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My question is

Is any CW complex with only finitely many nonzero homology groups homotopic to a finite dimensional CW complex?

(My thoughts on this which might not be useful at all.) Since an infinite dimensional CW complex could have only finitely many nontrivial homology groups($S^\infty$ for example), it seems to me that the relation between the dimension and the zeroness of the CW complexes is not very strong. On the other hand, we know that Moore spaces are unique up to homotopy equivalence and any CW complex with prescribed homology groups can be construct by taking the wedge sum of Moore spaces. If this statement above is true, then it means every CW complex with only finitely many nonzero homology groups is essentially built up in this way...

Please note that I meant finite dimensional CW complex instead of finite CW complex. Otherwise the infinite dimensional discrete space will serve the purpose. Also, I really appreciate it if someone can point it out whether the statement can become true buy adding some small conditions(One of my professors said we need $X$ to be simply connected).

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    $\begingroup$ No, there is an $\pi_1$ issue. Let $G$ be an acyclic group, then the classifying space BG is a space whose homology vanishes, but it is not contractible and if G is infinite than it is not homotopy equivalent to a finite complex. An example of such an infinite acyclic group is the group G of all bijections of an infinite set. $\endgroup$
    – Chris Schommer-Pries
    Commented Apr 26, 2017 at 17:49
  • $\begingroup$ @ChrisSchommer-Pries I think you should write this up as an answer. $\endgroup$
    – Gabriel C. Drummond-Cole
    Commented Apr 26, 2017 at 19:22
  • $\begingroup$ @ChrisSchommer-Pries: Perhaps you meant to say "if $G$ has infinite geometric dimension", since there are certainly infinite acyclic groups for which $BG$ may be taken to be a finite complex. Also, is the group of bijections of an infinite set the same thing as the infinite symmetric group $\Sigma_\infty$? I guess not, since the latter is not acyclic. $\endgroup$
    – Mark Grant
    Commented Apr 26, 2017 at 19:38
  • $\begingroup$ @MarkGrant You are right. What I meant to say was something like "not finitely generated". The infinite symmetric group $\Sigma_\infty$ is the subgroup which fixes all but finitely many elements of the (countable) infinite set. $\endgroup$
    – Chris Schommer-Pries
    Commented Apr 26, 2017 at 22:01

3 Answers 3

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In the simply connected case, the answer is yes.

In the general case, the theory was worked out in complete detail by Wall in the paper:

Wall, C. T. C. Finiteness conditions for CW-complexes. Ann. of Math. (2) 81 1965 56–69.

See "Theorem E."

Here is a description of the result:

In the non-simply connected case, to get that your complex $X$ is equivalent to an $n$-dimensional one, you need to know that Wall's condition $D_n$ is satisfied:

$D_n$: $H_i(\widetilde X) =0$ for $i > n$ and $H^{n+1}(X;\mathscr B) = 0$.

In the above, we $\widetilde{X}$ is the universal covering and $\mathscr B$ denotes any local coefficient system on $X$.

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  • $\begingroup$ Thank you! Is it easy to prove the simply connected case? Any reference for this if not? $\endgroup$
    – No One
    Commented Apr 28, 2017 at 0:55
  • $\begingroup$ @NoOne: The standard types of arguments use Whitehead moves. You presumably need something a little more subtle when the space isn't simply-connected. $\endgroup$
    – Ryan Budney
    Commented Dec 25, 2022 at 21:21
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As requested I am writing this as an answer.

No there are spaces with vanished homology which are not homotopy equivalent to finite CW-complexes.

For example if $G$ is an acyclic group, then the classifying space will be a space with vanishing homology. It is not contractible as $\pi_1 BG = G$.

Now a finite CW-complex will always have $\pi_1$ a finitely generated group (in fact finitely presentable). So if $G$ is not finitely generated, then $BG$ is not homotopy equivalent to a finite CW-complex.

There are many examples of large infinitely generated acyclic groups. For example can take the group of bijections of a countably infinite set. This is an uncountably infinite group and so it cannot be finitely generated by size considerations.

[edit: It was pointed out in the comments that I misread the question. I thought the question was asking if the space is homotopy equivalent to a finite CW-comples, when actually it asks for a finite dimensional CW-complex. Nevertheless BG where G is the group of bijections of an infinite set is still a counter example to the asked question, see Mark Grant's answer]

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    $\begingroup$ The question is about homotopy equivalence to a finite dimensional complex not a finite one. $\endgroup$
    – Igor Belegradek
    Commented Apr 27, 2017 at 3:09
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    $\begingroup$ The mentioned group of bijections is also "large" in the sense of having infinite (cohomological, and therefore geometric) dimension. See the very last sentence of the reference in my answer. So it does work as a counter-example to the stated question. $\endgroup$
    – Mark Grant
    Commented Apr 27, 2017 at 8:30
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Supplementing Chris's answer, you could take $G$ to be any acyclic group of infinite cohomological dimension. Such groups exist, e.g. binate groups. See

Berrick, A.J., The acyclic group dichotomy., J. Algebra 326, No. 1, 47-58 (2011). ZBL1253.20055.

Then any model of $BG$ will have trivial homology but must be infinite-dimensional. Taking wedge sums or cartesian products of $BG$ with a finite complex will give plenty more counter-examples.

For the (positive) simply-connected statement mentioned by John Klein, search for "minimal CW structures".

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