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 $. Can you give some hints or references?
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asked Jul 5, 2023 at 11:15
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2$\begingroup$ Helgason, Sigurdur. Differential geometry, Lie groups, and symmetric spaces. Pure and Applied Mathematics, 80. Academic Press, Inc. [Harcourt Brace Jovanovich, Publishers], New York-London, 1978. $\endgroup$– Mikhail KatzCommented Jul 5, 2023 at 11:21
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$\begingroup$ @MikhailKatzM Can you tell me the precise page? $\endgroup$– Luis Yanka AnnaliscCommented Jul 5, 2023 at 11:31
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2$\begingroup$ For $O(3)$, as a Riemannian manifold it consists of two copies of the real projective space, with metric the quotient of the metric of the 3-sphere, so the geodesics, which are exactly the one parameter subgroups, are exactly the real projective lines, the quotients of the great circles. So now you know about $n=0,1,2,3$. $\endgroup$– Ben McKayCommented Jul 5, 2023 at 19:44
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1$\begingroup$ For $n=4$, $O(4)$ is isometric as a Riemannian manfold to two copies of $SO(4)$, which is a 2-1 quotient of $S^3\times S^3$ with the usual metric. I think all of this is in many books on Lie groups. Maybe try Stillwell, Naive Lie Theory, although I don't have it with me at the moment. $\endgroup$– Ben McKayCommented Jul 5, 2023 at 19:46
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$\begingroup$ Is it obvious that the riemannian metric that O(n) inherits from R^(n×n) is the same as (or proportional to) the usual bi-invariant metric on O(n) ? $\endgroup$– Daniel AsimovCommented Jul 6, 2023 at 0:16
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1 Answer
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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.
