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Is it true that the homology of a manifold with field coefficients determines the homology over the integers? I know that by the universal coefficient theorem that $H_k(X; \mathbb{F}) \cong H_k(X; \mathbb{Z}) \bigotimes \mathbb{F}$ since the Tor part vanishes for fields. Then I think we can use the structure theorem of abelian groups (since $H_k(X; \mathbb{Z})$ is an abelian group). Is this true? Is there a reference for this claim?

Bob
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    I don't think that $H_\ast(X;\mathbf{Z})$ is finitely generated for any manifold. – Mister Benjamin Dover Feb 01 '15 at 22:24
  • @Later: it's finitely generated for any compact manifold. – Qiaochu Yuan Feb 01 '15 at 22:34
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    @Bob: do you mean the homology with coefficients in all fields or in a particular field? If the latter, the answer is no because of the possibility of torsion in the homology that tensoring with a field may kill. If the former, your claim is only true if $F$ has characteristic zero; if $F$ has positive characteristic then the Tor part of the universal coefficient theorem need not vanish. And of course if $F$ has characteristic zero then tensoring with $F$ does not detect torsion. – Qiaochu Yuan Feb 01 '15 at 22:36
  • @QiaochuYuan: I know (it is homotopy equivalent to a finite CW complex), but I doubt that it is so for any manifold – Mister Benjamin Dover Feb 01 '15 at 22:37
  • Yes, of course. – Qiaochu Yuan Feb 01 '15 at 22:37
  • @QiaochuYuan Thank you. I mean all fields, not a particular field. – Bob Feb 02 '15 at 00:03

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No. More generally, let $X$ be a space whose integral homology is levelwise finitely generated. This is true for any compact manifold but not true for general noncompact manifolds, e.g. a surface of infinite genus. Then the integral homology groups $H_k(X, \mathbb{Z})$ are determined by their rank and by the number of times the summand $\mathbb{Z}_{p^k}$ shows up in their torsion subgroups.

The rank, and only the rank, can be detected by taking homology with coefficients in $\mathbb{Q}$, or more generally any field of characteristic $0$. Some information about the $p$-torsion can be detected by taking homology with coefficients in $\mathbb{F}_p$, or more generally any field of characteristic $p$, although less directly than you suggest: universal coefficients gives a short exact sequence

$$0 \to H_i(X, \mathbb{Z}) \otimes \mathbb{F}_p \to H_i(X, \mathbb{F}_p) \to \text{Tor}(H_{i-1}(X, \mathbb{Z}), \mathbb{F}_p) \to 0.$$

The Tor term here does not vanish in general; its rank is the rank of the $p$-torsion in $H_{i-1}(X, \mathbb{F}_p)$. The rank of the leftmost term is the rank of $H_i(X, \mathbb{Z})$ plus the sum of the number of times $\mathbb{Z}_{p^k}$ shows up over all $k$.

So the problem is that taking homology with coefficients in a field cannot tell you all of the information in the torsion: in particular it cannot distinguish between an integral homology group with torsion of the form $\mathbb{Z}_{p^2}$ and torsion of the form $\mathbb{Z}_{p^3}$, for example (and manifolds with torsion of this form are not hard to write down, e.g. using lens spaces). You can detect this by taking homology with coefficients in $\mathbb{Z}_{p^k}$ for all $k$, or more cleanly by taking homology with coefficients in the $p$-adic integers, which to avoid collision with my previous notation I'll write $\widehat{\mathbb{Z}}_p$. A keyword to look up here is fracture theorem.

Qiaochu Yuan
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  • Thanks again. This shows that just considering $\mathbb{F}p$ is not enough. What about coefficients in the finite field $\mathbb{F}{p^k}$? – Bob Feb 02 '15 at 00:05
  • I guess it doesn't help very much since we know that $H_i(X, \mathbb{F}p) \bigotimes{\mathbb{F}p} \mathbb{F}{p^k} \cong H_i(X, \mathbb{F}_p)^k$. – Bob Feb 02 '15 at 00:20
  • @Bob: right. Like I said, any field of characteristic $p$ will just give you the same information as $\mathbb{F}_p$. – Qiaochu Yuan Feb 02 '15 at 00:29