Multiplication Operators on a Commutative Algebra

Introduction

Every element $a$ of an algebra $A$ carries the operator of multiplication by it, $x \mapsto ax$. When $A$ is commutative this single family of operators is closed under composition and under addition, and the assignment $a \mapsto L_a$ is an algebra homomorphism

$$ L : A \longrightarrow \operatorname{End}_R(A), \qquad L_a(x) = ax , $$

whose image is a copy of $A$ inside the ambient operator algebra. This image is the regular representation of $A$, and in the commutative case it is the whole story: the left multiplication and the right multiplication by $a$ coincide, the image is its own centralizer, and every derivation of $A$ is read on it as an operator.

The first purpose of the article is to assemble the multiplication operators and to record their algebra. The second is to read the Gelfand transform as an operator. A character of $A$ — a unital algebra homomorphism into the ground ring — is a common eigenvector of the whole family $L(A)$: on it, $L_a$ acts by the scalar the character assigns to $a$. Collecting the scalars over all characters gives the Gelfand transform $a \mapsto (\chi \mapsto \chi(a))$, and the statement that $L_a$ is multiplication by that function is the algebraic content of the transform. The topological reading of the character set — its topology, its compactness, the fact that the transform is an isometry for a norm — belongs to Topological Algebras and the Gelfand Transform in Part II and Banach Algebras in Part III, and is named here only as a forward reference: no distance, no norm and no open set occurs in this article.

Throughout, $R$ is a commutative ring with identity $1 \neq 0$ and $A$ is a commutative, associative and unital $R$-algebra. The unital hypothesis is used where an operator is recovered from its value at $1$; it is stated at each such point. The non-commutative case, where $L_a$ and $R_a$ differ, is Left and Right Multiplication in a Ring; the general ambient space of all operators is The Operators on an Algebra; the derivations are Derivations of a Ring, and their commutative-algebra form is the article The Derivations of a Commutative Algebra of this category.

The Multiplication Operators

Definition

Definition. For $a \in A$ the multiplication operator by $a$ is the $R$-linear map

$$ L_a : A \longrightarrow A, \qquad L_a(x) = ax . $$

Because $A$ is commutative, $L_a$ is also the right multiplication $x \mapsto xa$, so there is one family and not two. The assignment is $R$-linear in the subscript, and the following identities are immediate from associativity and from the commutativity of $A$.

Proposition. For all $a, b \in A$ and $r \in R$,

$$ L_{a+b} = L_a + L_b, \qquad L_{ab} = L_a \circ L_b = L_b \circ L_a, \qquad L_{ra} = rL_a, \qquad L_1 = \mathrm{id}_A . $$

Proof. Each identity is checked on an element $x$: $L_{a+b}(x) = (a+b)x = ax + bx$, $L_{ab}(x) = (ab)x = a(bx) = L_a(L_b x)$, $L_{ra}(x) = (ra)x = r(ax) = (rL_a)(x)$, and $L_1(x) = x$. Commutativity gives $L_aL_b = L_{ab} = L_{ba} = L_bL_a$. $\square$

Corollary. The image $L(A) = \{L_a : a \in A\}$ is a commutative subalgebra of $\operatorname{End}_R(A)$ with identity $L_1$, and $L : A \to L(A)$ is a surjective $R$-algebra homomorphism. If $A$ is unital then $L$ is injective, because $L_a = 0$ forces $a = L_a(1) = 0$; hence

$$ A \cong L(A) $$

as $R$-algebras, and the regular representation $L$ is a faithful representation of $A$ by multiplication operators.

The injectivity is exactly the point at which the unit is used, and it is what makes the regular representation a copy of the algebra rather than a quotient of it. Without a unit the kernel is the annihilator $\{a : aA = 0\}$, which may be nonzero.

Example. For $A = R$ the single operator $L_a$ is multiplication by the scalar $a$ in the one-dimensional space $R$, so $L(R) = \operatorname{End}_R(R) \cong R$.

Example. For $A = R[x]$, the operator $L_x$ is the shift $x^k \mapsto x^{k+1}$ on the monomial basis and $L_f = f(L_x)$ for a polynomial $f$; the map $L$ is the evaluation $f \mapsto f(L_x)$. Every $L_f$ raises the degree by $\deg f$, so each $L_f$ has no nonzero eigenvalue on the polynomial algebra and has infinite order of growth on the monomials. The operator $L_x$ is the shift, and the regular representation is the classical realisation of a polynomial algebra by a shift.

The Centralizer and the Double Centralizer

The multiplication operators are exactly the operators that commute with all of them.

Theorem. In $\operatorname{End}_R(A)$, the centralizer of $L(A)$ is $L(A)$ itself:

$$ \{T \in \operatorname{End}_R(A) : T L_a = L_a T \ \text{for all } a \in A\} = L(A) . $$

Equivalently, every $A$-linear endomorphism of the regular module $A$ is a multiplication operator, and $A \cong \operatorname{End}_A(A)$.

Proof. An operator $T$ commutes with every $L_a$ precisely when $T(ax) = a\,T(x)$ for all $a, x$, which is $A$-linearity of the regular module; putting $a = x$ gives $T(a) = a\,T(1) = L_{T(1)}(a)$, so $T = L_b$ with $b = T(1)$. Conversely every $L_b$ is $A$-linear by associativity. Here the unit is again used, in the evaluation at $1$. $\square$

Corollary. When $A$ is free of finite rank over the field $k$, the subalgebra $L(A)$ is a maximal commutative subalgebra of $\operatorname{End}_k(A)$: it equals its own centralizer and is commutative. The double centralizer theorem for the regular module therefore terminates at the first step.

Remark. For a non-commutative algebra the two families $L(A)$ and $R(A)$ are different, they commute with one another, and each is the other's centralizer in the appropriate sense; the coincidence above is peculiar to the commutative case, and it is why one representation suffices here.

The Interaction with Derivations

A derivation of $A$ is intertwined with the multiplication operators in a single identity, and the identity says that a derivation is determined by how it moves the parameters.

Proposition. Let $\delta \in \operatorname{Der}_R(A)$ be a derivation, $\delta(ab) = \delta(a)b + a\delta(b)$. Then for every $a \in A$,

$$ \delta \circ L_a = L_{\delta(a)} + L_a \circ \delta, \qquad \text{equivalently} \qquad [\delta, L_a] = L_{\delta(a)} . $$

Proof. Apply both sides to $x$: $\delta(ax) = \delta(a)x + a\delta(x) = L_{\delta(a)}(x) + L_a(\delta x)$. $\square$

Corollary. The $R$-linear map $\operatorname{Der}_R(A) \to L(A)$, $\delta \mapsto L_{\delta(a)}$ for a fixed $a$, is the restriction of the adjoint action of $\operatorname{Der}_R(A)$ on the operator algebra; on the whole of $L(A)$ the derivation acts by $L_b \mapsto L_{\delta(b)}$, and this is the derivation of the commutative algebra $L(A) \cong A$ carried across the isomorphism. In particular the derivation space depends only on the algebra structure of $A$, as it must.

Example. For $A = R[x]$ the derivations form the free module $R[x]\partial_x$ of rank one, and $[\partial_x, L_f] = L_{f'}$ where $f' = \partial_x f$ is the formal derivative. The operator identity $[\partial_x, L_f] = L_{f'}$ is the operator form of the product rule.

Characters and the Gelfand Transform

Characters

Definition. A character of $A$ over $R$ is a unital $R$-algebra homomorphism $\chi : A \to R$, so that $\chi(1) = 1$, $\chi(a+b) = \chi(a)+\chi(b)$ and $\chi(ab) = \chi(a)\chi(b)$. The character set is

$$ \mathfrak X(A) = \operatorname{Hom}_{\mathsf{CAlg}_R}(A, R) . $$

It is a set, and it is written here as a set: the topology it carries in the analytic theory is not used.

Proposition. Every character is a common eigenvector of the family $L(A)$: for every $a \in A$,

$$ \chi \circ L_a = \chi(a)\,\chi , $$

that is, $\chi$ is a linear functional on $A$ that is an eigenvector of the transposed operator $L_a^{\mathsf T}$ with eigenvalue $\chi(a)$. Equivalently, the ideal $\ker\chi$ is a maximal ideal of $A$ and $A/\ker\chi \cong R$.

Proof. $\chi(L_a x) = \chi(ax) = \chi(a)\chi(x) = \chi(a)\,\chi(x)$; the quotient statement is the first isomorphism theorem for algebras, and $\ker\chi$ is maximal because $A/\ker\chi$ is a subalgebra of the field $R$ (or of $R$ itself) containing $1$. $\square$

The eigenvalue assigned to $a$ by the character is forced: a character is a multiplicative functional, and the eigenvalues of the family are exactly the values of the characters.

Proposition (simultaneous triangularisation). Suppose $A$ is a finitely generated free $R$-module and that the characters separate the elements of $A$, in the sense that $\chi(a) = 0$ for all $\chi$ forces $a = 0$. Then the family $L(A)$ is simultaneously diagonalisable in a suitable extension of scalars: there is a basis of $A \otimes_R S$ over a ring $S$ in which every $L_a$ is diagonal, with the diagonal entries the values $\chi(a)$.

Proof. The hypothesis exhibits $A$ as a subalgebra of the algebra of functions $\mathfrak X(A) \to R$ and hence of a product of copies of $R$, one for each character; in the corresponding idempotent decomposition $A \otimes_R S \cong \prod_\chi S$, the operator $L_a$ is multiplication by the scalar $\chi(a)$ in the $\chi$-th coordinate. $\square$

For a finite-dimensional reduced commutative algebra over an algebraically closed field $k$, the characters are exactly the $k$-algebra homomorphisms into $k$, they separate points, and the proposition reduces to the classical diagonalisation of a commuting family of diagonalisable operators.

The Gelfand Transform

Definition. The Gelfand transform of $A$ is the map

$$ \Gamma : A \longrightarrow \operatorname{Fun}(\mathfrak X(A), R), \qquad \Gamma(a)(\chi) = \chi(a), $$

into the $R$-algebra of functions on the character set with pointwise operations.

Theorem. $\Gamma$ is a unital $R$-algebra homomorphism, and under it the multiplication operator $L_a$ becomes multiplication of functions by $\Gamma(a)$:

$$ \Gamma(L_a x) = \Gamma(a)\,\Gamma(x), \qquad \text{that is} \qquad \widehat{ax} = \hat a\,\hat x , $$

where $\hat a = \Gamma(a)$.

Proof. For $a, b \in A$ and a character $\chi$, $\Gamma(a+b)(\chi) = \chi(a+b) = \chi(a)+\chi(b)$ and $\Gamma(ab)(\chi) = \chi(ab) = \chi(a)\chi(b) = \Gamma(a)(\chi)\Gamma(b)(\chi)$, so $\Gamma$ is additive and multiplicative; $\Gamma(1)(\chi) = 1$. The second display is the defining property $\chi(ax) = \chi(a)\chi(x)$. $\square$

The kernel of $\Gamma$ is the Jacobson radical — the intersection of the maximal ideals, that is, the elements annihilated by every character — and $\Gamma$ is injective exactly when the characters separate points, which is the hypothesis of the triangularisation proposition. For a reduced algebra of finite type over a field the radical vanishes and the transform is injective.

Remark (what is deferred). The transform is treated here as an algebraic homomorphism into a ring of functions. Its analytic theory — that when $R = \mathbb{C}$ and $A$ carries a norm the character set becomes a compact space, the transform is an isometry, and it is an isomorphism onto the algebra of continuous functions for a suitable class of algebras — needs the distance of Part II and the limit of Part III. The reader will find it in Topological Algebras and the Gelfand Transform and Banach Algebras; nothing of it is used here.

Worked Examples

The Split-Complex Numbers

Let $A = \mathbb D = R[x]/(x^2-1)$ over $R = \mathbb R$, with basis $1, j$ and $j^2 = 1$. In this basis

$$ L_1 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \qquad L_j = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad L_{a+bj} = \begin{pmatrix} a & b \\ b & a \end{pmatrix} , $$

so $L(A)$ is the algebra of matrices of the form $\binom{a\ b}{b\ a}$, a two-dimensional commutative subalgebra of $M_2(\mathbb R)$. The characters are the two homomorphisms $\chi_\pm$ with $\chi_\pm(j) = \pm 1$, and the eigenvalues of $L_j$ are $1$ and $-1$, so $L_j$ is a reflection, of order two. The Gelfand transform sends $a+bj$ to the pair $(\chi_+,\chi_-) = (a+b, a-b)$, which is the isomorphism $\mathbb D \cong \mathbb R \times \mathbb R$ of the idempotent decomposition.

A Split Polynomial Algebra

Let $A = k[x]/(f)$ with $f$ monic of degree $n$ splitting into $n$ distinct linear factors over $k$: $f = (x-r_1)\cdots(x-r_n)$. Then $A \cong k^n$ by the Chinese remainder theorem and the characters are the evaluations $\chi_i(x) = r_i$. The multiplication operator $L_x$ is the companion-type matrix acting on the basis $1, x, \dots, x^{n-1}$,

$$ L_x = \begin{pmatrix} 0 & 0 & \cdots & 0 & -c_0 \\ 1 & 0 & \cdots & 0 & -c_1 \\ 0 & 1 & \cdots & 0 & -c_2 \\ \vdots & & \ddots & & \vdots \\ 0 & 0 & \cdots & 1 & -c_{n-1} \end{pmatrix}, \qquad f(x) = x^n + c_{n-1}x^{n-1} + \cdots + c_0 , $$

and the Cayley–Hamilton theorem is the statement that its characteristic polynomial is $f$. Its eigenvalues are $r_1, \dots, r_n$, the values of the characters, and the Gelfand transform is the diagonalisation by the Vandermonde matrix, which is invertible exactly because the roots are distinct.

The Trace Pairing

Definition. Let $A$ be free of rank $n$ over $R$. The trace pairing is

$$ T : A \times A \longrightarrow R, \qquad T(a,b) = \operatorname{tr}(L_{ab}) = \operatorname{tr}(L_a L_b) . $$

It is $R$-bilinear and symmetric, because $L_{ab} = L_aL_b = L_bL_a$ and the trace is symmetric. It satisfies $T(a, 1) = \operatorname{tr}(L_a)$, the regular trace of $a$, and it is associative,

$$ T(ab, c) = T(a, bc), $$

because both sides are $\operatorname{tr}(L_{abc})$. The pairing is the coefficient of $a \mapsto \operatorname{tr}(L_a)$ read as a bilinear form.

Proposition. Every multiplication operator is self-adjoint for the trace pairing:

$$ T(L_a x, y) = T(x, L_a y) \qquad \text{for all } a, x, y \in A . $$

Proof. $T(L_a x, y) = \operatorname{tr}(L_{(ax)y}) = \operatorname{tr}(L_{a(xy)}) = T(x, ay) = T(x, L_a y)$, using associativity of $A$ in the middle. $\square$

Corollary. The family $L(A)$ is a commutative subalgebra of the space of operators self-adjoint for $T$. When $T$ is non-degenerate the adjoint operation makes $\operatorname{End}_R(A)$ an algebra with an involution, and $L(A)$ lies in its self-adjoint part; the involution is developed in the * Operator Theory group of this category, where the pairing and the adjoint are read with the involution of the elements.

Example. For $A = k^n$ the multiplication operator $L_a$ with $a = (a_1, \dots, a_n)$ is the diagonal matrix $\operatorname{diag}(a_1, \dots, a_n)$, and $T(a,b) = \sum_i a_i b_i$; the trace pairing is the standard bilinear form on $k^n$. For $A = \mathbb D$ above, $T(a+bj, c+dj) = 2(ac+bd)$ by the matrix form of $L$, the factor $2$ being the rank.

Summary

For a commutative unital $R$-algebra $A$, the multiplication operators $L_a(x) = ax$ satisfy $L_{ab} = L_aL_b = L_bL_a$ and $L_{a+b} = L_a + L_b$, so $L : A \to \operatorname{End}_R(A)$ is a faithful $R$-algebra homomorphism; its image is the regular representation $L(A)$, a commutative subalgebra isomorphic to $A$. The family is its own centralizer in the operator algebra: every operator commuting with all $L_a$ is a multiplication operator, which is the statement $\operatorname{End}_A(A) \cong A$ for the regular module. A derivation interacts with the family by $[\delta, L_a] = L_{\delta(a)}$.

The characters of $A$ are the unital homomorphisms $\chi : A \to R$, and each is a common eigenvector of the family, with $\chi \circ L_a = \chi(a)\chi$. Collecting the eigenvalues gives the Gelfand transform $\Gamma(a)(\chi) = \chi(a)$, a unital algebra homomorphism into the algebra of functions on the character set, under which $L_a$ is multiplication of functions by $\hat a$. The transform is injective exactly when the characters separate points, and its kernel is the Jacobson radical. The topological and analytic theory of the transform — the compact character space, the norm, the isometry — needs the distance of Part II and the limit of Part III and is deferred to Topological Algebras and the Gelfand Transform and Banach Algebras. The trace pairing $T(a,b) = \operatorname{tr}(L_a L_b)$ is symmetric and associative, and every multiplication operator is self-adjoint for it.

Summary of Notation

Symbol	Meaning
$R$	Commutative ring with identity $1 \neq 0$
$A$	Commutative unital associative $R$-algebra
$L_a$, $L_a(x) = ax$	Multiplication operator by $a$
$L : A \to \operatorname{End}_R(A)$	The regular representation
$L(A)$	Image of $L$, a commutative subalgebra of $\operatorname{End}_R(A)$
$[\delta, L_a] = L_{\delta(a)}$	A derivation acts on the parameters
$\mathfrak X(A) = \operatorname{Hom}_{\mathsf{CAlg}_R}(A,R)$	Character set of $A$
$\chi$	A character, $\chi(ab) = \chi(a)\chi(b)$, $\chi(1) = 1$
$\Gamma$, $\hat a = \Gamma(a)$	Gelfand transform; the function $\chi \mapsto \chi(a)$
$\ker\Gamma$	Jacobson radical, the intersection of the maximal ideals
$T(a,b) = \operatorname{tr}(L_aL_b)$	Trace pairing; symmetric, associative
$\operatorname{tr}(L_a)$	Regular trace of $a$