Timeline for Lyndon–Hochschild–Serre spectral sequence for a non-normal subgroup
Current License: CC BY-SA 4.0
18 events
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| Feb 16, 2022 at 16:09 | history | edited | YCor | CC BY-SA 4.0 |
fixed English, added top-level tag
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| Feb 16, 2022 at 16:05 | answer | added | Joshua Mundinger | timeline score: 1 | |
| Jan 5, 2018 at 9:15 | answer | added | Mark Grant | timeline score: 8 | |
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Feb 1, 2015 at 22:25 | comment | added | Tyler Lawson | This works when analyzing $(X \times EG)/G \to X/G$ for any contractible $X$. For example, when you have a group $G$ that acts on a tree $T$ sufficiently nicely, you get a method for computing the cohomology of $G$ in terms of some kind of sheaf cohomology of $G/T$ where the stalks are the cohomology of the stabilizers. | |
| Feb 1, 2015 at 22:17 | comment | added | Tyler Lawson | @QiaochuYuan I don't think so. In that case, the fiber is the discrete set $G/H$ and so the Serre spectral sequence degenerates. I believe that it may be the Leray spectral sequence associated to the map $(E{\cal F} \times EG)/G \to E{\cal F}/G$ where $E{\cal F}$ is a certain contractible $G$-space (the classifying space for families that I mentioned earlier). | |
| Feb 1, 2015 at 22:09 | comment | added | Tyler Lawson | @quinque You are thinking of the boundary operator on inhomogeneous chains, e.g. $d g_0(g_1,g_2) = g_0 g_1(g_2) - g_0(g_1g_2) + g_0(g_1)$. I am talking about the boundary operator on homogeneous chains: $d[g_0,\ldots,g_n] = \sum (-1)^i [g_0,\ldots,\widehat{g_i},\ldots,g_n]$, where $G$ acts diagonally. | |
| Feb 1, 2015 at 19:50 | comment | added | Qiaochu Yuan | @Tyler: in more topological language, is this equivalent to taking the Serre spectral sequence associated to the fiber sequence $F \to BH \to BG$, where $F$ is the homotopy fiber of the natural map $BH \to BG$? | |
| Feb 1, 2015 at 19:48 | comment | added | quinque | @Tyler Lawson , I do not belive in complex $ \mathbb{Z} [ (G/H)^k ] $ . You have to have group structure on $G/H$ to explain what is "the same boundary operator". | |
| Feb 1, 2015 at 17:30 | comment | added | Tyler Lawson | (In answer to your explicit question, no, I do not know an immediate reference for this. I do know that this technique, from the topological point of view, appears in calculations of homological stability (I think it's used in Quillen's calculations for number fields). The chain complex I just described is the simplicial chain complex of a "classifying space for the family of subgroups of $H$".) | |
| Feb 1, 2015 at 17:27 | comment | added | Tyler Lawson | From the point of view of filtered complexes, this is a little easier to do because there is a nonprojective resolution of $\Bbb Z$ by a complex with terms $\Bbb Z[(G/H)^k]$, with the same boundary operator as on homogeneous chains. You can apply $\Bbb RHom_G(-,M)$ to this resolution and get a filtered complex, and this gives you the spectral sequence too. | |
| Feb 1, 2015 at 17:24 | comment | added | Tyler Lawson | The shortest way to say it is the following. For a group $G$, the category of $G$-modules is equivalent to the category of quasicoherent (etale) sheaves on the classifying stack $BG$, and the global section functor is the fixed-point functor. There is a faithfully flat cover $BH \to BG$, and so there is a Cech-to-derived / descent spectral sequence. But to compute effectively with it, you need to know the iterated fiber products of $BH$ over $BG$, which correspond to $G$-orbits in $(G/H)^k$. | |
| Feb 1, 2015 at 7:56 | comment | added | quinque | It is very interesting. Can you provide references? | |
| Feb 1, 2015 at 7:47 | history | edited | quinque | CC BY-SA 3.0 |
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| Feb 1, 2015 at 1:22 | comment | added | Tyler Lawson | The main difficulty is that, if $H < G$, there is not really a functor that will take the $H$-fixed points and produce the $G$-fixed points unless you include extra data. There are methods to get the cohomology of $G$, but almost all of them will require as extra input the cohomology of intersections of conjugates of $H$. | |
| Jan 31, 2015 at 21:46 | history | edited | quinque | CC BY-SA 3.0 |
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| Jan 31, 2015 at 21:41 | answer | added | Qiaochu Yuan | timeline score: 4 | |
| Jan 31, 2015 at 21:10 | history | asked | quinque | CC BY-SA 3.0 |