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Let $(S,g)$ be an orientable Riemannian 2-manifold having constant Gaussian curvature $K=-1$ and $\gamma$ a loop on $S$. Is $\gamma$ freely homotopic to a geodesic? Note the lack of completeness assumption on $S$.

By the uniformization theorem $S$ is conformally equivalent to a complete Riemannian manifold of constant curvature $K=-1$.

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  • $\begingroup$ This is false even for complete hyperbolic surfaces. $\endgroup$ Jan 4, 2023 at 23:50
  • $\begingroup$ What about the hyperbolic plane itself? $\endgroup$
    – Didier
    Jan 5, 2023 at 19:55

1 Answer 1

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  1. The simplest incomplete counter-example is the hyperbolic plane (in the unit disk model) minus the origin, let's call it $S_1$. Then the loop $\gamma=\{z: |z|=\frac{1}{2}\}$ is not freely homotopic to a closed geodesic in $S_1$.

  2. If you want a complete counter-example, take any noncompact complete hyperbolic surface of finite area. A bit simpler example (of infinite area) is obtained as follows:

I will work with the upper half-plane model $U\subset {\mathbb C}$ of the hyperbolic plane. Consider the strip $$ P=\{z\in U: 0\le Re(z)\le 1\}\subset U. $$ Identify the boundary lines of $P$ by the translation $z\mapsto z+1$. The quotient has a natural complete hyperbolic structure. I will call this hyperbolic surface $S_2$. Now, take the loop $\gamma$ in $S_2$ obtained from the Euclidean line segment between the points $i, i+1\in P$ by identifying the end-points (via the same translation as above, of course). Then $\gamma$ is not freely homotopic to any geodesic loop.

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