## What are the shapes of rational functions?

I would like to understand and compute the shapes of rational functions, that is, holomorphic maps of the Riemann sphere to itself, or equivalently, ratios of two polynomials, up to Moebius transformations in both domain and range. For degree 1 and 2, there is only one equivalence class. For degree 3, there is a well-understood one-complex-parameter family, so the real challenge is for higher degrees.

Given a set of points to be the critical values [in the range], along with a covering space of the complement homeomorphic to a punctured sphere, the uniformization theorem says this Riemann surface can be parametrized by $S^2$, thereby defining a rational function. Is there a reasonable way to compute such a rational map?

Given a set of $2d-2$ points on $CP^1$ to be critical points [in the domain], it has been known since Schubert that there are Catalan(d) rational functions with those critical points. Is there a conceptual way to describe and identify them?

In the case that all critical points are real, Eremenko and Gabrielov, Rational functions with real critical points and the B. and M. Shapiro conjecture in real enumerative geometry. Annals of Mathematics, v.155, p.105-129, 2002 gave a good description. They are determined by $f^{-1}(R)$, which is $R$ together with mirror-image subdivisions of the upper and lower half-plane by arcs. These correspond to the various standard things that are enumerated by Catalan numbers. Is there a global conceptual classification of this sort? And, is there a way to find a rational map with given critical points along with some kind of additional combinatorial data?

Note that for the case of polynomials, this is very trivial: the critical points are zeros of its derivative, so there is only one polynomial, which you get by integrating its the derivative.

Is there a complete characterization of the Schwarzian derivative for a rational map, starting with the generic case of $2d-2$ distinct critical points?

Cf. the recent question by Paul Siegel. The Schwarzian $q$ for a generic rational map has a double pole at each critical point. As a quadratic differential, it defines a metric $|q|$ on the sphere - critical points which is isometric to an infinitely long cylinder of circumference $\sqrt 6 \pi$ near each. Negative real trajectories of the quadratic differential go from pole to pole, defining a planar graph.

What planar graphs occur for Schwarzian derivatives of rational functions? What convex (or other) inequalities do they satisfy?

The map from the configuration space of $(2d-2)$ points together with branching data to the configuration space of $2d-2$ points, defined by mapping (configuration of critical values plus branched cover data) to (configuration of critical points) is a holomorphic map, which implies it is a contraction of the Teichmuller metric.

Is this map a contraction for other readily described metrics?

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Perhaps the "na" tag was meant to be "na.numerical-analysis". – Gjergji Zaimi Sep 10 2010 at 8:36
Yes, na-numerical-analysis was what I meant and what I actually typed. I surmise the software truncated it, and I don't see how to fix it. – Bill Thurston Sep 10 2010 at 12:35
The first question for three critical values seems to be the problem of "dessin d'enfant". A reasonable amount of calculation (some by hand, some by computer) has been done but the number of explicit examples is not large. – Torsten Ekedahl Sep 10 2010 at 17:02
The first question (which, as TE says, is the problem of computing equations for dessins d'enfant when there are three branch points) appears to be hard. One person who's done a lot of work on it is Jean-Marc Couveignes; see math.univ-toulouse.fr/~couveig/publi/volk.pdf for a representative piece of work. – JSE Sep 12 2010 at 2:19
Thanks TE and JSE for the pointer. I'm not yet convinced that the computation should be hard in principle, even if nobody has yet implemented an efficient process. The map from {rational functions/up to precomposition with Moebius} to {critical values, branching data} is biholomorphic, so at worst the inverse function theorem should be efficient once implemented, although annoying to implement because of the complexity of tracking the branch data, the braid group action, degeneracy relationships and orbifold singularities well enough to get good local coordinates, especially in the range. – Bill Thurston Sep 12 2010 at 18:13

The algorithmic unsolvability of the general polynomial equation of degree greater than 4 implies that this is not possible in the general case.

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Shows that what is not possible? – Gerald Edgar Jan 14 at 18:24
If "algorithmic unsolvability" here is intended to mean unsolvability by radicals, then I don't see its relevance. If it's intended to mean something else, then I'd appreciate more information about the intended meaning. – Andreas Blass Jan 14 at 18:32
1. There is a characterization of Schwarzian derivatives of rational maps: section 3 in the text: http://www.math.purdue.edu/~eremenko/dvi/schwarz.pdf There is something similar also in arXiv:math/0512370, chapter 2. All these descriptions are various systems of algebraic equations. One of them, the "Bethe ansatz equations for the Gaudin model", proved to be very useful, see Mukhin, Tarasov and Varchenko, The B. and M. Shapiro conjecture in real algebraic geometry and the Bethe ansatz, Ann. Math. 170, 2 2009, 863-15.

2. There is some cell decomposition of the sphere which can be intrinsically related to a ratonal function. It is described in the paper Bonk, Eremenko, Schlicht regions of entire and meromorphic functions, J. d'Analyse, 77, 1999, 69-104, Sections 7.8. For a given cell decomposition, a rational function can be recovered using an algorithm similar to Thurston's circle packing algorithm. However, with this description, critical points or critical valued cannot be prescribed, and the cell decomposition does not determine the rational function completely.

Alex Eremenko.

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 Prof. Eremenko, welcome to MO! – David Speyer Aug 3 at 20:04 Thanks, David. I should have posted my comment as a "comment", not an "answer", sorry, have not mastered the rules completely yet:-( – Alexandre Eremenko Aug 3 at 20:17