15
$\begingroup$

What is known about spaces $X$ with the property that $K^*(\text{point})\to K^*(X)$ is an isomorphism?

The same question for $K$-homology $K_*(X)\to K_*(\text{point})$; I don't even know whether these conditions are equivalent.

Note that replacing $K$-theory with integral homology one gets very interesting (I think) class of spaces, studied in "Acyclic spaces" by E. Dror.

Improve this question
$\endgroup$
11
  • 5
    $\begingroup$ All acyclic spaces have trivial K-theory (it's a simple argument with the Atiyah-Hirzebruch spectral sequence), but I think you get more things $\endgroup$
    – Denis Nardin
    Commented Dec 30, 2018 at 7:34
  • 3
    $\begingroup$ I don’t know a characterization, but it’s bigger than you might expect: K(Z/p, 2) is such a space. $\endgroup$
    – Eric Peterson
    Commented Dec 30, 2018 at 11:28
  • 3
    $\begingroup$ @მამუკაჯიბლაძე A space is acyclic iff its suspension spectrum is contractible (in fact iff its second suspension is contractible), hence all (co)homology theory have trivial values on acyclic spaces $\endgroup$
    – Denis Nardin
    Commented Dec 30, 2018 at 11:39
  • 2
    $\begingroup$ @მამუკაჯიბლაძე you can either prove this directly (a good exercise using the bar spectral sequence), or note that it follows from a result of Ravenel-Wilson: their results (using the bar spectral sequence) give that $(KU^\wedge_p \wedge \Sigma^\infty K(\mathbf{Z}/p^i, n))^\wedge_p$ vanishes for n>1 and all i. Coupled with the arithmetic fracture square and the fact that $KU_\mathbf{Q} \wedge \Sigma^\infty K(\mathbf{Z}/p^i, n)$ is contractible (the rationalization of $K(\mathbf{Z}/p^i, n)$ is contractible), you can conclude $KU$-acyclicity for $K(\mathbf{Z}/p^i, n)$ with n>1 and all i. $\endgroup$
    – skd
    Commented Dec 31, 2018 at 0:19
  • 1
    $\begingroup$ Sure, just take any rational object. $\endgroup$
    – skd
    Commented Dec 31, 2018 at 18:58

1 Answer 1

Reset to default
13
$\begingroup$

I'll just give an answer for finite complexes $X$. The condition $\widetilde{K}^*(X)=0$ only depends on the suspension spectrum of $X$ so this is naturally regarded as a question in stable homotopy theory. The condition also implies that $\widetilde{H}^*(X;\mathbb{Q})=0$ and thus that $n.1_X=0$ as a stable map for some $n>0$. From this it follows that $X$ splits stably as a wedge of finitely many $p$-torsion finite spectra for different primes $p$, so we are really looking at a question in $p$-local stable homotopy theory. The condition $\widetilde{K}^*(X)=0$ then says that $X$ has chromatic type at least two. For the general theory of chromatic type of finite spectra you can read Ravenel's book "Nilpotence in stable homotopy theory". For type two, it is possible to be a little more explicit than for higher types. For example, Adams constructed a certain self-map of the mod $p$ Moore spectrum which induces an isomorphism in $K$-theory, so the cofibre of that map has type two.

Improve this answer
$\endgroup$
2
  • 3
    $\begingroup$ It might be worth mentioning the work about $v_n$-local unstable homotopy theory, e.g. arxiv.org/abs/1803.06325 $\endgroup$
    – Denis Nardin
    Commented Dec 30, 2018 at 11:44
  • 4
    $\begingroup$ Is there a non-acyclic closed manifold with vanishing reduced K-theory? For oriented, this is clearly not possible because of rational homology, but there might be a non-oriented example? $\endgroup$
    – Lennart Meier
    Commented Dec 30, 2018 at 14:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .