10
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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

1 Answer 1

10
$\begingroup$

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 $K_0(\mathbb{Z}[\pi])/K_0(\mathbb{Z})$.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks Ian, that's exactly what I was seeking. $\endgroup$
    – John Klein
    Commented Oct 6, 2022 at 2:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.