Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 3 / 3 / c

Keep row-vector action and the block order $(W^{\prime },U,W)$. Let $R$ be the reversal matrix of size $k$, arising from the original reversed order of $f_1,\dots ,f_k$, and let $J_U$ be the alternating bilinear form matrix on $U$. ThenFor an element of $Q$, the equation $gJ{g}^T=J$ givesThus $D$, a $k\times (2m-2k)$ matrix, is arbitrary and uniquely determines $E$. The right side of the second equation is alternating, including in characteristic two: its diagonal entries vanish because $J_U$ represents an alternating bilinear form.

For any alternating matrix $K$, the equation $T-T^T=K$ has exactly $q^{k(k+1)/2}$ solutions. For each pair $i<j$, choose one entry freely and solve for the opposite entry; each diagonal entry is free. This works in characteristic two as well as odd characteristic. Taking $T=FR$ therefore givesThis is the unipotent radical count for a symplectic parabolic subgroup. It does not incorrectly replace the alternating constraint by division by two.

For a block-diagonal element, form preservation saysGiven any $C\in \mathrm{GL}(W)$, the first equation uniquely determinesWith dual bases ordered in matching rather than reversed order, the same relation is simply $A=C^{-T}$. This is the contragredient action on the paired space $W^{\prime }$.

The second equation independently allows every $B\in \mathrm{Sp}(U)$. The map taking a block-diagonal element to $(C,B)$ is a group isomorphism, with inverse $(C,B)\mapsto (R{C}^{-T}R,B,C,0,0,0)$. Thus

Taking diagonal blocks is a homomorphism $P\to L$. Its kernel is $Q$, and block-diagonal inclusion is a section. For any $g\in P$, remove its diagonal element $\ell \in L$ by $g{\ell }^{-1}\in Q$. Also $Q\cap L=1$. ThusUsing $|{\mathrm{GL}}_k(q)|=q^{k(k-1)/2}{\prod }_{i=1}^k(q^i-1)$ and part (a), the product simplifies toThe exponent simplification is $2k(m-k)+k(k+1)/2+k(k-1)/2+(m-k{)}^2=m^2$.

For an independent orbit-stabilizer theorem count, choose an ordered independent isotropic tuple $f_1,\dots ,f_k$. After $r$ choices, their span has $q^r$ elements and its orthogonal complement has dimension $2m-r$. The next choice therefore has $q^{2m-r}-q^r$ possibilities. The number of tuples isEvery totally isotropic $k$-space has $|{\mathrm{GL}}_k(q)|$ ordered bases, so the number of such spaces isEach tuple extends to a symplectic basis by successively choosing paired partners and taking orthogonal complements. Consequently the symplectic group is transitive on these spaces. The stabilizer subgroup of $W$ also preserves $W^{\perp }$, and so is exactly $P$. Dividing $|{\mathrm{Sp}}_{2m}(q)|$ by $N_k$ reproduces the boxed answer. Dividing by $T_k$ instead would count the pointwise stabilizer subgroup of the ordered tuple, a different subgroup.