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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 Hermitian form is a sesquilinear form on a complex vector space satisfying conjugate symmetry,Hence $q(v)=H(v,v)$ is a real number for every vector. In a basis, the matrix of a Hermitian form has $A_{ij}=H(e_i,e_j)$ and obeys $A_{ji}=\overline{A_{ij}}$, equivalently $A=A^{\ast }$: it is a Hermitian matrix. The coordinate rule remains $H(x,y)=x^{T}A\overline{y}$. Conjugate symmetry does not imply positivity.^[ref]

The diagonal contains more information than might be expected. For any complex sesquilinear form $B$, put $q(v)=B(v,v)$. Expanding four expressions gives the complex polarization identity in the linear-first convention:This is an algebraic identity, with no positivity assumption. If every $q(v)$ is real, interchanging $u,v$ preserves the first difference and negates the second: use $q(\alpha v)=|\alpha {|}^{2}q(v)$ for $\alpha =-1,i$. The identity then gives $B(v,u)=\overline{B(u,v)}$. Thus a complex sesquilinear form is a Hermitian form if and only if all its diagonal values are real. Testing only the basis vectors is insufficient: the off-diagonal example in sesquilinear form has zero diagonal entries in its matrix but a nonreal value at $(1,i)$.

Consider the indefinite Hermitian formIts values at $e_1,e_2$ are $1,-1$. Moreover $u=(1,1)$ and $v=(1,-1)$ are nonzero isotropic vectors: $q_{-}(u)=q_{-}(v)=0$, yet $H_{-}(u,v)=2$. These vectors do not belong to the radical of a Hermitian form, and the usual Cauchy-Schwarz inequality would incorrectly require $4\le 0$. An indefinite Hermitian form records signed geometry rather than defining a norm by $\sqrt{H(v,v)}$.

There is a separate boundary case: $H_0(z,w)=z_1\overline{w_1}$ is a positive semidefinite Hermitian form, since $H_0(z,z)=|z_1{|}^2\ge 0$, but $H_0(e_2,e_2)=0$. Its radical of a Hermitian form is the entire second coordinate line. Positive definiteness is exactly the additional condition in a Hermitian inner product. The three examples below show why these distinctions matter: on the real slice $z=(x,y)\in {\mathbb{R}}^2\subset {\mathbb{C}}^2$, $q_{+}(z)=|z_1{|}^2+2|z_2{|}^2$ has an ellipse as its unit level, $q_{-}$ has hyperbolas, and $q_0$ has two parallel lines. These are slices of the complex examples, not drawings of their full four-dimensional domains.

A Hermitian inner product is a Hermitian form $H$ on a complex vector space which has positive definiteness: $H(v,v)>0$ for every nonzero vector $v$. It is precisely a complex inner product. Only with this definiteness condition doesdefine a norm rather than a seminorm or an expression that may not be real. The standard Hermitian inner product on ${\mathbb{C}}^n$ is $H(z,w)={\sum }_j{z}_j\overline{w_j}$; positive weights $c_j$ give ${\sum }_j{c}_j{z}_j\overline{w_j}$, with squared norm ${\sum }_j{c}_j|z_j{|}^2$.^[ref]

A Hermitian inner product need not have a real matrix. For instance, with the linear-first coordinate convention,The last expression proves positive definiteness without a lengthy calculation of eigenvalues. For $z=(1,i)$ it gives $6$, while for $z=(1,-i)$ it gives $2$. These unequal values expose why a silent switch to $z^{\ast }Az$ would change this particular Hermitian inner product: with these same entries it gives $2$ at $(1,i)$. All entries must keep their roles from the matrix of a Hermitian form.

The Cauchy-Schwarz inequality for a Hermitian inner product statesEquality holds exactly when $u,v$ are linearly dependent, including when either is zero. For a nonzero $u$, outline the proof by setting $a=H(v,u)/H(u,u)$ and $r=v-au$. Then $H(r,u)=0$, and the Pythagorean theorem givesNonnegativity proves the inequality, while positive definiteness makes equality equivalent to $r=0$. The scalar $a$ has no extra conjugate because the first argument is linear.^[ref]

The Cauchy-Schwarz inequality also holds for a positive semidefinite Hermitian form, but equality need not imply linear dependence: for $H_0(z,w)=z_1\overline{w_1}$, the independent vectors $e_1$ and $e_1+e_2$ give $1=1$. Equality then describes dependence after passing to the quotient vector space by the radical of a Hermitian form. The indefinite Hermitian form in Hermitian form shows that conjugate symmetry alone does not even guarantee the inequality.

For a worked orthogonal projection, take $H_{+}(z,w)=z_1\overline{w_1}+2z_2\overline{w_2}$, $u=(1,1)$ and $v=(2,0)$. Then $a=2/3$, $au=(2/3,2/3)$ and $r=(4/3,-2/3)$, with $H_{+}(r,u)=0$ and $4=4/3+8/3$ for the squared norms. The residual is perpendicular in this Hermitian inner product even though its ordinary real dot product with $u$ is $2/3$. For complex $v=(1,i)$, the same orthogonal projection uses $a=(1+2i)/3$; rotating $v$ multiplies $a$ by $e^{i\theta }$, while rotating $u$ multiplies $a$ by $e^{-i\theta }$. This combines geometric decomposition with sensitivity to the relative complex arguments of the multipliers.

A Hermitian inner product therefore supplies more than a positive diagonal: orthogonality, orthogonal projection onto a span, and the triangle inequality for its norm make approximation and distance meaningful. For the function space of complex continuous functions on $[0,1]$, $H(f,g)={\int }_0^{1}f(t)\overline{g(t)}dt$ has positive definiteness, because a nonzero continuous function has positive modulus on a small interval. The same construction turns overlap under integration into lengths and best approximation in finite-dimensional vector subspaces.

A Hermitian manifold is a complex manifold whose tangent bundle carries a smoothly varying positive-definite Hermitian form. Its real part is a Riemannian metric invariant under the complex structure $J$.

A Hermitian metric on a holomorphic vector bundle $E\to X$ is a smoothly varying positive-definite Hermitian form $h_x$ on every complex fiber $E_x$.

Such a metric makes the underlying smooth bundle a Hermitian vector bundle.

This is a smoothly varying positive-definite Hermitian form on each fiber, without requiring a holomorphic base or bundle. A partition of unity glues local positive metrics into a global one. Applying smooth Gram-Schmidt to local frames then makes the transitions unitary, reducing the structure group of a vector bundle to $U(k)$. For a line bundle its unit vectors form a principal bundle for the circle group.

A bounded operator $T:H\to H$ on a real or complex Hilbert space is Hermitian, or self-adjoint, when $T=T^{\ast }$. Equivalently,Its directional values $q_T(x)=⟨Tx,x⟩$ are real, since this identity makes $q_T(x)$ equal its complex conjugate. Reality allows either sign; a positive operator additionally requires nonnegative values in every direction.

The diagram compares Hermitian quadratic forms along paths of unit vectors. The Hermitian matrix $D=\mathrm{diag}(1,-1)$ takes the values $1$ and $-1$ on the standard basis vectors; $P=\mathrm{diag}(1,0)$ takes the values $1$ and $0$; andsatisfies $H{e}_1=e_1$ and $H{e}_3=3e_3$. These eigenvectors form an orthonormal basis, and every $x=a{e}_1+b{e}_3$ has $q_H(x)=|a{|}^2+3|b{|}^2$. Thus $H$ is a positive operator, with eigenvalues $1,3$ and operator norm $3$.

Three conditions describe different geometry. A Hermitian operator satisfies $T=T^{\ast }$; a normal operator satisfies $T^{\ast }T=T{T}^{\ast }$; a unitary operator satisfies $T^{\ast }T=T{T}^{\ast }=I$, where $I$ is the identity operator. Every Hermitian operator and every unitary operator is a normal operator. Concrete matrices separate the conditions:

The spectrum of a bounded Hermitian operator is real. In finite dimension, the spectral theorem supplies an orthonormal basis of eigenvectors, turning its action into independent real scalings. This makes the geometry easy to interpret: each eigenvalue records a signed response along an orthogonal direction. An arbitrary bounded Hermitian operator in infinite dimension need not have any eigenvectors; the general spectral theorem uses a projection-valued measure. MIT's lecture 21 develops the real spectrum and the orthogonality of eigenspaces.^[ref]

A historical connection is the study of forms with conjugate variables. In his 1901 account of Charles Hermite's work, Camille Jordan describes these forms and emphasizes their role in reducing forms with real or complex coefficients. In today's notation, $x^{\ast }Hx$ is the diagonal value of a Hermitian form; the conjugate variables make a real-valued expression possible even with complex coefficients.^[ref]

For a densely defined operator outside this bounded setting, the displayed pairing identity alone defines a symmetric operator. A self-adjoint operator must also have the same operator domain as its adjoint operator.

A bounded Hermitian operator determines a Hermitian form $b_T(x,y)=⟨Tx,y⟩$ and its real-valued diagonal expression $q_T(x)=b_T(x,x)$. On a complex Hilbert space,Thus it is a quadratic form on the underlying real vector space; it is generally not a quadratic form over the complex field, whose scaling rule would involve ${\alpha }^2$. This distinction explains why replacing $x^{\ast }$ by $x^{\mathrm{T}}$ loses the geometry of a complex inner product.

For a finite-dimensional Hermitian operator with an orthonormal basis of eigenvectors $e_j$ and corresponding eigenvalues ${\lambda }_j$, the spectral theorem givesFor a unit vector, this is an average of the eigenvalues with nonnegative weights summing to one. It is also the Rayleigh quotient, so its minimum and maximum are the smallest and largest eigenvalues.

In the Hermitian operator diagram, take $x(\theta )=(\cos \theta ,\sin \theta {)}^{\mathrm{T}}$ for $D,P$, and $x_H(\theta )=\cos \theta e_1+\sin \theta e_3$ for $H$. The computed curves areThese paths illustrate the extremal values; the spectral theorem supplies the bounds for every complex unit vector, including those outside the plotted paths.

A Hermitian vector bundle is a complex vector bundle with a smoothly varying positive definite Hermitian form on each fibre. When the bundle is holomorphic, this is a Hermitian metric on a holomorphic vector bundle. Orthogonal projection to a smooth subbundle produces its induced metric and the quotient Hermitian metric.

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.

An indefinite Hermitian form takes both positive and negative values on $⟨v,v⟩$ for nonzero vectors. For example, $-|v_0{|}^2+|v_1{|}^2+|v_2{|}^2+|v_3{|}^2$ is indefinite. It is not a positive inner product defining a Hilbert space, and a nonzero vector can have zero norm without being orthogonal to all vectors. A positive semidefinite restricted form becomes a positive inner product only after quotienting its radical of a Hermitian form.

On $\mathbb{C}/(\mathbb{Z}+\mathbb{Z}i)$, a Hermitian form linear in its first argument has integral imaginary part on the period lattice exactly when its real scalar coefficient is integral. Positivity is equivalent to $t>0$. The associated holomorphic line bundle has degree $t$ because its First Chern class is represented by $(it/2)dz\wedge d\overline{z}$. In particular the oriented Chern pairing is minus the imaginary part under this convention.

A linear bijection between spaces carrying compatible bilinear forms, Hermitian forms or quadratic forms is an isometry of a space with a form when it preserves the specified form. Preserving the polar form alone need not preserve a quadratic form in characteristic two. Subspaces may have degenerate restricted forms even when the ambient form is nonsingular.

For a basis $(v_i)$, the matrix $A$ of a Hermitian form has entries $A_{ij}=H(v_i,v_j)$. With the convention that $H$ is linear in its first argument, coordinate columns $x,y$ satisfy $H(x,y)=x^{T}A\overline{y}$.

With a Hermitian form linear in its first slot, $H(s\otimes \bar{t})=h(s,t)$ is a complex-linear dual tensor. The original connection, its conjugate connection, the tensor product connection and the dual connection induce $D_0$. Its derivative is $(D_{0,V}H)(s\otimes \bar{t})=Vh(s,t)-h(D_{V}s,t)-h(s,D_{V}t)$. Consequently $D_{0}H=0$ is exactly metric compatibility.

Integrate a polynomial times the conjugate of another over the product of unit circles, with normalized product angular measure. Fourier orthogonality makes the monomials $z_1^{{\alpha }_1}\cdots z_d^{{\alpha }_d}$ an orthonormal basis of the polynomial vector space. The Hermitian form is positive definite because the squared norm is the sum of squared absolute values of the finitely many coefficients. No completion or infinite series is needed to obtain this algebraic basis.

For a Hermitian form or a Hermitian form over a quadratic finite field, the displayed subspace is the orthogonal complement of $W$. If the form restricts nondegenerately to a finite-dimensional $W$, its invertible Gram matrix supplies, for each $v$, a unique $w_0\in W$ with $B(v-w_0,W)=0$. Consequently $V=W\oplus W^{\perp }$. When the form on $V$ is nondegenerate, its restriction to $W^{\perp }$ is too: a radical vector there is orthogonal to both summands and hence is zero.

Set $B(x,y)=⟨Tx,y⟩$. It is a positive semidefinite Hermitian form. Write $A=B(x,x)$, $C=B(y,y)$ and $c=B(x,y)$. For every complex $t$,If $C>0$, choosing $t=-c/C$ gives $A-|c{|}^2/C\ge 0$. If $C=0$ and $c\ne 0$, choosing $t=-sc$ for large positive $s$ makes the expression negative, a contradiction. Thus $c=0$ in that case. In all cases,This is the Cauchy-Schwarz inequality for a positive operator.

Use the realization $R_3=W\subset {\mathbb{C}}^4$ from solution. Put $x=({\varphi }_1,{\varphi }_2,{\varphi }_3,-{\varphi }_1-{\varphi }_2-{\varphi }_3)$, which identifies the three-dimensional scalar field with $W$. The polynomial invariants of the standard S4 representation include the symmetric power sumsbecause every permutation merely rearranges their summands. In these coordinates,Neither polynomial vanishes identically: at $({\varphi }_1,{\varphi }_2,{\varphi }_3)=(1,1,1)$ they equal $12$ and $-24$. Hence coefficients multiplying $I_2$ and $I_3$ provide nonzero invariant quadratic and cubic terms. A change of basis transports these polynomial invariants to any equivalent realization of $R_3$. For a real scalar field, both are real. For a complex scalar field, a real scalar potential may contain ${\lambda }_k{I}_k+{\lambda }_k^{\ast }{I}_k^{\ast }$; the quadratic Hermitian form ${\sum }_i|x_i{|}^2$ is another invariant.

As an independent character check, the symmetric square of a group representation has character $[\chi (g{)}^2+\chi (g^2)]/2$, and the symmetric cube of a group representation has character $[\chi (g{)}^3+3\chi (g)\chi (g^2)+2\chi (g^3)]/6$. With $\chi ={\chi }_3$ these are $(6,2,2,0,0)$ and $(10,2,-2,1,0)$. Their character inner products with ${\chi }_0$ are both one, so each degree has exactly one independent polynomial invariant.