Opposite Algebras and Anti-Isomorphisms

Introduction

Every algebra has a mirror image, the opposite algebra $A^{\mathrm{op}}$, obtained by keeping the additive group and the scalars and reversing the order of every product. The mirror is not a curiosity: it is the algebra in which the right modules of $A$ become left modules, and it is the algebra that a right action secretly uses. The maps that compare an algebra with its mirror are the anti-isomorphisms, the bijections that reverse products, and the involutions of Involutive Linear Algebras are exactly the anti-isomorphisms of order two from an algebra to itself.

This article develops the opposite algebra as a construction, with its functoriality and its behaviour under the tensor product, the group algebra and the matrix algebra; the anti-isomorphisms onto it, with the calculus of composition that turns two anti-maps into a map; and the place of the involution as the special anti-isomorphism of order two. The notion of an involution, the symmetric and the skew elements and their Lie and Jordan structures are the subject of Involutive Linear Algebras, and they are cited rather than restated.

Throughout, $k$ is a field, $A$ and $B$ are unital associative $k$-algebras, and the unit is $1$; the opposite algebra is written $A^{\mathrm{op}}$, and the element $a$ of $A$, viewed in $A^{\mathrm{op}}$, is written $a^{\mathrm{op}}$ when the two algebras are being compared. The tensor products are over $k$, and $\operatorname{End}$ means $\operatorname{End}_k$. The algebras, the ideals and the centre are those of Algebras, Ideals and Quotients of Algebras and Centre, Units, Zero Divisors and Division Algebras; the modules are those of Modules over an Algebra and the tensor products those of Tensor Products of Algebras.

The Opposite Algebra

Definition

Definition. The opposite algebra $A^{\mathrm{op}}$ of $A$ is the $k$-algebra whose additive group and scalar action are those of $A$ and whose product is

$$ a^{\mathrm{op}} \cdot b^{\mathrm{op}} = (ba)^{\mathrm{op}} . $$

The identity of $A^{\mathrm{op}}$ is $1^{\mathrm{op}}$, and the opposite map

$$ \iota_A : A \longrightarrow A^{\mathrm{op}}, \qquad \iota_A(a) = a^{\mathrm{op}} $$

is $k$-linear and bijective, and reverses products, $\iota_A(ab) = \iota_A(b)\iota_A(a)$; it is therefore an anti-isomorphism.

The opposite algebra is introduced with the object in Algebras; the definition is repeated here because the whole article is the reading of the construction. The map $\iota_A$ is the identity on the underlying set and the reversal of the product, so it is an anti-isomorphism and not an isomorphism unless $A$ is commutative.

Elementary Properties

Proposition. Let $A$ be a unital associative $k$-algebra. Then

(a) $(A^{\mathrm{op}})^{\mathrm{op}} = A$, with $\iota_{A^{\mathrm{op}}} \circ \iota_A = \mathrm{id}_A$;

(b) $A^{\mathrm{op}}$ is associative with the same unit, and $Z(A^{\mathrm{op}}) = Z(A)$;

(c) $A$ is commutative if and only if $A^{\mathrm{op}} = A$;

(d) $(A^{\mathrm{op}})^\times = A^\times$ as a set, and the group $(A^{\mathrm{op}})^\times$ is the opposite group of $A^\times$;

(e) for ideals $I \subseteq A$ the set $I$ is a two-sided ideal of $A^{\mathrm{op}}$, and $(A/I)^{\mathrm{op}} = A^{\mathrm{op}}/I$.

Proof. (a) Reversing the product twice restores it, and the two maps are inverse bijections. (b) The associativity of $A^{\mathrm{op}}$ is the associativity of $A$ read in the reverse order, and the unit is unaffected; $z$ commutes with every $a$ in $A^{\mathrm{op}}$ exactly when $az = za$ in $A$, which is the centre. (c) $A^{\mathrm{op}} = A$ says $ab = ba$ for every pair, which is commutativity. (d) A bijective map reverses the product exactly when it carries the unit to the unit and preserves invertibility with the inverse reversed; the group of units of the opposite algebra is the opposite group of $A^\times$ because the product is reversed. (e) An ideal is closed under multiplication on both sides by every element, a property symmetric in the order of the two factors, so it is again an ideal in $A^{\mathrm{op}}$; the quotient product is reversed.

Corollary. Every algebra is anti-isomorphic to its opposite, and the anti-isomorphism $\iota_A$ is canonical. The algebras $A$ and $A^{\mathrm{op}}$ therefore have the same dimension, the same centre and the same lattice of two-sided ideals, and are either equal (when $A$ is commutative) or distinct as subalgebras of a common ambient algebra (when $A$ is not).

The corollary is the reason the opposite algebra is a construction and not a new object: it is the same underlying structure with the product read backwards, and every property of $A$ that is symmetric in the order of a product passes to $A^{\mathrm{op}}$.

Functoriality

Proposition. The assignment $A \mapsto A^{\mathrm{op}}$ on the objects extends to a functor: a $k$-algebra homomorphism $f : A \to B$ induces a $k$-algebra homomorphism $f^{\mathrm{op}} : A^{\mathrm{op}} \to B^{\mathrm{op}}$ by $f^{\mathrm{op}}(a^{\mathrm{op}}) = f(a)^{\mathrm{op}}$, the identities $\mathrm{id}^{\mathrm{op}} = \mathrm{id}$ and $(g \circ f)^{\mathrm{op}} = g^{\mathrm{op}} \circ f^{\mathrm{op}}$ hold, and an isomorphism $f$ induces an isomorphism $f^{\mathrm{op}}$.

Proof. For $a, b \in A$, $f^{\mathrm{op}}(a^{\mathrm{op}} b^{\mathrm{op}}) = f^{\mathrm{op}}((ba)^{\mathrm{op}}) = f(ba)^{\mathrm{op}} = (f(b)f(a))^{\mathrm{op}} = f(a)^{\mathrm{op}} f(b)^{\mathrm{op}} = f^{\mathrm{op}}(a^{\mathrm{op}})f^{\mathrm{op}}(b^{\mathrm{op}})$. The functorial identities follow from the definition and the bijectivity from that of $f$.

Tensor Products and Standard Instances

Proposition. For $k$-algebras $A$ and $B$ the swap $\tau(a \otimes b) = b \otimes a$ is an isomorphism of algebras

$$ (A \otimes_k B)^{\mathrm{op}} \cong A^{\mathrm{op}} \otimes_k B^{\mathrm{op}}, \qquad (a \otimes b)^{\mathrm{op}} \longmapsto a^{\mathrm{op}} \otimes b^{\mathrm{op}} . $$

Proof. Both sides are spanned by the decomposable elements and both products reverse the two factors and keep their order: $(a \otimes b)^{\mathrm{op}}(a' \otimes b')^{\mathrm{op}} = ((a' \otimes b')(a \otimes b))^{\mathrm{op}} = (a'a \otimes b'b)^{\mathrm{op}}$, which corresponds to $a^{\mathrm{op}}a'^{\mathrm{op}} \otimes b^{\mathrm{op}}b'^{\mathrm{op}}$, the product of the two images in $A^{\mathrm{op}} \otimes B^{\mathrm{op}}$. The map is $k$-bilinear on the factors, hence defined on the tensor product, and it is a bijection with inverse the corresponding map for the opposite algebras.

Corollary (the standard instances). The following hold.

(1) For a finite-dimensional $k$-linear space $V$, the transpose is an isomorphism $\operatorname{End}_k(V)^{\mathrm{op}} \cong \operatorname{End}_k(V)$: a matrix acts on the opposite algebra by acting on the transposed matrix, $(X^{\mathsf{T}})^{\mathrm{op}} = X^{\mathsf{T}}$.

(2) The matrix algebra is isomorphic to its opposite, $M_n(k)^{\mathrm{op}} \cong M_n(k)$, by the transpose; the isomorphism is an anti-automorphism of $M_n(k)$ of order two, hence an involution, and it is the canonical example of Involutive Linear Algebras.

(3) For a group $G$, the inversion $g \mapsto g^{-1}$ is an isomorphism $k[G]^{\mathrm{op}} \cong k[G]$, because it reverses products; the group algebra is therefore isomorphic to its opposite for every group.

(4) For a quiver $Q$, the reversal of the arrows is an isomorphism $kQ^{\mathrm{op}} \cong kQ^{\mathrm{rev}}$, where $kQ^{\mathrm{rev}}$ is the path algebra of the reversed quiver; it is an isomorphism onto $kQ$ itself exactly when the quiver is isomorphic to its reversal, which is the subject of Involutions of a Path Algebra.

Proof. (1) and (2): the transpose reverses the order of a product, $(XY)^{\mathsf{T}} = Y^{\mathsf{T}}X^{\mathsf{T}}$, and has order two, so it is an anti-automorphism of order two; the identification of matrices with endomorphisms is The Operators on an Algebra. (3) $(gh)^{-1} = h^{-1}g^{-1}$ and $(g^{-1})^{-1} = g$. (4) A path is a word in the arrows and its reversal is the word in the reversed arrows, and concatenation is reversed; the path algebra is that of Quiver Representations and Representation Type, and the reversal construction is developed in Involutions of a Path Algebra.

The list is the reason the opposite algebra is invisible in the commutative examples and in the matrix algebra: the transpose, the inversion and the arrow reversal are the anti-isomorphisms that identify each of these algebras with its opposite. In each case the anti-isomorphism has order two and is an involution, and it is the map that the rest of the category studies.

Anti-Isomorphisms

Definition and Calculus

Definition. Let $A$ and $B$ be $k$-algebras. An anti-homomorphism is a $k$-linear map $f : A \to B$ with

$$ f(ab) = f(b)f(a) \quad \text{for all } a, b \in A, $$

an anti-isomorphism is a bijective anti-homomorphism, and an anti-automorphism is an anti-isomorphism $A \to A$. The set of anti-homomorphisms is written $\operatorname{Anti}(A,B)$ and the set of anti-isomorphisms $\operatorname{AntiIso}(A,B)$.

Proposition. Composition of maps gives the following table: the composite of two anti-homomorphisms is a homomorphism; the composite of a homomorphism with an anti-homomorphism, in either order, is an anti-homomorphism; and the composite of two anti-isomorphisms is an isomorphism. In particular $\operatorname{Anti}(A,A)$ is closed under composition, and it is a monoid whose square lies in the endomorphism monoid of $A$.

Proof. If $f$ and $g$ both reverse products, then $f(g(ab)) = f(g(b)g(a)) = f(g(a))f(g(b))$, so the composite preserves products; if exactly one reverses, the composite reverses; bijectivity is preserved by composition.

Theorem (the dictionary). For $k$-algebras $A$ and $B$ there is a bijection

$$ \operatorname{Anti}(A, B) \longrightarrow \operatorname{Hom}_k(A, B^{\mathrm{op}}), \qquad f \longmapsto \iota_B \circ f, $$

and it restricts to a bijection $\operatorname{AntiIso}(A,B) \to \operatorname{Iso}_k(A, B^{\mathrm{op}})$. Under the dictionary the anti-automorphisms of $A$ correspond to the isomorphisms $A \to A^{\mathrm{op}}$, and the involutions of $A$ to those isomorphisms of Involutive Linear Algebras whose square is the identity.

Proof. The composite $\iota_B f$ preserves products, because $f$ reverses them and $\iota_B$ reverses them back; the correspondence is inverted by $\rho \mapsto \iota_B \rho$, since $\iota_B^2 = \mathrm{id}$. The statements about anti-automorphisms and involutions are the dictionary read with $B = A$.

The dictionary is the reason anti-isomorphisms need no separate theory: an anti-isomorphism onto $B$ is an isomorphism onto the opposite of $B$, and everything about homomorphisms applies to it through the mirror. In particular the set $\operatorname{AntiIso}(A,B)$ is nonempty exactly when $A \cong B^{\mathrm{op}}$, and the anti-automorphisms of $A$ are nonempty exactly when $A \cong A^{\mathrm{op}}$.

The Coset of the Anti-Automorphisms

Theorem. Let $A$ carry an anti-automorphism $\rho_0$. Then the map $\alpha \mapsto \rho_0 \circ \alpha$ is a bijection

$$ \operatorname{Aut}_k(A) \longrightarrow \operatorname{Anti}(A,A), \qquad \alpha \longmapsto \rho_0 \alpha, $$

and the anti-automorphisms of $A$ form a coset of the automorphism group in the monoid of all bijective $k$-linear maps of $A$. Two anti-automorphisms compose to an automorphism, and the products of an odd number of anti-automorphisms are anti-automorphisms.

Proof. For an automorphism $\alpha$ the composite $\rho_0\alpha$ reverses products, so the map lands in the anti-automorphisms; it is inverted by $\rho \mapsto \rho_0^{-1}\rho$, which is the composite of two anti-automorphisms and hence an automorphism, and the two composites are the identity by the associativity of composition and $\rho_0^{-1}\rho_0 = \mathrm{id} = \rho_0\rho_0^{-1}$. The statement that the anti-automorphisms form a coset is the definition of a coset under composition with the fixed element $\rho_0$.

Remark. When $A$ carries an involution the coset theorem is that of Involutive Linear Algebras, and the present article adds the case in which the fixed anti-automorphism $\rho_0$ is not of order two; the coset structure is the same and only the inverse map changes. The reason the coset is the right language is that the anti-automorphisms are never a group under composition, because the product of two of them leaves the coset.

Involutions as Isomorphisms to the Opposite

The Correspondence

Definition. An involution of $A$ is an isomorphism $\sigma : A \to A^{\mathrm{op}}$ with $\sigma^2 = \mathrm{id}$, equivalently a $k$-linear map with $\sigma(ab) = \sigma(b)\sigma(a)$, $\sigma(1) = 1$ and $\sigma \circ \sigma = \mathrm{id}$.

The definition is the one of Involutive Linear Algebras, where the involution, its symmetric and skew elements, the Lie algebra of the skew elements and the Jordan algebra of the symmetric ones are developed; the present article records only the place of the involution in the calculus of the opposite algebra.

Proposition. A $k$-linear map $\sigma : A \to A$ is an involution exactly when $\iota_A \circ \sigma : A \to A^{\mathrm{op}}$ is an isomorphism of algebras and $\sigma \circ \sigma = \mathrm{id}_A$. Under the dictionary the involutions of $A$ correspond bijectively to the anti-automorphisms of $A$ of order two, that is, to the elements of order two in the coset $\operatorname{Anti}(A,A)$.

Proof. The composite $\iota_A\sigma$ preserves products, because $\sigma$ reverses them and $\iota_A$ reverses them back, so $\sigma$ is an anti-homomorphism exactly when $\iota_A\sigma$ is a homomorphism; $\sigma$ is bijective exactly when $\iota_A\sigma$ is, and the condition $\sigma^2 = \mathrm{id}$ is the order-two condition on $\sigma$ itself. The final statement is the dictionary read with $B = A$.

Corollary. An algebra carries an anti-automorphism exactly when it is isomorphic to its opposite, and it carries an involution exactly when there is an isomorphism $\rho : A \to A^{\mathrm{op}}$ whose corresponding anti-automorphism has order two. An algebra with no anti-automorphism has no involution, and an algebra with an anti-automorphism of infinite order need not have an involution.

Example. On $M_n(k)$ the transpose is an involution. On $k[G]$ the inversion $g \mapsto g^{-1}$ extends to an anti-automorphism, and it is an involution exactly when every element of $G$ has order dividing two; otherwise it is an anti-automorphism of infinite order, and the group algebra may or may not carry an involution of order two.

The Centre, the Units and the Ideals of the Opposite

Proposition. The centre, the group of units, the radical and the lattice of two-sided ideals of $A^{\mathrm{op}}$ are those of $A$; the simple quotients of $A^{\mathrm{op}}$ are the opposites of the simple quotients of $A$, and $A$ is simple if and only if $A^{\mathrm{op}}$ is.

Proof. The centre and the ideals were settled above; the units form the opposite group. A two-sided ideal $I$ is maximal exactly when $A/I$ is simple, and $(A/I)^{\mathrm{op}} = A^{\mathrm{op}}/I$; a ring is simple exactly when its opposite is, because the lattice of two-sided ideals is the same.

Corollary (central simple algebras). If $A$ is central simple over $k$, then so is $A^{\mathrm{op}}$, and the two are related by the Brauer group: the class of $A^{\mathrm{op}}$ is the inverse of the class of $A$ in $\operatorname{Br}(k)$, as in Central Simple Algebras and the Brauer Group. In particular $A \cong A^{\mathrm{op}}$ for every central simple algebra, and the isomorphism can be chosen of order two when $A$ is a matrix algebra.

The corollary is the reason the opposite algebra is a construction of the theory and not a separate family: over a field every central simple algebra is anti-isomorphic to itself, and the anti-isomorphism is what the transpose realises for the matrix algebra.

Summary

The opposite algebra $A^{\mathrm{op}}$ is $A$ with the product reversed, $a^{\mathrm{op}}b^{\mathrm{op}} = (ba)^{\mathrm{op}}$, the opposite map $\iota_A : A \to A^{\mathrm{op}}$ being an anti-isomorphism; $(A^{\mathrm{op}})^{\mathrm{op}} = A$, $A^{\mathrm{op}}$ has the same centre, unit, units, radical and lattice of two-sided ideals as $A$, and $A^{\mathrm{op}} = A$ exactly when $A$ is commutative. The construction is functorial, it commutes with the tensor product by the swap, $(A \otimes B)^{\mathrm{op}} \cong A^{\mathrm{op}} \otimes B^{\mathrm{op}}$, and it is realised by the transpose for $\operatorname{End}_k(V)$ and for $M_n(k)$, by the inversion for a group algebra, and by the arrow reversal for a path algebra.

An anti-homomorphism $f : A \to B$ satisfies $f(ab) = f(b)f(a)$; the composite of two anti-maps is a map, and the dictionary $f \mapsto \iota_B f$ is a bijection $\operatorname{Anti}(A,B) \cong \operatorname{Hom}_k(A,B^{\mathrm{op}})$ restricting to $\operatorname{AntiIso}(A,B) \cong \operatorname{Iso}_k(A,B^{\mathrm{op}})$. The anti-automorphisms of $A$ form the coset $\rho_0\operatorname{Aut}_k(A)$ for any fixed anti-automorphism $\rho_0$, and they are a group only when empty or when $A$ is commutative. An involution is an isomorphism $\sigma : A \to A^{\mathrm{op}}$ of order two, that is an anti-automorphism of order two; the involutions, their symmetric and skew elements and the Lie and Jordan structures they carry are the subject of Involutive Linear Algebras, the involutions of the tensor algebra, the free algebra, the graded algebras and the path algebra are the subject of Involutions of the Tensor Algebra, Involutions of a Free Algebra, Involutive Graded Algebras and Involutions of a Path Algebra, and the adjoints built from an involution are the subject of the group - * Operator Theory of this category.

Summary of Notation

Symbol	Meaning
$k$	the field of scalars
$A, B$	unital associative $k$-algebras
$A^{\mathrm{op}}$	the opposite algebra, product $a \cdot b = ba$
$\iota_A : A \to A^{\mathrm{op}}$	the opposite map, an anti-isomorphism
$\operatorname{Anti}(A,B)$	the anti-homomorphisms $A \to B$
$\operatorname{AntiIso}(A,B)$	the anti-isomorphisms $A \to B$
$\operatorname{Anti}(A,A)$	the anti-automorphisms, a coset of $\operatorname{Aut}_k(A)$
$\sigma : A \to A^{\mathrm{op}}$	an involution, an isomorphism of order two
$M_n(k)^{\mathrm{op}} \cong M_n(k)$	the transpose
$k[G]^{\mathrm{op}} \cong k[G]$	the inversion
$kQ^{\mathrm{op}} \cong kQ^{\mathrm{rev}}$	the reversal of the arrows