Question about the exponential map and parallel transport I have a question about the connection between parallel transport and the exponential map. Consider and Riemannian manifold $(M,g)$ $z \in M$ two points $x,y\in B_{r}(z)$, a geodesic ball.
So if for $v\in T_x M$ $exp_x v=y$ and $w \in T_yN$ $exp_y v=x$.
Is there a way to somehow compare $v$ and $w$?
If I take the curve $ c(t)=exp_x t v$, $t \in[0,1]$ and parallel transport along c. Is it true that
$$
w=-T_{t,1}v
$$
holds?
 A: I believe this should hold by uniqueness of geodesics. Let $v$ be the initial tangent vector to the geodesic $\gamma : [0, 1] \to M$ with $\gamma(0) = x, \gamma(1) = y$ (so then $\text{exp}_x(v) = y$, $\dot \gamma(0) = v$). Since $\gamma$ is a geodesic, its velocity $\dot \gamma$ is parallel transported along itself. Therefore $Z := \dot \gamma(1)$ is equal to the parallel transport of $v$ along $\gamma$ up to time 1. By reversing the traversal of $\gamma$, $\text{exp}_y(-Z) = x$. This is because, if we define $\eta(t) = \gamma(1 - t)$, then $\eta(0) = y, \eta(1) = x$, and $\dot \eta(0) = -\dot \gamma(1) = -Z$. In particular this implies that $w = -Z$ and therefore your proposed equation holds.
Warning. This only holds if $B_r(z)$ is a totally normal neighborhood (so that each point in it is connected to every other point via exactly one geodesic lying in the neighborhood), and if you're requiring that the geodesics generated by the vectors you're talking about lie in this neighborhood. Otherwise (think of the sphere) you can get $w$ to be $+Z$ (by traveling in the opposite direction, the "long way around").
