Integral of a closed form over an compact manifold

Question

If $w$ is closed ($dw=0$), for a compact manifold $M$, why the integral $\int_M w$ is topological invariant instead of 0? Because from Storke theorem, I would expect that $\int_M w=\int_{Volume~enclosed~by~M} dw =0$.

Answer 1

Probably what you are looking for is as MarkM wrote in his answer.

However, the question remains why your argument doesn't show that the integral is 0. Note that when applying Stokes, you assume that $\omega$, or at least $d\omega$ is defined on the volume $V$ enclosed by $M$, so $\partial V = M$. Strictly speaking in Stokes' theorem you should write

$$\int_Md\omega = \int_{\partial M}i^\ast\omega$$

where $i:\partial M\hookrightarrow M$ is the inclusion.

What you showed is not that the integral is 0. How you should interpret the formula that you wrote

$$\int_{\partial V = M} \omega = \int_{V}d\omega$$

is that if $M$ can be written as the boundary of a manifold $V$, and $\omega$ can be extended to a form on the whole of $V$, then the integral of $\omega$ over $M$ is equal to the integral of $d\omega$ over the whole of $V$. If the extension of $\omega$ to $V$ is closed, then the original integral was 0.

Answer 2

That isn't quite the correct statement of Stoke's theorem. If $M$ is a compact orientable manifold with boundary $\partial M$ and dimension $n$, then $$ \int_{M} d\omega = \int_{\partial M} \omega $$ for any form $\omega \in \Omega^{n-1}(M, \mathbb{R})$. If $\partial M = \emptyset$, i.e. $M$ is closed, then Stokes' theorem asserts that that $$ \int_M d\omega = 0. $$ In other words, the integral of any exact form is zero. A (necessarily closed, by considerations of degree) form $\eta \in \Omega^n(M, \mathbb{R})$ need not be exact, so its integral $\int_M \eta$ need not be zero.