The Graded Adjoint Action on a Module over an Algebra

Introduction

On a graded module over a graded algebra the adjoint operation acquires a sign: passing an operator of degree $|S|$ past one of degree $|T|$ past the pairing costs the Koszul sign $(-1)^{|S||T|}$, and the adjoint of a product is

$$ (ST)^{*}=(-1)^{|S||T|}\,T^{*}S^{*}, $$

the graded involution of the operator algebra, ordinary on the even operators and signed on the odd ones. The graded adjoint and the sign-free adjoint $\dagger$ of the earlier entries of the group differ by the grade involution alone, $T^{*}=\alpha_M^{\lvert T\rvert}T^{\dagger}$, so the grading, the sign and the involution are three readings of one operation. The graded action $\rho : A \to \operatorname{End}_k(M)$, $a\mapsto\rho(a)$, carries a homogeneous element of the algebra to an operator of the same degree, and its adjoint is computed by the same sign rule. The graded adjoint action is the graded commutator $\mathrm{ad}_x(y)=xy-(-1)^{|x||y|}yx$, the inner derivation of the graded algebra; as an operator it reads $\mathrm{ad}_x=L_x-R_x\alpha^{|x|}$, the grade involution acting on the argument, and its adjoints are explicit: $\mathrm{ad}_x^{\dagger}=R_x-\alpha^{|x|}L_x$ for the sign-free pairing and $\mathrm{ad}_x^{*}=\alpha^{|x|}\mathrm{ad}_x^{\dagger}$ for the graded one, so that an even $x$ gives $-\mathrm{ad}_x$ for both, while an odd $x$ produces the mirrored one-sided operators $R_x-\alpha L_x$ and $\alpha R_x-L_x$, which are not again adjoint actions.

This article fixes the graded pairing on a graded module, computes the adjoint of a homogeneous operator and the Koszul sign rule, relates the graded adjoint to the graded involution $\alpha$ of the algebra and the module, records the adjoint of the action map, and treats the graded adjoint action with its degree and its two adjoints. It assumes The Graded Action on a Module over an Algebra for the graded module, the action map $\rho$ and the sign rule of the graded commutator, The Operators on an Algebra and The Sandwich Operator on an Algebra for $L_a$, $R_a$ and $T_{a,b}$, The Commutator Operator for the inner derivations, Involutions of the Operator Algebra, The Adjoint in an Involutive Algebra and Unitary Operators of an Involutive Algebra for the adjoint operation, the two pairings and the unitary operators, The Adjoint of the Left Multiplication on an Algebra, The Signed Adjoint Sandwich on an Algebra and The Signed Adjoint of the Left Multiplication on an Algebra for the one-sided and the signed adjoints, Involutive Linear Algebras, Involutive Graded Algebras and The Self-Adjoint Part of an Algebra for the involutions, the grading and the fixed parts, and Superalgebras and Graded Structures for the Koszul rule and the symmetric monoidal structure. The module case over a graded algebra is The Graded Adjoint Action on a Module over a Graded Algebra of the later category Symmetric Linear Algebras; the analytic case is Part II. This article stays inside Part I: no distance, norm, form with a norm, topology or limit.

Throughout, $k$ is a field of characteristic not two, $A$ is a finite-dimensional unital associative $k$-algebra with an involutive automorphism $\alpha$ and the induced grading $A=A^0\oplus A^1$ of The Signed Sandwich on an Algebra, and $\tau$ is a trace with nondegenerate product pairing on $A$. The graded module is $M=M^0\oplus M^1$ over $A$ in the sense of The Graded Action on a Module over an Algebra, the action map is $\rho(a)(m)=a\cdot m$, the grade involution of the module is the operator $\alpha_M(m)=(-1)^{|m|}m$, homogeneous elements carry the parity $|x|\in\{0,1\}$ and the sign $(-1)^{|x||y|}$ is the Koszul sign, and the pairing on $M$ is graded, supersymmetric and nondegenerate. The sign-free adjoint of the earlier entries is written $T^{\dagger}$, the graded adjoint $T^{*}$, the one-sided operators are $L_a$, $R_a$ on $A$, and $\mathrm{ad}_x(y)=xy-(-1)^{|x||y|}yx$.

The Graded Pairing and the Koszul Adjoint

Definition. A pairing $\langle\cdot,\cdot\rangle : M \times M \to k$ on the graded module is graded when it is biadditive and homogeneous of degree zero, $\lvert\langle x,y\rangle\rvert=\lvert x\rvert+\lvert y\rvert$, so that it vanishes on a pair of opposite parity; it is supersymmetric when

$$ \langle y,x\rangle=(-1)^{\lvert x\rvert\lvert y\rvert}\langle x,y\rangle , $$

and nondegenerate when $\langle x,M\rangle=0$ and $\langle M,y\rangle=0$ force $x=0$ and $y=0$. A graded, supersymmetric, nondegenerate pairing is a graded pairing of the category.

Definition. Let $\langle\cdot,\cdot\rangle$ be a graded nondegenerate pairing on $M$ and let $T$ be a homogeneous operator of degree $\lvert T\rvert$. The graded adjoint of $T$ is the operator $T^{*}$ defined by

$$ \langle Tx,y\rangle=(-1)^{\lvert T\rvert\lvert x\rvert}\langle x,T^{*}y\rangle $$

for homogeneous $x$ and extended by linearity; dropping the sign gives the sign-free adjoint $T^{\dagger}$ of the earlier entries of the group,

$$ \langle Tx,y\rangle=\langle x,T^{\dagger}y\rangle . $$

Theorem. Both adjoints exist and are unique, they are homogeneous of the same degree, $\lvert T^{*}\rvert=\lvert T^{\dagger}\rvert=\lvert T\rvert$, and for homogeneous $S,T$

$$ (ST)^{*}=(-1)^{\lvert S\rvert\lvert T\rvert}\,T^{*}S^{*}, \qquad (ST)^{\dagger}=T^{\dagger}S^{\dagger}, \qquad (T^{*})^{*}=T, \qquad \mathrm{id}^{*}=\mathrm{id} . $$

The graded adjoint is therefore an anti-automorphism of order two up to the Koszul sign, and the sign-free adjoint one of order two without a sign: the two rules differ exactly on a pair of odd operators and agree on everything else, which is the graded involution of the operator algebra.

Proof. Existence and uniqueness are those of Involutions of the Operator Algebra applied to each homogeneous degree: the functional $x\mapsto(-1)^{|T||x|}\langle Tx,y\rangle$ is linear and is represented by a unique $T^{*}y$, and the representing element is homogeneous of degree $|T|+|y|$ because the pairing has degree zero. For the sign rule, move $S$ and then $T$ across the pairing: $\langle STx,y\rangle=(-1)^{|S||Tx|}\langle Tx,S^{*}y\rangle=(-1)^{|S|(|T|+|x|)}(-1)^{|T||x|}\langle x,T^{*}S^{*}y\rangle=(-1)^{|S||T|}(-1)^{(|S|+|T|)|x|}\langle x,T^{*}S^{*}y\rangle$, which is the defining identity of $(-1)^{|S||T|}T^{*}S^{*}$; the sign-free rule is the same computation with both signs dropped; the order-two and the unit statements are the defining identity read twice and at the identity.

Corollary. On the even operators the Koszul sign is $1$ and the graded adjoint is the sign-free adjoint of Involutions of the Operator Algebra; on products of odd operators the sign survives, and the composite of two odd operators is adjointed with the sign $-1$. The self-adjoint operators satisfy $T^{*}=T$ and the skew-adjoint ones $T^{*}=-T$ on each degree, and the decomposition $T=\tfrac12(T+T^{*})+\tfrac12(T-T^{*})$ is the one of The Self-Adjoint Part of an Algebra, taken degree by degree.

Proof. The sign $(-1)^{|S||T|}$ is $1$ unless both operators are odd, which gives the first two statements; the decomposition is the splitting of a degree-zero operator into fixed and negated parts, and each homogeneous component is carried to itself because the adjoint preserves the degree.

The Graded Involution and the Adjoint

Definition. The grade involution of the module is the operator

$$ \alpha_M(m)=(-1)^{\lvert m\rvert}m ; $$

it is an operator of order two, it is self-adjoint for every graded pairing, and it is the module analogue of the grade involution $\alpha$ of the algebra.

Theorem. The graded adjoint and the sign-free adjoint differ by the grade involution alone,

$$ T^{*}=\alpha_M^{\lvert T\rvert}\,T^{\dagger}, $$

where $\alpha_M^{\lvert T\rvert}$ is the identity on the even operators and the grade involution applied to the values on the odd ones; consequently the two adjoints agree exactly on the even operators. The pairing twisted by the grade involution,

$$ \{x,y\}_\alpha=\langle x,\alpha_My\rangle , $$

is graded and supersymmetric, and the graded adjoint for the twisted pairing is the conjugate of the graded adjoint for the plain one,

$$ T^{*_\alpha}=\alpha_M\,T^{*}\,\alpha_M=c_{\alpha_M}(T^{*}), $$

so the graded adjoint is moved by the grade involution when the pairing is.

Proof. For homogeneous $T$ the two defining identities read $(-1)^{|T||x|}\langle x,T^{*}y\rangle=\langle Tx,y\rangle=\langle x,T^{\dagger}y\rangle$ for every homogeneous $x$ and $y$; comparing the even $x$ and the odd $x$ separately, and using that each graded piece of $M$ is nondegenerate for a pairing of degree zero, the even parts of $T^{*}y$ and $T^{\dagger}y$ agree and their odd parts are opposite when $|T|=1$, while for $|T|=0$ the two identities coincide; that is $T^{*}=\alpha_M^{|T|}T^{\dagger}$ on homogeneous $y$, and hence on all of $M$. Supersymmetry of $\{\cdot,\cdot\}_\alpha$ is the computation $\{y,x\}_\alpha=\langle y,\alpha_Mx\rangle=(-1)^{|x||y|}\langle\alpha_Mx,y\rangle=(-1)^{|x||y|}\langle x,\alpha_My\rangle$, using the supersymmetry of $\langle\cdot,\cdot\rangle$ and the self-adjointness of $\alpha_M$; the conjugation identity is the dictionary $T^{*_\sigma}=c_{\sigma}(T^{*})$ of The Adjoint in an Involutive Algebra with the grade involution as the twist.

Corollary. The invariants of the grade involution — the even part of the module — are the elements on which $\alpha_M=\mathrm{id}$, and the twisted pairing restricts to the plain one there; the odd part is the negated part, on which the twisted pairing differs from the plain one by the sign $-1$.

Proof. The two eigenspaces of $\alpha_M$ are the graded pieces, and the twisted pairing is $\{x,y\}_\alpha=\langle x,y\rangle$ for even $y$ and $-\langle x,y\rangle$ for odd $y$.

The Graded Action and its Adjoint

Theorem. The action map $\rho : A \to \operatorname{End}_k(M)$ is a homomorphism of unital algebras, it carries a homogeneous element to a homogeneous operator of the same degree, and

$$ \rho(a)\bigl(M^j\bigr) \subseteq M^{\,j+\lvert a\rvert} ; $$

the adjoint of the action is computed by the Koszul rule, $(\rho(a)\rho(b))^{*}=(-1)^{\lvert a\rvert\lvert b\rvert}\rho(b)^{*}\rho(a)^{*}$, and $\rho(\alpha(a))=\alpha_M\rho(a)\alpha_M$ for the grade involutions.

Proof. The homomorphism property, the degree and the containment are The Graded Action on a Module over an Algebra; the Koszul rule is the theorem of the first section applied to the operators $\rho(a)$ and $\rho(b)$, whose degrees are $\lvert a\rvert$ and $\lvert b\rvert$; the last identity is the multiplicativity of $\alpha$ read through the action, $\alpha(a)\alpha_M(m)=\alpha_M(am)$, which is the degree rule $|am|=|a|+|m|$ in operator form.

Theorem (the one-sided operators). For homogeneous $x$, on the algebra read as a module over itself,

$$ L_x^{*}=\alpha^{\lvert x\rvert}R_x, \qquad R_x^{*}=\alpha^{\lvert x\rvert}L_x , $$

so the graded adjoint preserves the side and inserts the grade involution when the element is odd, while the sign-free adjoint exchanges the sides, $L_x^{\dagger}=R_x$ and $R_x^{\dagger}=L_x$, as in The Adjoint of the Left Multiplication on an Algebra; the two statements are the identity $T^{*}=\alpha_M^{|T|}T^{\dagger}$ at $T=L_x$ and $T=R_x$.

Proof. The sign-free identity is the one-line computation $\langle xu,v\rangle=\tau(xuv)=\tau(uvx)=\langle u,vx\rangle$ of the cited article, which gives $L_x^{\dagger}=R_x$, and the same computation on the other side gives $R_x^{\dagger}=L_x$; applying $T^{*}=\alpha_M^{|T|}T^{\dagger}$ to the two gives the graded statements.

Corollary. For an even element the graded and the sign-free adjoints coincide on the one-sided operators, $L_x^{*}=L_x^{\dagger}=R_x$; for an odd element the graded adjoint of $L_x$ is the right multiplication twisted by the grade involution, $\alpha R_x=L_{\alpha(x)}\alpha$. In particular $L_x$ is self-adjoint for the graded pairing exactly when $\lvert x\rvert$ is even and $x$ is central.

Proof. The coincidence is $T^{*}=\alpha_M^{|T|}T^{\dagger}$ at $|T|=0$; for odd $x$ the identity $\alpha R_x=L_{\alpha(x)}\alpha$ is the multiplicativity of $\alpha$; self-adjointness needs $L_x^{*}=L_x$, which by the theorem is $R_x=L_x$ for even $x$, that is centrality, and for odd $x$ would require the grade involution to fix every element of the algebra, which it does only when the grading is trivial.

The Graded Adjoint Action

Definition. For $x \in A$ the graded adjoint action of $x$ is

$$ \mathrm{ad}_x(y)=xy-(-1)^{\lvert x\rvert\lvert y\rvert}yx , $$

the graded commutator; for homogeneous $x$ it is a linear operator on $A$ and, through the action, on every graded module over $A$.

Theorem. For homogeneous $x$,

$$ \mathrm{ad}_x=L_x-R_x\alpha^{\lvert x\rvert}, $$

the grade involution acting on the argument; the operator is a graded derivation of degree $\lvert x\rvert$,

$$ \mathrm{ad}_x(yz)=\mathrm{ad}_x(y)z+(-1)^{\lvert x\rvert\lvert y\rvert}y\,\mathrm{ad}_x(z) , $$

it satisfies the graded antisymmetry and the graded Jacobi identity, so the graded commutator makes $A$ a graded Lie algebra and $\mathrm{ad}$ a representation of it with kernel the graded centre.

Proof. On homogeneous $y$ the second term is $(-1)^{|x||y|}yx=(R_x\alpha^{|x|})y$, which is the operator identity and then, by linearity, the general one; the graded Leibniz rule, the graded antisymmetry and the graded Jacobi identity are The Graded Action on a Module over an Algebra and The Commutator Operator with the Koszul signs inserted, and the kernel is the graded centre because $\mathrm{ad}_x=0$ says exactly that $x$ graded-commutes with every element.

Theorem (the adjoints). For homogeneous $x$, with respect to the sign-free and the graded pairings,

$$ \mathrm{ad}_x^{\dagger}=R_x-\alpha^{\lvert x\rvert}L_x, \qquad \mathrm{ad}_x^{*}=\alpha^{\lvert x\rvert}\mathrm{ad}_x^{\dagger} . $$

Consequently, for even $x$ both adjoints are the negative of the operator itself,

$$ \mathrm{ad}_x^{\dagger}=\mathrm{ad}_x^{*}=-\mathrm{ad}_x , $$

so the graded adjoint action of an even element is skew-adjoint for both pairings; for odd $x$ the two adjoints are the mirrored one-sided operators

$$ \mathrm{ad}_x^{\dagger}=R_x-\alpha L_x, \qquad \mathrm{ad}_x^{*}=\alpha R_x-L_x , $$

which are not of the form $-\mathrm{ad}_z$: the sign of the graded commutator sits on the argument, and moving it through the adjoint moves it to the other side of the second term.

Proof. Adjoint the two terms of $\mathrm{ad}_x=L_x-R_x\alpha^{|x|}$ for the sign-free pairing: $L_x^{\dagger}=R_x$, and $(R_x\alpha^{|x|})^{\dagger}=(\alpha^{|x|})^{\dagger}R_x^{\dagger}=\alpha^{|x|}L_x$ because the grade involution is self-adjoint and of degree zero; this gives the first identity, and the second is $T^{*}=\alpha_M^{|T|}T^{\dagger}$ at $\lvert T\rvert=\lvert x\rvert$. For even $x$ the power of the grade involution is the identity, so both adjoints are $R_x-L_x=-\mathrm{ad}_x$. For odd $x$ the sign-free adjoint is $R_x-\alpha L_x$, whose second term carries the involution to the left of the argument, while $-\mathrm{ad}_z=-L_z+R_z\alpha$ carries it to the right; the two agree for no $z$ unless $\alpha$ fixes every element, which is the trivial grading.

Corollary. The graded adjoint action of an even element is the ordinary inner derivation $\mathrm{ad}_x=L_x-R_x$ of The Commutator Operator, and it is skew-adjoint for both the graded and the sign-free pairings; the adjoint action of an odd element is the graded derivation $L_x-R_x\alpha$ of The Graded Action on a Module over an Algebra, whose adjoints are the mirrored one-sided operators above. In particular the graded adjoint action of an even element is self-adjoint for no nonzero $x$.

Proof. Both statements are the theorem written for each parity; the even case is the inner derivation of the cited article, and self-adjointness would require $-\mathrm{ad}_x=\mathrm{ad}_x$, that is $2\,\mathrm{ad}_x=0$.

Examples

(a) The exterior algebra. For $A=\Lambda(V)$ with the degree grading and the natural graded pairing, the graded adjoint action of a vector $v$ is $\mathrm{ad}_v(\omega)=v\omega-(-1)^{\lvert\omega\rvert}\omega v$, a graded derivation of degree one; its sign-free adjoint is $R_v-\alpha L_v$ and its graded adjoint is $\alpha R_v-L_v$, the two mirrored odd operators of the theorem.

(b) The Clifford algebra. With the Clifford product and the grading by degree, the adjoint action of a vector is the commutator on the even part and the anticommutator on the odd part; the even part of the algebra acts by skew-adjoint derivations for the plain pairing, and the odd part by the operators of the odd case above. The metric reading is Part II.

(c) The matrix superalgebra. $A=M_{p|q}$ with the transpose and the grading by blocks: the adjoint action is the graded commutator of matrices, the Koszul sign appears the moment two odd matrices are multiplied, and the adjoint of a product is $(ST)^{*}=(-1)^{|S||T|}T^{*}S^{*}$.

(d) The trivial grading. If $\lvert a\rvert=0$ for all $a$ the Koszul sign is $1$, the graded pairing is ordinary, the graded adjoint action is the plain commutator $\mathrm{ad}_x=L_x-R_x$, and the article reduces to Involutions of the Operator Algebra, The Adjoint of the Left Multiplication on an Algebra and The Commutator Operator.

Summary

On a graded module over a graded algebra a graded nondegenerate supersymmetric pairing $\langle y,x\rangle=(-1)^{|x||y|}\langle x,y\rangle$ gives every homogeneous operator a homogeneous adjoint of the same degree with the Koszul sign rule $(ST)^{*}=(-1)^{|S||T|}T^{*}S^{*}$; dropping the sign gives the sign-free adjoint $\dagger$ with $(ST)^{\dagger}=T^{\dagger}S^{\dagger}$, and the two are related by the grade involution alone, $T^{*}=\alpha_M^{|T|}T^{\dagger}$, so they agree exactly on the even operators and the adjoint operation is a graded involution, ordinary on the even operators and signed on the odd ones. The grade involution $\alpha_M$ of the module gives the twisted pairing $\{x,y\}_\alpha=\langle x,\alpha_My\rangle$, whose graded adjoint is the conjugate $T^{*_\alpha}=c_{\alpha_M}(T^{*})$. The graded action $\rho$ sends a homogeneous element to an operator of the same degree and is compatible with the grade involution, $\rho(\alpha(a))=\alpha_M\rho(a)\alpha_M$. The one-sided operators have $L_x^{*}=\alpha^{|x|}R_x$ and $R_x^{*}=\alpha^{|x|}L_x$, so the graded adjoint preserves the side up to the involution, against $L_x^{\dagger}=R_x$ for the sign-free one. The graded adjoint action reads $\mathrm{ad}_x=L_x-R_x\alpha^{|x|}$, a graded derivation of degree $|x|$; its sign-free adjoint is $R_x-\alpha^{|x|}L_x$ and its graded adjoint is $\alpha^{|x|}$ times that, so an even $x$ gives $-\mathrm{ad}_x$ for both, while an odd $x$ gives the mirrored one-sided operators $R_x-\alpha L_x$ and $\alpha R_x-L_x$. When the grading is trivial the Koszul sign is $1$, the graded adjoint is the sign-free adjoint, and the whole article returns the sign-free adjoint involution and the ordinary inner derivation.

Summary of Notation

Symbol	Meaning
$\lvert a\rvert$	parity of a homogeneous element
$(-1)^{\lvert a\rvert\lvert b\rvert}$	Koszul sign
$\langle y,x\rangle=(-1)^{\lvert x\rvert\lvert y\rvert}\langle x,y\rangle$	supersymmetric graded pairing
$\alpha$, $\alpha_M$	grade involution of the algebra and of the module
$T^{}$, $\langle Tx,y\rangle=(-1)^{\lvert T\rvert\lvert x\rvert}\langle x,T^{}y\rangle$	graded adjoint
$T^{\dagger}$, $\langle Tx,y\rangle=\langle x,T^{\dagger}y\rangle$	sign-free adjoint of the earlier entries
$(ST)^{}=(-1)^{\lvert S\rvert\lvert T\rvert}T^{}S^{*}$, $(ST)^{\dagger}=T^{\dagger}S^{\dagger}$	Koszul sign rule and its sign-free case
$T^{*}=\alpha_M^{\lvert T\rvert}T^{\dagger}$	the two adjoints differ by the grade involution
$\{x,y\}_\alpha=\langle x,\alpha_My\rangle$, $T^{_\alpha}=c_{\alpha_M}(T^{})$	the twisted pairing and its adjoint
$\rho$, $\rho(\alpha(a))=\alpha_M\rho(a)\alpha_M$	the action map
$L_x^{}=\alpha^{\lvert x\rvert}R_x$, $R_x^{}=\alpha^{\lvert x\rvert}L_x$	graded adjoints of the one-sided operators
$L_x^{\dagger}=R_x$	sign-free adjoint of the left multiplication
$\mathrm{ad}_x(y)=xy-(-1)^{\lvert x\rvert\lvert y\rvert}yx$	the graded adjoint action
$\mathrm{ad}_x=L_x-R_x\alpha^{\lvert x\rvert}$	the operator form
$\mathrm{ad}_x^{\dagger}=R_x-\alpha^{\lvert x\rvert}L_x$, $\mathrm{ad}_x^{*}=\alpha^{\lvert x\rvert}\mathrm{ad}_x^{\dagger}$	the sign-free and the graded adjoints
even $x$: $\mathrm{ad}_x^{\dagger}=\mathrm{ad}_x^{*}=-\mathrm{ad}_x$	the even case
odd $x$: $\mathrm{ad}_x^{\dagger}=R_x-\alpha L_x$, $\mathrm{ad}_x^{*}=\alpha R_x-L_x$	the odd case
$\lvert a\rvert=0$	trivial grading; the sign-free case