Timeline for Where can I find basic "computations" of equivariant stable homotopy groups?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 31, 2018 at 13:11 | comment | added | Mikhail Bondarko | Possibly this is a stupid question; sorry! I have to think more about this. | |
| Jul 31, 2018 at 10:11 | comment | added | Mikhail Bondarko | Thank you very much indeed! It turns out that the the Burnside rings is not quite what I want. I am rather interested in the direct sum of all $Mor_{SH_G} (\Sigma_G (G/H_1)_+, \Sigma_G (G/H_2)_+)$ for $H_i$ running through all subgroups of $G$. This is a ring (with a unit if $G$ is finite) with multiplication given by compositions of morphisms, and I wonder whether anybody has considered it previously. | |
| Jul 4, 2018 at 3:06 | comment | added | Peter May | @Mikhail Bondarko. This is discussed in Section V.2 of the reference in my answer. Just click! For finite $G$, Segal first noticed that $\pi_0(S)$ is the Burnside ring of $G$. For compact Lie groups, tom Dieck defined the Burnside ring of $G$ in such a way that the same conclusion holds. For finite $G$, much more is true. Running through $\pi_0$ of the suspension $G$-spectra of orbits $G/H$ (with disjoint basepoints), one obtains the Burnside ring Mackey functor, and more recent work shows that there are multiplicative norms so that one actually obtains the Burnside ring Tambara functor. | |
| Jun 20, 2018 at 8:32 | comment | added | Mikhail Bondarko | So, I have had a look at you book, and I have found references that give the vanishing in question. However, I believe that for $i=0$ the corresponding groups should be related to the Burnside ring of $G$; still I wasn't able to find any reference that gives this statement. I would be deeply grateful for any advice here! | |
| Aug 1, 2016 at 1:06 | comment | added | Peter May | Yeah, I'm sorry it is so lengthy, and pre tex. | |
| Jul 31, 2016 at 14:41 | comment | added | Mikhail Bondarko | Thank you very much! So, everything "works in the way I want it to"; I only have to read you book thoroughfully. | |
| Jul 31, 2016 at 14:39 | vote | accept | Mikhail Bondarko | ||
| Jul 30, 2016 at 23:28 | history | answered | Peter May | CC BY-SA 3.0 |