The singular homology and cohomology of manifolds vanishes in high dimensions

Question

Let $M$ be an $n$-manifold. It seems that there are two results that the $p$-th singular homology and cohomology of $M$ are zero if $p>n$.

But I can not find them in my books of algebraic topology. Can anyone tell me some reference about them. If the proof is not difficult, is it OK to show them to me?

Answer 1

In Algebraic Topology by Allen Hatcher:

Theorem 3.26. Let $M$ be a closed, connected $n$-manifold. Then:

(a) If $M$ is $R$-orientable, the map $H_n(M;R) \to H_n(M | x; R) \approx R$ is an isomorphism for all $x \in M$.

(b) If $M$ is not $R$-orientable, the map $H_n(M; R) \to H_n(M | x; R) \approx R$ is injective with image $\{ r \in R | 2r = 0 \}$ for all $x \in M$.

(c) $H_i(M; R) = 0$ for $i > n$.

Here $H_k(X | A; G) := H_k(X, X \setminus A; G)$. The proof is somewhat technical and I suggest you read the book to get a full idea of how it works.

The basic idea is that you want to show $H_i(M | A) = 0$ for $i > n$ when $A$ is a compact subset, and then take $A = M$. First you show that if it holds for $A$, $B$ and $A \cap B$ then it holds for $A \cup B$. Now you use that $M$ is a manifold and the previous property to reduce to the case $M = \mathbb{R}^n$. You reduce then to the case of a simplicial complex $A = K$, and then it follows by induction on the number of simplices.

For cohomology, now use the universal coefficient theorem, and the explicit description of $H_n(M;\mathbb{Z})$ given by the theorem depending on whether $M$ is $\mathbb{Z}$-orientable or not. For $k > n$, you get an exact sequence: $$0 \to \operatorname{Ext}^1_\mathbb{Z}(H_{k-1}(M; \mathbb{Z}), R) \to H^k(M; R) \to \hom_\mathbb{Z}(H_k(M; \mathbb{Z}), R) \to 0.$$

Since $H_{k-1}(M; \mathbb{Z})$ is either $0$ (when $k > n+1$) or a subgroup of $H_n(M; \mathbb{Z})$ (when $k = n+1$), and is therefore free abelian (because $H_n(M; \mathbb{Z})$ is either $\mathbb{Z}$ or zero depending on the orientability of $M$), it is always projective as a $\mathbb{Z}$-module, so the $\operatorname{Ext}$ vanishes. Meanwhile $H_k(M; \mathbb{Z})$ is zero ($k > n$). In the end the left and right terms are zero so the middle term $H^k(M; R)$ is zero too.

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For noncompact manifolds, there is the following proposition:

Proposition 3.29. If $M$ is a connected noncompact $n$-manifold, then $H_i(M; R) = 0$ for $i \ge n$.

The proof is basically a corollary of the compact case (I'll just give the general idea here). Let $[z] \in H_i(M; R)$ is represented by a cycle $z$, then $z$ has compact image in $M$ (because $\Delta^i$ is compact). So you can find a relatively compact open set $U$ containing the image of $z$; let $V = M \setminus \overline{U}$.

For $i > n$ you can simply apply the long exact sequence of the triple $(M, U \cup V, V)$ to get that $[z] = 0$. For $i = n$ it's more complicated and you have to reuse the proof of the previous theorem, and use that since $M$ is noncompact there are points not in the image of $z$. See the book for more details.

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Now for a "dumb" proof: use Poincaré duality, then if $i > n$, you find $H^i(M) \cong H_{n-i}(M) = 0$ because $n-i < 0$. I say "dumb" because the proof of the Poincaré duality theorem depends on the fact that $H_*$ vanishes in dimensions higher that $n$, so the proof is circular... It could still be a helpful mnemonic, and it gives a hint that for extraordinary cohomology theories, the result isn't necessarily true (indeed $K^{\mathbb{C}}_{2n}(*) \neq 0$ for example).

Answer 2

You do need to decide what homology/cohomology theory you use and then what theorems you assume as tools.

One result that is not that easy to prove, but intuitive is the following: a manifold of dimension $n$ can be covered by a family of open subsets so that any nonvoid intersection of these open sets is homeomorphic to a ball and at most $n+1$ of these subsets have non-void intersection. Now your statement about $H^{\le n}$, $H_{\le n}$ would follow for any standard theory. You can explicitely produce such a cover for $S^2$, or even for $S^n$. I guess Maier-Vietoris would help with the proof after you have the cover.