Let $ O(n) $ be the manifold of orthornormal matrix, i.e.
$$
O(n)=\{A\in\mathbb{R}^{n\times n}:A^TA=I\}.
$$
Then $ O(n) $ is a submanifold of $ \mathbb{R}^{n\times n} $. On $ O(n) $, there is a Riemannian metric on $ O(n) $ induced by Euclidean metric of $ \mathbb{R}^{n\times n} $. I want to consider the geodesics on it. I know that if $ n=2 $, then $ O(n) $ can be seen as $ \mathbb{S}^1 $. However the case is much more hard if $ n\geq 3 $. Can you give some hints or references?



Geodesics on orthogonal matrix
A direct computation shows that for each $T\in O(n)$, the map $L\_T : \mathbb{R}^{n\times n} \to \mathbb{R}^{n\times n}$ given by
$$
 A \mapsto TA,
$$  

is an isometry. This isometry preserves the submanifold $O(n)$, thus it's also an isometry of $O(n)$ when the latter is endowed with the induced metric. The same applies for the map $R\_T : \mathbb{R}^{n\times n} \to \mathbb{R}^{n\times n}$,
$$
 A \mapsto AT.
$$
It follows that the induced metric on $O(n)$ is *bi-invariant.* For such metrics on Lie groups it is well-known that geodesics through the identity are given by orbits of one-parameter subgroups: see for instance problem 3 in Ch.3 of Do Carmo's Riemannian geometry (2nd Ed.), which includes some hints.


