# Stokes theorem implies that $\partial^2=0$?

First of all, is this true? Can you use Stokes' theorem plus $d^2 = 0$, or some other relatively easy to establish fact to prove that the boundary of a boundary is empty? If so, how exactly does one do this?

I've seen the argument (in Rudin p. 276, for example, see above) that, since $\int_{\partial\partial\Omega} \omega =\int_{\partial\Omega} d\omega = \int_\Omega d^2\omega$ for any smooth (or at least twice continuously differentiable) form $\omega$, and we know that $d^2\omega=0$, we get $\int_{\partial\partial\Omega}\omega = 0$. Since this is true for any $\omega$, that implies that $\partial\partial\Omega=\emptyset$.

Why is that true? Couldn't $\partial\partial\Omega$ be some nonempty but "small" domain such that you wind up integrating over a set of measure zero whenever you evaluate $\int_{\partial\partial\Omega}\omega$?

EDIT: I should clarify that I originally saw this in the context of integration over chains. I am not certain how this relates to integration over manifolds. I guess these are different questions (too much of a noob...). I don't see, for instance, how $\partial^2c=0$ for any $n$-chain $c$ would imply that $\partial^2\Omega=0$ for $n$-dimensional manifolds.

• Topologically the boundary operation is idempotent so the double boundary is not empty. Presumably the second boundary operation has to be understood in a different subspace. For example the boundary of $\partial \Omega$ as a subset of itself is empty. – Ian Sep 11 '17 at 4:52
• I believe you're right, but in this case, I think I should clarify that by ∂, I mean the manifold boundary and not the topological boundary. – WMe6 Sep 11 '17 at 5:02
• MathJax works in question titles and in the comment section as well. – gen-z ready to perish Sep 11 '17 at 6:46

The fact that $\partial\partial\Omega=\emptyset$ is trivial from the definition of the boundary of a smooth manifold and using Stokes' theorem is ridiculous overkill. In this context, $\Omega$ is a smooth $n$-manifold with boundary, meaning it has coordinate charts that make it locally diffeomorphic to $[0,\infty)\times\mathbb{R}^{n-1}$ or $\mathbb{R}^n$ at each point. The boundary of $\partial\Omega$ is defined as the union of the images of $\{0\}\times\mathbb{R}^{n-1}$ under the charts with domain $[0,\infty)\times\mathbb{R}^{n-1}$, and identifying $\{0\}\times\mathbb{R}^{n-1}$ with $\mathbb{R}^{n-1}$ these become the charts of an $(n-1)$-manifold structure on $\partial\Omega$. Since all of these charts have domain $\mathbb{R}^{n-1}$, not $[0,\infty)\times\mathbb{R}^{n-2}$, the boundary of $\partial\Omega$ is empty by definition.
That said, the Stokes' theorem argument does work. If $M$ is an oriented $m$-manifold and $\int_M \omega=0$ for all smooth $m$-forms $\omega$ on $M$, then $M$ must be empty. Indeed, if $M$ is nonempty, you can just use a bump function in any coordinate chart (times $dx_1\wedge dx_2\dots\wedge dx_m$ where the $x_i$ are the coordinate functions) to get an $m$-form whose integral is nonzero. You don't have to worry about $M=\partial\partial\Omega$ being some weird "small domain" as you suggest since $\partial\partial\Omega$ is by definition a manifold.
• If you are talking about chains then the Stokes' Theorem argument doesn't work to prove $\partial\partial c=0$ either, since you can have a nonzero smooth $n$-chain over which the integral of any $n$-form is trivial (for instance, just take a constant $n$-simplex for $n>0$). – Eric Wofsey Sep 11 '17 at 5:31
• That's interesting, because the claim comes from Nickerson, Spencer, and Steenrod, which states the theorem in terms of simplicial chains: $c_q$ – WMe6 Sep 11 '17 at 22:25
• Yeah, I think that's just an error in the book. The converse works, because it's easy to show that if a form vanishes on every chain, it is $0$. But there are nonzero chains on which every form vanishes, at least if you allow arbitrary smooth maps and not just embedded simplices (which I would assume is the case if they are described as "singular chains"). – Eric Wofsey Sep 11 '17 at 22:42