Lifted free homotopic closed Curves have same Endpoints at Infinity

I'm studying the book of Peter Buser, "Geometry and Spectra of Compact Riemann Surfaces". I want to understand the proof he gives in Lemma 1.6.5.

If $\gamma$ and $c$ are free homotopic closed curves on a hyperbolic surface, you can lift them uniquely (up to choosing a start point) to it's universal (Riemannian) covering, $\tilde{\gamma},\tilde{c}$. That's clear. Then he uses a "standard argument" which I don't understand:

"The cyclic subgroup of the covering transformation group which leaves $\tilde{\gamma}$ invariant also leaves $\tilde{c}$ invariant."

My idea so far: I lift the (free) homotopy $F$ beween $\gamma$ and $c$ and show that it behaves like $f(\tilde{F}(s,t))=\tilde{F}(s,t+\omega)$ under the deck transformation $f$ which leaves $\tilde{\gamma}$ invariant, where $\omega$ is the displacement of the geodesic $\tilde{\gamma}$ under $f$. Is that the right approach? If so, how can i proove that?

From this it should follow that $d(\gamma(t),c(t)) < d$ for all $t \in \mathbb{R}$ and some $d$. Since all deck transformation are isometries and the universal covering is a local isometry aswell, it should work somehow, but i don't get the details here.

Thank you for your help! :)

The equation $f(\tilde F(s,t)) = \tilde F(s,t+\omega)$ should follow from uniqueness of lifting: check that each side is a lift of $F$, and check that they have one common value by plugging in your favorite value of $s$ and of $t$.

For each fixed value of $t \in \mathbb{R}$, as $s \in [0,1]$ varies we obtain a path $p_t(s) = \tilde F(s,t)$. Since $[0,1]$ is compact and $p_t$ is continuous, the image $p_t[0,1]$ is compact and therefore has finite diameter which I will denote $D(t) = \text{diam}\left(p_t[0,1]\right) \in [0,\infty)$. The function $D : \mathbb{R} \to [0,\infty)$ varies continuously with $t$, which is not too hard to check. It follows from compactness that $D(t)$ is bounded for $t \in [0,\omega]$. Also $D(t) = D(t+\omega)$, because the deck translation $f$ is presumed to be an isometry (or, using your word, isotropy) of the hyperbolic plane, and so $D$ is a bounded function, $D(t) \le d$ for some $d > 0$. So $$d(\tilde \gamma(t), \tilde c(t)) = d(\tilde F(0,t), \tilde F(1,t)) = d(p_t(0),p_t(1)) \le D(t) \le d$$

• Yeah, of course, i meant "isometry". :D Thank you so far! :) By the way: Don't I need existence of such a covering transformation, acting on $\tilde{\gamma}$ invariantly? How do I get this? And for the first part: Unfortunately I've already got so far, but have no idea how to show that $\tilde{F}(s,t+\omega)$ is a lift of $F$, yet. But if you say the approach is right, I will think about it any further. – YoungMath Jun 18 '18 at 7:23
• Your subsidiary questions might be settled if you take a closer look at covering space theory, and at the lifting lemmas for covering spaces, including both existence and uniqueness. These concepts would then be applied to the universal covering map $p : \mathbb{H} \to S$ from the hyperbolic plane $\mathbb{H}$ to the Riemann surface $S$. – Lee Mosher Jun 18 '18 at 14:05
• For example, one uses covering space theory to prove that for any closed path $\gamma : [0,1] \to S$ with $\gamma(0)=\gamma(1)=p$, and any lift $\tilde p \in \mathbb{H}^2$ of $p$, there is a unique lift $\tilde\gamma : [0,1] \to S$ such that $\tilde\gamma(0)=p$. – Lee Mosher Jun 18 '18 at 14:08
• If $T_\gamma : \mathbb{H}^2 \to \mathbb{H}^2$ is the corresponding deck transformation, then lifting lemmas apply to prove that one can extend the domain of $\tilde\gamma$ uniquely to the entire real line so that $\tilde\gamma : \mathbb{R} \to \mathbb{H}^2$ so that $T_\gamma^k(\tilde\gamma(x)) = \tilde\gamma(x+k)$, and so that $p(\tilde\gamma(x+k)) = p(\tilde\gamma(x)) = \gamma(x)$ for all $x \in [0,1]$. That $\tilde\gamma$ is "the lift" of $\gamma$. If this is not familiar to you, then you really do need to buckle down and learn the covering space theory. – Lee Mosher Jun 18 '18 at 14:09
• I'm quite new to that topic, that's right. So I'm grateful for your patience. $\mathbb{R} \to S^1$ is given via universal covering, is it? I guess that was the missing part. So $\tilde{F}(s,t+k)$ goes down (via this covering) to $F(s,t)$ and since the resulting diagram commutes (since $\tilde{F}$ is a lift of $F$ by definition) we see that $\tilde{F}(s,t+k)$ is indeed a lift of $F(s,t)$. By construction $T_\gamma^k(\tilde{\gamma}(t))=\tilde{\gamma}(t+k) = \tilde{F}(0,t+k)$ and the statement $\tilde{F}(s,t+k) =T_\gamma^k(\tilde{F}(s,t))$ follows by uniqueness. Did I get it right? – YoungMath Jun 18 '18 at 18:00

Let us remark about your title "Lifting of free homotopic closed curves have same endpoints at infinity". For example, if a closed curve is freely-homotopic to a non-simple closed geodesic, say $\gamma$, there is an intersecting point $x_0$ of $\gamma$. If $\tilde{x}_0 \in \pi^{-1}(x_0)$ we can find two different lifting of $\gamma$ passing through $\tilde{x}_0$ having different endpoints at infinity.

I'm also new in this subject, and I prefer the following statement:

We start with a lifting $\tilde{c}$ of $c$. Consider the isometry $T_c \in \pi_1(M)$ such that $T_c(\tilde{c}(0))=\tilde{c}(1)$. Define $\tilde{\gamma}$ the only $T_c$-invariant geodesic line. Then, $\tilde{\gamma}$ and $\tilde{c}$ have the same endpoints at infinity.

We want to prove

$$d(\tilde{x}, \tilde{\gamma}) < \infty \ \forall \tilde{x} \in \tilde{c}.$$

Let $\tilde{\gamma}_0$ and $\tilde{c}_0$ be the restrictions of $\tilde{\gamma}$ and $\tilde{c}$ to $[0,1]$. For all $\tilde{x} \in \tilde{c}$, we have $\tilde{x} = T_{c}^n(\tilde{x}_0)$ for some $n \in \mathbb{Z}$ and $\tilde{x}_0 \in \tilde{c}_0$. We have

$$d(\tilde{x}, \tilde{\gamma}) < d(\tilde{x}, T^n_c(\tilde{\gamma}_0)) = d(T_{c}^n(\tilde{x}_0), T^n_c(\tilde{\gamma}_0)) = d(\tilde{x}_0, \tilde{\gamma}_0) = \text{a constant} < \infty.$$

Reference: Lectures on Hyperbolic Geometry.

• "Let us remark that [it] is only true if the closed curves are homotopic to a simple closed geodesic." Indeed, it seems like i forgot to mention it. $\gamma$ is a closed geodesic. But notice that $\gamma$ doesn't need to be simple if you require the lifts to be homotopic, like Buser does. If $\gamma$ isn't simple, $c$ isn't simple aswell. Even though, as you mentioned: Arbitrary lifts of these, which are both invariant under a cyclic subgroup of deck transformations, still have the same endpoints. – YoungMath Jul 21 '18 at 12:45