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?



Flat Riemannian manifold
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. 


