# 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?

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.
• 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