Let $M$ be a compact Riemannian manifold with negative sectional curvature and let $\tilde M$ be its universal cover.
Let $\gamma:\mathbb{R}\to\tilde M$ be the piecewise geodesic obtained by lifting the composition of two geodesic loops, say $c_1*c_2$, in $M$. Let $\gamma_1$ be the (unique) geodesic joining the end points at infinity of $\gamma$.
Assume that $\gamma$ and $\gamma_1$ are within bounded distance one from the other.
I would claim that $c_1*c_2$ and the projection of $\gamma$ (by the covering map) in $M$ are free homotopic.
My attempt: The projection of $\gamma$ in $M$ is a closed geodesic, since $M$ is compact. The free homotopy class of $c_1*c_2$ contains a unique closed geodesic whose lift to $\tilde M$, say $\hat{\gamma}$, is a geodesic with same end points at infinity of $\gamma$. Since a geodesic between two (distinct) points at infinity of $\tilde M$ is unique, it must be $\hat{\gamma}=\gamma_1$. Then $c_1*c_2$ and the projection of $\gamma$ are free homotopic.
Would that be correct? It bothers me that $c_1*c_2$ intersects itself, while the projection of $\gamma$ is a simple closed curve.
Best Answer
The claim you are trying to prove is false. As an example, consider $M=S^1$, $c_1=c_2$. (Note also that it is, in general, not true that a lift of $c=c_1*c_2$ is unbounded: You need to add the assumption that $c$ is not null-homotopic. As an example, consider the case when $c_2$ is the reverse of $c_1$.) The correct statement is that, up to conjugation, the element $[c]\in \pi_1(M)$ belongs to the infinite cyclic subgroup generated by the projection of $\gamma_1$ to $M$.