Timeline for Moving one family of commuting self-adjoint operators to another without losing commutativity on the way
Current License: CC BY-SA 2.5
18 events
| when toggle format | what | by | license | comment | |
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| S Oct 25, 2020 at 0:49 | history | suggested | CommunityBot |
Removed tag.
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| Oct 24, 2020 at 21:58 | review | Suggested edits | |||
| S Oct 25, 2020 at 0:49 | |||||
| May 24, 2013 at 16:09 | answer | added | Sean Lawton | timeline score: 5 | |
| Jan 11, 2013 at 11:17 | answer | added | Rami | timeline score: -3 | |
| Jan 10, 2013 at 22:30 | comment | added | Rami | I think that the proof the you gave for the f.d. case works also for compact operators. But you probably know that. | |
| Sep 4, 2011 at 8:53 | comment | added | jjcale | @Denis: $d_{m}\leq\pi$, since every unitary Matrix U can be written as $U=e^{iL}$, where L is self adjoint and $\|L\|\leq\pi$ | |
| Jan 8, 2011 at 17:06 | comment | added | Kate Juschenko | @Martin: $C_k(t)$ should commute for each t. It seems that if norms of $A_k$ and $B_k$ are bounded then the statement is true, just taking homotopy of $A_k$ to $0$ and then $0$ to $B_k$, i.e $C_k(t)=(1-2t) A_k$ if $t\in [0,1/2]$ and $C_k(t)=(2t-1)B_k$ if $t\in[1/2,1]$ – Kate Juschenko 2 mins ago | |
| Jan 8, 2011 at 16:47 | comment | added | Martin Argerami | I assume I'm missing something, but why wouldn't $C_k(t)=tA_k+(1-t)B_k$ work? | |
| Jan 7, 2011 at 7:28 | comment | added | Denis Serre | @Igor: But then the length of the path can be as large as the diameter $d_m$ of $SU_m$, $m$ the dimension of $H$. If the answer to the question is yes, we must have $d_m\le M(n)$, and therefore $\sup_md(m)<\infty$. Is this true ? | |
| Jan 6, 2011 at 23:46 | comment | added | Igor Rivin | @Andrey: I assume that in finite dimension you just take the orthonormal frame of eigenvectors of all $A$s and connect them by a geodesic (in the orthogonal/unitary group) to the orthogonal frame of eigenvectors of the $B$s. I did not do the computation, I confess, but it seems reasonably clear that the conditions are satisfied. This seems less obvious in the infinite dimensional case. | |
| Jan 6, 2011 at 22:46 | comment | added | Andrey Rekalo | Is this known when $H$ is finite dimensional? | |
| Jan 6, 2011 at 22:43 | history | edited | fedja | CC BY-SA 2.5 |
edited body
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| Jan 6, 2011 at 22:43 | comment | added | fedja | @ Kate: For $n=1$, you can just go straight: $C(t)=(1-t)A+tB$. A linear combination of self-adjoint operators is self adjoint and everything commutes with itself. @Survit Yes, that's one more way to put it and the difficulty is exactly as you said. @Ben I'll fix it now changing $C$ to $M$. | |
| Jan 6, 2011 at 22:40 | comment | added | Suvrit | @Kate: isn't $C1 = t*B1 + (1-t)*A1$ ok in this case? what am I missing? | |
| Jan 6, 2011 at 22:11 | comment | added | Kate Juschenko | @fedja: is your question clear for n=1? if A1 and B1 are orthogonal projections of different ranks, why do we have C1? maybe you mean $||A_k-B_k||<\epsilon<1$? | |
| Jan 6, 2011 at 21:56 | comment | added | Suvrit | In other words, $C_k(t)$ for $t \in [0,1]$ is a homotopy between $A_k$ and $B_k$ such that it has "bounded variation", right? I guess the difficulty stems from the $C_k$'s having to commute too? | |
| Jan 6, 2011 at 21:56 | comment | added | Ben Webster♦ | That was unfortunate overloading of C. | |
| Jan 6, 2011 at 21:39 | history | asked | fedja | CC BY-SA 2.5 |