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Feb 28, 2020 at 1:17 answer added Harshit Yadav timeline score: 1
May 8, 2018 at 14:15 comment added Rene Recktenwald @tj_: Could you explain what the inverse is?
Jan 9, 2017 at 11:36 comment added Mark Wildon A related question: mathoverflow.net/a/256208/7709.
Jan 9, 2017 at 5:21 review Close votes
Jan 9, 2017 at 11:04
Jan 9, 2017 at 2:17 comment added tj_ It's more interesting to define the inverse and to show that they compose on cocycle level to the identity up to a coboundary.
Jan 9, 2017 at 2:15 comment added tj_ Let $\pi: \text{Coind}^G_H(M) = Hom_H(\mathbb{Z}G,M)\to M,\, x \mapsto x(1)$. The isomorphism $\varphi: H^\ast(G,\text{Coind}^G_H(M)) \to H^\ast(H,M)$ from Shapiro's lemma can be described in the following way: Let $f: G \times \cdots \times G \to Hom_H(\mathbb{Z}G,M)$ be a cocycle. Then $\varphi([f])$ is represented by the cocycle $\pi \circ f: H \times \cdots \times H \to M$ (see Brown, Cohomology of Groups, III.8, exercise 2).
Jan 9, 2017 at 0:47 review First posts
Jan 9, 2017 at 1:09
Jan 9, 2017 at 0:44 history asked Shawn Cui CC BY-SA 3.0