Timeline for Point-ultraweak limit of *-homomorphisms/cpc order zero maps
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Sep 16, 2015 at 20:39 | history | edited | Aaron Tikuisis | CC BY-SA 3.0 |
Changed = to := for consistency.
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| Sep 16, 2015 at 20:39 | comment | added | Aaron Tikuisis | Alessandro: adding that hypothesis doesn't help. Let $U$ be the projection onto $e_1$ and $V$ the projection onto $e_2$ (so that $U,V$ are orthogonal to $S_i,T_i$ for $i>2$), and then define $S_i' := U + \epsilon S_i$ and $T_i':= V + \epsilon T_i$. | |
| Sep 16, 2015 at 17:19 | vote | accept | Alessandro Vignati | ||
| Sep 16, 2015 at 17:19 | comment | added | Alessandro Vignati | Uhg.You're right. And if I need them to be positive I can just take the projections onto $e_0+e_i$ and $e_0-e_i$. So even if $A$ is $\mathbb C^2$ that wouldn't work (answering Yemon). Thanks! Naively, what if there are operators $U$ an $V$, and $\epsilon>0$ (small enough) such that for all $i$ we have $||S_i-U||, ||T_i-V||<\epsilon$ (and everything has norm $1$!) | |
| Sep 16, 2015 at 16:22 | comment | added | Yemon Choi | Naive question: for which $A$ would Alessandro's question have a positive answer? (There is something in the original question which reminds me of AMNM, although it isn't actually the same) | |
| Sep 16, 2015 at 14:08 | history | answered | Aaron Tikuisis | CC BY-SA 3.0 |