Timeline for Is there a map of spectra implementing the Thom isomorphism?
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 13, 2016 at 19:23 | answer | added | Peter May | timeline score: 4 | |
| Nov 23, 2010 at 15:27 | answer | added | Robert Bruner | timeline score: 20 | |
| Nov 21, 2010 at 21:19 | comment | added | Oscar Randal-Williams | @skupers: No, it is not a theorem. | |
| Nov 21, 2010 at 19:33 | vote | accept | skupers | ||
| Nov 21, 2010 at 19:33 | comment | added | skupers | @Oscar. Is it a theorem that there is no such construction or is it simply too hard to give one when our models of $K(\mathbb{Z},n)$ get more complicated as $n$ increases? | |
| Nov 21, 2010 at 10:52 | comment | added | Oscar Randal-Williams | If we have the same standard of what is meant by "explicit", then there is not an explicit construction of such an equivalence of spectra. This is because in particular it would give an explicit construction of the Thom class $X^\mu \to K(\mathbb{Z},n)$, for which there is none. | |
| Nov 21, 2010 at 10:50 | answer | added | Johannes Ebert | timeline score: 23 | |
| Nov 21, 2010 at 0:44 | comment | added | Eric Peterson | You'd probably be interested in (at least the introduction of) arxiv.org/abs/0810.4535 , along with its lengthy bibliography. | |
| Nov 20, 2010 at 23:43 | history | asked | skupers | CC BY-SA 2.5 |