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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
up vote 1 down vote accepted

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.

EDIT The above answer is for hyperbolic surfaces. Obviously, if the fundamental group is abelian (as for the torus or the sphere or the projective plane), the answer is no.

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i had hyperbolic surfaces in mind when i wrote the question – student Jan 9 '12 at 9:46

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).

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