\section{Introduction}

In several talks I gave about my paper \cite{DM16} (e.g. at the meeting \it  Variet\'a reali e complesse: geometria, topologia e analisi armonica \rm, SNS, Pisa (Italy) 2014, at the Simons Semester \it Symmetry and Geometric Structures \rm at the IMPAN, Warsaw (Poland) 2017) I was asked about the proof of the following proposition:

\begin{prop}\label{main} Let $\pi:E \to M$ be a vector bundle with a connection $\nabla$. Let $(x_1,\cdots,x_n)$ be local coordinates centered at a point $p \in M$ and let $\xi_p \in E_p := \pi^{-1}(p)$. Then there exists a neighborhood $U$ of $p$ and a smooth section $\xi : U \to E$ such that $\xi(p) = \xi_p$ and \begin{equation} \label{rp} \nabla_{\frac{\partial}{\partial \rho}} \xi \equiv 0 \end{equation}
where $\frac{\partial}{\partial \rho} := \sum_{j=1}^n x_j \frac{\partial}{\partial x_j}$ is the radial vector field. That is to say, there exists a radial $\nabla$-parallel smooth section with a given initial condition $\xi_p \in E_p$.
\end{prop}

I learnt by heart the above proposition from C. Olmos when I was a PhD. student.

The standard theorem of existence and uniqueness of solutions of a system of ODE implies that the function $\xi : U \to E$ satisfying (\ref{rp})  is uniquely determined by its initial condition $\xi_p$.
People who asked me about $\xi$ were concerned about the smoothness of $\xi$ at the point $p$. The goal of this short note is to show that $\xi$ is indeed smooth at $p$.

\subsection{The proof.}

Let $k = \mathrm{rank}(E)$ and let $(\sigma_1,\cdots,\sigma_k)$ be a local smooth frame of $\pi: E \to M$ defined in a neighborhood $U$ of $p$. We assume that $U$ is contained in the domain of the coordinates $x_1,\cdots,x_n$ centered at $p$.
Let $\Gamma_{ij}^s \in C^{\infty}(U)$ be the connection symbols defined by
\[ (\nabla_{\frac{\partial}{\partial x_i}} {\sigma_j} )_{|_{p}}= \sum_{s = 1}^k \Gamma_{i j}^s(p) \cdot \sigma_s(p) \, \, .\]
For each $z \in U$ fixed let $\xi(t,z)$ be the $\nabla$-parallel section defined on the radial segment $t:[0,1] \to t \cdot z $  with initial condition $\xi_p$. So $\xi(t,z) = \sum_{j=1}^k y_j(t,z) \cdot \sigma_j(t \cdot z)$ and the coefficients $y_k(t,z)$ satisfies the ODE system:
\begin{equation} \label{ode}  \sum_{i=1}^n z_i \cdot \mathrm{M}_i \cdot \mathrm{y} + \frac{\partial \mathrm{y}}{\partial t} = 0
\end{equation}
where $\mathrm{y}$ is the column $\left[
                                 \begin{array}{c}
                                   y_1(t,z) \\
                                   \vdots \\
                                   y_k(t,z) \\
                                 \end{array}
                               \right]$ and $\mathrm{M}_i$, $(i=1,\cdots,n)$ are the $k \times k$ matrices
                               $\mathrm{M}_i := (\Gamma_{i j}^s(t \cdot z))$, $(s,j = 1,\cdots,k)$.\\

Set $f(t,\mathrm{y},z) := -\sum_{i=1}^n z_i \cdot \mathrm{M}_i \cdot \mathrm{y}$. Observe that $f(t,\mathrm{y},z)$ is a $C^{\infty}$ function of $(t,\mathrm{y},z)$ and that $\mathrm{y}(t,z)$ solves the ODE system \[ \frac{\partial \mathrm{y}}{\partial t} = f(t,\mathrm{y},z) \, .\]

According to \cite[Corollary 4.1., page 101]{H64} $\mathrm{y}(t,z)$ is a $C^{\infty}$ function of $(t,z)$. Hence $\mathrm{y}(1,z) \in C^{\infty}(U)$ and the section $\xi : U \to E$ defined as \[\xi(z) := \sum_{j=1}^k y_j(1,z) \cdot \sigma_j(z) \]
is smooth on $U$. Since $\mathrm{y}(1,t \cdot z) = \mathrm{y}(t,z)$ and $\mathrm{y}(t,z)$ satisfies (\ref{ode}) we get that $\xi$ satisfies equation (\ref{rp}). This complete the proof of our proposition. $\Box$

\subsection{Geometric interpretation} Let $\mathrm{I} = [0,1]$ be unit interval, let $\mu: I \times \mathbb{R}^n \to \mathbb{R}^n $ defined by $\mu(t,x):= t \cdot x $ and let $\pi:E \to \mathbb{R}^n$ be a vector bundle with a linear connection $\nabla$. Let $\nabla^*$ be the pull-back connection of the pull-back vector bundle $\pi^*: E^* \to \mathrm{I} \times \mathbb{R}^n$ defined by the commutative diagram
\[ \begin{tikzcd}
E^* \arrow{r}{\mu^*} \arrow[swap]{d}{\pi^*} & E \arrow{d}{\pi} \\%
\mathrm{I} \times \mathbb{R}^n \arrow{r}{\mu}& \mathbb{R}^{n}
\end{tikzcd}
\]

The usual existence, uniqueness, and smooth dependence on initial conditions of solutions to ODE applies directly to get a smooth section $\xi^*$ (of $E^*$) which is
$\nabla^*$-parallel in the direction of the factor $\mathrm{I}$ with the initial conditions given on the hyperplane $H := \{0\} \times \mathbb{R}^n$.
The restriction of $\xi^*$ to the hyperplane  $\{1\} \times \mathbb{R}^n$ gives the radial parallel section $\xi$ of $E$.

\begin{rem}Let $\mathrm{B}: \tilde{\mathbb{R}^n} \to \mathbb{R}^n$ be the blow-up at the origin and let $\tilde{\nabla}$ be the pullback connection on the pullback vector bundle $\tilde{E} \to \tilde{\mathbb{R}^n}$. The initial condition $\xi_p$ lifts to a initial condition $\tilde{\xi_p}$ on the exceptional divisor $D := \mathrm{B}^{-1}(0)$. Now the usual existence, uniqueness, and smooth dependence on initial conditions of solutions to ODE implies that there is smooth section of $\tilde{\xi}$ defined on neighborhood $\tilde{U}$ of $D$ which is $\tilde{\nabla}$-parallel along the transverse foliation of $\tilde{\mathbb{R}^n}$. In this setting Proposition \ref{main} implies that the pulldown (or pushforward) section $B^{*}(\tilde{\xi})$ of $\pi: E \to U$ is smooth at the origin and $\nabla$-parallel along the radial directions. Actually, Proposition \ref{main} is equivalent to the statement that the pulldown (or pushforward) section $B^{*}(\tilde{\xi})$ of $\pi: E \to U$ is smooth at the origin. But I could not show this directly. The problem is that, at difference as what happen in Algebraic Geometry (or under analyticity hypothesis), in the smooth setting the constancy of a regular function on the exceptional divisor does not implies that regular function is the pullback of a regular function down stairs.
\end{rem}







