6
$\begingroup$

Suppose that $X$ is a space whose suspension spectrum $\Sigma_+^\infty(X)$ is dualizable in the stable homotopy category. I believe this is equivalent to saying that $\Sigma_+^\infty(X)$ is (weakly) homotopy equivalent to a finite cell spectrum. What does this imply about $X$? In particular, does it imply that $X$ is weakly equivalent to a finite cell complex, as a space?

$\endgroup$

2 Answers 2

11
$\begingroup$

No. In the stable homotopy category a retract of a finite cell spectrum is again a finite cell spectrum, but in the weak homotopy category of spaces a retract of a finite cell complex is not necessarily a finite cell complex; there is an obstruction in the kernel of $K_0\mathbb Z[\pi_1(X)]\to K_0\mathbb Z$.

EDIT:

For a simply connected space, finite generation of the direct sum of its integral homology groups implies that it is equivalent to a finite complex. Thus a connected space must become finite after one suspension if its suspension spectrum is finite. The same then follows without assuming connected.

Therefore finiteness of $\Sigma^\infty X$ is equivalent to finiteness of $\Sigma X$, and (as shown by Fernando's answer) this is strictly weaker than finiteness of $X$.

$\endgroup$
9
$\begingroup$

Take an acyclic group, i.e. a group $G$ with trivial homology. Choose $G$ which is not finitely presented, so that its classifying space $X=BG=K(G,1)$ cannot be finite. Since $\Sigma X\simeq \star$ is contractible, $\Sigma^{\infty}_{+}(X)=S$ is the sphere spectrum (dualizable).

$\endgroup$

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.