Integrating over a interesting Manifold.

Question

I am writing a small paper on stokes generalised theorem namely: $\int_M d \omega=\int_{\partial M} \omega$. And I want an example problem. And not just a calculus 3 problem. Maybe over a Klien Bottle or something? I don't want a solution, just a problem.

Answer 1

This problem was found here. On $\mathbb{R}^4$, with coordinates $x_1,x_2,x_3,x_4$, consider the form \begin{equation} \alpha=x_1 dx_2\wedge dx_3\wedge dx_4 - x_2dx_1\wedge dx_3\wedge dx_4 + x_3 dx_1\wedge dx_2\wedge dx_4 - x_4 dx_1\wedge dx_2\wedge dx_3. \end{equation} Let $\iota\colon S^3\hookrightarrow \mathbb{R}^4$ be the inclusion map of the unit sphere into $\mathbb{R}^4$ and comptue $\int_{S^3}\iota^*\alpha$.

Notice that Stokes' tells us that $\int_{S^3}\iota^*\alpha=\int_{B^4}d\alpha$, where $B^4$ is the closed unit ball in $\mathbb{R}^4$. When you compute $d\alpha$, you'll see that $\int_{B^4}d\alpha$ is a much easier integral to evaluate than the original.

Answer 2

Here is an interesting result proved with Stokes' theorem:

A compact orientable smooth manifold M without boundary is not contractible.

Examples:
a) No sphere $S^n (n\geq1)$ is contractible.
b) No compact holomorphic manifold is contractible. In particular $\mathbb P^n(\mathbb C)\; (n\geq 1) $is not contractible.

Proof of result:
Let $n$ be the dimension of $M$ and let $\omega\in \Omega^n(M)$ be a smooth form with positive integral: $$\int_M\omega\gt0 \quad (\bigstar)$$ The existence of such an $\omega$ is easy to see: take an oriented chart $M\supset U\stackrel {\phi}{\to}\widetilde U\subset \mathbb R^n$, choose a non-negative function $u\in C_c^\infty(\widetilde U)$ with compact (but non-empty !) support and just take $\omega =\phi^*(udx_1\cdots dx_n)$ (extended by zero on $M\setminus U$).
Then $\omega$ is closed (like all $n$-forms on $M$!) but not exact: indeed if we had $\omega=d\eta$, we would deduce by Stokes $$\int_M\omega =\int_M d\eta=\int_{\partial M}\eta= \int_\emptyset\eta =0 $$ which contradicts $(\bigstar)$.
Since $\omega $ is closed but not exact its class $[\omega] \in H^n_{DR}(M)$ in the $n$-th De Rham cohomology group is not zero, so that $H^n_{DR}(M)\neq 0$ and thus $M$ is not contractible (because a contractible manifold has all $H^i_{DR}(M)=0$ for $i\gt0$).