Timeline for Semi-simple matrices over fields of finite characteristic
Current License: CC BY-SA 2.5
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 17, 2011 at 21:19 | vote | accept | Andreas Thom | ||
| Feb 3, 2011 at 19:58 | vote | accept | Andreas Thom | ||
| Feb 3, 2011 at 19:58 | |||||
| Jan 14, 2011 at 18:32 | answer | added | BS. | timeline score: 4 | |
| Jan 13, 2011 at 20:52 | comment | added | Andreas Thom | Pete, you are right. | |
| Jan 13, 2011 at 20:51 | comment | added | Andreas Thom | Ben, you are right. The reason I am interested in those questions is that there is a lot of theory about limits of large integer matrices in terms of their local statistics. A consequence of a theorem of Lück is that the normalized dimension of the kernel of a large matrix depends continuously on its local statistics. All the proofs are starting by assuming that the matrix is symmetric and by looking at the characteristic polynomial and its zero distribution on $\mathbb C$. I am looking for an analogue of Lück's results in finite characteristic. Maybe I'll ask a more precise question later. | |
| Jan 13, 2011 at 19:28 | comment | added | Ben Webster♦ | @Andreas- Saying "I am stuck on this problem" seems to imply that there's more going on here. Maybe saying more about your ultimate aims will get better answers? | |
| Jan 13, 2011 at 19:26 | comment | added | Ben Webster♦ | @Pete-I had the exact same thought. | |
| Jan 13, 2011 at 19:17 | answer | added | Anatoly Kochubei | timeline score: 1 | |
| Jan 13, 2011 at 17:53 | comment | added | ndkrempel | @Pete: Yes, I think the "i.e." in the question was bad choice of phrasing, that's all. (Replacing it with "a fortiori" would work.) As the question itself is over an algebraically closed field, there shouldn't be any confusion. | |
| Jan 13, 2011 at 17:20 | comment | added | Pete L. Clark | I dispute the definition (!) of semisimple: it should mean diagonalizable over the algebraic closure. This is equivalent to the $k$-algebra generated by the matrix $M$ to be semisimple and to the minimal polynomial being separable. Or am I wrong? | |
| Jan 13, 2011 at 16:46 | answer | added | David E Speyer | timeline score: 8 | |
| Jan 13, 2011 at 16:11 | comment | added | ndkrempel | @BS: That's interesting, I didn't know about that correspondence. It doesn't seem to be connected to the Lie correspondence as that would give you anti-symmetric matrices... | |
| Jan 13, 2011 at 15:48 | answer | added | ndkrempel | timeline score: 2 | |
| Jan 13, 2011 at 15:42 | comment | added | BS. | @ndkrempel : indeed this is more a hint than a proof. I was thinking to the cases of (not necessarily definite) orthogonal and unitary groups over $R$, which have only semisimple elements exactly when this is the case for the corresponding (pseudo-) symmetric or hermitian matrices. There must be a link, but I don't know enough of algebraic group theory to elaborate on this. | |
| Jan 13, 2011 at 15:22 | comment | added | ndkrempel | @BS: Wouldn't that only be an issue if the "shape" was required to be closed under multiplication? | |
| Jan 13, 2011 at 14:11 | comment | added | Andreas Thom | Jim, since I am stuck with this problem, I fear that any concrete question will have a negative answer. That is why I am trying to open up the realm of possible answers. Anything non-trivial and positive would certainly count as an answer. (Rich is only supposed to exclude the class of diagonal matrices.) | |
| Jan 13, 2011 at 14:06 | comment | added | Jim Humphreys | @Andreas: It seems unlikely that your loosely formulated question has an interesting answer. What's true is that the semisimple matrices form a Zariski-dense subset in the space of all square matrices; so there are plenty of them. But you are probably looking for a set of semisimple matrices given by polynomial conditions on entries, which would give a (proper) closed set in the Zariski topology. Still, "rich" is a flexible term. | |
| Jan 13, 2011 at 12:01 | comment | added | BS. | I doubt any shape (other than diagonal) implies semisimplicity, as all finite Chevalley groups in characteristic $p$ have order divisible by $p$, hence have nontrivial unipotent elements. | |
| Jan 13, 2011 at 11:22 | history | edited | Andreas Thom | CC BY-SA 2.5 |
added 20 characters in body; edited body
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| Jan 13, 2011 at 11:06 | history | asked | Andreas Thom | CC BY-SA 2.5 |