# Integration by parts on manifold with corners

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.

• I think the second term on the RHS should read $\int_Mf\text{div}X d\text{vol}$. – S.Surace May 16 '19 at 13:58
• Thanks, I fixed it. – Δημήτρης Ο May 16 '19 at 14:02
• 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. – Deane Yang Mar 4 '20 at 13:51
• ^Especially since corners are measure 0 points. – Chris Gerig Jul 2 '20 at 6:40

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 $$x \mapsto \sum_i x_i$$ is a Lipschitz map which is positive inside $$R$$ and negative outside it.
• The last sentence needs work; clearly if $n=2$ then at $x=(2,-1)$ that map is positive. – Ben McKay Jul 2 '20 at 7:34