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When $G$ acts trivially on $M$, the first homology group is just the abelianisation of $G$ tensored with $M$, i.e. $H_1(G;M)=(G/[G,G])\otimes_\mathbb Z M$.

Is there any similar statement when $G$ acts not trivially on $M$? Where does the abelianisation come from?

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You will find the answer in Brown's book "Cohomology of groups". When the action is not trivial you have to work with twisted coefficients. – berl13 Apr 18 '12 at 12:18
In Brown's book I can only find the statement with trivial coefficients... Do you have a page reference? Or are you suggesting I should calculate $H_1$ via the bar resolution, which I can't imagine has a straight-forward answer for an arbitrary action. – Earthliŋ Apr 18 '12 at 12:42
You'll get a certain quotient of $\mathbb Z[G]\otimes_{\mathbb Z[G]}M$ by the boundary operator. In the case of twisted coefficients the two parts of the tensor will be mixed up, so you won't have a nice factorization. – Jim Conant Apr 18 '12 at 12:48
But $\mathbb Z[G]\otimes_{\mathbb Z[G]}M$ is just $M$, no? A quotient of $M$ does sound not too bad. Thanks for your answers, I'll ask my pencil for more insights. – Earthliŋ Apr 18 '12 at 12:59
There is a nice description of the twisted first cohomology group as crossed homomorphisms modulo principal crossed homomorphisms. But there really isn't a particularly nice description of the twisted first homology group. – Andy Putman Apr 18 '12 at 17:06
up vote 8 down vote accepted

In the trivial case the abelianization comes from the short exact sequence

$0\to J\to \mathbb{Z}G\to \mathbb{Z}\to 0$

where $J$ is the augmentation ideal. The homology $H_*(G,-)$ are just derived functors and give a long exact sequence in homology, which since $H_1(\mathbb{Z}G,\mathbb{Z})$ is always trivial, gives a four term exact sequence which looks like

$0\to H_1(G,\mathbb{Z})\to J_G\to (\mathbb{Z}G)_G\to\mathbb{Z}\to 0$

Here the subscript $-_G$ just means the coinvariant functor from which the homology is derived: $M_G$ is the quotient of $M$ by the submodule generated by $gm - m$ for all $m\in M$. It is not hard to verify that $(\mathbb{Z}G)_G\to\mathbb{Z}$ is an isomorphism so that $H_1(G,\mathbb{Z})$ is isormorphic to $J_G$, which is in turn the abelianization. All of this is classical and in Brown, Weibel, or probably even Mac Lane, which you can consult for more details.

Alternatively, topologically it isn't hard to prove that $H_1(G,\mathbb{Z})$ is just the homology of a topological classifying space with $G$ as its fundamental group and whose higher homotopy groups are trivial, and from algebraic topology we know that $H_1$ of that space is $G/[G,G]$.

If $M$ is not a trivial $G$ module then you probably won't get as nice a description as for cohomology because the tensor product is not as nice as this case as the Hom set. Andy's comment on derivations/principal derivations comes from (or at least morally) from the bar resolution, which for any group $G$ is a free $\mathbb{Z}G$ resolution of $\mathbb{Z}$. This is great because homology is just the derived functor of $\mathbb{Z}\otimes_{\mathbb{Z} G}-$, which makes sense because $-_G$ is right exact and commutes with arbitrary direct sums.

So you can still work out a kind of analogue to the description for cohomology. If you actually use the bar resolution, you get that $H_1(G,M)$ for an arbitrary right $G$-module $M$ is just the homology $\mathrm{ker} (d)/\mathrm{im} (d)$ (abuse of notation, using the same $d$) of the complex

$M\otimes B_2\xrightarrow{1\otimes d} M\otimes B_1\xrightarrow{1\otimes d} M\otimes \mathbb{Z}G$

where the tensor is taken over $\mathbb{Z}G$. The module $B_1$ is the free $G$ module on the symbols $[g_1]$ for $g_1\in G$ and $g_1\not=1$. If we let $[]\in\mathbb{Z}G$ be the identity, then $d([g_1]) = g_1[] - []$. Similarly, $B_2$ is the free $G$ module on $[g_1|g_2]$ where neither are the identity, and $d([g_1|g_2]) = g_1[g_2] - [g_1g_2] + [g_1]$. Here $[g_1g_2] = 0$ if $g_1g_2 = 1$ in the group. You can do the same for left-$G$ modules by forming the analogous right free $G$-module resolution of $\mathbb{Z}$.

We could also have used the unnormalized bar resolution; regardless, it is now true that this is just a complex with each term a direct sum of copies of $M$, but the difficulty is determining what exactly the boundary formulas turn out to be. For a specific $M$ often it is not too hard to make a computation. For cyclic groups you can use the usual period resolution of $\mathbb{Z}$ to get that

$H_1(\mathbb{Z}/k,M)\cong M^{\mathbb{Z}/k}/NM$

where $N$ is the norm element, defined for a general (finite!) group to be $\sum_{g\in G}g$.

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Using the short exact sequence $0\to I\to \mathbb Z G \to \mathbb Z \to 0$ and the associated long exact sequence for ${\rm Tor}_*^{\mathbb ZG}(-,M),$ we get the following exact sequence $$0 \longrightarrow H_1(G,M) \longrightarrow I \otimes_{\mathbb ZG} M \longrightarrow M \longrightarrow H_0(G,M) \longrightarrow 0.$$ In particular, we obtain $$H_1(G,M)={\rm Ker}(I \otimes_{\mathbb ZG} M \longrightarrow M).$$ If $G$ acts trivially on $M,$ the map $I \otimes_{\mathbb ZG} M \longrightarrow M$ is trivial and we get $H_1(G,M)=I\otimes_{\mathbb ZG}M=I/I^2 \otimes M=G_{\rm ab}\otimes M.$ Hence, it is a generalisation of your formula.

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