Let $G$ be a locally compact group. Then we can define a modular action of $L^1(G)$ on $L^\infty(G)$ by
$$
(f.u)(t)=\int f(s)u(st) ds
$$
and
$$
(u.f)(t)=\int f(s)u(ts) ds
$$
for $f\in L^1(G)$ and $u\in L^\infty(G)$.


On the other hand we know that $L^\infty(G)$ is a sub von Neumann algebra of $\mathcal{B}(L^2(G))$ by multiplication operatot, $f\mapsto M\_f$.


My question:


Can we transfer the modular action of $L^1(G)$ on $L^\infty(G)$ to $M(L^\infty(G))$?



modules over group algebras
This is by no means a complete answer, but just to get the ball rolling while I have a moment spare, let me write some things down. **However**, I strongly encourage you to add some background to your question, explaining why you are asking it, where you came across the problem, and so on. This seems like a basic courtesy when asking questions, although perhaps I am overly touchy on this issue.




---


Anyway.


Consider the case of $G$ discrete. For each $x\in G$ let $\lambda\_x \in VN(G) \subset B(\ell^2(G))$ be the usual operator of left translation. Define $S:\ell^1(G) \to B(B(\ell^2(G))$ by


$$ S(f)(T) = \sum\_{x\in G} f(x) \lambda\_x^{-1} T \lambda\_x \qquad (f\in\ell^1(G), T\in B(\ell^2(G)). $$


Then you can check that $S$ defines a right action of $G$ on $B(\ell^2(G))$. Moreover, if $u\in \ell^\infty(G)$ and $M\_u\in B(\ell^2(G))$ is the corresponding diagonal multiplication operator, a direct calculation shows that
$$ S(\delta\_x)(M\_u) = M\_{x\cdot u}$$
where $(x\cdot u)(t) = u(xt)$. So $M(\ell^\infty(G))$ becomes a right sub-module for this action, and as a right module it is isomorphic to $\ell^\infty(G)$ with the right action you defined at the start.


A similar argument, using the right von Neumann algebra rather than the left one, will get you a left action of $G$ on $B(\ell^2(G))$ for which $M(\ell^\infty(G))$ is again a submodule, isomorphic to $\ell^\infty(G)$ with the left action that you defined at the start.


(IIRC, the first place I saw this representation was in an article of Bunce, where it is used in the proof that for $G$ discrete, amenability of $C\_r^\*(G)$ implies $G$ is amenable. I suspect it was already part of the experts' folklore by then, so I don't know who first noticed this. Similar ideas are surely there in the purely algebraic setting, this is one of the canonical ways to make End(V) into a G-module when V is a G-module.)


I think the argument should go through without difficult for unimodular locally compact groups, so I leave it to you. In the non-unimodular case, I am not sure without further checking.


