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Suppose that $M$ is a compact manifold with corners, where each boundary hypersurface is an embedded submanifold. Then, do we have an integration by parts identity? i.e. \begin{align*} \int_M g(\nabla f, X) \,d\textrm{vol} = \int_{\partial M} f\left\langle X,N \right\rangle \,d\textrm{vol}_{\tilde{g}} - \int_M f\cdot (\operatorname{div} X ) \,d\textrm{vol} \end{align*} with $f\in C^\infty(M), X\in \Gamma(M,TM)$. I know that this identity hold for domains with lipschitz boundary, but it is not very clear to me if a domain with corners is a special case of a lipschitz domain.

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    $\begingroup$ Proving this identity on a manifold with corners is quite straightforward using a partition of unity and local coordinates. Citing a theorem about domains with Lipschitz boundary seems unnecessary to me. $\endgroup$
    – Deane Yang
    Mar 4, 2020 at 13:51
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    $\begingroup$ ^Especially since corners are measure 0 points. $\endgroup$ Jul 2, 2020 at 6:40

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Yes, you can treat manifolds with corners as Lipschitz domains.

By definition of a manifold with corners $M$, for any $p \in \partial M$, there is a smooth coordinate map sending a neighbourhood of $p$ to a subset of $R:=(\mathbb{R}_{\ge 0})^n$. Then $x \mapsto \min_i x_i$ is a Lipschitz map which is positive inside $R$ and negative outside it.

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  • $\begingroup$ I found a theorem statement in Mikhail S. Agranovich's Sobolev Spaces, Their Generalizations, and Elliptic Problems in Smooth and Lipschitz Domains. There may be better ones out there, please leave a comment if you know of any. $\endgroup$
    – user7868
    Feb 3, 2020 at 4:02
  • $\begingroup$ I've corrected it now. $\endgroup$
    – user7868
    Jun 23, 2022 at 3:25

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