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I was trying to understand group (co)homology from a homological algebra point of view. Namely, given a group, $G$, one considers the category of (left) $\mathbb{Z}[G]$-modules, $\mathrm{Mod}_{\mathbb{Z}[G]}$. Then one can define the group homology $\mathrm{H}_n(G;M)$ as $\mathrm{Tor}^{\mathbb{Z}[G]}_k(\mathbb{Z},M)$ and the group cohomology $\mathrm{H}^n(G;M)$ as $\mathrm{Ext}^n _{{\mathbb{Z}[G]}}(\mathbb{Z},M)$. Of course, there is also the homotopical point of view: for each group $G$ we concoct a classifying space $\mathrm{B}G$ and a $G$-module $M$ can be thought of as a twisted coefficient system over this space; if we take the homology or cohomology of $\mathrm{B}G$ with the twisted coefficient system $M$ we recover $\mathrm{H}_n(G;M)$ and $\mathrm{H}^n(G;M)$.

The notion of the transfer comes to us naturally in the homotopical setting. Given an inclusion $H\subset G$, the map $\mathrm{B}H\rightarrow \mathrm{B}G$ can be arranged to be a finite covering map, so we can take the regular transfer map $\mathrm{H}_n(\mathrm{B}G;\mathbb{Z})\rightarrow \mathrm{H}_n(\mathrm{B}H;\mathbb{Z})$. Now my questions are:

$1$. How can we get this transfer map by homological means?

$2$. Can we replace the trivial coefficients in the transfer map by more general coefficients? If so, how would that look like?

The inclusion $H\subset G$ generates a finite free extension $\mathbb{Z}[H]\rightarrow \mathbb{Z}[G]$.

$3$. If $A\hookrightarrow B$ is ring homomorphism and $B$ is a finite $A$-module, are there analogs of transfers between $\mathrm{Tor}^A _{\ast}$ and $\mathrm{Tor}^B _\ast$?

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  1. Let $M$ be a $G$-module. Then there's an obvious injection $$M\rightarrow {\mathbb Z}G\otimes_{{\mathbb Z}H}M$$

  2. Taking homology gives $$H_\bullet(G,M)\rightarrow H_\bullet(G,{\mathbb Z}G\otimes_{{\mathbb Z}H}M)$$

  3. To compute the group on the right, start with a projective resolution $P$ of ${\mathbb Z}$ over ${\mathbb Z}G$, then take the homology of
    $$P\otimes_{{\mathbb Z}G}{\mathbb Z}G{\otimes_{\mathbb ZH}}M=P\otimes_{{\mathbb Z}H}M$$

  4. Notice that $P$ is also a projective resolution of ${\mathbb Z}$ over ${\mathbb Z}H$, so the homology you've just computed is in fact equal to $H_\bullet(H,M)$.

  5. Thus the map in 2) is actually a map $H_\bullet(G,M) \rightarrow H_\bullet(H,M)$ and is in fact the transfer map.

  6. To make the same argument work for a ring inclusion $A\rightarrow B$ you'd need to know that a projective $B$-resolution is also a projective $A$-resolution. This works, for example, if $B$ is $A$-free.

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  • $\begingroup$ By obvious you mean $m$ goes to $1\otimes m$? $\endgroup$
    – nerses
    Commented Mar 12, 2013 at 4:18
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    $\begingroup$ nerses: I'm not sure what you mean by "1". The element $m$ maps to $t\otimes m$ where $t$ is a sum of coset representatives for $G$ over $H$. (It doesn't matter how you pick the representatives.) $\endgroup$
    – Steven Landsburg
    Commented Mar 12, 2013 at 4:22
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    $\begingroup$ PS: Along with the map in 1), there is also an obvious surjection going the other way. Apply all of the same reasoning to that surjection, and you'll get the usual map $f:H(H,M)\rightarrow H(G,M)$. If you first do the obvious injection and then the obvious surjection, the induced map on $M$ is multiplication by $(G:H)$. This in turn tells you what happens when you compose the transfer with $f$. $\endgroup$
    – Steven Landsburg
    Commented Mar 12, 2013 at 4:27
  • $\begingroup$ Come to think of it, there is a bit of a problem: I don't understand why the map $m$ sending $t\otimes m$ is a $G$-module and how we get the map between homologies. It is if we have trivial $G$-module $M$, I think, we do recover the right transfer map in this case. I check out the case, $*\hookrightarrow C_2$ with $M=\mathbb{Z}[C_2]$ and it does not seem to work. $\endgroup$
    – nerses
    Commented Mar 12, 2013 at 17:29
  • $\begingroup$ Apologies; an earlier version of this comment mixed up rights and lefts. I've repaired that and am deleting the original. I'm not sure what's confusing you. The G-module structure on ${\mathbb Z}G\otimes_{{\mathbb Z}H}M$ is given by multiplication on the left. (Were you maybe thinking it was on the right?) So ${\mathbb Z}G\otimes_{{\mathbb Z}H}M$ is a direct sum of $(G:H)$ copies of M, permuted by the elements of $G$. The element $t\otimes m$ corresponds to the element $(m,m,\ldots m)$. (This does assume $(G:H)$ is finite, but I don't think that's where you're having trouble.) $\endgroup$
    – Steven Landsburg
    Commented Mar 13, 2013 at 17:04

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