Timeline for Do representations of finite groups of Lie type preserve diagonalizable elements?
Current License: CC BY-SA 2.5
21 events
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| Jun 14, 2010 at 22:30 | history | edited | Jim Humphreys | CC BY-SA 2.5 |
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| Jun 14, 2010 at 22:15 | comment | added | Jim Humphreys | @Someone: This didn't get my full attention, but I hope the latest edit refines the answer enough. Anyway, the original question got answered earlier in the negative, which is probably as far as it needs to go. Sorry to add to confusion along the way. | |
| Jun 14, 2010 at 22:13 | history | edited | Jim Humphreys | CC BY-SA 2.5 |
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| Jun 14, 2010 at 13:58 | comment | added | Someone | @Jim: Sorry, but the statement "So it can be diagonalized over k provided k has at least q elements (and thus contains all roots of unity of order dividing q−1)." is still wrong: the field with 8 elements does not contain a 3rd root of unity (q=4) as its multiplicative group has order 7. | |
| Jun 14, 2010 at 12:34 | history | edited | Jim Humphreys | CC BY-SA 2.5 |
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| Jun 14, 2010 at 7:28 | comment | added | Someone | @Jim: Thanks, your inequality is exactly where I got stuck for a little while in one direction of your proof. Victor's comment refers to the other direction ("Conversely ..."). Here the multiplicative group $k^\times$ must have order divisible by $q-1$. | |
| Jun 11, 2010 at 19:31 | comment | added | Jim Humphreys |
@Someone: Maybe so, but that doesn't seem relevant here(?) Actually, a more careful formulation and proof by me in the first place might have avoided some of this follow-up. The original question was (for me) out of focus. By the way, I didn't mention the easy inequality for $p>2$ needed in the "it follows" part of the argument: $$(p^e-1)/2 > p^{e-1}-1$$
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| Jun 11, 2010 at 15:11 | comment | added | Someone | @Jim: Victor referred to the fact that $F_{p^n}$ contains $F_{p^m}$ if and only if $m$ divides $n$. | |
| Jun 10, 2010 at 22:28 | comment | added | Jim Humphreys |
@Victor: Yes, that's equivalent. I just wanted to emphasize that $k$` might be smaller if the given matrix doesn't have to be sent to a matrix which is diagonalizable over $k$. The original question tends to equate "diagonal" with "diagonalizable" over a given ground field, which makes exact wording of an answer tricky. I would prefer the terms "semisimple" and "diagonalizable over $k$" (rather than some larger field) for clarity when talking about specific matrices.
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| Jun 10, 2010 at 22:00 | comment | added | Victor Protsak | "the field k must have at least q elements": I think you really mean that $k$ contains $F_q$. | |
| Jun 10, 2010 at 20:04 | comment | added | Jim Humphreys |
@Someone: Thanks for the refinement. In this rank 1 case, $q=2,3$ must be excluded (ditto for some rank 2 cases). Usually I don't think of a group like $S_3$ as being of "Lie type", to avoid making special statements about a few small groups. But conventions vary. For the question at hand, some use has to be made of the special feature of (most) finite linear groups of Lie type: such a group is almost simple, which limits the behavior of its semisimple elements in representations.
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| Jun 10, 2010 at 16:06 | comment | added | Someone | Just for completeness: you should exclude some small $q$ when claiming that $G/Z(G)$ is simple. | |
| Jun 10, 2010 at 14:40 | history | edited | Jim Humphreys | CC BY-SA 2.5 |
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| Jun 10, 2010 at 14:00 | history | undeleted | Jim Humphreys | ||
| May 20, 2010 at 13:08 | history | deleted | Jim Humphreys | ||
| May 20, 2010 at 13:07 | comment | added | Jim Humphreys | @Xandi Tuni: Sorry, my version was too hasty. I'll cancel it to you can post your comment as a negative answer to the question. | |
| May 20, 2010 at 12:22 | comment | added | Xandi Tuni | That's even simpler... | |
| May 20, 2010 at 12:17 | comment | added | Someone | Take $k=F_p$, $q=p^e$ and define $SL_2(q)\to GL_{2n}(p)$ by interpreting $F_q^2$ as $2e$−dimensional $k$-vector space. | |
| May 20, 2010 at 12:03 | comment | added | Xandi Tuni | Why must $k$ have $\geq q$ elements? Any finite group has a faithful representation over any field with say $p$ elements, e.g. permutation representations (given by permuting elements of a basis). | |
| May 20, 2010 at 11:08 | history | edited | Jim Humphreys | CC BY-SA 2.5 |
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| May 20, 2010 at 11:00 | history | answered | Jim Humphreys | CC BY-SA 2.5 |