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 $M(L^\infty(G))$?

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    $\begingroup$ Is this being downvoted and closed because of misspellings? Or is this just a trivial question? In either case it would at least be courteous to the OP if somebody would comment to indicate why the question is about to be closed. $\endgroup$ – MTS Jun 28 '12 at 1:55
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    $\begingroup$ I'm a little puzzled that the question has been down voted twice. The question seems to be: is there a sensible action of $L^1(G)$ on $B(L^2(G))$, for which the diagonal copy of $L^\infty(G)$ is a sub module, and moreover looks like the usual module as defined in the question? That seems to me a question whose answer is well known to experts, but not immediately obvious, and as such seems reasonable for MO. $\endgroup$ – Yemon Choi Jun 28 '12 at 2:02
  • $\begingroup$ The short answer: for discrete groups you can do it using conjugation, I would have to think harder in the setting of general locally compact groups to make sure things like non-unimodularity aren't an issue $\endgroup$ – Yemon Choi Jun 28 '12 at 2:04
  • $\begingroup$ Dear Yemon Choi, Sorry for poor english. I want to know that if we can define a modular action of $L^1(G)$ on $M(L^\infty(G))$ as a submodule of $B(L^2(G))$? If yes, how? $\endgroup$ – user24572 Jun 28 '12 at 3:01
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    $\begingroup$ Please add these clarifying remarks to the question in a coherent way. $\endgroup$ – S. Carnahan Jun 28 '12 at 7:32

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.


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.

  • $\begingroup$ The same thing works on any locally compact group (probably just measurable group is enough). It shouldn't matter that the Haar measure is being used, so "unimodular" shouldn't matter. All that matters is that $d\mu(s)\equiv f(s)ds$ is a finite signed measure, and $S(\mu)(T)=\int \lambda_s^{-1}T\lambda_s\,d\mu(s)$. $\endgroup$ – George Lowther Jun 28 '12 at 19:48
  • $\begingroup$ George, certainly you can define the action for any LC group (in view of Weil's theorem I tend to elide measurable groups with LC ones) but I didn't have time - or, to be frank, much inclination - to check that the diagonal operators still form a submodule, and that the restriction of the conjugation action is still the translation action on $L^\infty(G)$. (Keeping track of the modular function is something I still occasionally trip over.) $\endgroup$ – Yemon Choi Jun 28 '12 at 20:59
  • $\begingroup$ BTW, the map you wrote down in your comment turns out to be an isometric homomomorphism from $M(G)$ to $B(B(L^2(G))$. I think this was first noticed for abelian groups by Stormer and then for all LC groups by Ghahramani, but I may be misremembering. $\endgroup$ – Yemon Choi Jun 28 '12 at 21:01

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