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Let $M$ be a smooth connected oriented without boundary non-compact manifold of dimension n.

Let $k$ be a principal ideal, e. g. the integers $Z$

Let $H_n(M)$ and $H^n(M)$ be the homology and cohomology in degree n of $M$ with coefficients in $k$.

It is well-known that $H_n(M)=0$.

Is $H^n(M)$ also trivial ?

By the universal coefficient theorem for cohomology, $H^n(M)=Ext(H_{n-1}(M),k)$. Therefore if $k$ is a field, $H^n(M)=0$. I would like to know this answer over a principal ideal: $Z$.

It is also known (Bredon's book) that $H_{n-1}(M)$ is without torsion. But I believe that $H_{n-1}(M)$ in the non-compact case, is a not a finitely generated $k$-module. Therefore I don't know if $H_{n-1}(M)$ is free, i. e. projective. Since there exists abelians groups without torsion, non-free, e. g. the rationals $Q$

I suppose that this must be well-known. But I could not find a reference.

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3 Answers 3

up vote 3 down vote accepted

By a version of Poincaré duality, $\mathrm H^n(M, \mathbb Z)$ is isomorphic to the $0$-th Borel-Moore homology group $\mathrm H_0^{\mathrm BM}(M, \mathbb Z)$, (see, for example, IX-4 in Iversen's book on cohomology of sheaves). On the other hand $\mathrm H_0^{\mathrm BM}(M, \mathbb Z) = 0$; this can easily be seen considering the one-point compactification $\overline M$, and the exact sequence $$ 0 \longrightarrow \mathrm H_0^{\mathrm BM}(\overline M \smallsetminus M, \mathbb Z) \longrightarrow \mathrm H_0^{\mathrm BM}(\overline M, \mathbb Z) \longrightarrow \mathrm H_0^{\mathrm BM}(M, \mathbb Z) \longrightarrow \mathrm H_1^{\mathrm BM}(\overline M \smallsetminus M, \mathbb Z) = 0 $$ since $\mathrm H_0^{\mathrm BM}(\overline M, \mathbb Z) = \mathbb Z$.

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Could you give more details please? For instance how do you know that the last map is surjective? –  Mark Grant Jul 18 '11 at 10:36
    
You can see that $H_0^{BM}(M;\mathbb{Z})=0$ directly using that $H_*^{BM}$ is the homology of the complex of locally finite singular chains (for a nice enough space, and manifolds are very nice). For any generating $0$-simplex, just run a simple curve from it off to infinity and then triangulate it as you're triangulate a half-line. That's a Borel-Moore 1-simplex whose boundary is your generating $0$-simplex. –  Greg Friedman Jul 18 '11 at 12:54
    
(of course in Borel-Moore homology you might have $0$-chains with an infinite number of $0$-simplices, but the local finiteness condition says they can't have an accumulation point, so then you can run an infinite number of curves off to infinity to handle them) –  Greg Friedman Jul 18 '11 at 12:54
    
The exact sequence above incorrect, I fixed it. I like the sheaf-theoretic version of Borel-Moore homology, as in Iversen's book. With this definition $\mathrm H_)^{\rm BM}(M, \mathbb Z)$ is the dual of the $0$-th compactly supported cohomology group, which is clearly $0$. –  Angelo Jul 18 '11 at 14:35

I may be mistaken, I speak from memory, but I think that $H^i(M,k)\cong H_{n-i}^\infty(M,k)$ is the homology of infinite chains, in particular $H^n(M,k)\cong H_0^\infty(M)$. This homology is computed from a (locally finite) triangulation replacing $\bigoplus$ in the usual definition of the simplicial chain complex by $\prod$, i.e. we allow infinite linear combinations of simplices. I think one can argue $H_0^\infty(M)=0$ in the following way, at least if $\dim M\geq 2$, subdividing, or whatever. Any infinite $0$-cycle is a possibly infinite linear combination of vertices. For each vertex in the linear combination take a divergent simplicial half line ending on it. Take all of them in such a way that no edge is contained in infinitely many half lines. Then take the $1$-chain formed by the sum of the edges of all half lines coherently oriented. The boundary of this $1$-chain should be the original $0$-cycle.

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Maybe too late, and close to the Fernando's argument. Let $c$ be an infinite path going through each center of the $n$-simplex of a triangulation. Let $U$ be a neighbourhood of $c$, we can assume $U$ has the homotopy type of $[0,+\infty[$ and $\partial U$ is homeomorphic to $\mathbb R^{n-1}$. Then $V= \overline{X\setminus U}$ retracts on the $n-1$-skeleton, so $H^{n-1}(V)=0$ and we conclude by Mayer Vietoris

$$H^{n-1}(U\cap V) \to H^n(X) \to H^n(U)\oplus H^n(V)$$

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