Unit Quaternion 3D Rotation Representation

Introduction

In my previous article “Axis/Angle 3D Rotation Representation”, we have learned the axis/angle 3D rotation representation, there is another commonly used representation which is called unit quaternion 3D rotation representation.

In this blog post, I would like to discuss the quaternion 3D rotation representation and derive some of its properties mathematically.

Axis/Angle Rotation Representation

Here is a quick recovery from my previous article “Axis/Angle 3D Rotation Representation” on the axis/angle rotation representation.

$$\mathbf{R}(\hat{\mathbf{n}},\theta )=\mathbf{I}+[\hat{\mathbf{n}}{]}_{\times }\sin \theta +[\hat{\mathbf{n}}{]}_{\times }^2(1-\cos \theta )$$

where the unit rotation vector $\hat{\mathbf{n}}=({\hat{n}}_1,{\hat{n}}_2,{\hat{n}}_3)$ and $\Vert \hat{\mathbf{n}}\Vert =\sqrt{{\hat{n}}_1^2+{\hat{n}}_2^2+{\hat{n}}_3^2}=1$ and

$$[\hat{\mathbf{n}}{]}_{\times }=\left[\begin{array}{ccc}0& -{\hat{n}}_3& {\hat{n}}_2\\ {\hat{n}}_3& 0& -{\hat{n}}_1\\ -{\hat{n}}_2& {\hat{n}}_1& 0\end{array}\right]$$

Quaternion Rotation Representation

The quaternion rotation representation is closely related to the axis/angle rotation representation. The two representations could be converted from each other.

$$\begin{array}{rl}\mathbf{q}& =(x,y,z,w)\\ & =(\mathbf{v},w)\\ & =(\hat{\mathbf{n}}\sin \frac{\theta }{2},\cos \frac{\theta }{2})\\ & =({\hat{n}}_1\sin \frac{\theta }{2},{\hat{n}}_2\sin \frac{\theta }{2},{\hat{n}}_3\sin \frac{\theta }{2},\cos \frac{\theta }{2})\end{array}$$

where

$$\begin{array}{rl}x& ={\hat{n}}_1\sin \frac{\theta }{2}\\ y& ={\hat{n}}_2\sin \frac{\theta }{2}\\ z& ={\hat{n}}_3\sin \frac{\theta }{2}\\ w& =\cos \frac{\theta }{2}\end{array}$$

It is easy to verify that $\mathbf{q}$ is a unit vector.

$$\begin{array}{rl}\Vert \mathbf{q}\Vert & =\sqrt{\Vert \hat{\mathbf{n}}{\Vert }^2{\sin }^2\frac{\theta }{2}+{\cos }^2\frac{\theta }{2}}\\ & =\sqrt{{\sin }^2\frac{\theta }{2}+{\cos }^2\frac{\theta }{2}}\\ & =1\end{array}$$

Quaternion

Because $\sin \theta =2\sin \frac{\theta }{2}\cos \frac{\theta }{2}$ and $1-\cos \theta =2{\sin }^2\frac{\theta }{2}$,

$$\begin{array}{rl}\mathbf{R}(\hat{\mathbf{n}},\theta )& =\mathbf{I}+[\hat{\mathbf{n}}{]}_{\times }\sin \theta +[\hat{\mathbf{n}}{]}_{\times }^2(1-\cos \theta )\\ & =\mathbf{I}+[\hat{\mathbf{n}}{]}_{\times }(2\sin \frac{\theta }{2}\cos \frac{\theta }{2})+[\hat{\mathbf{n}}{]}_{\times }^2(2{\sin }^2\frac{\theta }{2})\\ & =\mathbf{I}+2w([\hat{\mathbf{n}}{]}_{\times }\sin \frac{\theta }{2})+2{([\hat{\mathbf{n}}{]}_{\times }\sin \frac{\theta }{2})}^2\end{array}$$

We define a new matrix $[\hat{\mathbf{v}}{]}_{\times }$,

$$\begin{array}{rl}[\hat{\mathbf{v}}{]}_{\times }& =[\hat{\mathbf{n}}{]}_{\times }\sin \frac{\theta }{2}\\ & =\left[\begin{array}{ccc}0& -{\hat{n}}_3\sin \frac{\theta }{2}& {\hat{n}}_2\sin \frac{\theta }{2}\\ {\hat{n}}_3\sin \frac{\theta }{2}& 0& -{\hat{n}}_1\sin \frac{\theta }{2}\\ -{\hat{n}}_2\sin \frac{\theta }{2}& {\hat{n}}_1\sin \frac{\theta }{2}& 0\end{array}\right]\\ & =\left[\begin{array}{ccc}0& -z& y\\ z& 0& -x\\ -y& x& 0\end{array}\right]\end{array}$$

Thus,

$$\begin{array}{rl}[\hat{\mathbf{v}}{]}_{\times }^2& =\left[\begin{array}{ccc}0& -z& y\\ z& 0& -x\\ -y& x& 0\end{array}\right]\left[\begin{array}{ccc}0& -z& y\\ z& 0& -x\\ -y& x& 0\end{array}\right]\\ & =\left[\begin{array}{ccc}-y^2-z^2& xy& xz\\ xy& -x^2-z^2& yz\\ xz& yz& -x^2-y^2\end{array}\right]\end{array}$$

Therefore, the rotation matrix could be expressed using $\mathbf{q}=(x,y,z,w)$.

$$\begin{array}{rl}\mathbf{R}(\hat{\mathbf{n}},\theta )& =\mathbf{I}+2w([\hat{\mathbf{n}}{]}_{\times }\sin \frac{\theta }{2})+2{([\hat{\mathbf{n}}{]}_{\times }\sin \frac{\theta }{2})}^2\\ & =\mathbf{I}+2w[\hat{\mathbf{v}}{]}_{\times }+2[\hat{\mathbf{v}}{]}_{\times }^2\\ & =\left[\begin{array}{ccc}1-2(y^2+z^2)& 2(xy-zw)& 2(xz+yw)\\ 2(xy+zw)& 1-2(x^2+z^2)& 2(yz-xw)\\ 2(xz-yw)& 2(yz+xw)& 1-2(x^2+y^2)\end{array}\right]\end{array}$$

Formally, the rotation matrix $\mathbf{R}(\mathbf{q})$ could be expressed as

$$\begin{array}{rl}\mathbf{R}(\mathbf{q})& =\left[\begin{array}{ccc}1-2(y^2+z^2)& 2(xy-zw)& 2(xz+yw)\\ 2(xy+zw)& 1-2(x^2+z^2)& 2(yz-xw)\\ 2(xz-yw)& 2(yz+xw)& 1-2(x^2+y^2)\end{array}\right]\end{array}$$

Notice that

$$\begin{array}{rl}1-2(y^2+z^2)& =(1-(y^2+z^2))-(y^2+z^2)\\ & =(1-{\sin }^2\frac{\theta }{2}({\hat{n}}_2^2+{\hat{n}}_3^2))-(y^2+z^2)\\ & =(1-{\sin }^2\frac{\theta }{2}(1-{\hat{n}}_1^2))-(y^2+z^2)\\ & =(1-{\sin }^2\frac{\theta }{2}+{\sin }^2\frac{\theta }{2}{\hat{n}}_1^2)-(y^2+z^2)\\ & =({\cos }^2\frac{\theta }{2}+{\sin }^2\frac{\theta }{2}{\hat{n}}_1^2)-(y^2+z^2)\\ & =w^2+x^2-y^2-z^2\end{array}$$

Similarly,

$$\begin{array}{rl}1-2(x^2+z^2)& =w^2+y^2-x^2-z^2\end{array}$$

$$\begin{array}{rl}1-2(x^2+y^2)& =w^2+z^2-x^2-y^2\end{array}$$

So the rotation matrix $\mathbf{R}(\mathbf{q})$ could also be expressed as

$$\begin{array}{rl}\mathbf{R}(\mathbf{q})& =\left[\begin{array}{ccc}{w}^2+x^2-y^2-z^2& 2(xy-zw)& 2(xz+yw)\\ 2(xy+zw)& w^2+y^2-x^2-z^2& 2(yz-xw)\\ 2(xz-yw)& 2(yz+xw)& w^2+z^2-x^2-y^2\end{array}\right]\end{array}$$

Quaternion Algebra and Properties

Identity Quaternion

It is easy to find that the identity quaternion is $\mathbf{q}=(\mathbf{0},1)=(0,0,0,1)$.

Antipodal Quaternion

Quaternion $\mathbf{q}=(\mathbf{v},w)=(x,y,z,w)$ and its antipodal quaternion $-\mathbf{q}=(-\mathbf{v},-w)=(-x,-y,-z,-w)$ represent the same rotation, because it is easy to find that the rotational matrices $\mathbf{R}(\mathbf{q})$ and $\mathbf{R}(-\mathbf{q})$ are exactly the same, i.e., $\mathbf{R}(\mathbf{q})=\mathbf{R}(-\mathbf{q})$.

Quaternion Multiplication

Suppose ${\mathbf{q}}_0=({\mathbf{v}}_0,w_0)=(x_0,y_0,z_0,w_0)$, ${\mathbf{q}}_1=({\mathbf{v}}_1,w_1)=(x_1,y_1,z_1,w_1)$, ${\mathbf{q}}_2=({\mathbf{v}}_2,w_2)=(x_2,y_2,z_2,w_2)$, quaternion multiplication is defined as follows

$$\begin{array}{rl}{\mathbf{q}}_2& ={\mathbf{q}}_0{\mathbf{q}}_1\\ & =({\mathbf{v}}_0\times {\mathbf{v}}_1+w_0{\mathbf{v}}_1+w_1{\mathbf{v}}_0,w_0{w}_1-{\mathbf{v}}_0{\mathbf{v}}_1)\end{array}$$

It carries the property that $\mathbf{R}({\mathbf{q}}_2)=\mathbf{R}({\mathbf{q}}_0)\mathbf{R}({\mathbf{q}}_1)$, which means quaternion multiplication is just applying two rotations ${\mathbf{q}}_1$ and ${\mathbf{q}}_0$ sequentially (not ${\mathbf{q}}_0$ and ${\mathbf{q}}_1$!). The verification of this property is skipped in this article as the algebra is too complex.

Concretely,

$$\begin{array}{rl}{\mathbf{q}}_2& ={\mathbf{q}}_0{\mathbf{q}}_1\\ & =({\mathbf{v}}_0\times {\mathbf{v}}_1+w_0{\mathbf{v}}_1+w_1{\mathbf{v}}_0,w_0{w}_1-{\mathbf{v}}_0{\mathbf{v}}_1)\\ & =((y_0{z}_1-z_0{y}_1,z_0{x}_1-x_0{z}_1,x_0{y}_1-y_0{x}_1)+(w_0{x}_1,w_0{y}_1,w_0{z}_1)\\ & +(w_1{x}_0,w_1{y}_0,w_1{z}_0),w_0{w}_1-(x_0{x}_1+y_0{y}_1+z_0{z}_1))\\ & =(x_2,y_2,z_2,w_2)\end{array}$$

where

$$\begin{array}{rl}{x}_2& =y_0{z}_1-z_0{y}_1+w_0{x}_1+w_1{x}_0\\ y_2& =z_0{x}_1-x_0{z}_1+w_0{y}_1+w_1{y}_0\\ z_2& =x_0{y}_1-y_0{x}_1+w_0{z}_1+w_1{z}_0\\ w_2& =w_0{w}_1-x_0{x}_1-y_0{y}_1-z_0{z}_1\end{array}$$

Notice that quaternion Multiplication is not communicative, i.e., usually ${\mathbf{q}}_0{\mathbf{q}}_1\ne {\mathbf{q}}_1{\mathbf{q}}_0$, simply because the rotation axis of the two rotations could be different. In addition, quaternion multiplication with identity quaternion does not change the quaternion, i.e., $\mathbf{q}{\mathbf{q}}_0={\mathbf{q}}_0\mathbf{q}=\mathbf{q}$, when ${\mathbf{q}}_0=(\mathbf{0},1)$.

Inverse Quaternion

Quaternion $\mathbf{q}=(\mathbf{v},w)=(x,y,z,w)$ and its inverse quaternion ${\mathbf{q}}^{-1}$ cancel each other.

From my previous article “Axis/Angle 3D Rotation Representation”, we already know that $(\hat{\mathbf{n}},-\theta )$ or $(-\hat{\mathbf{n}},\theta )$ cancels $(\hat{\mathbf{n}},\theta )$. Because the axis/angle representation $(\hat{\mathbf{n}},\theta )$ corresponds to the quaternion representation $(\mathbf{v},w)=(x,y,z,w)$, $(\hat{\mathbf{n}},-\theta )$ corresponds to $(-\mathbf{v},w)=(-x,-y,-z,w)$, $(-\hat{\mathbf{n}},\theta )$ corresponds to $(-\mathbf{v},w)=(-x,-y,-z,w)$, we could say $(-\mathbf{v},w)$ cancels $(\mathbf{v},w)$. Therefore, ${\mathbf{q}}^{-1}=(-\mathbf{v},w)=(-x,-y,-z,w)$.

In fact, $(\hat{\mathbf{n}},2\pi -\theta )$ also cancels $(\hat{\mathbf{n}},\theta )$, the axis/angle representation $(\hat{\mathbf{n}},2\pi -\theta )$ corresponds to the quaternion representation $(\mathbf{v},-w)=(x,y,z,-w)$. Therefore, ${\mathbf{q}}^{-1}=(\mathbf{v},-w)=(x,y,z,-w)$.

Taken together, there could be at least two expressions for the inverse quaternion ${\mathbf{q}}^{-1}$, ${\mathbf{q}}^{-1}=(\mathbf{v},-w)=(x,y,z,-w)$ or ${\mathbf{q}}^{-1}=(-\mathbf{v},w)=(-x,-y,-z,w)$.

Suppose we used the inverse quaternion expression ${\mathbf{q}}^{-1}=(\mathbf{v},-w)=(x,y,z,-w)$, let’s verify if rotation ${\mathbf{q}}^{-1}=(x,y,z,-w)$ cancels rotation $\mathbf{q}=(\mathbf{v},w)=(x,y,z,w)$.

$$\begin{array}{rl}\mathbf{R}({\mathbf{q}}^{-1})& =\left[\begin{array}{ccc}1-2(y^2+z^2)& 2(xy+zw)& 2(xz-yw)\\ 2(xy-zw)& 1-2(x^2+z^2)& 2(yz+xw)\\ 2(xz+yw)& 2(yz-xw)& 1-2(x^2+y^2)\end{array}\right]\\ & =\mathbf{R}(\mathbf{q}{)}^{\mathrm{\top }}\end{array}$$

From my previous article “Axis/Angle 3D Rotation Representation”, we can know that $\mathbf{R}(\mathbf{q})$ is orthonormal and $\mathbf{R}(\mathbf{q}{)}^{\mathrm{\top }}=\mathbf{R}(\mathbf{q}{)}^{-1}$. Therefore,

$$\mathbf{R}({\mathbf{q}}^{-1})=\mathbf{R}(\mathbf{q}{)}^{\mathrm{\top }}=\mathbf{R}(\mathbf{q}{)}^{-1}$$

So,

$$\mathbf{R}({\mathbf{q}}^{-1})\mathbf{R}(\mathbf{q})=\mathbf{I}$$

Similarly, we could also show that rotation ${\mathbf{q}}^{-1}=(-\mathbf{v},w)=(-x,-y,-z,w)$ cancels rotation $\mathbf{q}=(\mathbf{v},w)=(x,y,z,w)$.

Notice that $\mathbf{q}{\mathbf{q}}^{-1}={\mathbf{q}}^{-1}\mathbf{q}=(\mathbf{0},1)=(0,0,0,1)$.

Quaternion Division

Suppose ${\mathbf{q}}_0=({\mathbf{v}}_0,w_0)=(x_0,y_0,z_0,w_0)$, ${\mathbf{q}}_1=({\mathbf{v}}_1,w_1)=(x_1,y_1,z_1,w_1)$, ${\mathbf{q}}_2=({\mathbf{v}}_2,w_2)=(x_2,y_2,z_2,w_2)$, quaternion division is defined as follows

$$\begin{array}{rl}{\mathbf{q}}_2& =\frac{{\mathbf{q}}_0}{{\mathbf{q}}_1}\\ & ={\mathbf{q}}_0{\mathbf{q}}_1^{-1}\\ & =({\mathbf{v}}_0\times {\mathbf{v}}_1+w_0{\mathbf{v}}_1-w_1{\mathbf{v}}_0,-w_0{w}_1-{\mathbf{v}}_0{\mathbf{v}}_1)\\ & =(-{\mathbf{v}}_0\times {\mathbf{v}}_1-w_0{\mathbf{v}}_1+w_1{\mathbf{v}}_0,w_0{w}_1+{\mathbf{v}}_0{\mathbf{v}}_1)\end{array}$$

Spherical Linear Interpolation

In some scenarios, we would like to compute a rotation that is in the pathway of two rotations. This procedure is called spherical linear interpolation or slerp for short.

Suppose we have two rotations using unit quaternion representation, ${\mathbf{q}}_0$ and ${\mathbf{q}}_1$, we would like to compute the rotation ${\mathbf{q}}_2$ that is $\alpha$ partway of the two rotations. The key to this procedure is to compute the rotation angle ${\theta }_r$ between the two rotations.

Suppose ${\mathbf{q}}_1={\mathbf{q}}_r{\mathbf{q}}_0$, i.e., ${\mathbf{q}}_1$ is applying ${\mathbf{q}}_0$ followed by ${\mathbf{q}}_r$, and ${\mathbf{q}}_2={\mathbf{q}}_{\alpha }{\mathbf{q}}_0$. Because

$$\begin{array}{rl}{\mathbf{q}}_1{\mathbf{q}}_0^{-1}& ={\mathbf{q}}_r{\mathbf{q}}_0{\mathbf{q}}_0^{-1}\\ & ={\mathbf{q}}_r(\mathbf{0},1)\\ & ={\mathbf{q}}_r\end{array}$$

Using quaternion division, we could compute ${\mathbf{q}}_r$,

$$\begin{array}{rl}{\mathbf{q}}_r& ={\mathbf{q}}_1{\mathbf{q}}_0^{-1}=({\mathbf{v}}_r,w_r)\end{array}$$

We want to make sure that $-\frac{\pi }{2}\le \frac{{\theta }_r}{2}\le \frac{\pi }{2}$, i.e., $w_r\ge 0$. There are chances $w_r<0$, we could just apply the antipodal quaternion property and flip the quaternion ${\mathbf{q}}_r$, i.e., ${\mathbf{q}}_r\leftarrow -\mathbf{q}$. After all ${\mathbf{q}}_r$ and $-{\mathbf{q}}_r$ represent the same rotation.

Because $\Vert {\mathbf{v}}_r\Vert =\sin \frac{{\theta }_r}{2}$ and $w_r=\cos \frac{{\theta }_r}{2}$

$$\begin{array}{rl}\tan \frac{{\theta }_r}{2}& =\frac{\sin \frac{{\theta }_r}{2}}{\cos \frac{{\theta }_r}{2}}\\ & =\frac{\Vert {\mathbf{v}}_r\Vert }{w_r}\end{array}$$

${\theta }_r$ could be computed using

$$\begin{array}{r}{\theta }_r=2\arctan \left(\frac{\Vert {\mathbf{v}}_r\Vert }{w_r}\right)\end{array}$$

The fraction of the rotation angle ${\theta }_{\alpha }$ is therefore

$${\theta }_{\alpha }=\alpha {\theta }_r$$

The unit rotation axis could also be computed using

$$\begin{array}{rl}{\hat{\mathbf{n}}}_r& =({\hat{n}}_{r,1},{\hat{n}}_{r,2},{\hat{n}}_{r,3})\\ & =\frac{({\hat{n}}_{r,1}\sin \frac{{\theta }_r}{2},{\hat{n}}_{r,2}\sin \frac{{\theta }_r}{2},{\hat{n}}_{r,3}\sin \frac{{\theta }_r}{2})}{\sin \frac{{\theta }_r}{2}}\\ & =\frac{{\mathbf{v}}_r}{\Vert {\mathbf{v}}_r\Vert }\end{array}$$

Thus, by the definition of quaternion,

$${\mathbf{q}}_{\alpha }=({\hat{\mathbf{n}}}_r\sin \frac{{\theta }_{\alpha }}{2},\cos \frac{{\theta }_{\alpha }}{2})$$

$${\mathbf{q}}_2={\mathbf{q}}_{\alpha }{\mathbf{q}}_0$$