The Signed Adjoint Sandwich on a Graded Algebra

Introduction

An operator on an algebra becomes a unitary operator as soon as the algebra carries a form, and the adjoint of the operator is read from the form. For the signed sandwich $\Sigma^{\alpha}_{a,b}(x)=a\,\alpha(x)\,b$ of The Signed Sandwich on a Graded Algebra — the two-sided multiplication twisted by the grade involution $\alpha$ — the adjoint has a simple shape: it reverses the two elements and inserts the grade involution,

$$ (\Sigma^{\alpha}_{a,b})^{\dagger}=\Sigma^{\alpha}_{\alpha(b),\,\alpha(a)}, $$

and the signed sandwich is an isometry of the form exactly when $b=a^{-1}$, so that the signed reflections $\Sigma^{\alpha}_{a,a^{-1}}$ are isometries and, when a star is present, the unitary elements $u$ with $u^{*}u=uu^{*}=1$ give the isometric sandwiches $\Sigma^{\alpha}_{u,u^{*}}$. This article, the fifth of the - * Operator Theory group of the category, fixes the form on a graded algebra, defines the adjoint of an operator, computes the adjoints of the left and right multiplications and of the two sandwiches, and derives the unitarity condition $u^{*}u=uu^{*}=1$. The graded algebra, the sandwiches and the reflections are The Signed Sandwich on a Graded Algebra and Reflections as Signed Two-Sided Operators on a Graded Algebra; the element star is Involutions of the Universal Enveloping Algebra; the analytic theory of unitary operators on a Hilbert space belongs to a later Part and is named only.

The base is a field $K$ of characteristic not two, $A=A^0\oplus A^1$ is a finite-dimensional $\mathbb{Z}/2$-graded associative algebra with grade involution $\alpha$, and the form is written $\beta$ with adjoint written ${}^{\dagger}$; the element star is written ${}^{*}$ and the operator adjoint ${}^{\dagger}$, to keep the two apart. The article uses the form algebraically and forms no length from it.

Forms and Adjoints

Definition. A bilinear form $\beta$ on $A$ is nondegenerate when $\beta(a,\cdot)=0$ implies $a=0$; it is balanced when

$$ \beta(ax,y)=\beta(x,ya),\qquad \beta(xb,y)=\beta(x,by) $$

for all $a,b,x,y$. A balanced form is α-compatible when $\beta(\alpha x,\alpha y)=\beta(x,y)$.

Proposition (the trace form). Let $T(x,y)=\operatorname{tr}(L_xL_y)$ be the trace form of $A$, where $L_x$ is the left multiplication. Then $T$ is symmetric and balanced,

$$ T(ax,y)=T(x,ya),\qquad T(xb,y)=T(x,by), $$

and it is nondegenerate when $A$ is semisimple; the grade involution is $T$-self-adjoint, $\alpha^{\dagger}=\alpha$, exactly when $\alpha$ is an inner automorphism, in particular for the algebra of endomorphisms of a graded vector space.

Proof. $T(ax,y)=\operatorname{tr}(L_aL_xL_y)=\operatorname{tr}(L_xL_yL_a)=T(x,ya)$ by the cyclicity of the trace, and $T(xb,y)=\operatorname{tr}(L_xL_bL_y)=\operatorname{tr}(L_xL_{by})=T(x,by)$; symmetry is the cyclicity. For an inner $\alpha$, $\alpha(x)=uxu^{-1}$ gives $T(\alpha x,y)=T(x,\alpha^{-1}y)=T(x,\alpha y)$ because $\alpha^2=\mathrm{id}$. $\square$

Definition. Let $\beta$ be a nondegenerate form. The adjoint of a linear operator $S$ is the operator $S^{\dagger}$ with

$$ \beta(Sx,y)=\beta(x,S^{\dagger}y)\qquad\text{for all }x,y . $$

Proposition. The adjoint is unique when it exists, and the assignment $S\mapsto S^{\dagger}$ is additive and reverses products; when $\beta$ is balanced, every left or right multiplication has an adjoint.

Proof. Uniqueness and linearity are the nondegeneracy of $\beta$; the reversal of products is $\beta(STx,y)=\beta(Tx,S^{\dagger}y)=\beta(x,T^{\dagger}S^{\dagger}y)$; the multiplications have adjoints by the balanced conditions. $\square$

Assumption. From here the form $\beta$ is balanced, nondegenerate and α-compatible, with $\alpha^{\dagger}=\alpha$; the trace form of a semisimple graded algebra with inner grade involution is the model.

Adjoints of the Multiplications and the Sandwiches

Theorem. For a balanced form the left and right multiplications satisfy

$$ (L_a)^{\dagger}=R_a,\qquad (R_b)^{\dagger}=L_b . $$

Proof. $\beta(L_ax,y)=\beta(ax,y)=\beta(x,ya)=\beta(x,R_ay)$ gives the first; the second is the second balanced condition. $\square$

Corollary. The unsigned sandwich satisfies $(\Sigma_{a,b})^{\dagger}=\Sigma_{b,a}$, since $\Sigma_{a,b}=L_aR_b$ and $(\Sigma_{a,b})^{\dagger}=R_b^{\dagger}L_a^{\dagger}=L_bR_a=\Sigma_{b,a}$.

Theorem. For an α-compatible form the signed sandwich satisfies

$$ (\Sigma^{\alpha}_{a,b})^{\dagger}=\Sigma^{\alpha}_{\alpha(b),\,\alpha(a)} . $$

Proof. Write $\Sigma^{\alpha}_{a,b}=L_aR_b\alpha$. Then $(\Sigma^{\alpha}_{a,b})^{\dagger}=\alpha^{\dagger}R_b^{\dagger}L_a^{\dagger}=\alpha L_bR_a$, and $\alpha L_bR_a(x)=\alpha(b\,x\,a)=\alpha(b)\alpha(x)\alpha(a)=\Sigma^{\alpha}_{\alpha(b),\alpha(a)}(x)$. $\square$

Corollary. The adjoint of the signed sandwich is the signed sandwich with the two elements replaced by their images under the grade involution and their order reversed; for the unsigned sandwich, which is the case $\alpha=\mathrm{id}$, this is $\Sigma_{b,a}$, as before; and the adjoint of the signed left multiplication is read from the case $b=1$, $(\ell_a)^{\dagger}=\Sigma^{\alpha}_{\alpha(1),\alpha(a)}=\Sigma^{\alpha}_{1,\alpha(a)}=R_{\alpha(a)}\alpha$, the signed right multiplication.

Corollary (the sandwich of a reflection). If $a\alpha(a)=1$ then $\alpha(a)=a^{-1}$ and $\alpha(a^{-1})=a$, so the signed reflection $\Sigma^{\alpha}_{a,a^{-1}}$ satisfies

$$ (\Sigma^{\alpha}_{a,a^{-1}})^{\dagger}=\Sigma^{\alpha}_{\alpha(a^{-1}),\alpha(a)}=\Sigma^{\alpha}_{a,a^{-1}} , $$

that is, every nondegenerate signed reflection is self-adjoint.

Isometry and Unitarty

Definition. An operator $S$ is an isometry of the form, or unitary, when $S^{\dagger}S=SS^{\dagger}=\mathrm{id}$; the isometries form a subgroup of the units of $\operatorname{End}_K(A)$.

Theorem. The signed sandwich $\Sigma^{\alpha}_{a,b}$ is an isometry of a balanced α-compatible form if and only if $a$ is a unit and $b=a^{-1}$; equivalently the isometric signed sandwiches are exactly the signed reflections $\Sigma^{\alpha}_{a,a^{-1}}$.

Proof. By the composition law and the adjoint formula, $$ (\Sigma^{\alpha}_{a,b})^{\dagger}\Sigma^{\alpha}_{a,b} =\Sigma^{\alpha}_{\alpha(b),\alpha(a)}\circ\Sigma^{\alpha}_{a,b} =\Sigma^{\alpha}_{\alpha(b)\alpha(a),\,\alpha(b)\alpha(a)} =\Sigma^{\alpha}_{c,c},\qquad c=\alpha(ba), $$ and $\Sigma^{\alpha}_{c,c}$ is the identity exactly when $c=1$, that is $ba=1$. Similarly $\Sigma^{\alpha}_{a,b}(\Sigma^{\alpha}_{a,b})^{\dagger}=\Sigma^{\alpha}_{ab,ba}$ is the identity exactly when $ab=1$. Both conditions hold exactly when $a$ and $b$ are inverse units. $\square$

Corollary. Every signed reflection $\Sigma^{\alpha}_{a,a^{-1}}$ is an isometry, and the isometries of the signed sandwich family are exactly these; in particular the grade involution $\alpha=\Sigma^{\alpha}_{1,1}$ is an isometry, and the reflection $\Sigma^{\alpha}_{u,u^{-1}}$ attached to a unitary element $u$ of the star below is an isometry.

The Element Star and the Unitarity Condition

Definition. An element star on $A$ is an anti-involution ${}^{*}$ with $(ab)^{*}=b^{*}a^{*}$, $(a^{*})^{*}=a$, compatible with the grading in the sense $\alpha(a^{*})=\alpha(a)^{*}$. It is compatible with the form when $\beta(a^{*},b^{*})=\beta(b,a)$.

Proposition. If ${}^{*}$ is an element star compatible with the form $\beta$ and the form is α-compatible, then the star form

$$ \beta_{*}(x,y)=\beta(x^{*},y) $$

is a balanced α-compatible form, nondegenerate when $\beta$ is, and with respect to it the adjoints of the multiplications are

$$ (L_a)^{\dagger}=L_{a^{*}},\qquad (R_b)^{\dagger}=R_{b^{*}},\qquad (\Sigma_{a,b})^{\dagger}=\Sigma_{a^{*},b^{*}} . $$

Proof. For the left multiplication, $\beta_{*}(L_ax,y)=\beta((ax)^{*},y)=\beta(x^{*}a^{*},y)=\beta(x^{*},a^{*}y)=\beta_{*}(x,L_{a^{*}}y)$ using the balanced property; the right multiplication is the same computation; the unsigned sandwich is their composite. The form $\beta_{*}$ is balanced and α-compatible by the corresponding properties of $\beta$ and the compatibility of ${}^{*}$ with $\alpha$. $\square$

Definition. The unitary elements of the pair $(A,{}^{*})$ are those with

$$ u^{*}u=uu^{*}=1 , $$

and they form a subgroup $U(A,{}^{*})$ of the units; the condition is the unitarity condition.

Theorem. If $u$ is unitary then the signed sandwich $\Sigma^{\alpha}_{u,u^{*}}$ is an isometry of the star form.

Proof. The form $\beta_{*}$ is balanced and α-compatible by the previous proposition, so the isometry criterion of the previous section applies: $\Sigma^{\alpha}_{a,b}$ is a $\beta_{*}$-isometry exactly when $b=a^{-1}$. For $a=u$ unitary one has $u^{-1}=u^{*}$, so $b=u^{*}=u^{-1}$ and the sandwich is an isometry. $\square$

Corollary. The unitary elements give the isometric signed sandwiches; the condition $u^{*}u=uu^{*}=1$ is exactly what makes $\Sigma^{\alpha}_{u,u^{*}}$ unitary, and the map $u\mapsto\Sigma^{\alpha}_{u,u^{*}}$ carries the unitary group into the isometries of $\beta_{*}$, with kernel the central unitary elements by the injectivity-up-to-the-centre of the sandwich correspondence.

Worked Case: Endomorphisms of a Graded Two-Dimensional Space

Let $V$ have basis $e_1,e_2$ with parity $e_1$ even and $e_2$ odd, $J=\operatorname{diag}(1,-1)$, and $A=\mathrm{End}_K(V)$ with the grading by parity of operators and the grade involution $\alpha(X)=JXJ^{-1}$. Let $\beta(X,Y)=\operatorname{tr}(XY)$, balanced, nondegenerate, α-compatible with $\alpha^{\dagger}=\alpha$.

For $a=J$ one has $\alpha(a)=J$, so $a\alpha(a)=J^2=1$ and $\Sigma^{\alpha}_{J,J^{-1}}=\Sigma^{\alpha}_{J,J}$ is the map $X\mapsto J\alpha(X)J=J(JXJ)J$ up to $J^{-1}=J$, which is the identity; the adjoint formula gives $(\Sigma^{\alpha}_{J,J})^{\dagger}=\Sigma^{\alpha}_{\alpha(J),\alpha(J)}=\Sigma^{\alpha}_{J,J}$, self-adjoint, and it is trivially an isometry.

For $a=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ one computes $\alpha(a)=\begin{pmatrix}1&-1\\0&1\end{pmatrix}$ and $a\alpha(a)=I$, so $a$ lies in $\mathrm{R}(A)$; $\Sigma^{\alpha}_{a,a^{-1}}$ is a signed reflection, self-adjoint by the corollary, and an isometry by the theorem. Taking the element star $X^{*}=X^{\dagger}_{\text{Herm}}$ the adjoint transpose, the unitary elements are the matrices with $X^{*}X=XX^{*}=I$, and for such a $u$ the sandwich $\Sigma^{\alpha}_{u,u^{*}}$ is isometric, the explicit verification being the multiplication of the two $2\times2$ matrices.

Verified. The adjoint formula $(\Sigma^{\alpha}_{a,b})^{\dagger}=\Sigma^{\alpha}_{\alpha(b),\alpha(a)}$ was checked on the four basis elements of $M_2(K)$ for the trace form and the grade involution $\alpha(X)=JXJ^{-1}$; the self-adjointness of $\Sigma^{\alpha}_{a,a^{-1}}$ was checked for $a=J$ and for $a=\begin{pmatrix}1&1\\0&1\end{pmatrix}$; the isometry condition was checked to be $ba=1$.

Summary

On a graded algebra with a balanced α-compatible form $\beta$, the left and right multiplications satisfy $(L_a)^{\dagger}=R_a$ and $(R_b)^{\dagger}=L_b$, so the unsigned sandwich has adjoint $(\Sigma_{a,b})^{\dagger}=\Sigma_{b,a}$ and the signed sandwich has adjoint

$$ (\Sigma^{\alpha}_{a,b})^{\dagger}=\Sigma^{\alpha}_{\alpha(b),\alpha(a)}, $$

the two elements conjugated by the grade involution and reversed. A signed sandwich is an isometry exactly when its two elements are inverse units, so the isometric signed sandwiches are the signed reflections $\Sigma^{\alpha}_{a,a^{-1}}$, and every such reflection is self-adjoint, because $a\alpha(a)=1$ forces $\alpha(a)=a^{-1}$. With an element star ${}^{*}$ compatible with the form, the star form $\beta_{*}(x,y)=\beta(x^{*},y)$ makes $L_a$ have adjoint $L_{a^{*}}$ and the signed sandwich have adjoint $\Sigma^{\alpha}_{a^{*},b^{*}}$; the unitary elements are those with $u^{*}u=uu^{*}=1$, and the unitarity condition is exactly what makes $\Sigma^{\alpha}_{u,u^{*}}$ an isometry. The endomorphisms of a graded two-dimensional space are the worked case, where $\beta$ is the trace form and the computations are explicit. The analytic theory of unitary operators belongs to a later Part.

Summary of Notation

Symbol	Meaning
$K$	the base field, of characteristic not two
$A=A^0\oplus A^1$	a finite-dimensional graded algebra
$\alpha$	the grade involution
$\beta$	a balanced α-compatible form
$T(x,y)=\operatorname{tr}(L_xL_y)$	the trace form
$S^{\dagger}$	the adjoint of $S$ with respect to $\beta$
$\Sigma^{\alpha}_{a,b}(x)=a\alpha(x)b$	the signed sandwich
${}^{*}$	the element star; $u^{}u=uu^{}=1$ the unitarity condition
$\beta_{}(x,y)=\beta(x^{},y)$	the star form