Fraction of SO(3) within an angle

Question

SO(3) describes a space of rotations. These rotations can be described in axis-angle representation. I would like to know what fraction of SO(3) has an angle less than 30 degrees.

Answer 1

In order to answer the question, we need to decide how to measure sets of rotations. There is really only one reasonable candidate: Haar measure. Fortunately, there is a double cover $\def\SO{\operatorname{SO}}S^3\to\SO(3)$ obtained by taking $S^3$ to be the unit quaternions, with a unit quaternion $\omega$ acting on the imaginary quaternions by $z\to\omega^{-1}z\omega$.

Haar measure on the unit quaternions is just ordinary three-dimensional surface measure on $S^3$, since this measure is invariant under rotations, and multiplication by a unit quaternion from either side is a rotation on $S^3$; which is a defining property of Haar measure. Moreover, if we project this measure onto $\SO(3)$ we get Haar measure on $\SO(3)$ as well.

We can take the quaternion $\omega$ to be in the upper hemisphere of $S^3$, since $\omega$ and $-\omega$ induce the same rotation.

Let us compute the volume (“area” uf you wish) of the upper hemisphere: Representing a unit quaternion as $\omega=\cos\theta+\sin\theta\cdot u$ where $u$ is an imaginary unit quaternion, we arrive at the formula $$\int_0^{\pi/2}4\pi\cdot\sin^2\theta\,d\theta=2\pi\Bigl[\theta-\frac{\sin2\theta}2\Bigl]_{\theta=0}^{\pi/2}=\pi^2$$ for the volume (of a unit hemisphere in four dimensions). The technique here is familiar from the similar computation in three space dimensions: A narrow band corresponding to the interval $[\theta,\theta+d\theta]$ is practically the direct product of a 2-sphere of radius $\sin\theta$, and hence of surface area $4\pi\cdot\sin^2\theta$, and an interval of length $d\theta$.

The quaternion $\omega=\cos\theta+\sin\theta\cdot u$ induces a rotation of angle $2\theta$. So the stated question concerns $2\theta<30^\circ=\pi/6$, i.e.i $\theta<15^\circ=\pi/12$. Thus the answer to the stated question is $$\frac1{\pi^2}\int_0^{\pi/12}4\pi\cdot\sin^2\theta\,d\theta=\frac2\pi\Bigl(\frac\pi{12}-\frac{\sin(\pi/6)}2\Bigr)=\frac16-\frac1{2\pi}.$$