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.

Conjugation Distributes

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,

Conjugation is Additive

Suppose that

p, q \in H

then

\bar{p + q} = \bar{p} + \bar{q}

Suppose that

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

and

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

then

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

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 := i + j + k

Conjugate Quaternion

Given the quaternion

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

, it's conjugate is given by

\bar{q} := a - b i - c j - d k

Vector part of the Conjugate is minus 1 Times the Original

For any

p \in H

we have

v (\bar{p}) = - 1 v (p)

Suppose that

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

v (\bar{p}) = - b i + - c j + - d k = (- 1) v (p)

as needed.

Real part of the Conjugate doesn't Change

For any

p \in H

we have

r (\bar{p}) = r (p)

Suppose that

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

r (\bar{p}) = a = r (p)

as needed.

Conjugate Doesn't change the Real Part

For any

p \in H

, such that

v (p) = 0

, then

\bar{p} = p

Conjugate only Applies to Vector Part

For any

p \in H

we have

\bar{p} = r (p) - v (p)

\bar{p} = \bar{r (p) + v (p)} = r (p) - v (p)

Pure Quaternion

A quaternion

p \in H

is said to be pure, deiff

r (p) = 0

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)

Product Commutes in the Real Part

For any

p, q \in H

we have

r (p q) = r (q p)

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

Dot Product of Two Quaternions

Suppose that

p, q \in H

, then we define

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

Conjugation is a Homomorphism in the Real Part

For any

p, q \in H

we have

r (\bar{p q}) = r (\bar{p} \bar{q})

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

r (\bar{p q}) = r (p q) = r (p) r (q) - v (p) \cdot v (q)

and then we know that

r (\bar{p} \bar{q}) = r (\bar{p}) r (\bar{q}) - v (\bar{p}) \cdot v (\bar{q})

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

r (\bar{p}) r (\bar{q}) - v (\bar{p}) \cdot v (\bar{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 (\bar{p q}) = r (\bar{p} \bar{q})

Conjugation Swaps Order in the Vector Part

For any

p, q \in H

we have

v (\bar{p q}) = v (\bar{q} \bar{p})

We recall that, and so

v (\bar{p q}) = v (r (p q) - v (p q)) = - v (p q)

and then we know that

\begin{aligned} v (\bar{q} \bar{p}) & = r (\bar{q}) v (\bar{p}) + r (\bar{p}) v (\bar{q}) + v (\bar{q}) \times v (\bar{p}) \\ = - r (q) v (p) - r (p) v (q) + v (\bar{q}) \times v (\bar{p}) \\ = - r (q) v (p) - r (p) v (q) + (- 1) (- 1) v (q) \times v (p) \\ = - r (p) v (q) - r (q) v (p) - v (p) \times v (q) \\ = - v (q p) \end{aligned}

as needed.

Conjugation Distributes by Swapping

For any

p, q \in H

we have

\bar{p q} = \bar{q} \bar{p}

We show they are equal:

r (\bar{p q}) = r (\bar{p} \bar{q}) = r (\bar{q} \bar{p})

and then we recall

v (\bar{p q}) = v (\bar{q} \bar{p})

, so that

\bar{p q} = \bar{q} \bar{p}

Dot Product and Regular Product Commute

Suppose that

p, q \in H

, then we define

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

Quaternion times its Conjugate Equation

Suppose that

p \in H

, then we have:

q \bar{q} = (q \cdot q) e