Equivariant maps of "higher order"

Question

Given a group $G$, a ring $R$ and two $R[G]$-modules $M,N$. Then one can consider $Hom_R(M,N)$ and define inductively submodules $A_0,A_1,...$ via

$A_0:=0$

$A_{n+1}:=\{ \;f\; |\; \forall\; g\in G: (x\mapsto f(gx)-gf(x))\in A_n\}$

$A_1$ is then precisely $Hom_{R[G]}(M,N)$. So I am wondering, what the modules $A_2,..$ might be good for. A guess is, that this should be related to group cohomology in some way (, which I don't see).

For example with $G=\mathbb{Z}$ with generator $t$ and $M:=\mathbb{Z}[G],N:=\mathbb{Z}$ (with trivial $G$ action) one gets: $A_n=\{f \in Hom_\mathbb{Z}(M,N) | $ There are $ a_0,...,a_n \in \mathbb{Z} : f(t^n) =\sum_{i=0}^n a_i n^i\}$.

Answer 1

Let $\Lambda=R[G]$. Consider $L=Hom_R(M,N)$ as a $\Lambda$-module with the "conjugation" action: $(gf)(x)=gf(g^{-1}x)$. Let $I\subset \Lambda$ be the kernel of the augmentation $\Lambda\to R$, i.e. the ideal generated by $1-x$ for all $x\in G$. Then $A_n$ is the largest submodule of $L$ annihilated by $I^n$, or in other words $A_n=Hom_{\Lambda}(\Lambda/I^n,L)$.

I don't know what it's good for.

Answer 2

As you have not asked a question, I don't have to give any answer :-))

The group is irrelevant, you can do all the same with a group algebra $A=RG$, and then forget about group algebra and talk about any $R$-algebra $A$. If both $A$ and $R$ are commutative, then you have reinvented differential operators from $M$ to $N$. Congratulations! Needless to say that they are useful.

If $A$ or $R$ are no longer commutative then it is some noncommutative version of differential operators, whose usefulness is greatly diminished...