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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!

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  • $\begingroup$ Yes it is, and there is more: en.wikipedia.org/wiki/Gauss's_lemma_(Riemannian_geometry) $\endgroup$
    – Thomas Rot
    Commented Aug 6, 2014 at 13:48
  • $\begingroup$ If I am right, concerning the differentiability properties of the exponential map, Gauss'lemma tells you only that the exponential map has first order derivatives at the origine(that is: the identity), but it tells you nothing about high order derivatives. $\endgroup$
    – Julien Bernard
    Commented Aug 6, 2014 at 14:22
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    $\begingroup$ Why doesn't the normal argument for smooth dependence on initial conditions work at zero? $\endgroup$
    – Thomas Rot
    Commented Aug 6, 2014 at 14:42
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    $\begingroup$ Please use TeX on this site. $\endgroup$
    – GH from MO
    Commented Aug 6, 2014 at 15:10
  • $\begingroup$ I think that the usual argument for smooth dependence on initial conditions does not work at zero just because the notion of a geodesic with the zero vector as initial tangent vector has no meaning. It is not a legitimate initial condition for the equation of geodesics. $\endgroup$
    – Julien Bernard
    Commented Aug 6, 2014 at 15:26

1 Answer 1

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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 $v = 0$.

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  • $\begingroup$ Please, cf. my last answer to Thomas Rot. I think that this argument is correct for every vector in Tp, except the zero vector. Because you cannot write a geodesic equation with zero as initial condition for the tangent vector. $\endgroup$
    – Julien Bernard
    Commented Aug 6, 2014 at 16:23
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    $\begingroup$ @Julien: Why not? The ODE in local coordinates is perfectly well-defined, also for $0 \in T_p M$. Actually, the solution is immediately seen to be the constant curve, or point, $(p,0) \in T M$, hence $\exp_p(0) = p$. $\endgroup$
    – Jaap Eldering
    Commented Aug 6, 2014 at 16:45
  • $\begingroup$ Thank you Jaap Eldering. It did not see before that a constant curve could be considered as a degenerate geodesic curve. $\endgroup$
    – Julien Bernard
    Commented Aug 7, 2014 at 7:34

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