Is there explicit formula for the smallest enclosing ball in $SO_3$

244 Views Asked by At

In Euclidean Space, given some points, we can find the smallest enclosing ball which contains all the points. And the center is known as the Chebyshev Center. But now I encounter a problem related to rotation and I would like to find the smallest enclosing ball on the manifold $SO_3$, using the metric of geodesic distance (1):$$ d(R, Q)=\operatorname{acos}\left(\frac{\operatorname{tr}\left(R^T Q\right)-1}{2}\right) $$ To make this problem make sense, we may consider the points contained in a big ball with radius $\pi$.

I'm thinking about this question for quite a while, but I can't find much progress. Now I can solve this problem with some iterative algorithms, but due to this problem's simple form, I'm wondering if there is an explicit solution. I'm wondering if I can use something like stereographic projection to simply the problem.

Or, if I only consider the circumstances of 3 or 4 points on the $SO_3$, can I hope to get an explicit solution to this Chebyshev center problem?(To verify my iterative method). Any hints and help will be appreciated, thanks a lot in advance!

Edit 2023/6/3:

After some searchings, I find that the geodesic distance above is the bi-invariant metric defined on $SO_3$ and the angular distance in $S^3$ is also bi-invariant. The bi-invariant metric is unique up to scaling if the Lie group is compact and Lie algebra is simple. We can conclude that both these two bi-invariant metrics are unique up to scaling.

Now if I consider a diffeomorphism $F: SO_3 \to RP^3$ and locally the $RP^3$ looks the same as $S^3$, maybe we can consider the homomorphism: $\rho: SO_3 \to S^3$ and somehow prove that after the homomorphism, the new metric is still bi-invariant (I'm not sure if I'm right). If this holds, then maybe I can just study about the problem in $S^3$ ? But I'm still not sure if I'm right and how to interprete these in a precise way.

Also I think my previous thought of using the stereographic projection to simply the problem is not valid, for the smallest ball in $R^3$ after stereographic projection isn't the smallest on the $S^3$. So I'm still wondering how to tackle this.

1

There are 1 best solutions below

8
On BEST ANSWER

The key point is that there is a 2-to-1 mapping $S^3\to SO(3)$ where $S^3$ is viewed as the set of unit quaternions (see this wiki page for more on this map). This map preserves the standard geodesic distance (at least up to a constant scale factor depending on your convention). Using this covering map, we can associate each rotation $R\in SO(3)$ with a pair of antipodal points $\pm q\in S^3$. Given a set of rotations $R_1,\cdots,R_n$, there is a corresponding set of pairs $\pm q_1,\cdots,\pm q_n$, and the problem is equivalent to finding the smallest geodesic ball in $S^3$ which contains at least one member of each pair.

Since every geodesic ball in $S^3$ is the restriction of a geodesic ball in $\mathbb{R}^4$, you can find the minimal ball in $S^3$ by finding the minimal ball in $\mathbb{R}^4$ and computing its restriction to $S^3$