Timeline for symmetric 2-cocycle / many projective representations
Current License: CC BY-SA 3.0
18 events
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| S Feb 19, 2015 at 10:19 | history | bounty ended | CommunityBot | ||
| S Feb 19, 2015 at 10:19 | history | notice removed | CommunityBot | ||
| Feb 14, 2015 at 20:03 | history | edited | Simon Lentner | CC BY-SA 3.0 |
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| Feb 14, 2015 at 20:00 | vote | accept | Simon Lentner | ||
| Feb 14, 2015 at 14:27 | history | edited | Simon Lentner | CC BY-SA 3.0 |
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| Feb 13, 2015 at 10:32 | answer | added | Ofir Schnabel | timeline score: 3 | |
| S Feb 11, 2015 at 9:12 | history | bounty started | Simon Lentner | ||
| S Feb 11, 2015 at 9:12 | history | notice added | Simon Lentner | Draw attention | |
| Feb 6, 2015 at 19:57 | history | edited | Simon Lentner | CC BY-SA 3.0 |
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| Feb 6, 2015 at 18:09 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
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| Feb 5, 2015 at 2:33 | comment | added | S. Carnahan♦ | user63850 wanted to post the following comment. @SimonLentner Thanks, you're right. In the language of extensions the statement that worried me is: Every extension $\mathbb{C}^\times \to E \to G$ of abelian groups splits. That's true, because $\mathbb{C}^\times$ is a divisible (and hence injective) $\mathbb{Z}$-module. | |
| Feb 5, 2015 at 0:09 | comment | added | Simon Lentner | I understand you wonder about the 2-cocycle without extension interpretation: In case $C_2\times C_2=\langle g,h\rangle$ there is indeed a unique nontrivial cohomology class. Say the 2-cocycle $\sigma(g,g)=\sigma(h,h)=1, \sigma(g,h)=1,\sigma(h,g)=-1$; any other $\sigma$ in this class will have $\sigma(g,h)\sigma(h,g)=-1$, so all are nonsymmetric. | |
| Feb 4, 2015 at 23:58 | comment | added | Geoff Robinson | Well, I think the point is (at least the way I look at it) that a $2$-cocycle for a finite group can be normalized to take values which are finite order roots of unity ( basically, each finite group has a finite Schur multiplier). I agree that the Klein 4-group has a non-split central extension (by all of $\mathbb{C}^{\times}$ if you like), that's what I said above. | |
| Feb 4, 2015 at 23:47 | comment | added | user63850 | @Geoff: The question is concerned with central extensions $\mathbb{C}^\times \to E \to G$. So $E$ quaternion or dihedral doesn't fit here. What worries me is: The OP said there is no nontrivial symmetric 2-cocycle for $G$ abelian. To my understanding such a symmetric 2-cocycle corresponds to an extension with $E$ abelian. And my comment shows that there is such an abelian extension for $G=C_2 \times C_2$ (and many other groups) that dosn't split. So I wonder if this statement from the OP is really true. | |
| Feb 4, 2015 at 23:01 | comment | added | Geoff Robinson | I'm not sure what worries you. The Klein $4$-group certainly has proper central extensions ( either quaternion of order $8$ or dihedral of order $8$), but in either case, there is only one irreducible representation for the non-trivial associated twisted group algebra, compared to $4$ irreducible representations for the untwisted group algebra, | |
| Feb 4, 2015 at 22:41 | comment | added | user63850 | This is a comment (haven't enough reps to comment directly). Why are there no nonrivial symmetric 2-cocycles for abelian $G$ ? Note that for $G$ abelian, each 2-cocyle is symmetric (trivial coefficients!). Then, since $H^2(G,\mathbb{C}^\times)=H^2(G,S^1)= H^3(G,\mathbb{Z})$, there is, for instance, a nontrivial symmetric 2-cocycle for $G=C_2 \times C_2$, Or am I missing something ? | |
| Feb 4, 2015 at 12:03 | answer | added | Geoff Robinson | timeline score: 3 | |
| Feb 4, 2015 at 11:08 | history | asked | Simon Lentner | CC BY-SA 3.0 |