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?).

  • $\begingroup$ Do I understand correctly that you are interested only in closed geodesics on surfaces of constant curvature? If so, multiple geodesic loops are never contractible in any space of nonpositive curvature (including torus and double torus), let alone in the unit tangent bundle. $\endgroup$ – Sergei Ivanov Sep 14 '10 at 15:53
  • $\begingroup$ No, I'm interested in geodesics of other metrics as well. These metrics may posses "bumps" that give homotopically trivial closed geodesics. $\endgroup$ – Pablo Lessa Sep 14 '10 at 16:03
  • 1
    $\begingroup$ I added the algebraic topology tag since the metrics and geodesics are mostly irrelevant. $\endgroup$ – Sergei Ivanov Sep 14 '10 at 18:52

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 $M$).

  • $\begingroup$ Looks good! Thank you. Although I must admit I'm a little fuzzy on long exact sequences of homotopy groups. In particular, what is the canonical map from $\pi_2(M)$ to $\pi_1(S^1)$ in this case? Also: Any references? $\endgroup$ – Pablo Lessa Sep 14 '10 at 17:55
  • $\begingroup$ See e.g. Hatcher, "Algebraic topology", chapter 4. The map from $\pi_2(base)$ to $\pi_1(fiber)$ is the following: an element of $\pi_2(M)$ is a map $D^2\to M$ that sends the boundary circle to the marked point $p\in M$. Lift that map to the total space (using homotopy lifting property). The boundary circle gets lifted to an element of $\pi_1$ of the fiber. This is the canonical image of the original element of $\pi_2(M)$. $\endgroup$ – Sergei Ivanov Sep 14 '10 at 18:29
  • $\begingroup$ Excelent. That was exactly what I was looking for. Thanks. $\endgroup$ – Pablo Lessa Sep 14 '10 at 23:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.