# Triviality of the cohomology group of compact connected manifolds

Let $$M$$ be a compact connected $$n$$-manifold with non-empty boundary $$\partial M$$. Is its $$n$$-th cohomology group $$H^n(M)$$ always trivial?

I can prove it if the manifold is orientable as follows: by Lefschetz duality, $$H^n(M)=H_0(M,\partial M)=0.$$

• $M$ ought to be homotopic to a CW complex of dimension $n-1$ although I wouldn't know how to prove it; maybe Morse theory? Commented Jul 27, 2022 at 17:23
• In connection with your comment: compact manifolds are ENRs, and ENRs are retracts of finite simplicial complexes, however I don't see how to prove that it they should be of dimension $n-1$. Commented Jul 27, 2022 at 17:40
• Can you prove it in the case of empty boundary? If so, you can the result in this case via Mayer -Vietoris applied to the double $M\cup_{\partial M} M$. Commented Jul 28, 2022 at 1:18
• I think if the boundary is empty, then $H^n(M)=\mathbb{Z}$ if $M$ is orientable and $H^n(M)=\mathbb{Z}/2\mathbb{Z}$ otherwise. Commented Jul 28, 2022 at 3:07
• By the way, I didn't realize you responded to my comment until the question was bumped by Balarka's edit. If you want to notify me that you responded, you'll need to type @JasonDeVito in the comment. (There are some situations where someone is automatically notified, e.g., I think you'll be automatically notified by this comment. But I don't know all the rules.) Commented Aug 1, 2022 at 23:51

$$M$$ and $$M \setminus \partial M$$ have the same homology since by existence of a collar neighborhood of the boundary, $$M \setminus \partial M$$ deformation retracts to a properly embedded copy of $$M$$ inside $$M$$. Therefore, let us reassign $$M$$ to be the noncompact manifold $$M \setminus \partial M$$ from here on.

By compactly supported Poincare duality with twisted coefficients, $$H_n(M; \Bbb Z) \cong H^0_c(M; \Bbb Z_w)$$ where $$\Bbb Z_w$$ is the orientation sheaf on $$M$$. I do not know a reference for an elementary derivation, but this follows from, for example, Verdier duality. Since $$M$$ is noncompact, $$H^0_c(M; \Bbb Z_w) = 0$$.

EDIT: It is also true that $$H^n(M; \Bbb Z) = 0$$ for a noncompact $$n$$-manifold $$M$$. There is a dual version of Poincare duality with twisted coefficients, which says $$H^n(M; \Bbb Z_w) \cong H_0^{lf}(M; \Bbb Z)$$ where $$H_*^{lf}$$ is locally-finite homology; this also follows from Verdier duality. Since $$M$$ is noncompact, any $$0$$-chain (finitely many points) is a boundary of the locally finite $$1$$-chains given by a collection of disjoint proper PL-rays to infinity in $$M$$ starting at these points.

Unfortunately, the twisting of the coefficients is in the cohomology side this time, so it's not immediately clear why $$H^n(M; \Bbb Z) = 0$$ for $$M$$ nonorientable. Nevertheless, this proves $$H^n(M; \Bbb Z) = 0$$ for orientable noncompact $$n$$-manifold $$M$$. For nonorientable $$M$$, consider the cohomology transfer sequence $$\cdots \to H^n(M; \Bbb Z_w) \to H^n(\widetilde{M}; \Bbb Z) \to H^n(M; \Bbb Z) \to H^{n+1}(M; \Bbb Z_w) \to \cdots$$ Where $$\widetilde{M}$$ is the orientation cover of $$M$$. We know $$H^n(\widetilde{M}; \Bbb Z) = 0$$ by orientability and $$H^{n+1}(M; \Bbb Z_w) = 0$$ as $$M$$ is $$n$$-dimensional. This forces $$H^n(M; \Bbb Z) = 0$$.

• Thank you for your answer, but shouldn't it be cohomology rather than homology there? Commented Aug 1, 2022 at 12:30
• @Djamel I didn't notice you were asking for $H^n = 0$, and instead proved $H_n = 0$. I'll make an edit. Commented Aug 1, 2022 at 16:19
• Thank you @Balarka, do you know some references to understand more this proof? Commented Aug 1, 2022 at 17:13
• @Djamel Hatcher ch 3 has twisted coefficients and transfer homomorphism as an additional topic. I learnt the twisted Poincare duality from Goresky's notes here. Commented Aug 2, 2022 at 6:50
• OK thank you @BalarkaSen Commented Aug 2, 2022 at 15:23

Here's the answer I alluded to in the comments (which assumes you know how compute $$H^{\dim M}(M)$$ for a closed manifold $$M$$). We define $$N = M\coprod M/\sim$$ where $$\sim$$ identifies the boundary of each copy of $$M$$ via the identity map. Since $$\partial M$$ has a collar, $$N$$ is a closed manifold, called the double of $$M$$. Note that $$N$$ is non-orientable because it contains the non-orientable manifold $$M\setminus \partial M$$ as an open subset.

Set $$n =\dim M = \dim N$$ and consider the Mayer-Vietoris cohomology sequence for $$N$$ from the decomposition into copies of $$M$$. A portion of it looks like:

$$...\rightarrow H^n(N)\rightarrow H^n(M)\oplus H^n(M)\rightarrow H^n(\partial M)\rightarrow ...$$

Now, $$H^n(\partial M) = 0$$ since $$\partial M$$ has dimension $$n-1 < n$$, so we see that $$H^n(N)$$ must surject onto $$H^n(M)\oplus H^n(M)$$. Since $$N$$ is non-orientable, $$H^n(N)\cong \mathbb{Z}/2\mathbb{Z}$$.

Thus, we have a surjection $$\mathbb{Z}/2\mathbb{Z}\rightarrow H^n(M)\oplus H^n(M)$$. This obviously implies $$H^n(M) = 0$$, as claimed.

• By the way, if $M$ is non-compact, but has a boundary, almost the the same argument works to show that $H^n(M) = 0$: instead of $H^n(N)\cong \mathbb{Z}/2\mathbb{Z}$, we get $H^n(N) = 0$, so the argument is even easier after that point. Commented Aug 1, 2022 at 23:48
• Nice, thank you for your answer @JasonDeVito Commented Aug 2, 2022 at 15:25