Let $H$ be a separable, complex Hilbert space and let $\mathcal{B}(H)$ denote the algebra of bounded linear operators on $H$. Let $T \in \mathcal{B}(H)$. We define $$ A = \{ p(T,T^\*) : p \in \mathbb{C}[z\_1,z\_2] \}.$$ $A$ is the subalgebra generated by $T,T^\*$ - or the $\*$-algebra generated by $T$. If we denote by $\overline{A}^\sigma$ its weak (operator) closure, I would like to know whether can the equality $$\overline{A}^\sigma = \mathcal{B}(H)$$ holds.


Thank you very much.



Weak closure of subalgebra generated by an operator and its adjoint
Yes, the unilateral shift $S$ on $l^2(\mathbb{N})$ generates $B(l^2(\mathbb{N}))$ as a von Neumann algebra. This is a consequence of the double commutant theorem and the fact that the only bounded operators which commute with both $S$ and $S^\*$ are scalars.


(To see this, suppose $T$ commutes with both $S$ and $S^\*$. Then $S^\*Te\_0 = TS^\*e\_0 = 0$ so $Te\_0$ must be a scalar multiple of $e\_0$, say $Te\_0 = \lambda e\_0$. Then $Te\_n = TS^ne\_0 = S^nTe\_0 = \lambda e\_n$ for all $n$, and therefore $T = \lambda I$.)


