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Let $H$ be a Hilbert space. In Murphy's book on operator algebras, the weak topology on $B(H)$ is the locally convex topology determined by the seminorms $u\mapsto\langle ux,y\rangle$ for all $x,y\in H$. The ultraweak topology is defined by the seminorms $u\mapsto|\text{tr}(au)|$ for the trace class operators $a\in L^{1}(H)$. One can show that if a net converges ultraweakly, then it converges weakly. Hence the ultraweak topology is stronger or finer than the weak topology. However, the word "ultra" is latin for something like "far beyond". I think this is inconsistent with how the ultraweak and weak topology relate.

To motivate my thoughts, there is also the strong operator topology on $B(H)$ which is induced by the seminorms $u\mapsto\|u(x)\|$ for $x\in H$. And indeed, the strong topology is stronger than the weak topology. So this terminology is actually consistent with the behaviour of the topologies.

Is there an explanation for why people call it ultraweak? I find this very confusing.

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  • $\begingroup$ Is there an ultra strong topology for which this is analogous to? $\endgroup$ – Alfred Yerger Apr 21 at 18:55
  • $\begingroup$ @AlfredYerger Honestly, I don't know. I haven't encountered it yet. I don't think it is in Muprhy's book. $\endgroup$ – Calculix Apr 21 at 18:57
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    $\begingroup$ There is indeed an ultrastrong topology. There is a nice diagram on page 12 of "Lectures on von Neumann algebras" by Stratila and Zsido (sorry for not accenting I have no clue how). The weak and strong topologies are weaker than the ultraweak and ultrastrong respectively, and the ultreaweak is stronger than weak, but weaker than ultrastrong. The ultrastrong coincides with the strong on the unit ball of B(H). (see en.wikipedia.org/wiki/Ultrastrong_topology). $\endgroup$ – PStheman Apr 21 at 23:39
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    $\begingroup$ Not really an answer, but I do recall a professor once saying in a lecture that weak vs. strong for topologies go the opposite ways when talking with topologists vs. functional analysts. I assume you want to know how we ended up this way, but it sounds like it's a known fact of life. Kind of like how physicists and mathematicians have a different default base for their logarithms. $\endgroup$ – JonathanZ supports MonicaC Apr 22 at 2:21
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I don't have an authoritative answer, but here is my take.

As you say, the weak (operator) topology is the locally convex topology given by the seminorms $$\tag1 u\longmapsto |\langle ux,y\rangle|,\ \ \ x,y\in H. $$ Equivalently, you can use the seminorms $$\tag2 u\longmapsto |\langle ux,x\rangle|,\ \ \ x\in H. $$

The ultraweak topology is the locally convex topology given by the seminorms $$\tag3 u\longmapsto \left|\sum_j\langle ux_j,x_j\rangle\right|,\ \ \text{ where } \sum_j\|x_j\|^2=1. $$ Because of $(3)$, a common (and more reasonable) name for the topology is $\sigma$-weak topology. Similar to $(3)$, the $\sigma$-strong (ultrastrong) topology is given by the seminorms $$\tag4 u\longmapsto \sum_j\|ux_j\|^2,\ \ \text{ where } \sum_j\|x_j\|^2=1.. $$ So maybe the prefix ultra means to imply that in $(3)$ you need to use more than a single for the seminorms in $(2)$ to define the topology. Another possibility is that it is meant to imply that the ultraweak and ultrastrong topologies, when considering a von Neumann algebra, are intrinsic (as opposed to the weak and strong).

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