I am trying to understand how to construct rational functions whose analytic continuation corresponds to given finite-index subgroups of the free group $$ \pi_1(\mathbb C\setminus\{0,1\}). $$ Equivalently, I want rational maps $f:\widehat{\mathbb C}\to\widehat{\mathbb C}$ that restrict to unbranched finite-sheeted covers $$ f^{-1}(\mathbb C\setminus\{0,1,\infty\})\longrightarrow \mathbb C\setminus\{0,1,\infty\}. $$
For example, a $2$-fold cover corresponds to data of how the two sheets permute when looping around $0$ and $1$. Up to conjugation, there are three cases (Using polynomials so $\infty$ is automatically a branch value and the ramified preimage above $\infty$ is $x=\infty$):
- trivial at $0$ (nontrivial at $1,\infty$)$$f(w) = w^2$$
- trivial at $1$ (nontrivial at $0,\infty$)$$f(w)=1-(1-w)^2$$
- trivial at $\infty$ (nontrivial at $0,1$)$$f(w)=\frac{w^2}{2w-1}$$
each realizing a distinct index-2 subgroup of $F_2$.
Example: 3-fold covers
For 3-covers, I tried the same idea: choose a rational function of degree $3$ and solve for coefficients so that its critical points map to $\{0,1,\infty\}$ with specified local multiplicities.
The simplest non-abelian case with cycle data
$$(2,1)\mid(2,1)\mid(3)$$
admits a concrete formula. Setting
$$
f(w)=c\,\frac{w^2(w-1)}{(4w-1)^3}
$$
and solving for $c$ so that $f(-2)=1$ gives $c=\tfrac{243}{4}$. Then
$$
f(w)=\frac{243}{4}\frac{w^2(w-1)}{(4w-1)^3}
$$
indeed has two simple branch points over $0,1$ and a triple one over $\infty$.
The cycle structure $(2,1)|(2,1)|(2,1)$ is impossible
It can be seen without attempting to construct the rational function at all. $sgn((2,1)) = (-1) \cdot (+1) = -1$. All three desired permutations $σ_0, σ_1, σ_\infty$ must be odd.
For any rational map with critical values $\{0, 1, \infty\}$, the monodromy permutations must satisfy the relation: $$σ_∞ \cdot σ_1 \cdot σ_0 = id$$ Substituting the signs for our desired cycle structures, we get: $$(-1) \cdot (-1) \cdot (-1) = +1$$ This is a contradiction. It is impossible to find three odd permutations that compose to the identity. Therefore, no rational function can realize the cycle data $(2,1)|(2,1)|(2,1)$.
Question
For a given combinatorial pattern of local degrees at $\{0,1,\infty\}$, what are the necessary and sufficient conditions for existence of a rational function realizing it?