Rudin's *Principles of Mathematical Analysis*, p. 276 excerptFirst 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.

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    $\begingroup$ 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. $\endgroup$ – Ian Sep 11 '17 at 4:52
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    $\begingroup$ 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. $\endgroup$ – WMe6 Sep 11 '17 at 5:02
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    $\begingroup$ MathJax works in question titles and in the comment section as well. $\endgroup$ – 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.

  • $\begingroup$ Ah -- that's an interesting argument. I've only even seen texts prove ∂^2 = 0 using brute force computation by applying the definition of the boundary of a cube or simplex, which (I guess) would then extend to a chain and then to a manifold. $\endgroup$ – WMe6 Sep 11 '17 at 5:23
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    $\begingroup$ That is in a different context than the usual statement of Stokes' Theorem, where you are talking about some sort of chains rather than smooth manifolds. The definition of boundary there is different (as are the objects you are taking the boundaries of) and so the trivial argument does not apply. $\endgroup$ – Eric Wofsey Sep 11 '17 at 5:28
  • $\begingroup$ 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 $n>0$). $\endgroup$ – Eric Wofsey Sep 11 '17 at 5:31
  • $\begingroup$ That's interesting, because the claim comes from Nickerson, Spencer, and Steenrod, which states the theorem in terms of simplicial chains: $c_q$ $\endgroup$ – WMe6 Sep 11 '17 at 22:25
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    $\begingroup$ 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"). $\endgroup$ – Eric Wofsey Sep 11 '17 at 22:42

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