Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 3 / 6

Let $q$ be a prime power and let $V={\mathbb{F}}_q^{2m}$ carry a nondegenerate alternating bilinear form $B$. The symplectic group over a finite field isIn a symplectic basis its matrix description is $g^{\mathsf{T}}Jg=J$, where $J=\left(\begin{array}{cc}0& I_m\\ -I_m& 0\end{array}\right)$. The definition is valid in characteristic two: alternating means $B(v,v)=0$, and then the form is also symmetric.

Count ordered symplectic bases $(e_1,f_1,\dots ,e_m,f_m)$ with $B(e_i,f_j)={\delta }_{ij}$ and all $e$-$e$ and $f$-$f$ pairings zero. There are $q^{2m}-1$ choices for $e_1\ne 0$. Nondegeneracy makes $v\mapsto B(e_1,v)$ a nonzero linear functional, so there are $q^{2m-1}$ choices for $f_1$ satisfying $B(e_1,f_1)=1$.

The span $P=⟨e_1,f_1⟩$ is nondegenerate. Thus $V=P\oplus P^{\perp }$, and its bilinear orthogonal complement is nondegenerate of dimension $2m-2$. For completeness, a vector in the radical of a bilinear form of $P^{\perp }$ is orthogonal both to $P$ and to $P^{\perp }$, hence to all $V$ and therefore zero. Recursing constructs and counts every symplectic basis. If $b_m$ is their number, $b_m=(q^{2m}-1)q^{2m-1}{b}_{m-1}$ with $b_0=1$.

The symplectic group over a finite field acts freely and transitively on symplectic bases: there is a unique linear map sending one ordered basis to another, and its equality of basis pairings ensures it preserves $B$. Its order is consequently $b_m$, givingThe exponent is $1+3+\cdots +(2m-1)=m^2$. The counting is valid for every prime power $q$, including $q=2$.

Identify a subset $A\subseteq \Omega$ with its indicator vector $a=(a_1,\dots ,a_n)\in {\mathbb{F}}_2^n$. Symmetric difference becomes coordinatewise addition modulo two, so the subsets form a vector space of dimension $n$, with the singleton subsets as a basis. The proposed form becomes the dot productDistributivity proves bilinearity, and commutativity proves symmetry. Its Gram matrix in the singleton basis is $I_n$, so it is nondegenerate. A permutation merely permutes coordinates and preserves the dot product, proving invariance under $S_n$.

Write $u=(1,\dots ,1)$ for the full subset. Its bilinear orthogonal complement $E=u^{\perp }$ consists exactly of even-cardinality subsets. If $n\ge 2$ is even, $B(u,u)=n=0$ in ${\mathbb{F}}_2$, so $⟨u⟩\le E$. For any $v\in E$, $B(v,v)={\sum }_i{v}_i^2={\sum }_i{v}_i=0$, so the restricted form is alternating.

Nondegeneracy of the full form gives $E^{\perp }=⟨u⟩$; one can also see this directly by testing against all vectors $e_i+e_j$, which forces all coordinates of an element of $E^{\perp }$ to be equal. Hence the radical of a bilinear form of the restricted form isDefine the quotient form by $\overline{B}(v+⟨u⟩,w+⟨u⟩)=B(v,w)$. Adding $u$ to either representative does not change the pairing with $E$, so it is well-defined. It is alternating, and a class pairing to zero with all classes has a representative in $\mathrm{rad}(B{|}_E)=⟨u⟩$, hence is zero. ThereforeBoth spaces are $S_n$-invariant, so the coordinate action descends and preserves the quotient form. This is the symplectic quotient of the binary subset module. The evenness assumption is essential because otherwise $u$ does not lie in its own bilinear orthogonal complement.

Set $n=6$ in part (b). The quotient $W=u^{\perp }/⟨u⟩$ has dimension four and a nondegenerate alternating form. The coordinate action therefore definesProve it is faithful before comparing orders. If a permutation lies in the kernel of a group homomorphism, it fixes the quotient class of each two-element subset $A$. That class has exactly the two representatives $A$ and $A+\Omega =\Omega \setminus A$. Their cardinalities are two and four. A coordinate permutation preserves cardinality, so it must fix $A$ itself, not exchange it with its complement. Thus it fixes every two-element subset. Intersecting the fixed sets $\{i,j\}$ and $\{i,k\}$ shows it fixes each singleton $\{i\}$, so it is the identity. Hence $\rho$ is injective.

By part (a),An injection between finite groups of equal order is surjective. Choosing a symplectic basis of $W$ identifies the form with the standard one, soThis realizes the exceptional isomorphism through a concrete binary permutation module. Equality of orders alone would not establish it; the faithful form-preserving action supplies the required group homomorphism.