Quaternion Involutions and Projections

Introduction

The quaternion algebra carries a family of order-two maps that the corpus has met in several places and never collected: the three coordinate maps of Quaternion Augmented Statistics, the half-turn example of Quaternion Automorphisms and Derivations, and the three conjugations that Worked Examples in the Quaternion Algebra and The Scalar and Vector Subspaces of $\mathbb{H}$ call the involutions of the algebra. This article collects them. A real-linear map of $\mathbb{H}$ that is its own inverse and is multiplicative is an involution, and it is shown below that the involutions are the identity and the maps $\tilde q\mapsto-\nu\tilde q\nu$, one for each axis line $\mathbb{R}\nu$. The family is therefore infinite, parametrised by the axis lines, that is by the projective plane. The conjugations, which reverse the order of a product, are anti-involutions and form a separate family of the same size, glued to the involutions by the conjugate.

The article turns on one computation, the multiplicativity of $-\nu\tilde q\nu$, which uses $\nu^2=-1$. From it follow the geometry of an involution — it fixes the scalar part and the axis line and reflects the vector part in the axis, so that it is the reflection in the line and not in the plane — the composition rules of a mutually perpendicular triple, and the two formulas that name the article: the component of a vector parallel to an axis is half the vector plus its involution, and the perpendicular component is half the difference. The article closes with the resolution of a quaternion in an orthonormal frame, where the involutions supply the coefficients and no coordinate system need be chosen.

The treatment is mathematical throughout. No physical object is introduced, no state of a physical system is named, and no physical interpretation is invoked. A unit vector is an element of the imaginary subspace with unit quaternion norm; it is a direction in $\mathbb{R}^3$ and nothing else.

The quaternion algebra and its basis $e_0 = 1, e_1, e_2, e_3$ with $e_k^2 = -e_0$, the conjugate, the quaternion norm and the absence of zero divisors are from Quaternion Algebra and Quaternion Norm and Invertibility; the inner automorphism $\iota_u(\tilde q) = u\tilde q u^{-1}$ and the theorem that every automorphism of $\mathbb{H}$ is inner are from Quaternion Automorphisms and Derivations; the adjoint action, the plane reflection $\rho_v$, the half-angle formula and the composition $\rho_{v_1}\rho_{v_2} = \operatorname{Ad}_{v_1v_2}$ are from Quaternion Rotations and Reflections; the three coordinate involutions, the sign matrix, the Klein group and the recovery of the conjugate from the three coordinate involutions are from Quaternion Augmented Statistics; the sphere of unit vectors, identified with $Sp(1)/U(1)$, is from Quaternion Roots of Minus One.

Throughout, a quaternion is $\tilde q = q_0e_0+q_1e_1+q_2e_2+q_3e_3$ with scalar part $q_0 = \mathrm{Sc}\,\tilde q$ and vector part $\mathbf q = \mathrm{Vect}\,\tilde q = q_1e_1+q_2e_2+q_3e_3\in\operatorname{Im}\mathbb{H}$, the conjugate is $\tilde{q}^{\natural} = q_0e_0-q_1e_1-q_2e_2-q_3e_3$, the quaternion norm is $N(\tilde q) = \tilde q\tilde{q}^{\natural} = |\tilde q|^2$, the real inner product on $\operatorname{Im}\mathbb{H}$ is $\langle\mathbf{q},\mathbf{p}\rangle = \mathrm{Sc}(\mathbf{q}\mathbf{p}^{\natural}) = \sum_{k=1}^{3}q_kp_k$ with $|\mathbf{q}|^2 = \langle\mathbf{q},\mathbf{q}\rangle$, and a unit vector is an element $\nu\in\operatorname{Im}\mathbb{H}$ with $N(\nu) = 1$. For a unit vector $\nu$ one has $\nu^2 = -e_0$ and $\nu^{-1} = \bar\nu = -\nu$. The units of $\mathbb{H}$ are the non-zero elements, and $Sp(1)$ is the unit sphere.

The source of the arrangement is the paper of Ell and Sangwine, Quaternion Involutions (arXiv:math/0506034): the axioms, the infinitude of the family, the reflection reading, the composition theorems, the recovery of the conjugate and the projection formulas are theirs. The notation is the corpus's. Where the source writes $\tilde q^{\nu}$ for the value $-\nu\tilde q\nu$, this article writes $\iota_\nu(\tilde q)$, the inner-automorphism notation of Quaternion Automorphisms and Derivations, to which the involution is a specialisation.

Involutions and Anti-Involutions

The Axioms

Definition. An involution of $\mathbb{H}$ is a map $f : \mathbb{H}\to\mathbb{H}$ with

$f(f(\tilde q)) = \tilde q$ for every $\tilde q$;
$f$ is $\mathbb{R}$-linear, $f(\tilde p+\tilde q) = f(\tilde p)+f(\tilde q)$ and $f(\lambda\tilde q) = \lambda f(\tilde q)$ for real $\lambda$;
$f$ is multiplicative, $f(\tilde p\tilde q) = f(\tilde p)f(\tilde q)$.

A map satisfying 1 and 2 together with the reversed product rule $f(\tilde p\tilde q) = f(\tilde q)f(\tilde p)$ is an anti-involution.

Remark (why all three axioms). Axiom 1 alone admits too much. The real-linear map that fixes $e_0$ and $e_1$ and exchanges $e_2$ and $e_3$ is its own inverse, so it satisfies Axioms 1 and 2, and it fails Axiom 3: it sends the product $e_2e_3 = e_1$, which it fixes, to $e_3e_2 = -e_1$. It is Axiom 3 that forces the shape found in the next section, and the failure is not removable by a change of sign.

Proposition. An involution fixes $e_0$, and is a unital algebra automorphism of order dividing two; conversely every unital algebra automorphism $\sigma$ with $\sigma^2 = \mathrm{id}$ is an involution.

Proof. For an involution $f$, put $\tilde e = f(e_0)$. Then $\tilde e = f(e_0) = f(e_0e_0) = f(e_0)f(e_0) = \tilde e^2$, so $\tilde e$ is idempotent. It is not zero: were $\tilde e = 0$, then $e_0 = f(f(e_0)) = f(\tilde e) = f(0) = 0$, which is false. An idempotent $\tilde e$ in $\mathbb{H}$ satisfies $\tilde e(\tilde e-e_0) = 0$, and $\mathbb{H}$ has no zero divisors, so $\tilde e = 0$ or $\tilde e = e_0$; hence $f(e_0) = e_0$. Axioms 2 and 3 then say exactly that $f$ is a unital $\mathbb{R}$-algebra endomorphism, and Axiom 1 makes it bijective with inverse $f$. The converse is immediate from the definition of a unital automorphism.

The Conjugate Is an Anti-Involution

Theorem. Quaternion conjugation is an anti-involution, and is not an involution.

Proof. Conjugation is $\mathbb{R}$-linear and satisfies $(\tilde p\tilde q)^{\natural} = \tilde{q}^{\natural}\tilde{p}^{\natural}$, the reversed product rule, and it is its own inverse; so it satisfies Axioms 1 and 2 and the reversed form of Axiom 3. It fails Axiom 3: $(e_1 e_2)^{\natural} = e_3^{\natural} = -e_3$, while $e_1^{\natural}e_2^{\natural} = (-e_1)(-e_2) = e_1e_2 = e_3$.

Corollary. The anti-involutions of $\mathbb{H}$ are the conjugate $\tilde q\mapsto\tilde{q}^{\natural}$ and the maps $\tilde q\mapsto-\nu\tilde{q}^{\natural}\nu$ for unit vectors $\nu$, that is the composite $\iota_\nu\circ{}^{\natural}$ of an involution with the conjugate.

Proof. Let $f$ be an anti-involution. From the reversed product rule at $\tilde p = \tilde q = e_0$ one gets $f(e_0) = f(e_0)^2$, and $f(e_0)\neq0$ because $f$ is its own inverse; as above $f(e_0) = e_0$, so $f$ fixes $e_0$. Then $g = f\circ{}^{\natural}$ is multiplicative, since both $f$ and conjugation reverse products, and it fixes $e_0$ and is bijective, so it is a unital automorphism; by Quaternion Automorphisms and Derivations it is $\iota_u$ for a unit $u$, and $f = \iota_u\circ{}^{\natural}$. Computing, $f(f(\tilde q)) = u^2\tilde q u^{-2} = \iota_{u^2}(\tilde q)$, so $f$ is an anti-involution exactly when $u^2$ is real, that is when $u$ is real or pure imaginary; a real $u$ gives the conjugate, and a pure $u$ is $u = t\nu$ with $t\neq0$ real and $\nu$ a unit vector, giving $f(\tilde q) = t\nu\tilde{q}^{\natural}(t\nu)^{-1} = \nu\tilde{q}^{\natural}\nu^{-1} = -\nu\tilde{q}^{\natural}\nu$.

Remark (the terminology of the corpus). The corpus calls ${}^{\natural}$, $-{}^{\natural}$ and $-\mathrm{id}$ the three involutions of the algebra in Worked Examples in the Quaternion Algebra and in The Scalar and Vector Subspaces of $\mathbb{H}$, and Quaternion Augmented Statistics defines an involution by $\iota^2 = \mathrm{id}$ alone. In the sense of this article only the conjugate is an anti-involution, and the maps $-{}^{\natural}$ and $-\mathrm{id}$ are neither involutions nor anti-involutions. For $-\mathrm{id}$ the product $(-\tilde p)(-\tilde q)$ equals $\tilde p\tilde q$, while multiplicativity would need it to equal $-\tilde p\tilde q$ and the reversed rule would need it to equal $-\tilde q\tilde p$; for $-{}^{\natural}$ the product $(-\tilde{p}^{\natural})(-\tilde{q}^{\natural})$ equals $\tilde{p}^{\natural}\tilde{q}^{\natural}$, which is neither $-(\tilde p\tilde q)^{\natural} = -\tilde{q}^{\natural}\tilde{p}^{\natural}$ nor its reverse $-\tilde{p}^{\natural}\tilde{q}^{\natural}$. The maps $\iota_k$ of Quaternion Augmented Statistics are involutions in the sense of this article as well, and are the coordinate instances of the family classified next; it is the multiplicative sense that carries the classification.

The Involutions of the Algebra

The Family Defined by a Unit Vector

Theorem (the involutions). For every unit vector $\nu$ the map

$$ \iota_\nu(\tilde q) = \nu\tilde q\nu^{-1} = -\nu\tilde q\nu $$

is an involution of $\mathbb{H}$, and conversely the involutions of $\mathbb{H}$ are $\mathrm{id}$ and the maps $\iota_\nu$ with $\nu$ a unit vector. Hence $\mathbb{H}$ has infinitely many involutions. Moreover $\iota_\nu = \iota_\omega$ exactly when $\omega = \pm\nu$, so the non-trivial involutions are parametrised by the axis lines $\mathbb{R}\nu$, that is by the projective plane of directions of $\mathbb{R}^3$.

Proof. (a) The map is the inner automorphism of Quaternion Automorphisms and Derivations, hence multiplicative, and it is $\mathbb{R}$-linear; it is its own inverse because $\iota_\nu(\iota_\nu(\tilde q)) = \nu^2\tilde q\nu^{-2} = (-e_0)\tilde q(-e_0) = \tilde q$. The second form follows from $\nu^{-1} = -\nu$. (b) Conversely, let $f$ be an involution; by the proposition of the previous section it is a unital automorphism of order dividing two; by Skolem–Noether it is inner, $f = \iota_u$ for a unit $u$; and $\iota_u^2 = \iota_{u^2} = \mathrm{id}$ exactly when $u^2$ is central, that is real. Writing $u = u_0+\mathbf u$ with vector part $\mathbf u$, one has $u^2 = u_0^2-N(\mathbf u)+2u_0\mathbf u$, real precisely when $u_0 = 0$ or $\mathbf u = 0$. The second case is $u$ real, giving the identity; the first is $u$ pure imaginary, $u = t\nu$ with $t$ real and $\nu$ a unit vector, and $\iota_u = \iota_\nu$. (c) For the parametrisation, $\iota_\nu$ fixes the plane $\operatorname{span}(e_0,\nu)$ pointwise and no more, so $\iota_\nu = \iota_\omega$ forces $\operatorname{span}(e_0,\nu) = \operatorname{span}(e_0,\omega)$, hence $\omega = \pm\nu$; the converse is the identity $\nu\tilde q\nu^{-1} = (-\nu)\tilde q(-\nu)^{-1}$.

Corollary. The involutions $\iota_\nu$ of the coordinate axes are the three maps $\iota_k$ of Quaternion Augmented Statistics, and the involution about the axis $e_1$ is the half-turn of the example of Quaternion Automorphisms and Derivations.

Not Every Self-Inverse Sandwich Is an Involution

Proposition. For unit vectors $\nu_1,\nu_2$ the map $g(\tilde q) = \nu_1\tilde q\nu_2$ is its own inverse, $g^2 = \mathrm{id}$. It is multiplicative, hence an involution, exactly when $\nu_2 = -\nu_1$, and then it is $\iota_{\nu_1}$. In particular a map $\nu_1\tilde q\nu_2$ with $\nu_2\neq-\nu_1$ is self-inverse and is not an involution.

Proof. The square is $g(g(\tilde q)) = \nu_1^2\tilde q\nu_2^2 = \tilde q$, since $\nu_1^2 = \nu_2^2 = -e_0$. For multiplicativity, $g(\tilde p)g(\tilde q) = \nu_1\tilde p\nu_2\nu_1\tilde q\nu_2$ and $g(\tilde p\tilde q) = \nu_1\tilde p\tilde q\nu_2$; the two agree for all $\tilde p,\tilde q$ exactly when $\nu_2\nu_1 = e_0$ — set $\tilde p = \tilde q = e_0$ and cancel — that is exactly when $\nu_2 = \nu_1^{-1} = -\nu_1$, and then $g(\tilde q) = -\nu_1\tilde q\nu_1 = \iota_{\nu_1}(\tilde q)$.

The proposition is the reason Axiom 3 is not redundant: the self-inverse sandwiches of that shape form a four-dimensional family, the involutions a two-dimensional one, the projective plane.

The Geometry of an Involution

The Line Reflection

Theorem. Let $\nu$ be a unit vector and $\tilde q = q_0+\mathbf q$ its scalar and vector decomposition. Then

$$ \iota_\nu(\tilde q) = q_0-\nu\mathbf q\nu = q_0+2\langle\mathbf q,\nu\rangle\nu-\mathbf q . $$

Consequently $\iota_\nu$ fixes the scalar part, fixes the vector part parallel to $\nu$ and negates the vector part perpendicular to $\nu$: on the vector subspace $\operatorname{Im}\mathbb{H}$ it is the reflection in the line $\mathbb{R}\nu$, equivalently the rotation by $\pi$ about $\nu$. On $\mathbb{H}$ it fixes the plane $\operatorname{span}(e_0,\nu)$ pointwise and negates its orthogonal complement $\nu^{\perp}$ pointwise.

Proof. Since $\nu e_0\nu^{-1} = e_0$, the map fixes the scalar part. For pure $\mathbf q$ one has $\nu\mathbf q = \nu\times\mathbf q-\langle\nu,\mathbf q\rangle$ and $\mathbf q\nu = -\nu\times\mathbf q-\langle\nu,\mathbf q\rangle$ from the product formula of Quaternion Algebra, so $\nu\mathbf q\nu = (\nu\times\mathbf q)\nu-\langle\nu,\mathbf q\rangle\nu = \mathbf q-2\langle\nu,\mathbf q\rangle\nu$, using $(\nu\times\mathbf q)\nu = (\nu\times\mathbf q)\times\nu = \mathbf q-\langle\nu,\mathbf q\rangle\nu$; hence $-\nu\mathbf q\nu = 2\langle\mathbf q,\nu\rangle\nu-\mathbf q$. The right-hand side fixes $\mathbf q$ when $\mathbf q$ is parallel to $\nu$ and negates it when $\langle\mathbf q,\nu\rangle = 0$, so on $\operatorname{Im}\mathbb{H}$ the map is the reflection in the line $\mathbb{R}\nu$, equivalently the rotation by $\pi$ about $\nu$. Adding the fixed scalar line gives the fixed plane $\operatorname{span}(e_0,\nu)$ and the anti-fixed plane $\nu^{\perp}$.

Corollary (parallel and perpendicular elements). For a unit vector $\nu$, $\iota_\nu(\tilde q) = \tilde q$ when $\mathbf q$ is parallel to $\nu$, and $\iota_\nu(\tilde q) = \tilde{q}^{\natural}$ when $\mathbf q$ is perpendicular to $\nu$.

Proof. The involution fixes the scalar part and replaces $\mathbf q$ by its reflection in the line $\mathbb{R}\nu$, which is $\mathbf q$ in the parallel case and $-\mathbf q$ in the perpendicular case.

The Line Reflection and the Plane Reflection

Remark. The reflection of Quaternion Rotations and Reflections is the plane reflection

$$ \rho_\nu(\tilde q) = -\nu\tilde q\nu^{-1}, $$

which on $\operatorname{Im}\mathbb{H}$ equals $\tilde q-2\langle\tilde q,\nu\rangle\nu$ and is the reflection in the plane $\nu^{\perp}$; on $\mathbb{H}$ it also negates the scalar line, $\rho_\nu(e_0) = -e_0$. Since $\nu^{-1} = -\nu$, one has $\rho_\nu = -\iota_\nu$: the involution and the plane reflection are the two reflections attached to the axis $\nu$, and the sign between them is exactly the difference between reflecting in the line and reflecting in the plane. The plane reflection is orientation-reversing on $\operatorname{Im}\mathbb{H}$ and the involution is orientation-preserving there, the determinants being $-1$ and $+1$.

Composition of Involutions

The Composite Is an Inner Automorphism

Theorem. For unit vectors $\nu_1,\nu_2$,

$$ \iota_{\nu_1}\circ\iota_{\nu_2} = \operatorname{Ad}_{\nu_1\nu_2} , $$

the inner automorphism determined by the product $\nu_1\nu_2$.

Proof. $\iota_{\nu_1}(\iota_{\nu_2}(\tilde q)) = \nu_1(\nu_2\tilde q\nu_2^{-1})\nu_1^{-1} = (\nu_1\nu_2)\tilde q(\nu_2^{-1}\nu_1^{-1}) = (\nu_1\nu_2)\tilde q(\nu_1\nu_2)^{-1}$, since $(\nu_1\nu_2)^{-1} = \nu_2^{-1}\nu_1^{-1}$.

The composite is therefore the same map as the product $\rho_{\nu_1}\rho_{\nu_2}$ of two plane reflections of Quaternion Rotations and Reflections. It is the identity exactly when $\nu_1$ and $\nu_2$ are parallel; it is the involution $\iota_{\nu_1\times\nu_2}$ when they are perpendicular; and for a general pair of axes it is a rotation through an angle that is neither $0$ nor $\pi$.

Mutually Perpendicular Involutions

Theorem. Let $\nu_1,\nu_2$ be perpendicular unit vectors. Then $\nu_1\nu_2 = \nu_1\times\nu_2$ is a unit vector, the two involutions commute, and their composite is the involution about the common perpendicular:

$$ \iota_{\nu_1}\circ\iota_{\nu_2} = \iota_{\nu_2}\circ\iota_{\nu_1} = \iota_{\nu_1\times\nu_2} . $$

Proof. For perpendicular pure quaternions $\nu_1\nu_2 = -\langle\nu_1,\nu_2\rangle+\nu_1\times\nu_2 = \nu_1\times\nu_2$, of norm one because the two are unit and perpendicular; so the composite is $\operatorname{Ad}_{\nu_1\times\nu_2}$, the inner automorphism of the unit vector $\nu_1\times\nu_2$, which is $\iota_{\nu_1\times\nu_2}$. For the commutation, $\nu_2\nu_1 = -\nu_1\nu_2$ and $\operatorname{Ad}$ is unchanged by the sign of its argument, so $\operatorname{Ad}_{\nu_2\nu_1} = \operatorname{Ad}_{\nu_1\nu_2}$, and by the previous theorem the two composites agree.

Theorem (the Klein group of a perpendicular triple). Let $\nu_1,\nu_2,\nu_3$ be mutually perpendicular unit vectors. Then

$$ \{\mathrm{id},\iota_{\nu_1},\iota_{\nu_2},\iota_{\nu_3}\} $$

is a group under composition, isomorphic to $\mathbb{Z}/2\times\mathbb{Z}/2$. The product of two distinct members is the third — up to the sign of the axis, which the involution does not see — and the composition of all three is the identity.

Proof. Each member is its own inverse by the first theorem of the section. For the products, $\iota_{\nu_1}\iota_{\nu_2} = \iota_{\nu_1\times\nu_2}$ and $\nu_1\times\nu_2$ is one of $\pm\nu_3$, so the product is $\iota_{\nu_3}$; the other pairs are the same up to the order of the factors, which the commutation makes immaterial, and the sign of the axis is immaterial by the parametrisation theorem. For the triple, $(\iota_{\nu_1}\iota_{\nu_2})\iota_{\nu_3} = \iota_{\nu_3}\iota_{\nu_3} = \mathrm{id}$.

Remark. This is the general form of the sign matrix and of the Klein group of Quaternion Augmented Statistics, where the perpendicular triple is the coordinate one $e_1,e_2,e_3$ and the four maps are $\mathrm{id},\iota_1,\iota_2,\iota_3$; the four maps there are the four characters of the group, and the group is the same group.

The Composite Is a Rotation

Theorem. For unit vectors $\nu_1,\nu_2$ the composite $\iota_{\nu_1}\iota_{\nu_2} = \operatorname{Ad}_{\nu_1\nu_2}$ fixes the scalar part, preserves the quaternion norm, and rotates the vector part about the axis $\mathrm{Vect}(\nu_1\nu_2) = \nu_1\times\nu_2$ through twice the angle from $\nu_1$ to $\nu_2$.

Proof. The composite is $\operatorname{Ad}_p$ with $p = \nu_1\nu_2$, the product of two unit vectors, hence a unit quaternion; $\operatorname{Ad}_p$ is the inner automorphism, which fixes the centre and preserves the norm, and on $\operatorname{Im}\mathbb{H}$ it is the rotation of axis $\mathrm{Vect} p$ through twice the argument of $p$, by the half-angle formula and the rotation reading of Quaternion Rotations and Reflections. The product $\nu_1\nu_2$ has argument the angle from $\nu_1$ to $\nu_2$ and vector part $\nu_1\times\nu_2$, so the rotation is about $\nu_1\times\nu_2$ through twice that angle. The involution case is the perpendicular one, where $\nu_1\nu_2$ is already a unit vector and the angle is $\pi$.

The Conjugate from Three Involutions

The Sum of the Three Involutions

Lemma. Let $\nu_1,\nu_2,\nu_3$ be mutually perpendicular unit vectors and let $\mathbf q$ be a vector. Then

$$ \iota_{\nu_1}(\mathbf q)+\iota_{\nu_2}(\mathbf q)+\iota_{\nu_3}(\mathbf q) = -\mathbf q . $$

Proof. Resolve $\mathbf q = \eta_1+\eta_2+\eta_3$ with $\eta_i$ parallel to $\nu_i$. The involution about $\nu_i$ fixes $\eta_i$ and negates the other two components, by the corollary on parallel and perpendicular elements; the sum is therefore $(\eta_1-\eta_2-\eta_3)+(-\eta_1+\eta_2-\eta_3)+(-\eta_1-\eta_2+\eta_3) = -\eta_1-\eta_2-\eta_3 = -\mathbf q$.

The Conjugate Identity

Theorem. Let $\nu_1,\nu_2,\nu_3$ be mutually perpendicular unit vectors. For every quaternion $\tilde q$,

$$ \tilde{q}^{\natural} = \tfrac12\bigl(\iota_{\nu_1}(\tilde q)+\iota_{\nu_2}(\tilde q)+\iota_{\nu_3}(\tilde q)-\tilde q\bigr), \qquad q_0 = \tfrac14\bigl(\tilde q+\iota_{\nu_1}(\tilde q)+\iota_{\nu_2}(\tilde q)+\iota_{\nu_3}(\tilde q)\bigr). $$

Proof. Write $\tilde q = q_0+\mathbf q$. Each involution fixes $q_0$ and, by the lemma, the three together send $\mathbf q$ to $-\mathbf q$, so $\iota_{\nu_1}(\tilde q)+\iota_{\nu_2}(\tilde q)+\iota_{\nu_3}(\tilde q) = 3q_0-\mathbf q$. The first identity becomes $\tfrac12(3q_0-\mathbf q-q_0-\mathbf q) = q_0-\mathbf q = \tilde{q}^{\natural}$, and the second becomes $\tfrac14(q_0+\mathbf q+3q_0-\mathbf q) = q_0$.

Remark. The conjugate, which is the only anti-involution among the four maps, is a fixed integer combination of the three involutions: the anti-involution is generated by the involutions even though it is not one of them. For the coordinate triple this is the recovery formula of Quaternion Augmented Statistics, $\tilde{q}^{\natural} = \tfrac12(\iota_1\tilde q+\iota_2\tilde q+\iota_3\tilde q-\tilde q)$; the general form holds for every mutually perpendicular triple.

Projections

Projection of a Vector

Theorem. Let $\nu$ be a unit vector and $\mathbf q$ a vector. Then

$$ \mathbf q_{\parallel\nu} = \tfrac12\bigl(\mathbf q+\iota_\nu(\mathbf q)\bigr) = \langle\mathbf q,\nu\rangle\nu , \qquad \mathbf q_{\perp\nu} = \tfrac12\bigl(\mathbf q-\iota_\nu(\mathbf q)\bigr) = \mathbf q-\langle\mathbf q,\nu\rangle\nu , $$

and $\mathbf q = \mathbf q_{\parallel\nu}+\mathbf q_{\perp\nu}$, the first component parallel to $\nu$ and the second perpendicular to it.

Proof. By the line-reflection theorem $\iota_\nu(\mathbf q) = 2\langle\mathbf q,\nu\rangle\nu-\mathbf q$. Adding gives $2\langle\mathbf q,\nu\rangle\nu$ and subtracting gives $2\mathbf q-2\langle\mathbf q,\nu\rangle\nu$. The two components are the classical parallel and perpendicular projections, so the involution computes them without naming coordinates.

Projection of a Quaternion

Theorem. With $\tilde q = q_0+\mathbf q$ and $\nu$ a unit vector,

$$ \tfrac12\bigl(\tilde q+\iota_\nu(\tilde q)\bigr) = q_0+\langle\mathbf q,\nu\rangle\nu , \qquad \tfrac12\bigl(\tilde q-\iota_\nu(\tilde q)\bigr) = \mathbf q-\langle\mathbf q,\nu\rangle\nu . $$

The first component lies in the plane $\operatorname{span}(e_0,\nu)$, the Argand plane of the axis, and carries the scalar part; the second is a vector perpendicular to $\nu$.

Proof. The involution fixes $q_0$; applying the vector theorem to $\mathbf q$ gives the two expressions.

The Resolution in an Orthonormal Frame

Theorem. Let $\nu_1,\nu_2,\nu_3$ be mutually perpendicular unit vectors. Every quaternion has a unique resolution

$$ \tilde q = a+\nu_1\alpha+\nu_2\beta+\nu_3\gamma , \qquad a,\alpha,\beta,\gamma\in\mathbb{R} , $$

whose coefficients are computed by the involutions from $\tilde q$ alone:

$$ a = \tfrac14\bigl(\tilde q+\iota_{\nu_1}(\tilde q)+\iota_{\nu_2}(\tilde q)+\iota_{\nu_3}(\tilde q)\bigr), \qquad \alpha\nu_1+\beta\nu_2+\gamma\nu_3 = \tfrac12\bigl(\tilde q-\tilde{q}^{\natural}\bigr), $$

and, coefficient by coefficient, with $\mathbf q = \mathrm{Vect}\tilde q$ and $b_i = \tfrac12\bigl(\mathbf q+\iota_{\nu_i}(\mathbf q)\bigr)$,

$$ \alpha\nu_1 = b_1 , \qquad \beta\nu_2 = b_2 , \qquad \gamma\nu_3 = b_3 , \qquad \alpha = \langle\mathbf q,\nu_1\rangle,\ \ \beta = \langle\mathbf q,\nu_2\rangle,\ \ \gamma = \langle\mathbf q,\nu_3\rangle . $$

Proof. The scalar part is fixed by all four maps, so each of the four values has the same scalar part $a$, and their sum has scalar part $4a$; no involution contributes a further scalar, which gives the first formula. The vector part is $\mathbf q = \tfrac12(\tilde q-\tilde{q}^{\natural})$, resolved along the three axes by the parallel projection of the vector theorem, $b_i = \langle\mathbf q,\nu_i\rangle\nu_i$, so $\alpha = \langle\mathbf q,\nu_1\rangle$ and similarly for $\beta$ and $\gamma$. Uniqueness is the uniqueness of the decomposition of the vector part in the orthonormal basis.

The resolution is the computation the involutions were introduced for: the four maps produce the four coefficients in a frame the algebra chooses, and no numerical coordinates are fixed.

Summary

An involution of $\mathbb{H}$ is a real-linear self-inverse multiplicative map; multiplicativity is not implied by the other two axioms, and the map that exchanges $e_2$ and $e_3$ while fixing $e_0$ and $e_1$ shows it. Quaternion conjugation is an anti-involution and not an involution, and the anti-involutions are exactly the conjugate and its twists $\tilde q\mapsto-\nu\tilde{q}^{\natural}\nu$ by the involutions.

The involutions of $\mathbb{H}$ are the identity together with the maps $\iota_\nu(\tilde q) = \nu\tilde q\nu^{-1} = -\nu\tilde q\nu$, one for each unit vector $\nu$; since $\iota_\nu = \iota_{-\nu}$, the non-trivial involutions are parametrised by the axis lines, that is by the projective plane, and the family is infinite. The class contains the three coordinate involutions $\iota_k$ of Quaternion Augmented Statistics and the half-turn example of Quaternion Automorphisms and Derivations.

An involution fixes the scalar part and the axis line and negates the perpendicular plane; on the imaginary subspace it is the reflection in the axis line, equivalently the rotation by $\pi$ about the axis, and it is the negative of the plane reflection $\rho_\nu$ of Quaternion Rotations and Reflections. Two involutions compose to the inner automorphism $\operatorname{Ad}_{\nu_1\nu_2}$, which is a rotation about $\nu_1\times\nu_2$ through twice the angle between the axes; perpendicular involutions commute, the involutions of a mutually perpendicular triple form a Klein four-group, and the product of two perpendicular ones is the involution about the third axis.

Three mutually perpendicular involutions recover the conjugate and the scalar part, $\tilde{q}^{\natural} = \tfrac12(\iota_{\nu_1}\tilde q+\iota_{\nu_2}\tilde q+\iota_{\nu_3}\tilde q-\tilde q)$ and $q_0 = \tfrac14(\tilde q+\iota_{\nu_1}\tilde q+\iota_{\nu_2}\tilde q+\iota_{\nu_3}\tilde q)$; for the coordinate triple this is the recovery formula of Quaternion Augmented Statistics. The projection of a vector or a quaternion on an axis is $\tfrac12$ of the element plus or minus its involution, with the parallel component lying in the Argand plane of the axis, and the two formulas resolve any quaternion in any mutually perpendicular triple of axes, which is the constructive content of the family.

Summary of Notation

Symbol	Meaning
$\mathbb{H}$	Quaternion algebra
$e_0 = 1, e_1, e_2, e_3$	Quaternion basis, $e_k^2 = -e_0$
$\tilde q = q_0e_0+q_1e_1+q_2e_2+q_3e_3$	General quaternion, scalar part $q_0$, vector part $\mathbf q$
$\mathrm{Sc}, \mathrm{Vect}$	Scalar and vector part functionals
$\tilde{q}^{\natural}$	Quaternion conjugate, an anti-involution
$N(\tilde q) = \tilde q\tilde{q}^{\natural} = \lvert\tilde q\rvert^2$	Quaternion norm and modulus
$\operatorname{Im}\mathbb{H}$	Imaginary subspace, the space of vectors
$\langle\mathbf{q},\mathbf{p}\rangle = \mathrm{Sc}(\mathbf{q}\mathbf{p}^{\natural})$	Real inner product on $\operatorname{Im}\mathbb{H}$
$\nu$	Unit vector, $\nu\in\operatorname{Im}\mathbb{H}$, $N(\nu) = 1$
$Sp(1)$	Unit sphere and unit group of $\mathbb{H}$
$\iota_u(\tilde q) = u\tilde q u^{-1}$	Inner automorphism (from Quaternion Automorphisms and Derivations)
$\iota_\nu(\tilde q) = \nu\tilde q\nu^{-1} = -\nu\tilde q\nu$	Involution about the axis $\nu$, a unit vector
$\tilde q^{\nu} = -\nu\tilde q\nu$	The source's notation for the value of $\iota_\nu$
$\operatorname{Ad}_{\tilde q}(\tilde p) = \tilde q\tilde p\tilde q^{-1}$	Adjoint action and inner automorphism
$\rho_\nu(\tilde q) = -\nu\tilde q\nu^{-1} = -\iota_\nu(\tilde q)$	Plane reflection of the axis $\nu$
$\operatorname{span}(e_0,\nu)$	Argand plane of the axis $\nu$
$\nu_1,\nu_2,\nu_3$ mutually perpendicular	Orthonormal frame of axes
$\mathbb{R}\nu$	Axis line; the involutions are parametrised by the axis lines