Timeline for Strict topology on the multiplier algebra
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Aug 3, 2021 at 17:52 | history | bounty ended | Andromeda | ||
| S Aug 3, 2021 at 17:52 | history | notice removed | Andromeda | ||
| Aug 3, 2021 at 17:52 | vote | accept | Andromeda | ||
| Aug 2, 2021 at 15:45 | answer | added | Jamie Gabe | timeline score: 1 | |
| Aug 2, 2021 at 15:27 | comment | added | Andromeda | @JamieGabe Thank you. I didn't know this was true. In Lance's book on Hilbert C*-modules, the reader needed to use density of $A^*A$ into $A$ to prove this isomorphism, which gave me a false impression. If you want, you can make your comments into an answer. | |
| Aug 2, 2021 at 15:23 | comment | added | Jamie Gabe | Yes, and this is true, any element in a C*-algebra is a product of two elements. For instance, if $a\in A$ and $a = u|a|$ is the polar decomposition in the enveloping von Neumann algebra $A^{\ast \ast}$, then it is not hard to show that $a_1 := u|a|^{1/2} \in A$ (even though $u$ is not necessarily in $A$), and $a = a_1 |a|^{1/2}$, so $a$ is the product of two elements in $A$. | |
| S Aug 2, 2021 at 15:15 | history | bounty started | Andromeda | ||
| S Aug 2, 2021 at 15:15 | history | notice added | Andromeda | Draw attention | |
| Aug 2, 2021 at 14:56 | comment | added | Andromeda | @JamieGabe I think that is false? For example, what is $\theta_{a,b}+ \theta_{c,d}$? The isomorphism $A \cong \mathcal{K}(A)$ is given by $a^*b \mapsto \theta_{a,b}$. From your claim, it would follow that every element in a $C^*$-algebra can be written as $a^*b$ for some $a\in A$ and $b \in A$. | |
| Aug 2, 2021 at 14:19 | comment | added | Jamie Gabe | In the special case where the your Hilbert $A$-module is $A$, every element in $\mathcal K(A)$ is of the form $\theta_{a,b}$. | |
| Aug 1, 2021 at 21:19 | comment | added | Andromeda | @JamieGabe Yes, but what I am having problems with is the fact that $\mathcal{K}(A)$ is the closed linear span of the operators $\theta_{a,b}$, and it appears that we need to ask for boundedness of the nets to go from one to the other topology. Am I missing something? | |
| Aug 1, 2021 at 13:37 | comment | added | Jamie Gabe | Yes, these topologies agree basically by definition once you understand the isomorphism $M(A) \cong \mathcal L(A)$ (since $t\theta_{a,b}= \theta_{t(a),b}$ the canonical isomorphism $A \cong \mathcal K(A)$ takes this element to $t(a)b^\ast = t(ab^\ast ))$. | |
| Jul 31, 2021 at 19:38 | history | edited | Andromeda |
edited tags
|
|
| Jul 31, 2021 at 12:37 | history | asked | Andromeda | CC BY-SA 4.0 |