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Let $(M, g)$ be an $n$-dimensional smooth Riemannian manifold. Let $$ \Gamma_n = \{ [0,1] \ni t \mapsto \gamma_x^y(t) = (1-t)x + ty \mid x, y \in \mathbb{R}^n \}.$$ Let us choose $m \in M$. Can we always choose a map $\phi \colon U \to \mathbb{R}^n$, where $U$ is a neighbourhood of $m$ such that:

  • $\phi [U]$ is convex
  • for all $x, y \in \phi[U]$ the element $\gamma_x^y \in \Gamma_n$ satisfies: $\phi^{-1} \circ \gamma_x^y$ is a geodesic in $M$?
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    $\begingroup$ Don't you want $\phi$ to be a a diffeomorphism? If so, then you are asking whether there is a diffeomorphism that takes geodesics to lines (and vice versa). This requires $M$ to have constant curvature. (Essentially, this is a result of Cartan and Hadamard in dimension $\geq 3$ and of Blaschke in dimension 2.) $\endgroup$
    – Kapil
    Commented Sep 6, 2021 at 2:14
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    $\begingroup$ Yes, though, if possible, could you provide me with a reference to the result you mention? I tried to find it and I mostly find the Cartan-Hadamard theorem regarding the complete Riemannian manifolds with non-positive sectional curvature. Also, if that changes anything, I only want to map line segments from some neighbourhood to geodesics defined on $[0,1]$. $\endgroup$
    – Kacper Kurowski
    Commented Sep 6, 2021 at 7:55
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    $\begingroup$ Your condition is that the projective connection induced from the Levi-Civita connection is flat. You can probably find a proof of this in Projective Differential Geometry Old and New by Tabachnikov and Ovsienko. $\endgroup$
    – Ben McKay
    Commented Sep 6, 2021 at 11:13
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    $\begingroup$ One (somewhat convoluted) proof of the result for dimension $\geq 3$ can be found here on pages 12-13. The proof for dimension 2 is perhaps in the book by Sharpe; as Ben McKay has said it is about projectively flat connections. $\endgroup$
    – Kapil
    Commented Sep 6, 2021 at 17:02
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    $\begingroup$ Actually, I believe that the original proof that there is a local coordinate system in which the geodesics are straight lines if and only if the Gauss curvature is constant is due to Beltrami. The proof in higher dimensions follows immediately from the proof in dimension $2$. For a straightforward elementary proof in dimension $2$, you can see Exercise 13 in Chapter 4, Section 6 of do Carmo's Differential Geometry of Curves and Surfaces. $\endgroup$
    – Robert Bryant
    Commented Sep 6, 2021 at 17:38

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