Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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.)

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.