Action of $G/H$ on $H_n(H;M)$

Question

I'm currently studying group cohomology and have trouble with the Hochschild-Serre spectral sequence. My problem is this: Given a short exact sequence of groups

$$ 0 \to H \to G \to G/H \to 0$$

how do I compute the "natural" action of $G/H$ on the homology group $H_n(H;M)$, given a projective resolution of the $\mathbb{Z}[G]$-module $M$ in terms of free $\mathbb{Z}[H]$ modules? Note that I only have a resolution in terms of $\mathbb{Z}[H]$-modules, not $\mathbb{Z}[G]$-modules.

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For the sake of concreteness, let's consider the example

$$ 0 \to \mathbb{Z}_3 \to S_3 \to \mathbb{Z}_2 \to 0 .$$

A free resolution of $\mathbb{Z}$ in terms of $\mathbb{Z}[H] = \mathbb{Z}[X]/(X^3-1) =: \mathbb{Z}[t]$ (with $t^3=1$) modules is

$$ \dots \to \mathbb{Z}[t] \to^{(1-t)} \mathbb{Z}[t] \to^{(1+t+t^2)} \mathbb{Z}[t] \to^{(1-t)} \mathbb{Z}[t] \to^ε \mathbb{Z} \to 0$$

To calculate the homology, I drop the module that is being resolved and tensor with $\mathbb{Z}$ to get the chain complex

$$ \dots \to \mathbb{Z} \to^{0} \mathbb{Z} \to^{3} \mathbb{Z} \to^{0} \mathbb{Z} \to 0$$

But what is the action of $G/H \cong \mathbb{Z}_2$ on this complex, and why is it the correct/natural one?

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What I do understand is that given a resolution of $P_* \to M$ in terms of free $\mathbb{Z}[G]$ modules (G, not H!), I can define an action of $G/H$ on the modules $\mathbb{Z} \otimes_{\mathbb{Z}[H]} P_*$ by mapping $n \otimes m \mapsto n \otimes g·m$. Any two free resolutions are chain homotopic, but I don't see how I can transport this action to a free resolution that is only given in terms of $\mathbb{Z}[H]$ modules. (Maybe this action is even wrong.)

I also understand that given a $\mathbb{Z}[H]$-module $M$, I can define a new module structure $M_g$ on the same set by setting $h•m := g^{-1}hg·m$. Since homology is functorial, the map $m \mapsto g·m$ induces a map on homology

$$ H_n(H; M) \to^{g} H_n(H; M_g) $$

but the modules $M$ and $M_g$ are not the same, and I don't see how to get an action $H_n(H; M) \to^{g} H_n(H; M)$ from this.

Help?

Answer 1

It's helpful to first recall the construction of an induced map (cf. Brown, beginning of III/8 and II/6): Let $\varphi: H \to H'$ be a homomorphism of groups and let $F \to \mathbb{Z}\to 0$ resp. $F' \to \mathbb{Z}\to 0$ be a $H$- resp. $H'$-projective resolution. Via $\varphi$ any $H'$-module can be considered as a $H$-module. In this regard $F'$ is an acyclic complex of (not necessarily projective) $H$-modules. Hence by the fundamental lemma of homological algebra the identity $\mathbb{Z} \to \mathbb{Z}$ can be extended to a morphism $\varphi_\ast: F \to F'$ of complexes of $H$-modules ($\varphi_\ast$ is unique up to homotopy).

If $M$ resp. $M'$ is a $H$- resp. $H'$ module and $f: M \to M'$ is a homomorphism of $H$-modules (where $M'$ is considered a $H$-module via $\varphi$) then $\varphi_\ast \otimes f: F \otimes_H M \to F' \otimes_{H'} M'$ induces $(\varphi,f)_\ast: H_\ast(H;M) \to H_\ast(H';M')$.

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Now let $H$ be a normal subgroup of $G$. For $g \in G$ let $\varphi: H \to gHg^{-1} =: H'=H$ and let $f: M \to M,\;m \to gm$. Then, for $z \in H_\ast(H;M)$ the action of $gH$ is given by $(gH)\cdot z = (\varphi,f)_\ast(z)$.

So the action can be calculated by taking a $H$-projective resolution $F$ and by computing a morphism $\varphi_\ast: F \to F$ of complexes of abelian groups that extends the identity $\mathbb{Z} \to \mathbb{Z}$ and satisfies $\varphi_n(hx)=\varphi(h)\varphi_n(x) = ghg^{-1}\varphi_n(x)$ for $h \in H,x \in F_n$.

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Here a calculation for the concrete example $H_\ast(\mathbb{Z}_3;\mathbb{Z})$ and the automorphism $\varphi:\mathbb{Z}_3 \to \mathbb{Z}_3$, $\varphi(t)=t^2$ induced by the quotient of $S_3$.

In the following diagram

the maps $\varphi_k$ extend the identity on $\mathbb{Z}$ to . (Unfortunately, they are labelled $f_k$ in the diagram). The maps are determined by their action on the generator $1\in \mathbb{Z}[t]$. One possible choice is

$$\varphi_0(1) = 1,\quad \varphi_{2k+1}(1) = -t^2\varphi_{2k}(1),\quad \varphi_{2k}(1) = \varphi_{2k+1}(1), $$

Tensoring with $\mathbb{Z}$, we obtain maps as shown in the following diagram

In particular, we have

$$ \tilde\varphi_{4k+1}=\tilde\varphi_{4k+2}=-1, \quad \tilde\varphi_{4k}=\tilde\varphi_{4k+3}=+1$$

and for the action on homology, $\hat\varphi_\ast : H_\ast(H;\mathbb{Z}) \to H_\ast(H;\mathbb{Z})$, we obtain

$$ \hat\varphi_n = \begin{cases} 1 , &\text{ if } n=0 \\ -1 , &\text{ if } n \equiv 1 \mod 4 \\ +1 , &\text{ if } n \equiv 3 \mod 4 \\ 0 , &\text{ if } n=2,4,… \text{ even} .\end{cases} $$

Answer 2

This is a non-answer, but I don't think there is an answer. You want to compute the action directly from an $H$-module resolution, which as far as I know is not possible. (There is no reason it should be.)

Instead, you must find a larger resolution. A resolution by $\mathbb{Z}[G]$-modules is a priori a resolution by $\mathbb{Z}[H]$-modules, and we know that the homology doesn't depend on the choice of the free resolution. Given a resolution by $\mathbb{Z}[G]$-modules, after taking $H$-coinvariants, we get an action of $G/H$, which passes to homology precisely because the maps in the original resolution are $G$-equivariant and not just $H$-equivariant.

Now if for your explicit calculations you really need to understand the action on the homology as computed in terms of your original resolution, you are going to need to write down an explicit (sequence of?) quasi-isomorphisms between your original $H$-resolution and a $G$-resolution, and chase the action through the isomorphisms.

This is typical of derived functors -- what you get is "independent of the resolution," but only up to isomorphism. You still might have to use "special resolutions with extra properties" to get extra structure.

edit: for your example, I think the action is the non-trivial one, but did not calculate it out. Use the bar resolution for $S_3$.