How do I find the bisecting surface between a rotation R in SO(3) and the identity rotation? I.e. every point in the bisecting surface is the same distance from the identity rotation and from R. Lets say the metric between two rotations q1 and q2, is $|q_1^{-1} q_2|$, where |.| denotes the angle of its contained rotation.
Here's a better description of the metric I have in mind. Quaternion distance
Using the notation from that MSE answer, the task reduces to finding the locus $q\in S^3$ such that, for fixed $a,b\in S^3$, $$\arccos(2\langle q,a\rangle^2-1)=\arccos(2\langle q,b\rangle^2-1)\text{.}$$ Since $\arccos$ is a bijection between $[-1,1]$ and $[0,\pi]$, the equality holds iff $\langle q,a\rangle^2-\langle q,b\rangle^2=0$. This is last equality involves a difference of squares; consequently, equality holds iff $$\begin{align}\langle q,a+b\rangle&=0 &&\text{or} &\langle q,a-b\rangle&=0\end{align}\text{.}$$ Now, $\langle q,c\rangle=0$ is the equation of a hyperplane passing through the origin. The intersection of such a hyperplane with $S^3$ is a great sphere $S^2$, so in $S^3$ the solution locus is $S^{2}\coprod_{S^1}S^{2}$, two great spheres that intersect along the great circle determined by the intersection of the plane $\langle q,a\rangle=\langle q,b\rangle=0$ with $S^3$.
Now we use that $S^3$ is a double cover of $\mathbf{RP}^3\cong\mathrm{SO}(3)$ by identifying antipodal points. Thus, the solution locus in $\mathrm{SO}(3)$ is $\mathbf{RP}^{2}\coprod_{\mathbf{RP}^1}\mathbf{RP}^{2}$—note that the two great spheres are not identified.
