The Signed Adjoint of the Reflection on a Symmetry Group

Introduction

A reflection is an involutive isometry, and an operator that is both an isometry and an involution is its own adjoint: $\rho_u^{*} = \rho_u^{-1} = \rho_u$. The adjoint of the reflection is therefore the reflection itself, and the reflections are exactly the elements of the orthogonal group that the adjoint does not move — the self-adjoint unitaries of the group, among the involutions it contains. The property is what distinguishes a reflection from a rotation: a rotation is unitary but not self-adjoint, its adjoint being its inverse, and the two together generate the adjoint action on the whole orthogonal group. The property fails in the degenerate case: for an isotropic vector $u$, with $q(u) = 0$, the reflection formula has no meaning, the signed inner conjugation that realises it is not defined, and the substitute operator — the transvection — is not self-adjoint.

The article treats the reflection as an operator, its adjoint and self-adjointness, the failure of the property in the degenerate case, and the generation of the orthogonal group with its adjoint. The reflection and the Cartan–Dieudonné generation theorem are Isometries and Orthogonal Transformations, The Rotation Group and Orientation and Reflections as Signed Two-Sided Operators on a Symmetry Group; the adjoint of a symmetry operator, the unitarity theorem and the operator involution are The Adjoint of a Symmetry Operator, the first article of this group; the adjoint of the general signed sandwich is The Signed Adjoint Sandwich on a Symmetry Group, the previous article, of which this is the single-reflection case; the Clifford realisation of the reflection is The Clifford, Pin and Spin Groups with Signed Inner Conjugation. The two-structures question, whether an involution of the elements agrees with the adjoint of the operators, is The Adjoint of a Symmetry Operator and is not reopened.

The article has four sections: the reflection as an operator; its adjoint and self-adjointness; the degenerate case and the failure; and the worked cases. Throughout, $V$ is a finite-dimensional space over a field $F$ of characteristic not $2$ with a non-degenerate quadratic form $q$ and polar form $B$, $G = \operatorname{O}(V,q)$ the orthogonal group, $u \in V$ a vector with $q(u) \neq 0$ except in the degenerate section, and $\rho_u$ the reflection in $u^{\perp}$; the adjoint is taken with respect to $q$ as in The Adjoint of a Symmetry Operator.

The Reflection as an Operator

The Formula

Definition. For $q(u) \neq 0$ the reflection in the hyperplane $u^{\perp}$ is the linear operator

$$ \rho_u : V \longrightarrow V, \qquad \rho_u(v) = v - 2\,\frac{B(v,u)}{q(u)}\,u . $$

Proposition. The reflection is linear, fixes $u^{\perp}$ pointwise, sends $u$ to $-u$, is an involution $\rho_u^2 = \mathrm{id}$, is an isometry $q(\rho_u v) = q(v)$, is an element of $G$ of determinant $-1$, and is the negative of the transvection $v \mapsto v - \frac{B(v,u)}{q(u)}u$ composed appropriately; it is the operator whose signed two-sided form is the signed inner conjugation $\mathrm{Ad}^{\alpha}_u = \rho_u$ of Reflections as Signed Two-Sided Operators on a Symmetry Group.

Proof. These are the elementary properties of the reflection, Reflections as Signed Two-Sided Operators on a Symmetry Group, quoted.

The Reflection in the Clifford Algebra

Proposition. In the Clifford algebra $\mathrm{Cl}(V,q)$ the reflection is the restriction to the vectors of the signed inner conjugation by $u$,

$$ \rho_u(v) = u\,\alpha(v)\,u^{-1}, \qquad v \in V, $$

while the unsigned inner conjugation is its negative, $uvu^{-1} = -\rho_u(v)$; the vector $u$ satisfies $u^2 = q(u)$ and $u^{-1} = q(u)^{-1}u$, and the conjugation by $u$ is the reflection up to the sign of the grade involution.

Proof. The identity is the Clifford reflection formula, The Clifford, Pin and Spin Groups with Signed Inner Conjugation; it is quoted, not re-derived.

Remark. The reflection is at once an operator of the geometry, an element of the orthogonal group, and an inner conjugation of the Clifford algebra; the three descriptions coincide and the adjoint is the same in each.

The Adjoint of the Reflection

Self-Adjointness

Theorem (the reflection is self-adjoint). The adjoint of the reflection with respect to the form is the reflection itself,

$$ \rho_u^{*} = \rho_u^{-1} = \rho_u, \qquad \langle \rho_u v, w\rangle = \langle v, \rho_u w\rangle \ \text{for all } v, w, $$

so the reflection is both unitary and Hermitian (self-adjoint); in the notation of the operator involution it is a unitary element fixed by the adjoint.

Proof. An isometry of the form has adjoint its inverse, $g^{*} = g^{-1}$, by the unitarity theorem of The Adjoint of a Symmetry Operator; an involution has $g^{-1} = g$; combining, $\rho_u^{*} = \rho_u^{-1} = \rho_u$. Directly, $B(\rho_u v, w) = B(v,w) - 2q(u)^{-1}B(v,u)B(u,w)$ and $B(v, \rho_u w) = B(v,w) - 2q(u)^{-1}B(v,u)B(u,w)$ are equal by the symmetry of $B$.

The Involutions and the Reflections

Proposition. The self-adjoint unitary elements of $G$ are exactly the involutions of $G$,

$$ \{g \in G : g^{*} = g^{-1} = g\} = \{g \in G : g^2 = \mathrm{id}\}, $$

and a reflection is the involutive isometry whose fixed space is a hyperplane; an involution that fixes a subspace of codimension greater than one is a product of commuting reflections and is self-adjoint without being a single reflection.

Proof. $g^{*} = g$ and $g^{*} = g^{-1}$ together give $g^{-1} = g$, i.e., $g^2 = \mathrm{id}$; conversely an involutive isometry satisfies $g^{*} = g^{-1} = g$. The fixed-space statement is the classification of the involutions of an orthogonal group, Isometries and Orthogonal Transformations: a self-adjoint involution is orthogonally diagonalisable with eigenvalues $\pm1$, hence a product of reflections in the $(-1)$-eigenspaces.

The Rotations

Proposition. A rotation, an element of $\operatorname{O}^{+}(V,q)$ that is not an involution, is unitary but not self-adjoint: its adjoint is its inverse, $g^{*} = g^{-1} \neq g$. The adjoint therefore separates the reflections from the rotations inside the orthogonal group, and the Cartan–Dieudonné theorem reads $\rho_{u_1}\cdots\rho_{u_{2k}}$ for a rotation, whose adjoint is $\rho_{u_{2k}}\cdots\rho_{u_1} = g^{-1}$.

Proof. The adjoint of an isometry is its inverse; a rotation is not an involution, so it is not self-adjoint. The adjoint of a product reverses the order, $(g_1g_2)^{*} = g_2^{*}g_1^{*}$, and the reflections are self-adjoint, so the adjoint of a product of reflections is the product in reverse order, which for a rotation is its inverse. The generation theorem is Isometries and Orthogonal Transformations.

The Degenerate Case and the Failure

The Isotropic Vector

Proposition (the isotropic case). For $u$ with $q(u) = 0$ the reflection formula is undefined — the normalisation by $q(u)$ is lost — no operator of the reflection's shape exists in the orthogonal group, and the signed inner conjugation $\mathrm{Ad}^{\alpha}_u(x) = u\alpha(x)u^{-1}$ is likewise undefined, because an isotropic vector is a zero divisor in $\mathrm{Cl}(V,q)$ and does not lie in the Clifford group.

Proof. The reflection sends $u$ to $-u$ and fixes $u^{\perp}$; an isotropic $u$ lies in $u^{\perp}$, so the two requirements are contradictory and no such involution exists. For the Clifford statement, $u^2 = q(u) = 0$ makes $u$ a zero divisor with no inverse, The Clifford, Pin and Spin Groups with Signed Inner Conjugation.

The Failure of Self-Adjointness

Proposition. The operator that generalises the reflection to an isotropic vector is a transvection, a unipotent operator of the shape $\tau(v) = v + B(v,u)w$ with $w$ chosen so that $B(u,w) = 1$, or more generally $\mathrm{id} + N$ with $N$ of square zero; it fixes the isotropic line $Fu$, it is not an involution, $\tau^2 \neq \mathrm{id}$, and it is not self-adjoint, a unitary self-adjoint operator being an involution.

Proof. $\tau^2(v) = v + 2B(v,u)w \neq v$ for any $v$ with $B(v,u) \neq 0$, so $\tau$ is not an involution; a unitary self-adjoint operator satisfies $g^2 = \mathrm{id}$, so a non-involutive operator is not self-adjoint. In the degenerate case $\tau$ is not even an isometry, $q(\tau v) = q(v) + 2B(v,u)B(v,w) + B(v,u)^2q(w) \neq q(v)$ in general, so the self-adjointness fails on both counts. The degenerate-form theory is Quadratic Forms and Polarisation and Bilinear Forms.

Remark. The failure is the boundary of the theorem: the self-adjointness of the reflection is proved from two properties — isometry and involution — and the degenerate substitute has neither. The signed reflection $\mathrm{Ad}^{\alpha}_u$ is self-adjoint in the Clifford setting exactly when the vector satisfies $u^{*} = u$, by The Signed Adjoint Sandwich on a Symmetry Group; the two conditions, the non-degeneracy of the form and the reality of the vector, are the standing hypotheses of the self-adjointness.

Worked Cases

Example (the Euclidean reflection). Let $q$ be positive definite and $u \neq 0$; the reflection $\rho_u$ is an involution and an isometry, hence self-adjoint, $\rho_u^{*} = \rho_u$. In an orthonormal basis $\rho_u$ is a symmetric orthogonal matrix, and self-adjointness is the symmetry of the matrix; the example is the original case.

Example (the hyperbolic reflection). Let $q$ be of signature $(p,q)$ and $u$ a vector with $q(u) \neq 0$; the reflection is again self-adjoint, and it is an element of $O(p,q)$ with determinant $-1$. In the rank-one case $O(p,1)$ the reflection in $u^{\perp}$ is the geodesic symmetry of hyperbolic space when $u$ is the base vector, The Orthogonal Group and the Involutive Automorphism.

Example (the rotations are not self-adjoint). On the Euclidean plane with the rotation $r_{\theta}$ through an angle $\theta$ not $0$ or $\pi$, the adjoint is $r_{\theta}^{*} = r_{-\theta} \neq r_{\theta}$, so the rotation is unitary and not self-adjoint; on $\theta = \pi$ the rotation is an involution and is self-adjoint, which is the case where it is a product of two reflections in perpendicular lines. The example shows the separation of the reflections from the rotations by the adjoint.

Example (the transvection). Let $q$ be of signature $(1,1)$ with isotropic vectors $u = e_1+e_2$ in a basis with $q(e_1) = 1$, $q(e_2) = -1$; the transvection $\tau(v) = v + B(v,u)w$ with $B(u,w) = 1$ is unipotent, it fixes the isotropic line $Fu$, it is not an involution and it is not self-adjoint, and it generates the parabolic one-parameter subgroup of $O(1,1)$. The example is the degenerate substitute in the smallest indefinite case, and it is the boundary the self-adjointness theorem does not reach.

Summary

A reflection $\rho_u$ with $q(u) \neq 0$ is an involutive isometry, and its adjoint with respect to the form is itself: $\rho_u^{*} = \rho_u^{-1} = \rho_u$, so the reflection is both unitary and self-adjoint; the self-adjoint unitary elements of the orthogonal group are exactly its involutions, and a reflection is the involutive isometry with a hyperplane of fixed vectors, while a general involution is a product of commuting reflections. A rotation is unitary and not self-adjoint, its adjoint being its inverse, so the adjoint separates the reflections from the rotations and the Cartan–Dieudonné theorem gives the adjoint of a product of reflections as the reverse product. The property fails in the degenerate case: for an isotropic vector the reflection is undefined, the signed inner conjugation does not exist, and the substitute transvection is neither an isometry nor self-adjoint, so the self-adjointness is a theorem about the non-degenerate reflection alone. In the Clifford setting the signed reflection $\mathrm{Ad}^{\alpha}_u$ is self-adjoint exactly when $u^{*} = u$, by The Signed Adjoint Sandwich on a Symmetry Group. The adjoint of the one-sided signed operator is the subject of the next article.

Summary of Notation

Symbol	Meaning
$\rho_u$	the reflection in $u^{\perp}$, $q(u) \neq 0$
$\rho_u(v) = v - 2q(u)^{-1}B(v,u)u$	the reflection formula
$\rho_u^{*} = \rho_u^{-1} = \rho_u$	the reflection is self-adjoint
$\{g : g^{*} = g^{-1} = g\}$	the self-adjoint unitaries; the involutions
$g^{*} = g^{-1}$	a rotation is unitary, not self-adjoint
$\rho_{u_1}\cdots\rho_{u_{2k}}$	a rotation as a product of reflections (Cartan–Dieudonné)
$\mathrm{Ad}^{\alpha}_u(v) = \rho_u(v)$	the reflection as a signed inner conjugation
$\tau_u = \mathrm{id} - B(\cdot,u)w$	the transvection; the degenerate substitute
$u^{*} = u$	the reality condition for the signed self-adjointness