Difference between parallel transport and derivative of the exponential map

Question

Given a Riemannian manifold $M$, let $c(t) = \exp_p(tX)$ be the geodesic emanating from $p \in M$ with initial value $X$. Let $t_0$ be small enough, then we have to ways to map $T_pM$ to $T_{c(t_0)} M$ isomorphically. One is the parallel transport along $c$, let's call it $P_{c, 0, t_0}$ and the other is given by $$ d \exp_p|_{tX}: T_{tX}T_pM \cong T_p M \longrightarrow T_{\exp_p(tX)}M = T_{c(t_0)}.$$

My question is: What is the relation between those two?

The parallel transport is a linear isometry (in all dimensions), and the derivative of the exponential map is a radial isometry by the Gauss lemma (also in all dimensions), meaning $$ \langle d \exp_p|_{tX} \cdot Y, \dot{c}(t) \rangle = \langle Y, X \rangle $$ for all $Y \in T_pM$.

In two dimensions, this means that the two mappings coincide up to scaling, as there is only on orthogonal direction to the radial one. However, in higher dimensions, this is not true, i guess, though I find it hard to actually compute examples...

Are there formulas which relate the two concepts with curvature terms?

\Edit: Computed that on $S^3$, conincides with the parallel transport except that vectors orthogonal to the direction of parallel transport are multiplied by $\frac{\sin r}{r}$.

Answer 1

The relation between the differential of the exponential map $\text{d } \text{exp}$ and parallel transport $P$ (along geodesics) is expounded in the two papers (in the context of optimization on manifolds):

1 ''An accelerated first-order method for non-convex optimization on manifolds'' and

2 ''Curvature-Dependant Global Convergence Rates for Optimization on Manifolds of Bounded Geometry''

Both also provide bounds on $||\text{d } \text{exp}||$, see Proposition A.3 in 1 or Theorem 3.12 in 2. The following is from 1:

Let $$f_{K_{\text{low}}}(t) = \left\{ \begin{array}{lII} r^2\left(1 - \frac{\sin(t/r)}{t/r}\right) & \quad K_{\text{low}} = 1/r^2 > 0 \\ \frac{t^2}{6} & \quad K_{\text{low}} = 0 \\ r^2\left(\frac{\sinh(t/r)}{t/r} - 1\right) & \quad K_{\text{low}} = -1/r^2 < 0 \end{array} \right\}.$$

Let $(M, g)$ be a Riemannian manifold with sectional curvatures bounded below by $-K$ and above by $K$. Let $x \in M$, $s \in T_x M$, and $\gamma(t) = \text{exp}_x (t s)$. If $\gamma$ is defined and has no interior conjugate point on the interval $[0,1]$, then $$||(\text{d } \text{exp})_x (s) - P_s)[\dot{s}]|| \leq K ||\dot{s}_{\perp}|| f_{-K}(||s||) \quad \forall \dot{s} \in T_x M$$ where $\dot{s}_{\perp} = \dot{s} - \frac{\langle s, \dot{s} \rangle}{||s||^2} s$ is the component of $\dot{s}$ perpendicular to $s$, and $P_{t s}$ denotes parallel transport along $\gamma$ from $\gamma(0)$ to $\gamma(t)$.