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 $\text{Tor}_i(M,N)^G.$


1 Answer 1


$\newcommand{\Tor}{\mathrm{Tor}_1^\mathbb Z}$ 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 $x=0$.

So this was to adress your example; more generally :

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

  • $\begingroup$ Sorry, but I cannot see why $Tor_1^{\mathbb{Z}}(M,-)^G$ commutes with filtered colimit. $\endgroup$
    – Doug
    Jun 2, 2022 at 11:03
  • $\begingroup$ @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) $\endgroup$ Jun 2, 2022 at 11:10
  • $\begingroup$ 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 $\endgroup$ Jun 2, 2022 at 11:13
  • $\begingroup$ 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. $\endgroup$
    – Doug
    Jun 2, 2022 at 13:20
  • $\begingroup$ @DougLiu : it's ordinary fixed points here, so in fact $G$ finitely generated should be enough $\endgroup$ Jun 2, 2022 at 13:32

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