Reflexive operator algebra

Remarks

Always $\mathcal{A} \subseteq \operatorname{Alg} \operatorname{Lat} \mathcal{A}$, so reflexivity is the reverse inclusion. A reflexive algebra contains the identity and is closed in the weak operator topology, because each $\operatorname{Alg} \mathcal{L}$ has these properties. Example: $\mathcal{A} = B(H)$ is reflexive, since $\operatorname{Lat} B(H) = \set{\set{0}, H}$ and $\operatorname{Alg} \set{\set{0}, H} = B(H)$. Non-example: in $M_2(\C)$, let $\mathcal{A}$ be the matrices $\begin{pmatrix} a & b \\ 0 & a \end{pmatrix}$. Its invariant subspaces are those of $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, namely $\set{0}$, $\C e_1$ and $\C^2$, so $\operatorname{Alg} \operatorname{Lat} \mathcal{A}$ is the algebra of all upper triangular matrices, which is strictly larger. Every von Neumann algebra is reflexive: its invariant subspaces are the ranges of the projections in its commutant, and the bicommutant theorem gives the equality.