Timeline for Homotopy theoretic description of homotopy fixed points (and obstructions) for an action of group $G$ on a groupoid $X$
Current License: CC BY-SA 3.0
13 events
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| Sep 14, 2017 at 10:36 | comment | added | Saal Hardali | @YonatanHarpaz I've asked a followup question about this (oh-so-painful) topic: mathoverflow.net/questions/281125/… | |
| Sep 10, 2017 at 22:23 | vote | accept | Saal Hardali | ||
| Sep 10, 2017 at 22:09 | comment | added | Saal Hardali | @YonatanHarpaz I see. Thanks! (still it would be great to have a reference for this stuff if you know of one...) | |
| Sep 10, 2017 at 21:03 | comment | added | Yonatan Harpaz | -> This leads to the idea of breaking an action on an arbitrary space to its Postnikov pieces, which leads to the Bousfield-Kan spectral sequence. Of course the first Postnikov piece is always problematic when $\pi_1$ is not abelian. | |
| Sep 10, 2017 at 21:03 | comment | added | Yonatan Harpaz | -> You will then see that such affine linear actions are classified precisely by $H^2(G,A)$, that only the trivial one (i.e., the one which is just linear) has homotopy fixed points, and that the homotopy groups of the space of fixed points are the cohomologies $H^{1-n}(G,A)$. Then there is the observation that for abelian Eilenberg-Maclane objects the group of self equivalences of $BA$ is the same as the group of affine linear self equivalences. | |
| Sep 10, 2017 at 20:59 | comment | added | Yonatan Harpaz | It doesn't work like this. First the sequence $BA \to Aut_*(BA) \to Aut(BA)$ is not a principle fibration, and second this sequence doesn't lie over $BG$. If you want something in this kind of generality, the only thing I can suggest is what I wrote in the last edit. Take $A$ to be an $\mathbb{E}_{\infty}$-group with a fixed $G$-action and consider only affine-linear actions of $G$ on $BA$ whose linear part is the induced $\mathbb{E}_{\infty}$-action on $BA$. -> | |
| Sep 10, 2017 at 19:59 | comment | added | Saal Hardali | $H^0(\mathfrak{X},BA) \to H^0(\mathfrak{X},BAut_*(BA)) \to H^0(\mathfrak{X},BAut(BA))$ which for the case $\mathfrak{X} = \text{Spaces}_{/ BG}$ gives $" H^1(G,A) \to H^1(G,Aut_*(BA)) \to H^1(G, Aut(BA)) "$. Then you claim there's an $H^2$ which one can define which extends this exact sequence to the right? If there was then $H^2(G,A)$ would by this definition have to be the obstructions to existence of sections of our original $\mathcal{F}$ which determined an element in $H^1(G, Aut(BA))$. | |
| Sep 10, 2017 at 19:54 | comment | added | Saal Hardali | @YonatanHarpaz Is the following interpretation accurate? Lets work relatively in an (connected) $\infty$-topos $\mathfrak{X}$. Then the problem is basically obtaining an obstruction theory for determining when gerbes admit global sections (or just "points"). If $\mathcal{F}$ is any connected object of $\mathfrak{X}$ it must look locally like $BA$ for some $\infty$-group in $\mathfrak{X}$. But inside $\mathfrak{X}$ we can find a universal fibration $BA \to BAut_*(BA) \to BAut(BA)$. The long exact sequence of homotopy groups for this fibration only gets as far as... to be continued... | |
| Sep 9, 2017 at 11:51 | history | edited | Yonatan Harpaz | CC BY-SA 3.0 |
added 937 characters in body
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| Sep 9, 2017 at 9:14 | comment | added | Robert Furber | I would like to add that Giraud's definition of nonabelian $H^2$ (from "Cohomologie non abélienne") agrees with the usual one when $A$ is abelian. At the time I found out about this I couldn't work out if this was a mistake in the nLab or due to a difference in definitions. | |
| Sep 9, 2017 at 7:18 | comment | added | Yonatan Harpaz | Yes, it seems to be a different definition than the one I'm familiar with (and the one that appears, for example, in Galois descent or in the Bousfield Kan spectral sequence). nLab defines it as the set of homotopy classes of morphisms from $BG \to \mathbf{B}Aut(A) \simeq BAut(BA)$, while the $H^2$ I know fixes in advance the outer action and looks only at maps to $BAut(BA)$ which induce a given action (maybe this can even be takes as a definition, I'm not sure). | |
| Sep 9, 2017 at 4:41 | comment | added | Marc Hoyois | Your claim that nonabelian $H^2$ recovers the usual one when $A$ is abelian contradicts ncatlab.org/nlab/show/nonabelian+group+cohomology. Does the nLab have the wrong definition of nonabelian $H^2$? | |
| Sep 8, 2017 at 21:36 | history | answered | Yonatan Harpaz | CC BY-SA 3.0 |