Is it true that a Riemannian manifold is flat, if and only if a coordinate transformation $f$ exists, such that the geodesics after transformation is in linear form $\mathbf{y}_t=\mathbf{a}t+\mathbf{y}_0$, where $\mathbf{y}=f(\mathbf{x})$ are the transformed coordinates?

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    $\begingroup$ Yes. It is flat if and only if it is locally isometric to an open set in Euclidean space. $\endgroup$ – Peter Michor May 19 '15 at 19:02
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    $\begingroup$ This doesn't seem like a research-level question. It might have been more appropriate for math.SE. $\endgroup$ – Ben Crowell Jun 22 '15 at 17:34

Yes it is true. One direction is obvious as flat implies around each point there is a neighborhood which is isometric to an open set of standard Euclidean space. In that neighborhood if we use the co-ordinate given by that isometry, geodesics are of the prescribed form. Conversely if each point has a neighborhood where geodesics are linear w.r.t. some co-ordinate system then geodesic equation implies $a^i a^j \Gamma_{ij}^k=0$ for any reals $a^i,a^j$. That means $\Gamma_{ij}^k \equiv 0$. So Riemann Curvature Tensor vanishes identically.

  • $\begingroup$ An example that the OP might wish to think about would be a cylinder with a flat metric. That might help to clarify why the answer is framed in local terms. $\endgroup$ – Ben Crowell May 19 '15 at 23:23

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