TITLE: geodesic of $\rm SO(3)$ as a compact Lie group vs as a Riemannian symmetric space
QUESTION [6 upvotes]: 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?

REPLY [8 votes]: 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.