4
$\begingroup$

Let $H$ be a Hilbert space, and $B(H)$ its algebra of bounded operators. One of the reasons the Ultraweak topology is (in a way) more useful than the weak operator topology is that the Ultraweak topology preserves the crucial property of having a compact unit ball while being finer, so we have more to work with.

I think based on this, a natural question is to ask for objects like this that are maximal. So precisely, I would be asking for maximally fine locally convex topologies on $B(H)$ that are finer than the weak operator topology.

First, the Ultraweak topology is indeed such an object. It is the weak*-topology induced by the trace class operators, and by this MO post showing that weak* topologies are maximally fine like this, we are done.

Note that any such topology would have to agree on the unit ball with the WOT and the ultraweak topology, as a compact topology being finer than a Harsdorf one means they are equal.

So the question is, what are some other maximal objects like this (if any other even exist). Are any of them interesting? Could any of them be useful when our familiar topologies would not?

$\endgroup$
2
  • 3
    $\begingroup$ Perhaps the finest topology on the whole space which agrees with the weak or ultraweak one on the ball (or the finest l.c. topology which is the same). It has the same dual as the ultraweak topology (the nuclear operators) and has the advantage that it is complete. It can also be described as the topology of uniform convergence on the compact subsets of the latter. $\endgroup$
    – terceira
    Commented Mar 8, 2023 at 11:16
  • $\begingroup$ I should correct a mistake in the above comment. The following topologies on the space of operators coincide: 1. the finest locally (or even linear) topology which agrees with either of the two weak ones on the unit ball; 2. the finest topology which agrees with them on the dilations of the ball. $\endgroup$
    – terceira
    Commented Mar 12, 2023 at 16:28

0

You must log in to answer this question.