# Difference between parallel transport and derivative of the exponential map

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 $\frac{\sin r}{r}$.

• – Watson Jun 16 '17 at 8:33

This question has been asked and answered on MathOverflow. I have replicated the accepted answer by J. GE below.

To understand the realationship between $Q$ and $P$, it suffices to study how they act on an orthornormal(o.n.) basis. Here is the detailed argument:

Let $J_i(t)$ be the Jacobi field along $c$, with $J_i(0)=0$ and $J_i'(0)=e_i$, where $e_1, \cdots, e_{n-1}, X$ is an o.n basis of $T_p(M)$. Then one can show that $J_i(t)=Q_t(e_i)$, since $$Q_t(e_i)=dexp_{tX} (te_i).$$

Now extend $e_i$ to $E_i(t)$ along $c$ via parallel transport $P$. One can write the $J_i$ (i.e. $Q_t$) in terms of the o.n. basis $E_i(t)$: $$Q_t(e_i)=J_i(t)=\sum_j a_{ij}E_j(t)=\sum_j a_{ij}P_t(e_j).$$

Let $A=(a_{ij})$. Using Jacobi equation, one can show that $$A''(t)+R(t)A(t)=0,$$ where $R(t)=(R_{ij}(t))$ is the curvature term defined by $R_{ij}(t)=\langle R(\dot{c}, E_i) \dot{c}, E_i \rangle$.

This is essentially Jacobi equation. So roughly speaking, $Q$ and $P$ satisfy the Jacobi equation.

remark: It seems my definition of $Q$ differs by a rescaling fact $t$. So if we use $N$ to denote your map, then $Q_t(e_i)= N(t e_i)$.