# Why is the dual tree to a measured geodesic lamination in a compact hyperbolic surface not complete?

Let $$M$$ be a closed connected surface and $$\mathcal{F}$$ a minimal (every leaf is dense) measured foliation (as, for example, in Thurston's work on surfaces) on $$M$$. Let $$\tilde{M}$$ be the universal cover of $$M$$ and $$\tilde{\mathcal{F}}$$ the pullback of $$\mathcal{F}$$ to $$\tilde{M}$$.

Consider the leaf space of $$\tilde{\mathcal{F}}$$, $$\mathcal{T}=\tilde{M}/\tilde{\mathcal{F}}$$. Define the distance between two leaves of $$\tilde{\mathcal{F}}$$ as the minimum of the transverse measures of arcs joining the two leaves, we obtain a distance on $$\mathcal{T}$$ which turns $$\mathcal{T}$$ into a tree.

I have seen the following statement.

If the genus of $$M$$ is $$\geq 2$$, $$\mathcal{T}$$ is not complete for this distance.

Why is this true?

I also saw a notion "dual tree", which also shows up from time to time when I tried to search for related topics, from the book Hyperbolic manifolds and Discrete Groups (for geodesic laminations there). Is this indeed the concept that I am looking for?

Thanks in advance!

## 1 Answer

Here's a rough description of why $$\mathcal T$$ is incomplete.

Start at some point $$x_0$$ of $$\tilde M$$. Let $$\ell_0$$ be a lonnnng leaf segment of $$\tilde{\mathcal F}$$ with one endpoint on $$x_0$$, and let $$y_0$$ be the opposite endpoint. Let $$\tau_0$$ be a transverse arc of transverse measure $$2^{-0}=1$$ with one endpoint at $$y_0$$. Let $$x_1$$ be the opposite endpoint. Continue by induction: assuming $$x_n$$ has been defined, let $$\ell_n$$ be a lonnnng leaf segment with one endpoint on $$x_n$$, let $$y_n$$ be the opposite endpoint, let $$\tau_n$$ be a transverse arc of transverse measure $$2^{-n}$$, and let $$x_{n+1}$$ be the opposite endpoint. You get an infinite path of the form $$\ell_0 \tau_0 \ell_1 \tau_1 \ell_2 \tau_2 \cdots$$ The total transverse measure of the initial segment $$\ell_0 \tau_0 \cdots \ell_n \tau_n$$ that connects $$x_0$$ to $$x_n$$ is equal to $$2-2^{-n}$$, and this converges to $$2$$. By careful choice of the $$\tau$$'s in the induction step you can guarantee that this initial segment is "quasitransverse", implying that $$2-2^{-n}$$ is the minimum of the transverse measure of any arc from $$x_0$$ to $$x_n$$. So, in the dual tree $$\mathcal T$$, you get an isometric embedding $$[0,2) \mapsto \mathcal T$$. But, this embedding has no point to converge to as the parameter in $$[0,2)$$ approaches $$2$$; in fact, one can check that this infinite path escapes all compact subsets of $$\tilde M$$. So $$\mathcal T$$ is incomplete.

By the way, another place I know, where the dual tree is explained very clearly, is in Otal's book "Le théorème d'hyperbolisation pour les variétés fibrées de dimension 3", in the chapter on Skora's theorem.

• I think I can understand what you are saying, but what I am confused about is, after taking quotient, isn't a lonnng leaf got identified to a single point in the tree? – chikurin Nov 24 '18 at 16:14
• That's exactly correct. And therefore the infinite concatenation $\ell_0\tau_0\ell_1\tau_1\ell_2\tau_2\cdots$, after passing to the quotient $\mathcal T$, could (by abusing notation) be written as $\tau_0 \tau_1 \tau_2\cdots$, which is the image of the isometric embedding $[0,2) \mapsto \mathcal T$. – Lee Mosher Nov 24 '18 at 16:21
• Ohhh, I now see... Thanks a lot!! – chikurin Nov 25 '18 at 6:18
• Another thing, I tried to read the book "The Hyperbolization Theorem for Fibered 3-Manifolds" with my lame French for an hour and then found an English translation that day, in case there is anyone who does not read French and is not aware of the existence of an English translation. :) – chikurin Nov 30 '18 at 1:58