Final Exam

MA420 9:00-11:00am Analysis on Manifolds June 20, 2001

It is allowed to consult the textbook Differential Forms and Applications, Manfredo P. do Carmo, Springer-Verlag, 1994.

1. Let M be a bounded region in R^{n} with regular boundary ∂M and consider the (n − 1)-form

ω = ∑_{i=1}^{n}(−1)^{i−1}x_{i} dx_{1} ∧ ⋯ ∧ d̂x_{i} ∧ ⋯ ∧ dx_{n}.

Prove that

vol(M) = \frac{1}{n} ∫_{∂M} j^{∗}ω

where j : M → R^{n} is the inclusion map.

Solution: We have dω = ∑(−1)^{i−1}dx_{i} ∧ dx_{1} ∧ ⋯ ∧ d̂x_{i} ∧ ⋯ ∧ dx_{n} = n dx_{1} ∧ ⋯ ∧ d_{n}. Therefore, by Stokes Theorem,

vol(M) = ∫_{M} dx_{1} ∧ ⋯ ∧ dx_{n} = \frac{1}{n} ∫_{M} j^{∗}dω = \frac{1}{n} ∫_{M} d(j^{∗}ω) = \frac{1}{n} ∫_{∂M} j^{∗}ω.

2. Consider the the upper half-plane H^{2} = {(x, y) ∈ R^{2} | y > 0} with the following Riemannian metric:

⟨u, v⟩_{p} = \frac{u · v}{y^{2}} , for p = (x, y) ∈ H^{2} and u, v ∈ T_{p}H^{2}

where u · v is the canonical inner product of R^{2}. Prove that the Gaussian cuvature K is constantly −1.

Solution: The frame \{e_{1} = y \frac{∂}{∂x}, e_{2} = y \frac{∂}{∂y} \} has coframe \{ω_{1} = \frac{1}{y} dx, ω_{2} = \frac{1}{y} dy = d ln y\}. Since dω_{1} = −\frac{1}{y^{2}} dy∧dx = ω_{1}∧ω_{2} and dω_{2} = d^{2} ln y = 0, the defining relations ω_{12}(e_{1}) = dω_{1}(e_{1}, e_{2}) = 1 and ω_{12}(e_{2}) = dω_{2}(e_{1}, e_{2}) = 0 imply that w_{12} = ω_{1}. Therefore −K ω_{1} ∧ ω_{2} = dω_{12} = dω_{1} = ω_{1} ∧ ω_{2} and hence K = −1.

3. Prove that there exists no Riemannian metric on a torus T such that the Gaussian curvature K is nonzero and does not change sign on T.

Solution: Suppose, on the contrary, that there exists a Riemannian metric on T such that the Gaussian curvature K is nonzero and does not change sign on T. We may assume that K > 0 on T. Since T is compact there exists a positive number ϵ satisfying K > ϵ. Then ∫_{T} Kσ > ∫_{T} ϵσ = ϵ · area(T) > 0. On the other hand, by the Gauss-Bonnet Theorem, we have ∫_{T} Kσ = 2πχ(T) = 0, a contradiction.

4. Let M^{2} ⊂ R^{3} be a compact connected orientable surface without boundary embedded in R^{3} with the induced Riemannian metric. Suppose that the z-coordinate function z : M^{2} → R has only nondegenarate isolated critical points. Show that there exist points in M^{2} for which the Gaussian curvature is positive, negative and zero provided that M^{2} is not homeomorphic to a sphere. (Hint: The function z has a maximum and a minimum.)

Solution: Since M is an orientable surface not homeomorphic to a sphere, it is an n-torus for some n ≥ 1. Since M is compact, there exist only finitely many, say k, critical points of the z-coordinate function. Let I_{j} be the index of grad z at the j-th lowest critical point, j = 1, …, k. Then I_{j} = ±1 since the critical points are nondegenarate. By Gauss-Bonnet Theorem, we have (1/2π) ∫_{M} Kσ = ∑_{j=1}^{k} I_{j} = χ(M) = 2−2n ≤ 0. Since I_{1} and I_{k} correspond to the minimum and the maximum, we have I_{i} = I_{k} = +1. Hence, there must be some j’s with I_{j} = −1 and 1 < j < k, i.e., there must be some saddle among the critical points. The Gaussian curvature is positive at the extreme points and negative at the saddle points. By the intermediate value theorem there must be a point where the Gaussian curvature is zero.

5. (a) Let M^{n} be a compact orientable differentiable manifold without boundary. Construct a differentiable n-form ω_{M} which is nowhere zero on M^{n}.

(b) Let N^{m} be a compact orientable differentiable manifold with boundary ∂N ≠ ∅. Show that there exists no differentiable map f : N → ∂N such that f|_{∂N} = id_{∂N}.

Solution: (a) Let {(U_{α}, f_{α})} be a differentiable structure on M such that the change of coordinates have positive Jacobian. Since M is compact there is a finite subcover {U_{1}, … U_{k}} of {U_{α}} and a partition of unity {φ_{i}} subordinate to it. Let x_{1}^{(i)}, …, x_{n}^{(i)} denote the coordinates of U_{i} and define

ω_{i}(p) = \begin{cases}φ_{i}(p)dx_{1}^{(i)} ∧ ⋯ ∧ dx_{n}^{(i)} & p ∈ supp(φ) \\ 0 & p ≠ supp(φ) \end{cases}

Then ω_{M} = ∑_{i=1}^{k} ω_{i} is an n-form which is nowhere zero.

(b) By 5a, there is a nonvanishing (m − 1)-form ω = ω_{∂N} on ∂N. Since ∂N is compact, we may assume that there is a positive number ϵ and a continuous function g : ∂N → [ϵ, ∞) satisfying ω(p) = g(p)σ_{∂N}, p ∈ ∂N, where σ_{∂N} is the volume element of ∂N. Since dω = 0, we have d(f^{∗}ω) = f^{∗}(dω) = 0. Therefore, by Stokes Theorem, we get a contradiction:

0 = ∫_{N} d(f^{∗}ω) = ∫_{∂N} i^{∗}f^{∗}ω = ∫_{∂N} gσ_{∂N} ≥ ϵ · vol(∂N) > 0

(gtj2001/07/04)