Unitary Elements of an Involutive Algebra

Introduction

An involution of an algebra carries an element $u$ to its adjoint $u^* = \sigma(u)$, and the elements whose adjoint is their inverse, $u^*u = uu^* = 1$, are the unitary elements. They form a group under multiplication, they sit inside the group of units, and they are the algebraic group that an involution produces: the solutions of the quadratic equation $u^*u = 1$. Two companion objects live beside them — the self-adjoint elements, with $h^* = h$, and the skew elements, with $k^* = -k$ — and the three are tied together by the identities $u + u^* = 2h$ and $u - u^* = 2k$ for every unitary $u$.

This article develops the unitary elements, the group they form as the fixed subgroup of an order-two automorphism of the units, the skew elements as the algebraic tangent space at the identity and as a Lie algebra under the commutator, and the relation between the unitary and the self-adjoint parts. The involution on the elements, the decomposition $A = A^+\oplus A^-$, the Lie algebra of the skew elements and the Jordan algebra of the symmetric ones are the subject of Involutive Linear Algebras; the self-adjoint part is developed in The Self-Adjoint Part of an Algebra; and the forms, the adjoint involution that a form defines and the orthogonal and unitary groups of a form are Part II and Hilbert Algebras, which own them.

Throughout, $k$ is a field of characteristic not two, $A$ is a unital associative $k$-algebra, and $\sigma$ is an involution of $A$; the image of an element is written $u^* = \sigma(u)$, so that $(uv)^* = v^*u^*$ and $(u^*)^* = u$. The fixed elements $A^+ = \{h : h^* = h\}$ are self-adjoint and the negated elements $A^- = \{k : k^* = -k\}$ are skew; the group of units is $A^\times$, and the group of unitary elements is written $U(A) = U(A,\sigma)$.

The Unitary Elements

Definition

Definition. An element $u \in A$ is unitary for $\sigma$ if

$$ u^*u = 1 \quad \text{and} \quad uu^* = 1 . $$

The set of unitary elements is written $U(A) = U(A,\sigma)$, and an element of $U(A)$ is also called a unitary of $(A,\sigma)$.

Proposition. An element $u$ is unitary if and only if it is a unit and $u^* = u^{-1}$; every unitary is a unit, $1$ is unitary, and the two equations $u^*u = 1$ and $uu^* = 1$ are equivalent when $A$ is finite-dimensional.

Proof. If $u^*u = 1$ and $uu^* = 1$ then $u$ has the two-sided inverse $u^*$, so $u \in A^\times$ and $u^* = u^{-1}$. Conversely if $u$ is a unit with $u^* = u^{-1}$ then both products are $1$. The two equations are the two halves of invertibility, and in finite dimension a one-sided inverse of a linear map is two-sided, so either equation suffices.

The Group

Theorem. The unitary elements form a subgroup $U(A) \leq A^\times$. Precisely, the map

$$ \theta : A^\times \longrightarrow A^\times, \qquad \theta(g) = (g^*)^{-1} = \sigma(g)^{-1}, $$

is an automorphism of the group of units with $\theta^2 = \mathrm{id}$, and

$$ U(A) = \{g \in A^\times : \theta(g) = g\} $$

is its fixed subgroup.

Proof. The involution $\sigma$ restricts to an anti-automorphism of $A^\times$ and the inversion is an anti-automorphism of $A^\times$, so their composite $\theta$ is an automorphism; $\theta^2(g) = \sigma(\sigma(g)^{-1})^{-1} = \bigl(\sigma(\sigma(g))^{-1}\bigr)^{-1} = \bigl(g^{-1}\bigr)^{-1} = g$, so $\theta$ has order two. An element is fixed by $\theta$ exactly when $(g^*)^{-1} = g$, that is $g^* = g^{-1}$, which is the unitary condition. A fixed subgroup of an order-two automorphism is a subgroup, so $U(A)$ is a subgroup of $A^\times$, containing $1$, closed under products and inverses; the closure under products can also be checked directly, $(uv)^* = v^*u^* = v^{-1}u^{-1} = (uv)^{-1}$.

Corollary. $U(A,\sigma)$ is the fixed subgroup of the involutive group $(A^\times,\theta)$ in the sense of Involutive Groups, and the assignment $(A,\sigma)\mapsto(U(A,\sigma),\theta)$ is the passage from the involutive algebra to its involutive group of units. For the trivial involution $\sigma = \mathrm{id}$ the automorphism $\theta$ is the inversion, and $U(A,\mathrm{id})$ is the set of the elements $u$ with $u^2 = 1$, which is a subgroup of $A^\times$ exactly when those elements commute pairwise.

The Skew Elements and the Tangent Space

The Lie Algebra of the Skew Elements

Proposition. The skew elements $A^-$ are closed under the commutator,

$$ [A^-, A^-] \subseteq A^-, \qquad [A^-, A^+] \subseteq A^+, \qquad [A^+, A^+] \subseteq A^-, $$

so that $A^-$ is a Lie algebra under $[x,y] = xy - yx$, that $A^+$ is a module over it, and that $A^- \oplus A^+$ is a $\mathbb{Z}/2$-graded Lie algebra with the grading-compatible bracket of Graded Lie Algebras and Lie Superalgebras — the Koszul sign would give the anticommutator on two skew elements, which is self-adjoint, so the signed convention is not the one here; and since $[A^+,A^+]\subseteq A^-$ and $[A^-,A^+]\subseteq A^+$, the self-adjoint part is also a Lie triple system under $\{h_1,h_2,h_3\} = [h_1,[h_2,h_3]]$. The first statement is that of Involutive Linear Algebras.

Proof. For $x$ skew and $y$ skew, $(xy)^* = y^*x^* = (-y)(-x) = yx$, so $[x,y]^* = (xy-yx)^* = yx - xy = -[x,y]$ and $[x,y]$ is skew. The two mixed cases are the same computation with one sign, and the even-even case is the first with the signs cancelled.

The Tangent at the Identity

Theorem. The skew elements are the elements that make $1 + t x$ unitary to first order in $t$: expanding

$$ (1+tx)^*(1+tx) = 1 + t(x + x^*) + t^2 x^*x $$

shows that the coefficient of $t$ vanishes exactly when $x^* = -x$, that is exactly when $x$ is skew. Consequently $A^-$ is the algebraic tangent space of the unitary group at the identity, and the Lie bracket of two skew elements is the second-order shadow of the group commutator: the coefficient of the leading term in the expansion of $(1+tx)(1+ty)(1+tx)^{-1}(1+ty)^{-1}$ lies in the span of $[x,y]$.

Proof. The expansion displayed is the definition of the product; the linear term vanishes exactly for $x^* = -x$. For the second statement one expands the four factors to second order; the quadratic part involves $[x,y]$ and the identity $x^* = -x$, $y^* = -y$, and it is the standard identification of the bracket with the commutator of the group, whose totality belongs to the theory of Involutive Groups and of the linear groups of Part II.

Corollary. For every unitary $u$ the elements $u + u^*$ and $i(u - u^*)$ are self-adjoint when the square root of $-1$ exists in $k$, and $u - u^*$ is always skew. The unitary elements therefore decompose in the affine form

$$ u = \tfrac12(u + u^*) + \tfrac12(u - u^*), $$

the first summand self-adjoint and the second skew, which is the decomposition $A = A^+\oplus A^-$ applied to $u$.

The Relation to the Self-Adjoint Part

The Self-Adjoint Elements of a Unitary

Proposition. Let $u$ be unitary. Then $u + u^*$ and $u u^*$ are self-adjoint, $u - u^*$ is skew, and

$$ (u + u^*)^* = u + u^*, \qquad (u - u^*)^* = -(u - u^*), \qquad (u u^*)^* = u u^* = 1 . $$

Proof. Apply $^*$ and use $(uv)^* = v^*u^*$ and $u^* = u^{-1}$: $(u+u^*)^* = u^* + u = u+u^*$, $(u-u^*)^* = u^* - u = -(u-u^*)$, and $(uu^*)^* = u u^* = 1$.

The Symmetrised Product

Definition. The symmetrised product of two elements is $x \bullet y = \tfrac12(xy + yx)$, and the self-adjoint part $H(A) = A^+$ is a Jordan algebra under it, in the sense of Involutive Linear Algebras.

Proposition. The symmetrised product of two skew elements and of two self-adjoint elements is self-adjoint, while the symmetrised product of a self-adjoint and a skew element is skew:

$$ A^- \bullet A^- \subseteq A^+, \qquad A^+ \bullet A^+ \subseteq A^+, \qquad A^+ \bullet A^- \subseteq A^- . $$

Consequently $H(A)$ is closed under $\bullet$, and the skew part $A^-$ is a Jordan module over the Jordan algebra $H(A)$ under the same product.

Proof. For $x, y$ skew, $(xy)^* = y^*x^* = yx$ and $(yx)^* = xy$, so $\tfrac12(xy+yx)$ is fixed by $^*$; similarly for two self-adjoint elements. For $h$ self-adjoint and $k$ skew, $(hk)^* = k^*h^* = -kh$ and $(kh)^* = -hk$, so the half-sum is negated by $^*$.

The Cayley Transform

Proposition. Let $k$ be a skew element with $1 - k$ invertible, and set

$$ u = (1+k)(1-k)^{-1} . $$

Then $u$ is unitary, and its inverse is $(1-k)(1+k)^{-1}$. In particular the unitary group contains the image of the Cayley transform of the skew elements, and the transform is the algebraic substitute for the exponential of the Lie algebra: its first-order term at $k = 0$ is $2k$.

Proof. Compute $u^* = \bigl((1-k)^{-1}\bigr)^*(1+k)^*$. Since $(1-k)^* = 1 + k$ and $(1+k)^* = 1-k$, and since the involution reverses inverses, $u^* = (1+k)^{-1}(1-k)$. The factors $1+k$ and $1-k$ commute, so

$$ u^*u = (1+k)^{-1}(1-k)(1+k)(1-k)^{-1} = (1+k)^{-1}(1-k^2)(1-k)^{-1} = (1+k)^{-1}(1+k)(1-k)(1-k)^{-1} = 1, $$

and the same computation with the factors in the other order gives $uu^* = 1$. The derivative statement is the expansion $u = (1+k)(1 + k + k^2 + \cdots) = 1 + 2k + O(k^2)$ for nilpotent or small $k$.

The exponential $e^{k}$ is not available in a bare $k$-algebra; the formal series identifies the Lie algebra of the skew elements with the unitary group, as in the theory of the linear groups, and the convergent reading belongs to Hilbert Algebras and to Part II.

The Examples

The Trivial Involution

Let $\sigma = \mathrm{id}_A$. Then $u^* = u$, the group $U(A,\mathrm{id})$ is the set of the units $u$ with $u^2 = 1$, that is the involutions of the group of units, and the skew part $A^- = \{k : k = -k\}$ is zero in characteristic not two. The example is the degenerate one, and it shows that the unitary group measures the involution and not the algebra.

The Transpose on a Matrix Algebra

Let $A = M_n(k)$ with the transpose, $X^* = X^{\mathsf{T}}$, and $2 \neq 0$. Then $U(A) = \{X \in GL_n(k) : X^{\mathsf{T}}X = 1\} = O_n(k)$ is the orthogonal group of the standard symmetric form; the skew part is the space of the alternating matrices, the self-adjoint part the space of the symmetric matrices, and the decomposition $M_n = \mathrm{Sym}\oplus \mathrm{Alt}$ is the one of Involutive Linear Algebras. The form that makes $O_n(k)$ the orthogonal group, and its classification among the symmetric and the alternating forms, belong to Hilbert Algebras.

The Inversion on a Group Algebra

Let $A = k[G]$ with the involution $\sigma(g) = g^{-1}$; then $(uv)^* = v^*u^*$ because the inversion reverses products, so $\sigma$ is an involution, and each group element $g$ is unitary, $g^* = g^{-1}$. The map $G \to U(k[G],\sigma)$ is injective, and the unitary group contains the image of $G$; the rest of the unitary group consists of the invertible elements $u$ with $u^*=u^{-1}$ and depends on $G$ and on $k$.

The Exchange Involution

Let $A = B \times B^{\mathrm{op}}$ with the exchange involution $(b, c)^* = (c, b)$, as in Involutive Linear Algebras. An element $(b,c)$ is unitary exactly when $(c,b)(b,c) = 1$, that is when $cb = 1$ and $bc = 1$, so $c = b^{-1}$; hence

$$ U(B \times B^{\mathrm{op}}) = \{(b, b^{-1}) : b \in B^\times\} \cong B^\times , $$

and the unitary group of the exchange involution is the group of units of $B$.

The Conjugate Transpose

Let $A = M_n(\mathbb{C})$ with the $\varsigma$-semilinear conjugate transpose $X^* = \overline{X}^{\mathsf{T}}$. The unitary elements are the matrices with $X^*X = 1$, that is the unitary group $U_n$; the self-adjoint elements are the Hermitian matrices and the skew elements the skew-Hermitian ones. The conjugation $\varsigma$ of $\mathbb{C}$ is a semilinear involution of the scalars, the case of the second kind, and the form of the associated sesquilinear pairing belongs to Hilbert Algebras.

Summary

For an involution $\sigma$ of $A$, written $u^* = \sigma(u)$, the unitary elements are those with $u^*u = uu^* = 1$, equivalently the units with $u^* = u^{-1}$. They form the subgroup $U(A) \leq A^\times$, which is the fixed subgroup of the order-two automorphism $\theta(g) = (g^*)^{-1}$ of $A^\times$; this is the passage from the involutive algebra to the involutive group of units of Involutive Groups. The skew elements $A^- = \{k : k^* = -k\}$ are closed under the commutator, so they form a Lie algebra, and they are the algebraic tangent space of $U(A)$ at the identity: the elements $x$ with $1 + tx$ unitary to first order are exactly the skew elements. The self-adjoint elements $A^+ = H(A)$ form the Jordan algebra under the symmetrised product, and $A^-$ is a Jordan module over it; every unitary decomposes as $u = \tfrac12(u+u^*) + \tfrac12(u-u^*)$ into a self-adjoint and a skew part. The examples are the orthogonal group for the transpose, the image of $G$ for the inversion of a group algebra, the group of units for the exchange involution, and the unitary group for the conjugate transpose. The involution on the elements is Involutive Linear Algebras, the self-adjoint part is The Self-Adjoint Part of an Algebra, the forms and the linear groups belong to Hilbert Algebras and to Part II.

Summary of Notation

Symbol	Meaning
$k$	the field of scalars, of characteristic not two
$A$	a unital associative $k$-algebra
$\sigma$	an involution of $A$
$u^* = \sigma(u)$	the adjoint of an element
$A^+ = H(A)$	the self-adjoint elements, $h^* = h$
$A^-$	the skew elements, $k^* = -k$
$U(A) = U(A,\sigma)$	the unitary elements, $u^u = uu^ = 1$
$\theta(g) = (g^*)^{-1}$	the order-two automorphism of $A^\times$
$x \bullet y = \tfrac12(xy+yx)$	the symmetrised product
$O_n(k)$	the orthogonal group, for the transpose