# $G$-modules, Group invariants and Tor functor

Let $$M$$ be a $$G$$-module. Then functoriality induces a natural $$G$$-module structure on $$\text{Tor}_i(M,N)$$, where $$N$$ is any abelian group. My question is, what can we say about $$\text{Tor}_i(M,N)^G.$$ Is there some way to discribe this using $$M^G$$? For example, if we assume that $$M=\mathbb{Z}/n$$, then $$\text{Tor}_1(\mathbb{Z}/n,N)^G=\left(N[n]\right)^G=\left(N^G\right)[n]=N[n]=\text{Tor}_1(\mathbb{Z}/n,N).$$ I'm most interested in "nice" criteria for the vanishing of $$\text{Tor}_i(M,N)^G.$$

$$\newcommand{\Tor}{\mathrm{Tor}_1^\mathbb Z}$$ First, a comment about your example: note that in the isomorphism $$\mathrm{Tor}_1^\mathbb Z(\mathbb Z/n, N)^G\cong (N[n])^G$$, it is not the case that you can rewrite this as $$(N^G)[n]$$.

It would be the case if $$N$$ was a $$G$$-module, and the $$G$$-module structure on $$\mathrm{Tor}_1^\mathbb Z(\mathbb Z/n,N)$$ came from $$N$$, but that is not the case here : the $$G$$-action comes from $$\mathbb Z/n$$; in particular $$N[n]$$ has a $$G$$-action on its own that might now be the restriction to $$N[n]$$ of the trivial action on $$N$$.

Let me give a specific example to make this point clear : consider $$n=3$$, $$G=C_2$$ (the group with $$2$$ elements, and let $$\sigma \in C_2$$ be the nontrivial one), and $$C_2$$ acting on $$M=\mathbb Z/3$$ by $$x\mapsto -x$$.

Then on the projective resolution $$\mathbb Z\overset 3\to \mathbb Z$$ of $$M$$, $$C_2$$ acts simply by $$-1$$ too:

$$\require{AMScd}\begin{CD}\mathbb Z @>3>> \mathbb Z \\ @V-1VV @V-1VV \\ \mathbb Z @>3>> \mathbb Z\end{CD}$$

Therefore, on $$\Tor(M,N)$$, which you may identify as the kernel of $$N\otimes\mathbb Z \overset{id_N\otimes 3}\to N\otimes\mathbb Z$$, it is clear that $$C_2$$ also acts by $$x\mapsto -x$$ : just look at the diagram that defines the action of $$C_2$$ :

$$\begin{CD}\Tor(M,N)@>>> N\otimes\mathbb Z @>3>> N\otimes\mathbb Z \\ @VVV @V-1VV @V-1VV \\ \Tor(M,N) @>>> N\otimes\mathbb Z @>3>> N\otimes\mathbb Z\end{CD}$$

Therefore $$N[3]$$ here has a nontrivial $$G$$-action, and in fact $$N[3]^G=0$$ : if $$x=-x$$, then $$2x=0$$, but since $$x\in N[3]$$, $$3x=0$$, so $$x=0$$.

If you can describe $$\Tor(M,N)$$ in (functorial) terms of $$N$$, don't assume that this description has the $$G$$-action induced by $$N$$ : on the contrary, it will typically have a $$G$$-action induced by one on the functor you're using as a description.

That being said, let me move on to the more general question.

Over an arbitrary ring, the question can be extremely hard, in fact describing $$\Tor(M,N)^G$$ can involve some higher $$\mathrm{Tor}$$'s and some group cohomology. But we're lucky, because $$\mathbb Z$$ is not an arbitrary ring : it's a principal ideal domain.

Its higher $$\mathrm{Tor}$$'s therefore vanish, and so, note the following thing :

$$\Tor(-,N)$$ is a left exact functor.

Indeed suppose $$0\to A\to B \to C\to 0$$ is exact, then you get the long exact sequence for $$\mathrm{Tor}$$'s, but $$\mathrm{Tpr}_2^\mathbb Z= 0$$, so you're left with $$0\to\Tor(A,N)\to \Tor(B,N)\to\Tor(C,N)$$

(and of course $$\to A\otimes N\to B\otimes N\to C\otimes N\to 0)$$

In particular, note that we have the following exact sequence $$0\to M^G\to M\to \prod_{g\in G}M$$, where $$M\to \prod_{g\in G}M$$ is defined by $$m\mapsto (g\cdot m-m)_g$$

Assume temporarily that $$N$$ is finitely generated, so that $$\Tor(-,N)$$ commutes with products (if $$G$$ is finite, we don't need this - but anyway we'll get rid of it later)

It then follows that $$0\to\Tor(M^G,N)\to \Tor(M,N)\to \prod_{g\in G}\Tor(M,N)$$ is exact too (by left exactness), and since it's clear by the previous description that the map $$\Tor(M,N)\to \prod_{g\in G}\Tor(M,N)$$ is also $$g-id$$ on the $$g$$ coordinate, it follows that its kernel is $$\Tor(M,N)^G$$.

First conclusion:

If $$N$$ is finitely generated, the canonical map $$\Tor(M^G,N)\to \Tor(M,N)^G$$ is an isomorphism.

But now we have a natural map $$\Tor(M^G,N)\to\Tor(M,N)^G$$ which is an isomorphism for finitely generated $$N$$, and both sides commute with filtered colimits: it follows that it's an isomorphism for all $$N$$.

Second conclusion:

$$\Tor(M^G,N)\cong \Tor(M,N)^G$$

This gives you simple criteria for the vanishing of the RHS: whenever the LHS vanishes; for instance if $$M^G$$ is flat or projective, or of order coprime to $$N$$ (if both are finite, say).

I've seen you ask higher-level questions so let me just point out a way to generalize this and see why things get complicated over a general ring.

Work in the derived $$\infty$$-category of $$R$$-modules, then you have a canonical map $$M^{hG}\otimes_R^L N\to (M\otimes_R^LN)^{hG}$$. This map is an equivalence for $$N=R$$, and both sides commute with finite colimits, so it's an equivalence for $$N\in \mathrm{Perf}(R)$$.

The point is then that over $$\mathbb Z$$ or more generally a PID, it's quite easy to compute $$H_1$$ on each side, and compare them, you get the desired result (for $$N$$ finitely generated, but you can then extend to arbitrary $$N$$ as above - note, however, that this extension to arbitrary $$N$$ cannot be made in $$D(R)$$, as $$(-)^{hG}$$ need not commute with filtered colimits, so to extend to an arbitrary $$N$$, you need to go back to usual, discrete $$R$$-modules).

Over a general ring $$R$$, you will have spectral sequences controlling both sides of this equivalence, and the $$H_1$$'s will be related to $$\mathrm{Tor}_1^R(M^G,N)$$ and $$\mathrm{Tor}_1^R(M,N)^G$$, but the relation will be difficult because of some other input: $$\mathrm{Tor}_n^R(H^p(G,M),N)$$'s, as well as $$H^p(G,\mathrm{Tor}_n^R(M,N))$$'s

• Sorry, but I cannot see why $Tor_1^{\mathbb{Z}}(M,-)^G$ commutes with filtered colimit.
– Doug
Jun 2, 2022 at 11:03
• @DougLiu : It is because you can compute it as $H_1(P\otimes_\mathbb Z -)^G$ where $P$ is some (fixed) projective/flat resolution of $M$, and now it's a composite of functors which all preserve filtered colimits (homology, tensoring, and taking ordinary fixed points) Jun 2, 2022 at 11:10
• Although you're right that for this specific fact I'm using that $G$ is finite... Otherwise fixed points don't necessarily commute with filtered colimits Jun 2, 2022 at 11:13
• You can also require $G$ to be of type $FP_{\infty}$. I guess we can show finite group is $FP_{\infty}$ by using bar resolution.
– Doug
Jun 2, 2022 at 13:20
• @DougLiu : it's ordinary fixed points here, so in fact $G$ finitely generated should be enough Jun 2, 2022 at 13:32