Parallel Transport Along Radial Geodesics Yields a Smooth Vector Field? Let $M$ be a Riemannian manifold and $p$ be a point on $M$. Let $U$ be a normal neighborhood about $p$ (that is, the exponential map $\exp_p$ maps a neighborhood of the origin in $T_pM$ diffeomorphically onto $U$). Fix a vector $v_p\in T_pM$.
For each $q\in U$, let $v_q$ be the vector in $T_qM$ obtained by parallel transporting $v_p$ along the radial geodesic joining $p$ to $q$. So we get a map $X:U\to TU$ which takes $q$ to $v_q$. So $X$ is a vector field on $U$.

Question. Is $X$ necessarily smooth?

What I thought is that by definition of $X$, we have $\nabla_{\partial/\partial r}X=0$ at all point of $U\setminus\{p\}$, where $\partial/\partial r$ is the radial vector field (which is defined at all points of $U$ except $p$). So $X$ is a solution of a system of partial differential equations. I do not know if this guarantees that $X$ is smooth on $U\setminus\{p\}$. 
 A: Yes.
There's kind of a trick to it. Let $(x^i)$ be normal coordinates centered at $p$, so the radial geodesics are given by $\gamma(t) = (tx^1,\dots, tx^n)$.  The differential equation satisfied by $v = v^k \partial_k$ along such a geodesic is
$$
\dot v^k(t) = - \Gamma_{ij}^k(tx^1,\dots,tx^n)x^i v^j(t),\tag{$*$}
$$
with initial condition $v^k(0) = a^k$ (some arbitrary constants). (I'm using the summation convention here.)
The trick is to replace the $x^i$'s by new dependent variables $w^i$, and write this as a system of ODEs for the $2n$ functions $(v^i,\dots,v^n,w^1,\dots,w^n)$:
\begin{align*}
\dot v^k(t) &= - \Gamma_{ij}^k(tw^1(t),\dots,tw^n(t))w^i(t) v^j(t),\\
\dot w^k(t) &= 0,
\end{align*}
with initial conditions
\begin{align*}
v^k(0) &= a^k,\\
w^k(0) &= x^k.
\end{align*}
Because solutions to smooth ODEs depend smoothly on initial conditions as well as time, the solutions to this system can be written as smooth functions $v^k(t,a,x)$ and $w^k(t,a,x)$. It follows immediately from the form of the equation that $w^k$ is constant, $w^k \equiv x^k$, and therefore the vector field you're interested in is
$$
v(x) = v^k(1,a,x) \partial_k,
$$
which depends smoothly on $x$.
