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An alternating trilinear form is a scalar-valued map, linear in three arguments, which vanishes whenever two arguments agree. In characteristic zero, interchanging two arguments negates its value, while cyclic permutation preserves it. For a symmetric bilinear form $B$ on a Lie algebra, the tensor $B([x,y],z)$ is alternating exactly when $B$ is an invariant bilinear form on a Lie algebra.

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 map $B:V\times W\to Z$ between vector spaces is bilinear when it is linear in either argument with the other held fixed. A scalar-valued bilinear form is the case $V=W$ and $Z$ is the scalar field. Checking bilinearity is one of the algebraic steps in verifying a real inner product; being a symmetric bilinear form and having positive squared norm impose further conditions.

A bilinear form $B:X\times X\to \mathbb{R}$ on a real normed vector space is bounded when one constant $0\le C<\infty$ satisfiesThis is an upper bound on the magnitude of every pairing. It says nothing about the sign of $B(v,v)$ and does not require a symmetric bilinear form. The chosen norm is part of the hypothesis. Boundedness is equivalent to joint continuity, as shown by the boundedness criterion for a bilinear form.^[ref]

For example, on ${\mathbb{R}}^2$ with the Euclidean norm, the bilinear formsboth satisfy the upper bound with $C=1$, by the Cauchy-Schwarz inequality. Yet their quadratic forms are $B_{-}(v,v)=x^2-y^2$ and $B_0(v,v)=x^2$: the former takes negative values and the latter vanishes at the nonzero vector $(0,1)$. Thus neither is a coercive bilinear form. The latter is also a degenerate bilinear form, whereas $B_{-}$ is nondegenerate. Boundedness controls size, not positivity or loss of directions.

Every bilinear form on a real or complex finite-dimensional vector space equipped with any norm is bounded. In coordinates, write $B(u,v)=u^{T}Av$ using its matrix of a bilinear form. The Cauchy-Schwarz inequality gives $|u^{T}Av|\le \Vert A{\Vert }_2\Vert u{\Vert }_2\Vert v{\Vert }_2$, with $\Vert A{\Vert }_2$ the spectral norm; finite-dimensional equivalence of norms transfers this to the chosen norm. The conclusion concerns each fixed norm and fixed bilinear form: it does not supply a single constant for an arbitrary family of matrices or changing dimensions.

On an infinite-dimensional vector space, the upper bound is an additional condition. Equip the finitely supported sequences $c_{00}$ with the norm inherited from the l2 sequence space and setEach pairing is a finite sum, so this is a well-defined bilinear form. For the basis vector $e_n$, however, $\Vert e_n{\Vert }_2=1$ and $B(e_n,e_n)=n$; no finite $C$ works. It is even a coercive bilinear form with $\alpha =1$, since $\sum n{v}_n^2\ge \sum v_n^2$. Consequently coercivity alone does not imply boundedness on a normed vector space. This example uses $c_{00}$, which is not a Hilbert space; it does not define this pairing on all of the l2 sequence space.

A bounded bilinear form defines a bounded linear operator $T:X\to X^{\prime }$ by $(Tu)(v)=B(u,v)$, where $X^{\prime }$ is the continuous dual space. Its operator norm is the smallest admissible $C$:This converts an upper bound for pairings into an upper bound for a linear map taking an unknown vector to a linear functional.

Charles Weibel authored an exposition of bilinear forms over a field. It develops diagonalization and the classification of symmetric bilinear forms, including the real signature described by Sylvester's law of inertia. These classifications explain why changing a basis can simplify a form without removing invariants such as its numbers of positive and negative squares.^[ref]

A real bilinear form $B$ on a normed vector space $X$ is coercive when there is one constant $\alpha >0$ such thatThe lower bound uniformly controls every direction in the chosen norm. Unlike the upper bound defining a bounded bilinear form, it forces positive diagonal values at every nonzero vector. No symmetry is assumed. For complex Hilbert spaces, the usual analogue uses a sesquilinear form and $\mathrm{Re}B(v,v)\ge \alpha \Vert v{\Vert }^2$; it is not the same assertion for a complex bilinear form.

A concrete bounded and coercive example on ${\mathbb{R}}^2$, with the Euclidean norm, isIts quadratic form isThe best lower constant is $\alpha =2$, attained at $(1,0)$. Nevertheless, $B(e_1,e_2)=1$ and $B(e_2,e_1)=-1$, so coercivity does not imply a symmetric bilinear form. The symmetric part of a matrix $S=(A+A^T)/2=\mathrm{diag}(2,3)$ determines $q$, while the skew-symmetric matrix part changes mixed pairings without changing $q$. This is the symmetric and alternating parts of a bilinear form decomposition.^[ref]

For the upper constant, direct multiplication gives $A^{T}A=\left(\begin{array}{cc}5& -1\\ -1& 10\end{array}\right)$. Its eigenvalues are $(15\pm \sqrt{29})/2$, so the optimal upper constant, the spectral norm, is $C=\sqrt{(15+\sqrt{29})/2}$. The two constants answer different questions: how large a mixed pairing can be, and how small a self-pairing can be relative to squared norm.

The plotted quadratic forms show the mechanism: $x^2-y^2$ descends in one direction, $x^2$ leaves one direction uncontrolled, and $2x^2+3y^2$ controls both. These are self-pairings; their level sets cannot reveal the skew-symmetric matrix part. Calling $q$ a quadratic energy here refers to this diagnostic quantity, not to a claim that a nonsymmetric equation comes from minimizing an energy functional.

On a real finite-dimensional vector space with any norm, strict positivity $B(v,v)>0$ for every nonzero $v$ does imply coercivity. Indeed the quadratic form is continuous, and its positive minimum on the compact unit sphere supplies $\alpha$. If $B$ is also a symmetric bilinear form, this is positive definiteness. For general $B$, it is the symmetric and alternating parts of a bilinear form decomposition that isolates the positive part. Switching between equivalent norms preserves coercivity, but usually changes its constant.

The infinite-dimensional vector space distinction is explicit on the real l2 sequence space:The Cauchy-Schwarz inequality gives boundedness with $C=1$, and $B(v,v)>0$ for every nonzero vector. But $B(e_n,e_n)=1/n$ with $\Vert e_n{\Vert }_2=1$, so no positive $\alpha$ works. On the span of $e_1,\dots ,e_N$, the best lower constant is $1/N$: each finite restriction is coercive, but the control disappears as $N$ grows. Strict positivity is therefore weaker than a uniform lower bound.

The practical payoff is a stability estimate for a weak formulation. Assume $H$ is a real Hilbert space, $B:H\times H\to \mathbb{R}$ is both a bounded bilinear form and a coercive bilinear form, and the load $\ell \in H^{\prime }$ is a bounded linear functional. The Lax-Milgram theorem gives a unique $u\in H$ satisfying $B(u,v)=\ell (v)$ for all $v\in H$, including nonsymmetric $B$.^[ref] The dual norm $\Vert \ell {\Vert }_{H^{\prime }}={\sup }_{v\ne 0}|\ell (v)|/\Vert v\Vert$ measures load size. Testing with $v=u$ gives the short energy estimateFor $u=0$ the bound holds directly; otherwise divide by $\Vert u\Vert$. Subtracting the equations for two loads and using the same argument gives $\Vert u_1-u_2\Vert \le \Vert {\ell }_1-{\ell }_2{\Vert }_{H^{\prime }}/\alpha$. Thus a positive lower constant limits the response to load errors. The upper bound and completeness of the Hilbert space enter the existence theorem; the estimate itself uses the lower bound and the load's dual norm.

For the displayed matrix $A$, take $\ell (v)=f^{T}v$. Then $B(u,v)=\ell (v)$ for every $v$ means $A^{T}u=f$, not $Au=f$, with this ordering of the bilinear form. Since $\det A=7$,For $f=t(1,0{)}^T$, the exact response is $u=t(3,-1{)}^T/7$ and $\Vert u{\Vert }_2=\sqrt{10}|t|/7<|t|/2$. For $f=t(0,1{)}^T$, it is $u=t(1,2{)}^T/7$, with norm $\sqrt{5}|t|/7$. The actual maximal amplification is the operator norm $\Vert (A^T{)}^{-1}{\Vert }_2=\sqrt{2/(15-\sqrt{29})}<1/2$: the coercivity estimate is valid but need not be sharp for a nonsymmetric bilinear form.

An invertible matrix over the complex numbers preserving the symmetric bilinear form $v^{T}w$: $A^{T}A=I$. The transpose here does not include complex conjugation, unlike the defining condition for a unitary matrix. For example, $\left(\begin{array}{cc}\cosh t& i\sinh t\\ -i\sinh t& \cosh t\end{array}\right)$ is complex orthogonal for real $t$, but is not unitary when $t\ne 0$.

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 $B((1,1),(1,0))=1$.

A symmetric bilinear form on a tangent space is degenerate when some nonzero vector pairs to zero with every vector; equivalently its matrix has zero determinant. At such a point there is no inverse metric tensor, and the usual metric tensor construction of a unique Levi-Civita connection does not apply. A coordinate expression that degenerates must be checked to distinguish a removable coordinate failure from a genuinely excluded point of the geometric domain.

Over $\mathbb{C}$, the symplectic Lie algebra on a four-dimensional symplectic vector space acts faithfully on its five-dimensional primitive exterior square, preserving its symmetric bilinear form. Faithfulness follows because an operator killing all wedges $e_i\wedge e_j$ has all off-diagonal coefficients zero and satisfies $A_{ii}+A_{jj}=0$ for every pair, hence is zero. The source and target Lie algebras both have dimension ten, so this faithful map is an isomorphism to the Special orthogonal Lie algebra ${\mathfrak{so}}_5$.

For a complex vector space with a nondegenerate symmetric bilinear form $Q$, the map $u\wedge v\mapsto [w\mapsto Q(v,w)u-Q(u,w)v]$ identifies the exterior square of the defining representation with the Adjoint representation of a Lie algebra of $\mathfrak{so}(V)$. Equivariance follows from invariance of $Q$; an orthogonal basis verifies bijectivity. It is irreducible when $\mathfrak{so}(V)$ is simple. In dimension four it splits into the adjoint modules of the two commuting sl2 Lie algebras.

For a non-null hypersurface in a pseudo-Riemannian manifold, choose a unit normal vector $n$ and the ambient Levi-Civita connection $\nabla$. One sign convention for its extrinsic curvature isfor tangent vector fields $X,Y$. Reversing the normal reverses $K$; some authors use the overall negative convention.

Differentiating $g(n,Y)=0$ gives $K(X,Y)=-g(n,{\nabla }_{X}Y)$. The torsion-free property and tangency of $[X,Y]$ show that $K$ is a symmetric bilinear form. It measures the normal part of ambient acceleration along the hypersurface.

For a sphere of radius $R$ in Euclidean SPACE, the outward unit normal is $n(x)=x/R$, so ${\nabla }_{X}n=X/R$ and $K=g{|}_{TS}/R$. A plane has constant normal and $K=0$. The latter is the local totally geodesic case.

For a symmetric bilinear form $B$, the Gram-Schmidt process projection formula becomesIt produces orthogonal vectors whenever every denominator used is nonzero. A symmetric bilinear form can have nonzero vectors with zero self-pairing, so linear independence alone does not justify those divisions. For an inner product, positive definiteness excludes this obstruction.

For example, on ${\mathbb{R}}^2$ let $B((x_1,x_2),(y_1,y_2))=x_1{y}_2+x_2{y}_1$. Both standard basis vectors have zero self-pairing, so reordering them cannot provide the first pivot. Replacing them by $u=(1,1)$ and $v=(1,-1)$ works: $B(u,u)=2$, $B(v,v)=-2$ and $B(u,v)=0$. Changing linear combinations can therefore be necessary, rather than merely reordering the input. This example also separates the existence of an orthogonal basis from the success of an unmodified pivot order.

The inertia of a real symmetric bilinear form is the triple giving the numbers of positive, negative, and zero squares in a diagonalization. Sylvester's law of inertia makes it independent of the chosen basis.

For a closed manifold $M$ with an orientation and dimension $4k$, its integral intersection form is the symmetric bilinear form onobtained by pairing complementary middle-dimensional homology classes. Under Poincare duality, it is the cup product pairingon degree-$2k$ integral cohomology modulo its torsion subgroup. The even degree makes the pairing symmetric; integral duality makes it a unimodular intersection pairing. When representatives are transverse oriented submanifolds, the same value is the signed count of their intersection points.

The intersection form of a 4-manifold is the case $k=1$. On the Complex projective plane its matrix is $(1)$, whereas on $S^2\times S^2$ the factor spheres give $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$.

For a compact manifold with boundary, the perfect intersection pairing pairs absolute and relative homology, $H_{2k}(M)$ with $H_{2k}(M,\partial M)$; the induced absolute self-pairing can be degenerate. For instance the core sphere in $S^2\times D^2$ generates $H_2$ but has self-intersection zero.

For a closed manifold $M$ with an orientation and dimension four, its intersection form is the symmetric bilinear formon $H^2(M;\mathbb{Z})$ modulo its torsion subgroup. By Poincare duality it is the signed-intersection pairing on the free part of $H_2(M;\mathbb{Z})$, and it is a unimodular intersection pairing. Its real extension has positive and negative indices whose difference is its signature.

This is the four-dimensional case of the general intersection form. On a compact oriented four-dimensional manifold with boundary, the perfect intersection pairing uses one absolute and one relative homology class. An absolute self-pairing alone can be degenerate, so it should not be identified with the closed-manifold unimodular form without a boundary qualification.

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.

Over a field $F$ of characteristic different from two, an isometric structure in the knot-theoretic convention is a triple $(V,Q,T)$: $V$ is a finite-dimensional vector space, $Q$ is a nonsingular symmetric bilinear form, and $T$ is a linear automorphism satisfyingThus $T$ is a $Q$-isometry, with neither $1$ nor $-1$ as an eigenvalue. The structure is metabolic if $V$ has a half-dimensional $T$-invariant vector subspace on which $Q$ vanishes. The invariance requirement is part of this definition: an arbitrary isotropic subspace for $Q$ alone does not suffice.

For an invertible Seifert matrix $A$ with $A\pm A^T$ nonsingular, the associated matrices areDirect multiplication verifies $T^{T}QT=Q$, while $\det (T\pm I)=\det (A{)}^{-1}\det (A^T\pm A)$ verifies the excluded eigenvalues. Conversely $A=Q(I+T{)}^{-1}$ recovers the Seifert form in this convention. Quotienting metabolic structures gives the Witt group of isometric structures.

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.