## Pleated surfaces do not curl up too much

Hi!

Let $S$ be a hyperbolic surface with metric $\rho$ and $N$ a hyperbolic $3$-dimensional manifold with bounded geometry. Let $g\colon (S,\rho)\to N$ be an incompressible pleated surface, that is to say:

1. $g$ is a path-isometry (it maps paths of finite length to paths of the same length),
2. there exists a geodesic lamination $\lambda$ on $S$ such that $g$ maps leaves of $\lambda$ to geodesics in $N$ and is totally geodesic on $S\setminus \lambda$,
3. $g$ is $\pi_1$-injective.

In particular, $g$ is an isometry onto the image $g(S)$. On the other hand, $N$ has its own hyperbolic metric, so that for two points $x,y\in g(S)$ we may define two distances, say $d_\rho(x,y)$ and $d_N(x,y)$. I would like to prove that $\forall B > 0$ there exists $A > 0$ such that if $d_N(x,y)< B$ then $d_\rho(x,y)< A$.

I guess that this may be done using Lemma 4.4 in Minsky's paper "On rigidity, limit sets and end invariants of hyperbolic $3$-manifolds" which states:

Fix $S$ and $\varepsilon>0$. Given $B>0$ there exists $A$ such that if $g\colon (S,\rho)\to N$ is a pleated surface, $g_\ast$ is an isomorphism on $\pi_1$, and injectivity radii in $N$ are bounded below by $\varepsilon$, then the following holds: Let $\alpha\in S$ be a closed curve through $x\in S$, $\rho$-geodesic except possibly at $x$, and let $\beta$ denote the shortest curve in $N$ passing through $g(x)$ and homotopic to $g(\alpha)$. Then $$l_N(\beta)\le B\Rightarrow l_\rho(\alpha)\le A$$

This gives a uniform properness condition about inclusions of homotopy classes of loops from $S$ (identified with $g(S)$) into $N$. This sounds to me very similar to what I need, apart from the fact that what I need is a similar result which holds for inclusions of paths between points instead of loops. Do you think this is the right way? Could you help me with that? Thank you!

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 I fixed a typo in your transcription of Minsky's lemma (an inequality went the wrong way). – Dylan Thurston Apr 14 2012 at 15:25 Are you assuming a lower bound $\epsilon$ on injectivity radius of $N$? If you do then diameter of $g(S)$ is bounded above by some function $B$ of genus of the surface and $\epsilon$. Then you take your $B$ to be the function above. – Misha Apr 14 2012 at 23:20 I swapped $A$ and $B$ in your question to match the usage in Minsky. That is, $A$ is a bound in $(S, \rho)$ while $B$ is a bound in $N$. (This means that $B$ should be replaced by $A$ in Misha's comment. Sorry Misha!) – Sam Nead Apr 15 2012 at 8:58

This is an expansion of Misha's comment. Since $g : (S,\rho) \to N$ is a pleating map we have $g$ is $1$-Lipschitz. That is, for any $x, y \in S$ we have $d_N(g(x), g(y)) \leq d_\rho(x, y)$. In fact, if $\alpha$ is a geodesic arc in $(S,\rho)$ connecting $x$ to $y$, and if $\beta$ is a geodesic arc in $N$, connecting $g(x)$ to $g(y)$ in the relative homotopy class of $g(\alpha)$, then $\ell_N(\beta) \leq \ell_\rho(\alpha)$. Similarly, if $\alpha$ is a geodesic loop in $S$ then the geodesic representative $\beta$, of $g(\alpha)$, has $\ell_N(\beta) \leq \ell_\rho(\alpha)$.

It follows that $R = R_\rho$, the injectivity radius of $(S, \rho)$, is greater than or equal to the injectivity radius of $N$. Let $x, y \in S$ be any pair of points. Let $\alpha$ be the shortest geodesic segment in $S$ connecting $x$ to $y$. For any point $z$ of $\alpha$ let $D \subset S$ be the open ball of radius $R$ about $z$. It follows that $\alpha \cap D$ is a single arc, centered at $z$. Thus the $R/2$ neighborhood of $\alpha$ is an embedded strip with area less than the area of $(S, \rho)$. Since the area of $(S, \rho)$ is $-2\pi\chi(S)$, deduce an upper bound $A$ for the length of $\alpha$ and hence for the diameter of $S$.

So -

Reading the third to last sentence of your post, I think that you may be asking a different question from what you actually wrote in the first half. That is, instead of the distances $d_N(g(x), g(y))$ and $d_\rho(x,y)$, you are interested in upper bounds for certain geodesic arcs connecting the points.... Looking at Minsky's paper, I think that the proof with $x \neq y$ is basically the same compactness argument as his version with $x = y$.

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 @Sam: The upper bound on diameter of a pleated surface is due to Thurston and could be found in his Notes (it is the same proof that you gave). Also, once you know an upper bound on diameter, then your version of the question (with the relative homotopy classes), follows from Minsky's result, there is no need to go through his proof. Lastly, I think, it's time for Damiano Lupi to re-read his post and explain what question he actually has meant to ask. (Otherwise, we might edit it to a point where the original problem, whatever it was, is unrecognizable.) – Misha Apr 15 2012 at 12:12 Thank you everyone for your help. Actually what I meant is what I clearly asked in the first part of my question, i.e. that the inclusion $(S,\rho)\hookrightarrow (N,d_N)$ is uniformly proper. In the last part of the question I wrote something very confused: what I was trying to point out is that the result of Minsky holds for loops while I needed a statement about distance between points, that is length of shortest paths between points. Anyway, your explanation about the diameter of $S$ solved my problems. Thank you very much and sorry about my inaccuracy. – Damiano Lupi Apr 15 2012 at 16:34 @Misha - Of course, you are right: I didn't think to look in any references (all of my paper copies are at work). So... @Damiano, you should read Thurston's notes! Oh, and one remark about your comment: a pleating map $g: (S,\rho) \to N$ is generally not an inclusion. I'll leave it as an exercise (also answered in the notes) to give an example. – Sam Nead Apr 15 2012 at 18:48 @Damiano: One more remark: One can ask for uniform properness of the pleated map of the universal cover of $S$ to the hyperbolic space. You again get uniform bounds on uniform properness here, see Sam's and mine remarks on relative homotopy of paths in the surface. In my experience, uniform properness is most useful in the context of unbounded metric spaces, like the universal covers in this case. – Misha Apr 16 2012 at 1:11 @Sam: I will read Thurston's notes! Thanks! @Misha: actually what I was about to do was trying to figure out how to "lift" uniform properness to universal covers. Thank you for the hint! – Damiano Lupi Apr 16 2012 at 7:51