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I'm trying to figure out what the transfer map looks like in a specific case. Here's the set up

Let $G$ be a group and $H$ a subgroup of finite index, and let $h_{i}$ for $i=1,..,n$, be coset representatives.

Now following Weibel I know that if $A$ is a $G$ module, then $Tr_{0} :H_{0}(G,A) \rightarrow H_{0}(H,A)$ is the map that sends a 0-cycle $a$ to $\sum_{i} h_{i}.a$. Now theres two things:

the thing is I want to work out what $Tr_{1}: H_{1}(G,A) \rightarrow H_{1}(H,A)$ does specifically, and how does $G$ act on the 1-cycles, since I'm not sure how to define an action on 1-cycles. So I was wondering if I could get some help/hints or maybe some good references, so far I have looked at Weibels intro to homological algebra and a little of Rotmans Homological algebra.

Thank you

Thank you

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You should check out Ken Brown's textbook Cohomology of Groups, that's the definitive. A nice way to see the transfer is induced from the covering map of classifying spaces. Namely, if $\sigma$ is a cell of $BG$ then $\sum\tilde{\sigma}$ is a cell of $BH$ (i.e. all the lifts of $\sigma$). Then $Tr$ maps cochain $f$ to the cochain $\sigma\mapsto \sum f(\tilde{\sigma})$. – Chris Gerig Feb 14 '13 at 19:50

In case of trivial coefficients there are nice formulas: Let $G=\coprod_h Hh$ and
$$t: G \to H,\; g \mapsto \prod_h hgh_g^{-1}$$ where the representative $h_g$ is defined by $Hhg=Hh_g$. Under the identification $H_1(G,A)=G_{ab}\otimes A$ we get $$tr_1: G_{ab}\otimes A \to H_{ab}\otimes A,\; g[G,G]\otimes a \mapsto t(g)[H,H] \otimes a$$ (cf. Brown, Cohomology of Groups, III.9, Ex. 2). In case $A=\mathbb{Z}$ this is just the usual transfer homomorphism from group theory (cf. Robinson, Theory of Groups, 10.1).

In cohomology, using the identification $H^1(G,A)=Hom(G,A)$ we have $$tr^1: Hom(H,A) \to Hom(G,A),\; f \mapsto f\circ t.$$

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I thought that the identification $H_{1}(G,A)=G_{ab} \otimes A$ wasnt always true if $A$ is not $G$-trivial? – Chris Birkbeck Feb 14 '13 at 22:21
Yes, that's why the first sentence is "In case of trivial coefficients ...". – Ralph Feb 14 '13 at 22:37
Oh sorry I missed that :) – Chris Birkbeck Feb 15 '13 at 11:06

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