# Weak closure of subalgebra generated by an operator and its adjoint

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.

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 $$T = \lambda I$$.)
Yes, just take the unilateral shift on $$H=\ell^2({\bf N})$$. The norm closed ($${\rm C}^*$$-) algebra generated by this shift and its adjoint is the Toeplitz algebra, which contains all compact operators on $$H$$. Hence the WOT closure of the original $$*$$-algebra contains the WOT closure of $$K(H)$$, which is all of $$B(H)$$.