# Parallel transport on Riemannian symmetric spaces

What would be a reference for the following property of symmetric spaces?

Given a smooth curve $\gamma : [0,1] \rightarrow M$ on a symmetric space $(M,g)$, there exists an isometry $\varphi : M \rightarrow M$ taking $\gamma(0)$ to $\gamma(1)$ and such that the parallel transport map $$P_\gamma : T_{\gamma(0)}M \longrightarrow T_{\gamma(1)}M$$ equals the differential of $\varphi$ at the point $\gamma(0)$.

I guess one can get this from two particular cases: (1) the curve $\gamma$ is closed and we are looking at holonomy vs. isotropy groups, and (2) the curve $\gamma$ is a geodesic segment and is the projection of a translation of one-parameter subgroup of isometries, but I would love just to be able to cite something.

I looked up in Helgason and it does not seem to be there, at least in this guise.

• I would try Eberlein's book, though it might only cover non-positively curved symmetric spaces. – Benoît Kloeckner Mar 28 '17 at 11:31
• Looking at closed curves, my guess would be that this property is equivalent to $M$ being a rank one space. – user1688 Mar 28 '17 at 11:44
• @Antonius, no, the holonomy group at a point $p \in M$ is contained in the isotropy subgroup of the point for all symmetric spaces. – alvarezpaiva Mar 28 '17 at 18:18
• I know that you do not want a proof, but here is one :) --- take even number of evenly spaces points on your curve and compose their central symmetries --- you get an approximation for the parallel translation --- pass to the limit. – Anton Petrunin Mar 28 '17 at 18:30
• Yes, that's basically the proof you get if you look under the hood of the holonomy-isotropy result: use the result for geodesics by approximating general curves by broken geodesics. I just have paper-writing fatigue and would like to \cite this away. – alvarezpaiva Mar 28 '17 at 18:37