Timeline for Extensions of discrete groups by spectra
Current License: CC BY-SA 3.0
9 events
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| Jan 18, 2013 at 13:54 | comment | added | John Klein | @Greg: here's another way to see this classically. Let $A$ be a $G$-module. Then an extension gives a map $BG \to B\text{Aut}(BA)$, where $\text{Aut}(BA)$ is the monoid of unbased self-equivalences of $BA$. A $G$-module structure on $A$ is really a map $BG \to B\text{Aut}(A)$, where $\text{Aut}(A)=$ automorphisms of the abelian group. Applying $H_1$ to an automorphism gives a map $B\text{Aut}(BA) \to B\text{Aut}(A)$. I guess I'm looking for a version of this map in spectra: the hofiber of this map is identified with $B^2A$. By obstruction theory the lifts are classified by $H^2(G;A)$. | |
| Jan 18, 2013 at 13:41 | comment | added | John Klein | Greg, I'm confused by your answer. I think a bundle over $BG$ with fibers $A$ is really equivalent to specifying a $G$-action on the spectrum $A$ (i.e., the homotopy category of $G$-spectra is equivalent to the homotopy category of fiberwise spectra over $BG$). An "extension," whatever that might be, should have more structure than that: if we fix the given $G$-action on $A$, the extensions that induce the given $G$-action (up to some notion of equivalence) should be in bijection with $H^2(G;A)$, as Tyler points out. | |
| Jan 18, 2013 at 12:19 | comment | added | Gregory Arone |
John, it might help if you try to indicate what kind of answer you are hoping for. My first thought is something along the lines of a fiber bundle over $BG$ with fibers $A$'' (a bundle of spectra). But it is a variation of the same short and lazy'' answer.
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| Jan 18, 2013 at 11:47 | comment | added | John Klein | Yes, I was aware of the short and lazy answer---but it's not very satisfying. | |
| Jan 18, 2013 at 6:00 | comment | added | Tyler Lawson | A possible short + lazy answer by turning a classification into a definition: An extension is classified by an element of $H^2(G,A)$. You can give $K(A,2)$ a $G$-action, form the associated bundle $K(A,2) \to Y \to BG$, and then this $H^2$ can be identified with the set of space of sections of $Y \to BG$. Now replace $K(A,2)$ with $B^2 X$ for $X$ an infinite loop space. (A more modern thinker than me might say something about extensions of the $\infty$-groupoid $BG$ by a symmetric monoidal $\infty$-groupoid.) | |
| Jan 18, 2013 at 3:41 | history | edited | John Klein | CC BY-SA 3.0 |
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| Jan 18, 2013 at 2:45 | history | edited | John Klein | CC BY-SA 3.0 |
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| Jan 17, 2013 at 22:21 | history | edited | John Klein | CC BY-SA 3.0 |
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| Jan 17, 2013 at 22:05 | history | asked | John Klein | CC BY-SA 3.0 |