Take an equator on the two sphere $S^2$ and parametrize it by arc-length obtaining a closed loop $\alpha: S^1 \to S^2$. The curve $(\alpha,\alpha'):S^1 \to T^1S^2$ in the unit tangent bundle of $S^2$ is homotopically non-trivial.


However if you consider the concatenation $\beta$ of two copies of $\alpha$, you can take one copy and turn it about a diameter passing through two points of $\alpha$ so that it is now a copy of $\alpha$ but traversed in the opposite sense. In other words, the curve $(\beta,\beta') \in T^1S^2$ is homotopically trivial.


Does this occur on other compact orientable surfaces?


On the torus the answer is no.


What about on the double-torus (which is a hyperbolic surface)?


Also, does anybody have references for the above statements about the sphere and the torus? In particular, are they correct?


I've come across these problems while trying to picture the lift of a general geodesic flow to the universal covering space of $T^1M$ where $M$ is a compact orientable surface (in particular, what do homotopically trivial closed geodesics look like?).



What immersed closed curves on the double-torus are non-trivial when lifted to the unit tangent bundle?
If $M$ is not $S^2$ or $RP^2$, then $\pi\_1(T^1M)$ does not have elements of finite order
(in particular a double non-contractible loop is also non-contractible). Indeed, consider the long exact sequence of our fibration $E=T^1M\to M$:
$$
 \dots\to \pi\_2(M)\to \pi\_1(F)\to\pi\_1(E)\to\pi\_1(M)\to\dots
$$
where $F$ is a fiber (a circle).


Note that $\pi\_2(M)=0$ (since the universal cover of $M$ is the plane), hence the arrow $\pi\_1(F)\to\pi\_1(E)$ is injective. Hence the kernel of the arrow $\pi\_1(E)\to\pi\_1(M)$ is isomorphic to $\pi\_1(F)$ which is $\mathbb Z$. So this kernel does not contain elements of finite order. And if a non-kernel element has a finite order, then so does its image in $\pi\_1(M)$. But $\pi\_1(M)$ has no elements of finite order (e.g. due to existence of a nonpositively curved metric on $M$).


