I was reading about the weak operator topology in wiki, and I saw that: Norm-bounded sets in $B(H)$ are pre-compact in WOT.

I was wondering: Is there an analogous statement for von Neumann algebras? $C^*$-algebras?


1 Answer 1


The statement would have been that norm bounded sets are pre-compact, as being bounded does not guarantee that it is closed.

Other than that, yes. You would usually see a von Neumann algebra $M$ as embedded in $B(H)$. So a wot-closed bounded set is a closed subset of a compact set (a closed ball in $B(H)$) and thus compact.

In the case of a C$^*$-algebra, you would have to embed it in some $B(H)$ to even have a wot topolgy. In that case, bounded sets would be pre-compact.

  • $\begingroup$ By pre-compact, you mean has compact closure? $\endgroup$
    – chhro
    Mar 10, 2021 at 19:08
  • $\begingroup$ Yes. $ \ \ \ \ $ $\endgroup$ Mar 10, 2021 at 19:09
  • $\begingroup$ How about the following? Suppose $S\subset M$ is norm-bounded. Since $M=(M_*)^*$ (an isometric isomorphism), the unit ball in $M$ is $\sigma(M,M_*)$-compact (not sure if I'm saying it right). Then $S$ has $\sigma(M,M_*)$-compact closure. $\endgroup$
    – chhro
    Mar 10, 2021 at 19:21
  • $\begingroup$ Yes. That's the classic Banach-Alaoglu Theorem. $\endgroup$ Mar 10, 2021 at 19:40
  • $\begingroup$ $M$ is a von Neumann algebra in the argument I gave. In the argument you gave, the WOT for $M$ is induced by the containment $M\subseteq B(H)$. You get a non-equivalent topology if you use a different $B(K)$, no? $\endgroup$
    – chhro
    Mar 10, 2021 at 19:45

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