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Let $M$ be an oriented n dimensional manifold and let $\eta$ be an exact n-form on $M$ (say, $\eta=d\omega$).

If $M$ is compact, I know that Stokes theorem implies that $\int_M \eta=0$ (if I understand correctly, this is by "promoting" $M$ to a compact manifold with boundary where the boundary happens to be the empty set s.t one can then apply Stokes theorem, is that right?)

I am curious about the case where $M$ is not compact. Intuitively, I expect the integral to still vanish because if one can choose a compact submanifold with boundary $N \subset M$ (with orientation inherited from $M$) s.t. the support of $\eta$ lies within $N$ and $\omega$ vanishes on $\partial N$, then of course $\int_M\eta=\int_N\eta=0$ again, by Stokes theorem. However while intuitive, it is not entirely clear to me that such a submanifold with boundary $N$ always exists. $supp(\eta)$ seems like a candidate if it can be shown to be a submanifold with boundary.

Is my intuited answer correct and my approach to proving it sound? If so, how does one show that the required submanifold $N$ always exists? If not, how does one see the actual result?

Lostphysicist
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    Yes it is zero, and this should be established in the proof of Stokes theorem, even if it is not 'obvious' from one of the familiar versions of the statement of Stokes. You don't really need the $N$ either. It all boils down to the following observation: if $U\subset\Bbb{R}^n$ is open and $f:U\to\Bbb{R}$ is smooth and compactly supported, then for any $i\in{1,\dots, n}$, we have $\int_U\frac{\partial f}{\partial x^i},dV=0$ (use Fubini to integrate over $x^i$ first, then use FTC; you'll get $0$ at the 'endpoints' due to compact support). – peek-a-boo May 30 '22 at 23:21
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    Regardless, a very common way of stating Stokes is "Let $M$ be a smooth $n$-manifold with boundary (possibly empty), and $\omega$ a smooth $(n-1)$-form on $M$ which is compactly supported. Then, $\int_{\partial M}\omega=\int_Md\omega$", and this version clearly applies to your case (though again, the case when $\partial M=\emptyset$ is established in the proof of the theorem by the remark I made above). – peek-a-boo May 30 '22 at 23:25
  • Thanks, that's really helpful. Just to check: Every manifold is a manifold with boundary since R^n is diffeomorphic to the open upper half space H^n, right? Sorry this is really elementary but my manifolds course progressed a bit too fast for me. – Lostphysicist May 30 '22 at 23:54
  • yes, every smooth manifold is a smooth manifold-with-boundary, with empty boundary (lol that's a mouthful to say), and yes, it's because $\Bbb{R}^n$ is diffeomorphic to the open upper half space $H^n$ (or just because open balls in $\Bbb{R}^n$ can be considered as open balls in $H^n$ by translating up accordingly). – peek-a-boo May 31 '22 at 00:00
  • To add to peek-a-boo's first comment, the "standard" proof involves splitting the integral into a bunch of "nice" local coordinate charts and summing over interior charts (where the integral vanishes) and boundary charts (where we get portions of the boundary term). You never have to assume there are any boundary charts at all in this process. – Kajelad May 31 '22 at 00:17

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People are falling in a common trap here. You have said (if I read the title) that $\eta$ is a compactly supported exact $n$-form, with $\eta = d\omega$. The crucial question is this: Is $\omega$ compactly supported as well? If so, then the integral of $\eta$ is $0$. If not, it need not be.

For an example, take $f\colon\Bbb R\to\Bbb R$ with compact support and $\int_{-\infty}^\infty f(x)\,dx > 0$. Let $\eta = f(x)\,dx$ on $\Bbb R$. We note $\eta$ is compactly supported. If we set $F(x) = \int_{-\infty}^x f(t)\,dt$, then $\eta = dF$, so $\eta$ is exact, but $\int_{\Bbb R} \eta > 0$.

The moral of the story is this: One must be very careful with hypotheses! There's no reason the support of $\omega$ needs to be related to the support of $\eta$. :)

Ted Shifrin
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