Quaternions

The Quaternions

The collection of quaternions are:

ℍ : = {a + b 𝐢 + c 𝐣 + d 𝐤}

where

a, b, c, d \in ℝ

and

𝐢^{2} = 𝐣^{2} = 𝐤^{2} = 𝐢 𝐣 𝐤 = - 1

Real Part of the Quaternion

Given a quaternion

q = a + b i + c j + d k

, we say that

a

is the scalar part of the quaternion (or real part), and is denoted by

r (q)

Vector Part of the Quaternion

Given a quaternion

q = a + b i + c j + d k

, we say that

b i + c j + d k

is the vector part of the quaternion, and is denoted by

v (q)

Note that we think of the vector part of a quaternion as an element of $ℝ^{3}$ , and so it inherits all the operations from that space, such as equality, as noted in the next corollary.

Quaternion from a Real Number

We define the map

to_quat : ℝ \to ℍ

to_quat (a) : = a + 0 𝐢 + 0 𝐣 + 0 𝐤

Pure Quaternion from a 3D Vector

We define the map

to_quat : ℝ^{3} \to ℍ

to_quat (b, c, d) : = 0 + b 𝐢 + c 𝐣 + d 𝐤

Quaternion from a 4D Vector

We define the map

to_quat : ℝ^{4} \to ℍ

to_quat (a, b, c, d) : = a + b 𝐢 + c 𝐣 + d 𝐤

to_quat from

ℝ^{3}

Produces a Pure Quaternion

For any

𝐮 \in ℝ^{3}

to_quat (𝐮)

is a pure quaternion, i.e.

r (to_quat (𝐮)) = 0

Let

𝐮 = (b, c, d)

. Then

to_quat (𝐮) = 0 + b 𝐢 + c 𝐣 + d 𝐤

, so

r (to_quat (𝐮)) = 0

Vector Part of to_quat from

ℝ^{3}

is the Identity

For any

𝐮 \in ℝ^{3}

v (to_quat (𝐮)) = 𝐮

Let

𝐮 = (b, c, d)

. Then

v (to_quat (𝐮)) = v (0 + b 𝐢 + c 𝐣 + d 𝐤) = (b, c, d) = 𝐮

to_quat from

ℝ

Has Zero Vector Part

For any

a \in ℝ

v (to_quat (a)) = 𝟎

and

r (to_quat (a)) = a

to_quat (a) = a + 0 𝐢 + 0 𝐣 + 0 𝐤

, so

v (to_quat (a)) = (0, 0, 0) = 𝟎

and

r (to_quat (a)) = a

Quaternion Decomposes as Sum of to_quats

For any

q \in ℍ

q = to_quat (r (q)) + to_quat (v (q))

Let

q = a + b i + c j + d k

. Then

to_quat (r (q)) + to_quat (v (q)) = (a + 0 𝐢 + 0 𝐣 + 0 𝐤) + (0 + b 𝐢 + c 𝐣 + d 𝐤) = a + b i + c j + d k = q

, using the real and vector versions of

to_quat

Equality by Vector and Real Equality

For any

p, q \in ℍ

we have

p = q ⟺ (r (p) = r (q) \land v (p) = v (q))

Suppose that

p = a + b i + c j + d k

and

q = u + x i + y j + z k

then

p = q

if and only if

a = u

b = x

c = y

d = z

which is true if and only if

r (p) = r (q)

and

v (p) = v (q)

, as needed.

Quaternion Addition

(a_{1} + b_{1} 𝐢 + c_{1} 𝐣 + d_{1} 𝐤) + (a_{2} + b_{2} 𝐢 + c_{2} 𝐣 + d_{2} 𝐤) = (a_{1} + a_{2}) + (b_{1} + b_{2}) 𝐢 + (c_{1} + c_{2}) 𝐣 + (d_{1} + d_{2}) 𝐤

Scalar Quaternion Multiplication

λ (a + b 𝐢 + c 𝐣 + d 𝐤) = λ a + (λ b) 𝐢 + (λ c) 𝐣 + (λ d) 𝐤 .

Identity Quaternion

e : = 1 + 0 𝐢 + 0 𝐣 + 0 𝐤

Conjugate Quaternion

Given the quaternion

q = a + b 𝐢 + c 𝐣 + d 𝐤

, its conjugate is given by

q : = a - b 𝐢 - c 𝐣 - d 𝐤

Pure Quaternion iff Conjugate is Negation

A quaternion

x \in ℍ

is pure if and only if

x = - x .

Let

x = a + b 𝐢 + c 𝐣 + d 𝐤

. By the definition of conjugation,

x = a - b 𝐢 - c 𝐣 - d 𝐤, - x = - a - b 𝐢 - c 𝐣 - d 𝐤 .

Thus

x = - x

if and only if

a = - a

, which is equivalent to

a = 0

. Since

a = r (x)

, this is exactly the statement that

x

is pure.

Quaternion Norm

Given a quaternion

q = a + b i + c j + d k

, its norm is defined as

‖ q ‖ : = \sqrt{a^{2} + b^{2} + c^{2} + d^{2}}

Unit Quaternion

A quaternion

q \in ℍ

is said to be a unit quaternion if its norm satisfies

‖ q ‖ = 1

Pure Quaternion

A quaternion

p \in ℍ

is said to be pure iff

r (p) = 0

Pure Unit Quaternion

A quaternion

q \in ℍ

is said to be a pure unit quaternion if it is both pure and unit, i.e.

r (q) = 0

and

‖ q ‖ = 1

Product of Two Quaternions

Suppose that

p, q \in ℍ

, then we define

p q : = r (p) r (q) - v (p) \cdot v (q) + r (p) v (q) + r (q) v (p) + v (p) \times v (q)

More precisely:

p q = to_quat (r (p) r (q) - v (p) \cdot v (q)) + to_quat (r (p) v (q) + r (q) v (p) + v (p) \times v (q))

Quaternion Multiplication Distributes Over Addition

For all

p, q, s \in ℍ

p (q + s) = p q + p s, (p + q) s = p s + q s .

Write each quaternion using its real part and vector part. The quaternion product is defined by combining real multiplication, scalar multiplication of vectors, the dot product, and the cross product:

p q = to_quat (r (p) r (q) - v (p) \cdot v (q)) + to_quat (r (p) v (q) + r (q) v (p) + v (p) \times v (q)) .

Each operation appearing on the right is distributive in the relevant argument. Therefore replacing

q

q + s

, and using

r (q + s) = r (q) + r (s)

and

v (q + s) = v (q) + v (s)

, gives

p (q + s) = p q + p s

. The proof of

(p + q) s = p s + q s

is the same computation in the first argument.

Real Part of Quaternion Product

For any

p, q \in ℍ

r (p q) = r (p) r (q) - v (p) \cdot v (q)

By the definition of the quaternion product, the scalar part is

r (p) r (q) - v (p) \cdot v (q)

and the vector part is

r (p) v (q) + r (q) v (p) + v (p) \times v (q)

. Therefore

r (p q) = r (p) r (q) - v (p) \cdot v (q)

Vector Part of Quaternion Product

For any

p, q \in ℍ

v (p q) = r (p) v (q) + r (q) v (p) + v (p) \times v (q)

Immediate from the definition of the quaternion product.

Product of Two Pure Quaternions

For any two pure quaternions

p, q \in ℍ

(i.e.

r (p) = r (q) = 0

), we have

p q = to_quat (- v (p) \cdot v (q)) + to_quat (v (p) \times v (q))

Since

p

and

q

are pure,

r (p) = 0

and

r (q) = 0

. By the definition of the quaternion product:

\begin{matrix} p q & = to_quat (r (p) r (q) - v (p) \cdot v (q)) + to_quat (r (p) v (q) + r (q) v (p) + v (p) \times v (q)) \\ = to_quat (0 \cdot 0 - v (p) \cdot v (q)) + to_quat (0 \cdot v (q) + 0 \cdot v (p) + v (p) \times v (q)) \\ = to_quat (- v (p) \cdot v (q)) + to_quat (v (p) \times v (q)) \end{matrix}

Real Part of Product of Two Pure Quaternions

For any two pure quaternions

p, q \in ℍ

r (p q) = - v (p) \cdot v (q)

By the product of two pure quaternions, the real part is

- v (p) \cdot v (q)

Vector Part of Product of Two Pure Quaternions

For any two pure quaternions

p, q \in ℍ

v (p q) = v (p) \times v (q)

By the product of two pure quaternions, the vector part is

v (p) \times v (q)

Square of a Pure Unit Quaternion is Minus One

For any pure unit quaternion

\hat{n} \in ℍ

, we have

{\hat{n}}^{2} = - e

where

e

is the identity quaternion.

Since

\hat{n}

is pure and unit,

r (\hat{n}) = 0

and

‖ v (\hat{n}) ‖ = 1

. By the product of two pure quaternions:

{\hat{n}}^{2} = to_quat (- v (\hat{n}) \cdot v (\hat{n})) + to_quat (v (\hat{n}) \times v (\hat{n}))

Since

v (\hat{n}) \cdot v (\hat{n}) = ‖ v (\hat{n}) ‖^{2} = 1

and

v (\hat{n}) \times v (\hat{n}) = 𝟎

, we get:

{\hat{n}}^{2} = to_quat (- 1) + to_quat (𝟎) = - 1 + 0 𝐢 + 0 𝐣 + 0 𝐤 = - e

Perpendicular Pure Quaternions Anticommute

For any two pure quaternions

p, q \in ℍ

with

v (p) \cdot v (q) = 0

, we have

p q = - q p

By the product of two pure quaternions:

p q = to_quat (- v (p) \cdot v (q)) + to_quat (v (p) \times v (q))

Since

v (p) \cdot v (q) = 0

p q = to_quat (0) + to_quat (v (p) \times v (q)) = to_quat (v (p) \times v (q))

Similarly:

q p = to_quat (0) + to_quat (v (q) \times v (p)) = to_quat (v (q) \times v (p))

Since the cross product is anti-commutative,

v (q) \times v (p) = - (v (p) \times v (q))

, so

q p = - p q

, i.e.

p q = - q p

Quaternion Multiplication is Associative

For any

p, q, s \in ℍ

, we have

(p q) s = p (q s)

Write $p = a_{1} + b_{1} 𝐢 + c_{1} 𝐣 + d_{1} 𝐤$ , $q = a_{2} + b_{2} 𝐢 + c_{2} 𝐣 + d_{2} 𝐤$ , and $s = a_{3} + b_{3} 𝐢 + c_{3} 𝐣 + d_{3} 𝐤$ .

Since quaternion multiplication is defined to be distributive over addition and is determined by the multiplication rules on the basis elements ${1, 𝐢, 𝐣, 𝐤}$ , it suffices to verify associativity on all triples of basis elements. There are $4^{3} = 64$ such triples.

We verify the key non-trivial cases. Recall the basis multiplication rules derived from $𝐢^{2} = 𝐣^{2} = 𝐤^{2} = 𝐢 𝐣 𝐤 = - 1$ : $𝐢 𝐣 = 𝐤, 𝐣 𝐤 = 𝐢, 𝐤 𝐢 = 𝐣$ $𝐣 𝐢 = - 𝐤, 𝐤 𝐣 = - 𝐢, 𝐢 𝐤 = - 𝐣$

Case $(𝐢, 𝐣, 𝐤)$ : $(𝐢 𝐣) 𝐤 = 𝐤 \cdot 𝐤 = - 1$ and $𝐢 (𝐣 𝐤) = 𝐢 \cdot 𝐢 = - 1$ . ✓

Case $(𝐢, 𝐤, 𝐣)$ : $(𝐢 𝐤) 𝐣 = (- 𝐣) 𝐣 = - (𝐣^{2}) = 1$ and $𝐢 (𝐤 𝐣) = 𝐢 (- 𝐢) = - 𝐢^{2} = 1$ . ✓

Case $(𝐢, 𝐢, 𝐣)$ : $(𝐢 𝐢) 𝐣 = (- 1) 𝐣 = - 𝐣$ and $𝐢 (𝐢 𝐣) = 𝐢 𝐤 = - 𝐣$ . ✓

Case $(𝐣, 𝐤, 𝐢)$ : $(𝐣 𝐤) 𝐢 = 𝐢 \cdot 𝐢 = - 1$ and $𝐣 (𝐤 𝐢) = 𝐣 \cdot 𝐣 = - 1$ . ✓

All other non-trivial cases follow by similar computation. Any triple involving the identity $1$ is immediate since $1$ commutes with everything and acts as the identity. Since associativity holds on all basis triples and quaternion multiplication distributes over addition, associativity holds for all quaternions.

Product with a Real Quaternion

For any

q \in ℍ

and

a \in ℝ

to_quat (a) \cdot q = q \cdot to_quat (a) = a q

where

a q

denotes scalar multiplication.

Let

p = to_quat (a)

, so

r (p) = a

and

v (p) = 𝟎

. By the real to_quat definition and the quaternion product definition:

\begin{matrix} p q & = a \cdot r (q) - 𝟎 \cdot v (q) + a v (q) + r (q) \cdot 𝟎 + 𝟎 \times v (q) \\ = a r (q) + a v (q) = a (r (q) + v (q)) = a q \end{matrix}

Similarly

q p = r (q) \cdot a - v (q) \cdot 𝟎 + r (q) \cdot 𝟎 + a v (q) + v (q) \times 𝟎 = a r (q) + a v (q) = a q

Conjugation is Additive

Suppose that

p, q \in ℍ

then

p + q = p + q

Suppose that

p = a + b i + c j + d k

and

q = u + x i + y j + z k

then

\begin{matrix} p + q & = (a + u) + (b + x) i + (c + y) j + (d + z) k \\ = (a + u) - (b + x) i - (c + y) j - (d + z) k \\ = a - b i - c j - d k + u - x i - y j - z k \\ = p + q \end{matrix}

Vector Part of the Conjugate is Minus 1 Times the Original

For any

p \in ℍ

we have

v (p) = - v (p)

Suppose that

p = a + b i + c j + d k

. Then

v (p) = v (a - b i - c j - d k) = (- b, - c, - d) = - (b, c, d) = - v (p)

as needed.

Real Part of the Conjugate Doesn't Change

For any

p \in ℍ

we have

r (p) = r (p)

Suppose that

p = a + b i + c j + d k

. Then

r (p) = r (a - b i - c j - d k) = a = r (p)

as needed.

Conjugate Fixes Real Quaternions

For any

p \in ℍ

, such that

v (p) = 𝟎

, then

p = p

Since

v (p) = 𝟎

, we have

p = r (p) + 0 𝐢 + 0 𝐣 + 0 𝐤

. Then

p = r (p) - 0 𝐢 - 0 𝐣 - 0 𝐤 = p

Conjugate Only Applies to Vector Part

For any

p \in ℍ

we have

p = to_quat (r (p)) + to_quat (- v (p))

\begin{matrix} p & = r (p) - v (p) \\ = to_quat (r (p)) + to_quat (- v (p)) \end{matrix}

where the first equality follows from the definition of conjugation and the decomposition of a quaternion.

Conjugate of a Pure Quaternion is its Negation

For any pure quaternion

p \in ℍ

p = - p

Since

p

is pure,

r (p) = 0

. Then by the conjugate formula:

p = to_quat (0) + to_quat (- v (p)) = to_quat (- v (p)) = - p

Product Commutes in the Real Part

For any

p, q \in ℍ

we have

r (p q) = r (q p)

By the real part of a quaternion product,

r (p q) = r (p) r (q) - v (p) \cdot v (q) .

Using commutativity of real multiplication and commutativity of the dot product,

r (p) r (q) - v (p) \cdot v (q) = r (q) r (p) - v (q) \cdot v (p) .

By the same real-part formula, the final expression is

r (q p)

Quaternion Times its Conjugate

For any

q \in ℍ

q q = q q = to_quat (‖ q ‖^{2})

Let

q = a + b i + c j + d k

. Then

r (q) = a

v (q) = (b, c, d)

r (q) = a

v (q) = (- b, - c, - d)

\begin{matrix} r (q q) & = r (q) r (q) - v (q) \cdot v (q) \\ = a \cdot a - (b, c, d) \cdot (- b, - c, - d) \\ = a^{2} - (- b^{2} - c^{2} - d^{2}) \\ = a^{2} + b^{2} + c^{2} + d^{2} = ‖ q ‖^{2} \end{matrix}

\begin{matrix} v (q q) & = r (q) v (q) + r (q) v (q) + v (q) \times v (q) \\ = a (- b, - c, - d) + a (b, c, d) + (b, c, d) \times (- b, - c, - d) \\ = 𝟎 + 𝟎 = 𝟎 \end{matrix}

where

(b, c, d) \times (- b, - c, - d) = - (b, c, d) \times (b, c, d) = 𝟎

since the cross product of any vector with itself is zero. Therefore

q q = to_quat (‖ q ‖^{2})

. The proof that

q q = to_quat (‖ q ‖^{2})

is analogous.

Unit Quaternion Inverse is its Conjugate

For any unit quaternion

q \in ℍ

q^{- 1} = q

Since

‖ q ‖ = 1

, by the previous proposition,

q q = to_quat (1) = e

and

q q = e

. Therefore

q

is the multiplicative inverse of

q

Dot Product of Two Quaternions

Suppose that

p, q \in ℍ

, then we define

p \cdot q : = r (p) r (q) + v (p) \cdot v (q)

Conjugation is a Homomorphism in the Real Part

For any

p, q \in ℍ

we have

r (p q) = r (p q)

First of all, we know conjugation doesn't change the real part, so we have:

r (p q) = r (p q) = r (p) r (q) - v (p) \cdot v (q)

and then we know that

r (p q) = r (p) r (q) - v (p) \cdot v (q)

Since minus comes out of the vector part, and constants can get pulled out of the dot product we have that

r (p) r (q) - v (p) \cdot v (q) = r (p) r (q) - (- 1) (- 1) v (p) \cdot v (q) = r (p) r (q) - v (p) \cdot v (q)

therefore we have that

r (p q) = r (p q)

Conjugation Swaps Order in the Vector Part

For any

p, q \in ℍ

we have

v (p q) = v (q p)

We recall that, and so

v (p q) = - v (p q)

and then we compute:

\begin{matrix} v (q p) & = r (q) v (p) + r (p) v (q) + v (q) \times v (p) \\ = r (q) (- v (p)) + r (p) (- v (q)) + (- v (q)) \times (- v (p)) \\ = - r (q) v (p) - r (p) v (q) + v (q) \times v (p) \\ = - r (p) v (q) - r (q) v (p) - v (p) \times v (q) \\ = - (r (p) v (q) + r (q) v (p) + v (p) \times v (q)) \\ = - v (p q) \end{matrix}

where we used that

v (q) \times v (p) = - (v (p) \times v (q))

. Therefore

v (p q) = - v (p q) = v (q p)

Conjugation Distributes by Swapping

For any

p, q \in ℍ

we have

p q = q p

We show they are equal by checking both parts. For the real part:

r (p q) = r (p q) = r (q p) = r (q p) = r (q p)

where the first and third equalities use that conjugation preserves the real part, the second uses that the product commutes in the real part, and the last uses the conjugation homomorphism lemma. For the vector part, we have by the conjugation swaps order lemma that

v (p q) = v (q p)

. Therefore

p q = q p

Conjugation Preserves Norm

For any

q \in ℍ

‖ q ‖ = ‖ q ‖

Let

q = a + b i + c j + d k

. Then

q = a - b i - c j - d k

, so

‖ q ‖ = \sqrt{a^{2} + {(- b)}^{2} + {(- c)}^{2} + {(- d)}^{2}} = \sqrt{a^{2} + b^{2} + c^{2} + d^{2}} = ‖ q ‖

Quaternion Product is Norm-Multiplicative

For any

p, q \in ℍ

‖ p q ‖ = ‖ p ‖ \cdot ‖ q ‖

We compute:

\begin{matrix} ‖ p q ‖^{2} & = r (p q \cdot p q) \\ = r (p q \cdot q p) \end{matrix}

by conjugation distributing by swapping. By associativity:

p q \cdot q p = p (q q) p = p \cdot to_quat (‖ q ‖^{2}) \cdot p = ‖ q ‖^{2} (p p) = ‖ q ‖^{2} \cdot to_quat (‖ p ‖^{2})

where we used that

q q = to_quat (‖ q ‖^{2})

and that real quaternions commute with everything. Therefore

‖ p q ‖^{2} = ‖ p ‖^{2} ‖ q ‖^{2}

, and taking square roots gives

‖ p q ‖ = ‖ p ‖ \cdot ‖ q ‖

Conjugation of a Pure Quaternion by a Unit Quaternion is Pure

For any pure quaternion

p \in ℍ

and unit quaternion

q \in ℍ

, the quaternion

q p q

is pure, i.e.

r (q p q) = 0

We use the fact that a quaternion $x$ is pure if and only if $x = - x$ . We verify this for $x = q p q$ :

\begin{matrix} q p q & = q p q \\ = q (- p) q \\ = - (q p q) \end{matrix}

where we used conjugation distributes by swapping (applied twice), the fact that $q = q$ , and the fact that $p = - p$ since $p$ is pure. Since $q p q = - (q p q)$ , the quaternion $q p q$ is pure.

Conjugation by a Unit Quaternion Preserves the Norm of the Vector Part

For any pure quaternion

p \in ℍ

and unit quaternion

q \in ℍ

‖ q p q ‖ = ‖ p ‖

By norm-multiplicativity:

‖ q p q ‖ = ‖ q ‖ \cdot ‖ p ‖ \cdot ‖ q ‖ = 1 \cdot ‖ p ‖ \cdot 1 = ‖ p ‖

where we used that

‖ q ‖ = 1

and

‖ q ‖ = ‖ q ‖ = 1

Parallel and Perpendicular Decomposition

Given a vector

𝐰 \in ℝ^{3}

and a unit vector

\hat{n} \in ℝ^{3}

, we define the parallel and perpendicular components of

𝐰

with respect to

\hat{n}

as:

𝐰_{∥ \hat{n}} : = (\hat{n} \cdot 𝐰) \hat{n}, 𝐰_{⟂ \hat{n}} : = 𝐰 - 𝐰_{∥ \hat{n}}

so that

𝐰 = 𝐰_{∥ \hat{n}} + 𝐰_{⟂ \hat{n}}

, where

\hat{n} \cdot 𝐰_{⟂ \hat{n}} = 0

Pure Unit Quaternion Times a Parallel Pure Quaternion

Let

\hat{n}

be a pure unit quaternion and let

p_{∥ \hat{n}}

be a pure quaternion such that

v (p_{∥ \hat{n}}) = λ v (\hat{n})

for some

λ \in ℝ

. Then

\hat{n} p_{∥ \hat{n}} = to_quat (- λ)

In particular,

\hat{n} p_{∥ \hat{n}}

is a real quaternion.

Since both are pure, by the product of two pure quaternions:

\begin{matrix} \hat{n} p_{∥ \hat{n}} & = to_quat (- v (\hat{n}) \cdot v (p_{∥ \hat{n}})) + to_quat (v (\hat{n}) \times v (p_{∥ \hat{n}})) \\ = to_quat (- v (\hat{n}) \cdot λ v (\hat{n})) + to_quat (v (\hat{n}) \times λ v (\hat{n})) \\ = to_quat (- λ ‖ v (\hat{n}) ‖^{2}) + to_quat (λ (v (\hat{n}) \times v (\hat{n}))) \\ = to_quat (- λ) + to_quat (𝟎) \\ = to_quat (- λ) \end{matrix}

where we used

‖ v (\hat{n}) ‖ = 1

and

v (\hat{n}) \times v (\hat{n}) = 𝟎

Pure Unit Quaternion Times a Perpendicular Pure Quaternion

Let

\hat{n}

be a pure unit quaternion and let

p_{⟂ \hat{n}}

be a pure quaternion such that

v (\hat{n}) \cdot v (p_{⟂ \hat{n}}) = 0

. Then

\hat{n} p_{⟂ \hat{n}} = to_quat (v (\hat{n}) \times v (p_{⟂ \hat{n}}))

In particular,

\hat{n} p_{⟂ \hat{n}}

is a pure quaternion.

Since both are pure, by the product of two pure quaternions:

\begin{matrix} \hat{n} p_{⟂ \hat{n}} & = to_quat (- v (\hat{n}) \cdot v (p_{⟂ \hat{n}})) + to_quat (v (\hat{n}) \times v (p_{⟂ \hat{n}})) \\ = to_quat (0) + to_quat (v (\hat{n}) \times v (p_{⟂ \hat{n}})) \\ = to_quat (v (\hat{n}) \times v (p_{⟂ \hat{n}})) \end{matrix}

Pure Unit Quaternion Sandwich is a Reflection

Let

\hat{n} \in ℍ

be a pure unit quaternion, and let

p \in ℍ

be a pure quaternion. Then

\hat{n} p \hat{n} = to_quat (- v (p)_{∥ v (\hat{n})} + v (p)_{⟂ v (\hat{n})})

where the subscripts refer to the parallel and perpendicular decomposition with respect to

v (\hat{n})

. This is the reflection of

v (p)

through the plane perpendicular to

v (\hat{n})

Set $p_{∥ \hat{n}} = to_quat (v {(p)}_{∥ v (\hat{n})})$ and $p_{⟂ \hat{n}} = to_quat (v {(p)}_{⟂ v (\hat{n})})$ . Then:

$p_{∥ \hat{n}}$ and $p_{⟂ \hat{n}}$ are pure quaternions.
$v (\hat{n}) \cdot v (p_{⟂ \hat{n}}) = 0$ .
$v (p_{∥ \hat{n}}) = λ v (\hat{n})$ , where $λ = v (\hat{n}) \cdot v (p)$ .

Since $p = p_{∥ \hat{n}} + p_{⟂ \hat{n}}$ , by the associativity and distributive law: $\begin{matrix} \hat{n} p \hat{n} & = \hat{n} (p_{∥ \hat{n}} + p_{⟂ \hat{n}}) \hat{n} \\ = \hat{n} p_{∥ \hat{n}} \hat{n} + \hat{n} p_{⟂ \hat{n}} \hat{n} . \end{matrix}$ Thus it is enough to derive the values of $\hat{n} p_{∥ \hat{n}} \hat{n}$ and $\hat{n} p_{⟂ \hat{n}} \hat{n}$ .

By the parallel product lemma, $\hat{n} p_{∥ \hat{n}} = to_quat (- λ)$ . Since this is a real quaternion, by the real product property: $\hat{n} p_{∥ \hat{n}} \hat{n} = to_quat (- λ) \cdot \hat{n} = - λ \hat{n} = to_quat (- λ v (\hat{n})) = to_quat (- v {(p)}_{∥ v (\hat{n})})$

Since $v (\hat{n}) \cdot v (p_{⟂ \hat{n}}) = 0$ , the quaternions $\hat{n}$ and $p_{⟂ \hat{n}}$ anticommute: $\hat{n} p_{⟂ \hat{n}} = - p_{⟂ \hat{n}} \hat{n}$ . Therefore: $\hat{n} p_{⟂ \hat{n}} \hat{n} = (- p_{⟂ \hat{n}} \hat{n}) \hat{n} = - p_{⟂ \hat{n}} (\hat{n} \hat{n}) = - p_{⟂ \hat{n}} {\hat{n}}^{2}$ By the square of a pure unit quaternion, ${\hat{n}}^{2} = - e$ , so: $- p_{⟂ \hat{n}} (- e) = p_{⟂ \hat{n}} \cdot e = p_{⟂ \hat{n}} = to_quat (v {(p)}_{⟂ v (\hat{n})})$

Therefore: $\begin{matrix} \hat{n} p \hat{n} & = \hat{n} p_{∥ \hat{n}} \hat{n} + \hat{n} p_{⟂ \hat{n}} \hat{n} \\ = to_quat (- v {(p)}_{∥ v (\hat{n})}) + to_quat (v {(p)}_{⟂ v (\hat{n})}) \\ = to_quat (- v {(p)}_{∥ v (\hat{n})} + v {(p)}_{⟂ v (\hat{n})}) . \end{matrix}$

Unit Quaternion Factors into Two Pure Unit Quaternions

Let

q = \cos (\frac{θ}{2}) + \hat{n} \sin (\frac{θ}{2})

be a unit quaternion, where

\hat{n}

is a pure unit quaternion. Then there exist pure unit quaternions

\hat{a}, \hat{b} \in ℍ

with

v (\hat{a}) \cdot v (\hat{n}) = 0

and

v (\hat{b}) \cdot v (\hat{n}) = 0

, such that:

q = \hat{b} \hat{a}

Moreover, the acute angle between the two reflection planes perpendicular to

v (\hat{a})

and

v (\hat{b})

\frac{θ}{2}

, and the planes intersect along the axis

v (\hat{n})

Let $ϕ = \frac{θ}{2}$ . Choose a unit vector $𝐚 \in ℝ^{3}$ with $𝐚 \cdot v (\hat{n}) = 0$ , and define $𝐛 = - \cos (ϕ) 𝐚 - \sin (ϕ) (v (\hat{n}) \times 𝐚) .$ Set $\hat{a} = to_quat (𝐚)$ and $\hat{b} = to_quat (𝐛)$ .

Since $𝐚 ⟂ v (\hat{n})$ , the vector $v (\hat{n}) \times 𝐚$ is also perpendicular to $v (\hat{n})$ , so $𝐛 \cdot v (\hat{n}) = 0$ . Also $𝐚$ and $v (\hat{n}) \times 𝐚$ are perpendicular unit vectors, so $‖ 𝐛 ‖^{2} = \cos^{2} (ϕ) + \sin^{2} (ϕ) = 1.$ Therefore $\hat{a}$ and $\hat{b}$ are pure unit quaternions.

We now compute $\hat{b} \hat{a}$ using the product of two pure quaternions. First, $𝐛 \cdot 𝐚 = - \cos (ϕ) .$ Next, using $(v (\hat{n}) \times 𝐚) \times 𝐚 = - v (\hat{n})$ , $𝐛 \times 𝐚 = \sin (ϕ) v (\hat{n}) .$ Thus $\begin{matrix} \hat{b} \hat{a} & = to_quat (- 𝐛 \cdot 𝐚) + to_quat (𝐛 \times 𝐚) \\ = to_quat (\cos (ϕ)) + to_quat (\sin (ϕ) v (\hat{n})) \\ = \cos (ϕ) + \hat{n} \sin (ϕ) = q . \end{matrix}$ The normal vectors $𝐚$ and $𝐛$ have angle $π - ϕ$ , so the acute angle between the corresponding reflection planes is $ϕ = \frac{θ}{2}$ .

Unit Quaternion Rotation Formula

Let

q = \cos (\frac{θ}{2}) + \hat{n} \sin (\frac{θ}{2})

be a unit quaternion, where

\hat{n}

is a pure unit quaternion, and let

p \in ℍ

be a pure quaternion. Then

q p q

is the pure quaternion whose vector part is obtained by rotating

v (p)

around the axis

v (\hat{n})

by angle

θ

By the factorization of a unit quaternion into two pure unit quaternions, after renaming the two factors, there are pure unit quaternions $\hat{a}$ and $\hat{b}$ such that $q = \hat{a} \hat{b},$ and the angle between the reflection planes perpendicular to $v (\hat{a})$ and $v (\hat{b})$ is $\frac{θ}{2}$ . Since $\hat{a}$ and $\hat{b}$ are pure unit quaternions, we have $\hat{a} = - \hat{a}$ and $\hat{b} = - \hat{b}$ . Therefore: $\begin{matrix} q & = \hat{a} \hat{b} \\ = \hat{b} \hat{a} \\ = (- \hat{b}) (- \hat{a}) \\ = \hat{b} \hat{a} . \end{matrix}$

Using associativity: $\begin{matrix} q p q & = (\hat{a} \hat{b}) p (\hat{b} \hat{a}) \\ = \hat{a} (\hat{b} p \hat{b}) \hat{a} . \end{matrix}$

By the pure unit quaternion sandwich reflection theorem, $\hat{b} p \hat{b}$ is the reflection of $v (p)$ through the plane perpendicular to $v (\hat{b})$ . Applying the same theorem again, $\hat{a} (\hat{b} p \hat{b}) \hat{a}$ reflects the result through the plane perpendicular to $v (\hat{a})$ .

Thus $q p q$ is obtained by composing two plane reflections whose planes intersect along the axis $v (\hat{n})$ and whose angle is $\frac{θ}{2}$ . By the composition of two plane reflections theorem, this composition is rotation around $v (\hat{n})$ by angle $θ$ .