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):

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

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