Timeline for Genuine equivariant ambidexterity
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 17, 2016 at 2:24 | answer | added | Akhil Mathew | timeline score: 8 | |
| Jan 16, 2016 at 19:48 | vote | accept | Yonatan Harpaz | ||
| Jan 16, 2016 at 19:18 | answer | added | Jacob Lurie | timeline score: 8 | |
| Jan 16, 2016 at 19:05 | comment | added | Yonatan Harpaz | Of course, I realize that the definition itself is not symmetric, but I guess I thought about it as a technical thing, and that essentially, genuine equivariant is supposed to be some kind of a rigid action, to which you could in principle apply any construction you have for group actions. | |
| Jan 16, 2016 at 18:50 | comment | added | Yonatan Harpaz | @Jacob, oh, I missed this surprising point. How come the symmetry between quotient and fixed points is broken (maybe there isn't a good answer)? | |
| Jan 16, 2016 at 18:37 | comment | added | Jacob Lurie | @Yonatan Not to my knowledge. I believe that a $G$-spectrum (for $G$ cyclic of prime order) is determined by the data described above. | |
| Jan 16, 2016 at 18:27 | comment | added | Yonatan Harpaz | @Jacob, Isn't there also a genuine quotient space, through which the map $f$ factors? | |
| Jan 16, 2016 at 17:33 | comment | added | Jacob Lurie | @Yonatan I'm not sure what you mean by the "genuine norm map", unless you mean the map $f$. | |
| Jan 16, 2016 at 17:09 | comment | added | Yonatan Harpaz | In the setting you describe I guess my question would be if the genuine norm map from the genuine quotient to the genuine fixed points is a $K(n)$-local equivalence (something like in the Adams equivalence above). But I've been starting to think that this question is misguided, and that it should be whether there is some version of ambidexterity where finite groups are replaced by compact lie groups (but the action is still the "naive" one). | |
| Jan 16, 2016 at 16:32 | comment | added | Jacob Lurie | Let $G$ be cyclic of order $p$. To specify a genuine $G$-spectrum $X$, you need to specify a spectrum $Y$ with a "naive" $G$-action together with a factorization of the norm map of $Y$ as a composition $f: Y_{hG} \rightarrow Z$ with $g: Z \rightarrow Y^{hG}$; here $Z$ is the genuine fixed point spectrum. If $Y$ is $K(n)$-local, then the norm map $g \circ f$ exhibits $Y^{hG}$ as the $K(n)$-localization of $Y_{hG}$. I'm not sure if I understand the question: is it "if $Z$ is also $K(n)$-local, then does $f$ exhibit $Z$ as the $K(n)$-localization of $Y_{hG}$"? If so, the answer is no. | |
| Jan 15, 2016 at 18:58 | comment | added | Yonatan Harpaz | I did look at that paper before posting the question above, but I got the impression that it's not what I wanted because Rognes is not working in the genuine equivariant setting but rather in the ordinary equivariant one. However, looking at it now I'm starting to think that maybe my focus on genuine equivariance was somewhat misplaced. | |
| Jan 15, 2016 at 14:44 | comment | added | Sean Tilson | Have you looked at Stably Dualizable Groups by Rognes? | |
| Jan 15, 2016 at 9:28 | history | asked | Yonatan Harpaz | CC BY-SA 3.0 |