In page 79 of Bott-Tu, "Differential Forms in Algebraic Topology", they define the relative de Rham theory as follows:

Let $f:S\to M$ be a smooth map. Define the complex $\Omega^*(f)$ by $$\Omega^k(f):=\Omega^k(M)\oplus\Omega^{k-1}(S)$$ $$\underline{\mathrm{d}}(\alpha,\beta)=(\mathrm{d}\alpha,f^*\alpha-\mathrm{d}\beta)$$ It is easy to prove that $\underline{\mathrm{d}}^2=0$ which allows us to define the cohomology $H^*(f)$. As a particular case, one can consider a submanifold $\imath:N\hookrightarrow M$ and define $$\Omega^*(M,N):=\Omega^*(\imath)$$

My interest lies in the case when $N=\partial M$ and $M$ compact, where one can also define the integral of top forms as $$\int_{(M,\partial M)}(\alpha,\beta):=\int_M\alpha-\int_{\partial M}\beta$$ It is easy to check, using Stoke's theorem, that $$\int_{(M,\partial M)}\underline{\mathrm{d}}(\alpha,\beta)=0$$ Thus, we have a well defined map $$\tag{1}\label{one}\int_{(M,\partial M)}:H^n(M,\partial M)\to\mathbb{R}$$

If $\partial M=\varnothing$, then $H^n(M,\partial M)=H^n(M)$ and the previous integral is the standard one. The de Rham's theorem for top-forms then tells us that if $M$ has no boundary $$\tag{2}\label{two}\int_M:H^n(M)\to\mathbb{R}\quad \text{ is an isomorphism}$$ However, with boundary we have:

 1. It is surjective (applying \eqref{two} over the boundary and using elements of the form $(0,\beta)$).

 2. Its kernel is isomorphic to $H^n(M)$. Sketch of the proof: for every $[\alpha]\in H^n(M)$, build an element $[(\alpha,\beta)]$ such that $\int_{(M,\partial M)}(\alpha,\beta)=0$ using de Rham's theorem over the boundary. This map is well defined.

I have a heuristic argument to show that $H^n(M)$ is always zero: given $\alpha\in\Omega^n(M)$, take the double of $M$ along the boundary $\partial M$ and extend to some $\widetilde{\alpha}\in\Omega^n(M\sqcup_{\partial M}M)$ such that its integral is zero (using a tubular neighborhood over $\partial M$). Then using \eqref{two} (the double has no boundary) shows that $\widetilde{\alpha}$ is exact and, therefore, its pullback to $M$, which is $\alpha$, is also exact.

This seems a very strong result that I haven't found anywhere, while the proof seems very simple, thus I doubt if there are obstructions to the extension that invalidate the proof.

So the questions I have in mind (all of them are almost the same question) are:

  1. Is $H^n(M)=0$ if $M$ is compact with boundary?
  2. Is there a useful characterization of $H^n(M,\partial M)$ that can be used in this context?
  3. Is there a de Rham's theorem like \eqref{two} for manifolds with boundary (with no prescribed boundary conditions)?
  4. Is there a de Rham's theorem like \eqref{two} for relative cohomology?
  5. If $H^n(M)\neq 0$, is there another a map $G:H^n(M,\partial M)\to \mathbb{R}$ such that $(\int_{(M,\partial M)},G)\to\mathbb{R}^2$ is an isomorphism?

1 Answer 1


It is indeed true that $H^n(M)=0$ if $M$ is a compact manifold with boundary. In particular, $H^n(M,\partial M)\cong\mathbb{R}$ by Lefschetz duality (as Chris Gerig mentioned) and the integral (1) is an isomorphism.

The only reference I have found that states this results is:

Differential forms: theory and practice. Steven Weintraub. Academic Press (Elsevier) 2014.

  • Theorem 8.3.10 for compact manifolds with boundary.
  • Theorem 8.4.8 for non-compact manifold with boundary using compact support forms.

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