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Take a collection of pairwise disjoint, simple polygons in the plane. Identify pairs of sides between polygons by either a translation in the plane, or, a translation composed with a rotation by $\pi$. Do this in such a way that each edge gets glued to exactly one other edge, and in a way that the resulting surface is orientable.

We'll call the resulting object a half-translation surface. (You may also think of it as a point in Teichmueller space equipped with a quadratic differential, if you really want to.)

You may notice that a half-translation surface is a topological surface $S$ with some extra structure on it.

Question: Does there exist some constant $K=K(S)$ depending on $S$ such that for any half-translation surface with underlying topological surface homeomorphic to $S$, there exists an isometrically embedded metric cylinder of width at least $K$?

The motivation really comes from a result of Masur. He showed that there is some $K=K(S)$ so that we may always find an annulus (a priori, a wiggly thing that may contain singularities) of width at least $K$. I am finding it hard to find comments on whether we can promote this to cylinders (a union of closed geodesics).

(If your answer involves a reference, please make this citation include page number, or lemma, etc.)

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    $\begingroup$ Are the polygon edges the same lengths, or is the gluing less geometric than translation & rotation suggest? $\endgroup$ Commented Oct 1, 2014 at 14:43
  • $\begingroup$ @JosephO'Rourke: The gluing is actually accomplished by restricting the translation or the translation/$\pi$-rotation. That's what allows you to view the result as having a quadratic differential. $\endgroup$
    – Lee Mosher
    Commented Oct 17, 2014 at 16:22
  • $\begingroup$ You need to add a requirement - the areas of the polygons sum to one. (Otherwise you could just scale things down.) $\endgroup$
    – Sam Nead
    Commented Oct 17, 2014 at 16:45

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By passing to the branched double cover, branched over all singular points, it suffices to address your question for translation surfaces. (This gives up at worst a factor of two from the width.)

Some Googling finds the preprint (post 2002) "Periodic geodesics on translation surfaces", written by Yaroslav Vorobets. Here is the link http://arxiv.org/abs/math/0307249.

Define $\exp(x) = 2^x$. Theorem 1.3 of that paper is, basically, the following.

Suppose $g > 1$. Suppose that $S$ is a translation surface of genus $g$ and of area $1$. Then $S$ contains a flat annulus $A$ with

  • length at most $\exp(\exp(4(4g-4))$ and
  • area at least $1/(4g-4)$.

Taking their ratio gives a lower bound for width that depends only on $g$.

Two questions immediately suggest themselves. (A) Is the double exponential really necessary? Surely it isn't, but I guess you would have to understand the thick part of $Q(S)$ better to prove it... (B) How does this compare to the width of the annulus found by Masur? I believe that you can find explicit estimates in work of Bowditch. Of course those annuli are much, much wider. So why don't they contain wide-ish flat annuli? Presumably because these wide annuli are "expanding" in the language of Kasra Rafi: that is, they are modelled on the geometry of a round annulus in the plane, instead of on the geometry of a right cylinder in three-space.

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