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Caution: Written while trying to turn some vague confusion I was having into precise questions. A bit long and rambling. An appropriate way to answer may just be to point me toward references.

Let $G$ be a discrete group with identity $e$ and let $H$ be the Hilbert space $\ell^2(G)$. Let $g \mapsto \lambda_g$ and $g \mapsto \rho_g$ be the left and right regular representations of $G$ on $H$. These are determined on the canonical orthonormal basis $(\delta_x)_{x \in G}$ by the formulae: $$ \lambda_g \delta_x = \delta_{gx} \\ \rho_g \delta_x = \delta_{xg^{-1}} $$ Let $W$ be the set of operators $T \in B(H)$ such that $$\rho_x T \delta_x = T \delta_e$$ for all $x \in G$. Notice an operator $T \in W$ is completely determined by the vector $T \delta_e$ since the formula $T \delta_x = \rho_x^* T \delta_e$ tells $T$ what to do on an orthonormal basis. In fact, if $\eta = \sum_{x \in G} \eta_x \delta_x$ is any element of $\ell^2(G)$ and $T \in W$ has $T \delta_e = \xi = \sum_{x \in G} \xi_x \delta_x$, then $$ T\eta = \sum_{s \in G} \eta_s \rho_s^* \xi = \sum_{s \in G} \eta_s \sum_{t \in G} \xi_t \rho_s^* \delta_t = \sum_{t,s \in G} \xi_t \eta_s \delta_{ts} = \xi * \eta \in \ell^2(G)$$ where $\xi * \eta$ is the standard convolution product.

Remark: Concerning the above convolution, when $\xi, \eta \in \ell^2(G)$, we will have $\xi * \eta \in c_0(G)$. It is not generally the case that $\xi * \eta \in \ell^2(G)$.

It is simple to see that each of the identities $\rho_x T \delta_x = T \delta_e, x\in G$ which define $W$ is linear and strongly continuous in $T$. Thus $W$ is strongly closed subspace of $B(H)$.

Question 1: Is $W$ a von Neumann algebra?

Question 2: If $\xi \in \ell^2(G)$ is such that $\xi * \eta \in \ell^2(G)$ for all $\eta \in \ell^2(G)$, does $T \eta = \xi * \eta$ define a bounded operator (which is then in $W$)? Loosely, can $W$ be identified with the set of "convolvers" in $\ell^2(G)$?

To motivate the definition of $W$, note that, if $T \in B(H)$ commutes with the right representation of $G$ on $H$, then $$ \rho_x T \delta_x = T \rho_x \rho_x^* \delta_e = T \delta_e$$ so $T \in W$. In particular, since the left and right representations commute, $W$ contains the reduced group algebra $C^*_r(G)$ which we can define as the norm closure of the linear span of $\lambda(G)$.

Question 3: Does $W$ equal the strong-operator closure of $C^*_r(G)$?

Question 4: Does $W$ equal the commutant of $\rho(G)$? (it is clearly larger).

Thanks for reading.

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up vote 2 down vote accepted

It is not true in general that $\xi * \eta \in \ell^2(G)$ when $\xi, \eta \in \ell^2(G)$. There are easy examples to find when $G = \mathbb Z$.

The answer to all four of your questions is yes.

The fourth (and hence also first) question you basically answer yourself noting that $T \in W$ is completely determined by $T \delta_e$. From this it is not hard to see that $W = \{ \rho(G) \}'$: $T \rho(g) \delta_x = T \delta_{xg^{-1}} = \rho(gx^{-1}) T \delta_e = \rho(g) T \delta_x$, for each $x, g \in G$.

The easiest way to prove your second question is to first show that convolvers give rise to operators with closed graphs, and then apply the closed graph theorem.

Question 3 basically follows from the other questions. By symmetry from above we have that $\{ \lambda(G) \}'$ is the space of right convolvers $R_\xi \eta = \eta * \xi$. But left convolution commutes with right convolution hence $\{ \rho(G) \}' \subset \{ \lambda(G) \}'' \subset \{ \rho(G) \}''' = \{ \rho(G) \}'$. (Note: $\{ \lambda(G) \}''$ is the strong operator closure of $C_r^*(G)$ by von Neumann's double commutant theorem.)

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Thanks a lot! This is good timing because I was just returning to these sorts of questions. – Mike F Jul 15 '12 at 21:55

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