## Rotations in three dimensional space

It is well known that any ${A\in SO(3)=\{A\in M_3(\mathbb{R}): A^tA=I, \det A=1\}}$ is a rotation, i.e. there exists ${ v\in \mathbb{R}^3}$, ${|v|=1}$, such that ${Av=v}$ and ${A}$ acts on the 2-plane ${v^\perp}$ by a rotation of angle ${\theta}$. So we are free to call an element of ${SO(3)}$ a rotation. The picture below shows a continuous path in $SO(3)$:

Question: Suppose we are given two rotations ${A_1, A_2\in SO(3)}$ with axes ${v_1, v_2}$ and angles ${\theta_1, \theta_2}$ respectively. What are the axis and the rotation angle of their composition ${A_1 A_2}$?

After discussing with John Ma, WongTing and Zorn Leung, we have come up with the following

Theorem 1 (John Ma’s theorem, 2011) Suppose ${A_i}$ are two rotations with axis of rotation ${v_i}$ and rotation angles ${\theta_i}$ (see remarks below) for ${i=1,2}$. Then the new rotation ${A_1 A_2}$ have the axis of rotation in the direction of

$\displaystyle \sin\frac{\theta_1}{2}\cos\frac{\theta_2}{2}v_1+ \cos\frac{\theta_1}{2}\sin \frac{\theta_2}{2}v_2+ \sin\frac{\theta_1}{2}\sin\frac{\theta_2}{2}(v_1\times v_2)$

and the rotation angle ${\theta}$ is given by

$\displaystyle \cos \frac{\theta}{2} = \cos \frac{\theta_1}{2}\cos\frac{\theta_2}{2}-\sin \frac{\theta_1}{2}\sin\frac{\theta_2}{2}\;\langle v_1, v_2\rangle.$

Now we define more precisely what we mean by a rotation about ${v}$ with angle ${\theta}$. This means that ${|v|=1}$ and there exists an orthnormal basis ${\{v_1, v_2, v_3=v\}}$ with ${\det(v_1, v_2, v_3)=1}$ such that ${Av=v}$ and ${Av_1= \cos \theta v_1 + \sin \theta v_2, Av_2 =-\sin \theta v_1 + \cos \theta v_2}$. Note that the choice of $(v, \theta)$ is uniquely determined up to $\mathbb{Z}_2$: $(v, \theta)$ and $(-v, -\theta)$ defines the same rotation.

It turns that it is more convenient to work on the quaternions. (Well, originally I work on ${\mathbb{C}^2}$ and work with ${SU(2)}$ instead of ${Sp(1)}$ (see below), it is equivalent (as both are ${Spin(3)}$) but it turns out that ${Sp(1)}$ is even easier to work with, perhaps due to more built-in structure of ${\mathbb{H}}$ (cross product is natural here!), thanks John Ma for pointing this out!)

Let ${\mathbb{H}}$ be the space of quaternions, i.e. ${\mathbb{H}=\{x+yi+zj+wk: (x,y,z,w)\in \mathbb{R}^4\}}$ with ${i^2=j^2=k^2=ijk=-1}$. We denote the quaternions with unit norm by ${Sp(1)}$ (not to be confused with the real or complex symplectic group ${Sp(2n, F), F=\mathbb{C, R}}$). Clearly ${Sp(1)}$ is a (compact Lie) group. Let ${V=\{xi+yj+zk\}\subset \mathbb{H}}$ be the three dimensional real subspace of all imaginary quaternions. Naturally ${V}$ can be identified with ${\mathbb{R}^3}$. We define the action of ${Sp(1)}$ on ${V}$ by

$\displaystyle x\in V\mapsto qxq^{-1} \text{ for }q\in Sp(1).$

We will denote this action by ${\phi: Sp(1)\rightarrow GL(V)}$, where

$\displaystyle x\stackrel{\phi(q)}\longmapsto qxq^{-1}.$

The following proposition states that this is really a group action.

Proposition 2 ${q\mapsto \phi(q)}$ is a homomorphism ${\phi: Sp(1)\rightarrow GL(V)}$.

Proof: Clearly ${\phi(q)^{-1}=\phi(q^{-1})}$. Also, ${(qr)x(qr)^{-1}=q(rxr^{-1})q^{-1}}$, i.e. ${\phi(qr)=\phi(q)\phi(r)}$. $\Box$

Proposition 3 ${ \phi}$ defines a homomorphism ${Sp(1)\rightarrow SO(3)}$, where we have identified ${V}$ (under induced inner product from ${\mathbb{H}}$ ) with the ordinary ${(\mathbb{R}^3, \langle \;, \rangle)}$.

Proof: We have to check ${\phi(q)}$ preserves both the inner product and the orientation. But ${\langle qxq^{-1}, qyq^{-1}\rangle_{\mathbb{H}}=\langle x, y\rangle}$. So ${\phi: Sp(1)\rightarrow O(3)}$.

Also, ${Sp(1)}$ is connected, since it is the (three-dimensional) unit sphere inside ${\mathbb{H}}$, so ${\phi(Sp(1))}$ lies in the connected component of ${O(3)}$ containing identity, which is ${SO(3)}$. ($O(3)$ has precisely two (path) connected components corresponding to ${\det=\pm1}$. )

$\Box$

Remark 1 We can check that ${\phi}$ actually is the double cover of ${SO(3)}$ i.e. $\phi(Sp(1))=SO(3)$ (see also the proposition below), and it’s easy to see ${\phi(-q)=\phi(q)}$, so $\ker (\phi)=\pm 1$. This also shows that $SO(3)=Sp(1)/\{\pm1\}=\mathbb{S}^3/\{\pm id\}=\mathbb{RP}^3$, the three-dimensional projective space. Also ${Sp(1)}$ is simply connected, as it is the 3-sphere. But we don’t need these facts here.

Proposition 4 If ${A}$ is a rotation about ${v}$ with angle ${\theta}$. By regarding ${v\in V\subset \mathbb{H}}$, then for ${q=\cos \frac{\theta}{2} + \sin \frac{\theta}{2} v}$, ${\phi( q)=A}$ (and ${\phi(-q)=A}$).

Proof: It suffices to show that ${\phi(q)v=v}$ and ${\phi(q)v_1= \cos \theta v_1 + \sin \theta v_2, \phi(q)v_2 =-\sin \theta v_1 + \cos \theta v_2}$ where ${v_i}$ are defined as above.

Denote the conjugate of a quaternion by ${\overline z}$, so ${\overline v=-v=v^{-1 }}$. Consider

$\displaystyle \begin{array}{rcl} \phi(q)v=qvq^{-1}&=&(\cos \frac{\theta}{2} + \sin \frac{\theta}{2} v)v(\cos \frac{\theta}{2}+ \sin \frac{\theta}{2} \overline v)\\ &=&(\cos\frac{\theta}{2}v-\sin\frac{\theta}{2})(\cos \frac{\theta}{2} - \sin \frac{\theta}{2} v)\\ &=&v \end{array}$

where we have used ${v^2=-1}$ as it is purely imaginary.

On the other hand, since ${vv_1=v_2, v_1v=-v_2, v_2v=v_1}$ (see here),

$\displaystyle \begin{array}{rcl} \phi(q)v_1&=& (\cos \frac{\theta}{2}+ \sin \frac{\theta}{2} v)v_1(\cos \frac{\theta}{2}-\sin \frac{\theta}{2} v)\\ &=& (\cos \frac{\theta}{2}v_1 + \sin\frac{\theta}{2} v_2)(\cos \frac{\theta}{2}- \sin \frac{\theta}{2}v)\\ &=&(\cos^2 \frac{\theta}{2}-\sin ^2\frac{\theta}{2})v_1+ 2\sin \frac{\theta}{2}\cos\frac{\theta}{2}v_2\\ &=& \cos \theta v_1 + \sin \theta v_2. \end{array}$

As ${\phi(q)\in SO(3)}$, we conclude that ${\phi(q)v_2=-\sin \theta v_1 +\cos \theta v_2}$. Thus ${\phi(q)=A}$. $\Box$

We are now ready to prove John Ma’s theorem.

Proof of John Ma’s theorem:
Suppose ${v_1, v_2}$, ${\theta _1, \theta_2}$ are given as in the theorem. Then define ${q_i=\cos \frac{\theta_i}{2}+\sin \frac{\theta_i}{2} v_i}$ for ${i=1,2}$. By Proposition 4, ${\phi(q_i)=A_i}$ and by Proposition 3 ${A_1A_2=\phi(q_1q_2)}$, so we only have to compute ${q_1q_2}$ (see here):

$\displaystyle \begin{array}{rcl} q_1q_2&=&(\cos \frac{\theta_1}{2}+\sin \frac{\theta_1}{2} v_1)(\cos \frac{\theta_2}{2}+\sin \frac{\theta_2}{2} v_2)\\ &=& \left(\cos \frac{\theta_1}{2}\cos\frac{\theta_2}{2}-\sin \frac{\theta_1}{2}\sin\frac{\theta_2}{2}\langle v_1, v_2\rangle\right) +\\ && \left(\sin\frac{\theta_1}{2}\cos\frac{\theta_2}{2}v_1+ \cos\frac{\theta_1}{2}\sin \frac{\theta_2}{2}v_2+ \sin\frac{\theta_1}{2}\sin\frac{\theta_2}{2}(v_1\times v_2)\right)\\ &=& \cos\frac{\theta}{2}+ \sin \frac{\theta}{2} w \end{array}$

where ${\theta}$ and ${w}$ are the angle of rotation and the rotation axis of ${A_1A_2}$ respectively, by Proposition 4. Here ${\times}$ is the cross product in ${\mathbb{R}^3}$ and ${\langle \;, \;\rangle }$ is the inner product in ${\mathbb{R}^3}$ and we have used the fomula $(a+v)(b+w)=(ab-\langle v,w\rangle) +(aw+bv+ v\times w)$ if $a, b\in \mathbb{R}$ and $v, w$ are imaginary i.e. $v, w\in Im \mathbb({H})=V$. $\Box$