Let $(M, g)$ be a smooth Riemannian manifold, $p \in M$, and $\exp\_P$ the exponential map at the point $P$:
$\exp\_P: T(P) \to M$


It seems clear to me that $\exp\_P$ is smooth on $U \setminus \{0\}$, where $U$ is a neighborhood of the origin $0 \in T(P)$, because $\exp\_P$ is defined from the geodesics, which are solutions of a system of ordinary differential equations.


But is it true that $\exp\_P$ is also always smooth at the origin $0 \in T(P)$?


I am not interested for the moment on the ways to calculate these high order derivatives, but just on the theoretical question about their existence.


Thank you!



Smoothness of the exponential map at the origin
As Thomas Rot already suggested: this follows directly from smooth dependence of ODEs on initial conditions.


Let $p \in M$ and denote by $\Phi^t\colon TM \to TM$ the geodesic flow. Then the exponential map at $p \in M$ is defined as the time one geodesic flow, restricted to $T\_p M$ and projected onto $M$, i.e.
$$
 \exp\_p = \pi \circ \Phi^1|\_{T\_p M} \colon T\_p M \to M
$$
where $\pi$ is the tangent bundle projection.


In local coordinates around $p$ this amounts to an ODE on $\mathbb{R}^{2n}$ involving the Christoffel symbols, and these are smooth since $(M,g)$ was assumed smooth.


**Edit added:** In local coordinates the geodesic flow is given by
$$
 \dot{x}^i = v^i, \qquad \dot{v}^i = -\Gamma^i\_{jk}(x) v^j v^k
$$
with $(x,v) \in \mathbb{R}^{2n}$ induced local coordinates on $T M$.
This is well-defined, also for $v = 0$.


