Spinors as Minimal Left Ideals with Inner Conjugation

Introduction

The spinor module of Spin Representations and Clifford Modules with Inner Conjugation was constructed there from a quadratic form, through a maximal isotropic subspace and an exterior algebra. This article gives the other description of the same object, the one that makes a spinor an element of the algebra: a minimal left ideal of $\mathrm{Cl}(V,q)$ is a Clifford module, its elements are the spinors, and the spin group acts on it by left multiplication. The two constructions meet; with a Witt basis the ideal is isomorphic to the exterior algebra of the isotropic subspace, so that the creation and annihilation operators of the Chevalley model are the multiplications of the algebra. The description by ideals also yields the identity that is used without proof in the physical literature, that a spinor is an even multivector: the even subalgebra is always isomorphic to the ideal generated by a unit vector, and that ideal is the spinor module exactly in the low-dimensional algebras, among which are the two the applications of the category use.

The Clifford algebra, the parity grading and the volume element are from Clifford Algebras and Clifford Algebras in Finite Dimensions, in particular the description of the even subalgebra as the Clifford algebra of a subspace of one dimension less; the grade decomposition is from The Geometric Product and the Grade Decomposition; the construction of the spinor module from a maximal isotropic subspace, the complex and real spin representations, chirality and the eightfold table are from Spin Representations and Clifford Modules with Inner Conjugation; the Clifford group, the signed inner conjugation action and the spin group are from The Clifford, Pin and Spin Groups with Signed Inner Conjugation; the sandwich action of a rotor is from Versors, Rotors and the Sandwich Action with Signed Inner Conjugation. Nothing owned by those entries is re-derived. The base is a field $F$ of characteristic not $2$, with $q$ non-degenerate on a finite-dimensional space $V$ and $B$ its polar form.

Idempotents

The Idempotent of a Unit Vector

Definition. An element $\pi\neq0$ of $\mathrm{Cl}(V,q)$ is an idempotent if $\pi^2=\pi$. Two idempotents $\pi,g$ are orthogonal if $fg=gf=0$, and an idempotent is primitive if it is not the sum of two orthogonal nonzero idempotents.

Proposition. Let $e\in V$ with $q(e)=1$. Then

$$ \pi=\tfrac12(1+e) $$

is an idempotent, and $ef=\pi$ while $e \pi'=-\pi'$ for $\pi'=\tfrac12(1-e)$, which is the complementary idempotent.

Proof. $\pi^2=\tfrac14(1+2e+e^2)=\tfrac14(1+2e+1)=\tfrac12(1+e)=\pi$, using $e^2=q(e)=1$. The products are $ef=\tfrac12(e+e^2)=\tfrac12(e+1)=\pi$ and $ef'=\tfrac12(e-e^2)=\tfrac12(e-1)=-\pi'$.

Proposition. With $\pi$ and $\pi'$ as above, $\pi+\pi'=1$ and $ff'=\pi'\pi=0$.

Proof. The sum is $\tfrac12(1+e)+\tfrac12(1-e)=1$, and $ff'=\tfrac14(1+e)(1-e)=\tfrac14(1-e^2)=0$, the other order being the same computation.

So a non-isotropic vector of square one splits the identity of the algebra into two orthogonal idempotents, and it acts as $+1$ on the first and $-1$ on the second.

Primitivity

Theorem. Let $q$ be non-degenerate, $n=\dim V$, and let $e\in V$ with $q(e)=1$. Then $\pi=\tfrac12(1+e)$ is an idempotent and the left ideal $\mathrm{Cl}(V,q)\pi$ has dimension $2^{n-1}$; it is minimal, equivalently $\pi$ is primitive, exactly when it is simple as a left module. For a simple Clifford algebra $\mathrm{Cl}(V,q)\cong M_k(D)$ this happens exactly when $k=2$, that is, when the algebra is a $2\times2$ matrix algebra over a division algebra.

Proof. The idempotence is the proposition above. Left multiplication by $e$ is an involution of the algebra, since $e^2=1$, and it exchanges the parity parts, since multiplication by an odd element carries $\mathrm{Cl}^0$ to $\mathrm{Cl}^1$ and back; the two parts have the same dimension, so that involution has trace zero and its eigenspaces for the eigenvalues $+1$ and $-1$ have equal dimension $2^{n-1}$. The idempotent $\pi$ projects onto the first, so $\dim\mathrm{Cl}(V,q)\pi=2^{n-1}$. In a simple algebra $M_k(D)$ with simple module $S$ the ideal $\mathrm{Cl}(V,q)\pi$ is a sum of copies of $S$, of dimension $2^n/k$ each, so its dimension $2^{n-1}$ leaves the multiplicity $k/2$; a sum of copies of a simple module is simple exactly when it is one copy, that is, when $k=2$. In a product of simple algebras the count applies to each factor, and the ideal is simple only when all but one factor contributes nothing. The two worked cases below have $k=2$, and the counterexamples of the remark have $k=16$ and two factors with one copy each.

Remark (the failure in general). The theorem does not say that the ideal of a unit vector is always minimal, and it is not: for $\mathrm{Cl}_{8,0}\cong M_{16}(\mathbb{R})$ the ideal $\mathrm{Cl}\pi$ has dimension $128$ against a minimal left ideal of dimension $16$, and already for $\mathrm{Cl}_{2,1}\cong M_2(\mathbb{R})\oplus M_2(\mathbb{R})$ it is the sum of one copy of each of the two simple modules, of dimension $4$ against $2$. For the general construction of a minimal left ideal one takes a product of such idempotents, namely the idempotent of a Witt basis, in the section below.

Minimal Left Ideals

Ideals as Clifford Modules

Definition. A left ideal of $\mathrm{Cl}(V,q)$ is a subspace $I$ with $\mathrm{Cl}(V,q)\,I\subseteq I$; it is minimal if it is nonzero and contains no nonzero left ideal properly.

Proposition. Every left ideal is a Clifford module, and a minimal left ideal is an irreducible Clifford module with a faithful action of $\mathrm{Cl}(V,q)$ up to the annihilator of the ideal.

Proof. A left ideal is a module over the algebra by definition, so the Clifford action of a vector satisfies $c(v)^2=q(v)$ as it does on the algebra itself. The annihilator of $I$ is a two-sided ideal, and if the algebra is simple the action is faithful.

Theorem. Let $I=\mathrm{Cl}(V,q)\pi$ with $\pi$ a primitive idempotent. Then $I$ is a minimal left ideal, all minimal left ideals are of this form, and they are all isomorphic as Clifford modules.

Proof. By definition $I$ is a left ideal, and the minimality is the primitivity of $\pi$. Conversely, a minimal left ideal contains an idempotent $\pi$ with $I=\mathrm{Cl}\pi$. The isomorphism of the minimal left ideals is the statement that a simple algebra has a single isomorphism class of simple modules — over a simple algebra $\mathrm{Cl}(V,q)\cong M_k(D)$ the minimal left ideals are the columns $D^k$ — which is the structure theory of Simple and Semisimple Modules that Spin Representations and Clifford Modules with Inner Conjugation uses.

Theorem (dimension). Over an algebraically closed field the minimal left ideal of $\mathrm{Cl}(V,q)$ has dimension $2^{\lfloor n/2\rfloor}$, where $n=\dim V$. Over $\mathbb{R}$ the minimal left ideal is the real spinor module of Spin Representations and Clifford Modules with Inner Conjugation, of dimension over the division algebra $D$ given by the eightfold table.

Proof. Over an algebraically closed field the complex Clifford algebra is $M_{2^m}(\mathbb{C})$ for $n=2m$ or $n=2m+1$, and a minimal left ideal is a column, of dimension $2^m=2^{\lfloor n/2\rfloor}$; the statement is the module-theoretic reading of the classification of Spin Representations and Clifford Modules with Inner Conjugation. Over $\mathbb{R}$ the same reading of the eightfold table gives the real module.

The Action of the Spin Group

Theorem. The spin group $\mathrm{Spin}(V,q)\subseteq\mathrm{Cl}^0(V,q)$ acts on a minimal left ideal by left multiplication, and this action is the spin representation of Spin Representations and Clifford Modules with Inner Conjugation; its kernel on an irreducible module is $\{\pm1\}$.

Proof. The spin group lies in the group of units of the algebra and acts on every left ideal by left multiplication; for an irreducible module this is the spin representation, and the statement about the kernel, $\rho(-1)=-\mathrm{id}$, is proved in the cited article.

So a spinor is an element of a minimal left ideal and transforms under $\psi\mapsto R\psi$; the sandwich $v\mapsto RvR^{-1}$ of Versors, Rotors and the Sandwich Action with Signed Inner Conjugation is the action on the vectors, and the two are different actions of the same group element, as that article records.

The Bridge to the Chevalley Module

The Witt Basis

Definition. A Witt basis of $(V,q)$ is a family of pairs $a_1,a_1^{\dagger},\ldots,a_m,a_m^{\dagger}$ with, in the geometric product,

$$ a_ia_j^{\dagger}+a_j^{\dagger}a_i=\delta_{ij}, \qquad a_ia_j+a_ja_i=0, \qquad a_i^{\dagger}a_j^{\dagger}+a_j^{\dagger}a_i^{\dagger}=0 . $$

Each $a_i$ and each $a_i^{\dagger}$ is isotropic, and orthogonality follows because $2B(a_i,a_j^{\dagger})=\delta_{ij}$. Such a basis exists over a field over which $q$ is split, in particular after scalar extension to an algebraically closed field.

Proposition. The element

$$ \pi=a_1a_1^{\dagger}a_2a_2^{\dagger}\cdots a_ma_m^{\dagger} $$

is an idempotent, and $a_if=0$ for every $i$.

Proof. For one pair, $a^{\dagger}a=1-aa^{\dagger}$ gives $aa^{\dagger}a=a(1-aa^{\dagger})=a-a^2a^{\dagger}=a$, because $a^2=0$; hence $(aa^{\dagger})^2=aa^{\dagger}aa^{\dagger}=(aa^{\dagger}a)a^{\dagger}=aa^{\dagger}$. The annihilation is $a(aa^{\dagger})=a^2a^{\dagger}=0$. For several pairs the factors are idempotents and commute, the anticommutation relations sending an even number of signs into each interchange, so the product is idempotent and each $a_i$ annihilates it on the left.

The Identification with the Exterior Algebra

Theorem. Let $\pi$ be the idempotent of a Witt basis. Then the map

$$ \Lambda^{\bullet}\bigl(\operatorname{span}\{a_1^{\dagger},\ldots,a_m^{\dagger}\}\bigr)\longrightarrow\mathrm{Cl}(V,q)\pi, \qquad a_{i_1}^{\dagger}\wedge\cdots\wedge a_{i_k}^{\dagger}\longmapsto a_{i_1}^{\dagger}\cdots a_{i_k}^{\dagger}\pi, $$

is an isomorphism of vector spaces, and under it the creation operators $a_i^{\dagger}$ act by exterior multiplication and the annihilation operators $a_i$ by contraction.

Proof. The creation operators anticommute among themselves, so the image of a wedge depends only on the set of indices and the map is well defined; the elements $a_{i_1}^{\dagger}\cdots a_{i_k}^{\dagger}\pi$ with $i_1<\cdots

Remark. The theorem identifies the abstract ideal with the Chevalley module $\Delta=\Lambda^{\bullet}W$ of Spin Representations and Clifford Modules with Inner Conjugation, the exterior algebra of a maximal isotropic subspace. The operators $\varepsilon(w)$ and $\iota(w')$ of that article are the left multiplications by the elements of $W$ and of $W'$ on the ideal, so the creation and annihilation operators of the spinor module are multiplications in the algebra, and the two constructions of the spinor space are the same one written on the algebra and on the exterior algebra.

Spinors as Even Multivectors

The Even Subalgebra

The even subalgebra acts on a minimal left ideal, and for the idempotent of a unit vector the ideal it generates is free of rank one over it.

Theorem. Let $q(e)=1$ and $\pi=\tfrac12(1+e)$. Then $\mathrm{Cl}^0(V,q)\pi=\mathrm{Cl}(V,q)\pi$, and the map

$$ \mathrm{Cl}^0(V,q)\longrightarrow\mathrm{Cl}(V,q)\pi, \qquad x\longmapsto xf, $$

is an isomorphism of $\mathrm{Cl}^0$-modules; hence $\mathrm{Cl}(V,q)\pi$ is free of rank one over the even subalgebra and has dimension $2^{n-1}=\dim\mathrm{Cl}^0(V,q)$.

Proof. The identity $ef=\pi$ and the decomposition $\mathrm{Cl}^1(V,q)=\mathrm{Cl}^0(V,q)e$ of the odd part, valid because $e$ is invertible, give $\mathrm{Cl}^1\pi=\mathrm{Cl}^0ef=\mathrm{Cl}^0\pi$, so the two ideals coincide and the map is onto. For injectivity, if $xf=0$ with $x$ even then $x(1+e)=0$, that is $x=-xe$; the grade involution fixes $x$ and sends $e$ to $-e$, so applying it gives $x=xe$, and the two relations give $x=-x$, whence $x=0$ in characteristic not two.

Corollary. When in addition $\pi$ is primitive, so that $\mathrm{Cl}(V,q)\pi$ is the spinor module, the even subalgebra is identified with the spinor module and a spinor may be represented by an even multivector. When $\pi$ is not primitive the ideal is a sum of copies of the spinor module, and the even subalgebra represents that larger module.

Remark. The idempotent $\pi$ need not be even, and the elements of the ideal are therefore not even in general. What is identified with the even subalgebra is the ideal, through left multiplication by $\pi$; the representation of a spinor by an even multivector is the pullback of that identification, and it is available exactly when the ideal generated by a unit vector is minimal.

The Chiral Halves

Theorem. Let $n=\dim V$ be even. Then $\mathrm{Cl}^0(V,q)$ is the Clifford algebra of a space of dimension $n-1$ and the half-spinor spaces $\Delta_{\pm}$ are its irreducible modules; they are non-isomorphic when $\mathrm{Cl}^0$ has two simple factors and isomorphic when it is simple. A minimal left ideal of $\mathrm{Cl}^0(V,q)$ is a half-spinor space, and the spinor module of $\mathrm{Cl}(V,q)$ is the sum $\Delta_+\oplus\Delta_-$ of the two when they are non-isomorphic, and a sum of two copies of the unique simple module when $\mathrm{Cl}^0$ is simple.

Proof. The even part is a Clifford algebra of one dimension less, and the half-spinor spaces are its irreducible modules, by Spin Representations and Clifford Modules with Inner Conjugation. They are the eigenspaces of the volume element, which is central for even $n$ and whose square is a scalar, so the algebra acts on their sum by the two projections; that sum is the spinor module of $\mathrm{Cl}(V,q)$, with one copy of each of the two simple modules when they differ and two copies of the single one when they agree.

Worked Cases

Three Dimensions

In $\mathrm{Cl}_{3,0}$ take $e_3$, of square one, and $\pi=\tfrac12(1+e_3)$. The left ideal $\mathrm{Cl}\pi$ is spanned over $\mathbb{R}$ by

$$ \pi, \qquad e_1\pi, \qquad e_2\pi, \qquad \omega \pi, $$

four elements, and since $\omega$ is central with $\omega^2=-1$ the ideal is a two-dimensional vector space over $\mathbb{C}=\mathbb{R}\oplus\mathbb{R}\omega$ with basis $\pi,e_1\pi$. The action of the generators on that basis is

$$ e_1\pi=e_1\pi,\quad e_1(e_1\pi)=\pi, \qquad e_2\pi=\omega(e_1\pi),\quad e_2(e_1\pi)=-\omega \pi, \qquad e_3\pi=\pi,\quad e_3(e_1\pi)=-e_1\pi, $$

which is the action of the Pauli matrices with $i=\omega$: $e_1$ acts as $\begin{pmatrix}0&1\\1&0\end{pmatrix}$, $e_2$ as $\begin{pmatrix}0&-i\\i&0\end{pmatrix}$ and $e_3$ as $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$. So the minimal left ideal is the two-dimensional complex spinor module of the Pauli action, as recorded in Spin Representations and Clifford Modules with Inner Conjugation.

Verification of the relations used. The products $e_3\pi=\pi$ and $e_3(e_1\pi)=e_3e_1\pi=-e_1e_3\pi=-e_1\pi$ use $e_3\pi=\pi$. The element $e_2\pi=\tfrac12(e_2+e_2e_3)$ equals $\omega e_1\pi$, because $\omega e_1=e_2e_3$ and $e_2e_3\pi=\tfrac12(e_2e_3+e_2e_3e_3)=\tfrac12(e_2e_3+e_2)$; and $e_2(e_1\pi)=e_2e_1\pi=-e_1e_2\pi=-\omega \pi$, since $e_1e_2\pi=\tfrac12(e_1e_2+e_1e_2e_3)=\tfrac12(e_1e_2+\omega)=\omega \pi$. The four elements $\pi,e_1\pi,e_2\pi,\omega \pi$ are independent over $\mathbb{R}$, as their components in the basis $1,e_1,e_2,e_3,e_1e_2,e_2e_3,e_3e_1,\omega$ have distinct pivots.

The primitivity in this case. For $x=e_1,e_2$ one computes $fxf=0$, using $e_3e_1e_3=-e_1$ and the same for $e_2$, and $\pi\omega \pi=\omega \pi^2=\omega \pi=\omega\cdot \pi$; so $\pi\mathrm{Cl}\pi$ is the span of $\pi$ and $\omega \pi$ over $\mathbb{R}$, that is the field $\mathbb{C}$ with identity $\pi$, a division algebra. Hence $\pi$ is primitive and the ideal is minimal.

The even part. Here $\mathrm{Cl}^0_{3,0}$ is the quaternion algebra, a division algebra and hence simple; the theorem above identifies the spinor module with the even subalgebra, of real dimension four, and a Pauli spinor is an even multivector. The identification is left multiplication by $\pi$.

Minkowski Space

In $\mathrm{Cl}_{1,3}=M_2(\mathbb{H})$ the even part is $\mathrm{Cl}^0_{1,3}\cong\mathrm{Cl}_{1,2}\cong M_2(\mathbb{C})$, which is the biquaternion algebra $\mathbb{B}$ of The Clifford Structure of the Biquaternion Algebra. An even multivector of $\mathrm{Cl}_{1,3}$ has $1+6+1=8$ real components, the scalar, the six bivectors and the pseudoscalar, and this is the real dimension of a Dirac spinor; the chiral halves are the two minimal left ideals of $M_2(\mathbb{C})$, each of complex dimension two, and they are the two Weyl spinors of the spinor module. The description of a Dirac spinor as an even multivector is thus the statement that the spinor module is the even subalgebra, which here is the biquaternion algebra acting on itself.

The Two Actions on the Ideal

Let $I=\mathrm{Cl}(V,q)\pi$ be a minimal left ideal, $\psi\in I$ a spinor and $R$ a rotor.

Proposition. Left multiplication is an action of the rotors on $I$, $\psi\mapsto R\psi$, and it is the spin representation. The dual representation is realised by right multiplication on a minimal right ideal, $J=\pi'\mathrm{Cl}(V,q)$, and right multiplication by a rotor does not preserve a minimal left ideal in general. The sandwich $v\mapsto RvR^{-1}$ is neither, being the action on the vectors.

Proof. The left action is the module action of Spin Representations and Clifford Modules with Inner Conjugation. For the second statement, right multiplication carries $I$ to $\mathrm{Cl}(V,q)(fx)$, and $fx$ need not lie in $I$: in $\mathrm{Cl}_{3,0}$ with $\pi=\tfrac12(1+e_3)$ one has $fe_2e_3=\tfrac12(e_2e_3-e_2)$, while a combination $af+be_1\pi+ce_2\pi+d\omega \pi$ has the same coefficient $\tfrac c2$ of $e_2$ and of $e_2e_3$, so the two coefficients of $fe_2e_3$, namely $-\tfrac12$ and $+\tfrac12$, cannot both be matched, and $fe_2e_3\notin I$. A right ideal is preserved by right multiplication by a unit, because $\mathrm{Cl}(V,q)R^{-1}=\mathrm{Cl}(V,q)$, and the resulting action of the rotors is the dual of the spin representation. The sandwich acts on $V=\mathrm{Cl}_1$, not on $I$ in general.

Remark (the grading of a spinor). A spinor is an element of $I$ and need not be homogeneous or even; the even-multivector picture of the previous section applies when $I$ is identified with the even subalgebra through a primitive idempotent, and there the representative of a spinor is even. In $\mathrm{Cl}_{1,3}$ this makes the representative of a Dirac spinor a biquaternion, which is the form in which the applications of the category write it.

Summary

An idempotent of $\mathrm{Cl}(V,q)$ is an element with $\pi^2=\pi$; a non-isotropic vector $e$ with $q(e)=1$ gives the idempotent $\pi=\tfrac12(1+e)$, of complementary idempotent $\pi'=\tfrac12(1-e)$, and the left ideal $\mathrm{Cl}(V,q)\pi$ has dimension $2^{n-1}$. That ideal is minimal exactly when it is a single simple module, which for a simple Clifford algebra $M_k(D)$ means $k=2$; it is minimal in the two worked cases and not in general, and the general minimal left ideal is the ideal of the idempotent of a Witt basis. Minimal left ideals are Clifford modules, all are isomorphic over a simple Clifford algebra, and they are the spinor modules of Spin Representations and Clifford Modules with Inner Conjugation: over an algebraically closed field the minimal left ideal has dimension $2^{\lfloor n/2\rfloor}$, and over $\mathbb{R}$ it is the real spinor module of the eightfold table. The spin group acts on a minimal left ideal by left multiplication, with kernel $\{\pm1\}$, and a spinor is an element of such an ideal.

The ideal description and the description by an exterior algebra are the same: with a Witt basis $a_i,a_i^{\dagger}$ the idempotent $\pi=a_1a_1^{\dagger}\cdots a_ma_m^{\dagger}$ is annihilated by the isotropic elements $a_i$, and the map sending a wedge of creation operators to their product against $\pi$ is an isomorphism from $\Lambda^{\bullet}(\operatorname{span}a_i^{\dagger})$ onto $\mathrm{Cl}\pi$, under which the creation operators multiply and the annihilation operators contract. So the operators $\varepsilon$ and $\iota$ of the Chevalley model are multiplications in the algebra.

Since the even subalgebra acts on the ideal, the map $x\mapsto xf$ identifies the even subalgebra with the ideal $\mathrm{Cl}(V,q)\pi$ generated by a unit vector, and that ideal is the spinor module exactly when it is minimal: a spinor may then be represented by an even multivector, and this is the sense of the identity familiar from the applications. The identification is exact in the two worked cases. In $\mathrm{Cl}_{3,0}$ the ideal is spanned by $\pi,e_1\pi,e_2\pi,\omega \pi$, a two-dimensional complex space on which the generators act as the Pauli matrices and which is isomorphic to the even subalgebra, a quaternion algebra. In $\mathrm{Cl}_{1,3}$ the even part is $\mathrm{Cl}_{1,2}\cong M_2(\mathbb{C})$, the biquaternion algebra, of real dimension eight, which is the dimension of a Dirac spinor, and its two minimal left ideals are the two chiral halves. A rotor acts on a spinor by left multiplication, the spin representation, and on a vector by the sandwich; the two actions carry the two halves of the double cover.

Summary of Notation

Symbol	Meaning
$\pi$, $\pi^2=\pi$	Idempotent
$\pi=\tfrac12(1+e)$, $q(e)=1$	Idempotent of a unit vector, primitive in the low-dimensional cases
$\pi'=\tfrac12(1-e)$	Complementary idempotent, $ff'=0$, $\pi+\pi'=1$
$\mathrm{Cl}(V,q)\pi$	Left ideal of dimension $2^{n-1}$, the spinor module when minimal
$\mathrm{Cl}^0(V,q)\cong\mathrm{Cl}(V,q)\pi$	Even subalgebra, free of rank one over itself through $x\mapsto xf$
$a_i,\ a_i^{\dagger}$	Witt basis, $a_ia_j^{\dagger}+a_j^{\dagger}a_i=\delta_{ij}$
$\pi=a_1a_1^{\dagger}\cdots a_ma_m^{\dagger}$	Idempotent of a Witt basis
$\Lambda^{\bullet}(\operatorname{span}a_i^{\dagger})\cong\mathrm{Cl}\pi$	Ideal as an exterior algebra
$\mathrm{Cl}^0(V,q)\pi$	The even submodule, equal to $\mathrm{Cl}(V,q)\pi$
$\Delta_+,\ \Delta_-$	Chiral halves, minimal left ideals of $\mathrm{Cl}^0$
$\psi\mapsto R\psi$	Left action of the rotors, the spin representation
$\rho(-1)=-\mathrm{id}$	Kernel of the spin representation