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S Mar 19, 2020 at 16:41 history bounty ended as2457
S Mar 19, 2020 at 16:41 history notice removed as2457
Mar 15, 2020 at 14:09 answer added Denis Serre timeline score: 2
Mar 13, 2020 at 16:57 comment added Nick Gill The angled backed notation means "the group generated by $A$ and $B$" -- i.e. all matrices obtained by taking finite length products of $A$, $B$ and their inverses.
Mar 13, 2020 at 16:56 answer added Nick Gill timeline score: 2
S Mar 13, 2020 at 14:15 history bounty started as2457
S Mar 13, 2020 at 14:15 history notice added as2457 Draw attention
Mar 10, 2020 at 15:28 comment added as2457 Clearly if $AB$ is not unitarily similar to a permutation matrix then there is no such $U$. If $AB$ is similar to a permutation as well as $A$ and $B$ separately, can I say anything? Can I say something in the particular case that $A$ and $B$ are representations of the generators of the modular group?
Mar 10, 2020 at 8:58 comment added as2457 I don't know if it helps (or if I should add to the question), but in the case I'm interested in $A$ and $B$ are representations of the generators of the modular group (90deg rotation and Dehn twist). They are also symmetric in the original basis.
Mar 10, 2020 at 7:20 comment added YCor @MarkSapir and all the subgroup they generate. This is why I added the group tag. For instance one can wonder, if the product of, say all $2^n$ $n$-fold products ($n$ being the size) of $A,B$ are conjugate to permutation matrices, whether it's enough to ensure that they can be simultaneously conjugate into the subgroup of permutation matrices. (Already one can wonder about an algorithm to check if two matrices generate a finite group.)
Mar 9, 2020 at 17:48 comment added as2457 Yes, what I am imagining is being given two matrices and trying to determine whether there exists such a U. This could either be using the properties of A and B, or it could be determined through some algorithm designed to find U. Could you also clarify what you mean by the angled bracket notation please?
Mar 9, 2020 at 14:53 comment added Nick Gill When you say "how can I check", are you imagining that you have been given $A$ and $B$ explicitly, and you want to do a computation... Or are you asking for theoretical conditions that are equivalent? One obvious necessary condition is that $\langle A, B\rangle$ is isomorphic to a subgroup of $S_n$... I'm not sure how close it is to being sufficient.
Mar 9, 2020 at 14:19 comment added YCor Another restatement: is there a subset of $\mathbf{C}^n$ that forms an orthonormal basis, and which is invariant by both $A$ and $B$? (the existence of such a subset for each of $A$ and $B$ being the assumption)
Mar 9, 2020 at 14:16 history edited YCor
edited tags; edited tags
Mar 9, 2020 at 14:10 history asked as2457 CC BY-SA 4.0