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If the critical values are given, and the ramification degrees of critical points (I don't care about the locations of these points) are also given, does there exists a rational function on the complex plane (or Riemann sphere) with these information? Of course, it should satisfy the Riemann-Hurwitz formula. Thanks a lot.

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This is a group theoretical question. Indeed, by Riemann's existence theorem, it suffices to construct a topological ramified covering $S^2\to S^2$ with the desired properties; then, the complex structure would pull back from the target to the source (and the position of the critical values does not matter). Now, by Hurwitz, it suffices to solve in the symmetric group $S_n$ the equation $g_1\ldots g_k=1$ with the $g_i$'s in the prescribed conjugacy classes. If one of the ramification indices is $n$ (polynomials), a solution exists, but in general I do not know the answer. The number of solutions (in particular, zero or not) is given by the Frobenius formula, so I would suggest computer experiments to begin with. (Alternatively, one can use "generalized" dessins d'enfants.) And I would suggest to check early papers by S. Natanzon: he considered the real setting, which is more difficult, but he might have written as well something about the underlying complex problem.

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To complement the answer of Alex Degtyarev, the answer to the original question is "no", and a necessary and sufficient condition is quite complicated. It can be written in various forms, see, for example MR0754748
Mednykh, A. D. Nonequivalent coverings of Riemann surfaces with a prescribed ramification type, Sibirsk. Mat. Zh. 25 (1984), no. 4, 120–142.

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Let me expand on the answer by Alex a little bit, and give a specific example.

The question of whether such rational maps exist, and if yes, how many, goes back to Hurwitz ("Über Riemann’sche Flächen mit gegebenen Verzweigungspunkten", Math. Ann. 39 (1891) 1-60). In particular the counting problem, in certain cases where the answer is positive, is of particular importance in Algebraic Geometry today; the corresponding numbers are called Hurwitz numbers. (See http://en.wikipedia.org/wiki/ELSV_formula for a description of one of the key results in this context.)

As mentioend by Alex, this is fundamentally a combinatorial question, which can be formulated in a number of equivalent ways: One of these (the original one by Hurwitz) involves the symmetric group, but there are also graph-theoretical approaches (e.g. the generalized dessins d'enfants mentioned).

Finally, let me give a simple specific example where the answer is negative: There does not exist a rational map $R$ of degree $d=4$ with three critical values $0$, $1$ and $\infty$, having two simple critical points over each of $0$ and $1$, as well as one simple preimage and one preimage of degree three over $\infty$.

First note that this branch data satisfies the Riemann-Hurwitz formula: the number of critical points (counting multiplicity) is $2+2+2=6=2d-2$.

Now consider the preimage of the interval $[0,1]$ under the hypothetical map $R$. (This is a "dessin d'enfant".) This would have to be a connected bipartite planar graph with four vertices of degree 2 (corresponding to the preimages of 0 and 1). Clearly such a graph would have to be a simple cycle consisting of 4 edges, but then both faces of the graph would be bounded by exactly 4 edges, which contradicts the degree requirements of the preimages of $\infty$. (Each preimage of $\infty$ of degree $k$ must correspond to a face having $2k$ edges in its boundary.)

On the other hand, as Alex already pointed out, the answer is positive for polynomials, which may or may not answer your question about a "rational function on the complex plane".

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  • $\begingroup$ Couldn't the graph also consist of two 2-cycles, rather than just one 4-cycle? $\endgroup$
    – HJRW
    Jan 8, 2021 at 12:13
  • $\begingroup$ I guess it couldn't, because its complement is a cover of a punctured plane. But perhaps the easiest way to see this example is to use the group-theoretic criterion mentioned in Alex Degtyarev's answer, and note that the product of two (2,2)-cycles in S_4 is never a 3-cycle. That said, as one who loves (branched) covers of graphs, I really like this answer! $\endgroup$
    – HJRW
    Jan 8, 2021 at 18:13
  • $\begingroup$ @HJRW Yes, the graph will be connected, since each complementary component has to be simply connected - as you say. :) $\endgroup$ Jan 10, 2021 at 16:14

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