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Let $G$ be a group and $M$ a $G$-module. The basic definitions:

  • $H^0(G, M)$ will be the set of $G$-fixed points in $M$.
  • $Z^1(G, M)$ is the group of $1$-cocycles, i.e. the maps $f: G \rightarrow M$ such that $f(gg') = f(g) + g f(g')$ for all $g, g' \in M$.
  • $B^1(G, M)$ is the group of $1$-coboundaries, ie. maps $c_m : G \rightarrow M$ defined by $c_m(g) = gm - m$.
  • $H^1(G, M)$ is the quotient group $Z^1(G, M) / B^1(G, M)$.

Let $H < G$ be a subgroup of finite index. We have a map $$tr: H^0(H, M) \rightarrow H^0(G, M)$$

defined by $m \mapsto \sum_{g \in G/H} gm$. This can be extended to a map $H^*(H, M) \rightarrow H^*(G, M)$ of cohomological functors.

My question is:

Is there an explicit description of the corestriction map $H^1(H, M) \rightarrow H^1(G, M)$, e.g. in terms of $1$-cocycles?

For one idea, we have an isomorphism $\psi: \operatorname{Hom}_{\mathbb{Z}H}(\mathbb{Z}G, M) \rightarrow M$ defined by $$\psi(f) = \sum_{g \in G/H} gf(g^{-1}).$$ Hence there is an explicit map $H^1(G, \operatorname{Hom}_{\mathbb{Z}H}(\mathbb{Z}G, M)) \rightarrow H^1(G, M)$. By Shapiro's lemma we have $H^1(H, M) \cong H^1(G, \operatorname{Hom}_{\mathbb{Z}H}(\mathbb{Z}G, M))$, but I don't know if there is a way to make this isomorphism explicit.

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I believe it is the following. Let $f$ be a cocyle for $H$. Take a set of representatives $X$ of $G/H$ in $G$. Then $\operatorname{cor}(f)(g) = \sum_{x \in X} y\cdot f(y^{-1}gx)$ where $y\in X$ is the unique representative such that $gxH=yH$. Then $\operatorname{cor}(f)$ is a cocycle whose class is the well-defined corestriction of the class of $f$.

Cohomology of number fields, section I.5 has the formula for right cosets.

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  • $\begingroup$ 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 $H^n(H, M) \rightarrow H^n(G, M)$. $\endgroup$
    – spin
    Commented Oct 26, 2017 at 19:27
  • $\begingroup$ 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... $\endgroup$
    – Chris Wuthrich
    Commented Oct 27, 2017 at 11:50
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    $\begingroup$ 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 $gy_iH = xH$. $\endgroup$
    – spin
    Commented Oct 27, 2017 at 12:16

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