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Fix an integer $d > 1$ and $2d-2$ points $P_1, \ldots, P_{2d-2}$ in the Riemann sphere (not necessarily distinct). Thanks to the work of Eisenbud and Harris on limit linear series (Inventiones, 1983), we know that there are only a finite number of rational functions $\phi(z)$ of degree $d$ with complex coefficients that are ramified exactly at the points $P_i$, up to postcomposition by invertible rational functions. (The latter preserve the ramification points, so they must be taken into account.) Suppose that $\phi_1, \ldots, \phi_n$ are a maximal collection of such functions with the property that $\phi_i = \sigma \circ \phi_j$ for some $\sigma \in \mathrm{Aut}(\mathbb{P}^1)$ implies $i = j$.

Is there an invariant that allows one to distinguish between $\phi_1, \ldots, \phi_n$?

If $P_1, \ldots, P_{2d-2}$ are all distinct, then Goldberg (Advances in Math, 1991) showed that the number of postcomposition classes of rational functions ramified at exactly these points is positive and bounded above by the Catalan number $\rho(d) = \frac{1}{d}\binom{2d-2}{d-1}$. Moreover, if the $P_i$ are in general position, then the number of classes is exactly $\rho(d)$.

For example, when $d = 3$, we may normalize the ramified points to be at $0, 1, \infty$, and $c$, and we may further assume that $0, 1$, and $\infty$ are fixed after postcomposing by a suitable automorphism. Then Goldberg's result says there are precisely 2 rational functions of this shape for a general choice of $c \in \mathbb{C}$. How do we distinguish between them?

One can work out this example explicitly to see that $$ \phi(z) = \frac{\alpha z^3 + (1-2\alpha)z^2}{(2-\alpha) z - 1}.$$ The fourth critical point is $c = \frac{2\alpha - 1}{\alpha(2-\alpha)}$, so we require that $\alpha \neq 0, \pm 1,2^{\pm 1}$ in order to have four distinct critical points. A given $c$ generically determines two values of $\alpha$. But a priori, what data can I specify in order to nail down one or the other of these two functions?

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@Xander. Let $B$ be a finite set of points on the Riemann sphere $\mathbf{P}^1$. Then the fundamental group of $\mathbf{P}^1-B$ is finitely generated. Thus, for all $d\geq 1$, there are only finitely many branched covers of $\mathbf{P}^1$ of degree $d$ ramified only over $B$. In particular, the set of rational functions on $\mathbf{P}^1$ which ramify only over $B$ is finite. So I don't think you need to use Eisenbud-Harris. –  Ari Jul 19 '12 at 8:28
    
Continued: Also, branched covers of $\mathbf{P}^1$ of degree $d$ ramified only over $B$ correspond to subgroups of index $d$ of $\pi_1(\mathbf{P}^1-B)$. (This also gives you an upper bound on the number of rational functions. Of course, it's enormous because you're counting all branched covers.) And finally a question. Does this "topological/group-theoretical" point of view give you a way to distinguish between the functions? Or am I completely misunderstanding the question? –  Ari Jul 19 '12 at 8:28
    
@Ariyan - You're answering a slightly different question. I am not requiring that we fix the critical values ... only the critical points. Using the fact that $\phi$ and $\sigma \circ \phi$ have the same critical points for any $\phi$ of degree $d$ and $\sigma$ of degree $1$, observe that if there exists one map with a given set of critical points, then there is automatically a 3-dimensional family of them. –  Xander Faber Jul 19 '12 at 9:08
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2 Answers

up vote 5 down vote accepted

One answer is that there are NO rational expressions in $P_1, \dots, P_{2d-2}$ that allow you to single out one of the functions $\phi_i$, or that even allow you to single out a nonempty, proper subset of the set of all $\rho(d)$ functions. The proof of this is the usual argument of this type (also in Harris's paper on Galois groups of enumerative problems). The parameter space of degree $d$, $1$ dimensional linear systems on $\mathbb{P}^1$ is irreducible; it is just the Grassmannian of $2$-dimensional linear subspaces of $H^0(\mathbb{P}^1,\mathcal{O}(d))$. There is a dense, open subset $U$ parameterizing base point free linear systems with $2d-2$ distinct ramification points. As a dense open in an irreducible variety, also $U$ is irreducible. There is a morphism $f:U\to \text{Sym}^{2d-2}(\mathbb{P}^1)$, i.e., $f:U\to \mathbb{P}^{2d-2}$, sending such a linear system to the corresponding ramification divisor.

A rational expression in $P_1,\dots,P_{2d-2}$ is the same thing as a rational transformation defined on a dense open subset of $\mathbb{P}^{2d-2}$. So a rational expression choosing one $\phi_i$ would be the same thing as a rational section $s$ of $f$. Since $f$ is quasi-finite, the image of $s$ would be an irreducible component of $U$. As $U$ is irreducible, there can be no such rational section. More generally, a rational expression singling out a nonempty proper subset of size $\sigma$ is the same thing as a union of irreducible components of the domain which has degree $\sigma$ over the target. Once again, since $U$ is irreducible, this can only occur when $\sigma$ equals $\rho(d)$.

Of course you did not specify that you want your rule to be a "rational expression" in $P_1,\dots,P_{2d-2}$. Since $f$ is finite, \'etale over a dense open subset of $\mathbb{P}^{2d-2}$, obviously there do exist \'etale local sections, i.e., there do locally exist "algebraic" functions which pick out some $\phi_i$. I have not heard of any nice expression for such (e.g., in terms of other special functions such as modular functions).

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Jason: Do you know whether the Galois group is the full symmetric group on Catalan many elements? –  David Speyer Jul 18 '12 at 16:21
    
Dear David -- I don't know, although presumably the method in Harris's paper could be used: show that $f$ has simple branching in codimension 1, and then try to prove that the monodromy group acts doubly-transitively on a general fiber (so contains all transpositions). –  Jason Starr Jul 18 '12 at 17:13
    
Apparently, this is open. In his 2012 thesis math.tamu.edu/~sottile/advising/martindelcampo.pdf , del Campo-Sanchez shows that it is at least the alternating group and conjectures that it is the full symmetric group. He cites Leykin-Sotille who verify that it is the full symmetric group for $d \leq 9$ ams.org/mathscinet-getitem?mr=2501073 . –  David Speyer Jul 31 '12 at 19:44
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If all of the points $P_i$ lie on a circle $\gamma \subset \mathbb{CP}^1$, there is a beautiful description in

Eremenko and Gabrielov
Rational functions with real critical points and the B. and M. Shapiro conjecture in real enumerative geometry.
Ann. of Math. (2) 155 (2002), no. 1, 105–129.

Let $\phi: \mathbb{P}^1 \times \mathbb{P}^1$ be one of the rational maps. The circle $\gamma$ divides $\mathbb{P}^1$ into two hemispheres; call them $N$ and $S$. The image $\phi(\gamma)$ is contained in a circle; call it $C$. (This is NOT obvious.)

Consider $\phi^{-1}(C) \cap N$. (An earlier version of this answer wrote $\phi^{-1}(\phi(\gamma))$, but I want the inverse image of the whole circle $C$, which might be larger.) This is a collection of noncrossing arcs whose end points lie at the points $P_i$. Ermenko and Gabrielov show that there is precisely one rational function $\phi$ for each possible connectivity of these arcs. For example, there are $5$ possible ways to draw $3$ noncrossing arcs connecting $6$ points on the boundary of a disc, and there are $5$ degree $4$ maps with $6$ specified critical points.

I would be very interested in knowing a generalization of this result to the case where the points do not lie on a circle; I am fairly certain none is known.

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@David - Thanks for the pointer to Eremenko & Gabrielov. What a cool result! (And a beautiful proof, too.) This is especially helpful for me because I may have stumbled across an analogue in non-Archimedean geometry. The ramification locus for a rational function on the Berkovich projective line is a union of $\mathbb{R}$-trees, and in a number of cases I've looked at, the tree structure of the ramification locus distinguishes between different functions with the same set of critical points. –  Xander Faber Jul 19 '12 at 22:54
    
Thanks David! Very nice response. –  Kerry Jul 20 '12 at 2:27
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There is a much simpler proof of the same result in Eremenko and Gabrielov, Elementary proof of the B and M Shapiro conjecture for rational functions, arXiv:math/0512370. –  Alexandre Eremenko Aug 4 '12 at 8:12
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