Does a finite suspension spectrum make a space finite? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T08:58:35Z http://mathoverflow.net/feeds/question/67268 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67268/does-a-finite-suspension-spectrum-make-a-space-finite Does a finite suspension spectrum make a space finite? Mike Shulman 2011-06-08T15:18:02Z 2011-06-08T16:23:12Z <p>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?</p> http://mathoverflow.net/questions/67268/does-a-finite-suspension-spectrum-make-a-space-finite/67274#67274 Answer by Fernando Muro for Does a finite suspension spectrum make a space finite? Fernando Muro 2011-06-08T16:05:21Z 2011-06-08T16:05:21Z <p>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).</p> http://mathoverflow.net/questions/67268/does-a-finite-suspension-spectrum-make-a-space-finite/67276#67276 Answer by Tom Goodwillie for Does a finite suspension spectrum make a space finite? Tom Goodwillie 2011-06-08T16:09:18Z 2011-06-08T16:23:12Z <p>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 $K_0\mathbb Z[\pi_1(X)]\to K_0\mathbb Z$.</p> <p>EDIT: </p> <p>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.</p> <p>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 $X$.</p>