The Grading of a Hilbert Algebra with Signed Hermitian Adjoint

Introduction

A Clifford algebra carries a $\mathbb{Z}/2$-grading by the parity of the length of a blade, $\mathrm{Cl}(V,q)=\mathrm{Cl}^0\oplus\mathrm{Cl}^1$, and the grading is a structural fact and not a bookkeeping device: it is what makes the grade involution $\alpha$ an automorphism, it is what separates a rotation from a reflection, and it is the reason the signed Hermitian sandwich is a different operator from the Hermitian sandwich. This article is the Hermitian reading of that grading, in parallel with The Grading of the Clifford Algebra with Signed Inner Conjugation, where the same statements are made for the right factor $x^{-1}$.

The subject is the interaction of the grading with the dagger. The dagger preserves the parity, $x^{\dagger}\in\mathrm{Cl}^{|x|}$, because both reversion and the coefficient involution are even; consequently the signed Hermitian sandwich is even in the sense that it preserves the parity of its argument, and the sign that separates it from the unsigned member is the sign character of the grading itself,

$$ \Theta^{\alpha}_x=\varepsilon_x\,\Theta_x,\qquad \varepsilon_x=(-1)^{|x|}, $$

$\varepsilon$ being the algebra homomorphism $\mathrm{Cl}(V,q)\to\{\pm1\}$ that is $+1$ on the even part and $-1$ on the odd part.

Two consequences are developed. The first is that the odd part of the unitary group is a coset and not a subgroup, $U=U^0\sqcup uU^0$ with $U/U^0\cong\mathbb{Z}/2$, so that the odd elements are indexed by the even ones and no second group is needed; the same is the relation $\mathrm{Pin}/\mathrm{Spin}\cong\mathbb{Z}/2$. The second is that the family of signed Hermitian sandwiches is generated by a single odd operator together with the even ones, because the composite of two odd steps is an ordinary Hermitian sandwich, the two signs having cancelled.

The algebra, the grading and the three involutions are Clifford Algebras and Clifford Algebras in Finite Dimensions; the dagger is Hilbert Algebras; the two-sided family is Two-Sided Operators on a Clifford Algebra; the operator and its parity sign are Two-Sided Operators on a Hilbert Algebra with Signed Hermitian Adjoint; the inner-conjugation reading of the grading is The Grading of the Clifford Algebra with Signed Inner Conjugation; the slice is The Unitary Slice and the Compact Real Form with Hermitian Adjoint. Nothing of the inverse-conjugation article is reused except the comparison of the two maps, which is quoted and adapted.

Conventions. As in the companion articles: $A$ commutative with involution $\sigma$, $V$ free of rank $n$ with a non-degenerate $q$, $x^{\dagger}=\sigma(\alpha(x^{r}))$, $\Gamma$ the Clifford group, $U=\{x:x^{\dagger}x=1\}$ the unitary slice, $\varepsilon_x=(-1)^{|x|}$, and $\Theta^{\alpha}_x(y)=\alpha(x)\,y\,x^{\dagger}$, $\Theta_x(y)=x\,y\,x^{\dagger}$.

The Grading and the Sign

Proposition (the dagger is even). The dagger preserves the degree mod two, $(\mathrm{Cl}^k)^{\dagger}\subseteq\mathrm{Cl}^k$, and it is an anti-automorphism of the graded algebra: $(xy)^{\dagger}=y^{\dagger}x^{\dagger}$ with $|y^{\dagger}|=|y|$.

Proof. Reversion fixes each blade up to the sign $(-1)^{k(k-1)/2}$, hence preserves the degree; the grade involution $\alpha$ multiplies the degree-$k$ part by $(-1)^k$, hence preserves it as well; the coefficient involution acts on the coefficients and not on the degree. The anti-automorphism property is from Hilbert Algebras.

Proposition (the parity sign, restated as a statement about the grading). For homogeneous $x$ the twist is the sign character on the left factor, $\alpha(x)=\varepsilon_x x$, so

$$ \Theta^{\alpha}_x=\varepsilon_x\,\Theta_x . $$

The operator family is therefore a graded deformation of the Hermitian family: the two agree on $\mathrm{Cl}^0$, the even part of the algebra of operators, and differ by $-1$ on $\mathrm{Cl}^1$.

Proof. The grade involution is by definition multiplication by $(-1)^k$ on $\mathrm{Cl}^k$; substitute into the definition of $\Theta^{\alpha}_x$ and use centrality of the scalar.

Proposition (the sign is a homomorphism). The assignment $x\mapsto\varepsilon_x$ defined on homogeneous elements is the nontrivial character of the grading,

$$ \varepsilon_x\varepsilon_z=\varepsilon_{xz}, $$

so it extends to a monoid homomorphism on the whole algebra, and it induces an isomorphism $\mathrm{Cl}(V,q)/\mathrm{Cl}^0\cong\mathbb{Z}/2$ of the parity quotient.

Proof. Degrees add mod two, so the parities multiply; the kernel is the even part and every odd element is outside it.

Remark. The two propositions together are the whole content of the sign: it is a character, and the operator is the character times the unsigned operator. The parity of an element is thus an element of the dual group $\mathrm{Hom}(\mathbb{Z}/2,\{\pm1\})$, and the sign of the operator is that element evaluated at the parity.

The Composition of Two Signed Hermitian Sandwiches

Proposition. For all $x,z$ the family is exactly multiplicative,

$$ \Theta^{\alpha}_x\circ\Theta^{\alpha}_z=\Theta^{\alpha}_{xz}, $$

and written with the unsigned operator on the right the same statement reads

$$ \Theta^{\alpha}_x\circ\Theta^{\alpha}_z=\varepsilon_x\varepsilon_z\;\Theta_{xz}, $$

so for homogeneous $x,z$ the composite is the ordinary Hermitian sandwich exactly when the two parities agree,

$$ \Theta^{\alpha}_x\circ\Theta^{\alpha}_z= \begin{cases} \Theta_{xz}, & x,z \text{ both even or both odd},\\[2pt] -\,\Theta_{xz}, & x,z \text{ of opposite parity}. \end{cases} $$

Proof. Multiplicativity $\Theta^{\alpha}_{uv}=\Theta^{\alpha}_u\Theta^{\alpha}_v$ is Two-Sided Operators on a Hilbert Algebra with Signed Hermitian Adjoint; substituting the parity-sign identity $\Theta^{\alpha}_w=\varepsilon_w\Theta_w$ for $w=x,z,xz$ and using $\varepsilon_x\varepsilon_z=\varepsilon_{xz}$ gives the second form, and $\varepsilon_x\varepsilon_z=+1$ exactly when the parities agree.

Corollary (two odd steps give an ordinary step). If $x$ and $z$ are both odd then $xz$ is even, $\varepsilon_{xz}=+1$, and

$$ \Theta^{\alpha}_x\circ\Theta^{\alpha}_z=\Theta^{\alpha}_{xz}=\Theta_{xz}, $$

an ordinary Hermitian sandwich. The square of an odd step is ordinary as well, $(\Theta^{\alpha}_x)^{2}=\Theta_{x^{2}}$.

Proof. The product of two odd elements is even, so the twist is trivial on it; the rest is the multiplicativity.

Remark (the family is graded, and one odd generator suffices). Fix an odd $u$. Every odd $x$ is $x=us$ with $s$ even, so $\Theta^{\alpha}_x=\Theta^{\alpha}_u\circ\Theta^{\alpha}_s$: the odd operators are the composite of one fixed odd operator with the even ones, and the even ones are unsigned. The parity of the parameter is therefore the only new input, exactly as for the inverse member.

The Odd Part of the Unitary Group is a Coset

Proposition. Let $U=\{x:x^{\dagger}x=1\}$ be the unitary slice, and let $U^0=U\cap\mathrm{Cl}^0$ and $U^1=U\cap\mathrm{Cl}^1$. Then $U^0$ is a subgroup of $U$ of index at most two, and if $U^1$ is non-empty it is a single coset,

$$ U^1=u\,U^0=U^0u,\qquad u\in U^1\ \text{any}, \qquad U/U^0\cong\mathbb{Z}/2 . $$

Proof. The even part is closed under multiplication and inversion because the grading is multiplicative and the dagger preserves the parity; it is a subgroup. If $u$ is odd and $s$ even then $(us)^{\dagger}(us)=s^{\dagger}u^{\dagger}u\,s=s^{\dagger}s=1$, so $uU^0\subseteq U^1$ and $U^0u\subseteq U^1$; conversely an odd $v$ is $v=u\,(u^{-1}v)$ with $u^{-1}v$ even and in $U$ because $(u^{-1}v)^{\dagger}(u^{-1}v)=v^{\dagger}u\,u^{-1}v=v^{\dagger}v=1$ on the slice, where $u^{-1}=u^{\dagger}$. Hence $U^1=uU^0$, and the two cosets exhaust $U$.

Corollary (the double cover of the pin group). With the trivial involution and a definite form, $U=\mathrm{Pin}(V,q)$ (for an indefinite form $U\cap\Gamma=\{N=1\}$ is the double cover of the identity component, while $\Gamma_{\dagger}=\mathrm{Pin}$ always), the even part $U^0$ is the spin group, and

$$ \mathrm{Pin}=\mathrm{Spin}\sqcup(\text{odd coset}),\qquad \mathrm{Pin}/\mathrm{Spin}\cong\mathbb{Z}/2 , $$

whenever the odd part is non-empty, which over $\mathbb{R}$ always holds since a vector can be rescaled to square $-1$. Hence the odd part is a coset and it is the coset that carries the reflections; the signed Hermitian sandwich is constant on the two cosets up to the sign.

Proof. The trivial involution makes the dagger the Clifford conjugation, so $U=\{x:N(x)=1\}=\mathrm{Pin}$, and $U^0=\mathrm{Spin}$; the coset statement is the proposition, and the reflection statement is the vector proposition of the companion article.

Remark. The operator $\Theta^{\alpha}$ is not injective on $U$: on the slice it is the signed inner conjugation $\mathrm{Ad}^{\alpha}$, whose kernel is the scalars, so the odd coset is not separated by the operator from the even coset beyond the parity sign. What separates them is the determinant, in the next section but one.

The Two Components as Even Modules

Proposition. Let $e\in V$ with $q(e)\neq0$, an odd unit. Left multiplication by $e$ interchanges the two parity components and the Clifford algebra is a free left module of rank two over its even subalgebra,

$$ \mathrm{Cl}(V,q)=\mathrm{Cl}^0\oplus\mathrm{Cl}^1=\mathrm{Cl}^0\cdot1\ \oplus\ \mathrm{Cl}^0\cdot e , $$

with basis $\{1,e\}$. The dagger-form $\langle x,y\rangle_{\dagger}=\mathrm{Sc}(x^{\dagger}y)$ is even: it pairs $\mathrm{Cl}^0$ with $\mathrm{Cl}^0$ and $\mathrm{Cl}^1$ with $\mathrm{Cl}^1$, and the two components are orthogonal.

Proof. Multiplication by an odd element shifts the parity and is a bijection because $e$ is invertible, so $\mathrm{Cl}^1=\mathrm{Cl}^0e$ and the sum is direct; this is the module statement of The Grading of the Clifford Algebra with Signed Inner Conjugation. For the form, $x^{\dagger}y$ has parity $|x|+|y|$, and the scalar part is even; hence $\mathrm{Sc}(x^{\dagger}y)\neq0$ forces $|x|=|y|$, and the two components are orthogonal.

Remark (the copies are not independent). The two components are isomorphic as modules over $\mathrm{Cl}^0$, but the isomorphism is multiplication by the odd unit $e$, and the product of two odd elements is even and lives in the first component. The decomposition records that the algebra is a rank-two module over its even part; it does not say that the two components can be separated, and a pair of independent copies would lose the multiplication rule of $\mathrm{Pin}$, the action on $V$ and the twist $\alpha$ itself.

Remark (the Hermitian form is a form on each sector). By the second proposition the Hermitian structure of the algebra does not mix the two parities: it is the orthogonal direct sum of the two Hermitian forms on $\mathrm{Cl}^0$ and $\mathrm{Cl}^1$. The signed Hermitian sandwich is likewise even, and it acts on the two sectors separately; the chirality of the module of a representation, where the odd part of the algebra interchanges the two sectors, is the corresponding statement on a Clifford module and is Pin Representations and Hermitian Modules with Signed Hermitian Adjoint.

Remark (the full graded algebra is needed). If the odd part is removed and only $\mathrm{Cl}^0$ is kept, the twist $\alpha$ becomes the identity and the signed Hermitian sandwich collapses to the unsigned one: the sign is not a property of the even part. The correct object is the single graded algebra with the grading built in, and the odd component adjoined; the mod-two decomposition of the operator is written in that algebra and in no proper subalgebra of it. The biquaternion case of the same reading, with the grading the fermion parity and the odd generator the supercharge, is The Graded Algebra, Fermion Parity and the Two Sectors with Signed Inner Conjugation in Biquaternionic Form.

The Two Maps Compared

Proposition. For $x\in\Gamma(V,q)$ of parity $k$, the two Hermitian sandwiches act on the vectors as

$$ \Theta^{\alpha}_x\bigr|_{V}=\varepsilon_x\,\Theta_x\bigr|_{V}, $$

and on the unitary slice, where both reduce to conjugations, their determinants are

$$ \det\Theta^{\alpha}_x\bigr|_{V}=\varepsilon_x,\qquad \det\Theta_x\bigr|_{V}=(\varepsilon_x)^{n+1}= \begin{cases} \varepsilon_x, & n \text{ even},\\[2pt] +1, & n \text{ odd}. \end{cases} $$

Consequently for $n$ even the two members reach the same two cosets of $SO(V,q)$ in $O(V,q)$, while for $n$ odd the unsigned member misses every improper isometry and the signed member reaches them all.

Proof. On the slice the signed operator is $\mathrm{Ad}^{\alpha}_x$ and the unsigned one is $\mathrm{Ad}_x$; for the inverse member $\det\mathrm{Ad}^{\alpha}_x=\varepsilon_x$ and $\det\mathrm{Ad}_x=(\varepsilon_x)^{n+1}$, by The Grading of the Clifford Algebra with Signed Inner Conjugation. Off the slice both are scalar multiples of the same twisted conjugation, so the comparison of determinants is by the same scalar.

Remark (the sign is the determinant). The parity of the parameter is the determinant of the isometry on the slice: even parameters give rotations, odd parameters give improper isometries. This is the sense in which the sign is not a convention: the parity grading, the twist, and the determinant of the orthogonal transformation are one datum, and the signed Hermitian sandwich is the operator that reads it off.

Worked Cases

Two Odd Steps Give an Ordinary Step

In $\mathrm{Cl}_{0,3}(\mathbb{R})$ with $e_j^{2}=-1$ and $\sigma=\mathrm{id}$, let $u=e_1$ and $w=e_2$, both odd. Then $\varepsilon_u=\varepsilon_w=-1$ and

$$ \Theta^{\alpha}_{e_1}\circ\Theta^{\alpha}_{e_2}=\Theta^{\alpha}_{e_1e_2}=\Theta_{e_1e_2}, $$

the half-turn of the plane $\mathrm{span}(e_1,e_2)$ with $e_3$ fixed, of determinant $+1$: two signed Hermitian sandwiches, each the reflection $\rho_{e_1}$, $\rho_{e_2}$ of determinant $-1$, have composed into one ordinary Hermitian sandwich, a rotation.

The Odd Coset of the Unitary Group

In the same algebra $U=\mathrm{Pin}=\{x:N(x)=1\}$, with $\mathrm{Spin}=\{1,e_1e_2,e_2e_3,e_3e_1\}$ up to sign and the odd coset $e_1\,\mathrm{Spin}$. The element $e_1$ has $\varepsilon=-1$, and $\Theta^{\alpha}_{e_1}=\rho_{e_1}$ has determinant $-1$; the element $e_1e_2$ has $\varepsilon=+1$, and $\Theta^{\alpha}_{e_1e_2}=\Theta_{e_1e_2}$ has determinant $+1$. The sign of the operator is the determinant of the isometry, and it is constant along each coset.

The Hermitian Form is Even

In $\mathrm{Cl}_{0,3}$ the even part is spanned by $1,e_1e_2,e_2e_3,e_3e_1$ and the odd part by $e_1,e_2,e_3,e_1e_2e_3$. The blade form $\langle x,y\rangle_{\dagger}=\mathrm{Sc}(x^{\dagger}y)$ has $\langle e_1,e_1e_2\rangle_{\dagger}=\mathrm{Sc}(e_1^{\dagger}e_1e_2)=\mathrm{Sc}((-e_1)e_1e_2)=\mathrm{Sc}(e_2)=0$, while $\langle1,1\rangle_{\dagger}=1$ and $\langle e_1,e_1\rangle_{\dagger}=\mathrm{Sc}((-e_1)e_1)=\mathrm{Sc}(1)=1$: the two sectors are orthogonal and each is definite, the Hermitian form being the orthogonal sum of the two.

Summary

The grading $\mathrm{Cl}=\mathrm{Cl}^0\oplus\mathrm{Cl}^1$ interacts with the dagger in three ways. The dagger is even, so it preserves the parity; the twist is the sign character $\alpha(x)=\varepsilon_x x$, so the signed Hermitian sandwich is $\Theta^{\alpha}_x=\varepsilon_x\Theta_x$, agreeing with the unsigned member on the even part and differing by $-1$ on the odd part; and the sign is a homomorphism $\varepsilon_x\varepsilon_z=\varepsilon_{xz}$ of the parity quotient onto $\{\pm1\}$.

The composition is exactly multiplicative, $\Theta^{\alpha}_x\Theta^{\alpha}_z=\Theta^{\alpha}_{xz}$, equivalently $\Theta^{\alpha}_x\Theta^{\alpha}_z=\varepsilon_x\varepsilon_z\Theta_{xz}$: two operators with odd parameters compose to an ordinary Hermitian sandwich, $\Theta^{\alpha}_x\Theta^{\alpha}_z=\Theta_{xz}$, the two signs cancel, and the square of an odd step is ordinary. The odd part of the unitary group is a coset and not a subgroup, $U=U^0\sqcup uU^0$ with $U/U^0\cong\mathbb{Z}/2$, which for the trivial involution is $\mathrm{Pin}/\mathrm{Spin}\cong\mathbb{Z}/2$; the odd coset is what carries the reflections. The algebra is a rank-two module over its even part, $\mathrm{Cl}=\mathrm{Cl}^0\oplus\mathrm{Cl}^0e$, and the Hermitian form is even, the orthogonal sum of the forms on the two components, so the Hermitian structure does not mix the parities while the multiplication does. On the slice the two Hermitian sandwiches are the two conjugations, with determinants $\varepsilon_x$ and $(\varepsilon_x)^{n+1}$, so in odd dimension only the signed member reaches the improper isometries: the parity of the parameter is the determinant of the isometry.

Summary of Notation

Symbol	Meaning
$\mathrm{Cl}=\mathrm{Cl}^0\oplus\mathrm{Cl}^1$	The parity grading
$\varepsilon_x=(-1)^{\|x\|}$	The sign character, $\varepsilon_x\varepsilon_z=\varepsilon_{xz}$
$\Theta^{\alpha}_x=\varepsilon_x\Theta_x$	Signed vs unsigned Hermitian sandwich
$\Theta^{\alpha}_x\Theta^{\alpha}_z=\Theta_{xz}$ ($x,z$ odd)	Two odd signs cancel
$U=U^0\sqcup uU^0$, $U/U^0\cong\mathbb{Z}/2$	Odd part of the unitary group is a coset
$\mathrm{Pin}=\mathrm{Spin}\sqcup(\text{odd coset})$	The same for the pin group, $\sigma=\mathrm{id}$
$\mathrm{Cl}=\mathrm{Cl}^0\oplus\mathrm{Cl}^0e$	Rank-two module over the even part
$\langle x,y\rangle_{\dagger}=\mathrm{Sc}(x^{\dagger}y)$	Hermitian form, even, orthogonal on the two components
$\det\Theta^{\alpha}_x\|_V=\varepsilon_x$ on $U$	The sign is the determinant