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

The symplectic group consists of the invertible linear maps preserving the alternating bilinear form:We use row-vector action in this question, which is the convention compatible with its printed upper-triangular flag stabilizer subgroup. Thus matrices preserve a form matrix $J$ by $gJ{g}^T=J$.

Count ordered symplectic bases. There are $q^{2m}-1$ choices for the first nonzero vector $e_1$. Nondegeneracy makes the equation $(e_1,f_1)=1$ a nonzero linear-functional equation, with $q^{2m-1}$ solutions. Their span is a nondegenerate plane; its orthogonal complement is symplectic of dimension $2m-2$. Repeating there givesEach symplectic basis is the image of a fixed one under exactly one form-preserving map, justifying the count as a group order. If $q=p^a$, every factor $q^{2i}-1$ is prime to $p$, so the exact $p$-part is $q^{m^2}$.

For the block computation only, place the $f$-vectors in the order $f_1,\dots ,f_m$ after the $e$-vectors. This temporary reversal of the second block changes no transformations. The form matrix becomes $J=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)$.

Let $T$ be upper unitriangular of size $m$, and let $S$ be symmetric. The matricessatisfy $gJ{g}^T=J$ by direct block multiplication. The generator $x_{ij}(\lambda )$ is $d(I+\lambda E_{ij})$, because the inverse transpose adds $-\lambda f_i$ to $f_j$. The generators $y_{ij}(\lambda )$ and $z_i(\lambda )$ are respectively $n(\lambda (E_{ij}+E_{ji}))$ and $n(\lambda E_{ii})$. This proves all of them preserve the form, in every characteristic.

The $x$-generators generate every upper unitriangular $T$, by elimination of off-diagonal entries. The $y,z$ generators add all elementary symmetric entries, so they generate every $n(S)$. Moreoverwhich keeps symmetry. Therefore the generated group is exactlyThe two factors are uniquely determined by its diagonal and off-diagonal blocks. Their counts are $q^{m(m+1)/2}$ and $q^{m(m-1)/2}$, givingIt is a $p$-group, and this equals the full $p$-part found in part (a); hence $U_0$ is a Sylow $p$-subgroup. In the original reversed-$f$ ordering, these matrices are upper unitriangular throughout, exactly as the prescribed generators suggest. This is consistent with the finite symplectic group order. No factor of two was divided out, so characteristic two is included.

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.