Describing Levi factors and unipotent radicals of parabolic subgroups in classical groups

Question

I asked this question before at Math.SE (link) but got no answer.

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Let $G$ be an algebraic group over an algebraically closed field $k$ of characteristic $p \geq 0$. Then any parabolic subgroup $P$ of $G$ has factorization $P = Q \rtimes L$, where $Q$ is the unipotent radical of $P$ and $L$ is some Levi factor of $P$.

If $G = GL(V)$ with $\dim V = n$, then $$P = \{ \pmatrix{A_{n_1} & & & * \\ & A_{n_2} & & \\ & & \ddots & \\ 0 & & & A_{n_t} } \}$$

for some $n = n_1 + \cdots + n_t$ and $$Q = \{ \pmatrix{I_{n_1} & & & * \\ & I_{n_2} & & \\ & & \ddots & \\ 0 & & & I_{n_t} } \}$$ and for example $$L = \{ \pmatrix{A_{n_1} & & & 0 \\ & A_{n_2} & & \\ & & \ddots & \\ 0 & & & A_{n_t} } \}$$

So $L \cong GL_{n_1} \times GL_{n_2} \times \cdots \times GL_{n_t}$. What about when $G = \operatorname{Sp}(V)$ or $G = \operatorname{SO}(V)$? I suppose the answer will be different for $p = 2$ and $p \neq 2$.

I know that for $\operatorname{Sp}$ and $\operatorname{SO}$ any parabolic subgroup $P$ is a stabilizer of a flag of totally singular subspaces.

How does one describe $Q$ and $L$? Is there a good reference for this?

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I suppose if $P$ is the stabilizer of the flag $0 \subset V_1 \subset V_2 \subset \cdots \subset V_t$ of totally singular subspaces of $V$, then intuitively $Q$ should consist of those maps which stabilize each $V_i$ and act trivially on each $V_{i} / V_{i-1}$. But what do the Levi factors $L$ of $P$ look like?

Answer 1

If (say) $G = Sp(V) = Sp(V,\beta)$ for a non-degenerate symplectic form $\beta$ on $V$, and if $W \subset V$ is a totally singular subspace, the stabilizer $P$ of $W$ also stabilizes $W^\perp = \{v \in V \mid \beta(v,W) = 0\}$, so in fact $P$ stabilizes the chain $0 \subset W \subset W^\perp$. Note that the restriction of $\beta$ to $W^\perp$ determines a non-degenerate symplectic form $\overline{\beta}$ on the quotient $W^\perp/W$, and that as $P$-representations $V/W^\perp$ identifies with the dual of $W$. The reductive quotient $L$ of $P$ identifies with $GL(W) \times Sp(W^\perp/W,\overline{\beta})$.

The stabilizer of a chain of totally singular subspaces can be described using similar argumentation. And the case of an orthogonal group is not really different.

Answer 2

I'll assume here that $G$ is reductive (or even semisimple), since otherwise the description of unipotent radicals is more open-ended.

There is actually quite a bit of literature over the years, often concerned especially with questions involving smaller fields of definition (such as finite fields) or with the precise action of $L$ on $Q$. From the Borel-Tits conjugacy theorem, one can limit attention to the case of a standard parabolic subgroup arising from the choice of a set of simple roots.

In the case of classical groups a basic fact is that the semisimple derived group of $L$ itself (when nontrivial) only has as simple factors various classical groups: these are easily determined by removing nodes from the Dynkin diagram corresponding to simple roots whose negatives do not figure in the given parabolic subgroup. Of course, these simple algebraic groups can occur with varied isogeny types ranging from simply connected to adjoint. (On the Lie algebra level, this isn't a problem.)

Since the derived group of $L$ is only the almost-direct product of these simple factors, it's a somewhat more delicate task to examine each isogeny type of $G$ to describe the product more precisely. Similarly, $L$ is only an almost-direct product of the derived group with a central torus. It may help to keep in mind that when the given (reductive) group $G$ has a simply connected derived group, then the derived group of $L$ is also simply connected (a result going back to the work of Borel-Tits).

Concerning the way $L$ acts on $Q$ (or factors in a suitable series for it), there are explicit representation-theoretic descriptions, many by Gary Seitz and some of his collaborators or students: see for example Azad-Barry-Seitz, along with Roehrle and Anchouche. The special case when $Q$ is commutative gets special attention.