposition 4 of §9.9 that L = −N = −1 and M = 0, so the required solution ω of the Liouville equation is given by ω = log E. A routine calculation shows that 4E = (1 + u2 + v 2 )2 , so that

ω = 2 log (1 + u2 + v 2 )/2 .

9.13. The given parametrisation of the helicoid has L − N − 2iM = −2i. It follows from Lemma 1 of §9.13 that, in order to obtain an isothermal parametrisation x̃(ũ, ṽ) with L̃ = −Ñ = −1 and M̃ = 0, we should take ũ + iṽ = h(u, v) where (h0 )2 = i. So, taking h(z) = eiπ/4 z, we have √ √ ũ = (u − v)/ 2 , ṽ = (u + v)/ 2 .

Hence, our required isothermal parametrisation of the helicoid is given by x̃(ũ, ṽ) = (sinh v sin u, − sinh v cos u, u) , where

√ u = (ũ + ṽ)/ 2 ,

√ v = (−ũ + ṽ)/ 2 .

The corresponding Ẽ, F̃ and G̃ may be found either by direct calculation, or by using (3.32). Either way, we find that √ Ẽ = G̃ = cosh2 (−ũ + ṽ)/ 2 ,

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SOLUTIONS TO EXERCISES

and the required solution ω̃ of the Liouville equation is given by ω̃ = log Ẽ.

9.14. It follows from Exercise 9.6 that Lθ − iMθ = e−iθ (L − iM ) = −e−iθ .

Hence, from Lemma 1 of §9.13, we need a change of variable z̃ = h(z) such that 2 h0 = e−iθ , so we take z̃ = h(z) = ±e−iθ/2 z .

The final equation in the proof of Lemma 1 of §9.13 says that if x̃θ (z̃) = xθ (z) then

(xθ )u − i(xθ )v = ± (x̃θ )ũ − i(x̃θ )ṽ e−iθ/2 , so, in particular, since Eθ = E we see that Ẽθ (z̃) = E(z).

Hence, the required solution of the Liouville equation is ω̃(ũ, ṽ) = ω(u, v) where u + iv = eiθ/2 (ũ + iṽ).