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Oct 17, 2016 at 16:11 comment added YCor @SelimG OK; anyway I have little idea what you're asking in "What about the dynamic of such a flow?" (Also the foliation is well-defined but the flow depends on some choice of renormalization.)
Oct 17, 2016 at 7:25 comment added Selim G @BenMcKay Thank you very much for the reference :)
Oct 17, 2016 at 7:24 comment added Selim G @YCor, the completeness issue can be bypassed: the 'geodesic' flow defines a foliation on T^1(S) which is a compact manifold, and all dynamical questions can be rephrased in terms of this foliation. If you really insist on rephrasing the question using flow formalism, you can build a smooth vector field on T^1(S) whose integral curves are the leaves of the aforementioned foliation.
Oct 16, 2016 at 19:46 comment added YCor According to the classification given there, the only complete non-flat case is the case of the "transvection" quotient, namely the quotient of the affine plane by a lattice in the abelian group $G=\{u_{a,b},(a,b)\in\mathbf{R}^2\}$, where $u_{a,b}(x,y)=(x+ay+b,y+a)$.
Oct 16, 2016 at 15:27 comment added Ben McKay If the answer is known, it should be in Yves Benoist, Tores affines, Crystallographic groups and their generalizations (Kortrijk, 1999), Contemp. Math., vol. 262, Amer. Math. Soc., Providence, RI, 2000, pp. 1–37. But the examples of Nagano and Yagi in that paper seem to be too complicated to have explicit expressions for geodesic flows. So I don't think anyone knows.
Oct 16, 2016 at 15:23 comment added YCor And in general even a complete affine structure does not induce a geodesic flow on the unitary tangent bundle (which is not defined), but on the tangent bundle.
Oct 16, 2016 at 15:21 comment added YCor In the non-complete case (which can happen in spite of compactness) it can fail to be defined. For instance consider the quotient of $\mathbf{R}^n\smallsetminus\{0\}$ by the action of the cyclic subgroup generated by $x\mapsto 2x$ (for $n=2$ this is a 2-torus). Then start at the vector $v$ with tangent vector $-v$ (write $(v,-v)$). At time $1/2$ you're at $(v/2,-v)$ which equals in the quotient $(v,-2v)$. So at time $3/4$ you're at $(v,-4v)$ and so on, the flow is not defined at time 1.
Oct 16, 2016 at 14:30 history asked Selim G CC BY-SA 3.0