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Here are some definitions:

A space is homotopy finite if it is homotopy equivalent to a finite CW complex. A space finitely dominated if it is a retract of a homotopy finite space.

A space $X$ is a Poincaré duality space of dimension $d$ if there exists a pair $$ ({\mathscr L},[X]) $$ consisting of a rank one local system $\mathscr L$ on $X$ (i.e., a local coefficient system on $X$ which is locally isomorphic to $\Bbb Z$) and $[X] \in H_d(X;{\mathscr L})$ is a twisted homology class such that the cap product homomorphism $$ \cap [X]: H^\ast(X;{\mathscr E}) \to H_{d-\ast}(X;{\mathscr E} \otimes {\mathscr L}) $$ is an isomorphism in all degrees, where $\mathscr E$ runs over all local systems on $X$.

My question is this:

Question: Are there finitely dominated Poincaré duality spaces which are not homotopy finite?

(Note: I am a bit embarrassed about not knowing the answer to this question.)

Remarks:

(1). If $X$ is a finitely dominated space with finitely presented fundamental group $\pi$, then Wall's finiteness obstruction $w(X) \in K_0({\Bbb Z}[\pi])$ is defined. It is known that $X$ is homotopy finite if and only if $w(X)$ lies in the summand ${\Bbb Z} \cong K_0({\Bbb Z}) \subset K_0({\Bbb Z}[\pi])$.

(2). When $X= B\pi$ where $\pi$ is a discrete, finitely presented group and $X$ satisfies Poincaré duality, then $\pi$ is called a Poincaré duality group. In this case it automatically follows that $X$ is finitely dominated and $\pi$ is torsion free. It remains an open question as to whether $X$ is homotopy finite. However, it is known by work of Ian Leary that $w(X)$ is always a $2$-torsion element. Consequently, if $K_0({\Bbb Z}[\pi])$ contains no 2-torsion, it follows that $X$ is homotopy finite.

(3). It has been conjectured that for any torsion free group $\pi$, the class group $K_0(\Bbb Z[\pi])$ is torsion-free. If this conjecture holds, then every finitely presented Poincaré duality group $\pi$ will have the property that $B\pi$ is homotopy finite.

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  • $\begingroup$ I don't know the answer, but here is a suggestion : if $Y\to X$ is a finite covering space and $Y$ is PD, then so is $X$. So one could try to look for a (simply-connected ?) PD space $Y$ with a finite group action that realizes some nontrivial class in $\widetilde K_0(\mathbb Z[\pi])$. Maybe some manifold with an exotic action ? $\endgroup$ Commented Oct 5, 2022 at 8:15

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Corollary 5.4.2 of Wall's article `Poincaré complexes I', Ann. Math. 86 (1967) 213-245 gives examples of 4-dimensional Poincaré complexes $X$ with fundamental group of prime order $p\geq 23$ for which the Wall finiteness obstruction $\chi(X)$ is non-zero.

Incidentally, Theorem 1.3 of the same article is very close to the result that you attributed to me, except that I put in extra hypotheses that imply that (in Wall's notation) $\sigma(X)^*=\sigma(X)$. The thing I thought of as new in my article was considering PD groups over other rings such as $\mathbb{Q}$.

I think that Wall's $\sigma(X)$ is your $w(X)$ and Wall's $\chi(X)$ is the image of your $w(X)$ in the quotient $K_0(\mathbb{Z}[\pi])/K_0(\mathbb{Z})$.

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    $\begingroup$ Thanks Ian, that's exactly what I was seeking. $\endgroup$
    – John Klein
    Commented Oct 6, 2022 at 2:22

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