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I am learning some things about surfaces of genus greater than $1$, and I am trying to answer this question :

Let $S$ be a compact and orientable surface of genus $g \geq 2$, and $c$ a closed curve on $S$. Let $\widehat{S}$ denote the universal covering space of $S$. Given a deck transformation $\widehat{f}: \widehat{S} \rightarrow \widehat{S}$, is it true that there exists only a finite number of lifts $\widehat{c}$ of $c$ to $\widehat{S}$ that are $\widehat{f}-$invariant ?

(by "lift", I mean the image of a map $\widehat{c} : \mathbb{R} \rightarrow \widehat{S}$ that lifts the application $c : \mathbb{R}/\mathbb{Z} \rightarrow S$ defining the closed curve $c$).

I hope this is clear ! Thank you for any help :)

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Yes. If you pick a basepoint $b \in c$ then your curve represents an element of the fundamental group $\pi_1(S,b)$, and therefore also gives rise to a corresponding deck transformation $\hat f$. In particular, if you pick a basepoint $\hat b$ of the universal cover $\hat S$, then this deck transformation will map the lift $\hat c_{\hat b} \ni \hat b$ of $c$ to itself.

Conversely for any deck transformation $\hat f$ you can find a curve $c$ that realizes it in $\pi_1$.

Now some hyperbolic geometry is needed. You can endow $S$ with a hyperbolic metric (which lifts to the universal cover) and assume that $c$ was geodesic, that is to say that for any lift $\hat c$ of $c$, for any points $x,y$, the arc on $\hat c$ connecting $x,y$ is the shortest path between them. Unlike in Euclidean space where parallel lines stay close, hyperbolic geometry implies that no two distinct infinite geodesics can remain a bounded from one another.

Because the images of all lifts of $c$ correspond to distinct geodesics, there is only one of these geodesics that will be fixed by your deck transformation. This is viewing geodesics as subsets of the universal cover. Lifts, correspond to functions of $\mathbb R$ and it is possible to get multiple lifts being fixed by a deck transformation if $c$ is a power of a curve, but this number is exactly the power.

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  • $\begingroup$ Thank you very much for this precise answer ! However, there is still a point that puzzles me : why "the images of all lifts of $c$ correspond to distinct geodesics" ? Intuitively, I would say the following : if two lifts $\widehat{c}$ and $\widehat{g}(\widehat{c})$ correspond to the same geodesic, then they are homotopic, in the sense that there is a homotopy H between $c$ and $c$ such that the path of a basepoint of $c$ under this homotopy is a non-trivial loop. And this seems to be impossible. Is that true ? How can you write it properly ? Thanks again ! $\endgroup$
    – TheSilverDoe
    Commented Sep 26, 2018 at 10:10
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    $\begingroup$ It is possible if the path along which the basepoint $b$ travels is $c$ (or some power thereof). $\endgroup$
    – NWMT
    Commented Sep 26, 2018 at 12:48
  • $\begingroup$ I understand that (intuitively), but how can you prove rigorously that it is impossible in the other cases ? Or more generally, how do you prove the statement "the images of all lifts of c correspond to distinct geodesics" ? Thank you ! $\endgroup$
    – TheSilverDoe
    Commented Sep 26, 2018 at 12:55
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    $\begingroup$ Hi, you have to think about how deck transformations move the basepoint $\hat b$. We already chose it to lie in the specific lift $\hat c_{\hat b}$. Any deck transformation that sends $\hat b$ into $\hat c_{\hat b}$ will also fix $\hat c_{\hat b}$ (having the effect of shifting the curve along itself). A family of examples of such deck transformations will come from taking powers of the $\pi_1$-image of your original loop $c$. Other deck transformation that move $\hat b$ off $\hat c_{\hat b}$ will necessarily give a different lift. $\endgroup$
    – NWMT
    Commented Sep 28, 2018 at 0:46
  • $\begingroup$ Ok, thank you ! I think I begin to understand... but I have to learn a little bit of hyperbolic geometry. Thanks again ! $\endgroup$
    – TheSilverDoe
    Commented Sep 28, 2018 at 13:44

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