The Signed Adjoint of the Left Multiplication on a Group

Introduction

The signed left multiplication $x\mapsto a\,\alpha(x)$ is the one-sided operator built from the left translation and the grade involution, and its adjoint under the natural pairing is again a signed left multiplication, with parameter $\alpha(a)^{-1}$. The adjoint operation therefore acts on the set of signed left multiplications as an involution of the index set, and that involution is exactly the group involution $\sigma=\iota\alpha$ associated with $\alpha$; the self-adjoint signed left multiplications are those indexed by the fixed elements of $\sigma$. This article gives the adjoint, proves that the index assignment is an involution-preserving bijection, and records the relation to the signed sandwich. It is the sixth article of the * Operator Theory group; the signed left multiplication and its composition law are from The Signed Left Multiplication on a Group, the pairing and adjoint from Involutions on the Operator Layer, and the signed sandwich from The Signed Adjoint Sandwich on a Group.

Throughout, $(G,\alpha)$ is a group with an involutive automorphism $\alpha$, $\sigma=\iota\alpha$ is the associated involution of $G$, $\ell^{\alpha}_a(x)=a\,\alpha(x)=\Sigma^{\alpha}_{a,e}$ is the signed left multiplication, and the adjoint is taken with respect to the orthonormal pairing $\langle g,h\rangle=\delta_{g,h}$.

The Adjoint

Theorem (the adjoint of a signed left multiplication). For every $a\in G$,

$$ \bigl(\ell^{\alpha}_a\bigr)^{*}=\Sigma^{\alpha}_{\alpha(a)^{-1},\,e}=\ell^{\alpha}_{\alpha(a)^{-1}}=\ell^{\alpha}_{\alpha(a^{-1})}. $$

Hence the adjoint of a signed left multiplication is a signed left multiplication, and the set $\{\ell^{\alpha}_a:a\in G\}$ is closed under the adjoint.

Proof. The signed left multiplication is the signed sandwich with second parameter $e$, so the adjoint formula of The Signed Adjoint Sandwich on a Group gives $\bigl(\ell^{\alpha}_a\bigr)^{*}=\Sigma^{\alpha}_{\alpha(a)^{-1},\alpha(e)^{-1}}=\Sigma^{\alpha}_{\alpha(a)^{-1},e}$, because $\alpha(e)=e$. The last equality is $\alpha(a^{-1})=\alpha(a)^{-1}$.

Corollary (the adjoint inverts the index through the involution). The assignment $a\mapsto\alpha(a)^{-1}$ is an involution of the set $G$, and the adjoint on the signed left multiplications is that assignment on the indices:

$$ a\mapsto\alpha(a)^{-1}, \qquad a\mapsto\alpha(a)^{-1}\mapsto\alpha\bigl(\alpha(a)^{-1}\bigr)^{-1}=a . $$

It is the anti-automorphism $\sigma=\iota\alpha=\alpha\iota$ of Involutive Groups, read on the index set.

Proof. The double application returns $a$ because $\alpha^{2}=\mathrm{id}$, so the assignment is an involution of the set $G$. And $\alpha(a)^{-1}=\iota(\alpha(a))=\sigma(a)$; so the assignment is the group involution $\sigma$, which is an anti-automorphism of order two.

Proposition (the index map is a bijection). The map $a\mapsto\ell^{\alpha}_a$ is a bijection from $G$ onto the set of signed left multiplications; it intertwines the involution $\sigma$ of $G$ with the adjoint, and it carries the fixed set $G^{\sigma}$ onto the set of self-adjoint signed left multiplications.

Proof. Injectivity: $\ell^{\alpha}_a=\ell^{\alpha}_b$ means $a\alpha(x)=b\alpha(x)$ for all $x$, and evaluating at $x=e$ gives $a=b$; surjectivity is the definition of the set. The intertwining is the theorem read as $\ell^{\alpha}_{\sigma(a)}=(\ell^{\alpha}_a)^{*}$. The fixed set statement is the definition of a fixed point of $\sigma$ together with the same equation.

Self-Adjointness and Unitarity

Theorem (the self-adjoint signed left multiplications). $\ell^{\alpha}_a$ is self-adjoint if and only if $\alpha(a)=a^{-1}$, equivalently if and only if $a\alpha(a)=e$, equivalently if and only if $a$ is fixed by the involution $\sigma=\iota\alpha$. Hence the self-adjoint signed left multiplications are indexed exactly by the fixed set $G^{\sigma}$ of $\sigma$.

Proof. Self-adjointness is $\ell^{\alpha}_a=\ell^{\alpha}_{\alpha(a)^{-1}}$, and the index map is injective, so the condition is $a=\alpha(a)^{-1}$, that is $\alpha(a)=a^{-1}$. Applying $\iota$ gives $\sigma(a)=a$. The fixed set of $\sigma$ is precisely $\{a:\alpha(a)=a^{-1}\}$ by the identity $G^{\sigma}=I(\alpha)$ of Involutions and the Fixed-Point Subgroup.

Proposition (unitarity). Every signed left multiplication is unitary: $\bigl(\ell^{\alpha}_a\bigr)^{*}\ell^{\alpha}_a=\mathrm{id}=\ell^{\alpha}_a\bigl(\ell^{\alpha}_a\bigr)^{*}$.

Proof. It is the signed sandwich with second parameter $e$, and every signed sandwich is unitary by The Signed Adjoint Sandwich on a Group. Directly, $\ell^{\alpha}_a\ell^{\alpha}_{\alpha(a)^{-1}}=L_{a\,\alpha(\alpha(a)^{-1})}=L_{aa^{-1}}=\mathrm{id}$ by the composition law of The Signed Left Multiplication on a Group and $\alpha^{2}=\mathrm{id}$.

Example (the fixed and inverted elements). On the cyclic group $C_6$ with $\alpha=\iota$ the involution $\sigma=\iota\alpha=\mathrm{id}$ has fixed set all of $C_6$, so every signed left multiplication is self-adjoint; on $S_3$ with the conjugation by a transposition, the fixed set of $\sigma=\iota\alpha$ is the inverted set $I(\alpha)$ of the conjugation, which has four elements $\{e,(1\,2),(1\,2\,3),(1\,3\,2)\}$, and exactly four of the six signed left multiplications are self-adjoint.

Remark (the contrast with the unsigned case). The unsigned left multiplication satisfies $(L_a)^{*}=L_{a^{-1}}$, indexed by the inversion $\iota$; the signed one satisfies $(\ell^{\alpha}_a)^{*}=\ell^{\alpha}_{\alpha(a)^{-1}}$, indexed by $\sigma=\iota\alpha$. The insertion of the grade involution therefore replaces the inversion by the involution $\sigma$ on the index set, and the self-adjoint unsigned left multiplications are indexed by the elements of order at most two whereas the self-adjoint signed ones are indexed by the fixed set of $\sigma$. The two agree exactly when $\alpha=\mathrm{id}$.

The Relation to the Signed Sandwich

Proposition (the adjoint of the signed sandwich factors through the signed left multiplication). For every $a,b\in G$,

$$ \bigl(\Sigma^{\alpha}_{a,b}\bigr)^{*}=\bigl(\ell^{\alpha}_a\bigr)^{*}\circ(R_b)^{*}=\ell^{\alpha}_{\alpha(a)^{-1}}\circ R_{b^{-1}}. $$

Proof. The factorisation $\Sigma^{\alpha}_{a,b}=R_b\circ\ell^{\alpha}_a$ is from The Signed Adjoint Sandwich on a Group; the adjoint of a product reverses the order, and the adjoints of the two factors are the theorem above and $(R_b)^{*}=R_{b^{-1}}$.

Corollary (the two-sided adjoint from the one-sided one). The adjoint of the signed sandwich is determined by the adjoint of the signed left multiplication and the adjoint of the right translation; iterating, the adjoint of any word in the operators $\ell^{\alpha}_a$ and $R_b$ is the reversed word in the adjoints, and it remains in the algebra generated by the signed left multiplications and the right translations.

Proof. The adjoint is an anti-involution of the operator algebra, so it reverses words; the generators are mapped to generators by the theorem and by $(R_b)^{*}=R_{b^{-1}}$.

Remark (no $*$-representation). The map $a\mapsto\ell^{\alpha}_a$ is not a homomorphism, its composition law being $\ell^{\alpha}_a\ell^{\alpha}_b=L_{a\alpha(b)}$ of The Signed Left Multiplication on a Group; so the signed left multiplications do not form a $*$-representation of $G$. What they form is a set indexed by $G$ and closed under the adjoint, on which the adjoint acts as the involution $\sigma$ of the index set. The $*$-representation statement belongs to the unsigned left regular representation of The Adjoint of the Left Multiplication on a Group.

Summary

The adjoint of the signed left multiplication is the signed left multiplication with index sent through the involution,

$$ \bigl(\ell^{\alpha}_a\bigr)^{*}=\ell^{\alpha}_{\alpha(a)^{-1}}=\ell^{\alpha}_{\sigma(a)}, \qquad \sigma=\iota\alpha, $$

so the set $\{\ell^{\alpha}_a\}$ is closed under the adjoint and the index map $a\mapsto\ell^{\alpha}_a$ is a bijection intertwining $\sigma$ with the adjoint. The self-adjoint signed left multiplications are those with $\alpha(a)=a^{-1}$, that is those indexed by the fixed set $G^{\sigma}=I(\alpha)$ of the involution $\sigma$; the unsigned case is recovered at $\alpha=\mathrm{id}$, where $\sigma=\iota$ and the self-adjoint signed left multiplications are indexed by the $2$-torsion. Every signed left multiplication is unitary, being the signed sandwich with second parameter $e$.

The adjoint of the general signed sandwich factors through the signed left multiplication, $\bigl(\Sigma^{\alpha}_{a,b}\bigr)^{*}=\ell^{\alpha}_{\alpha(a)^{-1}}\circ R_{b^{-1}}$, so the two-sided adjoint is built from the one-sided adjoint and the adjoint of the right translation. The signed left multiplications do not form a $*$-representation, their composition law being that of The Signed Left Multiplication on a Group; they form a set indexed by $G$ on which the adjoint is the involution $\sigma$.

Summary of Notation

Symbol	Meaning
$\ell^{\alpha}_a(x)=a\,\alpha(x)$	the signed left multiplication
$\bigl(\ell^{\alpha}_a\bigr)^{*}=\ell^{\alpha}_{\alpha(a)^{-1}}$	the adjoint
$\sigma=\iota\alpha=\alpha\iota$	the involution of the index set
self-adjoint $\iff\alpha(a)=a^{-1}$	indexed by the fixed set $G^{\sigma}=I(\alpha)$
$\bigl(\ell^{\alpha}_a\bigr)^{*}\ell^{\alpha}_a=\mathrm{id}$	unitarity
$\bigl(\Sigma^{\alpha}_{a,b}\bigr)^{*}=\ell^{\alpha}_{\alpha(a)^{-1}}R_{b^{-1}}$	the adjoint of the sandwich