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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!

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  • $\begingroup$ 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. $\endgroup$ – Aweygan Apr 18 '18 at 14:28
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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 $T\not\in C^*(S)$.

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  • $\begingroup$ This is actually a pretty awesome idea! Thanks a lot! $\endgroup$ – Verdruss Apr 18 '18 at 21:35
  • $\begingroup$ Glad I could help. Just in case you are not aware, $C^*(S)$ is the very well-known Toeplitz algebra. $\endgroup$ – Martin Argerami Apr 18 '18 at 21:47
  • $\begingroup$ Yes, but looking for "not Toeplitz operators" didn't really get me something. Although I must admit that i didn't look into characterizations of Toeplitz operators, so maybe one could find something there, too. $\endgroup$ – Verdruss Apr 18 '18 at 22:10
  • $\begingroup$ I don't know how to get an explicit operator that way either. But the fact that the Toeplitz algebra is "abelian subalgebra plus the compacts" gives you the idea that it is fairly small. $\endgroup$ – Martin Argerami Apr 18 '18 at 22:11
  • $\begingroup$ Yeah, that's true. I mean, I knew about the "mod out the compacts, than one thing is abelian, the other one not" trick, but i wanted a nice explicit example. That you so kindly gave to me :D $\endgroup$ – Verdruss Apr 18 '18 at 22:21

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