computing monodromy of branched cover of sphere specified by a polynomial

We know by the Riemann Existence Theorem that any Riemann surface can arise holmorphically as the branched cover of a sphere:

Therefore, any branched cover of the sphere can be represented by a rational function polynomial $\frac{p (z)}{q(z)}: X \to \mathbb{C}$.

Given arbitrary polynomial $p$, do there exist ways to compute the monodromy around each critical points?

For example, $z \mapsto z^2 + \frac{1}{z^2}$ (not a polynomial) is 4-sheeted cover of the Riemann sphere $\hat{\mathbb{C}}$. It has 3 branch points, where $$p'(z) = 2z - \frac{2}{z^3} = 0$$ It seems to have 4 branch points at $z = 1, i, -1, -i$.

How do we calculate of which sheets were permuted given $p(z)$?

This is motivated by trying to understand a proof of the Riemann Existence Theorem in the case of genus 0 branched covers of the sphere $S^2$.

This question is the reverse direction of my previous question on branched covers: polynomial branched cover of the sphere with specified monodromy

Our example has a symmetry - and I am trying to avoid techniques which exploit them too much.

• It is not true that "any branched cover of the sphere can be represented by a polynomial". First of all, it has to be of genus $0$. Second, every $X$ of genus $0$ over the sphere can be represented by a rational function, not a polynomial. Your example is not a polynomial. – Alexandre Eremenko Nov 18 '13 at 20:08
• Essentially the same question as mathoverflow.net/questions/146971 – David E Speyer Nov 18 '13 at 22:17
• @DavidSpeyer Belyi's theorem says WLOG, we can assume the branch points are all at $0,1, \infty$. I had been drawing a tree where the leaves are the roots of $p(z)$ and the other vertices are the roots of $p'(z)$. So... I have been re-inventing the dessin? Maybe I can e-mail you, or post my entire train of thought in yet another question. – john mangual Nov 18 '13 at 22:31
• Your statement of Belyi's theorem is correct. Dessins are combinatorial devices that record the monodromy of a Belyi covering. All of the methods suggested in that answer work equally well for any branched covering, whether or not it is Belyi. – David E Speyer Nov 18 '13 at 23:32
• @DavidSpeyer and john: This is not exactly what Belyi's theorem says. Namely, if the branch points can be taken at 0,1,$\infty$, then up to a change of variables, the polynomial has its coefficients in $\overline{\mathbf Q}$ (the field of algebraic numbers). – ACL Nov 19 '13 at 7:51

In your example, everything is easy to calculate. Use the fact that your $f$ is a composition of $f(z)=z+1/z$ and $h(z)=z^2$. The plane is partitioned into $8$ regions by the coordinate axes and the unit circle. Each region corresponds to a sheet over upper or lower half-planes. The sheets are connected in the same way as the regions in the $z$-plane.

You found the critical points incorrectly: $0$ and $\infty$ are missing. In general, a rational function of degree $4$ has $6$ critical points.

The same method works in general (but in practice becomes complicated for functions of higher degree). You find critical points and critical values. Then make convenient cuts between critical values to break the sphere into simply connected regions. Then find out how components of the preimage of these regions tile the sphere of the independent variable. In your example, critical values are $\pm1$, $0$ and $\infty$. So it is reasonable to cut the image sphere into upper and lower half-planes. Finding the components of preimage of these half-planes is simplified by the fact that your function is a composition of two simpler functions. In general, an approximate calculation is sufficient to see how the components of preimage fit together.

EDIT. In the classical literature, they cut the image into two hemispheres by drawing a Jordan curve passing through all critical values. This defines a cell decomposition of the image sphere. Then they take the full preimage of this cell decomposition, or the full preimage of the dual call decomposition. In the second case one obtains an embedded bipartite graph (1-skeleton of the preimage cell decomposition). This graph is called the line complex; it permits to visualize the monodromy. It shows you how the half-sheets are pasted together. Of course everything depends on the choice of the Jordan curve, and there is no canonical way to choose it. Once it is chosen, the line complex can be computed if you can solve the equation $f(z)=a$ numerically. The line complex is what replaces the dessin d'enfant in the case of more than 3 critical values.

• So I am asking how to compute these "convenient" cuts of $\hat{\mathbb{C}}$. It is sounds like it will be tricky in general to decide how the pre-images fit together. – john mangual Nov 18 '13 at 22:20
• There is no general rule to choose the cuts in the image. It depends on the configuration of critical values. And you are right: in general it is difficult to decide how the preimages fit together. But if you know how to solve $f(z)=a$ at least numerically, this is possible. – Alexandre Eremenko Nov 19 '13 at 3:20
• Shouldn't it be 'dessin d'enfant' rather than 'design d'enfant' ? – brainjam Apr 15 '18 at 17:11
• @brainjam: yes, of course. I corrected. Thanks. – Alexandre Eremenko Apr 15 '18 at 23:39

David linked to some computational software implementing it; I just want to mention that this can also be done in Maple. I'd rather I was referencing either Sage because it's free and open source, or Bertini because it's free and Dan Bates here at CSU is one of the main authors, but I'm not actually much of a computing guy and this is what I figured out how to do in a few minutes with google.

For your example, "monodromy(y*z^2-z^4-1,y,z)" returns

[-3.60000000000, [-1.81567148865360 I, -0.550760424586247 I, 0.550760424586247 I, 1.81567148865360 I], [[-2., [[1, 2], [3, 4]]], [2., [[1, 3], [2, 4]]], [infinity, [[1, 4], [2, 3]]]]]

Where the output means the following: -3.6 is a regular value of your map, with four pure imaginary roots given. There are three critical values, -2,2, and infinity, and taking a path from -3.6 around your the given critical point (I'm not sure exactly which path) and back to -3.6, and lifting it to the cover, permutes the sheets in the following way.