Say that we have two Belyi maps, namely such that the composition is also a Belyi map. (For example, a sufficient condition here is that .) I am interested in computing the monodromy group of the compositon . To this end, I will show the following.

Proposition. Say that and are the monodromy groups of and , respectively, as subgroups of the symmetric groups and , respectively. Then is a subgroup of the wreath product of the symmetric groups.

Monodromy Groups of Belyi Maps

First, let me review how to compute the monodromy group of a Belyi map of degree .

Let be a complex number in the “thrice punctured sphere” . We know that the fundamental group , the group of closed loops satisfying modulo equivalence, is the free group on two generators. We will choose these generators to be the closed loops explicitly given by and as loops around 0 and 1, respectively.

Now the Belyi map restricts to an -fold cover whose domain is the punctured sphere . Let be those complex numbers which form the preimages of . For each preimage , there are unique paths such that

For instance, open differentiating, one can deduce that such paths are the unique solutions to the ordinary differential equations

Since , we see that for some . In particular, there exist permutations such that and for all . Hence we have a group homomorphism

The set , called the **monodromy group** of , is that subgroup of which is the image of this group homomorphism.

Consider the Belyi map given by of degree . We will show that is the Klein Vierergruppe.

Set and , so that is a 4-fold cover of punctured spheres. Choose , and write for some complex numbers and . Then in terms of the complex numbers

It is easy to check that the paths satisfy the differential equations

which have the exact solutions

In particular,

Hence while , so that the monodromy group is given by

Now let’s consider two Belyi maps such that the composition is also a Belyi map. Denote and so that . Continuing notation as above, set

- and choose ;
- and for ; and
- and for such that .

Hence is an -fold cover.

As before, we will choose the generators of the fundamental group to be the closed loops explicitly given by and . For each preimage , there are unique paths such that

As we know how to compute , we already know there exist such that

This means that we can find a -tuples and in such that

Hence we have a group homomorphism

The image of this group homomorphism is the monodromy group of . It is clear that there exists a subgroup such that we have a commutative diagram in the form

For any positive integer , consider the Belyi map given by which is the composition of the two Belyi maps and . It is easy to check that and are both cyclic groups, while is a dihedral group. In particular, is a proper subgroup of the wreath product , also known as the generalized symmetric group.

The Belyi map is the composition of the Belyi maps and . One checks that is the wreath product of and .

(I would like to thank Michael Musty and Sam Schiavone for reading a previous blog post, then sending me an e-mail with this result!)

The Belyi map is the composition of the Belyi maps and . It is easy to check that and . Curiously, the monodromy group is a proper subgroup because it has order . So what its the group ?

Question. Say that is a Belyi map which is the composition of and another Belyi map of degree . Since , we know that is a subgroup of the wreath product of and . Is it true that ?

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Let me try and set up how this would work by working through some explicit examples.

Say that we have a connected planar bipartite graph . The vertices consist of “black” vertices as well as “white” vertices , and the “midpoints” of the faces are . I am interested in the following

Problem.Given a connected planar bipartite graph , find a rational function which satisfies the following:

- Its only critical values are .
- are the “black” vertices.
- are the “white” vertices.
- are the “midpoints” of the faces.
- as the edges is the inverse image of the interval from to .

We say is a Belyi map, and is said to be a Dessin d’Enfant.

Computationally, I find working with connected planar bipartite graphs very difficult, so instead let me work with its valency list. Denote as the number of edges adjacent to each “black” vertex as indexed by , as the number of edges adjacent to each “white” vertex as indexed by , and as the number of edges adjacent to each face as indexed by . Define the degree sequence as a valency list:

and denote the positive integer . Define the polynomials in in variables by the determinants

for , each expressed in terms of the coefficients of the polynomials

each as elements of . The following is relatively trivial to prove.

Proposition 1.Continue notation as above.

- The Euler Characteristic and the Degree Sum Formula together imply

.- The following algebraic variety has dimension 0: In particular, Bezout’s theorem implies .
- Every point in gives rise to a Belyi map such that is its Dessin d’Enfant, and we have the explicit points Rather explicitly,

The basic idea behind this result is that we have the identity

Given a degree sequence , one can list the polynomials in very quickly. I would like to find the points in using the following

Algorithm. Denote as that extension of which is the field generated by the coordinates of the points .

- Using both the Homotopy Continuation and Newton’s Method, numerically find all of the points in .
- Using LLL, find a monic polynomial such that . Note that the roots of are in one-to-one correspondence with the points in .
- Also using LLL, find polynomials in such that for . Then for each root of .

I would like to use Bertini to complete Step #1. My graduate student Jacob Bond has code which can perform Steps #2 and #3 relatively quickly. I’ll say a few more words below when I discuss some examples.

Let me discuss a class of examples in some detail. We say that a connected planar graph is a tree if it has only one face.

I will view it as a bipartite graph by coloring its vertices “black” and placing “white” vertices at the midpoints of the edges. Assuming there are vertices and hence edges, its degree sequence is in the form

where we must have . Here , , and , so that is odd. The following result is closely related to the one above, but the notation is not quite the same.

Proposition 2.Continue notation as above. If we can find nontrivial complex numbers such that

is identically zero as a polynomial in , then the tree of interest is the Dessin d’Enfant of the Belyi map .

This Belyi map is not just a rational function of degree ; it is a polynomial. We say that such a Belyi map is a Shabat Polynomial. Note that there are equations in variables , so again we can construct an algebraic variety of dimension 0 and ask for all points .

Given a collection of positive integers such that , one can list the polynomials in very quickly. Remember that the goal is to find the variables such that we have the equations .

Say that so that . We have a degree sequence such that , so we must have . We have the polynomial where . Setting , we find . Hence the Shabat polynomial must be

Say that so that . We have a degree sequence such that , so up to permutation we must have and . We have the polynomial

where

Setting , we find two solutions: either or . We throw out the first solution because we will need to divide by , so from the second solution we find the Shabat polynomial

Consider the star graph on edges. It has degree sequence

We have already seen this when as in the two previous examples.

We are interested in an identity of the form for all . Let denote a primitive th root of unity. If we choose

then we recover the identity . Hence the corresponding Shabat polynomial is .

Now consider any tree with edges having the degree sequence

Since , we have the polynomial

in terms of

Question. Can Bertini find all solutions to ?

I only know that one solution corresponds to

, , , , , and . In particular, one set of identities which works is as follows:

Using the substitution we find a corresponding Shabat polynomial

However, this isn’t the only Shabat polynomial because this isn’t the only tree with this degree sequence! By using a slight permutation of the degree sequence above, say

it is easy to verify that a completely different set of identities is as follows:

Using the substitution , we find that a completely different polynomial:

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As considered in the 1994 paper by Georgii Borisovich Shabat and Alexander Zvonkin entitled *Plane trees and algebraic numbers*, the rational function

is a Shabat polynomial which happens to be the composition of two other Shabat polynomials. The first has monodromy group as a cyclic group, while the second has monodromy group as an alternating group. The monodromy group of the composition has order . Do we have as the wreath product of by ?

Let me begin by establishing some notation and making some remarks. A lot of this to begin with will be background information, so I apologize to those who already know this stuff pretty well.

Let be a compact connected Riemann surface. Remember that the only case we care about is when is the Riemann sphere. Indeed, we have a bijection which comes from stereographic projection:

A Belyi map is a morphism which is branched only above three points . When is a Riemann sphere, a Belyi map must be the ratio of two polynomials having complex coefficients. Rather implicitly:

Once we let be the number of times (, or , respectively) appears in the exponents (, or , respectively), we define the degree sequence of a Belyi map as a multiset of multisets:

and the passport of a Belyi map as shorthand notation for the degree sequence:

where

These quantities give combinatorial information about the critical points of , namely, the three sets , , and . For example, they correspond to partitions of the integer into , , and parts, respectively.

We say is a Shabat polynomial if is a Belyi map and consists of just point in the preimage. By pre- and post-composing with Moebius transformations, say

we may as well assume that a Shabat polynomial is a Belyi map branched only above such that and . In particular, we can factor a Shabat polynomial as

We often omit the information about infinity and write the degree sequence/passport of a Shabat polynomial as

where as before

There is a simple way to generate lots of examples of Shabat polynomials. Say that is one example of a Shabat polynomial. It is well-known that we can generate more Shabat polynomials as the composition of two maps and as long as is a Shabat polynomial which satisfies . One obvious such map is , although of course there are many others. I will return to this construction later.

Given a Belyi map , its Dessin d’Enfant is a bipartite graph which can be embedded on the Riemann sphere . It is defined as follows.

- The “black” vertices are ;
- The “white” vertices are ;
- The “midpoints of the faces” are ; and
- The “edges” are as the inverse image of the line segment .

The Dessin d’Enfant will have vertices, edges, and faces, so the Euler characteristic forces . When is a Shabat polynomial, there is only face, so the corresponding Dessin d’Enfant will be a tree. There is a nice post on MathOverflow which discusses this in a bit more detail.

In addition to Belyi maps/Shabat polynomials and degree sequences /passports , there is a third object of interest. The monodromy group is a transitive subgroup of $S_N$ which is constructed as follows.

- Label the edges of from 1 through .
- For each “black” vertex , read off the labels of the adjacent edges going counterclockwise, say as a -cycle in the symmetric group.
- For each “white” vertex , read off the labels of the adjacent edges going counterclockwise, say as a -cycle in the symmetric group.
- For each “midpoint of edge” , read off the labels of the adjacent edges going counterclockwise, say as a -cycle in the symmetric group.
- Form that subgroup of which is generated by the products of disjoint cycles , and .

If is a different set of permutations arising from this construction but from a different labeling of the edges, then there is a permutation so that these are simultaneously conjugate to , that is, , , and . This permutation must send the edge labels , , and .

It is not too difficult to recover the Dessin d’enfant from a triple . Indeed, the number (, respectively) of disjoint cycles in the decomposition of (, respectively) gives the number of “black” (“white”, respectively) vertices, and one can read off the cycle notation to find the labels of the edges which are adjacent to each “black” (“white”, respectively) vertex. One can connect the edges then slide around the vertices as necessary so that the faces match the disjoint cycle decomposition of .

Fix a positive integer , and let denote the symmetric group of degree , and let be a multiset of multisets such that .

Proposition. The following are equivalent:

- is the degree sequence of a Belyi map of degree on the Riemann sphere .
- There is a triple of elements in which can be expressed as a disjoint product of cycles , , and such that and is a transitive subgroup of .

The result was established by Adolf Hurwitz in his 1891 paper *Ueber Riemann’sche Flachen mit gegebenen Verzweigungspunkten*. I’ve explained how to construct a degree sequence from a Belyi map, but of course the opposite direction is an active area of research.

Proposition. All permutations of cycle type are conjugate, that is, for some . In particular, all triples of “cycle type ” are conjugate, that is, , , and for some .

This is an elementary fact about permutations.

Proposition. There is a one-to-one correspondence between:

- Belyi maps of degree on some compact connected Riemann surface .
- Triples of elements in , modulo simultaneous conjugation, such that and is a transitive subgroup of .

Proposition. Say that is a triple of elements in such that , and denote as the subgroup of generated by them. Denote as the centralizer of in . There is a one-to-one correspondence between:

- The set of all triples , modulo simultaneous conjugation, such that and , for some
- Double cosets in .

These correspondences between Belyi maps, triples , and double cosets were established by Michael Klug, Michael Musty, Sam Schiavone, and John Voight, in their 2013 paper *Numerical calculation of three-point branched covers of the projective line*. You can find the paper online at http://arxiv.org/abs/1311.2081v3. For the third correspondence, I’ve explained how to construct a triple of permutations from a Belyi map. Here’s a quick sketch of the fourth correspondence: say that and are two triples such that

for some , so denote and . Assume that they generate the same the double coset , that is, there exist and such that . We have the identity

in terms of so that the two triples are simultaneously conjugate. Conversely, assume that the triples are simultaneously conjugate. Then the elements and , so that and generate the same the double coset .

The monodromy group helps to count the number of Belyi maps of degree . Here’s an algorithm:

- Fix a positive integer .
- Compute all multisets of multisets
such that .

- For each , find a triple of elements in which can be expressed as a disjoint product of cycles , , and such that and is a transitive subgroup of . If no such triple exists, there is no Belyi map for which is a degree sequence.
- Compute and as the centralizers of and in . Given a double-coset representative in , the triple corresponding to and gives a unique Belyi map.

Let me give a relatively simple example. The rational is a Shabat polynomial of degree whose passport is , Dessin d’Enfant is the Star Graph with spokes, and whose monodromy group is as the cyclic group of order . To see why the latter is true, observe that

- the “black” vertex is ,
- the “white” vertices are for ,
- the “midpoint of the face” is , and
- the edges are the segments for .

If we label edge as the positive integer “”, then is an -cycle while is trivial. Hence the group generated by , , and is .

In fact, I can do even better: I claim this is the only Shabat polynomial associated with this passport! Indeed, if we have a triple associated with the passport , then must be the trivial permutation, and so the centralizer . This means there are no nontrivial double cosets in , so there can only be one Shabat polynomial.

As considered in the 1994 paper by Georgii Borisovich Shabat and Alexander Zvonkin entitled *Plane trees and algebraic numbers*, the rational function

is a Shabat polynomial of degree whose passport is . Observe that is the composition of the two rational functions and which are both Shabat polynomials, one of degree with passport , and the other of degree with passport . Since , we see that is the Star Graph and is the cyclic group of order 2. It is easy to see that the monodromy group of has the generators

so that is the alternating group. Here’s the strange part. The monodromy group of the composition has the generators

so that is a group of order .

Question. is a composition of Shabat polynomials, where and . Since , do we have as the wreath product of by ?

I’m really confused by this example because it seems to be part of a general phenomenon… I will post more on this later as I come up with more examples — and possibly a proof!

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Abstract.The purpose of this article is twofold. Firstly we show that the action of the absolute Galois group on certain classes of algebraic curves (Hurwitz curves, Hurwitz translation surfaces) and dessins d’enfants (regular dessins, classical and higher, with fixed signature) is faithful. Secondly we introduce a property of profinite groups, called Jarden’s property, and show this property for certain \’etale fundamental groups. Combining the latter result with faithfulness of the Galois action on these groups, which was known before, we obtain the results on curves and dessins.

You can download the paper here.

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Abstract.Branched covers of the complex projective line ramified over , and (Grothendieck’sDessins d’Enfant) of fixed genus and degree are effectively enumerated. More precisely, branched covers of a given ramification profile over and given numbers of preimages of and are considered. The generating function for the numbers of such covers is shown to satisfy a PDE that determines it uniquely modulo a simple initial condition. Moreover, this generating function satisfies an infinite system of PDE’s called the KP (Kadomtsev-Petviashvili) hierarchy. A specification of this generating function for certain values of parameters generates the numbers ofDessinsof given genus and degree, thus providing a fast algorithm for computing these numbers.

You can download the paper here.

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Abstract.We define and classify the analogues of the affine Kac-Moody Lie algebras for the ring corresponding to the complex projective line minus three points. The classification is given in terms of Grothendieck’s dessins d’enfants. We also study the question of conjugacy of Cartan subalgebras for these algebras.

You can download the paper here.

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Abstract.In this article, we survey methods to compute three-point branched covers of

the projective line.

You can download the paper here.

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This assignment is due Friday, October 18, 2013 at the start of class.

Homework Assignment 3 Download

**Problem 1.**

- Let denote the collection of those real matrices with nonzero determinant. Prove the Iwasawa Decomposition:

in terms of the real numbers , , , and . - Let denote those real matrices with positive determinant. Consider the map defined by . Show that is surjective, whereas

.

**Problem 2.** Show that . *Hint:* Use the Iwasawa Decomposition.

**Problem 3.** Define the the modular curve in terms of

The normalized hyperbolic area differential on the upper half plane is given by .

- Show that .
- Conclude that is a compact Riemann surface with finite volume.

**Problem 4.** Let be a complex number with . In the Poincare Disk Model, a “line” is given by .

- Identifying the affine complex line with the affine real plane , show that is a circle of radius centered at which intersects the boundary at the two distinct complex points .
*Hint:*Solve the simultaneous equations . - Show that is perpendicular to at the points where they intersect.
*Hint:*Consider the inner product of the vectors and which are normal to and , respectively.

**Problem 5.**Using the isomorphism

show that the intersection of the unit disk and the “line” is in bijection with the intersection of the hyperboloid and the plane . *Hint:* Show that points in map to .

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Abstract.We construct Belyĭ maps having specified behavior at finitely many points. Specifically, for any curve defined over , and any disjoint finite subsets , we construct a finite morphism such that ramifies at each point in , the branch locus of is , and is disjoint from . This refines a result of Mochizuki’s. We also prove an analogous result over fields of positive characteristic, and in

addition we analyze how many different Belyi maps are required to imply the above conclusion for a single and and all sets of prescribed cardinality.

You can download the paper here.

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Let’s review the main definitions and concepts from last time.

Consider a prime ideal in the polynomial ring . The set is an algebraic variety. The quotient ring is an integral domain; we define as the coordinate ring of . The map which sends a point to the maximal ideal is a bijection.

We say that is an algebraic curve if . In Lecture 14, we showed that the following are equivalent:

- For each , the matrix

yields an exact sequence:

That is, the Jacobian matrix has rank while the tangent space has dimension as a complex vector space. - The Zariski cotangent space has dimension as a complex vector space for each maximal ideal .
- For each , the localization is a discrete valuation ring.
- is a Dedekind Domain.

If any of these equivalent statements holds, we say that is a nonsingular algebraic curve.

Fix complex numbers , , , , and ; and consider the polynomial . Then the ideal in the polynomial ring is a prime ideal, so that the set is an algebraic curve over . We discuss when this is a nonsingular algebraic curve.

Define the complex numbers

Using the substitutions and we see that . Using the ideas in Lecture 5, we see that is non-singular if and only if . Such a curve is called an elliptic curve.

Now fix complex numbers , , , , and with ; and consider the polynomial in terms of the quartic polynomial . Again, the ideal in the polynomial ring is a prime ideal, so that the set is an algebraic curve over .

Define the complex numbers

Using the ideas in Lecture 5, we can find a substitution such that is equivalent to the curve . We see that is nonsingular if and only if .

Now that we have considered non-singular algebraic curves, we are motivated by the following result.

Theorem.

Given a prime ideal in such that , let be a non-singular algebraic curve over . Then can be given the structure of a compact, connected Riemann surface.

We will present a general proof later, but for now we sketch the ideas with a key example. You can read a proof in Proposition 0.4 on page 4 of J. S. Milne’s course notes *Modular Functions and Modular Forms*. There is an interesting discussion on this result at math.stackexchange.com with a thread entitled “Why are Riemann surfaces algebraic curves?”.

Here is one example. We have seen that the Riemann sphere is a compact, connected Riemann surface. Indeed, stereographic projection is the map defined by ; it is a bijection. The open sets

form an open cover of . You may wish to compare this with Lecture 10.

Here is another example with a historical approach. Fix complex numbers such that . We have seen that the set is a non-singular algebraic curve. Recall that is an elliptic curve. We explain why is a Riemann surface. (We’re cheating a little here: is not compact because we’re not including the “point at infinity” . This is a small point which we will ignore for now.)

Define a function implicitly as follows: For any , define as that extended complex number such that . We call the Weierstrass pae-function after the German mathematician Karl Weierstrass (1815 — 1897). Observe that . Unfortunately this concept doesn’t make a well-defined function. Indeed, the integrand has poles at the roots , , and of the cubic , so the integral is actually path dependent. If we define the lattice in terms of the periods and , then we find that the map which sends to is a bijection. In particular,

. Using this, we say that is a torus.

In fact, using this one shows that

This function is not as strange as it might seem at first. We explain how to construct a similar map for the unit circle . Define a map implicitly by . This is not well-defined; we can only determine the values up to multiples of . In particular, is a well-defined map . The map which sends to is a familiar bijection.

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