The Involutions of a Sesqualgebra

Introduction

A sesqualgebra is built from an associative algebra carrying a $\varsigma$-semilinear involution, and that involution is part of the datum: it is what makes the product linear in the first variable and $\varsigma$-semilinear in the second. This article reads the other involutions the same algebra carries, that is the $\varsigma$-semilinear anti-automorphisms $\alpha$ with $\alpha^{2} = \mathrm{id}$. They form a set $\operatorname{Inv}(A)$ that contains the datum $*$; it is closed under inversion, the composite of two of its elements is an $R$-linear automorphism with order two exactly when the two commute, it carries the action of the inner $*$-automorphisms of Units and the Unitary Elements, and each element gives a fixed subspace $H(A,\alpha)$ whose stability is governed by the commutator. The inner family $\sigma_{u}(x) = ux^{*}u^{*}$ is the family generated by the unitary elements, and the criterion for it to be an involution is that $u^{2}$ be central: the shape of the family is the shape of the unitary group, cut down by one centrality condition.

The subject is the order-two case of Anti-Automorphisms Twisted by the Base Involution, which treats the same maps at any order; there the parity of the order, the coset of the twisted anti-automorphisms and the Klein group of the four classes are developed, and the order-two case is referred to this article. Neither article repeats the other: the maps are the same, and what is added here is the order-two condition, the fixed subspace it produces, the criterion for the composite and the partial order those subspaces carry.

The setting is that of Sesqualgebras: $R$ is a commutative ring with $1$, $\varsigma$ is an involution of $R$, and $A$ is an associative unital $R$-algebra with a $\varsigma$-semilinear involution $*$, so that $x \mapsto x^{*}$ is additive, of order two, and reverses the product. The fixed and anti-fixed sets of the datum, the two halves, the scalar action and the decomposition of an element are Hermitian and Skew-Hermitian Elements; the unitary group, the unitary elements and the inner $*$-automorphisms $x \mapsto uxu^{*}$ are Units and the Unitary Elements; the stability of a subalgebra under an involution is Subalgebras and the Involution; the classical theory of the involutions of a central simple algebra, the two kinds and the inner reduction, is Involutions of a Central Simple Algebra; the involution as an isomorphism with the opposite algebra is Opposite Algebras and Anti-Isomorphisms; and the case $\varsigma = \mathrm{id}$, where the whole construction cuts down to the involutions of an ordinary algebra, is Algebras: A General Introduction.

The Set of the Involutions

The Definition

Definition. Let $A$ be an associative unital $R$-algebra with a $\varsigma$-semilinear involution $*$. An involution of $A$ over the datum $(R,\varsigma)$ is a $\varsigma$-semilinear anti-automorphism $\alpha$ of $A$ with $\alpha^{2} = \mathrm{id}$, that is a bijective additive map with

$$ \alpha(\lambda x) = \varsigma(\lambda)\alpha(x) , \qquad \alpha(xy) = \alpha(y)\alpha(x) , \qquad \alpha^{2} = \mathrm{id} . $$

The set of the involutions is written $\operatorname{Inv}(A)$, or $\operatorname{Inv}(A,\varsigma)$ when the datum has to be named.

Proposition. Every $\alpha \in \operatorname{Inv}(A)$ fixes the unit, $\alpha(1) = 1$, and it is its own inverse, $\alpha^{-1} = \alpha$.

Proof. From $\alpha^{2} = \mathrm{id}$ the map is the inverse of itself. For the unit, $\alpha(1) = \alpha(1 \cdot 1) = \alpha(1)\alpha(1)$, so $\alpha(1)$ is an idempotent; more precisely, for every $a \in A$ one has $\alpha(1)a = \alpha(1)\alpha(\alpha(a)) = \alpha(\alpha(a) \cdot 1) = \alpha(\alpha(a)) = a$, so $\alpha(1)$ is a left unit, and $A$ has the unit $1$.

Remark. The datum $*$ is an element of $\operatorname{Inv}(A)$, so the set is not empty, and every statement below is a statement about the datum and its neighbours. The definition is the same as that of the involutions of an algebra when $\varsigma = \mathrm{id}$, and it differs when $\varsigma \neq \mathrm{id}$ only through the first of the three displayed conditions: the twist is carried by the involution itself, not by the algebra.

The Datum and the Collapse

Proposition (the collapse at the identity). Let $\varsigma = \mathrm{id}$. Then the involutions of $A$ over $(R,\mathrm{id})$ are the $R$-linear anti-automorphisms $\alpha$ with $\alpha^{2} = \mathrm{id}$, that is the involutions of the algebra $A$ in the sense of Algebras: A General Introduction; the fixed ring is all of $R$ and the twist disappears from every formula.

Proof. With $\varsigma = \mathrm{id}$ the condition $\alpha(\lambda x) = \varsigma(\lambda)\alpha(x)$ reads $\alpha(\lambda x) = \lambda \alpha(x)$, which is $R$-linearity, and the other two conditions are unchanged.

Remark. The collapse is the reason the present article is a sesquilinear one and not merely a restatement of the classical theory: for $\varsigma = \mathrm{id}$ the set $\operatorname{Inv}(A)$ and the fixed subspace carry no twist, all the scalars are available, and the constructions of Algebras: A General Introduction apply without change. The content of the sesquilinear case is that the set of the involutions depends on $\varsigma$, that its scalars live in the fixed ring $R^{\varsigma}$, and that the inner family is cut by a centrality condition, as the next sections show.

The Composition of Two Involutions

The Criterion for the Composite

Theorem. Let $\alpha, \beta \in \operatorname{Inv}(A)$. Then the composite $\alpha\beta$ is an $R$-linear automorphism of $A$, and it has order two if and only if $\alpha$ and $\beta$ commute:

$$ \boxed{\;(\alpha\beta)^{2} = \mathrm{id} \iff \alpha\beta = \beta\alpha\;} . $$

Proof. The composite of two anti-automorphisms is an automorphism, and it is $(\varsigma \circ \varsigma)$-semilinear, that is $R$-linear, so $\alpha\beta$ is $R$-linear and multiplicative; it is an involution exactly when its square is $\mathrm{id}$. Now

$$ (\alpha\beta)^{2} = \alpha\beta\alpha\beta , $$

and this is $\mathrm{id}$ exactly when $\alpha\beta = (\alpha\beta)^{-1} = \beta^{-1}\alpha^{-1} = \beta\alpha$, using $\alpha^{2} = \beta^{2} = \mathrm{id}$. The composite is multiplicative and not anti-multiplicative, so it belongs to $\operatorname{Inv}(A)$ only when the two rules coincide, that is only when $A$ is commutative.

Corollary. $\operatorname{Inv}(A)$ is closed under inversion, and it is not closed under composition unless $A$ is commutative: the composite of two of its elements is an $R$-linear automorphism, and it lies in $\operatorname{Inv}(A)$ only when the two rules of multiplication coincide. In particular $\operatorname{Inv}(A)$ is not a group under composition.

Proof. The inverse of an element of $\operatorname{Inv}(A)$ is the element itself, by the proposition on the unit. The composite of two of its elements is multiplicative and not anti-multiplicative, so it fails the second of the three defining conditions of $\operatorname{Inv}(A)$; the two rules coincide exactly when $A$ is commutative.

Remark. The theorem is the reason the set of the involutions is a set and not a group: the composite of two of its elements leaves the set, and the commutation of the pair is exactly what makes its square the identity, while the products the set generates lie in the ambient group of the semilinear automorphisms and anti-automorphisms of $A$ over the twists $\{\mathrm{id},\varsigma\}$, which is Anti-Automorphisms Twisted by the Base Involution, where the two orientations and the two twists give the four classes. Here only the order-two part is kept.

Proposition (conjugation). For $\alpha, \beta \in \operatorname{Inv}(A)$ the conjugate $\alpha\beta\alpha^{-1} = \alpha\beta\alpha$ is again an involution, and $H(A,\alpha\beta\alpha) = \alpha H(A,\beta)$.

Proof. The conjugate of an anti-automorphism by an anti-automorphism is an anti-automorphism, it is $\varsigma$-semilinear, and its square is $\alpha\beta^{2}\alpha^{-1} = \mathrm{id}$, so it lies in $\operatorname{Inv}(A)$. For the fixed subspace, $(\alpha\beta\alpha)(\alpha x) = \alpha\beta x = \alpha x$ exactly when $x \in H(A,\beta)$, so the fixed subspace of the conjugate is the image of $H(A,\beta)$.

The Fixed Subspace

Definition. For $\alpha \in \operatorname{Inv}(A)$ the fixed subspace and the anti-fixed subspace are

$$ H(A,\alpha) = \{x \in A : \alpha(x) = x\} , \qquad S(A,\alpha) = \{x \in A : \alpha(x) = -x\} . $$

Proposition. For every $\alpha \in \operatorname{Inv}(A)$ the sets $H(A,\alpha)$ and $S(A,\alpha)$ are $R^{\varsigma}$-submodules of $A$, they are closed under addition, and when $2$ is invertible in $R$ the map $x \mapsto \bigl(\tfrac12(x + \alpha(x)), \tfrac12(x - \alpha(x))\bigr)$ is an isomorphism $A \cong H(A,\alpha) \oplus S(A,\alpha)$.

Proof. The computation is the one of Hermitian and Skew-Hermitian Elements, §The Scalar Action, with $\alpha$ in place of $*$: for $\lambda \in R^{\varsigma}$ and $x \in H(A,\alpha)$ one has $\alpha(\lambda x) = \varsigma(\lambda)\alpha(x) = \lambda x$, and the computation for $S(A,\alpha)$ carries the two signs. The decomposition is the standard one, and the two components are fixed and anti-fixed.

Remark. For the datum, $H(A,*) = H(A)$ is the Hermitian part and $S(A,*) = S(A)$ the skew-Hermitian part, so the definitions generalise the two halves from one involution to all of them. The generalised halves are $R^{\varsigma}$-submodules and not $R$-submodules, by the same argument as for the datum: a scalar with $\varsigma(\lambda) = -\lambda$ exchanges the two.

The Inner Involutions

The Family of a Unitary Element

Theorem. Let $u \in \mathrm{U}(A)$ be unitary and let $\sigma_{u}(x) = ux^{*}u^{*}$. Then $\sigma_{u}$ is a $\varsigma$-semilinear anti-automorphism of $A$ with $\sigma_{u} = \alpha_{u} \circ *$, where $\alpha_{u}(x) = uxu^{*}$ is the inner $*$-automorphism of Units and the Unitary Elements, and

$$ \boxed{\;\sigma_{u}^{2} = \alpha_{u^{2}} , \qquad \text{so that} \qquad \sigma_{u} \in \operatorname{Inv}(A) \iff u^{2} \in Z(A)\;} . $$

Proof. The map is additive, and it reverses products because

$$ \sigma_{u}(xy) = u(xy)^{*}u^{*} = u\,y^{*}x^{*}u^{*} = (uy^{*}u^{*})(ux^{*}u^{*}) = \sigma_{u}(y)\sigma_{u}(x) , $$

and it is $\varsigma$-semilinear because $\sigma_{u}(\lambda x) = u(\lambda x)^{*}u^{*} = u\,\varsigma(\lambda)x^{*}u^{*} = \varsigma(\lambda)\sigma_{u}(x)$. That $\sigma_{u} = \alpha_{u} \circ *$ is the identity $\alpha_{u}(x^{*}) = ux^{*}u^{*}$. For the square,

$$ \sigma_{u}^{2}(x) = u(u x^{*} u^{*})^{*}u^{*} = u\,(u x u^{*})\,u^{*} = u^{2}\,x\,(u^{*})^{2} = \alpha_{u^{2}}(x) , $$

so $\sigma_{u}^{2} = \mathrm{id}$ exactly when $u^{2}$ commutes with every $x$, that is $u^{2} \in Z(A)$.

Remark. The criterion is not $u \in \mathrm{U}(A)$ alone: the unitary element $u = \mathrm{diag}(i,1)$ of $M_{2}(\mathbb{C})$ has $u^{2} = \mathrm{diag}(-1,1)$, which is not central, and $\sigma_{u}$ is then not an involution, since $\sigma_{u}^{2}(E_{12}) = -E_{12}$; the unitary element $u = \mathrm{diag}(i,i)$ has $u^{2} = -I$ central and $\sigma_{u}$ is an involution. The condition $u^{2} \in Z(A)$ holds in particular for every unitary involution, $u^{2} = 1$, and for every central unitary.

Corollary. The family $\{\sigma_{u}\}$ is a family of involutions exactly over the unitary elements with $u^{2} \in Z(A)$, and $\sigma_{u} = *$ for every central unitary $u$.

Proof. The criterion above gives the first statement. For a central unitary $u$ one has $\sigma_{u}(x) = ux^{*}u^{*} = uu^{*}x^{*} = x^{*}$, by the centrality of $u$.

The Orbit of the Datum

Proposition. For $u, v \in \mathrm{U}(A)$ the maps $\sigma_{u}$ and $\sigma_{v}$ are equal exactly when $v^{*}u$ is central, and the assignment $u \mapsto \sigma_{u}$ identifies the family with the unitary elements modulo the central unitary ones.

Proof. From $ux^{*}u^{*} = vx^{*}v^{*}$ for every $x$, multiplying by $v^{*}$ on the left and by $v$ on the right, one gets $w x^{*} = x^{*} w$ with $w = v^{*}u$, for every $x$; this says exactly that $w$ is central. Conversely a central $w$ gives the equality.

Remark. The family is the coset of the inner $*$-automorphisms applied to the datum: $\sigma_{u} = \alpha_{u} \circ *$, so the family is the orbit of $*$ under the inner automorphism group, and its stabiliser is the group of the central unitary elements, the kernel computed in Units and the Unitary Elements, §Inner $*$-Automorphisms. The datum is therefore the base point of the family, and an involution outside the family is reached from it only by an automorphism that is not inner, as in Involutions of a Central Simple Algebra, §Comparing Two Involutions.

The Galois-Type Correspondence

Lemma (determination by the fixed subspace). Let $\alpha, \beta \in \operatorname{Inv}(A)$ and suppose that $H(A,\alpha)$ generates $A$ as an $R$-algebra and that $H(A,\alpha) = H(A,\beta)$. Then $\alpha = \beta$.

Proof. Both $\alpha$ and $\beta$ are additive and both reverse products, $\alpha(xy) = \alpha(y)\alpha(x)$ and $\beta(xy) = \beta(y)\beta(x)$; both fix the unit, so they agree on $1$, both fix $H(A,\alpha) = H(A,\beta)$ by hypothesis, and both carry a scalar $\lambda$ to $\varsigma(\lambda)$, so they agree on $R$ as well. On sums they agree because they agree on the summands, and on a product of two elements $h_{1}, h_{2}$ of $H(A,\alpha)$ they agree as well, since $\alpha(h_{1}h_{2}) = \alpha(h_{2})\alpha(h_{1}) = h_{2}h_{1} = \beta(h_{2})\beta(h_{1}) = \beta(h_{1}h_{2})$; the same computation on longer products shows that the two maps agree on the whole $R$-subalgebra generated by $H(A,\alpha)$ and $1$, which is $A$ by hypothesis, so $\alpha = \beta$.

Theorem (the stability criterion). Let $\alpha, \beta \in \operatorname{Inv}(A)$ and suppose that $H(A,\alpha)$ and $H(A,\beta)$ generate $A$. Then

$$ \boxed{\;\alpha H(A,\beta) = H(A,\beta) \iff \alpha\beta = \beta\alpha\;} . $$

Proof. By the conjugation proposition, $\alpha H(A,\beta) = H(A,\alpha\beta\alpha)$. Since $\alpha$ is bijective, $\alpha H(A,\beta)$ generates $A$ exactly when $H(A,\beta)$ does, so the determination lemma applies to the pair $\alpha\beta\alpha$, $\beta$: the subspace $\alpha H(A,\beta)$ is $H(A,\beta)$ exactly when $\alpha\beta\alpha = \beta$. This equality is $\alpha\beta\alpha\alpha = \beta\alpha$, that is $\alpha\beta = \beta\alpha$.

Corollary. For commuting involutions $\alpha, \beta$ the composite $\alpha\beta$ is an $R$-linear automorphism of order two with $H(A,\alpha) \cap H(A,\beta) \subseteq H(A,\alpha\beta)$, and $H(A,\alpha) \cap H(A,\alpha\beta) = H(A,\alpha) \cap H(A,\beta)$.

Proof. If $x$ is fixed by $\alpha$ and by $\beta$ then $\alpha\beta x = x$, so the first inclusion holds. For the second, $x \in H(A,\alpha) \cap H(A,\alpha\beta)$ gives $\alpha x = x$ and $\alpha\beta x = x$, hence $\beta x = \alpha^{-1}x = x$, so $x \in H(A,\beta)$.

Remark. The theorem is the Galois-type content of the set of the involutions: the involutions that preserve a given fixed subspace are exactly the involutions commuting with its involution. The determination lemma is the one hypothesis that is not automatic, and it holds in the matrix model, where the Hermitian matrices generate the whole algebra; the lemma is stated with the generation condition so that the correspondence is not asserted where it does not hold.

The Partial Order

Proposition. Define $\alpha \preceq \beta$ on $\operatorname{Inv}(A)$ by $H(A,\alpha) \subseteq H(A,\beta)$, under the hypotheses of the determination lemma for every pair. Then $\preceq$ is a partial order: it is reflexive and transitive because the inclusion of subspaces is, and it is antisymmetric because $\alpha \mapsto H(A,\alpha)$ is injective.

Proof. Reflexivity and transitivity are those of $\subseteq$. For antisymmetry, if $H(A,\alpha) \subseteq H(A,\beta)$ and $H(A,\beta) \subseteq H(A,\alpha)$ then the two subspaces are equal, so $\alpha = \beta$ by the determination lemma.

Remark. The order is not the order of a lattice in general, and it is not compatible with the composition: the criterion of the composite is commutation and not the order. Its role is to record the containment of the fixed subspaces, which is the inclusion of the two generalised Hermitian parts, and the datum is one element of the ordered set with $H(A,*)$ = $H(A)$.

The Derived Sesqualgebras

Theorem. Let $\alpha \in \operatorname{Inv}(A)$ and define $x \star_{\alpha} y = x\,\alpha(y)$. Then $(A,\star_{\alpha})$ is a sesqualgebra over $(R,\varsigma)$, the datum gives $\star_{*} = \star$, and the map $\alpha \mapsto \star_{\alpha}$ is injective. Conversely, if $x \star' y = x\,\alpha(y)$ is a sesqualgebra product on $A$ for some $\alpha \in \operatorname{Inv}(A)$, then $\alpha$ is determined by $\star'$.

Proof. The product is additive in each variable, and it need not be associative, which the definition of a sesqualgebra allows; for the scalars, $(\lambda x)\star_{\alpha} y = \lambda(x \star_{\alpha} y)$ and $x \star_{\alpha}(\lambda y) = x\,\alpha(\lambda y) = x\,\varsigma(\lambda)\alpha(y) = \varsigma(\lambda)(x\star_{\alpha} y)$, using the $\varsigma$-semilinearity of $\alpha$ and the $R$-bilinearity of the product of $A$. For the injectivity, $x \star_{\alpha} y = x\star_{\beta} y$ for all $x, y$ gives $\alpha(y) = \beta(y)$ on taking $x = 1$.

Remark. The theorem is the sense in which the involutions are the involutions of the sesqualgebra: each of them equips the same module with a sesqualgebra structure of the form $x\alpha(y)$, the datum gives the sesqualgebra one started from, and no two involutions give the same product. The map is the one of Sesqualgebras, §The Standard Example, read backwards: there the derived operation is built from a fixed involution, here the involution is recovered from the derived operation.

Worked Cases

The Matrix Algebra

Let $A = M_{n}(\mathbb{C})$ with the conjugate transpose and $\varsigma$ the complex conjugation. The involutions of the inner family are the $\sigma_{u}$ with $u$ unitary and $u^{2}$ central; the condition $u^{2} = cI$ with $|c| = 1$ says that $u = \lambda v$ with $\lambda^{2} = c$ and $v$ a unitary involution, $v^{2} = I$, so, the central unitary elements being absorbed in the kernel, the family is parametrised by the unitary involutions modulo the sign, two of them giving the same involution exactly when they are opposite. For $u = I$ and for $u = \lambda I$ the map is the conjugate transpose itself, by the corollary on the central unitaries; for the swap $u = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, which is a unitary involution, the involution $\sigma_{u}$ is the conjugate transpose followed by the interchange of the two coordinates. The map $x \mapsto x^{\mathsf{T}}$ is an anti-involution of the algebra, but it is $R$-linear and not $\varsigma$-semilinear, so it is not in $\operatorname{Inv}(A)$ for the datum $\varsigma$; it belongs to the layer $\varsigma = \mathrm{id}$ and is one of the four classes of Anti-Automorphisms Twisted by the Base Involution. The fixed subspace of the conjugate transpose is the space of the Hermitian matrices, which generates $M_{n}(\mathbb{C})$ as an algebra, so the determination lemma and the stability criterion apply.

The Field

Let $A = \mathbb{C}$ over $R = \mathbb{R}$ with $\varsigma = \mathrm{id}$. The anti-automorphisms of the field are its automorphisms, the identity and the conjugation, and both are involutions; the set is $\{ \mathrm{id}, \bar{\cdot} \}$, the fixed subspaces are $\mathbb{C}$ and $\mathbb{R}$, and the order of the proposition above is the inclusion $\mathbb{R} \subseteq \mathbb{C}$, so the conjugation is the smaller element. Let instead $A = \mathbb{C}$ over $R = \mathbb{C}$ with $\varsigma$ the conjugation. Then a $\varsigma$-semilinear map is determined on the scalars by $f(\lambda) = \varsigma(\lambda)f(1) = \varsigma(\lambda)$, so the only such map with $f(1) = 1$ is the conjugation, and $\operatorname{Inv}(A) = \{ \bar{\cdot} \}$ is a single point; its fixed subspace is $\mathbb{R}$, and the set of the involutions has no order beyond the trivial one. The two readings show how much of the set is visible: over $\mathbb{R}$ the twist is trivial and the automorphisms of the field are all available, over $\mathbb{C}$ the twist is the conjugation and it consumes the automorphism group.

Summary

The involutions of a sesqualgebra are the $\varsigma$-semilinear anti-automorphisms $\alpha$ with $\alpha^{2} = \mathrm{id}$. They contain the datum, they are closed under inversion, and the composite $\alpha\beta$, an $R$-linear automorphism, has order two exactly when $\alpha$ and $\beta$ commute. Each has a fixed subspace $H(A,\alpha)$ and an anti-fixed subspace $S(A,\alpha)$, both $R^{\varsigma}$-submodules, with $A = H(A,\alpha) \oplus S(A,\alpha)$ when $2$ is invertible, generalising the two halves of the datum.

The inner family $\sigma_{u}(x) = ux^{*}u^{*}$ is an involution exactly when $u^{2}$ is central, and the family is the orbit of the datum under the inner $*$-automorphisms, indexed by the unitary elements modulo the central unitary ones. The fixed subspace determines the involution when it generates the algebra, and with that hypothesis the involutions preserving $H(A,\beta)$ are exactly those commuting with $\beta$; the inclusion of the fixed subspaces is a partial order, and each involution equips the module with a sesqualgebra structure $x\alpha(y)$, injectively, of which the datum gives the original one.

Summary of Notation

symbol	meaning
$\operatorname{Inv}(A)$	the set of the $\varsigma$-semilinear anti-involutions of $A$
$\alpha$	an element of $\operatorname{Inv}(A)$, $\alpha^{2} = \mathrm{id}$
$\alpha(\lambda x) = \varsigma(\lambda)\alpha(x)$	the $\varsigma$-semilinearity
$\alpha(xy) = \alpha(y)\alpha(x)$	the reversal of the product
$\alpha\beta = \beta\alpha$	the criterion for the composite $\alpha\beta$ to have order two
$\alpha_{u}(x) = uxu^{*}$	the inner $*$-automorphism of a unitary $u$
$\sigma_{u}(x) = ux^{}u^{}$	the inner involution, $\sigma_{u} = \alpha_{u} \circ *$
$u^{2} \in Z(A)$	the criterion for $\sigma_{u} \in \operatorname{Inv}(A)$
$H(A,\alpha)$, $S(A,\alpha)$	the fixed and the anti-fixed subspace
$H(A,\alpha) \subseteq H(A,\beta)$	the partial order on the involutions
$x \star_{\alpha} y = x\alpha(y)$	the sesqualgebra product of the involution $\alpha$