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I think the answer to my question is known to many other people, but I'm still getting confused. Let $G$ be a (possibly infinite dimensional also) Lie group and $g$ be its Lie algebra. Consider the Lie group exponential map $exp:g\to G$ by defining $exp(tX):=\phi_t \forall t\in \mathbb{R}$, where $\phi:\mathbb{R}\to G$ is a Lie group homomorphism, and $\phi'(0)=X$.

Equip $g$ with an inner product and hence construct a left-invariant Riemannian metric on $G$.

1) It's NOT necessarily true that $\phi(t)=exp(tX)\in G$ is a Riemannian geodesic in $G$, right?

2) Can we find a condition on $X\in g$ so that $t \mapsto exp(tX)$is a geodesic in $G$?

3) Suppose we know that $\phi(t)$ is a Riemannian geodesic, then does it imply that the left invariant metric on $G$ is also right-invariant? If yes, what's an outline of proof? If no, what's a counter example where flow defined by exponential map is a geodesic but the metric on the group is not bi-invariant?

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    $\begingroup$ Look at this paper. $\endgroup$
    – Jorkug
    Commented May 14, 2015 at 6:20

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A great and complete answer to your question can be found in Arnold's ''Mathematical methods of classical mechanics'', in Appendix 2, plus a great physical motivation for the answer, from rigid body mechanics. If you read French, be sure to read Arnold's paper ''Sur la géométrie différentielle des groupes de Lie de dimension infinie et ses applications à l'hydrodynamique des fluides parfaits''. -- Salem

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    $\begingroup$ This is essentially a "link only" question, where the link in question here is a reference. $\endgroup$
    – amWhy
    Commented Aug 25, 2017 at 17:16

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