In a lecture of advanced calculus, my teacher made a very interesting remark about the generalized Stokes' Theorem (actually, he left it as an exercise!), such that, if I understood it right, is something like:

Let $\Omega \subseteq \mathbb{R}^n$ be an open set and $\tau:\mathrm{F}_{n-1}(\Omega) \to \mathrm{F}_n(\Omega)$ a linear operator satisfying $$ \int_{\Theta}\tau\omega = \int_{\partial\Theta} \omega,\quad \forall \omega\in\mathrm{F}_{n-1}(\Omega),\forall\Theta\in\mathrm{C}_{n}(\Omega),$$ where $\mathrm{C}_n(\Omega)$ is the set of singular $n$-chains in $\Omega$. Then $\tau = \mathrm{d}$, the exterior derivative operator.

My attempt: From Stokes' Theorem, we have that $\int_{\Theta} (\tau-\mathrm{d})\omega = 0,~ (\forall\omega,\forall \Theta)$, and we have to prove that this implies $(\tau-\mathrm{d})\omega = 0,~ (\forall\omega,\forall \Theta)$.

Suppose WLOG that $\Theta=\xi\in S_n(\Omega)$ (set of singular $n$-simplex in $\Omega$). Thus $$ \int_{\xi}(\tau-\mathrm{d})\omega = \int_{\sigma=[v_0,\cdots,v_n]}\left((\tau-\mathrm{d})\omega\right)_\xi,$$

but here there's a problem, as it's not clear whether I can commutate the operator $(\tau-\mathrm{d})$ with the pullback or not. Is this the wrong way?


As $(\tau - d)\omega \in F^n(\Omega)$, write $$ (\tau - d)\omega = f(x) dx^1\cdots dx^n,$$ where $f(x)$ is differentiable. If $(\tau - d)\omega \neq 0$, then there is $x\in \Omega$ so that $f(x) > 0$ (or $<0$, the argument are the same). Then $f(x) >0$ on an open set $\Omega' \subset \Omega$. Consider a singular chain $\Theta$ with image in this $\Omega'$, then you get $$\int_\Theta (\tau - d)\omega >0,$$ which is a contradiction.

(You just treat $(\tau - d)\omega$ as one whole thing and show that it is zero).

  • $\begingroup$ Hmm, sorry, but this doesn't create problems with the orientation of the chain? It feels like considering $\Theta$ to be a simplex fix that problem, but I'm not sure whether this is in fact a problem or I just don't understood this step at all haha $\endgroup$ – DylanOneida Dec 10 '15 at 21:12
  • $\begingroup$ @DylanOneida there won't be any problem with the orientation as $\Omega$ is just an open set in $\mathbb R^n$. Can you explain why you think there is issue with the orientation? $\endgroup$ – user99914 Dec 10 '15 at 21:45
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    $\begingroup$ Well, I thought that the chain could be something like $[v_0,v_1,...,v_n] + [v_n,v_1,...,v_(n-1),v_0]$ $\endgroup$ – DylanOneida Dec 10 '15 at 23:37
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    $\begingroup$ Ok I see what you mean. Yes we should choose $\Theta$ to be a single simplex at the end. @DylanOneida $\endgroup$ – user99914 Dec 10 '15 at 23:39
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    $\begingroup$ Copy that. Many thanks! $\endgroup$ – DylanOneida Dec 10 '15 at 23:55

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