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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 $\xi^\prime$?

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  • $\begingroup$ is the use of the word "surface" here meant to imply a constraint on the dimension of $M$? $\endgroup$ – hunter Jun 4 '15 at 18:04
  • $\begingroup$ 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.) $\endgroup$ – Andrew Bacon Jun 4 '15 at 18:08
  • $\begingroup$ got it, nice question. $\endgroup$ – hunter Jun 4 '15 at 18:30
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    $\begingroup$ 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? $\endgroup$ – Andrew D. Hwang Jun 4 '15 at 20:49
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    $\begingroup$ I wonder if your question has a positive answer assuming $(M,g)$ is a complete, simply connected Riemannian symmetric space. $\endgroup$ – Holonomia Jun 5 '15 at 20:52

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