Timeline for Non-trivial example of $H^2(G,M)$ where $M$ is a non-trivial G-representation
Current License: CC BY-SA 4.0
24 events
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| Dec 18, 2020 at 16:45 | vote | accept | A_Physicist. | ||
| Dec 18, 2020 at 9:20 | comment | added | Achim Krause | Maybe for additional context: I think $H^*(G;M) = H^*(\widetilde{G};\mathbb{F}_2)$, where $\widetilde{G}$ denotes the kernel of your homomorphism $G\to C_2$, and under that identification, your homomorphism $H^2(\widetilde{G};\mathbb{F}_2)\to H^2(G;\mathbb{F}_2)$ is the transfer map. | |
| Dec 18, 2020 at 6:45 | comment | added | A_Physicist. | I streamlined the question and checked another example. | |
| Dec 18, 2020 at 6:35 | history | edited | A_Physicist. | CC BY-SA 4.0 |
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| Dec 17, 2020 at 20:33 | comment | added | Qfwfq | (Well, now that I think if it was crystallography it'd be $H^2(G,\mathbb{Z}^n)$, not $H^2(G,\mathbb{Z}/2)$) | |
| Dec 17, 2020 at 20:32 | comment | added | Qfwfq | Just out of curiosity (assuming you're a physicist as your nickname suggests): why is a physicist interested in group cohomology of abstract groups? Is it crystallography? Or discrete symmetries like "parity" and stuff like that? :) | |
| Dec 17, 2020 at 19:27 | answer | added | Derek Holt | timeline score: 3 | |
| Dec 17, 2020 at 18:54 | history | edited | A_Physicist. | CC BY-SA 4.0 |
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| Dec 17, 2020 at 18:06 | comment | added | Benjamin Steinberg | I will delete my silly comments. | |
| Dec 17, 2020 at 18:06 | comment | added | Benjamin Steinberg | @DerekHolt, sorry I had it backward in my head. I shouldn't write comments late at night | |
| Dec 17, 2020 at 17:35 | history | edited | A_Physicist. | CC BY-SA 4.0 |
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| Dec 17, 2020 at 17:33 | comment | added | A_Physicist. | @DerekHolt Great point. I changed $H^2$'s to $Z^2$'s. | |
| Dec 17, 2020 at 17:32 | history | edited | A_Physicist. | CC BY-SA 4.0 |
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| Dec 17, 2020 at 17:29 | comment | added | Derek Holt | I don't believe that it is true in general that $H^2(G,Z/2)$ is (isomorphic to) a subgroup of $H^2(G,M)$. If $G$ is itself cyclic of order $2$, then $|H^2(G,Z/2)| = 2$, but $H^2(G,M)$ is trivial. | |
| Dec 17, 2020 at 15:46 | history | edited | YCor | CC BY-SA 4.0 |
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| Dec 17, 2020 at 15:35 | history | edited | A_Physicist. | CC BY-SA 4.0 |
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| Dec 16, 2020 at 21:10 | history | edited | A_Physicist. | CC BY-SA 4.0 |
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| Dec 16, 2020 at 8:36 | comment | added | Daniel Donnelly | @A_Physicist. yep that works :) | |
| Dec 16, 2020 at 6:15 | history | edited | A_Physicist. | CC BY-SA 4.0 |
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| Dec 16, 2020 at 6:03 | comment | added | A_Physicist. | @StudySmarterNotHarder. Thanks. Better? | |
| Dec 16, 2020 at 5:27 | comment | added | Daniel Donnelly | I just want to say please be more creative with your username... :| I upvoted. :D | |
| Dec 16, 2020 at 4:05 | history | edited | A_Physicist. | CC BY-SA 4.0 |
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| Dec 16, 2020 at 4:04 | review | Close votes | |||
| Dec 18, 2020 at 4:48 | |||||
| Dec 16, 2020 at 3:22 | history | asked | A_Physicist. | CC BY-SA 4.0 |