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I am trying to read the proof of the theorem:

On any closed hyperbolic $2$-manifold M there is a unique, closed geodesic in any non-trivial free homotopy class.

The reference I am using are these notes. I am unable to follow the argument here. He first considers the deck transformation associated to the homotopy class and then concludes that the transformation has to be a hyperbolic transformation. Then we know that a hyperbolic transformation leaves a geodesic invariant (the geodesic joining the two fixed points). He then says that the image of this geodesic is the require curve. But I am unable to see this. Thanks.

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To add a few details, suppose that your given non-trivial free homotopy class is represented by a closed path $f : S^1 \to M$.

Consider the universal covering maps $\mathbb{R} \mapsto S^1$ defined by $t \mapsto e^{2 \pi i t}$, choose a locally isometric universal covering map $\mathbb{H}^2 \mapsto M$, and choose a lift $F : \mathbb{R} \to \mathbb{H}^2$.

The hyperbolic transformation $T : \mathbb{H}^2 \to \mathbb{H}^2$ has the property that $T(F(t)) = F(t+1)$.

The geodesic invariant under $T$ has a constant speed parameterization $\gamma : \mathbb{R} \to \mathbb{H}^2$ such that $T(\gamma(t)) = \gamma(t+1)$.

Now simply define a $T$-equivariant homotopy between $F$ and $\gamma$, by moving each point $F(t)$ at constant speed along a geodesic segment to $\gamma(t)$.

This $T$-equivariant homotopy, when projected to $M$, produces a homotopy from $f$ to a constant speed reparameterization of a closed geodesic in $M$.

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  • $\begingroup$ Thanks a lot for the detailed answer $\endgroup$ – happymath Nov 18 '16 at 18:43

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