# Vectors that geodesically generate the same surface

Suppose that $\langle M,g \rangle$ is a complete, simply connected Riemannian symmetric space. The surface geodesically generated by a vector $\xi$ in $T_pM$ is the set of points lying on geodesics passing through $p$ that are orthogonal to $\xi$.

Suppose that $\xi$ and $\xi^\prime$ are vectors in $T_pM$ and $T_{p^\prime}M$ respectively that geodesically generate the same surface.

Does it follow that the vector obtained by parallel transporting $\xi$ along the geodesic connecting $p$ and $p^\prime$ is proportional to $\xi^\prime$?

• is the use of the word "surface" here meant to imply a constraint on the dimension of $M$? – hunter Jun 4 '15 at 18:04
• Oh, no it's not supposed to imply a constraint on the dimension of $M$. (Although I'm interested in the case of Lorentzian manifolds in particular. I don't think anything turns on this this though, so I posed the more general question.) – Andrew Bacon Jun 4 '15 at 18:08
• got it, nice question. – hunter Jun 4 '15 at 18:30
• The surface $S$ geodesically generated by a vector can be a fairly nasty subset of $M$; just view a "chaotic" geodesic $S$ in some surface $M$ as geodesically generated by a vector orthogonal to $S$ at one point. Particularly, what does "parallel transport along the geodesic connecting $p$ and $p'$ mean if $p = p'$ but $S$ is not a closed geodesic? – Andrew D. Hwang Jun 4 '15 at 20:49
• I wonder if your question has a positive answer assuming $(M,g)$ is a complete, simply connected Riemannian symmetric space. – Holonomia Jun 5 '15 at 20:52