# How should I think about the module of coinvariants of a $G$-module?

Let $$G$$ be a group, $$M$$ a $$G$$-module, then the group of coinvariants is the module $$M_G := M/I_GM$$, where $$I_G$$ is the kernel of the augmentation map $$\epsilon : \mathbb{Z}G\rightarrow \mathbb{Z}$$.

The dual notion of $$G$$-invariants is very simple, and for most $$G$$-modules, it's easy to tell whether or not for example $$M^G = 0$$.

On the other hand, I don't have a good intuition for $$M_G$$.

I'd appreciate a list of results that say something about this module of coinvariants, and criteria which might help recognize when it vanishes. Even facts in the case where $$G$$ is a finite, abelian, or even finite abelian group would be welcome.

• It may help to think in terms of the characterisation, rather than the definition: $M^G$ is the largest submodule of $M$ on which $G$ acts trivially, and $M_G$ is the largest quotient of $M$ on which $G$ acts trivially. Thus, for example, if $G$ acts as a commuting family of semisimple endomorphisms, then $M^G$ (the common $1$-eigenspace) is isomorphic to $M_G$ (the quotient by the non-$1$ eigenspaces); and, if $G$ acts by unipotent automorphisms, so that $M$ has a composition series with quotients which $G$ acts trivially, then $M^G$ is at the "bottom" of the composition series … (1/2) – LSpice Oct 11 at 2:22
• … and $M_G$ is at the "top", in the sense that it is the quotient of $M$ by the next-smallest term. (2/2) – LSpice Oct 11 at 2:23
• (In particular, if $G$ is finite Abelian and the module is over a field (I can't tell whether that's what you have in mind), then the action of $G$ is by commuting semisimple automorphisms, so $M^G \cong M_G$.) – LSpice Oct 11 at 2:24
• What about the modular case? Surely cyclic group of order $p$ can act by unipotent upper triangular $2\times 2$ matrices over $\mathbb{F}_p$, and tautologically this is an action by commuting non-semisimple automorphisms. – Victor Protsak Oct 11 at 3:11
• @LSpice Thanks for your comments, though I was definitely thinking more about the non-semisimple case (ie, $M$ is a more or less arbitrary $\mathbb{Z}G$-module) – stupid_question_bot Oct 11 at 4:08

I will say how to compute the module of coinvariants---this will give a criterion to say if it vanishes. We work over $$\mathbb{Z}$$, since this is the most general setting, but the following analysis works over any commutative ring.
Given a $$G$$-module $$M$$, pick generators $$x_1, \ldots, x_k \in M$$. These are elements so that every other $$m \in M$$ is a $$\mathbb{Z}$$-linear combination of the elements $$gx_i$$ where $$i \in \{1, \ldots, k\}$$ and $$g \in G$$. This is the same as saying that the $$x_i$$ generate $$M$$ as a left $$\mathbb{Z}G$$-module.
If $$M \cong \bigoplus_{i=1}^k \mathbb{Z}Gx_i$$, then $$M$$ is free as a left $$\mathbb{Z}G$$-module, and so the coinvariant module is just $$\bigoplus_{i=1}^k \mathbb{Z}x_i$$, the free $$\mathbb{Z}$$-module on the same generators. If $$M$$ is not free on the $$x_i$$, then there must be some relations. In other words, the surjection $$\bigoplus_{i=1}^k \mathbb{Z}Gx_i \twoheadrightarrow M$$ must have some kernel.
Let's suppose that $$G$$ is finite, or otherwise ensure that the kernel of this map is finitely generated. Pick generators for the kernel $$y_1, \ldots, y_l$$. We obtain an exact sequence
$$\bigoplus_{j=1}^l \mathbb{Z}Gy_j \xrightarrow{\varphi} \bigoplus_{i=1}^k \mathbb{Z}Gx_i \to M \to 0.$$
Since the map $$\varphi$$ is a map from one free module to another, it is given by a $$l \times k$$ matrix with entries in the ring $$\mathbb{Z}G$$. Let $$\epsilon\varphi$$ be the same matrix, but where we have replaced every group element $$g \in G$$ with the integer $$1 \in \mathbb{Z}$$. So $$\epsilon\varphi$$ is an integer matrix. Writing $$\mathbb{Z}$$ for the integers with the trivial right action of $$G$$, we have \begin{align} M_G &\cong \mathbb{Z} \otimes_G M \\ &\cong \mathbb{Z} \otimes_G \mathrm{coker} \, \varphi \\ &\cong \mathrm{coker} \left( \mathbb{Z} \otimes_G \varphi \right) \\ &\cong \mathrm{coker} \left( \epsilon \varphi \right). \end{align} In other words, the coinvariants are still generated by the same generating set; however, any group elements appearing in the relations are ignored. For example, any relation $$gx_1 = x_2$$ holding in $$M$$ just becomes $$x_1=x_2$$ in $$M_G$$. Similarly, the relation $$x_1 + gx_2 = hx_3$$ becomes $$x_1 + x_2 = x_3$$, $$x_1-gx_1=0$$ becomes $$x_1-x_1=0$$, etc. This gives a systematic way of obtaining a presentation for the $$\mathbb{Z}$$-module $$M_G$$ from a presentation for the $$\mathbb{Z}G$$-module $$M$$.