# group action on Tor groups of modules and smash product

I am trying to understand theorem 3.4.2 from the paper "Bernstein-Gelfand-Gelfand complexes and cohomology of nilpotent groups over $\mathbf Z_{(p)}$ for representations with $p$-small weights" by Polo and Tilouine.

Here's the (part of the) setup relevant for my question. $S$ is a noncommutative ring with a two-sided ideal $I$. $\Lambda$ is a group of automorphisms of $S$ preserving $I$. They form the smash product $S \Lambda$ defined as the ring $S \otimes_{\mathbf Z} \mathbf Z [\Lambda]$ with multiplication operation $$(s_1 \otimes \lambda_1) (s_2 \otimes \lambda_2) = s_1 \lambda_1(s_2) \otimes \lambda_1 \lambda_2$$

Let now $N$ be a $S \Lambda$-module (*). Then Theorem 3.4.2 gives (under some technical assumptions and more setup which I think is not relevant for my question) a filtered complex spectral sequence of $\Lambda$-modules converging to $$\mathrm {Tor}^S_{q-p} \left( N, S / I \right)_p$$

My question: I do not understand how the structure of $\Lambda$-modules is defined on these $\mathrm {Tor}$-groups.

First, let's address the (*): the statement of the theorem assumes that $N$ is a left $S \Lambda$-module, but $N$ also needs to be a right $S$-module for the notation they use to make sense - indeed they mention "with notation as above" in the statement of the theorem, and part of the previous notation is that $N$ is a right $S$-module.

I think that there should be a typo and that $N$ is only a right $S \Lambda$-module (rather than a $(S \Lambda, S)$-bimodule) since in the applications later in the paper the $N$'s they consider are definitely not bimodules. So I am assuming from now on that $N$ is a right $S \Lambda$-module.

My first attempt at understanding the $\Lambda$-action on the $\mathrm {Tor}$ group was for $N \otimes_S S /I$, where I guessed that $$\lambda. (n \otimes s) = n.\lambda^{-1} \otimes \lambda(s)$$ On the other hand, this does not really explain how to define a $\Lambda$-action on higher $\mathrm {Tor}$-groups. These are the homology groups of the complex obtained by applying $N \otimes_S -$ to a resolution of $L_{\bullet} \longrightarrow S/I \longrightarrow 0$ by finite, free $S$-modules. There is no reason why the differentials in this free resolution should be $\Lambda$-equivariant, and hence I do not see how the $\Lambda$-action on $N \otimes L_i$ could factor through to an action on homology.

Alternatively, one can think of the change of rings isomorphism $$\mathrm {Tor}^S_n (N, S/I) \cong \mathrm {Tor}^{S \Lambda}_n \left(N, S \Lambda \otimes_S S/I \right)$$ and then try to define $\Lambda$-actions on the modules $N \otimes_{S \Lambda} L'_i$ coming from tensoring a finite, free resolution of $S \Lambda \otimes_S S /I$ by $S \Lambda$-modules.

The formula above now defines a trivial action since we can 'move' $\lambda$ across the tensor product, but on the other hand this is the only $\Lambda$-action I can come up with that behaves well with the $S$-action - in the sense that a (left) $S \Lambda$-module is an $S$-module with a $\Lambda$-action such that $\lambda(s.m) = \lambda(s). \lambda(m)$, so a naive formula such as $\lambda. (n \otimes s) = n. \lambda^{-1} \otimes s$ is not well-posed.

How is this $\Lambda$-action defined?