Timeline for Extension of induced reps over Z: is it a sum of induced reps?
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Jan 10, 2010 at 20:23 | history | edited | Ben Webster♦ | CC BY-SA 2.5 |
added 29 characters in body
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| Jan 10, 2010 at 19:49 | comment | added | Leonid Positselski | For any finite group G, the first cohomology group H^1(G,Z) of the group G with constant integral coefficients Z vanishes, since there are no nonzero group homomorphisms G->Z. By the Shapiro Lemma, it follows that the first cohomology of G with coefficients in any integral permutational representation vanishes, too. Hence your argument is indeed correct and any extension of integral permutational representations splits. | |
| Jan 10, 2010 at 19:20 | vote | accept | Kevin Buzzard | ||
| Jan 10, 2010 at 19:20 | comment | added | Kevin Buzzard | Oh, but stop, the ext group clearly vanishes by Shapiro. So done! Thanks! | |
| Jan 10, 2010 at 19:12 | comment | added | Kevin Buzzard | I don't think I'm claiming that the ext group vanishes. I'm just saying that every element of the ext group gives rise to a sum-of-induceds representation, not that it's the sum of the two sum-of-induceds we're starting with. | |
| Jan 10, 2010 at 19:08 | history | answered | Ben Webster♦ | CC BY-SA 2.5 |