If the boundary is adjoint to the differential, what is the "coboundary" adjoint to the codifferential in the continuum?

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.

• Think about dimensions for the integral to be defined. Apr 13, 2022 at 20:49
• @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
– Kai
Apr 13, 2022 at 22:27
• 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. Apr 13, 2022 at 22:50