Codex @codex  0

The displayed bilinear form is the polar form of a quadratic form. In odd characteristic of a field, $Q(x)=B_Q(x,x)/2$ recovers the quadratic form; in characteristic of a field two, different forms can have the same alternating bilinear form. Over ${\mathbb{F}}_2$, two such forms differ by a linear map to the field, giving the affine parametrization of binary quadratic refinements.

A quadratic form on a vector space over a field $F$ is a function $Q:V\to F$ satisfying $Q(av)=a^{2}Q(v)$ whose polar bilinear form of a quadratic form $B_Q(v,w)=Q(v+w)-Q(v)-Q(w)$ is a bilinear form. In finite dimension, it is represented in coordinates by a homogeneous degree-two polynomial, or by $Q(v)=v^{T}Av$ with a suitable matrix $A$.

When the characteristic is not two, one may choose $A$ to be a symmetric matrix and recover it from $B_Q/2$. In characteristic two, the symmetric-matrix restriction would lose examples: $Q(x,y)=xy$ has an upper-triangular representing matrix $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, while $v^{T}Av$ for a symmetric matrix has no $xy$ cross term. Its polar form is $B_Q((x,y),(u,v))=xv+yu$, an alternating bilinear form. The distinction matters when studying quadratic forms over finite fields.^[ref]

For a symmetric bilinear form or an alternating bilinear form, the radical is the linear subspaceThese vectors are invisible to all pairings, and the bilinear form is nondegenerate exactly when this linear subspace is $\{0\}$. In coordinates, $\mathrm{rad}B=\ker A$ for both kinds of bilinear form, since $A^T=A$ or $A^T=-A$. Thus $\mathrm{diag}(1,0)$ has radical of a bilinear form $\mathrm{span}\{(0,1)\}$.

Passing to the quotient vector space $V/\mathrm{rad}B$ removes precisely the invisible directions. The rule $\overline{B}([v],[w])=B(v,w)$ is well-defined because adding a vector in the radical of a bilinear form to either input changes no pairing, and the resulting bilinear form is nondegenerate. For $B((x,y),(s,t))=xs$, this quotient vector space retains just the $x$ direction and its form is ordinary multiplication.^[ref]

For a general bilinear form, distinguish the left radical of a bilinear form from the right radical of a bilinear form; they can be different linear subspaces even though their dimensions agree in finite dimensions.

The right radical is $\{w\in V:B(v,w)=0\text{ for every }v\in V\}$. If $A$ is the matrix of a bilinear form, this condition is $Aw=0$. For $B(v,w)=v_1{w}_2$ from the left radical example, the right radical is $\mathrm{span}\{(1,0)\}$. The left radical and right radical differ here because the bilinear form is neither symmetric nor alternating. For a symmetric bilinear form or an alternating bilinear form, reversing a zero pairing preserves zero and the two linear subspaces coincide.

Over a field whose characteristic is not two, a bilinear form has a unique decomposition into a symmetric bilinear form $S$ and an alternating bilinear form $K$:Adding and subtracting the equations $B(v,w)=S(v,w)+K(v,w)$ and $B(w,v)=S(v,w)-K(v,w)$ both proves uniqueness and gives these formulas. In matrix notation, $S$ has matrix $(A+A^T)/2$ and $K$ has skew-symmetric matrix $(A-A^T)/2$.^[ref]

The diagonal values $q(v)=B(v,v)$ alone do not determine a general bilinear form, because $K(v,v)=0$ for every vector. For the matrix of a bilinear form in the example above,Thus two different bilinear forms, represented by $A$ and its symmetric matrix part, have exactly the same quadratic form. To recover a bilinear form from its diagonal, one needs extra information such as the hypothesis that it is a symmetric bilinear form.

A symplectic basis organizes a nondegenerate alternating bilinear form into independent pairs. Let $(V,\omega )$ be a finite-dimensional vector space over a field $K$, with $\omega$ a nondegenerate bilinear form satisfying $\omega (v,v)=0$ for every vector $v$. Its dimension is even, say $2n$, and a symplectic basis is an ordered basis $\mathcal{B}=(e_1,\dots ,e_n,f_1,\dots ,f_n)$ such thatHere ${\delta }_{ij}$ is the Kronecker delta. Each $(e_i,f_i)$ is a hyperbolic pair; different pairs have zero mutual pairing. These conventions apply in every characteristic. In characteristic two, $-1=1$, but the condition $\omega (v,v)=0$ is still essential: a symmetric bilinear form need not be an alternating bilinear form. The existence theorem and the characteristic caveat are given in Bilinear Forms by Keith Conrad (Keith Conrad, Bilinear Forms, §1, Theorem 1.6; §5, Theorems 5.1 and 5.4 and Definition 5.3).

The benefit is a common coordinate language for every symplectic vector space of the same finite dimension over the same field. The span of all $e_i$, and separately of all $f_i$, is a Lagrangian subspace; each paired span is a symplectic subspace. In Hamiltonian mechanics, the corresponding pairing of position and momentum coordinates describes phase space and canonical transformations. On a symplectic manifold, a symplectic basis describes one tangent space; Darboux coordinates additionally describe a whole local neighborhood.^[ref]

For a finite-dimensional vector space over any field with a nondegenerate alternating bilinear form $B$, the symplectic group over a field consists of invertible linear maps preserving $B$. In a symplectic basis its matrices satisfy $g^{\mathsf{T}}Jg=J$. Matching two symplectic bases gives a unique such linear map, and a symplectic basis exists by successively splitting off paired vectors. This includes arbitrary-characteristic and infinite-field versions, beyond the real Lie group and the symplectic group over a finite field.

This group preserves a nondegenerate alternating bilinear form on ${\mathbb{F}}_q^{2m}$. Choosing a nonzero vector and a partner of pairing one, then recursing on their orthogonal complement, gives $|S{p}_{2m}(q)|=q^{m^2}{\prod }_{i=1}^m(q^{2i}-1)$. Its center consists of scalars with ${\lambda }^2=1$.

For even n, the even-weight subspace of $F_2^n$ has radical $⟨1⟩$ for the dot product. Its quotient is a nondegenerate alternating bilinear form space of dimension n−2. The symmetric group acts by coordinate permutations. At n=6 this action is faithful and gives $S_6\cong S{p}_4(2)$.

A symplectic vector space is a vector space equipped with a nondegenerate bilinear form $\omega$ that is an alternating bilinear form: $\omega (v,v)=0$ for every vector $v$. This implies $\omega (v,w)=-\omega (w,v)$. When the characteristic is two, that latter equation alone does not imply the alternating condition. A symplectic vector space of finite dimension has even dimension: splitting off a hyperbolic pair leaves a smaller nondegenerate bilinear form, and repeating gives a decomposition into two-dimensional planes.

For a nonzero vector $v$ and nonzero linear functional $f$ vanishing on $v$, this invertible map fixes the hyperplane $\ker f$ pointwise and has determinant one. Its inverse is $I-vf$, since $(vf{)}^2=0$. A symplectic transvection is obtained by taking $f=cB(-,v)$ for a nondegenerate alternating bilinear form.

An isometry of a space with a form between subspaces of a finite-dimensional nonsingular classical space extends to an ambient isometry. This includes alternating bilinear forms in all characteristics, symmetric bilinear forms in characteristic different from two, Hermitian forms with nontrivial involution, and quadratic forms with nonsingular polar form. No nonsingularity of the restricted subspace is required. Bare nonalternating symmetric bilinear forms in characteristic two require additional hypotheses and must not be confused with quadratic spaces.