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