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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 finite-dimensional complex Lie algebra is semisimple if and only if its Killing form is nondegenerate.

A vector bundle $E$ with a fibrewise quadratic form $Q$ defines a bundle whose fibre over $x$ is the Clifford algebra $\mathrm{Cl}(E_x,Q_x)$. When the polar form is a nondegenerate bilinear form, local orthonormal frames identify its fibres with a fixed Clifford algebra, and orthogonal transition maps induce the algebra transition maps. A Clifford module bundle is a bundle of modules over these algebras, rather than the algebra bundle itself. For a Riemannian manifold, use its cotangent bundle and the metric quadratic form, with the sign convention specified.

The four-dimensional diamond Lie algebra is the semidirect product of Lie algebras of the three-dimensional Heisenberg Lie algebra with a derivation having eigenvalues zero, one and minus one. With basis $c,d,p,q$, its nonzero brackets are $[d,p]=p$, $[d,q]=-q$, $[p,q]=c$. Its derived series of a Lie algebra has dimensions four, three, one and zero. The symmetric form with $B(c,d)=B(p,q)=1$ and all other basis pairings zero is a nondegenerate bilinear form and invariant. The alternating tensor $B([x,y],z)$ is the volume form on $⟨d,p,q⟩$, verifying invariance.

If $T$ preserves a nondegenerate bilinear form $\omega$ that is alternating and satisfies $T^2=I$ on a finite-dimensional vector space over a field of characteristic different from two, thenThe two eigenspaces are orthogonal for $\omega$, since $\omega (v_{+},v_{-})=\omega (T{v}_{+},T{v}_{-})=-\omega (v_{+},v_{-})$. Each restricted form is nondegenerate: a vector annihilating its own eigenspace also annihilates the other and hence all of $V$. Both dimensions are even, say $2a$ and $2b$. Consequently $\mathrm{tr}T=2a-2b=\dim V-4b$ as an integer obtained from the eigenspace dimensions. Over the real numbers this gives the congruence $\mathrm{tr}T\equiv \dim V(\mathrm{mod}4)$.

A bilinear form $B$ on $V$ is invariant exactly when $v\mapsto B(v,-)$ is a $G$-homomorphism $V\to V^{\ast }$. On an irreducible representation, Schur's lemma makes a nonzero such form nondegenerate and unique up to scale.

In $\mathcal{L}=-g_{ab}{F}_{\mu \nu }^a{F}^{b\mu \nu }/4$, only the symmetric part of $g$ contributes. A constant symmetric bilinear form gives local gauge invariance precisely when it is an invariant bilinear form on a Lie algebra, because the gauge field strength transforms in the Adjoint representation. A nondegenerate kinetic term requires a nondegenerate bilinear form; positivity of the Hamiltonian is an additional condition. For a compact simple real Lie algebra, $-\kappa$ is positive definite, and a positive multiple supplies the usual positive-energy convention with Minkowski signature $(+---)$.

A smooth manifold carrying a nondegenerate bilinear form on each tangent space, with one negative and all remaining positive eigenvalues, or the overall reversed-sign convention. Its nonzero null vectors have vanishing metric quadratic form.

A metric tensor is a nondegenerate symmetric covariant tensor that defines squared line elements by $d{s}^2=g_{\mu \nu }d{x}^{\mu }d{x}^{\nu }$.

With the mostly-plus metric signature $(-,+,+,+)$ and coordinates $X^{\mu }=(ct,x,y,z)$, the Minkowski metric is the nondegenerate bilinear formIt is a symmetric bilinear form, but not a positive-definite bilinear form. For example, $U=(1,1,0,0)$ is nonzero yet $U\cdot U=0$, making it a null vector. This does not make the Minkowski metric degenerate: some other vectors have nonzero pairing with $U$.

Applied to infinitesimal displacements, the Minkowski metric givesFor a timelike curve, a clock records $d\tau =\sqrt{-d{s}^2}/c$. For a null curve, $d{s}^2=0$. For spacelike separation, the positive Minkowski interval measures a different kind of separation. A Lorentz transformation preserves this form, expressed in matrix notation by ${\Lambda }^T\eta \Lambda =\eta$, although it need not preserve ordinary Euclidean distance in the four-component coordinate list.

For nonzero $x\in K$, choose $y=x^{-1}$. Thenso the trace form is nondegenerate.

Let $E_{ji}$ be the matrix unit having its only nonzero entry, one, in position $(j,i)$. Direct multiplication gives $\mathrm{tr}(A{E}_{ji})=a_{ij}$. If $A\in R$, every such pairing vanishes, forcing every entry $a_{ij}$ to be zero. Conversely, zero pairs to zero with every matrix. ThusThe matrix trace pairing is therefore a nondegenerate bilinear form, despite its failure to be positive definite when $n\ge 2$.

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.

Put $b(v,w)=\phi (w)(v)$. Each $\phi (w)$ belongs to the dual space $V^{\prime }$, so it is a linear functional. For real numbers $a,b_0$ and vectors $v_1,v_2,w$, this givesThe linearity of $\phi$ gives linearity in the second argument as well:Thus $b$ is a bilinear form.

Its right radical of a bilinear form is exactly $\ker \phi$, becauseIf $b$ is a nondegenerate bilinear form, this kernel is zero. Since $V$ is a finite-dimensional vector space and $\dim V^{\prime }=\dim V$, the rank-nullity theorem makes $\phi$ surjective.

Conversely, if $\phi$ is surjective, the same dimension equality makes it injective, so the right radical is zero. If $v$ lies in the left radical, then $f(v)=0$ for every $f\in V^{\prime }$, because every $f$ is some $\phi (w)$. A nonzero $v$ can be included in a basis and detected by its coordinate functional, so this forces $v=0$. Both radicals vanish, proving

No. On the real vector space $V={\mathbb{R}}^2$, use the symmetric bilinear formIts matrix of a bilinear form is $\mathrm{diag}(1,-1)$, whose determinant is $-1\ne 0$, so it is a nondegenerate bilinear form. LetA vector $(s,t)$ belongs to the bilinear orthogonal complement $U^{\perp }$ exactly when $s-t=0$. HenceIn particular, the claimed direct sum fails. The isotropic vector $(1,1)$ has zero self-pairing, although it pairs nontrivially with other vectors of $V$.

The dimension identity in part (i) counts both vector subspaces but does not force their intersection of vector subspaces to be zero. For a symmetric bilinear form, $U\cap U^{\perp }$ is the radical of a bilinear form restricted to $U$. Thus the precise criterion isIndeed, trivial intersection of vector subspaces and part (i) give a sum of vector subspaces of dimension $\dim V$. A positive-definite bilinear form would guarantee this for every $U$; symmetry alone does not.

For the given nondegenerate bilinear form, define the linear mapBy part (a), $\Phi$ is a linear isomorphism. The paper's bilinear orthogonal complement uses $U$ in the first argument; with this convention,Here $U^0$ is the annihilator of a vector subspace $U$. Since $\Phi$ is a linear isomorphism, its restriction gives a linear isomorphism $U^{\perp }\to U^0$.

For completeness, choose a basis $u_1,\dots ,u_k$ of $U$ and extend it to a basis $u_1,\dots ,u_n$ of $V$. In its dual basis $u_1^{\ast },\dots ,u_n^{\ast }$, a linear functional vanishes on $U$ exactly when its first $k$ coefficients are zero. ThereforeConsequentlyThe argument includes $U=0$ and $U=V$, and requires no symmetry assumption on the bilinear form.

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.

For a bilinear form $\varphi$, define $F_{\varphi }:V\to V^{\ast }$ by $F_{\varphi }(v)(u)=\varphi (u,v)$. Bilinearity makes $F_{\varphi }(v)$ a linear functional and makes $F_{\varphi }$ linear in $v$. Conversely, every linear $F:V\to V^{\ast }$ defines a bilinear form by ${\varphi }_F(u,v)=F(v)(u)$. These constructions undo each other, giving the desired bijective correspondence.

For a nondegenerate bilinear form, $F_{\varphi }$ has zero kernel and is an isomorphism because $V$ and its dual space have the same finite dimension. For the two forms, putIt is an isomorphism and satisfies ${\varphi }_1(u,\alpha v)=F_{{\varphi }_1}(\alpha v)(u)=F_{{\varphi }_2}(v)(u)={\varphi }_2(u,v)$ for all $u,v$. This is the representation of a bilinear form relative to a nondegenerate bilinear form.

If both forms are symmetric, thenNondegeneracy makes the adjoint with respect to ${\varphi }_1$ unique; this equality proves $\boxed{{\alpha }^{†}=\alpha }$ in that bilinear-form sense. No positivity or Hermitian conjugation is assumed over the arbitrary field.

For a symplectic vector space $(V,\omega )$ and a vector subspace $W$, the symplectic orthogonal complement isNondegenerate bilinear form identifies $V$ with its dual space, so $\dim W^{\omega }=\dim V-\dim W$. If $W$ is a symplectic subspace, $W\cap W^{\omega }=0$, and hence $V=W\oplus W^{\omega }$; the restriction to $W^{\omega }$ is also nondegenerate.

Here is an inductive construction of a symplectic basis. Choose $e_1\ne 0$. Nondegenerate bilinear form supplies $f_1$ with $\omega (e_1,f_1)=1$. The plane $P=\mathrm{span}(e_1,f_1)$ is a symplectic subspace, and its symplectic orthogonal complement gives $V=P\oplus P^{\omega }$. Repeat on $P^{\omega }$. The process ends in dimension zero, since each step removes two dimensions and a one-dimensional alternating form is degenerate. We obtain $e_1,f_1,\dots ,e_n,f_n$ withThe linear map taking the coordinate vectors of ${\mathbb{R}}^{2n}$ to this basis pulls $\omega$ back to ${\sum }_{j}d{x}_j\wedge d{y}_j$, proving the asserted normal form.

Now split $V=C\oplus D$ with $D=C^{\omega }$, and choose symplectic bases on both planes. Parametrize ${\Gamma }_A$ by $v\mapsto (Av,v)$. Orthogonality removes the cross terms, and a two-dimensional alternating form transforms by the determinant, soConsequently the correct nondegenerate bilinear form criterion isThe strict inequality printed in the PDF requires an extra orientation condition: it says that the symplectic orientation on the graph agrees with the orientation transported from $D$. It is not equivalent to being a symplectic subspace. For example, $A=\mathrm{diag}(-2,1)$ has determinant $-2$ and gives a nondegenerate restriction $-{\omega }_D$. This is the orientation criterion for a graph of symplectic planes.

This same map supplies the requested obstruction. Orient $C$ and ${\Gamma }_A$ by their restricted symplectic forms. If $(d_1,d_2)$ is a positive basis of $D$, then $(A{d}_1+d_1,A{d}_2+d_2)$ is negatively oriented in ${\Gamma }_A$. The coordinate transformation from a basis of $C$ followed by these graph vectors to a basis of $C\oplus D$ has determinant $1$. Reversing the graph basis to give its positive orientation therefore makes the oriented direct sum $C\oplus {\Gamma }_A$ negative relative to the ambient symplectic orientation.

Two distinct complex lines in standard ${\mathbb{C}}^2$ have the opposite behaviour: complex bases of the two lines give a complex-linear isomorphism $\mathbb{C}\oplus \mathbb{C}\to {\mathbb{C}}^2$, whose real determinant is the positive number $|{\det }_{\mathbb{C}}{|}^2$. Their complex orientations are their symplectic orientations. A symplectic linear map preserves both the ambient and each plane's symplectic orientation, so it cannot change this sign. Thus the pair with $A=\mathrm{diag}(-2,1)$ cannot be the image of a pair of complex lines. Interchanging the planes does not change the sign, since both dimensions are even.

Define a symmetric bilinear form $B$ by pairing $c$ with $d$ and $p$ with $q$, each with value one, and setting all other pairings of basis vectors to zero. Its Gram matrix in the order $(c,d,p,q)$ isThe determinant is one, so this is a nondegenerate bilinear form.

For invariance, put $T(x,y,z)=B([x,y],z)$. The nonzero values on distinct basis triples are generated byand the reversed triples have value minus one. Every value involving $c$, or a repeated basis vector, vanishes. Thus $T$ is an alternating trilinear form, explicitly $d^{\ast }\wedge p^{\ast }\wedge q^{\ast }$. In particular $T(x,y,z)=T(y,z,x)$. By symmetry of $B$, this reads $B([x,y],z)=B(x,[y,z])$, the defining identity for an invariant bilinear form on a Lie algebra.

Suppose a finite-dimensional Lie algebra representation $\rho$ realized this form as $B(x,y)=\mathrm{tr}(\rho (x)\rho (y))$. The Lie theorem says that every finite-dimensional representation of a complex solvable Lie algebra admits a basis in which all representing matrices are upper triangular. A commutator of upper triangular matrices has zero diagonal, hence is strictly upper triangular. Therefore $\rho ([\mathfrak{g},\mathfrak{g}])$ is strictly upper triangular, and its product with any representing matrix still has zero diagonal. It follows thatThis is the general fact that trace forms of solvable Lie algebras annihilate the derived algebra. Here $c\in [\mathfrak{g},\mathfrak{g}]$, so such a trace form would have $B(c,d)=0$, contradicting the constructed value one. Thus the nondegenerate invariant form exists, but no finite-dimensional representation has it as its trace form. Finite dimensionality is implicit in the ordinary representation trace; no trace on arbitrary infinite-dimensional endomorphisms is being assumed.