# Flow of a left invariant vector field on a Lie group equipped with left-invariant metric and the group's geodesics

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?

• Look at this paper. May 14, 2015 at 6:20