Timeline for Which spaces have trivial K-theory?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 31, 2018 at 20:34 | comment | added | მამუკა ჯიბლაძე | @skd I see, thanks! | |
| Dec 31, 2018 at 19:42 | comment | added | skd | If a rationally trivial space has trivial p-completion for all p, then it is contractible. I'm a little worried that some connectivity assumption might be missing, but I'm fairly confident this is true in the simply-connected case. | |
| Dec 31, 2018 at 18:59 | comment | added | მამუკა ჯიბლაძე | @skd Well sorry I should then also add rationally trivial, since this is what happens in your situation. | |
| Dec 31, 2018 at 18:58 | comment | added | skd | Sure, just take any rational object. | |
| Dec 31, 2018 at 17:13 | vote | accept | მამუკა ჯიბლაძე | ||
| Dec 31, 2018 at 12:42 | comment | added | მამუკა ჯიბლაძე | @skd Thanks a lot, surely I would never figure this out! One question - in principle something might have trivial $p$-completion but nontrivial $p$-localization, no? Or is this irrelevant here? | |
| Dec 31, 2018 at 0:19 | comment | added | skd | @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. | |
| Dec 30, 2018 at 20:27 | comment | added | მამუკა ჯიბლაძე | @EricPeterson Sorry I don't have enough skills to compute this. Can you explain why? | |
| Dec 30, 2018 at 11:39 | comment | added | Denis Nardin | @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 | |
| Dec 30, 2018 at 11:28 | comment | added | Eric Peterson | I don’t know a characterization, but it’s bigger than you might expect: K(Z/p, 2) is such a space. | |
| Dec 30, 2018 at 10:13 | answer | added | Neil Strickland | timeline score: 13 | |
| Dec 30, 2018 at 8:59 | comment | added | მამუკა ჯიბლაძე | @DenisNardin I would rather think that all K-trivial spaces would be acyclic, while K-theory could distinguish more spaces than "ordinary" homology. I just don't know enough, but I think the Adams $e$-invariant detects K-theory classes killed by the Chern character, cannot one find something along these lines? | |
| Dec 30, 2018 at 7:34 | comment | added | Denis Nardin | 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 | |
| Dec 30, 2018 at 6:41 | history | asked | მამუკა ჯიბლაძე | CC BY-SA 4.0 |