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Consider a hyperbolic pair of pants with totally-geodesic boundaries of lengths $l_1,l_2,l_3$. Cap off boundary 1 with a conformal disk. The result is a conformal cylinder, which has a unique flat representative of the form $S^1 \times [0,h]$, where we choose the boundaries to have some fixed length $2 \pi$.

What is $h$ as a function of $l_1,l_2,l_3$?

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  • $\begingroup$ Do you need an exact formula for $h$, or would quantitative information suffice? The latter is much easier - for any fixed $l_1$, if $l_2$ or $l_3$ go to zero, $h$ goes to infinity. $\endgroup$ – Sam Nead Mar 2 '14 at 12:39
  • $\begingroup$ I need an exact formula, or at least an algebraic condition satisfied by $h,l_1,l_2,l_3$ that in principle determines $h$ exactly. $\endgroup$ – Jamie Vicary Mar 2 '14 at 12:56
  • $\begingroup$ There are many ways to cap off with a conformal disk. You get an open interval of possibilities for the modulus $h$, the lower bound corresponding to the disk squishing to a horizontal slit, and the upper bound corresponding to the disk squishing to a vertical slit. $\endgroup$ – Maxime Fortier Bourque Mar 4 '14 at 16:20
  • $\begingroup$ Thanks for your comment. This confuses me because there is only one conformal structure on the disk. So how can it matter which disk we choose as far as the conformal structure of the resulting cylinder is concerned? $\endgroup$ – Jamie Vicary Mar 4 '14 at 18:46
  • $\begingroup$ But there are many different ways to glue this unique disk. You need to pick a homeomorphism between the boundary of the disk and the cuff you're capping off. For every nice enough homeomorphism (e.g. quasisymmetric), you can realize the gluing, and in general different gluings give different annuli. $\endgroup$ – Maxime Fortier Bourque Mar 4 '14 at 22:39
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EDIT - Maxime has answer the question in the comments. The desired $h$ is not a function of the lengths $\ell_1, \ell_2, \ell_3$, but rather lies in an open interval. The minimum (resp. maximum) of $h$ is achieved by sewing the first circle shut so it lies along a concentric (resp. radial) arc of the resulting annulus. See Figure 46e in Nehari's book for a picture of the concentric case.

I wrote the following under the assumption that there is a unique best way to attach a disk. I'll leave it here as it seems like a particularly nice intermediate case.

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It suffices to answer the following question.

Fix six real numbers $A = \{a_0, a_1, \ldots a_5\}$ with $a_i < a_{i+1}$. We construct a right-angled hexagon $H_A$ in the complex plane by connecting $a_0$ to $a_1$, connecting $a_2$ to $a_3$, and connecting $a_4$ to $a_5$ by line segments, contained in the real axis. We then connect $a_5$ to $a_0$, connect $a_1$ to $a_2$, and connect $a_3$ to $a_4$ by arcs of circles orthogonal to the real axis, contained in the upper half plane.

$H_A$ appears to have six degrees of freedom, but these are reduced to just three after translating, scaling, and applying a Mobius transformation that preserves the real axis as well as $a_0$ and $a_5$. One must uniformize this hexagon (ie compute the hyperbolic lengths of the sides) to answer your question. I think this is a variant of the Schwarz-Christoffel transformation, but I failed to find a reference after looking for a bit. Perhaps look in Nehari's book "Conformal mappings".

That done, it is easy to double the hexagon across the real axis and apply a Mobius transformation to make the sides corresponding to $l_2$ and $l_3$ concentric. The log of the ratio of their radii now gives the desired modulus.

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  • $\begingroup$ The hyperbolic lengths of all six sides of the hexagon are definable straightforwardly in terms of the $l_1$, $l_2$, $l_3$: see mathoverflow.net/questions/159113/…. Is that what you mean? $\endgroup$ – Jamie Vicary Mar 2 '14 at 17:32
  • $\begingroup$ @JamieVicary: A hyperbolic right-angle hexagon has 3 parameters: the lengths of common perpendiculars to any three non-intersecting sides. This is what is meant above. $\endgroup$ – SashaKolpakov Mar 9 '14 at 1:02

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