ambidexterity?

Current License: CC BY-SA 4.0

15 events

when toggle format	what		by	license	comment
Jul 25, 2018 at 0:22	vote	accept	Tim Campion
Jul 24, 2018 at 16:35	answer	added	Dylan Wilson		timeline score: 4
Jul 24, 2018 at 16:01	comment	added	Tim Campion		For the record, we did move the discussion to chat. Link. Anybody should feel free to join the discussion there. I get the feeling that neither Dylan nor Denis is likely to write up an answer -- if nobody does, I'll probably try to write a CW response summarizing what I've learned.
Jul 24, 2018 at 15:13	comment	added	Denis Nardin		@TimCampion Contrary to the case of pointed $G$-spaces, the smash product in G-spectra is not computed fixed point-wise (i.e. $(E\wedge F)^G\neq E^G\wedge F^G$), so your proof doesn't work in that case If you want to discuss more these things may I suggest we move to chat? This comment chain is getting unwieldy.
Jul 24, 2018 at 14:46	comment	added	Tim Campion		@DenisNardin Thanks! I'm confused, though. If $E$ is just a pointed $G$-space, then $EG_+ \wedge E = (EG \times E) / (EG \times\{\ast\})$, and since $EG$ has a free $G$-action, $EG \times E$ has a free $G$-action. So I really don't see any fixed points other than the basepoint, unless things work very differently stably...
Jul 24, 2018 at 9:07	comment	added	Denis Nardin		@TimCampion No, $(EG_+\wedge E)^G=E_{hG}$ (it's a version of the Adams isomorphism). For your second question, you can find the most general version of that statement in this paper
Jul 23, 2018 at 21:14	comment	added	Tim Campion		By the way, in your last comment it sounds like you're saying that $(EG_+ \wedge E)^G = E_{hG}$ and $Fun(EG_+,E)^G = E^{hG}$ but only the second actually holds, right? It seems to me that $(EG_+ \wedge E)^G = 0$, it's $(EG_+ \wedge E)_G = E_{hG}$... Also I think I'm dimly remembering being told this before -- there's some range of groups / families where the data of a genuine $G$-spectrum is equivalent to supplying some factorizations of the Tate norms for each subgroup in the family, right?
Jul 23, 2018 at 20:57	comment	added	Tim Campion		@DylanWilson Thanks for sticking with me, I think I understand the construction now: Taking orbits of the transfer map $E \to E^G$, we obtain $E_{hG} \to E^G \wedge BG_+$. The map $BG^+ \to S^0$ then gives us $E^G \wedge BG_+ \to E^G$. Composing, we get $E_{hG} \to E^G$. Then we compose this with the inclusion of fixed points $E^G \to E^{hG}$ to get a map $E_{hG} \to E^{hG}$. Fantastic! Now why does this composite map agree with the norm map in the sense of (2)? (By the way, an explanation or a reference for this fact would make a great answer to the original question!)
Jul 23, 2018 at 20:21	comment	added	Dylan Wilson		the second map are indeed the homotopy fixed points and homotopy orbits, which can be done in various ways.
Jul 23, 2018 at 20:17	comment	added	Dylan Wilson		To produce a map from $X$ to a homotopy limit like $E^{hG}$ I need to produce a natural transformation from the constant diagram on $X$ to the BG-indexed diagram E, and I have such a thing as part of the data of a genuine G-spectrum, where X is $E^G$. If it's easier, you can work internally to the world of genuine G-spectra: there's a map $EG_+ \to S^0$ of $G$-spectra which induces natural transformations $EG_+ \wedge E \to E$ and $E \to F(EG_+, S^0)$. Now take genuine G-fixed points of both sides. You must show that the genuine fixed points of the source of the first map and target of
Jul 23, 2018 at 18:19	comment	added	Tim Campion		@DylanWilson I'm still not seeing it. Let $e$ be the trivial group. In (1), I have transfer and restriction maps between $E = E^e$ and $E^G$. When I take homotopy orbits I get maps between $E_{hG}$ and $E^G\wedge BG_+$, and when I take homotopy fixed points I get maps between $E^{hG}$ and $Fun(BG_+,E^G)$. So I think I'm missing something...
Jul 23, 2018 at 17:35	comment	added	Dylan Wilson		Take H to be trivial in your (1), but now note that the genuine restriction and transfer maps are equivariant for the trivial action on the fixed points, so they factor as indicated.
Jul 23, 2018 at 16:54	comment	added	Tim Campion		@DylanWilson Thanks! From what you said, I don't quite understand how the genuine equivariant transfer relates to the factorization $E_{hG} \to E^G \to E^{hG}$?
Jul 23, 2018 at 16:40	comment	added	Dylan Wilson		The transfer that appears for a genuine G-spectrum is a refinement of the Borel trace/norm in the sense that there is a factorization $E_{hG} \to E^G \to E^{hG}$. In the multiplicative case you also have something similar, but not every Borel G-spectrum comes with a multiplicative norm for free (unlike the additive one), and you'd want to replace the homotopy orbits by a multiplicative variant which I dunno a good name for in the non-fully-commutative case. (In the G-E_infty case, you'd like to take homotopy orbits in the category of E_infty rings, for example)
Jul 23, 2018 at 13:40	history	asked	Tim Campion	CC BY-SA 4.0