Understanding the construction of Riemann surface for $\xi^3-z\xi^2-(a^2-1)\xi+za^2=0$.

Question

I am reading an article which defines a Riemann surface by the following equation ($a>1$): $$\xi^3-z\xi^2-(a^2-1)\xi+za^2=0$$ The goal is to find the Riemann surface of $\xi(z)$.

What I know/understand:
The critical points satisfy $\frac{\partial z}{\partial \xi}=0$, which after some working out means that the branch points are $-z_1,-z_2,z_2,z_1$ ($0<z_2<z_1$) with $$z_1,z_2=\sqrt{\frac{1}{2}+a^2\pm\frac{1}{2}\sqrt{1+8a^2}}\frac{\sqrt{1+8a^2} \pm 3}{\sqrt{1+8a^2}\pm 1}.$$ Also, the solutions of the first equation have the following behaviour: $$\xi_1(z)=z-\frac{1}{z}+O\left(\frac{1}{z^3} \right),\xi_2(z)=a+\frac{1}{2z}+O\left(\frac{1}{z^2} \right),\xi_3(z)=-a+\frac{1}{2z}+O\left(\frac{1}{z^2} \right)$$ as $z\rightarrow \infty$.

What I don't understand:
Apparently, $\xi_1,\xi_2$ and $\xi_3$ can be analytically extended to $\mathbb{C}\backslash ([-z_1,-z_2]\cup [z_2,z_1])$, $\mathbb{C}\backslash [z_2,z_1]$ and $\mathbb{C}\backslash [-z_1,-z_2]$ respectively. On the cuts, it holds that $$\xi_{1+}(x)=\overline{\xi_{1-}(x)}=\xi_{2-}(x)=\overline{\xi_{2+}(x)},\quad z_2<x<z_1$$ and $$\xi_{1+}(x)=\overline{\xi_{1-}(x)}=\xi_{3-}(x)=\overline{\xi_{3+}(x)},\quad -z_1<x<-z_2$$ which determines the shape of the Riemann surface.

I understand that this likely requires some tedious working out, so just the idea of how to start will help a lot.

Answer 1

I think the easiest way to understand this Riemann surface is to view it as the graph of the function $$z(\xi) = \frac{\xi^3 - (a^2-1)}{\xi^2 - a^2}$$ since a function and its inverse have the same Riemann surface (which is essentially the graph of the relation defining those.) With your assumption $a>1$, the numerator and denominator do not have a common factor, so this is a rational functions of degree $3$. You can now find the critical points whose images are the branch points of the inverse $\xi(z)$, but basically this is it.

Answer 2

The branch points are the roots of the discriminant in the variable $\xi$, which is $$\Delta_\xi(P(z, \xi)) = \Delta_\xi(\xi^3 - z \xi^2 + a^2 z - a^2 + 1) =\\ (a^2 z - a^2 + 1)(4z^3 - 27 a^2 z + 27a^2 - 27),$$ they won't be quadratic irrationalities.

Consider the first branch point $z_1 = (a^2 - 1)/a^2$: $$P(z_1 + z, \xi) = \xi^3 - (z + z_1)\xi^2 + a^2 z.$$ $P(z_1, \xi)$ has a single root $\xi = z_1$ and a double root $\xi = 0$. The Puiseux expansions of $\xi(z)$ around $z = z_1$ are $\xi = z_1 + c_1(z - z_1) + \dots$ and $\xi = c_{1/2}(z - z_1)^{1/2} + \dots\,$.

The same computations show that at each of the other three branch points, $P(z_i, \xi)$ also has one single root and one double root, and the structure of the branches is the same, that is, there is one pair of sheets glued together and one single sheet, or, in terms of permutations, one cycle of length two.

At $z = \infty$, the Puiseux expansions are $\xi = z + c_{-1}z^{-1} + \dots$, with the leading term determined by the terms $\xi^3 - z \xi^2$ in $P$, and $\xi = \pm a + c_{-1}^{\pm}z^{-1} + \dots$, with the leading term determined by $-z \xi^2 + a^2 z$, and $\xi(z)$ is unbranched over $\infty$.

A loop around $\infty$ can be viewed as a composition of loops around the four branch points $z_i$. If there is a cycle that appears only once, the composition will not be equal to the identity permutation. Also, the group action is transitive because $P(z, \xi)$ is irreducible. Thus there have to be two cycles $(12)$ and two cycles $(23)$, and the structure looks like this:

A possible choice for the branch cuts is the segment connecting the branch points corresponding to the cycle $(12)$ together with the segment connecting the points corresponding to $(23)$.