Let $Y$ be a compact Riemann surface and $B$ a finite subset of $Y$. It is a standard fact that isomorphism classes of holomorphic ramified covers $f:X\rightarrow Y$ of degree $d$ with branch points in $B$ are in a correspondence with homomorphisms $\rho:\pi\_1(Y-B)\rightarrow S\_d$ with transitive image modulo conjugation by elements of the permutation group $S\_d$. Writing a formula for $f$ from the knowledge of $B\subset Y$ and $\rho$ is often hard, e.g. the task of recovering a Belyi map from its dessin where $|B|=3$. I am interested in the case of $X=Y=\Bbb{CP}^1$, and some points from $B$ moving in the Riemann sphere. Here is an example:


* Consider rational functions $f:\Bbb{CP}^1\rightarrow\Bbb{CP}^1$ of degree $3$ with four simple critical points that have $1,\omega,\bar{\omega}$ among their critical values $\left(\omega={\rm{e}}^{\frac{2\pi{\rm{i}}}{3}}\right)$, thus $B=\{1,\omega,\bar{\omega},\beta\}$ with $\beta$ varying in a punctured sphere. To fix an element in the isomorphism class, we can pre-compose $f$ with a suitable Möbius transformation so that $1$, $\bar{\omega}$ and $\omega$ are the critical points lying above $1$, $\omega$ and $\bar{\omega}$ respectively: $f(1)=1, f(\bar{\omega})=\omega, f(\omega)=\bar{\omega}$. A normal form for such functions is
$$
\left\{f\_\alpha(z):=\frac{\alpha z^3+3z^2+2\alpha}{2z^3+3\alpha z+1}\right\}\_\alpha.
$$
A simple computation shows that the fourth critical point is $\alpha^2$, and hence $\beta=\beta\_\alpha=:f\_\alpha(\alpha^2)=\frac{\alpha^4+2\alpha}{2\alpha^3+1}$. Here is my question:


**Why $\beta$ is not a degree one function of $\alpha?$ Shouldn't the knowledge of the branch locus and the monodromy determine $f\_\alpha(z)$ in the normalized form above? I presume the monodromy does not change because there are only finitely many possibilities for it and this is a continuous family.**


To monodromy of $f\_\alpha$ is a homomorphism
$$
\rho\_\alpha:\pi\_1\left(\Bbb{CP}^1-\{1,\omega,\bar{\omega},\beta\_\alpha\}\right)\rightarrow S\_3
$$
where small loops around $1,\omega,\bar{\omega},\beta\_\alpha$ generate the fundamental group, and are mapped to transpositions in $S\_3$ whose product is identity and are not all distinct. So I guess my question is how can such a discrete object vary with $\alpha$; and if it doesn't, why the assignment $\alpha\mapsto\beta(\alpha)$ is not injective. The degree of this assignment is four, and there are also four conjugacy classes of homomorphisms
$\rho:\langle\sigma\_1,\sigma\_2,\sigma\_3,\sigma\_4\mid\sigma\_1\sigma\_2\sigma\_3\sigma\_4=\mathbf{1}\rangle\rightarrow S\_3$ with ${\rm{Im}}(\rho)$ being a transitive subgroup of $S\_3$ generated by transpositions $\rho(\sigma\_i)$:
$$
\sigma\_1\mapsto (1\,2),\sigma\_2\mapsto (1,2),\sigma\_3\mapsto (1,3), \sigma\_4\mapsto (1,3);\\
\sigma\_1\mapsto (1\,2),\sigma\_2\mapsto (1,3),\sigma\_3\mapsto (1,2), \sigma\_4\mapsto (2,3);\\
\sigma\_1\mapsto (1\,2),\sigma\_2\mapsto (1,3),\sigma\_3\mapsto (1,3), \sigma\_4\mapsto (1,2);\\
\sigma\_1\mapsto (1\,2),\sigma\_2\mapsto (1,3),\sigma\_3\mapsto (2,3), \sigma\_4\mapsto (1,3).
$$  

Is it accidental that the degree of $\alpha\mapsto\beta(\alpha)$ is the same as the number of possibilities for the monodromy representations compatible with our ramification structure?



Recovering a family of rational functions from branch points
The reason for this phenomenon is that you are afflicted with a serious mathematical condition, that being:


Your monodromy has monodromy.




---


To be less cryptic, the key thing is that the fundamental group is not actually generated by small loops around these four points, except if by "small loop around $x$" you mean "a path from the base point to $x$, then a small loop around $x$, then following the same path back" - these do generate the fundamental group.


When you move the point $\beta$, you will have to continuously adjust these paths - the reason you need to adjust the path to $\beta$ is clear, but also the other paths may need to be adjusted if $\beta$ would otherwise pass through them.


If you bring $\beta$ around a loop and back to its starting value, these paths have no reason to be the same paths they were before, so while the conjugacy classes of the loops around the four points will be the same in your chosen map $\rho$ from $\pi\_1$ to $S\_3$, the actual values of the loops around the four points under $\rho$ could be different.


So, in the general case, what you get is a map from the moduli space of possible covers to the curve parameterizing values of $\beta$ (or a higher-dimensional version, if more points vary) which is locally constant but not (usually) constant, and where the fiber consists of all monodromy representations which have fixed local conjugacy classes.


So it is not an accident at all that the degree is the same as the number of possibilities - it just reflects the fact that this covering map is connected, which is a common occurrence but not universal - there are known invariants which sometimes separate different components.


