One-Sided Operators with the Signed Product

Introduction

The factors of the signed two-sided operator $a\alpha(y)b$ are one-sided, and they are of two different kinds. On the left the sign sits on the argument: the signed left multiplication is $y\mapsto a\,\alpha(y)$, the ordinary left multiplication composed with the grade involution. On the right the sign sits on the parameter: the signed right multiplication is $y\mapsto y\,\alpha(b)$, the ordinary right multiplication by the twisted parameter. The asymmetry comes from the same place as the anti-multiplicativity of the right family — an automorphism pushes forward through a product, an anti-automorphism pulls back — and it makes the two signed families behave differently under composition.

This article fixes the signed one-sided operators, their composition laws, their fixed elements, and the way a pairing of a signed factor with an ordinary one produces the reflections. The composition laws already show the asymmetry: two signed left multiplications compose to an ordinary left multiplication, the two grade involutions cancelling, while two signed right multiplications compose to a signed right multiplication, because there the parameter is twisted twice and the twist is an involution. The fixed elements are read off immediately: the signed left multiplication by a scalar is the identity on the even part, minus the identity on the odd part, or zero, according as the scalar is $1$, $-1$ or anything else. And the reflections arise exactly as in the ordinary theory, by pairing: the signed left multiplication by a vector with the ordinary right multiplication by the inverse of its twisted parameter is the reflection.

The one-sided operators and their composition are One-Sided Operators on a Clifford Algebra; the operators twisted by the grading are The Graded Multiplication Operators; the two-sided signed family is Two-Sided Operators with the Signed Product; the signed sandwich and the reflections are The Sandwich with the Signed Product; the one-sided action on a module is The One-Sided Action and the Spin Representation. Those are cited. The base is a field $F$ of characteristic not $2$, $q$ is non-degenerate and $\alpha$ is the grade involution.

The Signed One-Sided Family

Definition. For $a \in \mathrm{Cl}(V,q)$ the signed left multiplication and the signed right multiplication are

$$ \mathrm{L}^{\alpha}_a(y) = a\,\alpha(y) = L_a\bigl(\alpha(y)\bigr), \qquad \mathrm{P}^{\alpha}_a(y) = y\,\alpha(a) = R_{\alpha(a)}(y). $$

So $\mathrm{L}^{\alpha}_a = L_a\circ\alpha$ is the ordinary left multiplication with the argument twisted, and $\mathrm{P}^{\alpha}_a = R_{\alpha(a)}$ is the ordinary right multiplication with the parameter twisted. The two are different kinds of twist, and this is the source of the asymmetry in the composition laws.

Convention. The symbol $\Lambda^{\alpha}$ of One-Sided Operators on a Clifford Algebra and One-Sided Operators on a Clifford Algebra with Signed Inner Conjugation puts the grade involution on the parameter, $\Lambda^{\alpha}_x = L_{\alpha(x)} = \varepsilon_xL_x$, and it is that operator which is the left factor of the signed inner conjugation. The present family is the one-sided form of the signed product $T^{\alpha}_{a,b}(y) = a\alpha(y)b$, so its left factor twists the argument instead, and it is written $\mathrm{L}^{\alpha}$ to keep the two apart; the right factor agrees with the other convention, $\mathrm{P}^{\alpha}_a = R_{\alpha(a)}$.

Proposition (elementary properties). Both families are $F$-linear and bijective for $a$ a unit; $\mathrm{L}^{\alpha}_1 = \alpha = \mathrm{P}^{\alpha}_1$; $\mathrm{L}^{\alpha}_a$ coincides with $L_a$ on the even part and with $-L_a$ on the odd part; and for homogeneous $a$ one has $\mathrm{P}^{\alpha}_a = (-1)^{|a|}R_a$.

Proof. $\mathrm{L}^{\alpha}_a$ and $\mathrm{P}^{\alpha}_a$ are composites of $F$-linear bijections when $a$ is a unit; at $a = 1$ both reduce to $\alpha(1) = 1$ applied to the argument, that is to $\alpha$. The sign statements are $\alpha(a) = (-1)^{|a|}a$ for the right family and $\alpha(y) = (-1)^{i}y$ on $\mathrm{Cl}^{i}$ for the left: $\mathrm{L}^{\alpha}_a$ is $L_a\circ\alpha$, so it equals $(-1)^{i}L_a$ on $\mathrm{Cl}^{i}$ and hence $L_a$ on $\mathrm{Cl}^{0}$, $-L_a$ on $\mathrm{Cl}^{1}$; it is not a scalar multiple of $L_a$ unless $L_a$ is itself even, the argument twisting a different factor from the parameter.

Proposition (the parity). $\mathrm{L}^{\alpha}_a$ and $\mathrm{P}^{\alpha}_a$ have the parity of $a$: they carry $\mathrm{Cl}^{i}$ to $\mathrm{Cl}^{i+|a|}$.

Proof. $\mathrm{L}^{\alpha}_a = L_a\circ\alpha$ and $\alpha$ preserves the degree modulo two while $L_a$ shifts it by $|a|$; $\mathrm{P}^{\alpha}_a = R_{\alpha(a)}$ and $\alpha(a)$ has the parity of $a$, so the right multiplication by it shifts the degree by $|a|$.

The Composition Laws

Proposition (the left family). For all $a, c$

$$ \mathrm{L}^{\alpha}_a \circ \mathrm{L}^{\alpha}_c = L_{a\alpha(c)}, \qquad \mathrm{L}^{\alpha}_a \circ L_c = \mathrm{L}^{\alpha}_{a\alpha(c)}, \qquad L_a \circ \mathrm{L}^{\alpha}_c = \mathrm{L}^{\alpha}_{ac}. $$

So the composite of two signed left multiplications is an ordinary left multiplication, the two grade involutions cancelling; the composite of a signed and an ordinary left multiplication, in either order, is signed.

Proof. $\mathrm{L}^{\alpha}_aL_c(y) = a\alpha(cy) = a\alpha(c)\alpha(y) = \mathrm{L}^{\alpha}_{a\alpha(c)}(y)$; $L_a\mathrm{L}^{\alpha}_c(y) = ac\alpha(y) = \mathrm{L}^{\alpha}_{ac}(y)$; and composing two of the first kind gives $L_{a\alpha(c)}\alpha\circ\alpha = L_{a\alpha(c)}$.

Proposition (the right family carries no new operators). Since $\alpha$ is a bijection, $\{\mathrm{P}^{\alpha}_a : a \in \mathrm{Cl}(V,q)\} = \{R_b : b \in \mathrm{Cl}(V,q)\}$: the signed right multiplications are exactly the ordinary right multiplications, reparametrised by $b = \alpha(a)$. Consequently

$$ \mathrm{P}^{\alpha}_a \circ \mathrm{P}^{\alpha}_c = \mathrm{P}^{\alpha}_{ca}, \qquad \mathrm{P}^{\alpha}_a \circ R_c = \mathrm{P}^{\alpha}_{\alpha(c)\,a}, \qquad R_c \circ \mathrm{P}^{\alpha}_a = \mathrm{P}^{\alpha}_{a\,\alpha(c)}, $$

and the signed right family is a group, being the ordinary right family. The genuine asymmetry is that the signed left multiplications are new maps while the signed right multiplications are not.

Proof. $\mathrm{P}^{\alpha}_a = R_{\alpha(a)}$ by definition, and $\alpha$ is surjective, which gives the first sentence; the laws are the ordinary laws $R_bR_d = R_{db}$ under the substitution $b = \alpha(a)$, $d = \alpha(c)$. Explicitly, $\mathrm{P}^{\alpha}_aR_c(y) = yc\alpha(a) = y\alpha(\alpha(c))\alpha(a) = y\alpha(\alpha(c)a) = \mathrm{P}^{\alpha}_{\alpha(c)a}(y)$.

Corollary (the mirror of the coset structure). The signed left multiplications form, together with the ordinary ones, the coset $L(\Gamma)\alpha$ of the group $L(\Gamma)$ of invertible left multiplications; the signed right multiplications are the ordinary right multiplications themselves. The asymmetry is exactly the difference between an automorphism acting on the argument and the same automorphism acting on the parameter: on the left the twist changes the operator, and on the right it merely relabels it. The genuinely signed one-sided family is therefore the left one, and it is a coset and not a group.

Proof. The left statement is the first law, the right statement the first law of the right family, and the isomorphism is $\mathrm{P}^{\alpha}_a = R_{\alpha(a)}$.

Fixed Elements

Definition. The fixed space of an operator $T$ is $\mathrm{Fix}(T) = \{y : T(y) = y\}$.

Proposition. For $a \in \mathrm{Cl}(V,q)$,

$$ \mathrm{Fix}(\mathrm{L}^{\alpha}_a) = \{\, y : a\,\alpha(y) = y \,\}, \qquad \mathrm{Fix}(\mathrm{P}^{\alpha}_a) = \{\, y : y\,\alpha(a) = y \,\}. $$

In particular $\mathrm{Fix}(\mathrm{L}^{\alpha}_1) = \mathrm{Fix}(\alpha) = \mathrm{Cl}^{0}(V,q)$ is the even part, and for a scalar $\lambda$

$$ \mathrm{Fix}(L_{\lambda}\alpha) = \begin{cases} \mathrm{Cl}^{0}(V,q), & \lambda = 1, \\ \mathrm{Cl}^{1}(V,q), & \lambda = -1, \\ 0, & \lambda \neq \pm1. \end{cases} $$

Proof. Write $y = y_0 + y_1$. The equation $\lambda\alpha(y) = y$ is $\lambda y_0 = y_0$ and $-\lambda y_1 = y_1$; the two are satisfied independently, giving the three cases. The general formulae are the definitions.

Remark (fixed elements against the graded parts). The fixed space of the signed left multiplication by a scalar is a graded subspace, the even or the odd part; this is the operator form of the statement that the grade involution is the parity operator. For a general $a$ the fixed space is not graded, and it is a right or left translate of a graded subspace by the equation above.

The Reflections

Proposition (the pairing). For a unit $x$, the signed sandwich of The Sandwich with the Signed Product is the pairing of the signed left multiplication by $x$ with the ordinary right multiplication by the inverse of the twisted parameter:

$$ \Sigma_x = \mathrm{L}^{\alpha}_x \circ R_{\alpha(x)^{-1}} . $$

Proof. $\mathrm{L}^{\alpha}_x\bigl(R_{\alpha(x)^{-1}}(y)\bigr) = x\,\alpha\bigl(y\,\alpha(x)^{-1}\bigr) = x\,\alpha(y)\,\alpha\bigl(\alpha(x)^{-1}\bigr) = x\,\alpha(y)\,\alpha^{2}(x)^{-1} = x\,\alpha(y)\,x^{-1} = \Sigma_x(y)$, using that $\alpha$ is an automorphism, that $\alpha^{-1} = \alpha$, and that $\alpha^{2} = \mathrm{id}$. The inverse right multiplication of the other convention, $R_{x^{-1}}$, gives only $\varepsilon_x\Sigma_x$, because $\alpha(x)^{-1} = \varepsilon_x x^{-1}$; the two pairings agree exactly on the even part of the unit group, which is the reason the corpus keeps the two families apart.

Theorem (the reflections). Let $u \in V$ with $q(u) \neq 0$. Then $\Sigma_u = \mathrm{L}^{\alpha}_u\circ R_{\alpha(u)^{-1}}$ restricted to $V$ is the reflection $\rho_u$, and its fixed vectors in $V$ are the hyperplane $u^{\perp}$:

$$ \mathrm{Fix}\bigl(\Sigma_u\big|_V\bigr) = u^{\perp} = \{\, v \in V : B(u,v) = 0 \,\}. $$

Proof. The reflection statement is the theorem of The Sandwich with the Signed Product; for the fixed space, $\rho_u(v) = v$ exactly when $2B(v,u)q(u)^{-1}u = 0$, that is $B(v,u) = 0$.

Remark (why one side is not enough). A signed left multiplication by a vector never preserves $V$, for the same reason an ordinary left multiplication does not: $L_u\alpha(\mathrm{Cl}^{1})$ contains the scalars as well as the vectors, since $\alpha$ flips the parity. Only the pairing with an ordinary right multiplication brings the parity back and produces an operator of $V$. This is the one-sided form of the statement of One-Sided Operators on a Clifford Algebra that the scalar is the only one-sided operator preserving the quadratic space.

Worked Cases

A Scalar in $\mathrm{Cl}_{0,3}(\mathbb{R})$

With $e_j^{2} = -1$ let $\lambda = -1$. Then $L_{-1}\alpha$ is the identity on the odd part and minus the identity on the even part, so its fixed space is the four-dimensional odd part $\mathrm{span}(e_1,e_2,e_3,\omega)$, and $L_{-1}\alpha = -\alpha$ has square $-\mathrm{id}$; the example shows a signed one-sided operator whose fixed space is a full graded summand.

A Vector

In the same algebra let $a = e_1$. Write $y = y_0 + y_1$ with $y_0$ even and $y_1$ odd, so that $\alpha(y) = y_0 - y_1$ and the equation $e_1\alpha(y) = y$ reads $e_1y_0 - e_1y_1 = y_0 + y_1$. Comparing the even parts gives $y_1 = e_1y_0$, and the odd parts then give $-e_1y_1 = y_0$, that is $-e_1^{2}y_0 = y_0$, which holds identically. So $\mathrm{Fix}(\mathrm{L}^{\alpha}_{e_1}) = \{y_0 + e_1y_0 : y_0 \in \mathrm{Cl}^{0}(V,q)\}$, a four-dimensional space that is not graded, the graph of the left multiplication by $e_1$ in the even part.

The Reflection

With $u = e_1$ one has $\alpha(e_1)^{-1} = (-e_1)^{-1} = e_1$, and the signed sandwich $\Sigma_{e_1} = \mathrm{L}^{\alpha}_{e_1}\circ R_{e_1}$ acts on the basis of $V$ by $(-e_1, e_2, e_3)$ and fixes the hyperplane $e_1^{\perp} = \mathrm{span}(e_2, e_3)$, as a reflection should. The signed left factor alone does not preserve $V$; the pairing does.

Summary

The signed left multiplication $\mathrm{L}^{\alpha}_a = L_a\circ\alpha$ twists the argument and the signed right multiplication $\mathrm{P}^{\alpha}_a = R_{\alpha(a)}$ twists the parameter; both have the parity of $a$ and coincide with $\alpha$ at $a = 1$. The asymmetry between them is complete: $\mathrm{L}^{\alpha}_a = L_a\circ\alpha$ is a new operator for an odd $a$, while $\mathrm{P}^{\alpha}_a = R_{\alpha(a)}$ is the ordinary right multiplication relabelled, so the signed right family is just the ordinary right family and there are no new right operators. The composition laws record the rest: two signed left multiplications give an ordinary left multiplication, $L_{a\alpha(c)}$, because the two grade involutions cancel, so the signed left family is a coset of the ordinary left family and not a group, exactly the one-sided form of the torsor structure of Two-Sided Operators with the Signed Product. The fixed space of a signed operator is given by $a\alpha(y) = y$ on the left and $y\alpha(a) = y$ on the right; that of the scalar $\lambda$ is the even part, the odd part or nothing according as $\lambda = 1, -1$ or neither. Finally the reflections are pairings: $\Sigma_x = \mathrm{L}^{\alpha}_x R_{\alpha(x)^{-1}}$, a vector $u$ gives the reflection $\rho_u$, and its fixed vectors are the hyperplane $u^{\perp}$; a signed left multiplication alone never preserves $V$. The ordinary one-sided calculus is One-Sided Operators on a Clifford Algebra, the graded twisting is The Graded Multiplication Operators, and the two-sided signed family is Two-Sided Operators with the Signed Product.

Summary of Notation

Symbol	Meaning
$\mathrm{L}^{\alpha}_a = L_a\circ\alpha$	Signed left multiplication, argument twisted
$\mathrm{P}^{\alpha}_a = R_{\alpha(a)}$	Signed right multiplication, parameter twisted
$\mathrm{L}^{\alpha}_a\mathrm{L}^{\alpha}_c = L_{a\alpha(c)}$	Two signed lefts give an ordinary left
$\mathrm{P}^{\alpha}_a\mathrm{P}^{\alpha}_c = \mathrm{P}^{\alpha}_{ca}$	The signed right family is a group
$\mathrm{Fix}(T)$	Fixed space of $T$
$\mathrm{Fix}(L_{\lambda}\alpha)$	Even part, odd part or $0$ for $\lambda = 1, -1$ or else
$\Sigma_x = \mathrm{L}^{\alpha}_x R_{\alpha(x)^{-1}}$	The signed sandwich as a pairing
$\rho_u$, $u^{\perp}$	Reflection and its fixed hyperplane