Bilinear Operators on a Hermitian Module with Hermitian Adjoint

Introduction

A bilinear operator on a Clifford module is an operator built from two spinors rather than one. The space of such operators is finite-dimensional, and its structure is dictated by two facts: the Hermitian form of the module fixes the adjoint of every operator, and the Clifford algebra, acting on the module, generates all the endomorphisms of the module when the module is irreducible over a simple algebra. The second fact is the completeness of the Clifford action, and its bilinear reading is the classical Fierz identity: the bilinear covariants $(s,\Gamma^{A}t)$, indexed by a basis of the Clifford algebra, span the space of bilinear forms on a spinor module, and every endomorphism is a Clifford-linear combination of the covariant operators.

This article sets up the bilinear operators — the forms $S\times S\to A$, the equivariant ones, and the maps $S\to S$ they induce — and proves the completeness identity with the adjoint structure carried along: the adjoint of the operator $\Gamma^{A}$ is the operator $\Gamma^{A\dagger}$, the covariant form $(s,\Gamma^{A}t)$ has an adjoint covariant, and the types of the covariants are read off from the dagger. The two worked cases are the complex two-dimensional module, where the four blades are the four matrix units of $\mathrm{End}(\mathbb{C}^{2})$ and the Fierz identity is the completeness of a basis, and the quaternion algebra, where the invariants are the quaternion units.

The Clifford action and the module form are Hermitian Modules over a Hilbert Algebra with Hermitian Adjoint; the adjoint of the action is The Adjoint of the One-Sided Action with Hermitian Adjoint; the spinor adjoint and the Dirac adjoint are Spinor Adjoints and the Dirac Adjoint with Hermitian Adjoint; the forms of the dagger are Hermitian Forms on a Hilbert Algebra with Hermitian Adjoint; the classification of the invariant forms is Hermitian Forms over an Involution Ring and the Unitary Witt Group with Hermitian Adjoint; the trace form and the adjoint of multiplication are The Blade Form and the Hilbert Structure with Hermitian Adjoint; the completely positive maps and the operator systems are Completely Positive Maps of a Hilbert Algebra with Hermitian Adjoint; and the spinor module and its idempotents are Spinors as Minimal Left Ideals with Inner Conjugation.

Bilinear Forms and Bilinear Operators

Definitions

Definition. Let $S$ be a Clifford module over $A$. A bilinear form on $S$ is a map $b : S\times S\to A$ that is $A$-linear in each argument; it is Hermitian if $b(t,s) = \sigma(b(s,t))$, and the adjoint form $b^{*}$ is defined by $b^{*}(s,t) = \sigma(b(t,s))$, so that $b$ is Hermitian exactly when $b^{*} = b$. A bilinear operator is the $A$-linear map $S\to S^{*}$, $s\mapsto b(s,\cdot)$, induced by a bilinear form; the two descriptions carry the same data, by the adjunction $\mathrm{Bil}_A(S,S;A)\cong \mathrm{Hom}_A(S,S^{*})$.

Definition. A bilinear form $b$ is Clifford-equivariant (or invariant) if

$$ b(x\cdot s, t) = b\bigl(s, x^{\dagger}\cdot t\bigr) \qquad \text{for all } x \in \mathrm{Cl}(V,q), $$

which is the adjointness condition of Hermitian Modules over a Hilbert Algebra with Hermitian Adjoint. So a Hermitian Clifford module is exactly a Clifford module with a Hermitian equivariant bilinear form, and the spinor adjoint $J$ of Spinor Adjoints and the Dirac Adjoint with Hermitian Adjoint is the bilinear operator of that form.

Remark (the two-sided action on the forms). The space $\mathrm{Bil}_A(S,S;A)$ carries the action of the operator algebra: for $T\in \mathrm{End}_A(S)$ the form $b\circ(T\times T)$ is again bilinear, and the equivariant forms are the fixed points of the subspace of operators of the form $\rho(x)\otimes\rho(x^{\dagger}) - \mathrm{id}$. This is the module-level form of the two-sided operators of Two-Sided Operators on a Clifford Algebra.

The Adjoint of a Bilinear Operator

Proposition. With respect to the module form $(\cdot,\cdot)$ and the induced adjoint on $\mathrm{End}_A(S)$, the adjoint of the bilinear operator $T_b : s\mapsto b(s,\cdot)$ is $T_{b^{*}}$, where $b^{*}(s,t) = \sigma(b(t,s))$; equivalently, the assignment $b\mapsto T_b$ commutes with the adjoint:

$$ (T_b)^{*} = T_{b^{*}} . $$

Proof. $\langle T_bs, t\rangle = b(s,t)$ and $\langle s, T_{b^{*}}t\rangle = b^{*}(t,s) = \sigma(b(s,t))$; comparing with the defining property $\langle T_bs,t\rangle = \sigma(\langle s, T_b^{*}t\rangle)$ of the adjoint under the sesquilinear convention gives the identity.

Corollary. A bilinear form is Hermitian iff its bilinear operator is self-adjoint; a form is alternating iff its operator is skew-adjoint; and the Hermitian forms are the fixed points of the involution $b\mapsto b^{*}$ on $\mathrm{Bil}_A(S,S;A)$, so the Hermitian forms form the symmetric part of the space of bilinear forms under this involution.

The Completeness of the Clifford Action

The Theorem

Theorem (completeness, the Fierz identity). Let $V$ be even-dimensional over an algebraically closed field, so that $\mathrm{Cl}(V,q)\cong M_{2^m}(A)$ with $m = \dim V/2$, and let $S$ be the irreducible (spinor) module. Then the action map

$$ \rho : \mathrm{Cl}(V,q)\longrightarrow \mathrm{End}_A(S) $$

is an isomorphism of algebras, and the $2^n$ blade operators $\rho(e_{i_1}\cdots e_{i_k})$ form a basis of $\mathrm{End}_A(S)$; equivalently every endomorphism of the spinor module is a Clifford-linear combination of the blades. In bilinear form, the bilinear covariants

$$ b_A(s,t) = \bigl(s, \rho(\Gamma_A)\,t\bigr), \qquad \Gamma_A \text{ a basis of } \mathrm{Cl}(V,q), $$

span the space of bilinear forms on $S$, and every bilinear form is a linear combination of the covariants.

Proof. The Clifford algebra is simple and $\dim \rho(\mathrm{Cl}) = \dim \mathrm{Cl} = 2^n$ because the module is faithful; and $\dim \mathrm{End}_A(S) = (\dim_A S)^2 = (2^m)^2 = 2^n$. An injective linear map between spaces of the same dimension is an isomorphism, so the blades, being a basis of the algebra, map to a basis of the endomorphisms. The bilinear statement is the isomorphism $\mathrm{Bil}_A(S,S;A)\cong \mathrm{End}_A(S)$ read in the dual basis.

Corollary (the adjoint of a covariant). The operator $\rho(\Gamma_A)$ has adjoint $\rho(\Gamma_A)^{\dagger} = \rho(\Gamma_A^{\dagger})$, so the covariant $b_A$ has adjoint covariant $b_{A^{*}}$ with $\Gamma_{A^{*}} = \Gamma_A^{\dagger}$; the covariant is Hermitian or skew-Hermitian for the module form exactly as the blade $\Gamma_A$ is self-adjoint or skew-adjoint under the dagger. In particular, since every vector is skew-adjoint, the vector covariants $b_{e_j}(s,t) = (s,\rho(e_j)t)$ are skew-Hermitian, and the scalar covariant $b_1(s,t) = (s,t)$ is the Hermitian module form itself.

The Bilinear Covariants of a Spinor Module

Proposition (the classification by grade). For a basis of $\mathrm{Cl}(V,q)$ by grade, the covariants organise by degree: degree $0$ gives the scalar covariant $b_1$ (the module form); degree $1$ the vector covariants $b_{e_j}$; degree $2$ the bivector covariants $b_{e_ie_j}$ with $i

Proof. The dagger acts on a blade by $\Gamma_A^{\dagger} = \sigma(\alpha(\Gamma_A^{r})) = (-1)^{|A|(|A|-1)/2}(-1)^{|A|}\sigma(\Gamma_A) = (-1)^{T_{|A|}}\sigma(\Gamma_A)$ with $T_{|A|} = |A|(|A|+1)/2$ the triangular number, which is even exactly for $|A|\equiv0,3\pmod4$; the covariant inherits the sign by the corollary above. This was checked on the blades of $\mathrm{Cl}_{2,0}(\mathbb{C})$, $\mathrm{Cl}_{0,3}(\mathbb{R})$ and $\mathrm{Cl}_{3,0}(\mathbb{R})$: degrees $0$ and $3$ self-adjoint, degrees $1$ and $2$ skew-adjoint.

Remark (the Fierz coefficients). The expansion of a bilinear form in the covariants $b_A$ is the Fierz rearrangement. Because the blades are orthogonal for the trace form of The Blade Form and the Hilbert Structure with Hermitian Adjoint, the coefficient of $b_A$ in a form $b$ is $\mathrm{Tr}(\rho(\Gamma_A)^{*}T_b)$ up to the normalisation of the trace form, so the expansion is obtained by orthogonality and is a finite computation on the algebra.

The Invariance Under the Slice

Theorem. For $u$ in the unitary slice $U$, the bilinear form $b\circ(u\times u)$ has the same expansion coefficients as $b$ in the covariants:

$$ b_A(u\cdot s, u\cdot t) = b_{\,u^{\dagger}\Gamma_Au}(s,t) , $$

so the slice permutes the covariants by the conjugation $\Gamma_A\mapsto u^{\dagger}\Gamma_A u$ of the Clifford algebra; in particular the scalar covariant is invariant, the space of degree-$k$ covariants is preserved, and the expansion of any form in the covariants is transformed by the adjoint representation of the slice.

Proof. $b_A(us,ut) = (us, \rho(\Gamma_A)ut) = (s, \rho(u^{\dagger}\Gamma_A u)t)$ by the adjointness of the action, $= b_{u^{\dagger}\Gamma_Au}(s,t)$, and $u^{\dagger}\Gamma_Au$ is again a blade combination of the same degree.

Corollary (invariance and its failure). The slice acts on the space of covariants, and the invariants are the forms whose expansion is fixed by all $u$, which on the definite module is the span of the scalar covariant alone; off the slice, the conjugation is the two-sided operator of Two-Sided Operators on a Clifford Algebra, and the exponent reads the sign of the general operator.

Worked Cases

The Complex Two-Dimensional Module

Let $A = \mathbb{C}$ with the conjugation, $\dim V = 2$ and $e_1^{2} = e_2^{2} = 1$, so that $\mathrm{Cl}\cong M_2(\mathbb{C})$ and $S = \mathbb{C}^{2}$. The four blades $1,e_1,e_2,e_1e_2$ give four action matrices, and a direct computation of their action on a minimal left ideal shows that they are linearly independent and hence a basis of the four-dimensional $\mathrm{End}_{\mathbb{C}}(\mathbb{C}^{2})$: the rank of the span is $4$, equal to $(\dim S)^{2}$. The Fierz identity is then the statement that the four covariants $b_1, b_{e_1}, b_{e_2}, b_{e_1e_2}$ span the four-dimensional space of bilinear forms on $\mathbb{C}^{2}$: $b_1$ is the Hermitian module form, and $b_{e_1}, b_{e_2}, b_{e_1e_2}$ are the three skew-Hermitian covariants of degrees $1,1,2$. The matching of the counts — $2^n = 4$ blades, $(\dim S)^{2} = 4$ endomorphisms — is the content of the theorem.

The Quaternion Algebra

Let $\mathrm{Cl}_{0,2}(\mathbb{R}) = \mathbb{H}$ with the dagger positive and the frame $e_1,e_2$ of square $-1$. The module is the algebra itself over the division algebra $\mathbb{H}$, and the covariants are $b_1$ (the positive definite form $\mathrm{Sc}(x^{\dagger}y)$), the two skew-Hermitian vector covariants $b_{e_1}, b_{e_2}$, and the skew-Hermitian bivector covariant $b_{e_1e_2}$ of degree $2$. The completeness fails in the form stated above because $\mathbb{H}$ is a division algebra and the module is not the simple module of a full matrix algebra; instead the invariant forms are the multiples of the quaternion bilinear $\mathrm{Sc}(x^{\dagger}y)$ and of its left translations, and the coherent statement is the uniqueness theorem of Hermitian Modules over a Hilbert Algebra with Hermitian Adjoint rather than the Fierz basis. The example shows the hypothesis of the completeness theorem — a simple algebra that is a full matrix algebra over the base field, with its irreducible module — and what replaces it when the module is over a division algebra.

Summary

A bilinear operator on a Clifford module is the operator $s\mapsto b(s,\cdot)$ induced by a bilinear form $b$, the two carrying the same data up to the adjunction $\mathrm{Bil}_A(S,S;A)\cong \mathrm{Hom}_A(S,S^{*})$. The adjoint of a bilinear operator is the operator of the adjoint form $b^{*}(s,t) = \sigma(b(t,s))$, so the Hermitian forms are the self-adjoint operators and form the symmetric part of the space of bilinear forms. A form is Clifford-equivariant exactly when it satisfies the adjointness axiom, and the Hermitian Clifford module is the case of a Hermitian equivariant form, of which the spinor adjoint is the bilinear operator.

The completeness of the Clifford action — the Fierz identity — says that for an even-dimensional algebra over an algebraically closed field the action is an isomorphism $\mathrm{Cl}(V,q)\cong \mathrm{End}_A(S)$, so the $2^n$ blades are a basis of the endomorphisms and the $2^n$ bilinear covariants $(s,\Gamma_A t)$ span the space of bilinear forms; it was checked for the complex two-dimensional module, where the rank of the span is $4 = (\dim S)^{2}$. The covariants are classified by grade, the covariant $b_A$ being Hermitian for $|A|\equiv0,3\pmod4$ and skew-Hermitian for $|A|\equiv1,2\pmod4$, so the scalar and volume covariants are Hermitian and the vector and bivector covariants are skew-Hermitian; the adjoint of a covariant is the covariant of the dagger $\Gamma_A^{\dagger}$, so the scalar covariant is the module form and the vector covariants are skew-Hermitian. The unitary slice permutes the covariants by $\Gamma_A\mapsto u^{\dagger}\Gamma_Au$, fixes the scalar covariant, preserves the degree, and the invariants on the definite module are the scalar covariant alone; the expansion coefficients are the Fierz coefficients, computed by orthogonality of the blades for the trace form.

Summary of Notation

Symbol	Meaning
$b : S\times S\to A$	Bilinear form
$b^{*}(s,t) = \sigma(b(t,s))$	Adjoint form, $b$ Hermitian iff $b^{*}=b$
$T_b : s\mapsto b(s,\cdot)$	Bilinear operator, $T_b\in\mathrm{Hom}_A(S,S^{*})$
$(T_b)^{} = T_{b^{}}$	Adjoint of a bilinear operator
$b(xs,t) = b(s,x^{\dagger}t)$	Clifford-equivariance
$\rho : \mathrm{Cl}(V,q)\cong \mathrm{End}_A(S)$	Completeness (Fierz identity)
$b_A(s,t) = (s,\rho(\Gamma_A)t)$	Bilinear covariants, spanning the forms
$b_A(us,ut) = b_{u^{\dagger}\Gamma_Au}(s,t)$	Slice action on the covariants