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.
Best Answer
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.