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?



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


