Dimension of Unipotent Radicals

Question

A parabolic subgroup of a linear algebraic group $G$ defined over a field $k$ is a subgroup $P\subseteq G$, closed in the Zariski topology, for which the quotient space $G/P$ is a projective algebraic variety. A subgroup $P\subseteq G$ is a parabolic subgroup if and only if it contains some Borel subgroup of the group $G$. Each parabolic subgroup $P$ of a group $G$ is the semi-direct product of its unipotent radical $U$ and the Levi subgroup of the group $P$.

I am interesting in knowing about the dimension of $U$. In particular, for what $G$ can one find a parabolic $P$ whose unipotent subgroup $U$ has dimension a multiple of 4? The answer for $G=\mathrm{SL}_n(\mathbb{R})$ is clear. Is there a discussion of the computations for $U$ in a reference book? Is this something that LiE or Sage can help me determine?

(Apologies if this question has been asked before.)

Answer 1

I have to assume you are looking at a semisimple or simple group, since otherwise you can just take the direct product of such a group with a vector group of any dimension to enlarge $U$.

In the semisimple case, when $U$ is the unipotent radical of a Borel subgroup, its dimension is just the number of positive roots. All of this is laid out clearly as a consequence of the Borel-Chevalley structure theory, which itself takes a lot of work but is exposed in standard textbooks with a common title Linear Algebraic Groups. For parabolic subgroups in general, the dimension of $U$ is just the difference between the total number of positive roots and the number of positive roots in a Levi factor. So the same data is involved.

Since the root system ideas have been axiomatized by Bourbaki apart from Lie groups or Lie algebras or algebraic groups, their data in Chapter VI is easy to check. For instance, 4 is the dimension of $U$ for a Borel subgroup of type $B_2=C_2$, etc. Divisibility by 4 is easy enough to track down, though it doesn't seem to have any special significance theoretically.

P.S. Technically one might be dealing with reductive groups (or Lie algebras), but a central torus makes no difference here either for the big group or for a Levi subgroup.

Also, a concise table of the numbers of positive roots for irreducible root systems appears on page 66 of my Lie algebra textbook. Probably the algorithm I've sketched is easiest to do by hand, since for a given simple type you can enumerate first all possible root systems of Levi subgroups/subalgebras via the Dynkin diagram and use the number of positive roots for each irredudible component occurring. The alternative description by pranavk involves a similar amount of computation, but requires having at hand the full lists of positive roots (not all given explicitly by Bourbaki for exceptional types).

Answer 2

@Jim: Thank you for your help on this. With respect to your discussion above, I understand the $A_n$ case but wonder if I can run my $C_2$ calculation by you. In this case there are four parabolics $P_I$ of $Sp(4)$, corresponding to the subsets $I$ of the basis $\{\alpha, \beta\}$. If $u_I$ denotes the dimension of the unipotent radical and $\ell_I$ denotes the dimension of the Levi complement, then write $u_I+\ell_I=r_I$, the number of positive roots of the associated root system.

If $I=\{\alpha\}$ or $\{\beta\}$, then $u_I=3$ and $\ell_I=4$, so $r_I=7$.

If $I=\{\alpha, \beta\}$, then $u_I=0$ and $\ell_I=r_I=10$.

It seems that $r_I$ can be independently computed by $r_I=2^2+2+|I|^2$. I'm curious to know if this is the right combinatorics.

If $I=\emptyset$, this would give $r_I=2^2+2+0^2=6$. Then $\ell_I=2$ and subsequently $u_I=4$. Is this right?

The last computation confuses me a little because $I=\emptyset$ should correspond to the collection of upper triangular matrices in $Sp(4)$. Calculating the general form for this subgroup, I find a dimension of 5, and I'm not sure where the 5 fits into the $6=2+4$.

Thanks in advance for any thoughts.