Von Neumann bicommutant theorem

Remarks

Every commutant is weakly closed, so 1 implies 2, and 2 implies 3 because the strong topology is finer. For 3 implies 1, let $T \in M''$ and $x \in H$. The orthogonal projection $P$ onto the closure of $Mx$ commutes with $M$ because $M$ is closed under adjoints, so $T$ commutes with $P$ and $Tx \in \overline{Mx}$, using $x = Ix \in Mx$. Applying this to the diagonal action of $M$ on $H^n$ handles $n$ vectors at once, which shows that $T$ is a strong limit of elements of $M$.

The identity is needed: the algebra $M = \set{0}$ is closed under adjoints and weakly closed, but $M' = B(H)$ and $M'' = \C I \ne M$. More generally, without the identity the theorem holds in the form $M'' = \overline{M}$ only when $M$ acts nondegenerately.

For example, on $\C^n$ the diagonal matrices $D$ satisfy $D' = D$, so $D'' = D' = D$, as the theorem predicts for a finite-dimensional, hence closed, algebra.

Going deeper

For a student

The theorem lets one produce von Neumann algebras for free. The commutant of any self-adjoint set $S$ of operators is one, and so is the bicommutant $S''$, the von Neumann algebra generated by $S$. It turns questions about topological closure into algebraic questions about commuting operators. This is how one shows, for example, that the spectral projections of a self-adjoint element of $M$ lie in $M$. Together with the Kaplansky density theorem, it is the basic tool for approximating elements of $M''$ by elements of $M$.