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Is there a topological space $X$ such that $H^i(X; \mathbb{Z}) = 0$ for all $i > 0$, but $H_n(X; \mathbb{Z}) \neq 0$ for some $n > 0$?

In his answer to the question Is homology determined by cohomology?, Qiaochu Yuan points out that for a Moore space $X = M(A, n)$,

$$\begin{align*} H^n(X; \mathbb{Z}) &\cong \operatorname{Hom}(A, \mathbb{Z})\\ H^{n+1}(X; \mathbb{Z}) &\cong \operatorname{Ext}^1(A, \mathbb{Z}) \end{align*}$$

and $H^i(X; \mathbb{Z}) = 0$ for all $i > 0$. As $\operatorname{Hom}(\mathbb{Q}, \mathbb{Z}) = \operatorname{Hom}(\mathbb{Q}\oplus\mathbb{Q}, \mathbb{Z}) = 0$ and $\operatorname{Ext}^1(\mathbb{Q}, \mathbb{Z}) \cong \mathbb{R} \cong \mathbb{R}\oplus\mathbb{R} \cong \operatorname{Ext}^1(\mathbb{Q}\oplus\mathbb{Q}, \mathbb{Z})$, we see that $M(\mathbb{Q}, n)$ and $M(\mathbb{Q}\oplus\mathbb{Q}, n)$ have the same cohomology, but different homology because $\mathbb{Q} \not\cong \mathbb{Q}\oplus\mathbb{Q}$.

The above shows that, in general, cohomology does not determine homology, but my question is about the specific case where the cohomology is trivial.

Note, if there is an example of a topological space $X$ with trivial cohomology but non-trivial homology, then there is also an example among the Moore spaces. More precisely, if $H_n(X; \mathbb{Z}) \cong A \neq 0$, then $M(A, n)$ also has trivial cohomology but non-trivial homology. Therefore, my initial question is equivalent to the following algebraic question:

Is there a non-trivial abelian group $A$ such that $\operatorname{Hom}(A, \mathbb{Z}) = 0$ and $\operatorname{Ext}^1(A, \mathbb{Z}) = 0$?

If an example of such a group exists, it must not be finitely generated. As such, if there is an $X$ with trivial cohomology but non-trivial homology, it cannot be homotopy equivalent to a compact manifold.

Community
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Michael Albanese
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  • So far I can show that we can assume WLOG that $A$ is torsion-free, and from here $A$ must also be reduced (have no nontrivial divisible subgroups). – Qiaochu Yuan Apr 09 '16 at 05:33

1 Answers1

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No such groups exist.

Suppose $\operatorname{Ext}(A,\mathbb{Z})=\operatorname{Hom}(A,\mathbb{Z})=0$. First, note that the functor $\operatorname{Ext}(-,\mathbb{Z})$ turns injections into surjections (this follows from the long exact sequence and the fact that $\operatorname{Ext}^2$ always vanishes for abelian groups). So we must have $\operatorname{Ext}(B,\mathbb{Z})=0$ for all subgroups $B\subseteq A$. In particular, this means every finitely generated subgroup of $A$ is free, so $A$ is torsion-free.

Now fix a prime $p$ and consider the short exact sequence $0\to pA\to A\to A/pA\to 0$. Since $A$ is torsion-free, $pA\cong A$. The Ext exact sequence $$\operatorname{Hom}(pA,\mathbb{Z})\to \operatorname{Ext}(A/pA,\mathbb{Z})\to \operatorname{Ext}(A,\mathbb{Z})$$ now tells us that $\operatorname{Ext}(A/pA,\mathbb{Z})=0$. But $A/pA$ is just a direct sum of copies of $\mathbb{Z}/p$, so this implies $A/pA=0$. Thus $pA=A$ for all primes $p$.

This means that $A$ is a divisible torsion-free group, and hence actually a $\mathbb{Q}$-vector space. Since $\operatorname{Ext}(\mathbb{Q},\mathbb{Z})\neq 0$, this implies $A=0$.

Eric Wofsey
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    Do you know if there is such a group $A$ if we drop the abelian hypothesis? – Michael Albanese Apr 10 '16 at 02:22
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    What do you mean by $\operatorname{Ext}(A,\mathbb{Z})$ if $A$ is nonabelian? – Eric Wofsey Apr 10 '16 at 02:34
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    Maybe @MichaelAlbanese meant modules over a ring in place of Abelian groups? By the way, just to make sure I got this straight, if we apply the long exact sequence of homology/cohomology and the mapping cylinder construction, is it true we can conclude that a map $f:X\rightarrow Y$ induces isomorphisms on $H_\bullet(-,\mathbb Z)$ iff it does so on $H^\bullet(-,\mathbb Z)$? – user46652 Apr 10 '16 at 23:11
  • @user46652: That's correct. – Eric Wofsey Apr 10 '16 at 23:29
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    @EricWofsey: Ah, completely forgot $A$ needs to be a $\mathbb{Z}$-module. Thanks. – Michael Albanese Apr 11 '16 at 01:46
  • @EricWofsey: Dear Eric, Thank you for this fantastic answer. I have similar kind of question in the category of spectra. Let E be a ring spectrum and $f : X \to Y$ be a map between two spectra and the cohomology map $E^(f)$ is an isomorphism. Does it imply that the homology map $E_(f)$ is also an isomorphism? In the other words, Does $E^(X) =0$ imply $E_(X) =0.$ – Surojit Jan 31 '19 at 14:21
  • @math: No. For instance, when $E$ is the sphere spectrum, by a theorem of Lin, $E^*(H\mathbb{F}_p)=0$. – Eric Wofsey Jan 31 '19 at 16:23
  • @EricWofsey: Thank you for your wonderful counter example. Could you please send me a reference for Lin's theorem? – Surojit Jan 31 '19 at 20:16
  • @math: Here's Lin's paper: https://www.jstor.org/stable/2041622?seq=1#page_scan_tab_contents. – Eric Wofsey Jan 31 '19 at 20:39
  • @EricWofsey: Is it possible to claim the following? Let $HF_p^(X) =0.$ Then $M^(X) =0$ for every $HF_p$-module spectrum $M.$ – Surojit Feb 01 '19 at 06:20
  • @math: Yes, since every $H\mathbb{F}_p$-module is a wedge sum of shifted copies of $H\mathbb{F}_p$. – Eric Wofsey Feb 01 '19 at 06:23
  • @EricWofsey: Why $(\bigvee_{i \in I} \sum^i HF_p)^*(X) =0?$ I can't see it. It is not a finite wedge. – Surojit Feb 01 '19 at 06:35
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    @math: I guess this is not quite obvious if there are infinitely many summands. Here's another way to think about it. A map $X\to M$ is the same thing as an $H\mathbb{F}_p$-module map $H\mathbb{F}_p\wedge X\to M$. So we wish to show that if there are no nonzero $H\mathbb{F}_p$-module maps $H\mathbb{F}_p\wedge X\to \Sigma^i H\mathbb{F}_p$ for all $i$, there are also no nonzero $H\mathbb{F}_p$-module maps $H\mathbb{F}_p\wedge X\to M$ for any $M$. – Eric Wofsey Feb 01 '19 at 06:49
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    But if all module maps $H\mathbb{F}_p\wedge X\to \Sigma^i H\mathbb{F}_p$ are trivial this implies $H\mathbb{F}_p\wedge X=0$, since $H\mathbb{F}_p\wedge X$ is a wedge of shifted copies of $H\mathbb{F}_p$. – Eric Wofsey Feb 01 '19 at 06:50
  • @EricWofsey: Since $HF_p^(X) =0$ this gives $[HF_p \wedge X, \sum^i HF_p]{HF_p-module}=0$ for all i. So, we get $HF_p \wedge X =0.$ This impies $(\bigvee{i \in I} \sum^i HF_p)^(X) = M^*(X) =0$. Am I understanding correctly? Still there is a question? Why First line implies $HF_p \wedge X =0?$ – Surojit Feb 01 '19 at 08:03
  • @EricWofsey: Say $HF_p \wedge X \cong \bigvee_{j \in J} \sum^j HF_p$. then $[HF_p \wedge X , \sum^i HF_p] =[ \bigvee_{j \in J} \sum^j HF_p, \sum^i HF_p] = \bigoplus_{j \in J}[\sum^j HF_p, \sum^i HF_p] =0$ for all i. So, if $HF_p \wedge X$ is non zero then considering i=j for some j gives a contradiction. I think this argument is fine. – Surojit Feb 01 '19 at 08:14
  • @EricWofsey: So, we get that if $HF_p^(X) =0$ then $HF_p \wedge X =0.$ I know this also holds if we replace $F_p$ by $\mathbb{Z}.$ Can we expect a similar thing for any subring of rationals? More generally, Can we classify also such spectra $E$ for which $E^(X)=0$ implies $E \wedge X=0$ for all spectra $X?$ – Surojit Feb 02 '19 at 08:53
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    @Surojit: I think it is true for any subring of $\mathbb{Q}$ but have not worked through the details. I don't know a general classification. I would suggest posting a new question if you want to know more about the topic. – Eric Wofsey Feb 02 '19 at 16:13
  • @EricWofsey: Thank you Eric. I'll post a new question. – Surojit Feb 02 '19 at 16:30
  • @EricWofsey: I have posted a new question and tried a lot by myself but can't able manage the subring of rational case. I also think that may be this paper is helpful https://pdfs.semanticscholar.org/6c77/86bf40128bce8d1c02176589810f55773d1d.pdf. Can you tell me how to proceed this case? – Surojit Feb 06 '19 at 11:59