The Pin and Spin Groups with Signed Hermitian Adjoint

Introduction

The classical construction of the covering groups of the orthogonal group uses the inverse: the Clifford group is the set of units $x$ for which $x\,v\,x^{-1}$ is again a vector, the norm $N(x)=x x^{\natural}$ cuts it down to $\mathrm{Pin}$, and the parity cuts $\mathrm{Pin}$ down to $\mathrm{Spin}$. The right factor in all of it is $x^{-1}$, and the operator carrying it is the signed inner conjugation $\mathrm{Ad}^{\alpha}_x(v)=\alpha(x)vx^{-1}$.

This article makes the same construction with the right factor replaced by the dagger, that is with the signed Hermitian sandwich of Two-Sided Operators on a Hilbert Algebra with Signed Hermitian Adjoint,

$$ \Theta^{\alpha}_x(v)=\alpha(x)\,v\,x^{\dagger}, $$

and it isolates what changes. The dagger needs an involution $\sigma$ of the base and is defined on the whole algebra; the inverse needs invertibility and nothing else. Three objects must therefore be compared rather than one. The Clifford group $\Gamma(V,q)$ is defined by the condition $\mathrm{Ad}^{\alpha}_x(V)\subseteq V$; the unitary slice $U=\{x:x^{\dagger}x=1\}$ is defined by the dagger; and the new object is the set of parameters for which the Hermitian sandwich is an isometry of $V$,

$$ \Gamma_{\dagger}=\{\,x\in\Gamma(V,q) : \sigma(N(x))^{2}=1\,\}, $$

the Hermitian Clifford group. The three are related by two facts: the Hermitian sandwich preserves $V$ exactly for $x\in\Gamma$ and is an isometry exactly for $x\in\Gamma_{\dagger}$, by the action proposition of the companion article; and on the slice $U$ the dagger is the inverse, so there $\Theta^{\alpha}_x=\mathrm{Ad}^{\alpha}_x$ and the whole inverse theory transfers verbatim.

The payoff is a single clean statement. With the trivial involution the Hermitian Clifford group is the pin group, $\Gamma_{\dagger}=\mathrm{Pin}$, and the hermitian reading of $\mathrm{Pin}$ is the set of elements whose Hermitian sandwich is an isometry; the odd part of it gives the reflections and the even part the rotations, and the map to $O(V,q)$ is two-to-one exactly as for the inverse member. The general involution $\sigma$ widens the scalars and the norm condition but changes no group-theoretic statement, because everything is a scalar modification of the inverse member on the slice.

The groups $\Gamma$, $\mathrm{Pin}$, $\mathrm{Spin}$, the norm and the exact sequences are The Clifford, Pin and Spin Groups with Signed Inner Conjugation; the dagger and the involution of the base are Hilbert Algebras; the unitary slice, its compactness and the compact real form are The Unitary Slice and the Compact Real Form with Hermitian Adjoint; the operator and its action on $V$ are Two-Sided Operators on a Hilbert Algebra with Signed Hermitian Adjoint; the unsigned member and its Clifford-group identities are The Hermitian Sandwich on a Hilbert Algebra with Hermitian Adjoint. Nothing owned by those entries is reproved.

Conventions. The base $A$ is a commutative ring with involution $\sigma$, $F$ is the field of scalars of the Clifford algebra, $V$ is free of rank $n$ with a non-degenerate quadratic form $q$, $x^{\dagger}=\sigma(\alpha(x^{r}))$, $x^{\natural}=\alpha(x^{r})$ is Clifford conjugation, $N(x)=x x^{\natural}$, $\rho_u(v)=v-2g(v,u)q(u)^{-1}u$, and $\Theta^{\alpha}_x$, $\Theta_x$, $\mathrm{Ad}^{\alpha}_x$ are the three sandwiches named above.

The Three Conditions

Proposition (the slice is a group). $U=\{x:x^{\dagger}x=1\}$ is a subgroup of the unit group. It contains $1$; if $x\in U$ then $x^{\dagger}=x^{-1}$ and $x^{\dagger}\in U$; and $U$ is closed under multiplication.

Proof. This is the slice proposition of The Hermitian Sandwich on a Hilbert Algebra with Hermitian Adjoint: $(xy)^{\dagger}(xy)=y^{\dagger}x^{\dagger}xy=y^{\dagger}y=1$.

Proposition (the Hermitian Clifford group is a group). $\Gamma_{\dagger}=\{x\in\Gamma(V,q):\sigma(N(x))^{2}=1\}$ is a subgroup of $\Gamma(V,q)$, and it is the set of parameters for which the Hermitian sandwiches restrict to isometries of $V$,

$$ \Theta^{\alpha}_x(V)\subseteq V\iff x\in\Gamma(V,q), \qquad \Theta^{\alpha}_x\bigr|_{V}\in O(V,q)\iff x\in\Gamma_{\dagger}. $$

Proof. $N$ is multiplicative on $\Gamma$ and $\sigma$ is a ring homomorphism, so $\sigma(N(xy))=\sigma(N(x))\sigma(N(y))$ and the condition $\sigma(N)^{2}=1$ is closed under multiplication and inversion; it holds at $1$. The two equivalences are the action proposition of the companion article, where the operator acts on $V$ as $\varepsilon_x\sigma(N(x))\chi(x)$ with $\chi(x)\in O(V,q)$, so it is an isometry exactly when the scalar has square $1$.

Proposition (the slice meets the Clifford group in the norm-one subgroup). With the trivial involution, $\sigma=\mathrm{id}$, the dagger is the Clifford conjugation, $x^{\dagger}=x^{\natural}$, and on the Clifford group $x^{\natural}\,x=N(x)$. Hence

$$ U\cap\Gamma(V,q)=\{\,x\in\Gamma:N(x)=1\,\}, \qquad \Gamma_{\dagger}=\{\,x\in\Gamma:N(x)=\pm1\,\}=\mathrm{Pin}(V,q). $$

So the Hermitian Clifford group is the pin group, and the slice meets it in the norm-one subgroup, of index at most two. For a definite form in the convention of the corpus, $q$ negative definite, the value $N(x)=\prod_i(-q(v_i))$ is positive for every $x\in\Gamma$, so the two coincide, $U\cap\Gamma=\Gamma_{\dagger}=\mathrm{Pin}(V,q)$, and the slice recovers the pin group as in The Unitary Slice and the Compact Real Form with Hermitian Adjoint; for an indefinite form $U\cap\Gamma=\{N=1\}$ is the double cover of the identity component $SO^{+}(V,q)$.

Proof. For $\sigma=\mathrm{id}$ the dagger is $x^{\natural}$, so $U=\{x:x^{\natural}x=1\}$; on $\Gamma$ one has $x^{\natural}=N(x)x^{-1}$, hence $x^{\natural}x=N(x)$ and $U\cap\Gamma=\{x\in\Gamma:N(x)=1\}$, while $\Gamma_{\dagger}=\{x\in\Gamma:N(x)^{2}=1\}=\{x\in\Gamma:N(x)=\pm1\}$, which is the pin group of the inverse article. For a negative definite form the factors $-q(v_i)$ are positive, so $N$ is positive on the versors and on $\Gamma$, and $N(x)=\pm1$ is $N(x)=1$.

Remark (the equality is the point). The proposition says that in the trivial-involution case the three conditions — "the dagger is the inverse", "the Hermitian sandwich is an isometry", "the norm is $\pm1$" — cut $\Gamma$ down to the same group, the pin group being the isometry locus and the slice its norm-one subgroup. The Hermitian sandwich therefore needs no new group: it reconstructs $\mathrm{Pin}$ from the adjunction rather than from the inversion, and the identity $\Theta^{\alpha}_x=\mathrm{Ad}^{\alpha}_x$ on the slice is what makes the two constructions agree.

Remark (for a general involution the three differ). The slice $U$ and the Hermitian Clifford group $\Gamma_{\dagger}$ are different subsets of the unit group in general. The slice condition is $x^{\dagger}x=1$ and it does not involve the quadratic form; the isometry condition is $\sigma(N(x))^{2}=1$ and it does. An element can be a unit of norm $\pm1$ without being on the slice, and an element of the slice can have $\sigma(N(x))^{2}\neq1$; in the trivial-involution case the two reduce to $U\cap\Gamma=\{N=1\}$ and $\Gamma_{\dagger}=\{N=\pm1\}=\mathrm{Pin}$, which coincide for a definite form and differ by an index-two subgroup otherwise. The scalar widening by $\sigma$ is the only trace of the coefficient involution in the group theory.

The Reflections and the Covering

Proposition (the reflections are the odd elements). Let $u\in V$ with $q(u)\neq0$. Then $u$ is odd, $u^{\dagger}=-\sigma(u)$, and

$$ \Theta^{\alpha}_u(v)=u\,v\,\sigma(u)=-q(u)\,\rho_u(v),\qquad v\in V . $$

On a vector of square $-1$ the signed Hermitian sandwich is the reflection, $\Theta^{\alpha}_u=\rho_u$, and $\Theta^{\alpha}_u\in\Gamma_{\bar{\cdot}}$; the unsigned Hermitian sandwich is $-\rho_u$ there. So the odd elements of the Hermitian Clifford group realise the reflections, and the even ones realise the rotations.

Proof. The vector identity and the reflection statement are the propositions of the companion article; $\Theta^{\alpha}_u$ is an isometry for a unit vector by the determinant-and-scalar computation: $-q(u)$ has square $1$ when $q(u)=\pm1$. That an even element gives a rotation follows from $\Theta^{\alpha}_x=\Theta_x=\mathrm{Ad}_x$ on the slice and the parity of the determinant.

Theorem (the double cover, in the Hermitian formulation). Suppose that $-1$ is not a square in $F$ and that every element of $F^{\times}$ differs from $\pm1$ by a square; both hold over $\mathbb{R}$. Let the involution be trivial and the form definite in the corpus's convention, $q$ negative definite, so that every element of $\mathrm{Pin}(V,q)$ lies on the slice and $\Gamma_{\dagger}=U\cap\Gamma=\mathrm{Pin}(V,q)$. Then

$$ 1\longrightarrow\{\pm1\}\longrightarrow \mathrm{Pin}(V,q)\xrightarrow{\ \Theta^{\alpha}\ } O(V,q)\longrightarrow 1, \qquad 1\longrightarrow\{\pm1\}\longrightarrow \mathrm{Spin}(V,q)\xrightarrow{\ \Theta^{\alpha}\ } SO(V,q)\longrightarrow 1 $$

are exact, with the same two-to-one identification $x\sim-x$ as in the inverse formulation.

Proof. On $U$ the operator is the signed inner conjugation, $\Theta^{\alpha}_x=\mathrm{Ad}^{\alpha}_x$; the two exact sequences are therefore those of The Clifford, Pin and Spin Groups with Signed Inner Conjugation, transported along the identity $U=\Gamma_{\dagger}$ of the previous section. The kernel is $\{\pm1\}$ by the slice corollary of the companion article.

Remark (the general involution). For a nontrivial $\sigma$ the sequence must be read with the scalars of the base: the kernel of $\Theta^{\alpha}$ on $\Gamma_{\dagger}$ is $F^{\times}\cap\Gamma_{\dagger}=\{\lambda\in F^{\times}:\sigma(\lambda^{2})^{2}=1\}$, which contains the scalars of fourth root of unity and, over a field with $-1$ not a square, still gives the two elements $\{\pm1\}$ when $\sigma$ fixes $F$. The image is $O(V,q)$ whenever the reflections can be rescaled to unit vectors, exactly as in the inverse case. So the covering statement is insensitive to $\sigma$ up to the scalars, and no new group-theoretic phenomenon appears.

Cartan–Dieudonné and the Reflection Length

Theorem (Cartan–Dieudonné). Let $(V,q)$ be a non-degenerate quadratic space of dimension $n$ over a field of characteristic not two. Every isometry of $V$ is a product of at most $n$ reflections, and it lies in $SO(V,q)$ exactly when it is a product of an even number of them.

Proof. Quoted from The Clifford, Pin and Spin Groups with Signed Inner Conjugation; the theorem is independent of which sandwich realises the reflections.

Corollary (the Hermitian reading). With $\sigma=\mathrm{id}$, every element of $O(V,q)$ is a product of signed Hermitian sandwiches of unit vectors,

$$ \gamma=\Theta^{\alpha}_{u_1}\circ\cdots\circ\Theta^{\alpha}_{u_k},\qquad q(u_i)=-1, $$

with $k$ even exactly when $\gamma\in SO(V,q)$; the word length $k$ is the reflection length of $\gamma$ and the algebraic length of the corresponding element of $\mathrm{Pin}$ in the generating set of unit vectors. The reflections are the Hermitian sandwiches of the odd elements, and the parity of the number of factors is the determinant.

Proof. The unit vectors are in $\mathrm{Pin}$ and $\Theta^{\alpha}_u=\rho_u$; Cartan–Dieudonné writes the isometry as a product of reflections; the parity and determinant statements are the parity proposition of the companion article.

Low-Dimensional Cases

Over $\mathbb{R}$ with the definite negative form, $e_j^{2}=-1$ and $\sigma=\mathrm{id}$, the constructions read as follows.

$n=1$: $\mathrm{Cl}_{0,1}\cong\mathbb{C}$, $\mathrm{Pin}=\{\pm1,\pm e_1\}$, and $\Theta^{\alpha}_{e_1}=\rho_{e_1}$ is the single nontrivial reflection; $\mathrm{Spin}=\{\pm1\}$.
$n=2$: $\mathrm{Cl}_{0,2}\cong\mathbb{H}$, $\mathrm{Pin}$ is the group of $24$ Hurwitz units up to sign, equal to $\{\pm1,\pm e_1,\pm e_2,\pm e_3\}$ with $e_3=e_1e_2$; the even part $\mathrm{Spin}=\{\pm1,\pm e_3\}$ is the circle group of the plane, and $\Theta^{\alpha}_{e_3}$ is the rotation through $\pi$.
$n=3$: $\mathrm{Cl}_{0,3}\cong\mathbb{H}\oplus\mathbb{H}$, $\mathrm{Pin}$ is the binary octahedral group of order $48$, $\mathrm{Spin}$ is the binary tetrahedral group of order $24$; the odd elements give the reflections and the even ones the rotations of $\mathbb{R}^{3}$, and the unit vectors $e_1,e_2,e_3$ give $\rho_{e_1},\rho_{e_2},\rho_{e_3}$.

In each case $\mathrm{Pin}$ is compact because the form is definite, the norm $N(x)=x x^{\natural}$ is a nonzero real, and the Hermitian sandwich of a unit vector is the reflection with no rescaling. The dictionary with the biquaternion algebra, where the Hermitian sandwich is the corpus sandwich $\tilde Q\,x\,\bar{\tilde{Q}}$ and the slice is the Lorentz group, is Biquaternion Versors and the Orthogonal Group in the physics corpus.

The Indefinite and Degenerate Cases

Remark (isotropic vectors). If $q$ is indefinite, a vector with $q(u)=0$ is not invertible and not on the slice; neither Hermitian sandwich is defined for it as an isometry, and the reflection $\rho_u$ does not exist. The construction is confined to the non-isotropic vectors in both formulations, and the Hermitian one has no advantage there.

Remark (the norm can be isotropic). On $\mathrm{Cl}_{1,1}\cong M_2(\mathbb{R})$ the norm is the determinant of signature $(2,2)$ and vanishes on the nonzero rank-one matrices. Those elements are zero divisors and not in $\Gamma$, so $\Gamma_{\dagger}\subseteq\Gamma$ excludes them; but an element of the slice can lie outside $\Gamma$, and the two conditions are genuinely independent off the trivial-involution case, as the next remark records.

Remark (the slice is not inside the Clifford group). In $\mathrm{Cl}_{1,1}\cong M_2(\mathbb{R})$ the element $1-\tfrac12e_2-\tfrac12e_1e_2$ has norm $1$ and is a unit, and its signed inner conjugation does not preserve $V$, by The Clifford, Pin and Spin Groups with Signed Inner Conjugation. The same element shows that a unit can satisfy a norm condition and fail the Clifford-group condition, which is why $\Gamma_{\dagger}$ is defined inside $\Gamma$ and not by a norm condition alone. In the trivial-involution case this cannot happen, because there the norm condition and the slice condition coincide with membership in $\Gamma$.

Remark (the degenerate case). If $q$ is degenerate the Clifford group need not act on $V$ by invertible maps, the map to $O(V,q)$ need not be onto, and the covering statement fails in both formulations; the Hermitian sandwich does not repair it. The degenerate theory belongs to Degenerate Clifford Algebras and the Radical.

Summary

The Hermitian formulation of the covering groups replaces the inverse by the dagger and therefore compares three conditions. The Clifford group $\Gamma$ is where the Hermitian sandwich preserves the vector space, and the Hermitian Clifford group

$$ \Gamma_{\dagger}=\{\,x\in\Gamma:\sigma(N(x))^{2}=1\,\} $$

is where it is an isometry; it is a subgroup of $\Gamma$ because $N$ is multiplicative and $\sigma$ is a homomorphism. The unitary slice $U=\{x:x^{\dagger}x=1\}$ is a group, and it is where the dagger is the inverse, so that on $U$ the Hermitian sandwich is the signed inner conjugation and the entire inverse theory transfers verbatim. With the trivial involution the two conditions reduce to

$$ \sigma=\mathrm{id}\ \Longrightarrow\ U\cap\Gamma=\{\,x\in\Gamma:N(x)=1\,\},\quad \Gamma_{\dagger}=\{\,x\in\Gamma:N(x)=\pm1\,\}=\mathrm{Pin}(V,q), $$

and the two coincide for a definite form in the corpus's convention, so that the Hermitian sandwich reconstructs the pin group from the adjunction and needs no new group: the odd elements give the reflections through $\Theta^{\alpha}_u(v)=u\,v\,\sigma(u)=-q(u)\rho_u(v)$, which for a unit vector is $\rho_u$ exactly, and the even ones give the rotations; the map to $O(V,q)$ is two-to-one, with kernel $\{\pm1\}$ on the slice, and the sequences $1\to\{\pm1\}\to\mathrm{Pin}\to O\to1$ and $1\to\{\pm1\}\to\mathrm{Spin}\to SO\to1$ are exact over $\mathbb{R}$ and a definite form. Cartan–Dieudonné then writes every isometry as a product of signed Hermitian sandwiches of unit vectors, with the parity of the number of factors equal to the determinant. For a general involution the three conditions differ, the scalars widen, and the covering statement is unchanged up to those scalars; the indefinite and degenerate cases fail in the same way as for the inverse member, and the Hermitian form brings no repair.

Summary of Notation

Symbol	Meaning
$\Gamma(V,q)$	Clifford group, $\Theta^{\alpha}_x(V)\subseteq V$
$\Gamma_{\dagger}=\{x\in\Gamma:\sigma(N(x))^{2}=1\}$	Hermitian Clifford group, the isometry locus
$U=\{x:x^{\dagger}x=1\}$	Unitary slice, a group, dagger = inverse
$\Theta^{\alpha}_x=\mathrm{Ad}^{\alpha}_x$ on $U$	The two theories agree on the slice
$\sigma=\mathrm{id}\Rightarrow\Gamma_{\dagger}=\mathrm{Pin}$, $U\cap\Gamma=\{N=1\}$	Hermitian reconstruction of the pin group; slice is its norm-one subgroup
$\Theta^{\alpha}_u=\rho_u$ for $q(u)=-1$	Odd elements give reflections
$1\to\{\pm1\}\to\mathrm{Pin}\xrightarrow{\Theta^{\alpha}}O\to1$	Double cover, $\sigma=\mathrm{id}$, definite form
$1\to\{\pm1\}\to\mathrm{Spin}\xrightarrow{\Theta^{\alpha}}SO\to1$	Spin double cover, $\sigma=\mathrm{id}$, definite form