Quaternions

In this article, we describe what Quaternions are, how to use them, and their applications in Computer Graphics.

Defining a Quaternion §

Mathematical Definition §

Recall a complex number $c$. All complex numbers can be represented as

$$c=a+bi$$

Where $a,b\in \mathbb{R}$, and $i=\sqrt{i}$, the imaginary number.

Quaternions were invented as an extension of these imaginary numbers $i$. Mathematically, a quaternion $\hat{q}$ is defined as

$$\begin{gathered} \hat{q}=i{q}_x+j{q}_y+k{q}_z+q_w \\ \hat{q}=q_v+q_w=(q_v,q_w)q_v=i{q}_x+j{q}_y+k{q}_z=(q_x,q_y,q_z) \end{gathered}$$

Note that $q_v$ is a vector, and can be used with normal vector operations!

Where $q_x,q_y,q_z,q_w\in \mathbb{R}$ are real components of our quaternion, and $i,j,k$ are imaginary units such that

$$\begin{gathered} i^2=j^2=k^2=-1 \\ jk=-kj=iki=-ik=jij=-ji=k \end{gathered}$$

Note that by this definition, multiplication with these units is not commutative.

Components of each of these imaginary units are closely related to the axes of rotation!

One convenient way of understanding these imaginary unit conversions is by imagining the following cycle

graph LR
    i -.-> j -.-> k -.-> i;

Where for the product of any two units, its result is the third unit, and its sign is determined by if we “follow the arrows” (positive) or not (negative) on this cycle.

Example: Product of Imaginary Units

Using our cycle above, we see that the product of $i\cdot k$ is j, and as the path from $i$ to $k$ goes against the arrow, our result is negative. Thus,

$$i\cdot k=-j$$

Quaternion Operations §

Suppose we have two quaternions $\hat{q}$ and $\hat{r}$.

$$\begin{gathered} \hat{q}=i{q}_x+j{q}_y+k{q}_z+q_w \\ \hat{r}=i{r}_x+j{r}_y+k{r}_z+r_w \end{gathered}$$

Then, the following operations are defined.

$$\begin{array}{rlr}& \text{Product}& \hat{q}\cdot \hat{r}=(q_v\times r_v+r_w{q}_v+q_w{r}_v,q_w{r}_w-q_v\cdot r_v)\\ & \text{Addition}& \hat{q}+\hat{r}=(q_v+r_v,q_w+r_w)\\ & \text{Conjugate}& {\hat{q}}^{\ast }=(-q_v,q_w)\\ & \text{Norm}& n(\hat{q})=||\hat{q}||=\sqrt{\hat{q}{\hat{q}}^{\ast }}=\sqrt{q_x^2+q_y^2+q_z^2+q_w^2}\\ & \text{Identity}& \hat{i}=(0,1)\end{array}$$

These operations can easily be defined by considering their normal meanings, and applying them to the quaternions as so. See below for an example derivation.

Example: Quaternion Product

We calculate the product of two quaternions below, by considering their definition, taking the product, and distributing.

$$\begin{array}{rl}\hat{q}\cdot \hat{r}& =(i{q}_x+j{q}_y+k{q}_z+q_w)(i{r}_x+j{r}_y+k{r}_z+r_w)\\ & =i(q_y{r}_z-q_z{r}_y+r_w{q}_z+q_w{r}_x)+j(q_z{r}_x-q_x{r}_x+r_w{q}_y+q_w{r}_y)\\ & +k(q_x{r}_y-q_y{r}_x+r_w{q}_z+q_w{r}_z)\\ & +(q_w{r}_w-q_x{r}_x-q_y{r}_y-q_z{r}_z)\\ & =(q_v\times r_v+r_w{q}_v+q_w{r}_v,q_w{r}_w-q_v\cdot r_v\end{array}$$

From these definitions, we can define a variety of other rules! They are given below.

$$\begin{array}{rlr}& \text{Multiplicative Inverse}& {\hat{q}}^{-1}=\frac{1}{n(\hat{q}{)}^2}{\hat{q}}^{\ast }\\ & \text{Scalar Multiplication}& s\hat{q}=(0,s)(q_v,q_w)=(s{q}_v,s{q}_w)\\ & \text{Conjugate Rules}& ({\hat{q}}^{\ast }{)}^{\ast }=\hat{q}\\ & & (\hat{q}+\hat{r}{)}^{\ast }={\hat{q}}^{\ast }+{\hat{r}}^{\ast }\\ & & (\hat{q}\hat{r}{)}^{\ast }={\hat{r}}^{\ast }{\hat{q}}^{\ast }\\ & \text{Norm Rules}& n({\hat{q}}^{\ast })=n(\hat{q})\\ & & n(\hat{q}\hat{r})=n(\hat{q})n(\hat{r})\\ & \text{Linearity of Multiplication}& \hat{p}(s\hat{q}+t\hat{r})=s\hat{p}\hat{q}+t\hat{p}\hat{r}\\ & & (s\hat{p}+t\hat{q})\hat{r}=s\hat{p}\hat{r}+t\hat{q}\hat{r}\\ & \text{Associativity of Multiplication}& \hat{p}(\hat{q}\hat{r})=(\hat{p}\hat{q})\hat{r}\end{array}$$

Unit Quaternions §

Definition §

A quaternion $\hat{q}=(q_v,q_w)$ is a unit quaternion when $||\hat{q}||=1$.

Now suppose we have any 3-dimensional unit vector $u_q$. Using this, we can write a unit quaternion as

$$\hat{q}=(\sin \theta u_q,\cos \theta )$$

And observe that

$$||\hat{q}||=\sqrt{{\sin }^2\theta (u_q\cdot u_q)+{\cos }^2\theta }=1$$

This definition of unit quaternions is, interestingly enough, extremely well-suited for rotations and orientations!

This is described in the next section.

Quaternion Transforms §

Consider a quaternion $\hat{p}=(p_v,p_w)$, where $p_v$ represents a 3D position in space, and consider a unit quaternion $\hat{q}=(\sin \theta u_q,\cos \theta )$.

We can prove that the quaternion product

$$\hat{q}\hat{p}{\hat{q}}^{-1}$$

Rotates $\hat{p}$ (and thus our point $p_v$) around the axis $u_q$ by the angle $2\theta$! Thus, this product is equal to us applying a rotation transform onto our point $p_v$!

Note that because $\hat{q}$ is a unit quaternion, $q^{-1}={\hat{q}}^{\ast }$.

By this, and our definition of quaternion product, we see that if we wanted to perform multiple transforms, we have

$$\begin{array}{rl}\hat{r}(\hat{q}\hat{p}{\hat{q}}^{\ast }){\hat{r}}^{\ast }& =(\hat{r}\hat{q})\hat{p}({\hat{q}}^{\ast }{\hat{r}}^{\ast })\\ & =(\hat{r}\hat{q})\hat{p}(\hat{r}\hat{q}{)}^{\ast }\\ & =\hat{c}\hat{p}{\hat{c}}^{\ast }\end{array}$$

Where $\hat{c}=\hat{r}\hat{q}$. In other words, the product of our unit quaternions gave us a new quaternion representing the transform of applying both!

This provides us a simple operation that gives us a very convenient way to apply rotations on points in 3D space!

Matrix Conversions §

It’s often the case that several different transforms need to be applied to a point, and most of these transforms are given in a matrix form.

It can be shown that for our unit quaternion $\hat{q}$, we can convert it to matrix form

$$M^q=\left[\begin{array}{cccc}1-2(q_y^2+q_z^2)& 2(q_x{q}_y-q_w{q}_z)& 2(q_x{q}_z+q_w{q}_y)& 0\\ 2(q_x{q}_y+q_w{q}_z)& 1-2(q_x^2+q_z^2)& 2(q_y{q}_z-q_w{q}_x)& 0\\ 2(q_z{q}_z-q_w{q}_y)& 2(q_y{q}_z+q_w{q}_z)& 1-2(q_x^2+q_y^2)& 0\\ 0& 0& 0& 1\end{array}\right]$$

Note that given this, we don’t need to compute any trigonometric functions to get our transform! This makes our matrix conversion process very efficient in practice.