TITLE: finite complex with non-finitely generated homology with local coefficients
QUESTION [10 upvotes]: 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.

REPLY [10 votes]: 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.
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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 .