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Are all homology spheres compact? Are all generalized homology spheres compact? By a homology sphere I mean an $n$-manifold $X$ with same homology as the $n$-sphere. By a generalized homology sphere I mean the same with the assumption "$n$-manifold" replaced by "homology $n$-manifold".

If that helps, assume further that the spaces under consideration are simply connected.

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If $M$ is a connected, non-compact $n$-manifold, then $H_i(M;R)=0$ for $i\geq n$. For a proof, see Proposition 3.29 in Hatcher's Algebraic Topology book.

So, if you are going to have $H_n(M;R)=R$, $M$ had better be compact.

EDIT (to answer about homology $n$-manifolds):

A homology $n$-manifold is a finite dimensional, locally contractible space $X$ whose local homology groups $H_*(X, X-\{x\})$ are the local homology groups for $\mathbb{R}^n$ for every $x\in X$. In particular, $\mathbb{R}^n$ is a non-compact homology $n$-manifold.

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Thanks! How about generalized homology spheres? – Kestutis Cesnavicius Apr 21 '10 at 13:25
I edited to include a response to the second question. Does it help? – Bill Kronholm Apr 21 '10 at 15:07
$R^n$ is non-compact but it does not have same homology as the $n$-sphere. – Kestutis Cesnavicius Apr 21 '10 at 20:08
So, do you want a homology $n$-manifold with the same local homology as $S^n$ or the same (non-local) homology as $S^n$? – Bill Kronholm Apr 21 '10 at 20:51
Are you sure that finite dimensional and locally contractible is part of the definition of homology manifold? What notion of dimension is being used here? I vaguely recall that the definition only required $H_i(X,X-x)=H_i(R^n,R^n-0)$ for all $x$, and that most non-manifold examples were not locally compact. – Paul Apr 21 '10 at 23:49

The top dimensional cohomology of a connected manifold is non-zero if and only if the manifold is compact, so the answer is "yes" for homology spheres.

EDIT: As Mariano remarks below, "compact" should read "compact and orientable".

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iff the manifold is compact and orientable, no? – Mariano Suárez-Alvarez Apr 20 '10 at 16:24
Yes, thanks! (So the homology sphere is also orientable.) – Emerton Apr 20 '10 at 16:39

By a paper of Martio and Ryazanov, every connected, compact and simply connected homology $n$-sphere is homeomorphic to $S^n$, if $ \ n \geqslant 4$. By the affirmative answer to the Poincaré conjecture, this follows on dimension $3$ as well.

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