Timeline for A cohomology class associated with a complex representation of a group
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Feb 27, 2016 at 0:13 | comment | added | Mariano Suárez-Álvarez | (Here LBG is the free loop space on BG) The first Chern class of this descended bundle is an element of $H^2(LBG,\mathbb Z)$, which is a direct sum of cohomologies of centralizers. Is that class the same as the one described? That would be an answer to "what is c(V)?". | |
| Feb 27, 2016 at 0:13 | comment | added | Mariano Suárez-Álvarez | @Andreas, well, I know that the class is a bunch of homomorphisms (which are coherent in some way): that is how it is defined! What I am asking is what the class is, in the sense of is it something? For example, from V you can construct a vector bundle $\tilde V=EG\times_GV$ on $BG$, which you can pull back along the evaluation map $LBG\times S^1\to BG$ to get a vector bundle on that product which descends onto $LBG\times_{S^1}S^1$, which is $LBG$. | |
| Feb 26, 2016 at 14:47 | comment | added | Andreas Thom | @MarianoSuárez-Alvarez: Your question was "What is this cohomology class?". Maybe I do not understand what a possible answer would be, but didn't appear the class in $H^1$ naturally from the representation. What are you looking for? | |
| Feb 26, 2016 at 2:28 | comment | added | Ehud Meir | $\beta$ will be a cocycle here. An easy way to prove this is the following: in the twisted group algebra $\mathbb{C}^{\alpha}G$ we have the equality, if $g$ and $h$ commute in $G$, $U_h^{-1}U_gU_h = \beta(g,h)U_g$. Now it is quite easy to prove that the resulting map $C_g\rightarrow \mathbb{C}^{\times}$ is a homomorphism, and thus a one cocycle, since the action on $\mathbb{C}$ is trivial. | |
| Feb 25, 2016 at 21:52 | comment | added | Mariano Suárez-Álvarez | I honestly do not understand what «that is all» means in this context. | |
| Feb 25, 2016 at 21:47 | comment | added | Andreas Thom | The identification $H^2(C_g,\mathbb Z)=H^1(C_g,\mathbb C^{\times})$ makes it look more obscure than it is. For each $g$, the centralizer $C_g$ fixes the subspace spanned by non-fixed eigenvectors and hence there is a determinant homomorphism $C_g \to \mathbb C^{\times}$, i.e., a class in $H^1(C_g,\mathbb C^{\times})$, that's all. | |
| Feb 25, 2016 at 21:16 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 3.0 |
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| Feb 25, 2016 at 18:30 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 3.0 |
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| Feb 25, 2016 at 18:14 | comment | added | Mariano Suárez-Álvarez | @EhudMeir, indeed, my example with the determinant was an example of something else! :-| I'll correct it. if I start with with a $2$-cocycle $\alpha:G\times G\to\CC^\times$, fix $g\in G$ and let $\beta:h\in G_g\mapsto \alpha(g,h)/\alpha(h,g)\in\CC^\times$, I am only being able to prove that $d\beta=\beta^2\smile\beta^2$, no that it is a $1$-cocycle on $G_g$; is that construction treated somewhere? (I can do it if $\alpha:G\times G^{\ad}\to\CC^\times$ is a $1$-cocycle giving an element of $\Ext^1(\ZZ G^\ad,\CC^\times)$, though: is that what you meant?) | |
| Feb 25, 2016 at 16:17 | comment | added | Ehud Meir | Another question: if $[\alpha]\in H^2(G,\mathbb{C}^{\times})$, then we get another element in the same group $\oplus_{c\in CG}H^2(G_c,\mathbb{Z})\cong \oplus_{c\in CG}H^1(G_c,\mathbb{C}^{\times})$, namely if $h$ commutes with $g$ then we have the one cocycle $h\mapsto \alpha(g,h)/\alpha(h,g)$. Is it possible to see any sort of connection between elements arising from two cocycles and from representations? | |
| Feb 25, 2016 at 15:54 | comment | added | Ehud Meir | Shouldn't $\lambda(g,1_G)=1$? That is: for the identity element $e=1_G$ it holds that $d(e) = 0$ because the only eigenvalue of $e$ is 1. I guess that in this case the interpretation of $\wedge^0 V_e$ should be $\mathbb{C}$, and the induced map should be the identity? | |
| Feb 25, 2016 at 7:31 | history | edited | Denis Serre | CC BY-SA 3.0 |
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| Feb 25, 2016 at 7:21 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 3.0 |
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| Feb 25, 2016 at 6:37 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 3.0 |
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| Feb 25, 2016 at 6:21 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 3.0 |
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| Feb 25, 2016 at 6:06 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 3.0 |
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| Feb 25, 2016 at 5:58 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 3.0 |
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| Feb 25, 2016 at 5:50 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 3.0 |
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| Feb 25, 2016 at 5:43 | history | asked | Mariano Suárez-Álvarez | CC BY-SA 3.0 |