Timeline for Is the sigma-strong topology generated by bounded sets?
Current License: CC BY-SA 4.0
15 events
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| May 12, 2023 at 8:11 | vote | accept | Sebastian Meyer | ||
| May 11, 2023 at 13:29 | comment | added | Yemon Choi | Hilbert spaces and modules have an $\ell^2$-flavour, and in particular $\ell^2(\ell^2) \cong \ell^2$, so that is why the sigma-strong definition works well in the original setting. Banach spaces can look very different from $\ell^2$ so while there is nothing "incorrect" with the definition you proposed, it doesn't seem intrinsic to the original $V$ | |
| May 11, 2023 at 12:36 | comment | added | Sebastian Meyer | @Yemon Choi Thank your for pointing it out. I have not noticed that this definition was not made for a general Banach space. However, I have seen this definition for operators on Hilbert modules. There, this definition is meaningful, because only the sigma-strong- (or sigma-strong-*) topology on the bounded operators is (to some extend) independent of the underlying space. I think it was in B. Blackadar: Operator Algebras. To find out if this is a natural definition on Banach spaces was also the aim of my question. | |
| May 8, 2023 at 14:02 | comment | added | Narutaka OZAWA | The second sentence in my previous comment is still confusing... | |
| May 8, 2023 at 13:17 | comment | added | Yemon Choi | Can you give a reference that defines this sigma-strong topology for Banach spaces other than $\ell_2$? It does not look natural for general Banach spaces. | |
| May 8, 2023 at 3:19 | answer | added | Narutaka OZAWA | timeline score: 6 | |
| May 8, 2023 at 0:04 | comment | added | Narutaka OZAWA | I confused myself. What I had in mind was an unbounded convergent net consisting of countable elements. In any case, uniform convergence on compact subsets is strictly stronger than $\sigma$-strong (e.g., on $B(\ell_2)$). The Mackey topology, as Onur Oktay suggests, on $B(\ell_2)$ is the strongest among the loc convex topologies that coincide with the $\sigma$-strong$^\ast$ topology on bounded subsets. | |
| May 5, 2023 at 10:34 | comment | added | Sebastian Meyer | @Narutaka OZAWA Sorry for the late respond. When I am not mistaken, then the description of bounded strictly converging nets is not inherited by a topology as the product $B(V) \times V \to V$ (take the norm topology on $V$) would be continuous and this implies that we already have the operator norm topology on $B(V)$. However, there is a convergent, unbounded net in $l^2(\mathbb{N})$ if that helps you. | |
| May 5, 2023 at 10:12 | history | edited | Sebastian Meyer | CC BY-SA 4.0 |
a definition added
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| Apr 21, 2023 at 7:20 | comment | added | Narutaka OZAWA | I don't know what ultrastrong topology means in the case $V$ is not a Hilbert space. However, the answer is clearly NO if that topology admits an unbounded convergent net, because one can strengthen the topology by insisting that every convergent net has to be bounded. Perhaps, in the present case, this topology would be given by the uniform convergence on compact subsets. | |
| Apr 20, 2023 at 15:56 | comment | added | terceira | A reference: for the Hilbert space case, you might want to consult the fourth chapter of the monograph "Saks Spaces and Applications to Functional Analysis". | |
| Apr 20, 2023 at 14:03 | history | edited | Sebastian Meyer | CC BY-SA 4.0 |
deleted 122 characters in body; edited tags
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| Mar 29, 2023 at 20:44 | comment | added | Onur Oktay | I do not know the answer for exactly which $V$ the Mackey topology and $\sigma$-strong$^*$ topology coincide on bounded sets on $B(V)$ generally. They do however when $V$ is a Hilbert space, when $V=\ell^p$ $1<p<\infty$. Let's note that Mackey top. is stronger than $\sigma$-strong$^*$ top., and $\sigma$-strong$^*$ top. is stronger than $\sigma$-strong top. on $B(V)$. | |
| Mar 29, 2023 at 20:03 | comment | added | Onur Oktay | For the general case, please search the literature/textbooks for the Mackey topology. For the second case, the multiplier algebra is a unital $C^*$-algebra. On unital $C^*$-algebras, the Mackey topology and the $\sigma$-strong$^*$ topologies coincide on bounded sets. | |
| Mar 29, 2023 at 13:07 | history | asked | Sebastian Meyer | CC BY-SA 4.0 |