Von Neumann Algebra built from the bilateral shift

Question

Let $T: \ell^2(\mathbb{Z}) \rightarrow \ell^2(\mathbb{Z})$ be the bilateral shift, and let $S= \{ I, T, T^* \}$.

Why do we have $S'= S'' = \{ W\in B(\ell^2(\mathbb{Z})) : \langle e_n, We_k \rangle = \langle e_{n-k}, We_0 \rangle\} $?

I know that if we think of it in terms of matrices then this is $w_{{n+j},{k+j}} = w_{n,k}$. But I still don't really see what's going on here. I wonder if there is way to see this with out the use of matrices.

Thank you in advance!

Answer 1

Let $\mathcal{W}:=\{W\in B(H):\langle e_n,We_k\rangle=\langle e_{n-k},We_0\rangle\}$.

Then $\mathcal{S}'=\mathcal{W}$.
Proof: If $W\in\mathcal{S}'$, that is, $WT=TW$, then $$\langle e_n, We_k\rangle=\langle e_n,WT^ke_0\rangle=\langle e_n,T^kWe_0\rangle=\langle T^{-k}e_n,We_0\rangle=\langle e_{n-k},We_0\rangle$$ Conversely, if $W\in\mathcal{W}$, then $$\langle e_n,T^{-1}WTe_k\rangle=\langle e_{n+1},We_{k+1}\rangle=\langle e_{n-k},We_0\rangle=\langle e_n,We_k\rangle$$ Since this is true of all $n,k$, then $T^{-1}WT=W$, so $W\in\mathcal{S}'$.

Since $T$ and $T^*=T^{-1}$ commute, it follows that $\mathcal{S}\subset\mathcal{S}'$ and thus $\mathcal{S}''\subseteq\mathcal{S}'$.

To show $\mathcal{S}''=\mathcal{W}$, let $W,V\in\mathcal{S}'$, and let $We_0=\sum_i\alpha_ie_i$. Then $W^*e_0=\sum_i\alpha_i^*e_{-i}$ since $$\langle W^*e_0,e_n\rangle=\langle e_0,We_n\rangle=\langle e_{-n},We_0\rangle=\alpha_{-n}^*$$

Then since $\mathcal{W}$ is an algebra $$\langle e_n,VWe_k\rangle=\langle e_{n-k},VWe_0\rangle=\sum_i\alpha_i^*\langle e_{-i},Ve_{k-n}\rangle=\langle W^*e_0,Ve_{k-n}\rangle=\langle e_n,WVe_k\rangle$$ so $VW=WV$ and thus $\mathcal{W}\subseteq \mathcal{S}''$.

Answer 2

A more suggestive way of describing the von Neumann algebra you are dealing with might be "all the bounded operators whose matrices with respect to the standard basis of $\ell^2(\mathbb{Z})$ are constant along diagonals". Note that the matrix of the bilateral shift itself has this property; it has one diagonal consisting of $1$s and the rest of the diagonals consist only of $0$s.

To avoid working with matrices, you can apply the Fourier transform. Then you are working on $L^2(\mathbb{T})$, where $\mathbb{T} \subseteq \mathbb{C}$ is the unit circle, and the generators become the coordinate functions $z$ and $\overline z$. The $C^*$-algebra generated by those functions is the commutative algebra $C(\mathbb{T})$ of continuous functions on the circle, represented as multiplication operators. The weak-star closure must also be commutative and indeed you get $L^\infty(\mathbb{T})$, likewise represented as multiplication operators.

The dual perspective is probably not that important at the moment because one is mainly interested in the von Neumann algebras of noncommutative groups, and the harmonic analysis of noncommutative groups is much more complicated. The matrix perspective is the right place to start, I think.