Geodesic distance on $\mathrm{SO}(n)$

Question

$\DeclareMathOperator\SO{SO}$Recently I came across this old MSE post or this paper (w.o. proof) discussing the geodesic distance on $\SO(n)$ when it is equipped with the left-invariant Riemannian metric given by $g_{I_n}(X,Y):=-\operatorname{tr}(XY)$ for $X,Y\in T_{I_n}\SO(n)\leftrightarrow \mathfrak{so}(n)$. The post states that, the length/geodesic metric induced by $g$ is $$ d(A,B)^2:= \sum_{i=1}^d \theta_i^2 $$ where $\theta_i=\ln(\lambda_{i}(A^{\top}B))$, for any $A,B\in \SO(n)$, where $\{\lambda_{i}(A^{\top}B)\}_{i=1}^d$ are the eigenvalues of $A^{\top}B$. Where can I find a proof, or a reference, to this fact?

Answer 1

The formula holds with a minus sign on the RHS and only for small enough distances. Here's a short proof: by invariance you may set $A = I_n$. Then for $B$ close enough to $I_n$, the unit-speed minimising geodesic from $I_n$ to $B$ is a one-parameter subgroup $\gamma(s) = \exp(s \Xi)$, for some $\Xi \in \mathfrak{so}(n)$ with ${\rm tr}\, \Xi^2 = -1$. We thus have $B = \exp(d \Xi)$, where $d = d(I_n, B)$, and the eigenvalues satisfy $\theta_i = \ln \lambda_i = d \xi_i$, where $\xi_i$ are the eigenvalues of $\Xi$. Your formula follows since $\sum \xi_i^2 = -1$. (Notice the minus sign on the RHS.)