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I am looking for an explicit example, if one exists, of a (pointed) finite connected CW-complex $X$ such that some homology group with local coefficients $H_n(X,{\mathbb Z}[\pi_1 X])$ is not a finitely generated ${\mathbb Z}[\pi_1 X]$-module.

Such an example would in particular give a finitely presented group $\pi$, and a chain complex of finitely generated free ${\mathbb Z}[\pi]$-modules whose homology groups are not all finitely generated over ${\mathbb Z}[\pi]$. Suggestions on finding an explicit example of such a chain complex (of length 3, without loss of generality) are also welcome.

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Something very similar was asked and answered in… – Andreas Thom Jul 11 '12 at 8:27
up vote 10 down vote accepted

As Ricardo points out in the comments, there's an error in my sketched calculation below. I also didn't notice the requirement that $X$ should be finite, so the natural $BK$ fails on two counts! However, it seems possible that a presentation complex for $K$ would do the job. Stallings shows that $\pi_2$ of any complex with $\pi_1=K$ is infinitely generated as a $K$-module.


I think you want to start with a famous example of Stallings, from the paper 'A finitely presented group whose 3-dimensional integral homology is not finitely generated'. Stallings constructs a finitely presented group $K$ with the property that `there is no projective resolution of $\mathbf{Z}$ over $\mathbf{Z}[K]$ which is finitely generated in dimension 3' (Corollary 1).

In fact, as observed by Bieri, $K$ can be realizes as an explicit subgroup of the direct product of three free groups, $G=F_2\times F_2\times F_2$: $K$ is the kernel of a map $G\to\mathbf{Z}$ that sends every generator to $1$. So $K$ defines an `explicit' 3-complex, namely a covering space of the natural $K(G,1)$. As $K$ is 3-dimensional and finitely presented, it follows from Corollary 1 that $H_3(K,\mathbf{Z}[K])$ is infinitely generated.

Stallings's paper was the starting point for many beautiful constructions. Highlights include the word of Bieri and Bestvina--Brady .

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Thanks for the suggestion. Nevertheless, and I hope I am not doing something completely silly, doesnt't $H_3(K,{\mathbb Z}[K])=H_3(BK,{\mathbb Z}[K])=H_3(\widetilde{BK},{\mathbb Z})=H_3(pt,{\mathbb Z})=0$ hold? – Ricardo Andrade Jul 10 '12 at 20:50
It's certainly possible that I've made a mistake---homological algebra isn't my strong point! But I don't understand your argument, as it doesn't seem to use anything about the group $K$. Where are you getting $H_3(BK,\mathbb{Z}[K])=H_3(\widetilde{BK},\mathbb{Z})$ from? – HJRW Jul 10 '12 at 21:27
@HW: It is just a property of homology with local coefficients: $H_n(X,{\mathbb Z}[\pi_1 X])=H_n(\tilde{X},{\mathbb Z})$ for any path connected space $X$ with a universal cover $\tilde{X}$. See, for example, the section on local coefficients in Hatcher's book (specifically, example 3H.2 on page 329). – Ricardo Andrade Jul 10 '12 at 21:51
Ricardo - you're right. When tensoring $C(\widetilde{X})$ with $\mathbb{Z}[K]$, I got $C(X)$ instead of $C(\widetilde{X})$. Your argument shows that no aspherical space is going to work. So it turns out that the theory of group homology with local coefficients is rather boring! I didn't know that. – HJRW Jul 11 '12 at 9:37
@HW: Thanks for your edit. I had not actually looked at Stallings' paper, and so had not noticed his corollary 2. I will restate it here for future convenience: if $X$ is any finite complex with $\pi_1 X=K$ ($K$ is the group Stallings constructs), then $\pi_2 X$ is infinitely generated as a module over ${\mathbb Z}[\pi_1 X] = {\mathbb Z}[K]$. Now we simply note that $\pi_2 X=\pi_2(\tilde{X})=H_2(\tilde{X})=H_2(X,{\mathbb Z}[\pi_1 X])$ using the Hurewicz theorem for $\tilde{X}$. So, as HW points out in his edit, any presentation complex for a finite presentation of $K$ will indeed do the job. – Ricardo Andrade Jul 12 '12 at 7:42

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