Timeline for A group 3-cocycle, trivial on a pair of generating subgroups?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| May 17, 2013 at 7:57 | comment | added | Ralph | Yes: Let $H$ resp. $K$ be any finite group those integral cohomology has a non-zero class x resp. y of degree 2. Then $xy\in H^4(H\times K;\mathbb{Z})=H^3(H\times K;\mathbb{C}^\times)$ is non-zero and restricts to zero on $H$ and $K$. If you take $H=K=\mathbb{Z}/p\oplus \mathbb{Z}/p$ then you get an example with $H^2(H;\mathbb{C}^\times)\neq 0$. | |
| May 17, 2013 at 2:34 | vote | accept | Kim Morrison | ||
| May 17, 2013 at 2:34 | comment | added | Kim Morrison | Ah. Do you know an example with $\mathbb k = \mathbb C$? | |
| May 16, 2013 at 22:06 | comment | added | Ralph | No, I'm afraid, but it doesn't: For $\mathbb{k}=\mathbb{C}$ the question is equivalent to finding $\omega \in H^4(G;\mathbb{Z})\cong H^3(G;\mathbb{C}^\times)$ that restricts to zero. Since the cohomology of $Q_8$ is 4-periodic, $H^4(Q_8;\mathbb{Z})$ is generated by a cohomology class z (of order 8) that restricts non-zero on both, H and K. | |
| May 16, 2013 at 15:57 | comment | added | Kim Morrison | Thanks Ralph, that's perfect. The same groups also work for $\mathbb k = \mathbb C$, right? | |
| May 16, 2013 at 10:15 | history | answered | Ralph | CC BY-SA 3.0 |