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Clifford multiplication is the action of a Clifford algebra on a module. The displayed relation uses the negative Riemannian metric convention; in a compatible Hermitian inner product, $c(v{)}^{\ast }=-c(v)$ for real $v$. Another convention writes $C(v)=ic(v)$, giving self-adjoint generators with positive anticommutator. Switching between these conventions switches a geometric Dirac operator between self-adjoint and skew-adjoint realizations.

For a matrix $A$ over the complex numbers with the standard Hermitian inner product, the linear system $Ax=b$ is solvable exactly whenThe star is the conjugate transpose; for matrices over the real numbers, use the transpose. Necessity follows from $y^{\ast }Ax=(A^{\ast }y{)}^{\ast }x=0$. For sufficiency, $(\mathrm{im}A{)}^{\perp }=\ker A^{\ast }$, and taking the orthogonal complement again recovers $\mathrm{im}A$ in finite dimension. This is the cokernel obstruction expressed as scalar equations. ^[ref]

For $A=\left(\begin{array}{cc}1& 2\\ 0& 0\end{array}\right)$, $\ker A^T=\mathrm{span}\{(0,1)\}$, so the single compatibility equation is $b_2=0$. If it holds, the kernel supplies the remaining input freedom; if it fails, changing the input cannot remove the obstruction.

A finite-dimensional complex representation of a compact group is a direct sum of irreducible representations. Average a positive Hermitian inner product over normalized Haar measure. The orthogonal complement of an invariant subspace is then invariant, so induction on dimension gives the decomposition. The same argument works for finite-dimensional real representations with a positive real inner product.

For a Hermitian manifold with the pointwise Hermitian inner product linear in the first variable, define the conjugate-linear star by $\alpha \wedge \ast \beta =⟨\alpha ,\beta ⟩{\omega }^n/n!$. It is the usual real Hodge star operator extended complex-linearly and then composed with complex conjugation. It has the displayed bidegree and satisfies ${\ast }^2=(-1{)}^{p+q}$ on $(p,q)$-forms.

Exchanging the arguments of a Hermitian inner product takes its complex conjugate. This property, together with sesquilinearity and positive definiteness, distinguishes a complex inner product from a general bilinear pairing.

The discrete L2 inner product is the cell-volume-weighted Hermitian inner product on a uniform grid. Its associated norm is the discrete L2 norm. It provides the finite-dimensional analogue of spatial integration by parts and continuous energy method identities.

For a finite-dimensional continuous complex representation $R$ of a compact group, normalize Haar measure to mass one and average any positive-definite Hermitian inner product:The resulting inner product is positive-definite and invariant. If $W$ is an invariant subspace, then $W^{\perp }$ is invariant as well, since $⟨R(g)v,w{⟩}_K=⟨v,R(g^{-1})w{⟩}_K$. Thus $V=W\oplus W^{\perp }$. Restriction to a compact real form, followed by integration of a Lie-algebra representation, proves the Weyl complete reducibility theorem.

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.

For a complex finite-dimensional group representation, the displayed matrices give another group representation: taking inverses and taking the conjugate transpose each reverse multiplication order, so their composition preserves it. This operation is involutive. With the standard Hermitian inner product, an invariant vector subspace $W$ for $D^{♯}$ has orthogonal complement invariant for $D$, because $D(g{)}^{†}W=D^{♯}(g^{-1})W\subseteq W$. Thus $⟨w,D(g)v⟩=⟨D(g{)}^{†}w,v⟩=0$ for $v\in W^{\perp },w\in W$. It follows that $D^{♯}$ is an irreducible representation exactly when $D$ is. No finiteness assumption on the group or unitarity assumption on the group representation is needed.

For $n\ge 2$, ordered orthogonal complex lines in ${\mathbb{C}}^n$ form a projective bundle over ${\mathbb{CP}}^{n-1}$, with fiber ${\mathbb{CP}}^{n-2}$. Here $L$ is the first tautological line and the complement is formed using the standard Hermitian inner product. Its real dimension is $4n-6$. It is also the partial complex flag space of a line contained in a two-plane: send an orthogonal pair to $({\ell }_1,{\ell }_1\oplus {\ell }_2)$, and recover the second line by orthogonal complement inside the two-plane.

A matrix represents a linear projection when applying it twice has the same effect as applying it once. For an orthogonal projection with the standard Hermitian inner product on ${\mathbb{C}}^m$, the additional condition is that the conjugate transpose equals the original matrix:Thus $P$ is both idempotent and a Hermitian matrix. Over the real numbers this says $P^2=P=P^T$. Its image of a linear map is the retained linear subspace, while its kernel of a linear map is the orthogonal complement: vectors in the image are unchanged, and vectors in the kernel become zero.

For a matrix $A$ with full column rank, projecting onto its column space givesThe invertibility of $A^{†}A$ follows from linear independence of the columns. If those columns are an orthonormal set, $A^{†}A=I$ and $P=A{A}^{†}$. These formulas and the conditions $P^2=P=P^T$ in the real case are developed in MIT 18.06SC, Projections onto Subspaces, pp. 1–3.

For the line spanned by $e=(1,2{)}^T$,In particular, $Pe=e$ and $P(2,-1{)}^T=0$. The eigenvalues are $1$ on the retained direction and $0$ on the removed direction. This exposes the meaning of an orthogonal projection as retaining selected components, rather than stretching them.

Symmetry refers to the specified inner product. In the real weighted setting, the formula becomes $P_M=A(A^{T}MA{)}^{-1}{A}^{T}M$ and satisfies $P_M^2=P_M$ and $P_M^{T}M=M{P}_M$. For the preceding weighted line example,It is not symmetric under the ordinary transpose, but it is an orthogonal projection for the weighted Euclidean inner product. The matrix criterion $P=P^{†}$ should therefore be used with the standard Hermitian inner product, not silently transferred to a different geometry.

The scalar fields commute, so a degree-$d$ expression in one doublet belongs to a symmetric power, rather than an arbitrary tensor product. Put $x={\varphi }_1$, $y={\varphi }_2$. The generators of a group $r={\tau }_1{\tau }_2$ and $s={\tau }_3$ act byA monomial $x^a{y}^b$ is fixed by $r$ precisely when $a-b$ is divisible by three. Imposing $s$ then pairs the exponents. Through degree four the resulting holomorphic polynomial invariants areIn particular, two explicit invariant terms are ${\varphi }_1{\varphi }_2$ and ${\varphi }_1^3+{\varphi }_2^3$. Their invariance follows directly from ${\alpha }^3=1$ and the interchange of the two components. A quartic example is $({\varphi }_1{\varphi }_2{)}^2$.

For a physical real scalar potential, complex conjugate fields are allowed as well. The Hermitian inner product ${\varphi }^{†}\varphi$ and its square are also invariant, while a holomorphic term can enter with its complex conjugate, for example $b({\varphi }_1^3+{\varphi }_2^3)+\bar{b}({\bar{\varphi }}_1^3+{\bar{\varphi }}_2^3)$. There is still no invariant linear term, because neither the doublet nor its conjugate representation contains a trivial representation. Thus the possible nonconstant total degrees areThe constant term is unrestricted.

For the fundamental representation of the special unitary group $SU(2)$, its central element $-I$ sends $\varphi \mapsto -\varphi$ and $\bar{\varphi }\mapsto -\bar{\varphi }$. Every odd-degree polynomial changes sign, excluding degrees one and three. Further, $SU(2)$ acts transitively on the unit sphere in ${\mathbb{C}}^2$: a unit vector $(u,v{)}^T$ is the first column ofAn invariant polynomial must therefore depend only on ${\varphi }^{†}\varphi$. Through quartic order the general real scalar potential isHence the possible nonconstant degrees are $2,4$; the cubic term permitted by $S_3$ is excluded by $SU(2)$.

If the notation $V(\varphi )$ is instead read as restricting to holomorphic polynomials alone, the distinction matters: a single commuting $SU(2)$ doublet has no nonconstant holomorphic polynomial invariant. The candidate contraction ${ϵ}_{ab}{\varphi }_a{\varphi }_b={\varphi }_1{\varphi }_2-{\varphi }_2{\varphi }_1$ vanishes identically. The quadratic and quartic $SU(2)$ terms above require complex conjugate fields. This also explains why counting a trivial representation in $2\otimes 2$ alone would give an incorrect holomorphic answer.

With conjugation in its first slot, a complex inner product has conjugate symmetry, linearity in its second slot, and positive definiteness. For the proposed Hermitian inner product, direct expansion givesThe last expression vanishes exactly for the zero vector. Conjugate symmetry and linearity also give antilinearity in the first slot: $⟨\alpha z+\beta w,y⟩=\bar{\alpha }⟨z,y⟩+\bar{\beta }⟨w,y⟩$. These verify all three axioms and the associated first-slot consequence.

Using the Hermitian inner product, the scalar product is $1-i+2i^2=-1-i$. The two squared norms are $6$ and $4$. The specified real angle in a complex inner product space therefore hasThe real part is essential: the complex scalar product itself need not be real.

The characteristic polynomial of the Hermitian matrix isSolving the two eigenvector equations and using the Hermitian inner product to normalize givesIndeed $v_1^{†}{v}_2=0$ and both norms equal one. Multiplication of either eigenvector by any unit phase factor gives an equally valid normalized choice.

Both expressions are the same scalar:This identity uses the ordinary transpose and remains true for complex entries. A Hermitian inner product uses a conjugate transpose instead and has conjugate symmetry, not this bilinear symmetry.

A unitary matrix preserves the Hermitian inner product and hence the norm. If $Uv=\lambda v$ with $v\ne 0$, thenDividing by the positive squared norm givesUnlike the Hermitian case, this constrains eigenvalues to the unit circle, not necessarily the real axis.

Let $Hv=\lambda v$ with $v\ne 0$ and $H^{\ast }=H$. Then $v^{\ast }Hv=\lambda v^{\ast }v$, while taking the complex conjugate gives $v^{\ast }Hv=\overline{\lambda }{v}^{\ast }v$. Since $v^{\ast }v>0$, every eigenvalue of a Hermitian matrix is real.

If $Hv=\lambda v$ and $Hw=\mu w$, thenDistinct eigenvalues therefore force $v^{\ast }w=0$: their eigenvectors are orthogonal in the Hermitian inner product. A degenerate eigenvalue occurs more than once in the spectrum, counted with multiplicity. For a Hermitian matrix, its algebraic multiplicity equals its geometric multiplicity, so degeneracy gives an eigenspace of dimension greater than one; an orthonormal basis can be chosen within it.

A unitary matrix has $A^{\ast }A=I$. For example,This example is invertible, has nonreal off-diagonal entries, and preserves the standard Hermitian inner product. Although it is also symmetric, that additional property is not required for unitarity.