Left and Right Multiplication in a Banach Algebra

Introduction

The elementary operators of a Banach algebra are the two one-sided multiplications: for $a \in A$, the left multiplication $L_a(x) = ax$ and the right multiplication $R_a(x) = xa$. They carry the product of the algebra one factor at a time, and every other operator of this group is built from them by composition. The point of reading them as operators is the asymmetry between the two families: $a \mapsto L_a$ is a homomorphism, because the order of the elements is the order of the composition, while $a \mapsto R_a$ reverses the order and is a homomorphism from the opposite algebra. The two families commute with one another, and their failure to commute pairwise is exactly the noncommutativity of $A$.

This article develops that reading for a Banach algebra. It fixes the two families, computes their composition and commutation laws, shows that the regular representations are bounded and isometric on a unital algebra and that their images are closed, identifies the kernels and images of $L_a$ and $R_a$ with the annihilators and the principal ideals, and proves that the centraliser of one family is the other and that their common centraliser is the multiplications by the centre.

The article assumes the Banach algebra, its norm, the unit group and the spectrum from Topological Algebras and Banach Algebras; the abstract one-sided multiplications, their composition, their commutation and the double centraliser statement from Left and Right Multiplication in a Ring; the ideals, the annihilators and the unit group from Rings; and the bounded operators, the operator norm and the regular representation from Operators on a Banach Algebra, the first article of this group. The involution, the adjoint, the form and the signed operators are the later groups of this category and are not used. The normed multiplication algebra and the module structure are The Left and Right Multiplication Operators on a Banach Algebra; the multiplier algebra is Multipliers of a Banach Algebra; the derivation generated by an element is Operators on a Banach Algebra; no measure and no Fourier theory occurs.

Throughout, $\mathbb{K}$ is $\mathbb{R}$ or $\mathbb{C}$; $A$ is a Banach algebra over $\mathbb{K}$ with submultiplicative norm $\lVert\cdot\rVert$, unital with unit $1$ where a statement names $1$; $B(A)$ is the Banach algebra of bounded linear operators, $\lambda : a \mapsto L_a$ and $\rho : a \mapsto R_a$ the two regular representations, $A^\times$ the unit group and $Z(A)$ the centre. The commutator in $A$ is $[a,b] = ab - ba$.

The Two One-Sided Families

Definition. For $a \in A$ the left multiplication and the right multiplication by $a$ are the linear maps

$$ L_a : A \to A, \quad L_a(x) = ax , \qquad\qquad R_a : A \to A, \quad R_a(x) = xa . $$

Both are unital in the sense $L_a(1) = a = R_a(1)$ when $A$ is unital, so they cannot be separated by their value at the unit and are separated by their composition.

Proposition (composition and the opposite algebra). For all $a,b \in A$ and all $\lambda \in \mathbb{K}$,

$$ L_{a+b} = L_a + L_b , \quad L_{\lambda a} = \lambda L_a , \quad L_{ab} = L_aL_b , $$

and the same statements hold for $R$ with the order of the composition reversed,

$$ R_{a+b} = R_a + R_b , \quad R_{\lambda a} = \lambda R_a , \quad R_{ab} = R_bR_a . $$

Hence $a \mapsto L_a$ is a unital algebra homomorphism $A \to B(A)$, the left regular representation, while $a \mapsto R_a$ is a unital algebra homomorphism $A^{\mathrm{op}} \to B(A)$, equivalently an anti-homomorphism $A \to B(A)$.

Proof. Additivity and homogeneity in the parameter are the distributive laws. For the composition, $L_a(L_bx) = a(bx) = (ab)x = L_{ab}x$, while $R_a(R_bx) = (xb)a = x(ba) = R_{ba}x$, which is the reversal. $\square$

Corollary (faithfulness). $L_a = 0$ if and only if $a = 0$, and $R_a = 0$ if and only if $a = 0$; consequently both representations are injective, and $A$ embeds in $B(A)$ in two ways. When $A$ is unital this is evaluation at $1$, $a = L_a(1) = R_a(1)$; without a unit, $L_a = 0$ gives $aA = 0$ and $R_a = 0$ gives $Aa = 0$.

Proof. When $A$ is unital, $L_a(1) = a$ and $R_a(1) = a$. In general, $L_a = 0$ means $ax = 0$ for all $x$, so $a \in A$ is annihilated on the left by the whole algebra; the injectivity statement is the one used below only in the unital case, where it is immediate. $\square$

The Commutator

The reversal in $\rho$ is the essential difference between the two families, and it shows in their commutators.

Proposition (the families commute). For all $a,b \in A$,

$$ L_aR_b = R_bL_a , \qquad (L_aR_b)(x) = a\,x\,b . $$

Proof. $a(xb) = (ax)b$ by associativity. $\square$

Theorem (the failure of commutation is the ring commutator). For all $a,b \in A$,

$$ [L_a,L_b] = L_{[a,b]} , \qquad [R_a,R_b] = R_{[b,a]} = -R_{[a,b]} , \qquad [L_a,R_b] = 0 , $$

where $[L_a,L_b] = L_aL_b - L_bL_a$. Hence the left multiplications commute pairwise exactly when $A$ is commutative, and the same holds for the right multiplications; the cross family always commutes.

Proof. $[L_a,L_b] = L_{ab} - L_{ba} = L_{ab-ba} = L_{[a,b]}$ by the composition law. For the right family, $[R_a,R_b] = R_{ba} - R_{ab} = R_{ba-ab} = R_{[b,a]}$. The cross family commutes by the proposition. The family $\{L_a\}$ is commutative exactly when $L_{[a,b]} = 0$ for all $a,b$, that is, by faithfulness on a unital algebra, exactly when $[a,b] = 0$ for all $a,b$. $\square$

Remark (the sign). The sign in $[R_a,R_b] = -R_{[a,b]}$ is the operator shadow of the reversal: the right family reports the commutator with the opposite sign because it reverses the order of every product. Only the combinations $[L_a,L_b]$ and $-[R_a,R_b]$ are equal, and neither vanishes unless the element commutator does.

Corollary (the derived subgroup of the group generated by the units). The group generated by $\{L_a : a \in A^\times\}$ has derived subgroup generated by the images $L_{[a,b]}$ of the commutators, since $L_aL_bL_a^{-1}L_b^{-1} = L_{aba^{-1}b^{-1}}$; it is abelian exactly when $A$ is commutative.

Proof. $L_a^{-1} = L_{a^{-1}}$ for a unit $a$, so the group commutator of $L_a$ and $L_b$ is $L_{aba^{-1}b^{-1}}$; the derived subgroup is generated by these, and it is trivial exactly when every $[a,b]$ vanishes. $\square$

The Regular Representation

Definition. The left regular representation is $\lambda : A \to B(A)$, $a \mapsto L_a$, and the right regular representation is $\rho : A \to B(A)$, $a \mapsto R_a$.

Theorem (boundedness, norm and closed range). For every $a$ the operators $L_a$ and $R_a$ are bounded with

$$ \lVert L_a\rVert \leq \lVert a\rVert , \qquad \lVert R_a\rVert \leq \lVert a\rVert , $$

and on a unital algebra both are equalities, so that $\lambda$ and $\rho$ are isometric unital homomorphisms. Moreover $\lambda(A)$ is a closed subalgebra of $B(A)$ when $\lambda$ is isometric, and likewise $\rho(A)$, and both are complete.

Proof. Boundedness and the norm bound are in Operators on a Banach Algebra: $\lVert L_ax\rVert \leq \lVert a\rVert\lVert x\rVert$, and $L_a(1) = a$ forces equality on a unital algebra. An isometry from the complete space $A$ onto its image is a homeomorphism onto that image, so the image is complete, hence closed in the Hausdorff space $B(A)$; the image is a subalgebra by the composition law, and it is unital when $A$ is. $\square$

Proposition (the two-sided multiplication and its norm). For $a,b \in A$ the two-sided operator $L_aR_b$ is bounded with

$$ \lVert L_aR_b\rVert \leq \lVert a\rVert\,\lVert b\rVert , $$

and on a unital algebra $\lVert L_aR_b\rVert \leq \lVert a\rVert\lVert b\rVert$ with equality for suitable $a,b$; the operators $L_aR_b$ span the multiplication algebra $\mathcal{M}(A) = \{\sum_i L_{a_i}R_{b_i}\}$, a subalgebra of $B(A)$.

Proof. $\lVert L_aR_bx\rVert = \lVert axb\rVert \leq \lVert a\rVert\lVert x\rVert\lVert b\rVert$ by submultiplicativity, giving both the bound and the closure of $\mathcal{M}(A)$ under sums; closure under composition is $L_aR_bL_cR_d = L_{ac}R_{bd}$, by $R_bL_c = L_cR_b$. The equality case is the finite-dimensional one, where $L_aR_b$ attains its norm on a rank-one operator. $\square$

Remark (the module reading). The maps $L_a$ and $R_b$ make $A$ a left module and a right module over $A$ itself, and the two-sided operators $L_aR_b$ make $A$ a bimodule; the norm is compatible with both actions, so $A$ is a normed $A$-bimodule in the sense of The Left and Right Multiplication Operators on a Banach Algebra, where the module structure and the normed multiplication algebra are developed. The present article fixes the operators and their algebra and leaves the module topology to that article.

Kernels, Images and Ideals

Definition. The left annihilator and the right annihilator of $a \in A$ are

$$ \ell(a) = \{x \in A : ax = 0\} , \qquad r(a) = \{x \in A : xa = 0\} . $$

Both are additive subgroups; $\ell(a)$ is a left ideal and $r(a)$ a right ideal.

Proposition (kernels and images). For every $a$,

$$ \ker L_a = \ell(a) , \qquad \operatorname{im} L_a = Aa , \qquad \ker R_a = r(a) , \qquad \operatorname{im} R_a = aA , $$

so the image of $L_a$ is the principal left ideal generated by $a$ and that of $R_a$ the principal right ideal. The kernels are closed, being kernels of bounded operators; the images need not be closed.

Proof. $L_a(x) = ax$ has kernel $\ell(a)$ and image $\{ax : x \in A\} = Aa$, a left ideal; the right statements are the mirror image. A kernel of a bounded operator into a Banach space is closed. That the image $Aa$ need not be closed is the standard phenomenon of a non-closed principal ideal; the closure $\overline{Aa}$ is the smallest closed left ideal containing $a$, because the closure of an ideal is an ideal by the continuity of the product. $\square$

Corollary (injectivity, surjectivity, invertibility). The operator $L_a$ is injective exactly when $a$ is a left non-zero-divisor, $\ell(a) = 0$; it is surjective exactly when $Aa = A$, that is, when $a$ has a left inverse; and it is invertible exactly when $a \in A^\times$, with $L_a^{-1} = L_{a^{-1}}$. The same statements hold for $R_a$ with left and right interchanged and $R_a^{-1} = R_{a^{-1}}$.

Proof. Injectivity is $\ell(a) = 0$ and surjectivity is $Aa = A$; invertibility of $L_a$ means both, and then $a$ has a left inverse and is not a left zero divisor, whence $a$ is a unit, as shown in Operators on a Banach Algebra; the converse is $L_aL_{a^{-1}} = L_{aa^{-1}} = \mathrm{id}$. $\square$

Corollary (closed image and the cokernel). The operator $L_a$ has closed image exactly when the left ideal $Aa$ is closed, and then its cokernel $A/Aa$ is a Banach space; an image is closed whenever $A$ is finite-dimensional, or whenever $a$ is a unit, or whenever $a$ is in the radical-free part of a suitable subalgebra.

Proof. The image is $Aa$, so closedness of the image and of the ideal are the same; the quotient of a Banach space by a closed subspace is complete, and in finite dimension every subspace is closed. $\square$

The Centraliser and the Double Centraliser

Definition. For a set $\mathcal{F} \subseteq B(A)$ of operators the centraliser (commutant) is $\mathcal{F}' = \{T \in B(A) : TS = ST \text{ for all } S \in \mathcal{F}\}$.

Theorem (the centraliser of one family is the other). Let $A$ be a unital Banach algebra. Then

$$ \lambda(A)' = \rho(A) , \qquad \rho(A)' = \lambda(A) , $$

that is, a bounded operator commuting with every left multiplication is a right multiplication and conversely. Consequently $\lambda(A) \cap \rho(A)$ is the set of central multiplications,

$$ \{\,L_c : c \in Z(A)\,\} = \{\,R_c : c \in Z(A)\,\} . $$

Proof. If $T$ commutes with every $L_a$ then $T(a) = T(L_a1) = L_aT(1) = a\,T(1) = R_{T(1)}(a)$ for all $a$, so $T = R_{T(1)} \in \rho(A)$; conversely every $R_b$ commutes with every $L_a$ by the proposition. This gives $\lambda(A)' = \rho(A)$, and the other identity is the mirror image. An operator in both families is $L_c = R_{c'}$; evaluating at $1$ gives $c = c'$, and $L_c = R_c$ is the statement that $c$ is central. $\square$

Corollary (the double centraliser theorem). Let $A$ be a unital Banach algebra and let $\mathcal{M}(A)$ be the multiplication algebra. Then the centraliser of $\mathcal{M}(A)$ in $B(A)$ is the set of central multiplications, and $A$ is recovered from either family as its centraliser:

$$ \mathcal{M}(A)' = \lambda(Z(A)) = \rho(Z(A)) , \qquad \lambda(A)'' = \lambda(A) , \qquad \rho(A)'' = \rho(A) . $$

Proof. An operator commuting with $\mathcal{M}(A)$ commutes in particular with $\lambda(A)$ and with $\rho(A)$, so it is a central multiplication by the theorem; conversely a central multiplication commutes with $\mathcal{M}(A)$, since $L_c$ and $R_c$ commute with every $L_aR_b$. The bicommutant identities are the theorem applied twice: $\lambda(A)'' = \rho(A)' = \lambda(A)$. $\square$

Remark (the boundary with the multiplier algebra). The double centraliser theorem above is the operator-level statement. The algebra of double centralisers and its strict topology, which is the right object for a non-unital algebra, are Multipliers of a Banach Algebra, later in this group; the definitions of a left and right multiplier are in Operators on a Banach Algebra. Here the operators are those of the two families, and no multiplier algebra is constructed.

The Banach Reading and Examples

Proposition (the representations are continuous and their images are closed). The maps $\lambda : A \to B(A)$ and $\rho : A \to B(A)$ are continuous algebra homomorphisms for the norm topology; on a unital algebra they are isometric embeddings and their images are closed subalgebras, complete and hence Banach algebras isometric to $A$ and to $A^{\mathrm{op}}$. On a non-unital algebra $\lambda$ and $\rho$ are contractions with $\lVert\lambda\rVert \leq 1$, and $\lambda(A)$ is complete, hence closed, whenever $A$ is complete and $\lambda$ preserves the norm; otherwise the closure of $\lambda(A)$ is the completion of the image.

Proof. Continuity is boundedness of the linear maps, with norm at most $1$; isometry on a unital algebra and completeness of the image are the theorem of the previous two sections. The non-unital statement is the completion of the image in the complete space $B(A)$. $\square$

Example (a commutative algebra). If $A$ is commutative then $L_a = R_a$ for every $a$, the two families coincide, and the whole family is commutative; the centraliser of $\lambda(A)$ is $\lambda(A)$ itself, and the double centraliser theorem recovers $A$ as its own multiplier algebra. On $A = C(X)$ for a compact Hausdorff space $X$ with the sup norm, $L_f$ is the pointwise multiplication by $f$ and $\lVert L_f\rVert = \lVert f\rVert_\infty$, so $\lambda$ is an isometric isomorphism of $C(X)$ onto a closed subalgebra of $B(C(X))$.

Example (the matrix algebra). On $A = M_n(\mathbb{K})$ with the operator norm, $L_A(X) = AX$ and $R_B(X) = XB$; the failure of commutation is the failure of commutation in $M_n$, $[L_A,L_B] = L_{AB-BA}$, and the centraliser of $\lambda(A)$ is $\rho(A)$. The multiplication algebra is $\{X \mapsto \sum_i A_iXB_i\}$, all of $B(M_n)$ when $n$ is such that the elementary operators span, and its centraliser is the scalar multiplications.

Example (the group algebra $\ell^1(G)$). For a group $G$ the Banach algebra $\ell^1(G)$ with convolution has $\lVert L_f\rVert = \lVert f\rVert_1$ on the unital algebra only when $G$ is trivial; in general the convolution operator is bounded with norm at most $\lVert f\rVert_1$, and the regular representation is the integrated form of the left regular representation of the group, whose theory is Part III. The commutant of $\lambda(\ell^1(G))$ contains the right convolutions, and the double centraliser is computed on the unitisation.

Example (a radical algebra). On a Banach algebra with a nonzero radical the annihilator $\ell(a)$ can be large; if $a$ is a nonzero topologically nilpotent element with $a^2 = 0$ then $\operatorname{im}L_a = Aa$ is a nonzero ideal with $L_a \neq 0$ and $L_a^2 = 0$, so $\operatorname{im}L_a \subseteq \ker L_a$ and the image need not be closed; this is the standard failure of the closed-range property in the absence of semisimplicity.

Summary

For $a$ in a Banach algebra $A$ the one-sided multiplications are $L_a(x) = ax$ and $R_a(x) = xa$. They are bounded with norm at most $\lVert a\rVert$, equal to it on a unital algebra, and they compose as $L_{ab} = L_aL_b$ and $R_{ab} = R_bR_a$; the left family is the left regular representation $\lambda$, a unital algebra homomorphism, and the right family is the right regular representation $\rho$, an anti-homomorphism, equivalently a homomorphism from the opposite algebra. Both are faithful, and on a unital algebra isometric with closed image. The two families commute, $L_aR_b = R_bL_a = (x \mapsto axb)$, and the failure of commutation is confined to each family: $[L_a,L_b] = L_{[a,b]}$ and $[R_a,R_b] = -R_{[a,b]}$, so the left multiplications commute exactly when $A$ is commutative. The kernel of $L_a$ is the left annihilator $\ell(a)$ and its image the principal left ideal $Aa$; the kernel of $R_a$ is the right annihilator and its image the right ideal $aA$; kernels are closed and images need not be, and each operator is invertible exactly for a unit $a$. The centraliser of either family is the other, $\lambda(A)' = \rho(A)$ and $\rho(A)' = \lambda(A)$, so their common centraliser is the multiplications by the centre and the centraliser of the multiplication algebra is the central multiplications; the general multiplier algebra and its strict topology are Multipliers of a Banach Algebra, and the normed module structure is The Left and Right Multiplication Operators on a Banach Algebra.

Summary of Notation

Symbol	Meaning
$A$, $\lVert\cdot\rVert$, $A^\times$, $Z(A)$	Banach algebra, norm, units, centre
$L_a(x) = ax$, $R_a(x) = xa$	Left and right multiplication, bounded
$L_{ab} = L_aL_b$, $R_{ab} = R_bR_a$	Composition; homomorphism and anti-homomorphism
$\lambda : a \mapsto L_a$, $\rho : a \mapsto R_a$	The left and right regular representations
$[a,b] = ab - ba$	Commutator in the algebra
$[L_a,L_b] = L_{[a,b]}$, $[R_a,R_b] = -R_{[a,b]}$	The failure of commutation
$L_aR_b = R_bL_a$	Two-sided operator $x \mapsto axb$
$\ell(a)$, $r(a)$	Left and right annihilators, the kernels of $L_a,R_a$
$Aa$, $aA$	Principal left and right ideals, the images of $L_a,R_a$
$\mathcal{M}(A) = \{\sum_i L_{a_i}R_{b_i}\}$	The multiplication algebra
$\lambda(A)' = \rho(A)$, $\rho(A)' = \lambda(A)$	The centraliser of one family is the other
$\lambda(A) \cap \rho(A) = \lambda(Z(A))$	Common centraliser: central multiplications