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!
Best Answer
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 $\mathcal{W}\subseteq \mathcal{S}''$.