---
abstract: |
    A pair is a holomorphic map from a Riemann surface to $S^2$ with additional properties. A dessin d’enfants is a bipartite graph with additional structure. It is well know that there is a bijection between pairs and dessins d’enfants.

    Vassiliev has defined a filtration on formal sums of isotopy classes of knots. Motivated by this, we define a filtration on formal sums of Belyĭ pairs, and another on dessin d’enfants. We ask if the two definitions give the same filtration.
author:
- |
    Jonathan Fine\
    Milton Keynes\
    England\
    `jfine@pytex.org`
date: 28 September 2009
title: A filtration question on Belyĭ pairs and dessins
---

Introduction
============

First, we recall some definitions [@Belyui; @Dessins]. A *Belyĭ pair* is a Riemann surface $C$ together with a holomorphic map $f:C \to S^2 = \C \cup \{\infty\}$ to the Riemann sphere, such that $f'(p)$ is non-zero provided $f(p)$ is not $0$, $1$ or $\infty$. (Belyĭ proved that given $C$ such an $f$ can be found iff $C$ can be defined as an algebraic curve over the algebraic numbers.)

A *dessin d’enfants*, or *dessin* for short, is a graph $G$ together with a cyclic order of the edges at each vertex, and also a partition of the vertices $V$ into two sets $V_0$ and $V_1$ such that every edge joins $V_0$ to $V_1$. Necessarily, $G$ must be a bipartite graph. Traditionally, the vertices in $V_0$ and $V_1$ are coloured black and white respectively.

It is easy to see that a Belyĭ pair gives rise to a dessin, where $V_0=f^{-1}(0)$, $V_1 = f^{-1}(1)$, and the edges are the components of the inverse image $f^{-1}([0,1])$ of the unit interval in $\C$. The cyclic order arise from local monodromy around the vertices.

A much harder result, upon which our definitions rely, is that up to isomorphism every dessin arises from exactly one Belyĭ pair, or in other words that there is a bijection between isomorphism classes of Belyĭ pairs and dessins.

Definitions
===========

A *Belyĭ object* $B$ consists of $((B_C, B_f), B_D)$ where $(B_C, B_f)$ is a Belyĭ pair and $B_D$ is the associated dessin (or vice versa for the dessin and the pair).

The *Vassiliev space* $V=V_\C$ (for Belyĭ objects) is the vector space over $\C$ which has as basis the isomorphism classes of Belyĭ objects.

Clearly, when an edge is removed from a dessin then it is still a dessin. Suppose $D$ is a dessin, and $T$ is a subset of its edges. We will use $D \setminus T$ to denote the dessin so obtained. This same operation can also be applied to a Belyĭ object $B$, even though computing the associated curve $(B\setminus T)_C$ from $B_D$ and $T$ might be hard.

We will now define one or two filtrations of $V$.

Let $D$ be dessin and $S$ a $d$-element subset of $D$. Each subset $T$ of $S$ determines a dessin $S\setminus T$ and hence a object $B_{S\setminus T}$. Let $|T|$ denote the number of edges in $T$. Use $$B_S = \sum\nolimits _{T\subseteq S} (-1)^{|T|}B_{S \setminus T}$$ to define a vector $B_S$ in $V$, which we call *the expansion of a dessin with $d$ optional edges*.

Let $V_{D,d}$ be the span of the expansions of all dessins with $d$ optional edges. The sequence $$V =
    V_{D, 0} \supseteq
    V_{D, 1} \supseteq
    V_{D, 2} \supseteq
    V_{D, 3} \ldots$$ is the *dessin filtration* of $V$.

We can also think of a Belyĭ object as a map $f:C\to S^2$ (with special properties). Let $(C_1, f_1)$ and $(C_2, f_2)$ be Belyĭ pairs. Then there is of course a map $$g: C_1 \times C_2 \to S^2 \times S^2 \>.$$

Let $\Delta \subset S^2 \times S^2$ denote the diagonal, and let $C$ denote $g^{-1}(\Delta)$, and $f$ the restriction of $g$ to $C$. In general $$f: C \to \Delta \cong S^2$$ will not be a Belyĭ pair. There are two possible problems. The first is that $C\subset C_1\times C_2$ might have self intersections or be otherwise singular. If this happens, we replace $C$ by its resolution, which is unique.

The second problem is more interesting. It might be that $f$ has critical points not lying above the special points $0$, $1$ and $\infty$. This problem cannot be avoided. However, the above discussion does show that there is product, which we will denote by ‘$\circ$’, on holomorphic branched covers of $S^2$.

Let $W$ be the vector space with basis isomorphism classes of branched covers of $S^2$. We set $W_n$ to be the span of all products of the form $$(A_1 - B_1) \circ
  (A_2 - B_2) \circ
  \ldots \circ
  (A_n - B_n)$$ for $A_i$ and $B_i$ basis vectors of $W$. Clearly, the $W_n$ provide a filtration of $W$.

The induced filtration of $V$ defined by $V_{B,n} = W_n \cap V$ is called the *filtration* of $V$.

Questions
=========

Are the two filtrations $V_D$ and $V_B$ equal?

If so, then we have also answered the next two questions.

The absolute Galois group acts on pairs, and preserves the filtration. Does this action also preserve the dessin filtration?

Because the dessins with $d$ edges, all of which are optional, span $V_d/V_{d+1}$, the dessin filtration has finite dimensional quotients. Does the filtration have finite dimensional quotients?

Investigating the last two questions might help us answer the first. They might also be of interest in their own right.

[9]{}

D. Bar-Natan, On the Vassiliev knot invariants, Topology 34 (1995), 423–472

G. V. , Another proof of the three points theorem, Subornik: Mathematics 193 (2002), 329–32.

Leila Schneps, ed, The Grothendieck Theory of Dessins d’Enfants, London Math. Soc. Lecture Note Ser., vol 200, Cambridge Univ. Press 1994.
