# Deriving explicit formulas for the transfer maps

Let $G$ be a group, $H \leq G$ a subgroup of finite index. The transfer map $\mathrm{tr} \colon H^*(H,-) \Rightarrow H^*(G,-)$ is uniquely determined as a morphism of $\delta$-functors by its behavior on the 0th cohomology groups: $\mathrm{tr}_A^0 \colon A^H \to A^G, a \mapsto \sum_{\overline{g} \in H \setminus G} ag$ ($A$ is any right $G$-module). Now there is an obvious approach to find explicit formulas for $\mathrm{tr}_A^n \colon H^n(H,A) \to H^n(G,A)$ for higher $n$, namely:

1. Pick any exact sequence $0 \to A \to \hat{A} \to A^* \to A$ of $G$-modules such that $H^k(H,\hat{A}) = 0$ for all $k$. E.g. take $\hat{A} = \mathrm{CoInd}^G(A) = \mathrm{Hom}_\mathbb{Z}(G,A)$.
2. Derive a formula for $\mathrm{tr}^n$ inductively by looking at the commutative diagram: $\begin{array}{ccc} H^{n-1}(H,A^*) & \xrightarrow{\delta'} & H^n(H,A)\\ \downarrow{\mathrm{tr}_{A^*}^{n-1}} & & \downarrow{\mathrm{tr}_A^n}\\ H^{n-1}(G,A^*) & \xrightarrow{\delta} & H^n(G,A) \end{array}$

However, this approach becomes very complicated. Are there more elegant ways to find these formulas? I would also be happy about a reference, where those formulas are given.

PS: I consider $H^n(G,A)$ as the cohomology of the standard complex here, i.e. co-cycles modulo co-boundaries.

## 1 Answer

Let $H$ be a subgroup of $G$ with finite index $[G:H]=\#(H\backslash G)$. For every right coset $c \in H\backslash G$ choose a representative $\bar c\in G$ (i.e. such that $H\bar c = c$).

Notice that $G$ acts on $H\backslash G$ on the left, and that for all $\sigma\in G$ and all $c\in H\backslash G$, the element $\bar c \sigma \overline{c\sigma}^{-1}$ is in $H$.

Let $X^n(G,A):=\operatorname{Maps}(G^{n+1},A)$ and let $C^n(G,A):=X^n(G,A)^G$ be the complex of homogeneous A-valued cochains.

Then the corestriction $\operatorname{cor}: C^n(H,A)\to C^n(G,A)$ is given by the formula $$(\operatorname {cor} u)(\sigma_0,\ldots,\sigma_n)=\sum_{c\in H\backslash G} \bar c^{-1} u(\bar c\sigma_0\overline{c\sigma_0}^{-1},\dots,\bar c\sigma_n\overline{c\sigma_n}^{1})\,.$$

If you are interested in the inhomogeneous cochains, the formula becomes (I think, please check): $$(\operatorname {cor} u)(\sigma_1,\ldots,\sigma_n)=\sum_{c\in H\backslash G} \bar c^{-1} u(\bar c\sigma_1\overline{c\sigma_1}^{-1},\overline{ c\sigma_1}\sigma_2\overline{c\sigma_1\sigma_2}^{-1}\dots,\overline{ c\sigma_1\dots\sigma_{n-1}}\sigma_n\overline{c\sigma_1\dots\sigma_n}^{-1})\,.$$

You can find this formula in [1, Chapter I.$\S$5, pag 48] or in [2, Proposition 2.5.2]

[1] Neukirch, Jürgen; Schmidt, Alexander(D-RGBGNS1); Wingberg, Kay(D-HDBG) Cohomology of number fields. Second edition. Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], 323. Springer-Verlag, Berlin, 2008. xvi+825 pp. ISBN: 978-3-540-37888-4

[2] Weiss, Edwin Cohomology of groups. Pure and Applied Mathematics, Vol. 34 Academic Press, New York-London 1969 x+274 pp.

• Source [2] has an approach of deriving a formula by induction, imposing the compatibility with $\partial$. Source [1] instead, defines two homomorphisms of complexes: one given by the formula, the other defined abstractly. Then proves they are equal, by induction, with dimension shifting. Commented Aug 10, 2017 at 23:12