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.

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?

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.


  • $\begingroup$ You don't want a $\mathbb{Z}H$-projective resolution of $M$ but of the trivial $\mathbb{Z}G$-module $\mathbb{Z}$. $\endgroup$
    – tj_
    Aug 3 '14 at 14:09
  • $\begingroup$ @tj_ Both are possible, the definition of homology is symmetric with respect to which module you resolve. $\endgroup$ Aug 3 '14 at 15:54
  • 1
    $\begingroup$ Yes, but a projective resolution of $\mathbb{Z}$ can be used for all coefficients. $\endgroup$
    – tj_
    Aug 6 '14 at 19:54

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 $(\varphi,f)_\ast: H_\ast(H;M) \to H_\ast(H';M')$.

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 $h \in H,x \in F_n$.

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

enter image description here

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

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

  • $\begingroup$ Many thanks for your answer! (I'll make a small edit and add the concrete calculation later.) Brown's book was not available in my university's library. A book by Lang mentioned that the action on the zeroth homology uniquely determines the action on higher homology groups via the long exact sequence / uniqueness of group homology, but a concrete calculation using this method seemed very daunting to me. $\endgroup$ Aug 6 '14 at 20:17
  • $\begingroup$ How did you get the relations in that example? $\endgroup$
    – Liddo
    May 29 '19 at 15:14
  • $\begingroup$ @Liddo: Which relations exactly do you mean? $\endgroup$
    – tj_
    May 29 '19 at 21:07
  • $\begingroup$ @tj In the example of $Z_{3}$ I am trying to do it in general for $Z_{n}$ for the $\varphi$ In that particular case of $n=3$ instead of multiplying by $-t^{2}$ I have $t+1$, that's why my question. $\endgroup$
    – Liddo
    May 31 '19 at 16:50
  • $\begingroup$ @tj In general I don't have any, I am trying to find an action on the homology $H_{*}(Z_{n},Z)$ $\endgroup$
    – Liddo
    Jun 2 '19 at 20:39

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 $S_3$.

  • $\begingroup$ Oh! But wouldn't this mean that the Hochschild-Serre spectral sequence has only limited utility? Right now, it seems to me that in order to calculate the G/H action, I have to look at the bar resolution of $M$ as a $\mathbb{Z}[G]$-module, which I wanted to avoid doing in the first place by decomposing $G$ into $H$ and $G/H$. (Of course, I need some information about how $H$ sits inside $G$.) I was hoping for some "functorial" way of defining the action on $H_n(H;M)$. $\endgroup$ Aug 2 '14 at 9:11
  • $\begingroup$ Of course it's possible to compute the action of $G/H$ on $H_\ast(H;M)$ by any $\mathbb{Z}H$-projective resolution of $\mathbb{Z}$. This follows right from the definition of the action. $\endgroup$
    – tj_
    Aug 3 '14 at 14:07
  • $\begingroup$ @tj_ Apparently, I'm missing the definition of the action. Could you elaborate it into a short answer? That would be much appreciated. $\endgroup$ Aug 3 '14 at 15:56

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