Timeline for Naive equivariant transfer
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Aug 2, 2017 at 20:51 | history | bounty ended | Omar Antolín-Camarena | ||
| S Aug 2, 2017 at 20:51 | history | notice removed | Omar Antolín-Camarena | ||
| Aug 2, 2017 at 1:11 | answer | added | Dylan Wilson | timeline score: 4 | |
| Jul 26, 2017 at 23:12 | comment | added | jdc | I see. Thank you. So it seems I should just include $RO(G)$-grading as a hypothesis. That said, just in case, all I truly want is the maximum generality under which a covering action of a finite cyclic group $C$ on a $G$-space $X$ (so the actions commute) induces an isomorphism $E^*_G (X/C) \to E^*_G (X)^C$. | |
| Jul 26, 2017 at 21:21 | comment | added | Dylan Wilson | So what I'm saying is: it's probably the case that what you're asking for is weaker than having an RO(G)-graded cohomology theory, but actual examples of non-RO(G)-graded-coh-theories which have these transfers would probably be artificially built and not naturally occurring. | |
| Jul 26, 2017 at 21:19 | comment | added | Dylan Wilson | an E_infty G-space and you want to give it a "G-H_infty" structure without requiring that it be "G-E_infty". This is a little different than the classical transfer story (which begins with just an arbitrary H_infty space and asks whether it's also E_infty) since you do have non-equivariant coherent multiplication. But I think I would still be a little surprised if it turned out that coherent addition + H_infty-style-transfers = coherent transfers. | |
| Jul 26, 2017 at 21:16 | comment | added | Dylan Wilson | If you ask for a structured enough version of these transfers, and G is finite, then I think this is equivalent to asking for an RO(G)-graded cohomology theory (basically by the equivariant version of Segal's machine.) If you drop down the structure, there might be counterexamples akin to the classical counterexamples of the "transfer conjecture" in nonequivariant homotopy theory. Basically, it's like measuring the difference between an H_infty-space and an E_infty space... there are examples of the former which aren't the latter, but they tend to be weird. In your case, it's like you have | |
| S Jul 26, 2017 at 20:01 | history | bounty started | Omar Antolín-Camarena | ||
| S Jul 26, 2017 at 20:01 | history | notice added | Omar Antolín-Camarena | Draw attention | |
| Jul 11, 2017 at 17:49 | history | edited | jdc | CC BY-SA 3.0 |
Asked a question that didn't assume the wrong result from the get-go.
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| Jul 9, 2017 at 20:56 | comment | added | jdc | @user51223, I believe it's the second meant. The filtration $F_p E^*(Y) = \ker(E^* Y \to E^* Y^{p-1})$ gets you $E_2^{p,q} = H^p(Y;E^q(*))$. In the Becker–Gottlieb proof, the additional assumption that $\chi(F)$ is invertible makes $p_! p^*\colon E^* Y \to E^* Y$ induce an automorphism of this $E_2$ page, so that $p_! p^*$ is an automorphism. | |
| Jul 9, 2017 at 6:21 | comment | added | user51223 | @OscarRandal-Williams Thank for this. I went back to Brumfiel and Madsen and Madsen-Tillmann to update my knowledge on this. That the equality holds up a filtration, I get it. But, I actually don't know the definition of the Atiyah-Hirzebruch filtration. Is this a filtration on the stable cohomotopy that you refer to or is it related to the fibre bundle that we use and just simply obtained by restriction over the finite subcomplexes of the base of the fibre bundle? | |
| Jul 8, 2017 at 19:51 | comment | added | Oscar Randal-Williams | @user51223: I don't mean equivariantly, the equation is false for the ordinary transfer. The push-pull formula is true for any multiplicative cohomology theory, but shows that $t \circ p$ is multiplication by $t(1) \in \pi^0(Y)$, which is why stable cohomotopy plays a distinguished role. Mapped to ordinary cohomology this is a scalar, namely $\chi(F)$, but that is not generally true. In a general cohomology theory it is $\chi(F)$ modulo higher Atiyah-Hirzebruch filtration, and in particular becomes a unit if $\chi(F)$ is inverted (at least if the base of the fibration is a finite complex). | |
| Jul 8, 2017 at 15:30 | comment | added | Omar Antolín-Camarena | Thanks, @OscarRandall-Williams, I've believed that false statement for a while now! (Luckily, I've never had to use it.) | |
| Jul 8, 2017 at 13:19 | comment | added | user51223 | @OscarRandal-Williams. I suppose you mean that in the equivariant case the equation needs higher filtration, right? Although, apart from Theorem 5.5, Becker and Gottlieb state their pull back-push forward formula for any orientable cohomology, which some useful theories do satisfy. The stable (co-) homotopy stable homotopy are very specific examples really! What confuses me is that in cohomology or homology, the multiplication by $2$ would imply a splitting away from $2$ taking place in a localised stable homotopy category; how this does match with the higher filtration you mention? | |
| Jul 8, 2017 at 9:05 | comment | added | Oscar Randal-Williams | @OmarAntolinCamarena: the equation is not true, see my answer at mathoverflow.net/a/270996/318 | |
| Jul 8, 2017 at 4:59 | comment | added | Omar Antolín-Camarena | @TomGoodwillie. Trivial $G$ is the non-equivariant case, right? I thought the transfer was implemented by a map $t : \Sigma^\infty_+ Y \to \Sigma^\infty_+ X$ satisfying $t \circ \Sigma^\infty_+ p = 2$. Doesn't that show the desired formula upon applying $Map(-,E)$ for an arbitrary spectrum $E$? | |
| Jul 8, 2017 at 3:51 | comment | added | jdc | Well, that would certainly answer my question! What's the example? | |
| Jul 7, 2017 at 21:24 | comment | added | Tom Goodwillie | I thought that the equation $p_!p^\ast=n$ was false already when $G$ is trivial and $n=2$ (for generalized cohomology theories). | |
| Jul 7, 2017 at 21:05 | comment | added | user51223 | I am not very familiar with the equivariant cohomology. But, couldn't this be derived from what is known about ordinary homology? For example, doesn't the Becker-Gottlieb together with Atiyah-Hizerbruch-Whitehead SS then gives the result for any cohomology theory? | |
| Jul 7, 2017 at 19:13 | history | asked | jdc | CC BY-SA 3.0 |