6
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ @SebastianGoette I think it should still be $\times$. $\tau$ is the transitive action by the group of displacement of $\rm SO(3)$. $\endgroup$
    – Troy Woo
    Nov 3 '15 at 14:19
  • $\begingroup$ @SebastianGoette Ok...I changed my application of $\tau$... $\endgroup$
    – Troy Woo
    Nov 3 '15 at 14:24
  • 1
    $\begingroup$ 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. $\endgroup$
    – YCor
    Nov 3 '15 at 15:30
8
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ First of all, you are redefining my action into the projection. This does not mean my equation is wrong. I know what you said from the texts. So you are saying the base point should be chosen as the identity. It can be chosen as any point in fact, since the action is transitive. $\endgroup$
    – Troy Woo
    Nov 3 '15 at 17:00
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 3 '15 at 17:12
  • 1
    $\begingroup$ Ok...I see, so the transvections on points other than the fixed point are not geodesics. I go back to check Helgason, and it is what he said too. Thanks for the counterexample. $\endgroup$
    – Troy Woo
    Nov 3 '15 at 17:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.