Timeline for What are the possible eigenvalues of these matrices?
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| Mar 9, 2018 at 17:09 | history | edited | Chris Woodward | CC BY-SA 3.0 |
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| Mar 9, 2018 at 16:58 | history | edited | Chris Woodward | CC BY-SA 3.0 |
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| Mar 9, 2018 at 16:34 | comment | added | Chris Woodward | Being a convex polyhedral cone of maximal rank, it could be much smaller than the full Weyl chamber. Because the cone is maximal rank, the facets of the cone you are looking for (relative to the Weyl chamber) are a subset of the hyperplanes corresponding to one-dimensional stabilizers. Taking A to be the identity gives a one-dimensional stabilizer and so one particular hyperplane, while taking A to be a different permutation matrix would give others. I wonder if the hyperplanes at the boundary of the cone are of the form (sum of four eigenvalues) = (sum of four other eigenvalues). | |
| Mar 9, 2018 at 12:34 | comment | added | Nik Weaver | As noted in the other answers, not all eigenvalue sequences are possible ... | |
| S Mar 9, 2018 at 7:20 | history | suggested | Hans | CC BY-SA 3.0 |
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| Mar 9, 2018 at 2:38 | review | Suggested edits | |||
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| Mar 8, 2018 at 22:54 | comment | added | Chris Woodward | Pick a point x in the space X being acted up on by G and let $G_x$ be all elements $g$ such that $gx = x$. There exists a dense open subset of $x$ such that $G_x$ is independent of $x$ up to isomorphism. | |
| Mar 8, 2018 at 22:52 | history | edited | Chris Woodward | CC BY-SA 3.0 |
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| Mar 8, 2018 at 22:45 | comment | added | Nik Weaver | I'm still baffled as to how "belonging to the generic stabilizer" becomes "stabilizing all antidiagonal matrices". What is the formal definition of generic stabilizer? | |
| Mar 8, 2018 at 22:26 | history | edited | Chris Woodward | CC BY-SA 3.0 |
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| Mar 8, 2018 at 22:02 | comment | added | Chris Woodward | Agreed - my "identity matrix" was not sufficiently generic. Once one takes a truly generic 4x 4 matrix, there are no solutions, which implies that the convex polyhedral cone is maximal rank. | |
| Mar 8, 2018 at 22:01 | history | edited | Chris Woodward | CC BY-SA 3.0 |
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| Mar 8, 2018 at 21:42 | comment | added | Kevin P. Costello | I'm also not sure about the specific linear relation mentioned in the post here -- I tried generating random matrices of the form in the original post, and they didn't satisfy any linear relationship between the eigenvalues with all coefficients in $\{-3,-2,-1,0,1,2,3\}$. | |
| Mar 8, 2018 at 21:36 | history | edited | Chris Woodward | CC BY-SA 3.0 |
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| Mar 8, 2018 at 20:55 | comment | added | jjcale | Counterexample : a = b = 1, c = -1, A=diag(2,1,1,1), B = 0 . | |
| Mar 8, 2018 at 19:42 | comment | added | Nik Weaver | I don't understand what you mean by "generic stabilizer" then. | |
| Mar 8, 2018 at 17:52 | comment | added | Chris Woodward | Right, you have to take B "generic" antidiagonal, then $\beta$ is zero. (I think you are also missing an inverse somewhere.) (In my first post I said I already knew the generic stabilizer was abelian; there is some abstract nonsense that implies this that I didn't explain.) | |
| Mar 8, 2018 at 16:47 | comment | added | Nik Weaver | Shoot, I'm still not getting it. For example, in the $4\times 4$ case if you take $B = \left[\begin{matrix}0&1\cr 1&0\end{matrix}\right]$ then for any special unitary $A_1 = \left[\begin{matrix}\alpha&-\bar{\beta}\cr \beta&\bar{\alpha}\end{matrix}\right]$ the special unitary $A_2 = \left[\begin{matrix}\alpha&-\beta\cr \bar{\beta}&\bar{\alpha}\end{matrix}\right]$ satisfies $A_1BA_2^{-1} = B$. So how is the stabilizer of $B$ just ${\rm diag}(t, t^{-1}, t^{-1}, t)$? What am I misunderstanding? | |
| Mar 8, 2018 at 15:59 | history | edited | Chris Woodward | CC BY-SA 3.0 |
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| Mar 8, 2018 at 15:01 | history | edited | Chris Woodward | CC BY-SA 3.0 |
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| Mar 8, 2018 at 14:45 | history | edited | Chris Woodward | CC BY-SA 3.0 |
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| Mar 8, 2018 at 14:40 | comment | added | Chris Woodward | Tried to clarify. - C | |
| Mar 8, 2018 at 14:34 | history | edited | Chris Woodward | CC BY-SA 3.0 |
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| Mar 8, 2018 at 6:25 | comment | added | Nik Weaver | I'm starting to understand some of this, but I'm afraid there's still a lot I don't follow. I suppose that "generic stabilizer" means that the stabilizers of a generic set of points are conjugate, but then (1) how do you know that the stabilizer of the specific $A$ and $B$ you chose is generic, and (2) once you have fixed $A$ and $B$, how can you go on to assume that their stabilizer is diagonal? So obviously I'm still pretty confused. | |
| Mar 8, 2018 at 2:30 | comment | added | Chris Woodward | see if above helps - C | |
| Mar 8, 2018 at 2:30 | history | edited | Chris Woodward | CC BY-SA 3.0 |
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| Mar 8, 2018 at 1:27 | history | edited | Chris Woodward | CC BY-SA 3.0 |
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| Mar 8, 2018 at 1:21 | history | edited | Chris Woodward | CC BY-SA 3.0 |
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| Mar 8, 2018 at 0:04 | comment | added | Nik Weaver | That's exactly the kind of condition I was hoping for! Is there any way you could give me some more hints about how you do this computation ... even a low-tech definition of the moment polytope in this case would help a lot. | |
| Mar 7, 2018 at 21:55 | history | edited | Chris Woodward | CC BY-SA 3.0 |
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| Mar 7, 2018 at 21:49 | history | edited | Chris Woodward | CC BY-SA 3.0 |
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| Mar 7, 2018 at 20:44 | comment | added | Nik Weaver | Thank you --- this is going to take some work for me to digest. | |
| Mar 7, 2018 at 20:17 | history | edited | Chris Woodward | CC BY-SA 3.0 |
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| Mar 7, 2018 at 20:11 | history | edited | Chris Woodward | CC BY-SA 3.0 |
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| Mar 7, 2018 at 20:04 | history | answered | Chris Woodward | CC BY-SA 3.0 |