How generalize the bicommutant theorem?

Question

Let $H$ be an infinite dimensional separable Hilbert space.

Bicommutant theorem : Let $\mathcal{S}$ be $*$-subset of $B(H)$, then $\mathcal{S}''$ is the strong closure $\overline{\langle \mathcal{S} \rangle}$ of the algebra $\langle \mathcal{S} \rangle$ generated by $\mathcal{S}$ (called the von Neumann algebra generated by $\mathcal{S}$).

• Now if $\mathcal{S}$ is a non-selfadjoint subset of $B(H)$, what can we say about $\mathcal{S}''$ ?

• Is it true that $\overline{\langle \mathcal{S} \rangle} \subset \mathcal{S}''$ ?

• What are the simplest known subsets $\mathcal{S}$ with $\overline{\langle \mathcal{S} \rangle} \subsetneq \mathcal{S}''$ ? Single operator subsets ?

How generalize the bicommutant theorem ?

Answer 1

First note, e.g. by net manipulations, that for any set $\mathcal{S}\subseteq B(H)$, the bicommutant $\mathcal{S}^{''}$ is a wot/sot-closed unital subalgebra of $B(H)$. So we want to consider the wot/sot-closures of the algebra $\langle \mathcal{S}\rangle$ of all polynomials in elements of $\mathcal{S}$. Since $\langle \mathcal{S}\rangle\subseteq \mathcal{S}^{''}$, and since the wot and sot closures of a convex set coincide, we always have $$ \overline{\langle \mathcal{S}\rangle}^{sot}=\overline{\langle \mathcal{S}\rangle}^{wot}\subseteq \mathcal{S}^{''}. $$ The bicommutant property is when the equality holds. I will only consider the case of a single operator $S\in B(H)$, i.e. $\mathcal{S}=\{S\}$. Then $\langle\mathcal{S}\rangle =\mathbb{C}[S]$ is the algebra of all polynomials in $S$.

• In finite dimension, we always have $\mathbb{C}[S]=\{S\}^{''}$.

• In the infinite-dimensional case, there are quite a few examples of non-normal operators $S$ for which the double commutant property $\overline{\mathbb{C}[S]}^{wot}= \{S\}^{''}$ holds. I'll just mention the unilateral shift operator (Brown-Halmos, "Algebraic properties of Toeplitz operators", Crelle 1964). See also the paper "Contractions with the bicommutant property", Proc.AMS 1985, by Takahashi. And see also, indeed, the recent paper by Marcoux and Mastnak for possibly non-singly-generated algebras with the bicommutant property.

• From now on, let us restrict to the case where $S$ is normal, in infinite dimension. Note that by Fuglede's theorem, the commutant of $\{S\}$ coincides with the commutant of the self-adjoint set $\{S,S^*\}$. Therefore, by von Neumann bicommutant theorem, $$\{S\}^{''}=\{S,S^*\}^{''}=\overline{\mathbb{C}[S,S^*]}^{wot}$$ is a von Neumann algebra. So $\overline{\mathbb{C}[S]}^{wot}= \{S\}^{''}$ if and only if $\overline{\mathbb{C}[S]}^{wot}$ is a $*$-algebra. The next key point is due to Sarason, "Invariant subspaces and unstarred operator algebras", Pacific JM 1966 (corollary p.511): $\overline{\mathbb{C}[S]}^{wot}$ is a $*$-algebra if and only if $S$ is reductive, i.e. every closed invariant subspace $F$ of $S$ is a reducing subspace (i.e. $F^\perp$ is invariant under $S$ as well or, equivalently, $F$ is invariant under $S^*$). In finite dimension, $S$ is reductive if and only if $S$ is normal. Recall that the wot/sot are not metrizable in infinite dimmension (although they are metrizable on norm bounded sets when $H$ is separable): in the 1952 Proc. AMS paper "On invariant subspaces of normal operators", Wermer constructed an example of reductive normal operator $S\in B(H)$ such that $S^*$ does not belong to the wot sequential closure of $\mathbb{C}[S]$, although it lies in the wot closure, and in sharp contrast with the finite-dimensional case where $S^*$ is a polynomial in $S$ for every normal operator $S$.

• Of course, every bounded (skew-)self-adjoint operator operator is reductive. So these give other examples where the bicommutant property holds. Now we finally reach what could be the simplest counterexample: the bilateral shift $$ S:\ell^2(\mathbb{Z})\longrightarrow\ell^2(\mathbb{Z})\qquad S:e_n\longmapsto e_{n+1} $$ is unitary, whence normal, and not reductive as $\ell^2(\mathbb{N})$ is $S$-invariant, but not $S^*$-invariant. Therefore $$\overline{\mathbb{C}[S]}^{wot}\subsetneq\{S\}^{''}.$$

Answer 2

Here is a preprint (Marcoux, Mastnak 2012) : Non-selfadjoint double commutant theorem.

I asked the authors for posting an answer here. While waiting here's the abstract of their article:

Abstract. The von Neumann Double Commutant Theorem states that if $\mathcal{N}$ is a unital and selfadjoint subalgebra of the set $B(H)$ of all bounded linear operators acting on a Hilbert space $H$, and if $\mathcal{N}':=\{T \in B(H) :TN=NT \text{ for all} N \in \mathcal{N} \}$ denotes the commutant of $\mathcal{N}$, then $\mathcal{N}'' = \overline{\mathcal{N}}$. In this paper, we begin the analysis of not necessarily selfadjoint subalgebras $\mathcal{S}$ of $B(H)$ whose second commutant $\mathcal{S}''$ agrees with $\mathcal{S}$. More specically, we examine the case where $\mathcal{S}= \mathcal{D}+ \mathcal{R}$, where $\mathcal{R}$ is a bimodule over a masa $\mathcal{M}$ in $B(H)$ and $\mathcal{D}$ is a unital subalgebra of $\mathcal{M}$.

If the authors don't respond to my call, I will take the time to browse their article in order to post here a more appropriate response.

Answer 3

What are the simplest known subsets S with ⟨S⟩⊊S′′

Since I was just considering this: A finite dimensional example (that works over all fields) of this going wrong (and probably the easiest example in general) is given by the algebra $T$ of upper triangular matrices in $M_n(K)$. It is an easy exercise to show that the commutant of this are just the scalar multiples of the identity and therefore $T''=M_n(K)$.