# Is every loop on a hyperbolic surface freely homotopic to a geodesic?

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 $$K=-1$$.

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

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 $$S_1$$.
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.