# The integral of an exact form over an orientable closed manifold is $0$ [closed]

A form $\beta^{p}$ is closed if $d\beta=0$.

A form $\beta^{p}$ is exact if $\beta^{p}=d\alpha^{p-1}$, for some form $\alpha^{p-1}$.

An orientable closed manifold is a compact manifold without boundary.

What is a compact manifold?

Why is an orientable closed manifold the same as a compact manifold without boundary?

If a manifold is compact, does it not, be definition, have a boundary?

How do you prove that the integral of an exact form over an orientable closed manifold is $0$?

## closed as too broad by Andrew D. Hwang, Willie Wong, Jack Lee, pjs36, MickGNov 10 '16 at 20:07

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Voting to close as "missing context": If you don't know what a compact manifold (such as a circle, a sphere, a torus...) is, explaining integration of forms is the subject of a short course or textbook, not a Math.SE post. – Andrew D. Hwang Nov 10 '16 at 11:40

3) No, a compact manifold could have no boundary. For example, consider the circle $S^1$. It is definitely compact and does not have a boundary.
4) As we have just concluded that this manifold has no boundary, we can just use Stoke's theorem for our manifold $M$: $$\int_{M} d\omega = \int_{\partial M} \omega.$$ We have an exact form, let's call it $\beta^p$. As you already said, this means that $\beta^p=d\alpha^{p-1}$. Also, because $M$ has no boundary, the integral over $\partial M$ will be zero: $$\int_M \beta^p = \int_M d\alpha^{p-1} = \int_{\partial M} \alpha^{p-1} =0.$$