For a smooth manifold, Stoke's theorem says that the differential/exterior derivative $\mathrm{d}$ is adjoint to the boundary operator $\partial$, i.e. $$\int_{\partial U} \omega = \int_{U} \mathrm{d}\omega.$$ where $\omega$ is a differential form. Given an inner product of forms (e.g. induced by a metric) $\langle\cdot,\cdot\rangle$, the codifferential $\delta$ can be defined as the adjoint of $\mathrm{d}$ with respect to the inner product, $$\langle \mathrm{d}\alpha,\beta \rangle = \langle \alpha, \delta \beta\rangle.$$ My question is, is there a coboundary $\partial^\dagger$ such that we could write a dual version of Stoke's theorem, $$\int_{\partial^\dagger U}\omega = \int_U \delta \omega.$$ I am wondering if there is a meaningful version of this in the continuum. Naively it seems like the answer is ``no'', but it feels like $\partial^\dagger$ ought to have some sort of meaning. One could certainly define such a thing for chains, say on a simplicial complex.

  • $\begingroup$ Think about dimensions for the integral to be defined. $\endgroup$ Apr 13, 2022 at 20:49
  • $\begingroup$ @TedShifrin I understand that $\partial^\dagger U$ would be one dimension higher. Very roughly, it would be something like an "infinitesimal thickening" of $U$. For chains on a simplicial complex, $\partial^\dagger$ of a single 0-simplex $(i)$ would be the sum of 1-simplices $\sum_{j@i} (ji)$ which terminate at $(i)$, but such a thing doesn't seem to have any continuous analogue $\endgroup$
    – Kai
    Apr 13, 2022 at 22:27
  • $\begingroup$ Yeah, it's not that simple. If you use $\delta = \pm{\star}d{\star}\omega$, you see that you're never going to be integrating $\omega$ itself over something. $\endgroup$ Apr 13, 2022 at 22:50


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