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Let $M$ be a smooth $d$-dimensional manifold. Let $\nabla$ be a flat non-symmetric connection on $TM$ (i.e., curvature=0 and torsion$\ne$0). Let $p\in M$ and denote by $\exp_p:\Omega\subset T_pM\to M$ the exponential map with respect to $\nabla$. Under what conditions (on the torsion) does $\exp_p$ map straight lines into $\nabla$-geodesics. In other words, under what conditions any curve of the form $$ \gamma(t) = \exp_p(v + tw), \qquad v,w\in T_pM $$ satisfies $\ddot{\gamma}(t) \parallel \dot{\gamma}(t)$. An answer for the particular case $d=2$ would also be of interest. (Note that this is always the case if the torsion vanishes).

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  • $\begingroup$ do you have an example for $d=2$ ? I ask you so because after some ugly computations (perhaps wrong) it seems to me that there are no non-symmetric example for $d=2$. $\endgroup$ – Holonomia Sep 22 '16 at 9:50
  • $\begingroup$ Yes. Take the punctured plane; use polar coordinates. The following property define a connection: $\{\partial_r,1/r\partial_\phi\}$ is a parallel frame. Note that both $\gamma(t) = (r + v t,\phi)$ and $\gamma(t) = (r, \phi + v t)$ are geodesic. This connection is flat and has non-zero torsion. $\endgroup$ – Raz Kupferman Sep 22 '16 at 15:11
  • $\begingroup$ The motivation for such questions can be found in ``The emergence of torsion in the continuum limit of distributed dislocations" by R. Kupferman and C. Maor, published in J. Geom. Mech. 7 (2015) 361-387 $\endgroup$ – Raz Kupferman Sep 22 '16 at 15:17
  • $\begingroup$ your connection is not defined at the origen. So the point $p$ in your question (where you use $exp_p$) can not be the origen. What are the coordinates of $p$ in your example? $\endgroup$ – Holonomia Sep 22 '16 at 19:34

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