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A bilinear form is alternating if $B(v,v)=0$ for every vector $v$. Expanding $B(v+w,v+w)=0$ gives $B(v,w)+B(w,v)=0$, hence $B(v,w)=-B(w,v)$. If the characteristic is not two, the converse follows from $B(v,v)=-B(v,v)$ by dividing by two. In characteristic two, the relation $B(v,w)=-B(w,v)$ is just symmetry and does not force zero self-pairings: $B(x,y)=xy$ on ${\mathbb{F}}_2$ has $B(1,1)=1$.^[ref]

The matrixgives an alternating bilinear form over every field. Over the real numbers, its value is the signed area of the parallelogram with sides $v,w$: $B((2,1),(1,3))=6-1=5$, and swapping the sides gives $-5$. Every self-pairing is zero, yet $\det J=1$, so this bilinear form is nondegenerate. Directly, pairing $(x,y)$ with $(1,0)$ gives $-y$, and pairing it with $(0,1)$ gives $x$; both vanish only when $(x,y)=0$.

In characteristic different from two, a skew-symmetric matrix of odd integer size has zero determinant, since $\det A=\det (-A^T)=(-1{)}^n\det A$. Thus an alternating bilinear form on a finite odd-dimensional vector space is a degenerate bilinear form. The stronger statement that the rank of an alternating bilinear form is even holds in all characteristics: splitting off a two-dimensional block with matrix $J$ repeatedly proves it. A nondegenerate alternating bilinear form defines a symplectic vector space, the linear setting for signed area rather than squared length.

A vector is anisotropic for a quadratic form if $Q(v)\ne 0$, and for a Hermitian form if $h(v,v)\ne 0$. The complementary zero-value condition is isotropy. For a symmetric bilinear form in odd characteristic of a field, this agrees with the associated quadratic form. A nondegenerate alternating bilinear form has all self-pairings zero, so nondegeneracy must not be confused with the existence of anisotropic vectors.

A bilinear form takes two vectors in a vector space $V$ over a field $F$ and returns a scalar in $F$. Its defining property is separate linearity: holding either input fixed gives a linear functional of the other. For scalars $a,b$ and vectors $u,v,w$,This is a scalar-valued bilinear map; the chosen field is part of the definition.^[ref]

Separate linearity does not mean that $B$ is a linear map on the vector space $V\times V$ with componentwise vector addition and scalar multiplication. Over the real numbers, $B(2v,2w)=4B(v,w)$, whereas a linear map on the pair would give $2B(v,w)$. The simplest example is $B(x,y)=xy$ on the one-dimensional vector space $\mathbb{R}$: fixing $y=3$ gives $x\mapsto 3x$, but doubling both $x$ and $y$ quadruples their product.

A bilinear form measures how two inputs interact. The dot product measures alignment; a symmetric bilinear form can describe a quadratic form; an alternating bilinear form can measure signed area. A nondegenerate bilinear form turns each vector into a unique linear functional and so connects geometry with the dual space. The same short definition therefore supports lengths, bilinear orthogonal complements, and symplectic vector spaces, depending on the additional properties imposed.

Over the complex numbers, a bilinear form is linear in both inputs without conjugating either scalar. A Hermitian form instead is a sesquilinear form and conjugates when its inputs are exchanged. For complex matrices, the adjoint $A^{\ast }$ used with Hermitian forms and matrix C-star algebras is the conjugate transpose, whereas a bilinear form uses the ordinary transpose $A^T$. For example, $B(z,w)=zw$ has $B(i,i)=-1$, while the Hermitian form $H(z,w)=z\overline{w}$ has $H(i,i)=1$.^[ref]

A binary quadratic refinement of a fixed nonsingular alternating bilinear form $B$ has polar form $B$. The difference of two refinements is linear, so nondegeneracy gives the displayed unique parametrization by $v$. For $\dim V=2m$, refinements of plus and minus Arf invariant of a quadratic form have respectively $2^{m-1}(2^m+1)$ and $2^{m-1}(2^m-1)$ elements. Completing the square in the binary character sum gives $ϵ(Q_v)=ϵ(Q_0)(-1{)}^{Q_0(v)}$.

A bilinear form is degenerate if some nonzero vector pairs to zero with every vector on at least one side. For a symmetric bilinear form or an alternating bilinear form, this says its radical of a bilinear form contains a nonzero vector. On a finite-dimensional vector space, it is equivalent to a singular matrix of a bilinear form, or $\det A=0$.

For $A=\mathrm{diag}(1,0)$, $B((x,y),(s,t))=xs$. The nonzero vector $(0,1)$ is invisible to every pairing, and changing $y$ never changes $B((x,y),w)$. The quadratic form $q=x^2$ therefore has two parallel lines $x=\pm 1$ as its unit level set. This is a lost direction, a stronger condition than being an isotropic vector: with $A=\mathrm{diag}(1,-1)$, $B((1,1),(1,1))=0$ but $B((1,1),(1,0))=1$.

For a closed oriented manifold of dimension $4r+2$, Poincare duality pairs all rational Betti numbers except the middle one, making their total contribution to the Euler characteristic even. The middle cohomology degree is odd, so its nondegenerate Poincare duality pairing is an alternating bilinear form and has even dimension. Hence the Euler characteristic is even. Nonorientable manifolds can have odd Euler characteristic, as Real projective space of even dimension demonstrates.

For $V=L({\omega }_1)$ with the G2 root system numbered short-first,In the chain basis of the crystal of the seven-dimensional G2 representation, normalize $v_1=f_1{v}_0$, $v_2=f_2{v}_1$, $v_3=f_1{v}_2$. Then $v_0\wedge v_1$ and $v_0\wedge v_3-2v_1\wedge v_2$ are nonzero highest-weight vectors of weights ${\omega }_2$ and ${\omega }_1$; the identities $e_1{v}_1=v_0$, $e_1{v}_3=2v_2$, $e_2{v}_2=v_1$ verify both raising conditions. The Weyl complete reducibility theorem and G2 dimension polynomial exhaust the twenty-one dimensions of the exterior square. In the symmetric square, $v_0^2$ generates the twenty-seven-dimensional summand. Self-duality supplies a Lie-invariant bilinear form, which is symmetric because a nondegenerate alternating bilinear form cannot have odd dimension; its inverse gives the remaining invariant line.

On $K^4$, use the standard basis $\mathcal{U}=(u_1,u_2,u_3,u_4)$ and the alternating bilinear form with matrix $J$ above for $n=2$. A second symplectic basis isFor instance, $\omega (f_1,f_2)=1-1=0$, while $\omega (e_1,f_1)=\omega (e_2,f_2)=1$ and the cross-pairings vanish. The change of basis matrix, whose columns are these new basis vectors in $\mathcal{U}$, isThe symmetric matrix $C$ makes $S^{T}JS=J$, so $S$ is a symplectic matrix. Explicitly, multiplying the blocks gives a lower-right block $C^T-C=0$; the other blocks agree with $J$. For a fixed vector, change of basis means $[v{]}_{\mathcal{U}}=S[v{]}_{\mathcal{B}}$, henceTheir symplectic form gives $3\cdot 0+2\cdot 1-1\cdot 1-1\cdot 0=1$ in $\mathcal{U}$, and $1\cdot 0+1\cdot 1-1\cdot 0-1\cdot 0=1$ in $\mathcal{B}$. The separate coordinate-plane contributions change, but their sum does not.

The same matrix also represents an active symplectic linear map: keeping one basis fixed, it sends $(q,p)$ to $(q+Cp,p)$. The following diagram shows this active interpretation over the real numbers. Its preserved total pairing is precisely the test used for linear canonical transformations.

Two vectors spanning a hyperbolic plane. For an alternating bilinear form, choose $B(e,f)=1$; then their span is nondegenerate. For a quadratic form with polar form $B$, require also $Q(e)=Q(f)=0$. The analogous definition for a Hermitian form uses two isotropic vectors of pairing one. Extending such pairs to adapted bases is useful for constructing form-preserving maps.

In the irreducible sl2 Lie algebra module $V(m)$, use $v_j=f^j{v}_0$, with $0\le j\le m$. The Lie-invariant bilinear form $B(v_j,v_k)=(-1{)}^j{1}_{\{j+k=m\}}$ is nondegenerate: its anti-diagonal is nonzero. Invariance follows from $h{v}_j=(m-2j)v_j$, $f{v}_j=v_{j+1}$ and $e{v}_j=j(m-j+1)v_{j-1}$. Its transpose equals $(-1{)}^{m}B$, so it is a symmetric bilinear form when $m$ is an even number, and an alternating bilinear form when $m$ is an odd integer. The Weyl complete reducibility theorem gives a nondegenerate invariant form on any finite-dimensional module by taking the orthogonal direct sum of these forms.

For trivial coefficients in a field $k$, a Lie algebra two-coboundary is a bilinear map $\beta (x,y)=t([x,y])$ for a linear functional $t$ in the dual space ${\mathfrak{g}}^{\ast }$. Some conventions put a minus sign in this formula; the resulting vector subspace $B^2(\mathfrak{g},k)$ is the same. Each such map is a Lie algebra two-cocycle: it is an alternating bilinear form, and its cyclic expression is $t([[x,y],z]+[[y,z],x]+[[z,x],y])=0$ by the Jacobi identity. Replacing a linear section of a central extension of a Lie algebra by $s^{\prime }(x)=s(x)+t(x)z$ changes its Lie algebra two-cocycle by $-\beta$. Thus $H^2(\mathfrak{g},k)=Z^2(\mathfrak{g},k)/B^2(\mathfrak{g},k)$ in Lie algebra cohomology records exactly the obstruction to a section that is a Lie algebra homomorphism.

A bilinear form on a Lie algebra representation is invariant when it satisfies the displayed identity. Equivalently, $T_B(v)=B(v,-)$ is an intertwining operator from the representation to its dual Lie algebra representation. For a finite-dimensional irreducible representation over an algebraically closed field, any nonzero such form is nondegenerate, and the Schur lemma makes all invariant forms proportional. In characteristic different from two, transposing twice then proves that the form is a symmetric bilinear form or an alternating bilinear form.

A bilinear form $B$ is nondegenerate when its left radical and right radical are both zero: every nonzero vector can be detected by some pairing, whichever input slot it occupies. On a finite-dimensional vector space, either condition implies the other, and the following tests are equivalent:Here $A$ is the matrix of a bilinear form, and $V^{\ast }$ is the dual space. The linear map $w\mapsto B(-,w)$ has matrix $A$ in the chosen basis and its dual basis, so its kernel is the right radical. The map $v\mapsto B(v,-)$ has matrix $A^T$. An invertible matrix has zero kernel on both sides, proving the finite-dimensional equivalence. In infinite dimensions, zero left radical and right radical imply these maps are injective, but do not generally make them surjective.^[ref]

For the real symmetric bilinear form $B((x,y),(s,t))=xs-yt$, the determinant is $-1$. Its isotropic vector $(1,1)$ has zero self-pairing but pairs to $1$ with $(1,0)$, so it is detected. The alternating bilinear form with matrix $J$ has determinant $1$ and detects every nonzero vector even though all self-pairings vanish. Neither zero self-pairings nor the absence of positive definiteness implies degeneracy.

Nondegeneracy is useful because it turns a linear functional into a unique vector. For $A=\mathrm{diag}(1,-1)$, the linear functional $\ell (s,t)=2s+3t$ is $B((2,-3),(s,t))$; solving for that representing vector amounts to solving an invertible system of linear equations. The analogous identification underlies bilinear orthogonal complements and the dual isomorphisms induced by a nondegenerate pairing.

The group of invertible linear maps preserving a specified quadratic form over a finite field. In characteristic two one must preserve the quadratic form itself, not merely its polar alternating bilinear form. For a nonsingular form, any two nonzero isotropic vectors can be incorporated into hyperbolic pairs and standard adapted bases, producing an isometry between them.

Since the general linear Lie algebra ${\mathfrak{gl}}_1(k)$ has zero Lie bracket, identify it with $kz$. A central extension of a Lie algebra is a short exact sequenceof Lie algebra homomorphisms with $\iota (kz)$ contained in the center of a Lie algebra $Z(\tilde{\mathfrak{g}})$. Thus $\pi$ is surjective, its kernel is precisely the indicated line, and that line commutes with all of $\tilde{\mathfrak{g}}$.

Choose a linear map $s:\mathfrak{g}\to \tilde{\mathfrak{g}}$ with $\pi s=\mathrm{id}$. Such a section exists by lifting a basis. Identify $\tilde{\mathfrak{g}}$ with the direct sum of vector spaces $\mathfrak{g}\oplus kz$ by $(x,tz)\mapsto s(x)+t\iota (z)$. The difference $[s(x),s(y)]-s([x,y])$ belongs to the kernel of $\pi$, so there is a unique bilinear map $\alpha :\mathfrak{g}\times \mathfrak{g}\to k$ such thatIt is an alternating bilinear form: $\alpha (x,x)=0$, equivalently $\alpha (x,y)=-\alpha (y,x)$ in characteristic zero. Taking the central component of the Jacobi identity gives the Lie algebra two-cocycle conditionConversely, any alternating bilinear form satisfying this identity defines a Lie algebra byIndeed its Lie bracket is an alternating multilinear map and a bilinear map, and the two components of its Jacobi identity are respectively the Jacobi identity in $\mathfrak{g}$ and the displayed Lie algebra two-cocycle condition. The line $kz$ is central, and projection to $\mathfrak{g}$ gives the required central extension of a Lie algebra.

For later use, replacing $s$ by $s^{\prime }(x)=s(x)+t(x)\iota (z)$, where $t\in {\mathfrak{g}}^{\ast }$ is in the dual space, changes the Lie algebra two-cocycle toThus changes of section add a Lie algebra two-coboundary, up to the conventional sign. A section that is a Lie algebra homomorphism has ${\alpha }^{\prime }=0$ and gives a direct sum of Lie algebras.

Choose a basis $x,y$ of the two-dimensional abelian Lie algebra $\mathfrak{a}$, and lifts with the same names. Every alternating bilinear form is determined by $c=\alpha (x,y)$, and every such form is a Lie algebra two-cocycle because every Lie bracket in $\mathfrak{a}$ is zero. Thus every central extension of a Lie algebra has a basis $x,y,z$ withIf $c=0$, all Lie brackets vanish, so this is the direct sum of Lie algebras $\mathfrak{a}\oplus kz\cong k^3$. If $c\ne 0$, replace the central basis vector by $z^{\prime }=cz$. Then $[x,y]=z^{\prime }$, which is the Heisenberg Lie algebra ${\mathfrak{h}}_3(k)$. This change of basis shows that every nonzero $c$ gives the same Lie algebra isomorphism class.

These two Lie algebras cannot be isomorphic: their derived algebras have dimensions zero and one, respectively, and a Lie algebra isomorphism preserves the derived algebra. Moreover, the Heisenberg Lie algebra cannot be a direct sum of Lie algebras $kz\oplus \mathfrak{a}$, since that direct sum of Lie algebras would have zero Lie bracket. ThereforeHere “up to isomorphism” means the Lie algebra isomorphism class of the middle Lie algebra, allowing a change of the identifications at the ends of the short exact sequence. If instead equivalence of central extensions of a Lie algebra must induce the identity on both $kz$ and $\mathfrak{a}$, the parameter $c$ cannot change: the formula in part (a) has $t([x,y])=0$. That stricter notion has one class for each $c\in k$, rather than two. This distinction explains the convention required by the question.

First put the nonabelian two-dimensional Lie algebra in a convenient form. For any basis $p,q$, let $u=[p,q]=ap+bq\ne 0$. Then $[p,u]=bu$ and $[q,u]=-au$. Since $(a,b)\ne (0,0)$, some linear combination $x$ of $p,q$ has $[x,u]=u$. In particular, $x,u$ are linearly independent. Taking $y=u$ gives a basis withThis is the affine Lie algebra of the line.

Every alternating bilinear form on this vector space isFor any alternating bilinear form $\alpha$, the expressionis an alternating trilinear form: it is trilinear, and setting any two arguments equal makes the terms cancel. Every alternating trilinear form on a two-dimensional vector space vanishes, since any triple is linearly dependent. Hence every ${\alpha }_c$ satisfies the Lie algebra two-cocycle condition, and there are no others:

We use the supplied weight decomposition $V={⨁}_{\lambda \in \Psi (\mathfrak{a})}{V}_{\lambda }$. There is a small distinction to justify: the definition of a toral subalgebra uses the Adjoint representation, while these weight spaces use the defining representation. For the symplectic Lie algebra, adjoint semisimplicity in the symplectic Lie algebra implies that each $a\in \mathfrak{a}$ is diagonalizable on $V$ as well. Here is a direct verification.

Define the symplectic adjoint $a^{\ast }$ by $\omega (av,w)=\omega (v,a^{\ast }w)$. The symplectic Lie algebra condition is $a^{\ast }=-a$. Write the Jordan–Chevalley decomposition in $\mathrm{End}(V)$ as $a=S+N$, with $S$ diagonalizable, $N$ a nilpotent linear map, and $SN=NS$. Uniqueness of the Jordan–Chevalley decomposition, applied to $a^{\ast }=-a$, gives $S^{\ast }=-S$ and $N^{\ast }=-N$. Thus $S,N\in \mathfrak{sp}(V)$. On $\mathrm{End}(V)$, $\mathrm{ad}S$ is diagonalizable by using a matrix unit basis in an eigenbasis of $S$, and $\mathrm{ad}N$ is a nilpotent linear map, sincevanishes for large $m$. These operators commute and preserve $\mathfrak{sp}(V)$, so their restrictions are the Jordan–Chevalley decomposition of $\mathrm{ad}a$. Because $\mathrm{ad}a$ is diagonalizable, its nilpotent part is zero: $\mathrm{ad}N=0$ on $\mathfrak{sp}(V)$.

To conclude that $N=0$ without an assumption on the center of a Lie algebra, observe that each map $T_v(w)=\omega (v,w)v$ lies in $\mathfrak{sp}(V)$. If $v\ne 0$, its image is $kv$, because $\omega$ is a nondegenerate bilinear form. Since $N$ commutes with $T_v$, it preserves $kv$. Hence $Nv\in kv$ for every $v$, forcing $N$ to be a scalar linear map: compare its action on two linearly independent vectors and their sum. A scalar nilpotent linear map is zero. For $V=0$ the assertion is immediate. This proves the needed diagonalizability. Commuting diagonalizable linear maps admit simultaneous diagonalization, giving the displayed weight decomposition.

Now let $v\in V_{\lambda }$ and $w\in V_{\mu }$. For every $a\in \mathfrak{a}$, preservation of the alternating bilinear form givesIf $\lambda +\mu \ne 0$ as a linear functional, choose $a$ on which it is nonzero; then $\omega (v,w)=0$. ThereforeFor $v\ne 0$ in $V_{\lambda }$, the fact that $\omega$ is a nondegenerate bilinear form supplies a $w\in V$ with $\omega (v,w)\ne 0$. Decomposing $w$ into weight spaces, only its $V_{-\lambda }$ component can contribute. Thus $V_{-\lambda }\ne 0$, and $\Psi (\mathfrak{a})=-\Psi (\mathfrak{a})$. The same argument on both sides shows that $\omega :V_{\lambda }\times V_{-\lambda }\to k$ is a perfect pairing, so $\dim V_{\lambda }=\dim V_{-\lambda }$. On $V_0$, the restriction of $\omega$ is a nondegenerate bilinear form that is an alternating bilinear form, so $\dim V_0$ is even.

Choose one representative ${\lambda }_i$ from each pair of nonzero weights of a representation, for $1\le i\le r$. HereThe linear map $\mathfrak{a}\to k^r$ sending $a$ to $({\lambda }_1(a),\dots ,{\lambda }_r(a))$ is injective: an element in its kernel acts by zero on every nonzero weight space, and also on $V_0$, so it is the zero element of $\mathfrak{a}\subseteq \mathrm{End}(V)$. Taking dimensions yieldsThe final inequality follows because each pair contributes at least two to $\dim V=2n$. Equivalently, the nonzero weights of a representation span the dual space ${\mathfrak{a}}^{\ast }$; they need not be linearly independent at this stage.

Writeon $V=W\times W^{\prime }$, where $W^{\prime }$ is the dual space. The dual pairing is linear separately in the vector and the linear functional. For example, for real numbers $a,b$,Thus $\omega$ is a bilinear form. It is also an alternating bilinear form, since

To prove that this is a nondegenerate bilinear form, suppose $(v,f)$ pairs to zero with every $(w,g)$. Setting $g=0$ gives $f(w)=0$ for every $w\in W$, hence $f=0$. Setting $w=0$ gives $g(v)=0$ for every $g\in W^{\prime }$. If $v\ne 0$, basis extension supplies a coordinate functional with $g(v)=1$, a contradiction. Thus $v=0$. The left radical is zero, and the displayed change of sign implies the right radical is zero as well. Hence $\omega$ is a nondegenerate bilinear form, making $V$ a symplectic vector space.

For the required example take the coordinate subspaceUsing the definition of the bilinear orthogonal complement in part (b),ThereforeThis is a Lagrangian subspace. Its dimension is $\dim W$, exactly half of $\dim V=2\dim W$, so part (b)(i) is satisfied.

There is no contradiction with part (b)(ii): that part asks whether a direct sum is guaranteed, and the answer is already no. Moreover, when $W\ne 0$, the present alternating bilinear form is not a symmetric bilinear form. Choose $v\ne 0$ and $f\in W^{\prime }$ with $f(v)=1$; thenIt therefore does not satisfy the symmetry hypothesis of part (b)(ii). If $W=0$, then $V=U=0$, and both $U=U^{\perp }$ and $V=U\oplus U^{\perp }$ hold trivially.

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 $q=2$.