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Dear Members of Mathoverflow,

I am interested about a Fact (if it is right) of the structure of parabolic subgroups of finite classical algebraic groups:

Let G be a classical algebraic group over the finite field of order $r^{e}$, and let $ P \leq G $ denote a maximal parabolic subgroup of G.

We regard the unipotent radical of P: $ R:=R_{U}(P) = O_{r}(P) $. Is it true, that R is a minimal normal subgroup of P? (For P being a parabolic subgroup which is not a maximal in G, it is not true. If we regard the PSL(3,$r^{e}$) and a Borel subgroup.)

If it is right for the maximal parabolic subgroups of G, are there any Theorems or Propositions about that?

Thank you much for your Answers.

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Thank you all a lot, now I see that it is quite rare that $R_{U}(P)$ is a minimal normal subgroup of P. For the $PSL(n,q)$ for example it holds for the stabilizer of a minimal Flag $ 0 \lneq W \lneq V $, with dim W = 1 or dim W = dimV – 1. Regards –  Little Lie Theory Mar 22 at 9:54

4 Answers 4

No, you can see the problem here already in type $B_2$. If the simple roots are called $\alpha$ (long) and $\beta$ (short), a maximal (proper) subgroup $P$ has as Levi factor a copy of $\mathrm{GL}_2(k)$ corresponding to the root $\alpha$, while its unipotent radical involves three root groups for $\beta, \alpha+\beta, \alpha+2 \beta$. But the last of these root groups is a minimal normal subgroup of $P$. (Here it also doesn't matter what the field $k$ is. And I don't see why you specify "classical" types in the question.)

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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? –  Nick Gill Mar 21 at 21:17
@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. –  Jim Humphreys Mar 21 at 22:29
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. –  Allen Knutson Mar 22 at 1:37
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). –  Nick Gill Mar 22 at 1:52

The unipotent radical $U$ of a maximal parabolic $P$ of a classical group is not always abelian - see Jim's answer to a specific example and the paper by Richardson, Rohrle and Steinberg for the general case. So in these cases $Z(U)$ is proper subgroup of $U$ that is normal in $P$, and you have a counterexample.

In the cases where $U$ is abelian, one needs to check whether the natural action of a Levi subgroup of $P$ on the unipotent radical $U$ is irreducible. There are lots of sources for this sort of thing, for instance Volume 3 of the series by Gorenstein, Lyons and Solomon.

Edit: As I mention in my comment above on Jim's answer, I don't know under what circumstances $Z(U)$ is minimal normal - again this would come down to studying the irreducibility of the action of the Levi on $Z(U)$.

(I should add, in case anyone thinks I was cheeky, that I initially phrased this answer speculatively... and adjusted it in light of the counter-example given in Jim's answer.)

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I think we can verify explicitly that this is true for classical groups. For $SL_n$, the Levi looks like $SL_a \times SL_b$ for $a+b=n$ and $U$ is just Homs from the standard representation of $SL_a$ to the standard representation of $SL_b$. For $SO_n$ the Levi is $Gl_a \times SO_{n-2a}$, and $U$ is an extension of homs between the standard representations by $\wedge^2$ of $GL_a$, with the second one the center unless the first is $0$. Same thing with $SP_{2n}$ and $\operatorname{Sym}^2$. –  Will Sawin Mar 22 at 14:27

The standard reference for this is Azad, Barry, Seitz. On the structure of parabolic subgroups. Comm. Algebra 18 (1990), no. 2, 551–562.

It gives explicit information on the structure of the unipotent radical for the algebraic group case together with all split Chevalley groups and twisted 'Steinberg' groups, and also gives details on what happens when the characteristic is 'special', e.g. p=3, G=G_2.

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It is a general fact that if $P$ is a parabolic $k$-subgroup of an arbitrary connected reductive group $G$ over an arbitrary field $k$ then $U := \mathscr{R}_u(P)$ has a canonical $P$-equivariant filtration $$U = U_0 \supset U_1 \supset \dots \supset U_m \supset U_{m+1} = \{1\}$$ by smooth connected $k$-subgroups such that each $U_i/U_{i+1}$ is a vector group that admits a unique $P$-equivariant linear structure. (This can be defined over $\mathbf{Z}$ as well when working with "Chevalley groups", so it is truly characteristic-free). In particular, the final stage of this canonical filtration is a vector group that is a normal $k$-subgroup of $P$. This must be well-known in the classical case, but the only reference I know in the literature is SGA3, Exp. XXVI, 2.1.

If $G$ is split semisimple of adjoint type and $T$ is a split maximal $k$-torus in $P$ then the resulting linear representation of $T$ on each $U_i/U_{i+1}$ is a direct sum of 1-dimensional weight spaces given by certain roots in $\Phi(G,T)$. I expect that when the root system is moreover irreducible then each $(U_i/U_{i+1})(k) = U_i(k)/U_{i+1}(k)$ is irreducible as a $k$-linear $P(k)$-representation, apart from perhaps some special cases with low rank and/or $k$ with size 2 or 3. In particular, the final stage $U_m(k)$ would then be minimal as a normal subgroup of $P(k)$ (since there are no divisible roots in the character lattice due to $G$ being of adjoint type). Maybe this is sufficient for your purposes? (No motivation was given for the posted question, beyond perhaps curiosity.)

Since you use the notation "PSL" which is a perennial source of confusion (do you mean ${\rm{SL}}_n(k)/\mu_n(k)$ or ${\rm{PSL}}_n(k)$ where ${\rm{PSL}}_n$ is meant as the quotient ${\rm{SL}}_n/\mu_n$ in the sense of algebraic groups?), I should note that although such confusion creates headaches with the tori, it does not with the "unipotent radicals", so the above should be applicable to various possible meanings of the phrase "finite classical group", but (as always) be careful.

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