Timeline for About structure of parabolic subgroups of finite classical algebraic groups
Current License: CC BY-SA 3.0
7 events
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| Mar 22, 2014 at 1:52 | comment | added | Nick Gill | Jim, I din't view your answer as undercutting mine at all! Mine was, in the first instance, idle speculation, and yours was a specific counterexample which was just what was needed. Although the original question is dealt with, I'm still interested about the minimality of the centre (from idle curiosity, nothing more). | |
| Mar 22, 2014 at 1:37 | comment | added | Allen Knutson | Let me spell out a framework I think is useful here. Given a set $S$ of simple roots (which will be the ones not in your parabolic), define the $S$-height of a root $\beta$ to be the sum of the coefficients on $S$, when expanding $\beta$ in simple roots. Let $m$ be the maximum $S$-height. Then for each $k\in [0,m]$ we can define a normal subgroup $J_k$ of $P$, using those roots of height $\geq k$, and $P = J_0 > J_1 > \ldots > J_m$. (Also $J_1 = Rad(P)$.) For your question, you want $m=1$. For $P$ maximal, $m$ is the coefficient of $P$'s missing simple root in the highest root. | |
| Mar 21, 2014 at 22:29 | comment | added | Jim Humphreys | @Nick: I didn't intend to undercut your tentative answer but rather to emphasize that elementary examples exist and don't require further references. Certainly there may be nuances for small fields, but I'm more puzzled by the way the original question is formulated. | |
| Mar 21, 2014 at 21:17 | comment | added | Nick Gill | in your counterexample you have $Z(U)$ as a minimal normal subgroup, as I suggested would happen in my answer. I guess the original question could be reposed to ask whether $Z(U)$ is always minimal normal, i.e. whether a Levi always acts irreducibly on $Z(U)$? I imagine that there will be exceptions for small fields in special cases, but I wonder if this is true most of the time? | |
| Mar 21, 2014 at 20:36 | history | undeleted | Jim Humphreys | ||
| Mar 21, 2014 at 20:34 | history | deleted | Jim Humphreys | via Vote | |
| Mar 21, 2014 at 20:32 | history | answered | Jim Humphreys | CC BY-SA 3.0 |