Timeline for Two densely defined traces on a $C^*$-algebra coinciding on a dense subalgebra are equal
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 31, 2021 at 5:19 | vote | accept | Nikita Safonkin | ||
| Oct 29, 2021 at 9:47 | answer | added | Jamie Gabe | timeline score: 5 | |
| Oct 29, 2021 at 9:40 | comment | added | Nikita Safonkin | @JamieGabe Thank you! I have edited the question. | |
| Oct 29, 2021 at 9:30 | history | edited | Nikita Safonkin | CC BY-SA 4.0 |
added 4 characters in body
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| Oct 29, 2021 at 9:21 | comment | added | Jamie Gabe | Do you mean that $t_1(b^\ast b) = t_2(b^\ast b)$ for all $b\in B$? Otherwise you might have a lot of elements $b\in B_+$ such that $t_i(b) = \infty$. If this is what you mean, I can provide a proof of the result. | |
| S Oct 28, 2021 at 16:37 | review | First questions | |||
| Oct 28, 2021 at 16:47 | |||||
| S Oct 28, 2021 at 16:37 | history | asked | Nikita Safonkin | CC BY-SA 4.0 |