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I am wondering about the following:

Suppose that $S$ is a non-compact hyperbolic surface of finite area. Suppose that $\lambda \subset S$ is a non-trivial, geodesic, measured lamination. Forget the transverse measure. Is there a (non-compact) geodesic lamination $\lambda'$ containing $\lambda$, and a sequence of ideal geodesic triangulations $T_i$, so that $T_i \to \lambda'$ in the Chabauty topology?

A little bit of context:

The Chabauty topology is a generalization of the Hausdorff topology to non-compact sets. It is characterized by the condition that every point of $\lambda'$ is the limit of a sequence of points $x_i \in T_i$, and conversely every convergent sequence $x_{i_n} \in T_{i_n}$ limits to a point of $\lambda'$. See, e.g. Notes on Notes of Thurston.
I am mainly interested in the answer in the setting where $\lambda$ is the pleating lamination on the boundary of the convex core of a quasifuchsian $3$-manifold. This places some additional hypotheses on $\lambda$: for example, it would have to be compact. But the question seems to be intrinsically $2$--dimensional, and it's not clear to me how to use compactness of $\lambda$ as a hypothesis.
If the ideal triangulations $T_i$ are replaced by simple closed curves, the result is well-known. So one approach would be to take a sequence of closed curves $C_i$, limiting to $\lambda' \supset \lambda$, approach each $C_i$ by triangulations (twisting more and more), and then take a diagonal sequence of triangulations. But it's not clear that this diagonal sequence even converges.

Anyway, either a reference or a way to argue would be much appreciated!

edited Jun 21 2011 at 2:07

asked Jun 20 2011 at 14:43

Dave Futer
986●4●12

	I suspect you mean something different than what you said. An ideal triangulation would need to have at least one edge going to each cusp of the surface, so the Chaubaty limit could not be a compactly supported lamination. Approximations by ideal triangulations exist only in the measure topology, unless you change the hypotheses. – Bill Thurston Jun 21 2011 at 0:48
	Bill: thank you for taking a look at the question. I would be perfectly happy to have the limit of ideal triangulations be $\lambda' \supset \lambda$, where of course (as you say) $\lambda'$ is non-compact, and has leaves going out the cusps. I edited the question to clarify that $\lambda'$ need not be compact. – Dave Futer Jun 21 2011 at 1:24
	Also, what I'm really after is a way to approximate the pleated boundary of the convex core by a sequence of surfaces $S_i$, with each $S_i$ pleated along an ideal triangulation. If the measure topology is more well-suited to this, then that's great too! – Dave Futer Jun 21 2011 at 1:29

Dave Futer · Accepted Answer · 2011-09-02 19:31:28Z UTC

It turns out that the answer to the question is "yes". Saul Schleimer and I needed this result for a paper that we just finished writing, so we ended up sorting it out. The full argument is written down in Lemma A.6 in the Appendix of this paper, so what follows below is an outline.

Take a sequence of closed curves $C_i$, which limits to $\lambda$ in the measure topology. This sequence has a subsequence (which I will still call $C_i$) limiting to $\lambda' \supset \lambda$ in the Chabauty topology. Now, each $C_i$ is contained in the Chabauty limit of a sequence of triangulations $T_{i,j}$. This means that one can take a representative triangulation $T_{i,j(i)}$ that is very close to $C_i$, where "very close" can be quantified (say, closer than distance $1/i$) because the Chabauty topology is metrizable. Now, the sequence $T_{i,j(i)}$ will converge to a lamination $\lambda'' \supset \lambda' \supset \lambda$.

Laminations as a limit of ideal triangulations

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