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Oct 27, 2017 at 12:16 vote accept spin
Oct 27, 2017 at 12:16 comment added spin Perhaps, since for $f \in H^n(H, M)$ we can define $\operatorname{cor}(f)(g_1, \ldots, g_n) = \sum_{x \in X} x f(x^{-1}g_1y_1, \ldots, x^{-1}g_ny_n)$ where $y_i \in X$ is such that $gy_iH = xH$.
Oct 27, 2017 at 11:50 comment added Chris Wuthrich That is the same thing. You just shuffled the sum by swapping the names of $x$ and $y$. Maybe your expression is a nicer way to write it...
Oct 26, 2017 at 19:27 comment added spin Thanks. But how about $\operatorname{cor}(f)(g) = \sum_{x \in X} x f(x^{-1} g y)$, where $y \in X$ is the unique rep. such that $gyH = xH$? That would also generalize to $H^n(H, M) \rightarrow H^n(G, M)$.
Oct 26, 2017 at 13:48 history answered Chris Wuthrich CC BY-SA 3.0