# Inclusion of $C^*(S)$ in $W^*(S)$ is proper

I have a little question concerning operator algebras.

Consider the unilateral shift $S\in \mathcal{B}(\ell^2)$ defined by $Se_n=e_{n+1}$, $e_i$ being the canonical orthonormal basis of $\ell^2$. Then take the $C^*$-algebra generated by $S$, that is the the completion of the algebra of all polynomials in $S$ and $S^*$ with respect to the operator norm, and the von-Neumann-Algebra $W^*(S)$ generated by $S$ which is the same but with respect to the strong operator topology.

It is easy to see that $C^*(S) \subset W^*(S)$ and that $W^*(S)$ is actually the whole space $\mathcal{B}(\ell^2)$. But is this inclusion proper? What would be a linear Operator that is not in $C^*(S)$? I tried to think of a property that all elements in $C^*(S)$ fulfill and then find a linear operator that does not have this property but i can't seem to find something fitting. Does someone know an example?

Thanks!

• Well I know it contains all finite-rank projections, hence all projections with finite-dimensional kernel. But what about projections with infinite-dimensional kernel and range? I imagine $C^*(S)$ might fail to contain a few of those. – Aweygan Apr 18 '18 at 14:28

The inclusion is "very" proper: $C^*(S)$ is separable, while $B(\ell^2)$ isn't.
To look explicitly for an operator not in $C^*(S)$, as Aweygan mentions, it is very likely that infinite projections $P$ with $I-P$ also infinite are not in $C^*(S)$. But I cannot come with a proof. So let's look somewhere else. Consider $$V_1=1,\ V_2=\begin{bmatrix} 0&1\\1&0\end{bmatrix},\ V_3=\begin{bmatrix} 0&0&1\\0&1&0\\1&0&0\end{bmatrix},\ \cdots$$ and form $$T=\bigoplus_n V_n$$ (that is, put the $V_n$ one after the other in the diagonal). The algebra $C^*(S)$ is the closed linear span of $\{S^nS^{*m}:\ n,m\in\{0\}\cup\mathbb N\}$. If $$\tag1 R=\sum_{j=0}^r \alpha_jS^{n_j}S^{*m_j}.$$ If we think of the matrix representation of $R$, all its entries will be zero outside of a certain diagonal band. Our operator $T$, on the other hand, has a $1$ as far from the diagonal as we want; thus $\|T-R\|\geq1$. This estimate will be preserved as we take limits of operators of the form $(1)$, so $T\not\in C^*(S)$.
• Glad I could help. Just in case you are not aware, $C^*(S)$ is the very well-known Toeplitz algebra. – Martin Argerami Apr 18 '18 at 21:47