# On geodesics of flat, non-symmetric connections

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).

• 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 $d=2$. – Holonomia Sep 22 '16 at 9:50
• 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. – Raz Kupferman Sep 22 '16 at 15:11
• 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 – Raz Kupferman Sep 22 '16 at 15:17
• 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? – Holonomia Sep 22 '16 at 19:34