The Signed Adjoint of the Left Multiplication on a Ring

Introduction

The signed left multiplication on a ring with a grade involution is the one-sided operator

$$ T_a(x) = a\,\alpha(x) = S_{a,1}(x), $$

the left multiplication composed with the sign rule; it is the one-sided companion of the signed sandwich and the elementary case of the reflection. Its signed adjoint with respect to the twisted pairing is again a signed left multiplication, $T_a^{*_\sigma} = T_{\delta(a)}$ with $\delta = \sigma\alpha$, so the signed adjoint does not switch the side, in contrast with the unsigned adjoint of The Adjoint of the Left Multiplication on a Ring, which sends $L_a$ to $R_a$. The operator is unitary, $T_a^{*_\sigma}T_a = \mathrm{id}$, exactly when $\delta(a)\alpha(a) = 1$, and it is an involution exactly when $a\alpha(a) = 1$; the relation with the signed sandwich is $S_{a,b} = T_aR_{\alpha(b)} = L_aT_{\alpha(b)}$, so the one-sided operators generate the two-sided ones.

This article computes the signed adjoint of the signed left multiplication, its inverse, its unitarity and involution conditions, and its relation to the signed sandwich and the reflection. It assumes The Signed Left Multiplication on a Ring for the operator, The Signed Adjoint Sandwich on a Ring for the general adjoint, and Involutions of the Endomorphism Ring for the twisted pairing. Throughout, $A$ is a ring with $1 \neq 0$ and an involution $\sigma$, $\tau$ is a $\sigma$-invariant trace, $\alpha$ is a grade involution commuting with $\sigma$, $\delta = \sigma\alpha$, the twisted pairing is $\{x,y\} = \tau(x\sigma(y))$, and $T_a(x) = a\alpha(x)$.

The Signed Adjoint

Theorem. The signed left multiplication is the sandwich $S_{a,1}$, and its signed adjoint is

$$ T_a^{*_\sigma} = S_{\delta(a),\delta(1)} = T_{\delta(a)} ; $$

the signed adjoint of the signed left multiplication by $a$ is the signed left multiplication by $\delta(a)$, and it stays on the same side.

Proof. $T_a = S_{a,1}$ by definition, and $S_{a,b}^{*_\sigma} = S_{\delta(a),\delta(b)}$ from The Signed Adjoint Sandwich on a Ring; with $b = 1$ and $\delta(1) = 1$ this is $S_{\delta(a),1} = T_{\delta(a)}$.

Proposition (inverse and composition). The operator $T_a$ is invertible exactly when $a$ is a unit, with

$$ T_a^{-1} = T_{\alpha(a)^{-1}} , $$

and $T_aT_b = L_{a\alpha(b)}\alpha$; the signed left multiplications form a semigroup under composition, closed under inverses on the units, and the two-sided sandwich is generated by them, $S_{a,b} = T_aR_{\alpha(b)} = L_aT_{\alpha(b)}$.

Proof. $T_{\alpha(a)^{-1}}(T_a(x)) = \alpha(a)^{-1}\alpha(a\alpha(x)) = \alpha(a)^{-1}\alpha(a)\,x = x$ for $a$ a unit; the composition is the direct computation $T_a(T_b(x)) = a\alpha(b\alpha(x)) = a\alpha(b)x = (L_{a\alpha(b)}\alpha)(x)$; and $T_aR_{\alpha(b)}(x) = T_a(x\alpha(b)) = a\alpha(x\alpha(b)) = a\alpha(x)b$.

Unitarity and Involution

Theorem. The signed left multiplication $T_a$ is unitary, $T_a^{*_\sigma}T_a = T_aT_a^{*_\sigma} = \mathrm{id}$, exactly when

$$ \delta(a)\alpha(a) = 1 , $$

and it is an involution, $T_a^2 = \mathrm{id}$, exactly when

$$ a\alpha(a) = 1 . $$

For a unitary $a$ the operator $T_a$ is an isometry of the twisted pairing, and for an involutive $a$ it is a signed involution of the ring.

Proof. $T_a^{*_\sigma}T_a(x) = T_{\delta(a)}(a\alpha(x)) = \delta(a)\alpha(a\alpha(x)) = \delta(a)\alpha(a)\,x$, so the product is the identity exactly when $\delta(a)\alpha(a) = 1$; the same formula with $\delta(a)$ in place of $a$ gives $T_aT_a^{*_\sigma}$. The square is $T_a(T_a(x)) = a\alpha(a\alpha(x)) = a\alpha(a)x$, which is $x$ for all $x$ exactly when $a\alpha(a) = 1$.

Corollary (the specialisations). For $\alpha = \mathrm{id}$ the conditions are $\sigma(a)a = 1$ for unitarity and $a^2 = 1$ for the involution: the unitary signed left multiplications are the $\sigma$-unitary elements $a^{*_\sigma}a = 1$, and the involutive ones are the involutions of the ring. The reflection $r_u = S_{u,u^{-1}}$ is the composite $T_uR_{\alpha(u)^{-1}}$, and its signed adjoint is $T_{\delta(u)}R_{\alpha(\delta(u))^{-1}}$; the self-adjointness condition of The Signed Adjoint of the Reflection on a Ring is the comparison of these two one-sided operators.

Proof. Put $\alpha = \mathrm{id}$ in the theorem: $\delta(a)\alpha(a) = \sigma(a)a$ and $a\alpha(a) = a^2$. The reflection formula is $S_{a,b} = T_aR_{\alpha(b)}$ with $b = u^{-1}$, and the adjoint is read off from $S_{\delta(u),\delta(u)^{-1}}$.

Examples

(a) The matrix case. $A = M_n$ with the transpose $\sigma$, the grade involution $\alpha$ of a $\mathbb{Z}/2$-grading and $T_X(Y) = X\alpha(Y)$: $T_X^{*_\sigma} = T_{\sigma\alpha(X)}$, unitary when $\sigma\alpha(X)\alpha(X) = I$, an involution when $X\alpha(X) = I$. For $\alpha = \mathrm{id}$, $T_X = L_X$ and the signed adjoint is $L_{\sigma(X)} = L_{X^{\mathrm t}}$.

(b) The group ring. $A = K[G]$ with the standard involution and the trivial grading: $T_g = L_g$, $T_g^{*_\sigma} = L_{g^{-1}}$, unitary always, and $T_g^2 = L_{g^2}$, an involution only for $g^2 = 1$. The signed adjoint of the signed left multiplication by a group element is the left multiplication by the inverse, the one-sided companion of the group inversion of The Group Inversion as an Adjoint.

(c) The graded algebra. For a superalgebra with $\alpha$ the grade involution and a homogeneous unit $u$ of odd degree, $T_u$ is not $\alpha$-linear but graded-linear, $T_u(uv) = \alpha(u)T_u(v)$ up to the sign; the unitarity condition $\delta(u)\alpha(u) = 1$ carries the sign of the degree, the sign rule of the graded case.

(d) The relation with the sandwich. Every signed sandwich is a product of a signed left multiplication and a right multiplication twisted by $\alpha$, $S_{a,b} = T_aR_{\alpha(b)}$; the signed left multiplications therefore generate the two-sided operators, and the signed adjoint of the sandwich is the composite of the signed adjoints by the anti-multiplicativity of the adjoint.

Summary

The signed left multiplication $T_a(x) = a\alpha(x) = S_{a,1}$ has signed adjoint $T_a^{*_\sigma} = T_{\delta(a)}$ with respect to the twisted pairing, where $\delta = \sigma\alpha$: the signed adjoint stays on the same side. Its inverse is $T_{\alpha(a)^{-1}}$ for a unit $a$, its composition is $T_aT_b = L_{a\alpha(b)}\alpha$, and every signed sandwich is generated by it, $S_{a,b} = T_aR_{\alpha(b)} = L_aT_{\alpha(b)}$. It is unitary exactly when $\delta(a)\alpha(a) = 1$ and an involution exactly when $a\alpha(a) = 1$; for $\alpha = \mathrm{id}$ these are the $\sigma$-unitarity $\sigma(a)a = 1$ and the involution $a^2 = 1$. The reflection is the composite $T_uR_{\alpha(u)^{-1}}$, so the self-adjointness of the reflections is the comparison of the signed adjoints of two one-sided operators.

Summary of Notation

Symbol	Meaning
$T_a(x) = a\alpha(x)$	Signed left multiplication; $T_a = S_{a,1}$
$\delta = \sigma\alpha$	Twist of the grade involution
$T_a^{*_\sigma} = T_{\delta(a)}$	Signed adjoint; same side
$T_a^{-1} = T_{\alpha(a)^{-1}}$	Inverse
$T_aT_b = L_{a\alpha(b)}\alpha$	Composition
$S_{a,b} = T_aR_{\alpha(b)} = L_aT_{\alpha(b)}$	Sandwich generated by one-sided operators
$\delta(a)\alpha(a) = 1$	Unitarity
$a\alpha(a) = 1$	Involution ($T_a^2 = \mathrm{id}$)
$\alpha = \mathrm{id}$: $\sigma(a)a = 1$, $a^2 = 1$	Unitary and involutive left multiplications