用户: Solution/ 习题: do Carmo

0Differentiable Manifolds

2.

Prove that the tangent bundle of a differentiable manifold is orientable (even though may not be).

Let be the projection from to . Cover with , then covers , and we need to find the trivialization associated to for each .

Given any smooth chart , define by For any two overlapping charts and , the transition map is given by where . Then the determinant is .

1Riemannian Metrics

2.

In terms of , which is,we have . So the induced metric on is

4.

(a) For and , That is to say, carries to , and thus As , we have (b) Check that is invariant under and .

6.

Exercise 2 provides a local isometry , but there is no local isometry . Indeed, write . mapping the compact set to attains its maximum at some , and consequently . Then is singular.

2Affine and Riemannian Connections

1.

Denote by instead. Since is an isometry. Then . Because is continuously dependent on and , we have is always , and in particular positive.

2.

Proved in [Chen] Chapter 2 Theorem 7.2 (pp.142-3).

4.

a) According to Exercise 3, is the tangential component of , i.e. the orthonormal projection from onto , where is the Riemannian connection of . is simply the usual derivative , so iff the projection of to vanishes.

b) The acceleration of the unit velocity field along a great circle, namely , points to the center of the sphere, perpendicular to the tangent space embedded in (or ).

7.

Method I. Consider the cone tangent to along , then the parallel transport of along is the same, whether taken relative to or to , as the tangent spaces along whether taken relative to or to are the same. Flatten the cone (locally, or at least minus a line) to a sector on the plane (in other words, is locally isometric to a sector in with being its arc). The parallel transport of on the plane is merely translation in . From the perspective of one walking along , appears to be rotating at constant speed. For details, refer to [dC] 4.4 Example 1 (p.246). 運命のルーレット廻して、ずっと君を見ていた.

Method II. Consider the cone tangent to along , parametrized as where . The parallel transport of along is the same, whether taken relative to or to , as the tangent spaces along whether taken relative to or to are the same. Now we carry out a bit of calculations to see the effect of parallel–translating along , written . Besides, let be the local expression for . First, and then Substituting into where , we get Hence, for two constants . We conclude that the vector rotates at angular velocity , if the angular velocity of along the parallel is taken to be .

3Geodesics; Convex Neighborhoods

1.

a) The induced metric is b) The inverse is Then the coefficients in these differential equations are c) Let then Thus by the first (which gives ) and the second equation subsequently, We have and , That is constant has been verified, and shows that is constant as well.

d) Click, or cf. [dC] 4-4 Example 6 (pp.261-2), or [Chen] Chapter 3 Exercise 2 (pp.465-6).

2.

a) The inner product being well–defined is because depends only on , not essentially on . To show that it is a Riemannian metric, properties like symmetry, bilinearity, non-negativity and smooth dependence are all obvious, and implies , which further implies by the expression of in local coordinates above.

b) We have the decomposition and iff , i.e. is parallel along .

c) Simply because is parallel along .

d) The hint is a solution indeed.

e) With respect to the decomposition given in b), if is horizontal, and if is vertical.

4.

a) Write for the convex neighborhood and for any of its points. By definition, for any , there is a unique geodesic joining and . Then is a homotopy between and .

b) For any pick a convex neighborhood. A finite union of convex neighborhoods is again convex (check by definition), hence contractible.

7.

Let be a normal neighborhood of in , that is, is a diffeomorphism. Define , where , , and are the standard Euclidean coordinates on . In other words, maps at to at .

For any , is a geodesic. Then , where . In particular let , we get where , and consequently by the arbitrariness of .

8.

a) The first is because For the second, for , the trace of the mapping is .

b) Apply a), as is a geodesic frame.

10.

Differentiate with respect to , obtaining we get

4Curvature

1.

a) Combine and

b) Using a) and the Jacobi identity,

c) By the equation (3) of Chapter 1 section 2 (p.40), we have

2.

a) .

b) Put to be the right side, that is to say, Using the Killing equation , the sum of first two terms becomes note that , implying , and then the third term becomes the fourth term is Adding together, we get Because of the Killing equation at and a), for any , it is inferred that , and therefore

3.

That is anti–symmetric is simply and that is an isomorphism is because by 2.b), As , taking determinant on both sides gives This implies is even, which is a contradiction to being even.

5.

Observe that, for all , at we have (where ) The first equality is because the tensor is tensorial in the second component and vanishes at , and the third equality is because .

6.

a) implies that for any , Substituting , we get which is .

b) For any curve , let be two orthonormal parallel unit vector fields along . Then we have To put it another way, is constant along . Therefore is constant over .

c) Recall that has constant sectional curvature means Then , i.e. becomes the identity

7.

In this frame , hence Note that though there is supposed to be one more term in the second line above, it actually vanishes, as long as we observe that In the second Bianchi Identity, stays invariant, while permutes cyclically. Summing these three equations yields where the last equality is simply since at .

5Jacobi Fields

3.

Assume that , then the existence of a non-trivial Jacobi field along the geodesic , with . Using the Jacobi equation, we have But at , then . Since , we get , and hence , a contradiction.

4.

From Example 2.3, the Jacobi field along satisfying , is given by In addition, from Corollary 2.5, It follows that In addition, since , we have which implies what was asserted.

6.

a) By Proposition 2.9, is a diffeomorphism onto an open subset of . To make the polar coordinates into a coordinate system indeed, an entire radius in must be removed, and correspondingly the ray in , forming the required .

b) is the point instead of a curve, so . Considering the unit–speed geodesic where varies and is fixed, we get , and for , the latter implying i.e. So , and consequently . is too simple to be explained.

c) On one hand, according to b) and Corollary 2.10, on the other hand, the Taylor expansion yields Comparing the coefficients, we get at , and then

d)

7.

Using Exercise 6 c), we have

6Isometric Immersions

1.

a) Trivially verified by definition or local expressions.

b) Suppose that is a geodesic of , naturally identified with which is a geodesic of . That is to say, . Then using a), we too have .

c) Easy to see that commute, and so do . Hence , and then .

2.

, given by is injective, and the composite ( is the projection) is also injective, since the vector is perpendicular to , and as a result, lies in for and then .

The zero sectional curvature is simply from and Exercise 1.c).

3.

Use the fact that is totally geodesic in iff all geodesics in are also geodesics in .

4.

For , from Exercise 1.a) we have Then for , we haveso vanish implies that vanishes.

5.

Let , then for all . The reason is that for all , and thus . Expand the curvature below, we have as .

Let be the inclusion from into an equator of . By Exercise 4, is totally geodesic.

7.

For any and , , which is to say, .

8.

a) Note that , and therefore . And check without effort that form an orthonormal basis of .

b) Extend to , and to . As for in , we have and , , . Then c) Since and , the mean curvature of is .

9.

Follow the hint, and a visualization is offered here.

10.

Differentiating with respect to , we obtain Using the fact that and the previous expression, we obtain that

11.

a) By definition, b) Use and c) Follow the hint, where .

d) We need to verify that . In fact, for any , let be a curve in with , and then . It follows that .

7Complete Manifolds; Hopf–Rinow and Hadamard Theorems

1.

Take to be a non–convex subset of .

7.

For any curve joining , is a curve joining , then Taking infimum on , we obtain that We are now equipped to verify that is complete as a metric space. Let be a Cauchy sequence in , that is, . Then too. As is complete, converges to some point in , written , for is surjective. Then converges to , as is a diffeomorphism.

8.

Were not injective, i.e. for two distinct points , then for a geodesic joining to , is a loop. Because is a local isometry, is a geodesic too. Take any point different from in the image of , then there are two geodesics joining to , a contradiction.

Next we show that is surjective. For any , let be the geodesic segment joining to , where is any point in the image of . Let be the geodesic starting from with initial speed such that . Then is precisely as both are geodesics that start from with the same initial speed, and it follows that , where owing to the completeness of , is defined on the entire , and in particular at .

Here we used the fact that a local isometry maps geodesics to geodesics (mentioned on p.73). The reason: The geodesic equation depends on the metric and its first derivatives in a coordinate chart. A local isometry preserves the metric and is a local diffeomorphism. So it induces coordinates on with the same metric coefficients. In particular, it must take geodesics to geodesics.

9.

The length is Note that is a geodesic that does not extend beyond , and therefore is not complete.

10.

Method I. As illustrated by Chapter 3 Example 3.10 (p.73), all the geodesics in the Lobatchevski plane are of the form or . Their lengths are respectively

Method II. Exhaust by a sequence of compact sets . Let , then for , it is not hard to show that

8Spaces of Constant Curvature

1.

a) Recall that by Chapter 4 Corollary 3.5 (p.96), the metric having constant curvature is equivalent to for all and in the other cases. From the calculation at the end of p.161 and we get from the formula (2) on the same page, taking , we get provided that .

b) i. for implies that depends only on for all . Then for , ii. Note that , denote , where each for coincides due to the second equation in a). Then we write where the means is a function independent of . Further, express as and make a comparison ( comparisons indeed), we find that does not depend on for each indeed. Therefore and substituting into the second equation in a) yields

c) This is a statement, instead of a problem.

d) For any , the distance between and is bounded by the length of the curve , which is Hence is bounded, and also closed as it is the whole space. But is non–compact (compactness is a topological property), violating b) of Hopf–Rinow Theorem. Then is not complete.

13.

It is clear that , as both are geodesics with the same starting point and initial speed . To prove that is parallel along , compare and we get the last term is , and consequently by the arbitrariness of (and that is surjective). Note that at the colored “”, is used (check it using (9) of Chapter 2 Theorem 3.6, p.55).

9Variations of Energy

1.

Suppose that is a sequence of points in such that , then it follows from the boundedness and thus compactness of that there is a cluster point , and as is closed. Let be a minimizing geodesic joining to . Consider the variational fields of variations keeping the starting point fixed yet with varying end points in . As and , we have then is perpendicular to all .

2.

Suppose, to the contrary, that on for a certain . As will be soon shown, , where is compact hence bounded. Then the whole is bounded and hence compact, a contradiction.

For any , join them by a minimizing geodesic and establish a uniform upper bound of its length. If , then ; if , then as the proof of Bonnet–Myers theorem (Theorem 3.1) indicates. Things are worse if , in which case ; the worst case is when intersects twice (had it crossed the boundary more than twice, it would not be the shortest), and .

3.

Suppose that the geodesic is of length , then So

4.

Suppose that is the parallel transport along from to , and denote by , where is assumed to be parametrized by (so ). preserves the inner product and is thus orthogonal; is orientable and is even, so . Applying Lemma 3.8 (p.203), leaves invariant some nonzero , that is, . Then is a vector field along the closed curve , and ensures that it is well–defined. Now Note that though a variation that generates is improper, the last four terms of (5) of this chapter (p.199) all offset as is a closed curve.

5.

a) Existence is from compactness, and using Exercise 1 twice we get .

b) Denote the Riemann connection on , and write for . We have Here we used where are the two endpoints of , and hence .

6.

a) (i) Since , (ii) Put . Then where we used as , and as is a geodesic. Suppose that , then combined with we get , so as a result of .

b) By Hadamard’s theorem (3.1 of Chapter 7), is a diffeomorphism. Hence every geodesic joining to any point of is a curve in the variation (where is assumed to range over the entire ). Use the property of convex functions.

c) When is dropped, let be a sphere, be the north pole and be its equator, then there are infinitely many perpendicular lines of minimal length.

When simple connectedness is dropped, let be an infinite cylinder and be any point, and be a straight line passing through the antipodal point of , then there are two lines reaching minimal length.

10The Rauch Comparison Theorem

2.

Take arbitrarily small in Klingenberg’s lemma and show that if is simply connected, there exists a unique geodesic joining the two points . This is because were there to exist two such geodesics and , they would be homotopic by the simple connectedness, and by Klingenberg’s lemma, there would exist a sequence of curves in this homotopy of lengths , which would be arbitrarily big; however, as a continuous function of is bounded, a contradiction.

This asserts that is a one-one map. That is, it has an inverse . It suffices to show that is a local diffeomorphism then, as this will imply that is continuous locally, and hence globally. This is a consequence of Rauch’s Theorem, in combination with Chapter 5 Proposition 3.5 (p.117) (and Chapter 0 Theorem 2.10, p.10).

3.

Note that is a geodesic on , and is a Jacobi field along ; further, is a Jacobi field along . As and , we have , which is .

5.

a) For all , integration by parts gives From the initial conditions and , is positive in a neighborhood of . Assume that for certain . Then take and get forcing . Then and satisfy exactly the same ode, with the same initial conditions and , and for this reason .

b) Again, examine It follows that , that is, . Conclude by For equality, forces as a).

To verify the Theorem of Rauch in dimension two as an application, first reduce the Jacobi fields in question to normal Jacobi fields as the proof in the book does, and hence suppose that where is the parallel transport of a unit vector with . Then , so the Jacobi equation is equivalent to and since spans . The first equation trivially holds, and the second is . Similarly for .

6.

Suppose that for all . Using the expression from the last exercise, This forces , as well as . Then the initial conditions imply that , contradicting that on .

7.

A correction: it should be

a)b) , and . .

c) , which is . Then the oscillation theorem of Sturm yields that has a zero on . But .

11The Morse Index Theorem

2.

Expand and into the Fourier series Then the inequality follows from Yet we need to justify that differentiating by terms does produce , which relies on the decay of the coefficients . Indeed, integrating by parts, we find Then yielding the uniform convergence of the series of , and consequently the justification.

3.

It follows that for . Let , we get . Then use Theorem 4.1 of Chapter 8 (p.163).

4.

implies that for some . Suppose that has no positive zero, then on . Hence for . But then as , a contradiction.

5.

Since , and with strict inequality at ,

6.

From Exercise 5, there exists and a vector field along the segment such that . Let be a proper variation that generates , then , which is, does not minimize the length between its endpoints.

If instead, violates the theorem.

Remark: Cf. [YS] 1.2 (p.13-7) for the proof of the splitting theorem (Cheeger–Gromoll, 1971) below.

Let be an -dimensional complete Riemannian manifold with . Suppose that has a line. Then is isometric to , where is an -dimensional Riemannian manifold with the product metric.

References

[Chen]

陈维桓, 李兴校. 黎曼几何引论 (上册) . 北京大学数学教学系列丛书. 北京大学出版社, 2002.

[dC]

M.P. do Carmo. Differential Geometry of Curves and Surfaces. Dover Publications Inc., 2016.

[YS]

丘成桐, 孙理察. 微分几何讲义. 高等教育出版社, 2018.