Commutant of a set of operators and norm topology.

Question

In the references I have it's remarked that the commutant $S'$ of a set $S$ in $B(H)$, where $H$ is a Hilbert space, is closed in the weak operator topology.

And this is true because if $\{A_\alpha\}_\alpha$ is a net in $S'$ converging to $A$ in the wot, then $\lim_\alpha\langle A_\alpha f,g \rangle=\langle A f,g \rangle$ for every $f,g$ in $H$. Thus if $B$ is in $S$ we have $\langle ABf,g\rangle=\lim_\alpha\langle A_\alpha B f,g \rangle = \lim_\alpha\langle BA_\alpha f,g\rangle = \lim_\alpha\langle A_\alpha f,B^\ast g\rangle = \langle A f,B^\ast g\rangle = \langle BA f, g\rangle$.

The commutant it is also closed in the strong operator topology because if $\{A_\alpha\}_\alpha$ is a net in $S'$ converging to $A$ in the sot: $\lim_\alpha A_\alpha f = Af$ for every $f$ in $H$. Thus if $B$ is in $S$ we have $ABf=\lim_\alpha A_\alpha B f = \lim_\alpha BA_\alpha f = BAf$.

I'm wondering what happens in the norm topology. If $\{A_\alpha\}_\alpha$ is a net in $S'$ converging to $A$ then $\lim_\alpha \|A_\alpha - A\|=0$ but it seems I can't conclude anything about $\|AB-BA\|$... or am I wrong?

Answer 1

$||AB-BA||=||AB-A_{\alpha}B+A_{\alpha}B-BA|| \leq ||AB-A_{\alpha}B||+||A_{\alpha}B-BA||$.

But $A_{\alpha}B=BA_{\alpha}$, and $||AB-A_{\alpha}B|| \leq ||B||||A-A_{\alpha}||$.

So $S'$ is closed in the norm topology as well.

A "high-level" explanation can be given:

$S'$ is a von-neumann algebra, and every von-neumann algebra is a $C^*$ algebra- and so $S'$ is norm closed.

Answer 2

Note the weak topology is weaker than the strong topology which in turn is weaker than the norm topology. So saying the commutant is WOT closed implies it is closed in the other two (this is in reference to the question in the comment by Benzio).