Symplectic parabolic subgroup

The stabilizer of a totally isotropic $k$-space in a finite symplectic space also stabilizes its orthogonal complement. Its block decomposition has a unipotent normal subgroup $Q$ and a block-diagonal complement $L\cong {\mathrm{GL}}_k(q)\times {\mathrm{Sp}}_{2m-2k}(q)$. The paired $k$-space carries the contragredient action. A row-vector convention gives upper-triangular blocks; a column-vector convention transposes the arrangement. The unipotent radical count for a symplectic parabolic subgroup yields its order.

With a Lagrangian subspace first in a paired symplectic basis, its stabilizer consists of $\left(\begin{array}{cc}A& AS\\ 0& A^{-\mathsf{T}}\end{array}\right)$ with $S$ symmetric. The additive symmetric-matrix unipotent radical has order $q^{m(m+1)/2}$; the Levi subgroup is $G{L}_m(q)$, acting by $S\mapsto AS{A}^{\mathsf{T}}$. Its index in $S{p}_{2m}(q)$ is ${\prod }_{i=1}^m(q^i+1)$, the number of Lagrangian subspaces.

For a line $⟨e⟩$, complete $(e,f)$ to a symplectic basis with complementary symplectic space $W$. The unipotent radical has elements $u(b,t)$ sending $e$ to $e$, $w$ to $w-B(w,b)e$ and $f$ to $f+b+te$. Its multiplication is $(b,t)(b^{\prime },t^{\prime })=(b+b^{\prime },t+t^{\prime }+B(b,b^{\prime }))$, giving order $q^{2m-1}$. Its commutator is central with parameter $2B(b,b^{\prime })$. It is Heisenberg-type in odd characteristic and abelian in characteristic two. The full stabilizer also scales $e,f$ inversely and acts symplectically on $W$.

In row-block order $(W^{\prime },U,W)$ with dual bases, a unit-diagonal symplectic flag stabilizer satisfies $E=J_U{D}^T$ and $F-F^T=D{J}_U{D}^T$. The matrix $D$ is arbitrary. The latter right side is alternating, so each off-diagonal pair of entries of $F$ gives one free choice and each diagonal entry is free. ConsequentlyReversed dual-basis order inserts a reversal matrix but does not change the count. The argument holds in characteristic two without division by two.