The Graded Adjoint Action on a Module over a Bimodule over an Algebra

Introduction

The graded action of a graded algebra on a graded module is even, and the Koszul sign rule governs every exchange of two homogeneous things. This article is about the adjoint action, the transpose of the graded action with respect to an $\alpha$-invariant pairing: the operator $L_a$ has adjoint $L_{\sigma(a)}$, so the algebra acts on the module a second time by $a\triangleright x=\sigma(a)x$; the article computes this action, shows that it is compatible with the grading, and derives the sign rule for the adjoint of a graded commutator.

The article is the seventh and last of the * Operator Theory group of this category. It assumes the graded action and the Koszul sign rule of The Graded Action on a Module over a Bimodule over an Algebra, the one-sided adjoints of The Signed Adjoint of the Left Multiplication on a Module over an Algebra and The Signed Adjoint of the Sandwich on a Bimodule over an Algebra, and the involution and unitarity of Module Operators with an Involution. The grading itself, superalgebras and their morphisms belong to Superalgebras and Graded Structures. The article stays inside Part I: no distance, norm, form with a norm, positivity, topology or limit. Throughout, $R$ is a commutative ring with $1 \neq 0$ in which $2$ is invertible, $(A,\sigma)$ is a graded involutive $R$-algebra with graded involution ($\sigma(A_i)\subseteq A_i$) and grade involution $\alpha$ commuting with $\sigma$, $\beta=\alpha\sigma$, ${}_A M_A$ is a graded bimodule with the Koszul right action and an $\alpha$-invariant balanced $\sigma$-sesquilinear pairing, and $L_a$, $L^{\alpha}_a$ are the one-sided operators.

The Adjoint Action

The transpose of the left action

Definition. The adjoint action of $A$ on $M$ is

$$ a \triangleright x = \sigma(a)\,x \qquad (a \in A,\ x \in M), $$

the one-sided operator adjoint read as an action: since $(L_a)^{*}=L_{\sigma(a)}$, the adjoint of the left multiplication by $a$ is the left multiplication by $\sigma(a)$.

Proposition. The adjoint action is a right action of $A$ on $M$,

$$ (a b)\triangleright x = b\triangleright(a\triangleright x), $$

and it satisfies $a\triangleright x = L_a^{*}(x)$; the original left action is recovered from the adjoint action by $a x=\sigma(a)\triangleright x$.

Proof. $(ab)\triangleright x=\sigma(ab)x=\sigma(b)\sigma(a)x=b\triangleright(\sigma(a)x)=b\triangleright(a\triangleright x)$, so it is a right action; the last two clauses are the definitions and $\sigma^{2}=\mathrm{id}$. $\square$

The pairing converts the left action of $A$ into the right adjoint action of $A$; the two actions together are the module-level form of the statement that the adjoint reverses the order, the content of $(L_{ab})^{*}=(L_b)^{*}(L_a)^{*}$.

Evenness of the adjoint action

Proposition. If the involution $\sigma$ is graded, $\sigma(A_i)\subseteq A_i$, then the adjoint action is even:

$$ a \in A_i \implies a\triangleright M_j \subseteq M_{i+j}, \qquad \deg(a\triangleright x)=\deg a+\deg x \quad (a,x \text{ homogeneous}). $$

Proof. $a\triangleright M_j=\sigma(a)M_j$, and $\deg\sigma(a)=\deg a$ because $\sigma$ is graded, so $a\triangleright M_j\subseteq M_{i+j}$. $\square$

Corollary. The operator $L_a^{*}=L_{\sigma(a)}$ has the same parity as $L_a$, and the adjoint operation preserves the $\mathbb{Z}/2$-degree of every homogeneous one-sided operator:

$$ \deg L_a^{*}=\deg L_a=\deg a . $$

Proof. $L_a$ has parity $\deg a$ by the evenness of the action, and $\deg\sigma(a)=\deg a$, so $L_{\sigma(a)}$ has the same parity. $\square$

The adjoint is therefore not only additive and anti-multiplicative but graded: it maps operators of parity $i$ to operators of parity $i$, so no sign appears in the parity alone.

The Sign Rule

The Koszul sign of the operators

Proposition. For homogeneous $a$ and for every homogeneous operator $T$ of parity $\deg T$ the grade involution acts on the operators by the Koszul sign:

$$ \alpha\,L_a\,\alpha^{-1}=(-1)^{\deg a}L_a, \qquad \alpha\,T\,\alpha^{-1}=(-1)^{\deg T}T . $$

Proof. This is the sign rule of The Graded Action on a Module over a Bimodule over an Algebra. $\square$

Corollary. The sign rule is compatible with the adjoint: for homogeneous $T$,

$$ \alpha\,T^{*}\,\alpha^{-1}=(-1)^{\deg T}T^{*}. $$

Proof. $T^{*}$ has the same parity as $T$ by the gradedness of the adjoint, so applying the sign rule to $T^{*}$ gives the identity. $\square$

Conjugation by the grade involution, the adjoint, and the grading are therefore mutually compatible: both operations multiply a homogeneous operator by the same sign $(-1)^{\deg T}$.

The graded commutator

Definition. For homogeneous operators $S$ and $T$ the graded commutator is

$$ [S,T]_{\epsilon}=ST-\epsilon\,TS, \qquad \epsilon=(-1)^{\deg S\,\deg T}. $$

When $\epsilon=1$ it is the ordinary commutator; when $\epsilon=-1$, the case of two odd operators, it is the ordinary anticommutator.

Proposition. The graded commutator of the left multiplications is a left multiplication,

$$ [L_a,L_b]_{\epsilon}=L_{ab-\epsilon\,ba}, \qquad \epsilon=(-1)^{\deg a\,\deg b}, $$

and it vanishes exactly when $L_{ab-\epsilon ba}=0$, that is, when $ab-\epsilon ba$ annihilates $M$.

Proof. $L_aL_b-L_bL_a\cdot\epsilon=\ldots$: since $L_aL_b=L_{ab}$, the expression is $L_{ab}-\epsilon L_{ba}=L_{ab-\epsilon ba}$; the last clause is the kernel statement for $L$. $\square$

The graded commutator is the graded version of the commutator, and the sign $\epsilon$ is the Koszul sign attached to the exchange of $a$ and $b$.

The Adjoint of the Graded Commutator

The sign rule for the bracket

Theorem. For homogeneous $a$ and a homogeneous operator $T$, with $\epsilon=(-1)^{\deg a\,\deg T}$,

$$ [L_a,T]_{\epsilon}^{*}=-\epsilon\,\bigl[L_{\sigma(a)},T^{*}\bigr]_{\epsilon}. $$

Thus the adjoint of the graded commutator is the graded commutator of the adjoints, up to the sign $-\epsilon$: the ordinary sign $-1$ for the commutator ($\epsilon=1$) and the sign $+1$ for the anticommutator ($\epsilon=-1$).

Proof. By the laws of the adjoint, $(L_aT)^{*}=T^{*}L_{\sigma(a)}$ and $(TL_a)^{*}=L_{\sigma(a)}T^{*}$, and $T^{*}$ has parity $\deg T$; hence

$$ [L_a,T]_{\epsilon}^{*}=T^{*}L_{\sigma(a)}-\epsilon\,L_{\sigma(a)}T^{*}=-\epsilon\bigl(L_{\sigma(a)}T^{*}-\epsilon\,T^{*}L_{\sigma(a)}\bigr)=-\epsilon\,[L_{\sigma(a)},T^{*}]_{\epsilon}. \qquad\square $$

Corollary. For two homogeneous left multiplications,

$$ [L_a,L_b]_{\epsilon}^{*}=L_{\sigma(ab-\epsilon ba)}=-\epsilon\,[L_{\sigma(a)},L_{\sigma(b)}]_{\epsilon}. $$

Proof. Apply the theorem with $T=L_b$, $T^{*}=L_{\sigma(b)}$, and use $[L_a,L_b]_\epsilon=L_{ab-\epsilon ba}$ and $\sigma(ab-\epsilon ba)=\sigma(b)\sigma(a)-\epsilon\,\sigma(a)\sigma(b)$. $\square$

The theorem is the sign rule the grading imposes on the adjoint action: for two odd operators the anticommutator has a self-adjoint bracket, $[L_a,T]_{-1}^{*}=[L_{\sigma(a)},T^{*}]_{-1}$, while for the ordinary commutator a sign is required. It is the graded refinement of the elementary fact that the adjoint of a composite reverses the order.

The sign rule for the sandwich

Proposition. The adjoint action on the signed sandwiches is compatible with the grading:

$$ \bigl(S^{\alpha}_{a,b}\bigr)^{*}=S^{\alpha}_{\beta(a),\beta(b)}, \qquad \bigl(S_{a,b}\bigr)^{*}=S_{\sigma(a),\sigma(b)}, $$

and it sends operators of parity $i$ to operators of parity $i$: the adjoint action preserves the degree and the sign rule of the graded operators.

Proof. This is The Signed Adjoint of the Sandwich on a Bimodule over an Algebra; the parity statement follows because $\sigma$ and $\alpha$ are graded and their composite $\beta$ is graded, so both parameters keep their degrees. $\square$

The Adjoint Action on the Operators

The action of the adjoint action

Proposition. The adjoint action of $A$ on $M$ induces an action on the operators by

$$ a \cdot T = L_{\sigma(a)}\,T, \qquad a \cdot T = T\,L_{\sigma(a)} , $$

the left and right regular actions of $A$ composed with the adjoint action, and the adjoint of $a\cdot T$ is computed by the rules above; in particular

$$ (L_a T)^{*}=T^{*}L_{\sigma(a)}, \qquad (T L_a)^{*}=L_{\sigma(a)}T^{*} . $$

Proof. The two formulas are the anti-multiplicativity of the adjoint with $(L_a)^{*}=L_{\sigma(a)}$. $\square$

Corollary. The adjoint action intertwines the involution of the algebra with the involution of the operators: the algebra element $a$ acts through $L_{\sigma(a)}$, so the adjoint action is the original action precomposed with $\sigma$, and applying the operator involution to an action applies $\sigma$ to the algebra element.

Proof. $L_a^{*}=L_{\sigma(a)}$ is the identity defining the adjoint action; the rest is its restatement. $\square$

Self-adjointness under the adjoint action

Proposition. Let $T$ be homogeneous and let $a$ be homogeneous with $\sigma(a)=a$. Then $T$ is self-adjoint if and only if the graded commutator $[L_a,T]_{\epsilon}$ is skew with respect to the adjoint in the graded sense,

$$ T^{*}=T \implies [L_a,T]_{\epsilon}^{*}=-\epsilon\,[L_a,T]_{\epsilon}, $$

and conversely $T^{*}=T$ follows when this holds for a separating family of such $a$.

Proof. By the theorem $[L_a,T]_{\epsilon}^{*}=-\epsilon[L_a,T^{*}]_{\epsilon}$, and with $\sigma(a)=a$ this is $-\epsilon[L_a,T^{*}]_{\epsilon}$. If $T^{*}=T$ it equals $-\epsilon[L_a,T]_{\epsilon}$, which is the displayed skewness. Conversely, if the skewness holds then $[L_a,T^{*}]_{\epsilon}=[L_a,T]_{\epsilon}$ for the family, that is $L_{a(T^{*}-T)-\epsilon(T^{*}-T)a}=0$, and for a separating family this forces $T^{*}=T$. $\square$

Examples

(a) The trivial grading. For $\alpha=\mathrm{id}$ the sign rule is vacuous, the graded commutator is the ordinary one, and the theorem reads $[L_a,T]^{*}=-[L_{\sigma(a)},T^{*}]$, the classical rule for the commutator with a left multiplication.

(b) The trivial involution. For $\sigma=\mathrm{id}$ the adjoint action is the original action, $a\triangleright x=ax$, and every operator is graded: the sign rule reduces to the graded action of The Graded Action on a Module over a Bimodule over an Algebra.

(c) The Clifford algebra. For a Clifford algebra with its canonical involution $\sigma$ and grade involution $\alpha$, the graded commutator $[L_a,L_b]_\epsilon=L_{ab-\epsilon ba}$ is the algebra bracket, and the theorem computes its adjoint with the Koszul sign; the anticommutator of two odd vectors is even and self-adjoint under the adjoint action.

(d) The exterior algebra. For $A=\Lambda(V)$ with the graded-commutative product, the graded commutator of two odd elements is the anticommutator, and its adjoint is the anticommutator of the adjoints with the sign $+1$, in agreement with the theorem.

(e) The matrix algebra with a diagonal grading. For $A=M_n(k)$ with the transpose involution and a diagonal grading by parity of the index, $\sigma$ is graded, and the adjoint action is $a\triangleright X=a^{\mathsf{T}}X$; the graded commutator's adjoint acquires the sign $-\epsilon$.

Summary

For a graded module over a graded involutive algebra with an $\alpha$-invariant balanced pairing, the transpose of the left action $L_a$ is the left multiplication by $\sigma(a)$, so the algebra acts on the module a second time by the adjoint action $a\triangleright x=\sigma(a)x$, which is a right action and is even when the involution is graded. The adjoint operation preserves the parity of every homogeneous operator, and it is compatible with the Koszul sign rule: both the grade involution and the adjoint multiply a homogeneous operator by $(-1)^{\deg T}$. The graded commutator $[S,T]_\epsilon=ST-\epsilon TS$ has adjoint $[L_a,T]_\epsilon^{*}=-\epsilon[L_{\sigma(a)},T^*]_\epsilon$, so the adjoint of the ordinary commutator is minus the commutator of the adjoints and the adjoint of the anticommutator is the anticommutator of the adjoints; for two left multiplications the bracket is again a left multiplication and the formula reads $[L_a,L_b]_\epsilon^{*}=L_{\sigma(ab-\epsilon ba)}$. The adjoint action on the signed sandwiches is $(S^{\alpha}_{a,b})^{*}=S^{\alpha}_{\beta(a),\beta(b)}$ with $\beta=\alpha\sigma$, and it preserves the degree and the sign rule. Nothing in the article uses a norm or a topology; the sign rule is the Koszul sign of the grading, and the adjoint action is the transpose of the graded action with respect to the pairing.

Summary of Notation

Symbol	Meaning
$R$, $(A,\sigma)$	base ring, graded involutive $R$-algebra, graded involution
$\alpha$, $\beta=\alpha\sigma$	grade involution, commuting with $\sigma$, and the composite
$M$	graded bimodule with α-invariant balanced σ-sesquilinear pairing
$a\triangleright x=\sigma(a)x$	the adjoint action, a right action
$L_a^{*}=L_{\sigma(a)}$	the adjoint of the left multiplication
$(-1)^{\deg T}$	the Koszul sign of the operator $T$
$\alpha T\alpha^{-1}=(-1)^{\deg T}T$	the sign rule
$[S,T]_\epsilon=ST-\epsilon TS$	the graded commutator, $\epsilon=(-1)^{\deg S\deg T}$
$[L_a,T]_\epsilon^{}=-\epsilon[L_{\sigma(a)},T^{}]_\epsilon$	the sign rule for the adjoint of the bracket
$(S^{\alpha}_{a,b})^{*}=S^{\alpha}_{\beta(a),\beta(b)}$	the adjoint action on the signed sandwich