Commutant built from the Unilateral Shift

Question

Let $T: \ell^2 \rightarrow \ell^2$ be the unilateral shift, and let $\mathcal{T}= \{ I, T, T^* \}$.

Show that $\mathcal{T}' = \{\lambda I_{\ell^2} : \lambda \in \mathbb{C} \}$ and $\mathcal{T}'' = B(\ell^2)$.

Attempt:

Since $T$ is the unilateral shift, we have that $Te_n = e_{n+1}$.

Also, by definition, we have $\mathcal{T}' = \{ S \in B(\ell^2) : ST =TS, \forall T \in \mathcal{T}\}$.

Let's look at the each element in $T$:

We know that the identity $I$ always has the property $IS=SI$ for $S \in B(\ell^2) $.

Also, let $S \in B(\ell^2)$. Then we have $STe_n= Se_{n+1} =e_{n+1 }S= Te_n S = TSe_{n}$.

For $T^*$, I know that $T^*e_n= e_{n-1}$, for $n>1$. So don't we get a similar result as above, i.e $ST^*e_n =T^* S e_n$?

So I'm a bit confused on how to get the results we want. I feel like I got the definitions of $\mathcal{T}'$ and $\mathcal{T}''$ mixed up somewhere...

Thank you for your help!

Answer 1

Observing that, for every $n\geq 1$, one has that $Te_n=e_{n+1}$, and $$ T^*e_n = \left\{\matrix{0, & \text {if } n=1, \cr e_{n-1}, & \text {if } n>1, }\right. $$ one quicky sees that $$ (I-TT^*)e_n = \left\{\matrix{e_1, & \text {if } n=1, \cr 0, & \text {otherwise.} }\right. $$ So we deduce that the operator $P$ defined by $$ P:=I-TT^* $$ is the orthogonal projection onto the space spanned by $e_1$.

If $S$ is any operator commuting with $T$ and $T^*$, then obviously $S$ also commutes with $P$, so $$ Se_1 = SPe_1 = PSe_1 = \lambda e_1, $$ for some scalar $\lambda $. We then claim that $S$ necessarily coincides with $\lambda I$, the reason being that for all $n>1$, $$ Se_n = ST^{n-1}e_1 = T^{n-1}Se_1 = \lambda T^{n-1}e_1 = \lambda e_n. $$

Answer 2

Hint: If an operator commutes with the the unilateral shift and its adjoint, then it commutes with the whole Toeplitz algebra. Furthermore, the Toeplitz algebra contains all the compact operators.

If you don't want to appeal to such facts, I would suggest showing by hand that the algebra generated by the unilateral shift and its adjoint contains every operator whose matrix with respect to the standard basis of $\ell^2(\mathbb{N})$ has only finitely many nonzero entries. Basically, figure out how to construct the matrix $E_{ij}$ which has a $1$ in the $(i,j)$ position and zeros elsewhere.

Answer 3

Let's rename $E=\{I,T,T^*\}$. If we show that $E'=\mathbb{C}I$, then the claim that $E''=\mathbb{B}(\ell^2)$ follows trivially, since $E''=(E')'=(\mathbb{C}I)'=\mathbb{B}(\ell^2)$.

Lemma: Let $H$ be a Hilbert space and $\mathbb{K}(H)$ denote the compact operators over $H$. Then $\mathbb{K}(H)''=\mathbb{B}(H)$.

Proof: $\mathbb{K}(H)$ is a $C^*$-algebra that has an approximate unit of the form $\{p_\lambda\}$ where $p_\lambda$ are finite rank projections and $p_\lambda\to\text{id}_{H}$ in the strong operator topology. In other words, the inclusion $\mathbb{K}(H)\subset\mathbb{B}(H)$ is a non-degenerate representation, so, by von Neumann's double commutant theorem we have that $$\mathbb{K}(H)''=\overline{\mathbb{K}(H)}^{sot}$$ But, in general, if $I$ is a closed, self-adjoint two sided ideal in $\mathbb{B}(H)$ then so is $\overline{I}^{sot}$: it is a $C^*$-algebra as we know that the strong closures of $C^*$-subalgebras are again $C^*$-algebras and it is an ideal since if $x\in \overline{I}^{sot}$ and $y\in\mathbb{B}(H)$ then we take a net $\{x_\mu\}\in I$ with $x_\mu\to x$ strongly and simply observe that $x_\mu y\in I$ for all $\mu$ and that $x_\mu y\to xy$ strongly, since multiplication is separately strongly continuous. Since $\mathbb{K}(H)$ is an ideal, its strong closure is an ideal and it contains the identity operator, since $p_\lambda\to\text{id}_{H}$. Thus it is equal to the entire algebra $\mathbb{B}(H)$, q.e.d.

Now it is elementary to check that $\mathbb{B}(\ell^2)'=\mathbb{C}I$, so $\mathbb{K}(\ell^2)'=\mathbb{K}(\ell^2)'''=\mathbb{B}(\ell^2)'=\mathbb{C}I$. It is also elementary to check that if $x\in E'$ then $x\in C^*(E)'$. So, if we show that $\mathbb{K}(\ell^2)\subset C^*(E)$, then $E'\subset C^*(E)'\subset\mathbb{K}(\ell^2)'=\mathbb{C}I$ and we will be done. Let's show this inclusion:

For $n\geq1$ define the projection $p_n\in\mathbb{B}(\ell^2)$ by $$p_n(\xi_1,\xi_2,\dots)=(\xi_1,\dots,\xi_n,0,0,\dots)$$ The projections $\{p_n\}$ form an approximate unit for $\mathbb{K}(\ell^2)$, so every $x\in\mathbb{K}(\ell^2)$ satisfies $x=\lim_{n\to\infty}p_nxp_n$. So it suffices to show that $p_nxp_n\in C^*(E)$ for all $x\in\mathbb{K}(\ell^2)$.

More generally, if $\{e_k\}$ is the canonical orthonormal basis of $\ell^2$, we define $y_{i,j}:\ell^2\to \ell^2$ by $y_{i,j}(\xi)=\langle \xi,e_j\rangle e_i$. Observe that $p_n=\sum_{i=1}^ny_{i,i}$. Moreover, $$p_1=y_{1,1}=I-TT^*\in C^*(E)$$ and also $$y_{i,j}=T^{j-1}\cdot p_1\cdot (T^*)^{i-1}\in C^*(E)$$ for all $i,j$. Now it is easy to verify that $p_n\cdot\mathbb{B}(\ell^2)\cdot p_n=\text{span}\{y_{i,j}: 1\leq i,j\leq n\}$, so $p_n\cdot\mathbb{B}(\ell^2)\cdot p_n\subset C^*(E)$ for all $n$. This shows that $p_nxp_n\in C^*(E)$ for all $x\in\mathbb{K}(\ell^2)$ and all $n\geq1$, so, since $C^*(E)$ is closed, we have that $x=\lim_{n\to\infty}p_nxp_n\in C^*(E)$ and the claim is proved.

Comment: My answer is basically the details to what Mike F describes in their answer. I am posting this because OP asks for "a more detailed answer".