Versors, Rotors and the Sandwich Action with Signed Hermitian Adjoint

Introduction

A versor is a product of invertible vectors, and a rotor is an even versor of unit norm. The geometric content of the Clifford algebra is that a versor acts on the quadratic space by an isometry, through the sandwich, and that a rotor acts by a rotation. In the inverse formulation the sandwich is the signed inner conjugation $v\mapsto\alpha(x)vx^{-1}$, and the theory is Versors, Rotors and the Sandwich Action with Signed Inner Conjugation.

This article is the Hermitian reading of the same geometry. The sandwich is now

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

and its form on a versor is computed once and for all by the identity

$$ \Theta^{\alpha}_x=\varepsilon_x\,\sigma\bigl(N(x)\bigr)\;\sigma\text{-conjugation by }x, \qquad \Theta^{\alpha}_x(v)=\varepsilon_x\,\sigma\bigl(N(x)\bigr)\,x\,v\,\sigma(x)^{-1}, \qquad x\in\Gamma(V,q), $$

with $\varepsilon_x=(-1)^{|x|}$ the parity sign and $\sigma$ the involution of the base. Three features are Hermitian and have no counterpart in the inverse formulation. The first is that the dagger of a versor is again a versor, of the same length and parity, so the Hermitian sandwich is a sandwich in a versor on both sides. The second is the extra factor $\sigma(x)^{-1}$: the right factor of the operator is the dagger, which on the Clifford group is $\sigma(N(x))\sigma(x)^{-1}$, and the coefficient involution therefore appears in the geometry. The third is that the two sides of a rotor sandwich agree only when the rotor is $\sigma$-real and of norm one, and otherwise the operator is a similarity: the unitary slice is exactly the condition that makes the Hermitian sandwich a rotation.

The theory of versors, of the Lipschitz group, of the rotor group and of the double cover is Versors, Rotors and the Sandwich Action with Signed Inner Conjugation; the dagger, the slice 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 Clifford Algebra with Signed Hermitian Adjoint; the Hermitian module and the adjoint of the Clifford action are The Adjoint of the One-Sided Action with Hermitian Adjoint; the biquaternion dictionary is Biquaternion Versors and the Orthogonal Group in the physics corpus. Nothing owned by those entries is reproved.

Conventions. The coefficient involution $\sigma$ is assumed compatible with the form, $\sigma(q(v))=q(\sigma(v))$, so that it preserves the subspace $V$ and the quadratic form, as in the Hermitian theory of Involutive Clifford Algebras. $N(x)=x x^{\natural}$ is the Clifford norm, $\Gamma(V,q)$ the Clifford group, $U$ the unitary slice, $\varepsilon_x=(-1)^{|x|}$, and $x^{\dagger}=\sigma(\alpha(x^{r}))$.

Versors and the Hermitian Adjoint

Definition. A versor is an element of $\mathrm{Cl}(V,q)$ of the form

$$ x=v_1v_2\cdots v_k,\qquad q(v_i)\neq0\ \text{ for every }i, $$

a product of non-isotropic, hence invertible, vectors. Its parity is the parity of $k$.

Proposition (the dagger of a versor is a versor). For a vector $v$ one has $v^{\dagger}=-\sigma(v)$, a non-isotropic vector, and for a versor

$$ (v_1v_2\cdots v_k)^{\dagger}=(-1)^{k}\,\sigma(v_k)\cdots\sigma(v_2)\sigma(v_1), $$

a versor of the same length and parity. Hence the set of versors is stable under the dagger up to the scalar $(-1)^{k}$ and up to $\sigma$, and the Hermitian sandwich $\Theta^{\alpha}_x$ of a versor has versors on both sides.

Proof. For $v\in V$ one has $v^{r}=v$ and $\alpha(v)=-v$, so $v^{\dagger}=\sigma(\alpha(v))=-\sigma(v)$; by compatibility of $\sigma$ with $q$, $\sigma(v)$ is a non-isotropic vector, and by compatibility with the Clifford relations $\sigma$ is a ring automorphism of the algebra, so the product formula holds. The scalar $(-1)^{k}$ is a versor of length two, as recorded in the inverse article.

Proposition (the form of the sandwich on a versor). Every versor lies in the Clifford group $\Gamma(V,q)$, and for a versor $x$

$$ \Theta^{\alpha}_x=\varepsilon_x\,\sigma\bigl(N(x)\bigr)\,x\,(\ )\,\sigma(x)^{-1}, \qquad \Theta^{\alpha}_x(v)=\varepsilon_x\,\sigma\bigl(N(x)\bigr)\,x\,v\,\sigma(x)^{-1},\quad v\in V . $$

On the unitary slice this is the signed inner conjugation, $\Theta^{\alpha}_x=\varepsilon_x x\,(\ )\,x^{-1}=\mathrm{Ad}^{\alpha}_x$; on the $\sigma$-real part of the slice with $\sigma(N(x))=1$ it is the plain sandwich $x\,(\ )\,x^{-1}$.

Proof. The Lipschitz group, generated by the non-isotropic vectors, is $\Gamma(V,q)$, so a versor lies in $\Gamma$ and its norm is a scalar. On $\Gamma$ one has $x^{\dagger}=\sigma(N(x))\sigma(x)^{-1}$, by the Clifford-group identity of The Hermitian Sandwich on a Clifford Algebra with Hermitian Adjoint; substituting into $\Theta^{\alpha}_x(v)=\alpha(x)vx^{\dagger}$ and using $\alpha(x)=\varepsilon_x x$ gives the displayed form. On the slice $x^{\dagger}=x^{-1}$, so $\Theta^{\alpha}_x=\alpha(x)(\ )x^{-1}=\mathrm{Ad}^{\alpha}_x$; if in addition $\sigma(x)=x$ and $\sigma(N(x))=1$ the scalar is $1$ and the sandwich is the plain one.

Corollary (the value at the unit and the norm). For a versor $x$,

$$ \Theta^{\alpha}_x(1)=\alpha(x)\,x^{\dagger}=\varepsilon_x\,\sigma\bigl(N(x)\bigr)\,x\,\sigma(x)^{-1}, $$

which is the identity on $1$ exactly for an even, $\sigma$-real versor of norm with $\sigma(N(x))=1$.

Proof. Immediate from the proposition at $v=1$.

Remark (why the slice is the geometric locus). The three corrections to the plain sandwich are the parity sign, the norm factor and the coefficient involution. All three are scalars or are trivial on the $\sigma$-real slice: the parity sign is $\pm1$, the norm factor is $\sigma(N(x))$, and $\sigma(x)^{-1}$ is $x^{-1}$ for $\sigma$-real $x$. So the Hermitian sandwich is the inverse sandwich up to scalars off the slice and exactly the inverse sandwich on it, and the geometry — rotors rotating, reflections reflecting — is the geometry of the slice.

Rotors

Definition. A rotor is an even versor $R$ with $\sigma(N(R))=1$ and $\sigma(R)=R$. For such an $R$ the Hermitian sandwich acts as the plain sandwich and produces a rotation,

$$ \Theta^{\alpha}_R(v)=\varepsilon_R\,\sigma\bigl(N(R)\bigr)\,R\,v\,\sigma(R)^{-1}=R\,v\,R^{-1},\qquad \varepsilon_R=+1 . $$

The rotor group is the group of all even versors of norm with $\sigma(N(R))=1$; over the reals with $\sigma=\mathrm{id}$ it is the even part of $\mathrm{Pin}$, that is $\mathrm{Spin}(V,q)$.

Proposition. The rotors form a subgroup of the unitary slice and of the Hermitian Clifford group, and the map $R\mapsto\Theta^{\alpha}_R$ is a homomorphism into $SO(V,q)$ with kernel $\{\pm1\}$ under the hypotheses on the field of The Clifford, Pin and Spin Groups with Signed Hermitian Adjoint.

Proof. A rotor is even, so $\Theta^{\alpha}_R=\Theta_R$; on a norm-one versor the dagger is inverse on the slice, so $R\in U$; and $\sigma(N(R))=1$ is the isometry condition, so $R\in\Gamma_{\dagger}$. The group statements are those of the inverse article, transported along the slice identity.

Example (the rotational rotor). Let $B$ be a bivector with $B^{2}=-1$ in the definite algebra $\mathrm{Cl}_{0,m}(\mathbb{R})$, $\sigma=\mathrm{id}$, and put

$$ R=\exp\Bigl(-\tfrac{\theta}{2}B\Bigr)=\cos\tfrac{\theta}{2}-B\sin\tfrac{\theta}{2}. $$

Then $R$ is even and $\bar R=\cos\tfrac{\theta}{2}+B\sin\tfrac{\theta}{2}=R^{-1}$, so $N(R)=1$, $R$ is real and the Hermitian sandwich is the plain sandwich $R\,v\,R^{-1}$, the rotation through $\theta$. The dagger is the inverse here, $R^{\dagger}=R^{-1}$, and the same rotation is obtained from $\Theta_R$ and from $\Theta^{\alpha}_R$, which agree because $R$ is even.

Example (the hyperbolic rotor). Let $B$ be a bivector with $B^{2}=+1$, as for a boost in $\mathrm{Cl}_{1,3}$, and put $R=\exp\bigl(-\tfrac{\eta}{2}B\bigr)=\cosh\tfrac{\eta}{2}-B\sinh\tfrac{\eta}{2}$, with $\eta$ the rapidity. Then $\bar R=\cosh\tfrac{\eta}{2}+B\sinh\tfrac{\eta}{2}=R^{-1}$, $N(R)=1$, and the Hermitian sandwich is the Lorentz boost $R\,v\,R^{-1}$. The example is the biquaternion one in the physics corpus, where the sandwich is written $\tilde Q\,x\,\tilde{Q}^{*}$; the Hermitian sandwich of this article is exactly that object, and the signed member is its parity-signed version.

The Two Actions

The sandwich on the vectors. For a versor $x$ the Hermitian sandwich maps $V$ to $V$ and is given by the formula of the second proposition; it is an isometry exactly when $\sigma(N(x))^{2}=1$ and otherwise a similarity of ratio $\sigma(N(x))^{2}$, and its value on the vectors is a scalar multiple of the sandwich of the inverse member. For a rotor it is the rotation, and the parameter inside the rotor is the half angle while the rotation is through the full angle.

The left action on spinors. Let $S$ be a Hermitian Clifford module over $\mathrm{Cl}(V,q)$, with the Hermitian form for which the Clifford action is self-adjoint, $(x\cdot s,t)=(s,x^{\dagger}\cdot t)$, of Hermitian Clifford Modules with Hermitian Adjoint. The spinor action of the rotors is left multiplication $\psi\mapsto R\psi$, the restriction to the rotors of the left action of the even subalgebra; the signed member of the family acts on the module as the scalar $\varepsilon_R$ times that action.

Proposition (the two actions and the double cover). Left multiplication by a rotor is a representation of the rotor group on $S$ with kernel $\{\pm1\}$ on an irreducible module, while the sandwich acts on $V$ with the same kernel; a full turn, $\theta=2\pi$, is the identity on $V$ and $-\mathrm{id}_S$ on the module. The two actions are the two halves of the double cover of $SO(V,q)$, in the Hermitian formulation.

Proof. The kernel statement and the full-turn statement are those of the inverse article and depend only on the group, not on the right factor; they are transported along the slice identity $\Theta^{\alpha}_x=\mathrm{Ad}^{\alpha}_x$, and the identification of the module action with the spin representation is Spin Representations and Clifford Modules with Inner Conjugation.

Remark (the Hermitian specificity in the module statement). The dagger enters the module side not in the rotation but in the adjointness: the form on $S$ is chosen so that the Clifford action satisfies $(x\cdot s,t)=(s,x^{\dagger}\cdot t)$, and it is this axiom that makes the Hermitian sandwich the adjoint of its parameter and the left action self-adjoint for the elements of the slice. The unsigned Hermitian sandwich is the adjoint of the Clifford action; the signed one is the same up to the parity sign. The analytic consequences are Dirac Operators with Hermitian Adjoint and are not part of this article.

Off the Slice: the Similarity

Proposition. Let $x\in\Gamma$ be a versor with $\sigma(N(x))^{2}\neq1$. Then $\Theta^{\alpha}_x$ preserves $V$, acts on it by a similarity of ratio $\sigma(N(x))^{2}$, and is not an isometry; the signed and the unsigned members differ only by the sign $\varepsilon_x$.

Proof. The action proposition of the companion article gives $\Theta^{\alpha}_x|_V=\varepsilon_x\sigma(N(x))\chi(x)$ with $\chi(x)\in O(V,q)$, and a scalar multiple of an orthogonal map by $\lambda$ scales $q$ by $\lambda^{2}$.

Remark (the split and biquaternion cases). In $\mathrm{Cl}_{1,1}$ the even element $x=2+3e_1e_2$ has $N(x)=-5$ and the Hermitian sandwich is a similarity of ratio $25$, as computed in The Hermitian Sandwich on a Clifford Algebra with Hermitian Adjoint; the signed member is the negative of it. In the biquaternion algebra the corpus sandwich $\tilde Q\,x\,\tilde{Q}^{*}$ is an isometry exactly on the slice, which is the Lorentz group there, and a similarity off it; that is the geometric form of the statement that the Hermitian sandwich needs the unitary condition to produce a transformation of the physical slices. The dictionary is Biquaternion Versors and the Orthogonal Group, and the slice itself is The Unitary Slice and the Compact Real Form with Hermitian Adjoint.

Worked Cases

The Plane Rotor

In $\mathrm{Cl}_{0,2}(\mathbb{R})$, $e_1^{2}=e_2^{2}=-1$, $\sigma=\mathrm{id}$, let $B=e_1e_2$, $B^{2}=-1$, and $R=\cos\tfrac{\theta}{2}-B\sin\tfrac{\theta}{2}$. Then $R^{\dagger}=R^{-1}$ and

$$ \Theta^{\alpha}_R(e_1)=R\,e_1\,R^{-1}=\cos\theta\,e_1+\sin\theta\,e_2, \qquad \Theta^{\alpha}_R(e_2)=-\sin\theta\,e_1+\cos\theta\,e_2 , $$

the rotation through $\theta$. The operator is even and its determinant is $+1$; the unsigned member gives the same, since $R$ is even.

The Reflection Read Unevenly

In $\mathrm{Cl}_{0,3}(\mathbb{R})$ let $u=e_1$, an odd versor of length one. Then $\Theta^{\alpha}_u(v)=e_1ve_1$ and $\Theta^{\alpha}_u=\rho_{e_1}$, while $\Theta_u=-\rho_{e_1}$: the odd versor gives the reflection for the signed member and the negative for the unsigned one. The dagger of the versor is $e_1^{\dagger}=-e_1$, again a versor of length one and the same parity.

A Similarity in the Split Algebra

In $\mathrm{Cl}_{1,1}(\mathbb{R})$, $e_1^{2}=1$, $e_2^{2}=-1$, let $x=e_1e_2$, an even versor with $N(x)=-1$ and $\sigma(N(x))^{2}=1$. Then $\Theta^{\alpha}_x=\Theta_x=\mathrm{Ad}_x$ is an isometry, not a similarity, even though $N(x)=-1$: the isometry condition is on the square of the norm, as in the companion article, and the sign of the norm is not the obstruction. For $x=2+3e_1e_2$ the norm is $-5$, its square is $25$, and the operator is the similarity of ratio $25$.

Summary

A versor is a product of invertible vectors, and the Hermitian sandwich of a versor has versors on both sides, because the dagger of a versor is again a versor of the same length and parity. On the Clifford group the operator has the explicit form

$$ \Theta^{\alpha}_x=\varepsilon_x\,\sigma\bigl(N(x)\bigr)\,x\,(\ )\,\sigma(x)^{-1}, \qquad x\in\Gamma(V,q), $$

with the parity sign $\varepsilon_x$, the norm factor $\sigma(N(x))$ and the coefficient involution $\sigma(x)^{-1}$ as the three corrections to the plain sandwich. A rotor is an even, $\sigma$-real versor of norm with $\sigma(N(R))=1$, and for it the sandwich is the plain one, $R\,v\,R^{-1}$, a rotation; the rotor group is the even part of the Hermitian Clifford group, it lies in the unitary slice, and it double-covers $SO(V,q)$ with kernel $\{\pm1\}$. The two actions are the sandwich on the vectors, where a full turn is the identity, and the left multiplication on a Hermitian Clifford module, where a full turn is $-\mathrm{id}$; the dagger enters the module side through the adjointness axiom $(x\cdot s,t)=(s,x^{\dagger}\cdot t)$, which is what makes the Hermitian sandwich the adjoint of its parameter. Off the slice the operator is a similarity of ratio $\sigma(N(x))^{2}$ and the signed member is the negative of the unsigned one; the geometry of rotations, reflections and boosts is the geometry of the slice, where the two formulations coincide.

Summary of Notation

Symbol	Meaning
$x=v_1\cdots v_k$, $q(v_i)\neq0$	Versor, $k$ its length, $k\bmod2$ its parity
$v^{\dagger}=-\sigma(v)$	The dagger of a vector, again a non-isotropic vector
$\Theta^{\alpha}_x=\varepsilon_x\sigma(N(x))\,x(\ )\sigma(x)^{-1}$	The sandwich on a versor, $x\in\Gamma$
$\Theta^{\alpha}_x=\mathrm{Ad}^{\alpha}_x$ on $U$	Slice reduction
Rotor: even, $\sigma$-real, $\sigma(N(R))=1$	Rotor; $\Theta^{\alpha}_R=R\,v\,R^{-1}$
$R=\exp(-\tfrac\theta2B)$	Rotational rotor ($B^{2}=-1$) and hyperbolic rotor ($B^{2}=+1$)
$(x\cdot s,t)=(s,x^{\dagger}\cdot t)$	Hermitian Clifford module, self-adjointness
$\Theta^{\alpha}_x\|_V=\varepsilon_x\sigma(N(x))\chi(x)$, $\chi\in O$	Similarity off the slice