geodesic of $\rm SO(3)$ as a compact Lie group vs as a Riemannian symmetric space

I got a little bit confused about the definition of geodesic for $\rm SO(3)$ as

• a compact Lie group
• a Riemannian symmetric space

In the former case, it is given by the usual matrix exponential: $$\exp_{g}tX=ge^{tX}\quad g\in{\rm SO(3)}, X\in\mathfrak{so}(3)$$ In the latter case, the geodesic is given by the transvections $\tau(\exp t(X',-X'))$: $${\rm Exp}_{g}~t(\frac{1}{2}X'g+\frac{1}{2}gX')=\tau(\exp\frac{t}{2}(X',-X'),g)=e^{tX'/2}ge^{tX'/2}$$ where $\tau$ is given by: \begin{aligned} \tau:{(\rm SO(3)\times SO(3))}\times{\rm SO(3)}&\to{\rm SO(3)}\\ ((g,h),x)&\mapsto gxh^{-1} \end{aligned} The two only coincide at the identity. Now my question is, we have two types of geodesics, all defined by the same bi-invariant metric (though in difference sense). Is there anything wrong with my calculation?

If the two types of geodesics are indeed different, which one will be shorter in length?

• @SebastianGoette I think it should still be $\times$. $\tau$ is the transitive action by the group of displacement of $\rm SO(3)$. Nov 3 '15 at 14:19
• @SebastianGoette Ok...I changed my application of $\tau$... Nov 3 '15 at 14:24
• Both "geodesic of $SO(3)$ as a compact Lie group" and "geodesic of $SO(3)$ as a Riemannian manifold" are not well-defined. On a Lie group there is no notion of geodesic, or it depends on the choice of a choice of Riemannian metric. Indeed in case a simple compact Lie group is endowed with a bi-invariant Riemannian metric, the latter is unique up to scalar multiplication, and the notion of geodesic is well-defined up to linear rescaling, and the geodesics are the 1-parameter subgroups.
– YCor
Nov 3 '15 at 15:30

I think the second formula is wrong. The map $\tau$ should be given as $$\tau\colon (SO(3)\times SO(3))/SO(3)\to SO(3)\;;\quad [(g,h)]\mapsto gh^{-1}\;.$$ A geodesic in this picture is then given by $$\tau(ge^{tX/2},he^{-tX/2})=ge^{tX}h^{-1}=gh^{-1}e^{t\mathrm{Ad}_hX}\;.$$ This also proves the equivalence of both constructions.
• The geodesics can start at any $g$. This example shows the problems with your formula: Let $g$ be the rotation by $\pi$ around the $x$-axis. Let $X$ be the velocity vector of a rotation around the $z$-axis. Then Ad$_gX=-X$. Your formula gives a constant curve $e^{tX/2}ge^{tX/2}=e^{tX/2}e^{-tX/2}g=g$. Hence, it does not work. If you take $g$ to be a rotation around the $x$-axis by a different angle, then you get curves on $SO(3)=\mathbb R P^3$ that lift to small circles on $S^3$, not to geodesics. Nov 3 '15 at 17:12