Let $\mathcal H, \mathcal K$ be Hilbert spaces and $M \subseteq B(\mathcal H)$ a (concrete) von Neumann algebra (Here, $B(\mathcal H)$ denotes the algebra of bounded operators on $\mathcal H$). Furthermore, let $$\pi \colon M \to B(\mathcal K)$$ be a *-Homomorphism. Note that $\pi$ is automatically a contractive, i.e. especially bounded, operator.

I want to prove that $\pi$ is strong-strong (SOT-SOT) continuous if and only if $\pi$ is weak-weak (WOT-WOT) continuous.

strong-strong means that $B(\mathcal H)$ as well as $B(\mathcal K)$ carry the strong operator topology (SOT). Same goes for the term weak-weak and the weak operator topology (WOT).

I know that the topological duals of $B(\mathcal H)$ w.r.t the strong/weak operator topologies coincide, i.e. a linear functional on $B(\mathcal H)$ is strongly continuous if and only if it is weakly continuous. The same property is inherited by the sub-von Neumann Algebra $M$.

Do you know how to prove this or can you share a reference on this matter?

EDIT. I have asked another more general question which could yield an answer to this concrete problem. See here. However, I am not conviced that there is a closed-graph theorem for the strong and weak operator topologies.

2nd EDIT (06.04.2017)

I think one can prove WOT-WOT implies SOT-SOT in a straigthforward way:

It suffices to prove that if $x_i \overset{\mathrm{SOT}}{\to} 0$ then $\pi(x_i) \overset{\mathrm{SOT}}{\to} 0$.

We know that $$x_i \overset{\mathrm{SOT}}{\to} 0 \iff x_i^* x_i \overset{\mathrm{WOT}}{\to} 0 \tag{$\ast$}.$$

Now if $x_i \overset{\mathrm{SOT}}{\to} 0$, since $\pi$ is WOT-WOT continuous, ($\ast$) gives that $$\pi(x_i)^* \pi(x_i) = \pi(x_i^* x_i) \overset{\mathrm{WOT}}{\to} 0$$ which, again by ($\ast$) implies that $\pi(x_i) \overset{\mathrm{SOT}}{\to} 0$.


1 Answer 1


I don't know a reference for your exact assertion, but you may want to look at 7.1.14-7.1.16 in volume 2 of Kadison-Ringrose.

  • $\begingroup$ Thank you. As you said, this is not the exact assertion. My lecture notes say that my statement is true due to the fact that SOT and WOT create the same topological dual space... Do you know if there may be a more general principle at work? $\endgroup$
    – el_tenedor
    Feb 6, 2017 at 7:06
  • $\begingroup$ I have asked a similar question, which would help with this one. You can take a look if you find the time: math.stackexchange.com/questions/2131594 $\endgroup$
    – el_tenedor
    Feb 6, 2017 at 10:14
  • $\begingroup$ I have found a proof for one implication (posted it under "2nd edit"). Maybe you have an idea for the other implication? $\endgroup$
    – el_tenedor
    Apr 6, 2017 at 10:35

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