-- $x_1,\ldots,x_n$ be $n$ distinct points on the complex plane $\mathbb C$.

-- $r_1,\ldots,r_n$ be $n$ real numbers.

Consider the map $$ z\mapsto u(z)=\int^z \frac{1}{(x-x_1)^{r_1}\cdots (x-x_n)^{r_n}} d\xi $$

It defines a multivalued holomorphic function on $\mathbb C\setminus (x_1,\ldots,x_n)$

Question: what is the monodromy of this map?

Remark: when the $x_i$'s are real and if the $r_i$'s verify $\sum_i r_i=2$, the map $u$ is nothing but a classical Schwarz-Chritoffel map that maps conformally the upper half-plane onto a closed $n$-gon.

In this classical case, it is well known that the monodromy is affine. But I have been unable to find any reference where the monodromy of $u$ is explicited.

Any help would be welcome!

  • $\begingroup$ +1. However, the function is not well defined on $\mathbb{C}$ minus a set of points. You have to choose branches of logarithms at each point or only consider it as a function on the upper halfplane. Can you elaborate what you mean by "the monodromy is affine" and how this conclusion arises (only for the sake of my education;)? $\endgroup$
    – Marc Palm
    Commented Apr 19, 2013 at 15:52
  • $\begingroup$ As said just before the Question, $z\mapsto u(z)$ is ${\bf multivalued}$ on the complement of the union of the $z_i$'s, noted by $U$. Let $z_0$ be a point distinct with $z_0\neq z_i$ for every $i$. For any loop $\gamma: [0,1]\rightarrow U$ centered at $z_0$ (ie. \gamma(0)=\gamma(1)=z_0$), the monodromy $\mathcal M_{\gamma} u$ is the germ at $z_0$ obtained by analytic continuation of the initial determinantion of $u$ at $z_0$. Affine monodromy means $\mathcal M_{\gamma}u=a_{\gamma} u+b_{\gamma}$ with $a_{\gamma},b_{\gamma}\in \mathbb C$ $\endgroup$
    – Babar
    Commented Apr 19, 2013 at 18:15
  • $\begingroup$ Computation of the local monodromies (up to conjugation) for the small loops around points $x_i$, is a nice undergraduate complex analysis exercise. $\endgroup$
    – Misha
    Commented Apr 19, 2013 at 19:09

1 Answer 1


Monodromy is affine in the general case. It is generated by finitely many elliptic transformations of the form $az+b$. Proof: your function satisfies the differential equation $u^{\prime\prime}=Ru',$ where $R$ is rational. The general solution of this equation is obtained from a particular solution by the formula $au+b$.

Computing this monodromy explicitly is as difficult as in the classical case: you need to compute the integrals of your $u$ from $x_j$ to $x_k$.


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