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.)

• @Stanley: You mention a real split semisimple Lie or algebraic group as an example, but it's usually best to focus first on the corresponding complex group or better yet on its Lie algebra. All the dimension information you want is found in the latter more elementary setting. In any case, no computer calculation is needed, just a lot of routine elementary arithmetic. Is there some motivation? – Jim Humphreys Jan 19 '13 at 22:04
• @Jim: What if we complicate this idea a bit further by asking for the Q-dimensions of Q-unipotent radicals associated to Q-parabolic subgroups? The number of Q-parabolic subgroups will depend on the Q-rank, which of course will at times differ from the real rank. Again, the goal is to find the cases in which the Q-dimension is divisible by 4. Thanks in advance. – user31353 Feb 11 '13 at 17:46
• @Stanley: Your newest comment raises a new question about working over the field of rational numbers. Keep in mind that all Lie groups in your first version have rational forms, so the list of parabolics defined over $\mathbb{Q}$ includes the ones originally discussed. For them it's easy to compute dimensions of unipotent radicals. For other isotropic but nonsplit forms, you have to dig more into the classification. – Jim Humphreys Feb 12 '13 at 1:57

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).

• Equivalently, since parabolic subgroups of a connected reductive group are classified up to conjugacy by subsets $I$ of a basis $\Delta$ of a positive system of roots $\Phi^+$, and more specifically ${\rm{Lie}}(R_u(P_I))$ supports the positive roots which are a non-negative $\mathbf{Z}$-linear combination of elements of $\Delta - I$ (with each such root space being 1-dimensional), the problem is purely combinatorial and can be settled via Bourbaki tables via inspection of the tables of reduced irreducible root system separately (as one varies $I$), as Jim suggests. This avoids computing Levis. – user30379 Jan 19 '13 at 18:16
• @pranvavk: Yes, what I've described is just based on elementary combinatorics of root systems. But note that your description of roots in the unipotent radical isn't quite right as stated. Also, Bourbaki doesn't list all positive roots in spite of leaving ample blank space in the "planches"; a later IHES paper by Springer does that for the exceptional types, arranged conveniently by heights. – Jim Humphreys Jan 19 '13 at 22:02
• @Jim Humphreys: Oops, I should have said that the roots on ${\rm{Lie}}(R_u(P_I))$ are the roots that are non-negative $\mathbf{Z}$-linear combinations of $\Delta$ that lie outside the $\mathbf{Z}$-span of $I$. My main point was just to say that one can view the problem as combinatorial without needing to compute Levi factors. (I also forgot that Bourbaki sometimes only lists the positive roots with a coefficient $\ge 2$.) – user30379 Jan 19 '13 at 22:48
• @pranavk: A concrete example is $F_4$, where there are 24 positive roots and $2^4= 16$ classes of parabolic subgroups. Here it's easy (without pencil and paper) to work out the 16 dimensions of unipotent radicals, which lie between 0 and 24, knowing just the Dynkin diagram and the number of positive roots for each smaller root system. The Levi types are visible in the diagram, with no "computation" required. Anyway, divisibility by 4 looks accidental in that list. – Jim Humphreys Jan 20 '13 at 18:41
• Thanks to both of you for these ideas. The divisibility by 4 issue has to do with some structure sets that I want to compute via the surgery exact sequence. – Stanley Chang Jan 21 '13 at 18:36

@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.

• @Stanley: Your approach is too complicated. Each of the intermediate Levi subgroups has derived group of simple type $\mathrm{SL}_2$ with a single positive root, so the unipotent radical has dimension $3=4-1$ in both cases. The dimensions of the four possible unipotent radicals for $C_2$ (up to conjugacy) are $0, 3, 3, 4$. Such dimensions are always bounded above by the total number of positive roots, the dimension of a maximal unipotent subgroup. In my example $F_4$`, you get 16 dimensions between 0 and 24, for example, with a few intermediate ones divisible by 4. – Jim Humphreys Jan 25 '13 at 20:01