# intersection of geodesiques

Let $(M,g)$ be a closed riemannian surface . let $\alpha$ be a simple closed geodesique . does there is exist a simple closed geodesic $\beta$ that intersect alpha at only 1 point p such that $[\alpha]$ and $[\beta]$ does not commute in $\pi_1(M,p)$

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I'm assuming you mean a surface with a Riemannian metric? Otherwise for a Riemann surface, presumably you mean a unique constant curvature metric with curvature $1,0$ or $-1$? – Ian Agol Jan 6 '12 at 16:11
You put about 10 spelling errors in 3 lines of text, I corrected them all. Please be more precise next time, we are mathematicians after all, and this is a professional forum. – GH from MO Jan 7 '12 at 1:55
It doesn't seem to me that you corrected all the spelling errors! – YangMills Apr 14 '12 at 3:18

Yes. Consider the punctured torus, then the $(1, 0)$ and $(0, 1)$ curves together generate the fundamental group (which is the free group on two generators), and so don't commute. Now, if you have a closed riemann surface, one of its handles is a punctured torus, so the above construction goes through without change.
No, this is false for any curve $\alpha$ on $M$ a torus, sphere, or projective plane (choosing any Riemannian metric on the surface). For a general surface with a Riemannian metric, you might have a simple closed $\mathbb{Z}/2$-homologically trivial geodesic ($\alpha$ bounds a subsurface), in which case there is no geodesic $\beta$ meeting it in a single point.
If the curve $\alpha$ is non-separating, then this will be true (if $\chi(M)<0$). There exists transverse curves $\alpha$ and $\beta$ such that $|\alpha\cap \beta|=1$ and $[\alpha]$ and $[\beta]$ not commuting in $\pi_1(M)$ (with the natural base point). Then minimal length representatives of $\alpha$ and $\beta$ will intersect transversely in a single point (see e.g. a paper of Hass-Scott for an elementary proof).