# Is there an inverse of the Stokes' Theorem for manifolds with boundary?

I would like to consider this question When does a null integral implies that a form is exact? (also related to Top deRham cohomology group of a compact orientable manifold is 1-dimensional), but for manifolds with boundaries.

## 1. Stokes' theorem for a manifold without boundary

Let $$M$$ be an oriented n-manifold with no boundary and let $$\omega\in\Omega^n_c(M)$$.

If $$\omega=\mathrm{d}\eta\$$ for some $$\eta\in\Omega^{n-1}_c(M)\quad\Longrightarrow\quad\displaystyle\int_M\omega=0$$

## 2. "Inverse" of Stokes' theorem without boundary

Let $$M$$ be an oriented and connected n-manifold with no boundary and let $$\omega\in\Omega^n_c(M)$$.

If $$\displaystyle \int_M\omega=0 \quad\Longrightarrow\quad\exists\eta\in\Omega^{n-1}_c(M) \ /\ \omega=\mathrm{d}\eta$$

This is a direct consequence of the isomorphism $$[\omega]\in H^n_c(M)\mapsto\int_M\omega\in\mathbb{R}$$, given by the De Rham theorem for top-graded forms.

## 3. The Stokes' theorem with boundaries

Let $$M$$ be an oriented n-manifold with boundary $$\partial M\overset{\imath}{\hookrightarrow} M$$ (with the induced orientation) and let $$\omega\in\Omega^n_c(M)$$

If $$\omega=\mathrm{d}\eta\$$ for some $$\eta\in\Omega^{n-1}_c(M)\quad\Longrightarrow\quad\displaystyle\int_M\omega=\int_{\partial M}\imath^*\eta$$

## 4. "Inverse" of Stokes' theorem with boundary?

Let $$M$$ be an oriented and connected n-manifold with boundary $$\partial M\overset{\imath}{\hookrightarrow} M$$ and let $$\omega\in\Omega^n_c(M)$$ and $$\alpha\in\Omega^{n-1}_c(\partial M)$$.

If $$\displaystyle \int_M\omega=\int_{\partial M}\alpha\quad\Longrightarrow\quad\begin{array}{l}\exists\eta\in\Omega^{n-1}_c(M)\\\exists\gamma\in\Omega^{n-2}_c(\partial M)\end{array} \ /\ \begin{array}{l}\omega=\mathrm{d}\eta\\\alpha=\imath^*\eta+\mathrm{d}\gamma\end{array}$$

My question is what additional hypotheses are required? Notice that once we prove that $$\omega$$ is exact, then the existence of $$\gamma$$ is a consequence of Stokes' theorem over $$\partial M$$ (which has no boundary).

Notice also that if $$\omega$$ is non-exact, then we can always find $$\alpha$$ (by de Rham's theorem) such that

$$\int_{\partial M}\alpha=\int_M\omega\in\mathbb{R}$$

so this question is actually equivalent to prove that $$H_c^n(M)=0$$ if $$M$$ has boundary.

No additional assumptions are needed. In fact, you need fewer assumptions: of $$\omega$$ is any $$n$$-form with compact support on a connected oriented smooth $$n$$-manifold $$M$$ with nonempty boundary, then $$\omega=d\eta$$ for some $$(n-1)$$-form $$\eta$$ with compact support. To prove this, consider the "double" $$N$$ of $$M$$ obtained by gluing together two copies of $$M$$ along $$\partial M$$ (giving the second copy the opposite orientation). We can extend $$\omega$$ to an $$n$$-form on $$N$$ whose integral is $$0$$ (just first extend along a collar neighborhood of the boundary, and then add some $$n$$-form supported inside the second copy of $$M$$ to cancel out the integral). Since $$\partial M$$ is nonempty, $$N$$ is connected, so by the converse of Stokes' theorem for manifolds without boundary, the extension of $$\omega$$ can be written as $$d\eta$$ for some $$\eta$$ on $$N$$ with compact support. Since $$M$$ is closed in $$N$$, $$\eta$$ still has compact support when restricted to $$M$$.
(The orientability assumption can also be dropped; if $$M$$ is not orientable, then $$N$$ is not either, and then $$H_c^n(N)$$ is automatically trivial by Poincaré duality for nonorientable manifolds.)