Timeline for Almost commuting unitary matrices
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14 events
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| Dec 3, 2019 at 9:26 | history | edited | YCor |
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| Nov 8, 2015 at 15:27 | vote | accept | Omid Hatami | ||
| Nov 4, 2015 at 15:12 | comment | added | Omid Hatami | No! $n$ can be very large. For example Lev Glebsky obtains a bound that doesn't depend on $n$ for any pair of matrices in the case that norm is the Hilbert-Schmidt norm. arxiv.org/pdf/1002.3082v1.pdf | |
| Nov 4, 2015 at 15:12 | comment | added | Victor Protsak | Are you fixing the order $n$ of the matrices? As I understand it, it is crucial in the question linked by Mikael de la Salle that the bound be uniform in $n$. | |
| Nov 4, 2015 at 14:58 | comment | added | Omid Hatami | Well as I understand two cases of unitary and self-adjoint matrices are different. For example there are two unitary matrices $A$ and $B$ such that $||[A,B]||_{op}$ is small but they can't be approximated with commuting unitary matrices. mtm.ufsc.br/~exel/papers/asympt.pdf But the answer to the other case is yes for any pair of almost commuting matrices and it is called Huaxin Lin's Theorem math.ku.dk/~rordam/manus/short.pdf | |
| Nov 4, 2015 at 14:52 | comment | added | Victor Protsak | You can use Cayley transform to pass between hermitian and unitary matrices. This preserves commutativity, but of course distorts the norm. (Cayley transform is only a birational map, it is undefined at unitary matrices with eigenvalue $-1$.) | |
| Nov 4, 2015 at 14:48 | comment | added | Omid Hatami | The question is that you know any two (or three or small number) of them can be approximated simultaneously by commuting unitary matrices and the question is to approximate all of them together (with a good bound such as $c\varepsilon$) with commuting unitary matrices. As I understand Miakel's answer is no for approximately commuting Hermitian matrices and doesn't apply here. | |
| Nov 4, 2015 at 14:36 | comment | added | Suvrit | I'm a bit confused. You're saying that you know that there exists a solution and the question is how to find it? Or am I misunderstanding (because Mikael's answer says "no" for one instance...). | |
| Nov 4, 2015 at 13:50 | history | edited | Omid Hatami | CC BY-SA 3.0 |
added 142 characters in body; edited title
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| Nov 4, 2015 at 8:16 | answer | added | Mikael de la Salle | timeline score: 14 | |
| Nov 4, 2015 at 1:32 | comment | added | Vectornaut | Now, suppose we can find commuting operators $A_i'$ and $A_j'$ within $\varepsilon$ of $A_i$ and $A_j$ respectively. Since $A_i'$ and $A_j'$ commute, they have the same eigenlines $L_1, \ldots, L_n$. If $\varepsilon$ is small enough, I think the eigenlines of $A_i$ and $A_j$ should have to be near $L_1, \ldots, L_n$, and therefore near each other. In other words, for small enough $\varepsilon$, your commutative approximation condition implies that the eigenlines of the operators $A_1, \ldots, A_k$ almost match. | |
| Nov 4, 2015 at 1:32 | comment | added | Vectornaut | Here's a thought. Suppose we're in a vector space $V$ over an algebraically closed field, and each of the operators $A_1, \ldots, A_k$ has distinct eigenvalues. Make $\varepsilon$ small enough that the operators in the $\varepsilon$-ball around each $A_i$ also have distinct eigenvalues. That means any operator $\varepsilon$-near one of the $A_i$ decomposes $V$ into eigenlines. | |
| Nov 4, 2015 at 0:15 | history | edited | Omid Hatami | CC BY-SA 3.0 |
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| Nov 4, 2015 at 0:07 | history | asked | Omid Hatami | CC BY-SA 3.0 |