$\DeclareMathOperator\SO{SO}$Recently I came across this old [MSE post](https://math.stackexchange.com/questions/4267430/measuring-distance-in-son) or [this paper (w.o. proof)](https://epubs.siam.org/doi/10.1137/S0895479801383877) 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?



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


