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Let's keep our groups finitely presented for the time being. All spaces in this post are path connected.

Background: By a standard construction (e.g., on p. 365 of Hatcher), there exists a $K(\pi, n)$ for any (Abelian for $n>1$) group $\pi$ and any positive integer $n$. We can then multiply these varying over $n$ to obtain a space with arbitrary homotopy groups (abelian for $n>1$ again). An easier construction shows that we can do the same for homology, at least for any sequence of finitely presented groups, by wedging together a bunch of spheres and adding $n$-cells to give the appropriate torsion.

Now, we can't do this simultaneoulsy in a completely arbitrary fashion: the Hurewicz theorem gives an isomorphism between (the abelianization of) the first nontrivial homotopy group and the homology in the same dimension. But are there any other obstructions to doing this?

Question: Fix $n>0$. Assume I have two sequences of (finitely presented) Abelian groups $G_{n+1}, G_{n+2}, \dots$ and $H_{n+1}, H_{n+2}, \dots$, and in addition a (finitely presented) group $G_n$ and $H_n \cong G_n^{ab}$.

Is there some path-connected space $X$ with $\pi_k(X) \cong G_k$ and $H_k(X) \cong H_k$ for $k\geq n$ and $\pi_k(X) \cong H_k(X) \cong 0$ for $k<n$?

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Sorry if the homotopy-theory tag isn't appropriate. Wasn't sure. –  Mike Miller Jun 22 at 4:43
    
You don't consider the homotopy/homology groups of degree $>n$? If so $X=K(G_n,n)$ will do. Also only $G_1, H_1\sim G_1^{\mathrm{ab}}, H_2,\dots$ are required for your question in its current state. –  Olivier Bégassat Jun 22 at 4:48
    
@OlivierBégassat Stupid typo. Thanks. –  Mike Miller Jun 22 at 4:52
    
In general, the first nontrivial homology groups and homotopy groups appear in the same dimension. If this is $1$, then $H_1$ is the abelianization of $\pi_1$. Otherwise $H_n\cong \pi_n$ for the first nontrivial group. This is the Hurewicz Isomorphism. –  Grumpy Parsnip Jun 22 at 4:52
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You might do a search on Postnikov Towers, which is what Zach L is probably referring to when he talks about twisted products. –  Grumpy Parsnip Jun 22 at 5:18

2 Answers 2

up vote 5 down vote accepted

There are relatively easy-to-state obstructions coming from Serre classes. In short, a Serre class of abelian groups is a collection of abelian groups closed under various natural constructions; examples include

  • finite abelian groups,
  • finitely generated abelian groups,
  • finite abelian $p$-groups for a fixed prime $p$.

Serre proved the following.

If $X$ is simply connected, then for any $n$, all of the $H_k(X), k \le n$ lie in some fixed Serre class iff all of the $\pi_k(X), k \le n$ lie in the same Serre class.

This has various corollaries, among the most important being that the homotopy groups of spheres are finitely generated.

More generally, if you know the homotopy groups of your space and they aren't too complicated you can try to analyze the structure of all possible Postnikov towers your space could have. The theory of Postnikov towers tells you that at least in nice cases, and in particular if a space is simply connected, then the only additional data aside from the homotopy groups you need to uniquely determine its homotopy type (and in particular to uniquely determine its homology) is a sequence of cohomology classes called the $k$-invariants of its Postnikov tower.

For example, the next most complicated case after Eilenberg-MacLane spaces is when exactly two of the homotopy groups are nontrivial. If $X$ is a space with $\pi_n$ and $\pi_m$ nontrivial, where $n \ge 2$ and $m \ge n + 1$, then $X$ fits into a fibration sequence ($X$ is a "twisted product") of the form

$$K(\pi_m, m) \to X \to K(\pi_n, n)$$

where $X \to K(\pi_n, n)$ induces an isomorphism on $\pi_n$, and applying the Serre spectral sequence to this fibration puts strong restrictions on the possible homology groups of $X$; in particular if everything is finitely generated then you can bound the ranks of the homology groups of $X$.

But you can say more: isomorphism classes of such fibrations are completely classified by their $k$-invariants, which are cohomology classes in $H^{m+1}(K(\pi_n, n), \pi_m)$, so if you can compute this group you can try to write down all of the possibilities for $X$ and so all of the possibilities for the homology of $X$, or at the very least you can count them. In particular, if this group vanishes then $X \cong K(\pi_n, n) \times K(\pi_m, m)$ and so its homology is uniquely determined.

Example. Suppose $X$ is a space whose only nontrivial homotopy groups are $\pi_2 \cong \mathbb{Z}$ and $\pi_m$, where $m \ge 4$ is even. The classifying space $K(\mathbb{Z}, 2)$ is the infinite complex projective space $\mathbb{CP}^{\infty}$, and in particular has no cohomology in odd dimensions with any coefficients. Hence in this case $H^{m+1}(\mathbb{CP}^{\infty}, \pi_m)$ vanishes and the homology of $X$ is uniquely determined.

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This is really very enlightening (as usual). Thank you. –  Mike Miller Jun 22 at 6:27
    
@Mike: You're welcome! For further reading one possibility is May and Ponto's More Concise Algebraic Topology (math.uchicago.edu/~may/TEAK/KateBookFinal.pdf). –  Qiaochu Yuan Jun 22 at 6:40
    
Regarding bounding the ranks of homology given bounds on the ranks of homotopy there is, for example, the rational Hurewicz theorem: en.wikipedia.org/wiki/… I think I know what the correct bound is in general but I'll wait until I'm more confident about the proof. –  Qiaochu Yuan Jun 22 at 6:58
    
I hadn't expected that the mentioned book would be accessible to me, but according to the introduction apparently I've got the requisite background. Thanks for the reference! –  Mike Miller Jun 22 at 7:04
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Okay, so in addition to the rational Hurewicz theorem here is a general bound for ranks, if I'm not mistaken. Let $X$ be simply connected with homotopy (equivalently homology) finitely generated in each degree. Then $\dim H_k(X) \otimes \mathbb{Q}$ is at most the dimension of the degree-$k$ component of the symmetric algebra on the graded vector space $\pi_{\bullet}(X) \otimes \mathbb{Q}$, where "symmetric algebra" is understood in the graded sense (so the tensor product of the symmetric algebra on the even components and the exterior algebra on the odd components). This bound is realized... –  Qiaochu Yuan Jun 22 at 7:22

For starters, an Eilenburg-MacClane space is uniquely determined up to homotopy by its $\pi_n$'s, and so the only possible homology groups that could go with the sequence $G,0,0,\ldots$ of homotopy groups are the homology groups $H_n(K(G,n)), H_{n+1}(K(G,n)),\ldots$. I know that there's a notion of building up spaces by "twisted products" of Eilenburg-MacClane spaces, and so I would imagine that knowing the homotopy groups of a space puts severe restrictions on what its homology groups can be.

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Sorry for retracting the acceptance - I've been seduced. –  Mike Miller Jun 22 at 5:58

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