# Polynomials for Bicubic Planar Dessin d'Enfants

A dessin d'enfant is a graph, with its vertices colored alternately black and white, embedded in ... a plane. For the coloring to exist, the graph must be bipartite. ... The ... embedding may be described combinatorially using a rotation system...

Any dessin can provide the surface it is embedded in with a structure as a Riemann surface. It is natural to ask which Riemann surfaces arise in this way.

The wikipage on dessin d'enfants, also gives an example:

The dessin d'enfant arising from the rational function $$f = −(x − 1)^3(x − 9)/64x$$. Not to scale.

Further a table, listing the degrees of vertices, is given. Later Shabat polynomials are mentioned, which are the corresponding ones for trees.

My question:

Is it possible to assign a rational function to any kind of bipartite planar cubic graph?

Or more specific: How to construct the polynomial for the chamfered cube?

• I have no idea about any of this, but I just want to say that I am very surprised no other users have answered or even commented on this post. I apologise that I cannot help you, and also on their behalf. Mar 21, 2018 at 6:50
• You may find Modular subgroups, dessins d’enfants and elliptic K3 surfaces relevant. Oct 16, 2021 at 9:16

Some introductory words. My interest in dessins d'enfants (dessins for short from now on) was related some years ago to (my speculation concerning) some connections with the K-theory, where the object "$$\Bbb P^1_{\Bbb Q}\setminus\{0,1,\infty\}$$" is a central object. (Symbols, Galois Groups over $$\Bbb Q$$, regulators in K-theory, ... are related to this object.) There was also an Arbeitsgemeinschaft (AG) on the subject, MFO-2000 centered on Dessins d'Enfants, and some very specific applications. (The report of the AG from loc. cit. shows a wide spectrum of related domains.) Often, the starting point is a given more or less complicated Belyi function. (E.g. related to quotients of the upper half plane by modular groups, where the Belyi function is a $$j$$-invariant.) Building the dessin gives a "compact information" on the situation.

In this case, the question of the OP is the converse one. We have the dessin, and need a realization. To obtain an answer to the question, it is natural to get a simplest situation of a map. Which is the situation of a rational map $$f=\frac PQ:\Bbb P^1(\Bbb C)\to\Bbb P^1(\Bbb C)$$, defined over $$\bar{\Bbb Q}$$, unramified above $$\Bbb P^1(\Bbb C)-\{0,1,\infty\}$$. If this is possible.

Existence theorems show it is possible. I found a good reference for this, the master thesis of Jeroen Sijsling.

Let us see the machinery at work for the case of the Chamfered Cube.

The planar graph for the ($$1$$-skeleton of the) cube is as follows:

To obtain its chamfered version, we draw inside of each (deformated) square a smaller one. For the outer square we must go outside. Every "old vertex" gets connected to the three new square vertices in its proximity.

The new picture is:

The circles (in black and/or blue) $$\circ$$ and $$\color{blue}{\circ}$$ are preimages of $$0$$ via the searched function $$f$$ (up to repositioning by hand, which does not destroy the dessin). The ramification index of each such point is $$3$$, i.e. locally around each such point, after a new choice of uniformizers, the map $$f$$ looks like $$z\to z^3$$. (Our graph is a cubic graph.)

The bullets (in black and/or blue) $$\bullet$$ and $$\color{blue}{\bullet}$$ are preimages of $$1$$ via the searched function $$f$$ (up to same repositioning). The ramification index of each such point is $$3$$.

The edges correspond to preimages via $$f$$ of the interval $$(0,1)\subset \Bbb C\subset \Bbb P^1(\Bbb C)$$.

In each "region" delimited by such edges there is exactly one preimage of $$\infty$$, we build the preimages of the three distinct paths between $$0,1,\infty$$ that build the projective real line $$\Bbb P^1(\Bbb R)$$ inside $$\Bbb P^1(C)=\Bbb H\cup\Bbb P^1(\Bbb R)\cup\Bbb H'$$. Here $$\Bbb H$$ is the upper half plane, and $$\Bbb H'$$ is its reflection w.r.t. the real affine line $$\Bbb R\subset \Bbb C\subset \Bbb P^1(C)$$. Imagine that the three paths are building a "triangle", and both $$\Bbb H$$ and $$\Bbb H'$$ may be considered to fill it, we give them two different colors. Preimages are then colored in the same way. Then the six "squares" and the twelve "hexagons" are colored with two white and two gray triangles as follows:

My paint dried short before painting the two "bounded" regions (those with one arc as boundary) outside the outer most square, please imagine them also as painted in gray. (The "triangle region", and the "unbounded region" are painted in white.)

Each preimage of $$0$$ and/or $$1$$ becomes thus a vertex of three white and three gray triangles, the corresponding ramification index is three. For the pink bullets $$\color{magenta}{\bullet}$$, preimages of $$\color{magenta}{\infty}$$ we have two different situations, there are:

• either two white and two gray "triangles" building a "square", and the corresponding ramification index is two,
• or three white and three gray "triangles" building a "hexagon", and the corresponding ramification index is three.

We can now make an Ansatz for $$f=P/Q$$:

• there are $$16$$ points $$a_1,a_2, \dots a_{16}$$ building $$f^{-1}(0)$$, each ramification index is three,
• there are $$12$$ points $$b_1,b_2, \dots b_{12}$$ with ramification index three in $$f^{-1}(\infty)$$, they correspond to the points $$\color{magenta}{\bullet}$$ in the "hexagons",
• there are $$6$$ points $$b'_1,b'_2, \dots b'_6$$ with ramification index two in $$f^{-1}(\infty)$$, they correspond to the points $$\color{magenta}{\bullet}$$ in the "squares",
• there are $$16$$ points $$c_1,c_2, \dots c_{16}$$ building $$f^{-1}(1)$$, each ramification index is three.

We need to exhibit the above constants, so that:

Ansatz: $$A\underbrace{\left(\prod_{1\le i\le 16}(x-a_i)\right)^3}_{P=p^3} - B\underbrace{\left( \prod_{1\le j\le 12}(x-b_j) \right)^3 \cdot\left( \prod_{1\le j\le 6}(x-b'_j) \right)^2 }_{Q=q^3s^2} = C\underbrace{ \left(\prod_{1\le k\le 16}(x-c_k) \right)^3 }_{R=r^3} \ .$$ (Each term has degree $$48$$, which is $$3\cdot 16=3\cdot 12+2\cdot 6$$.)

There is of course a homogenous version of the above polynomial relation, obtained by setting $$x=X/Y$$. The factors from above are $$(X-a_iY)$$, $$(X-b_jY)$$, etc. - and using Möbius isomorphisms, we may change the above, so that three points among the many points $$(a_i)$$, $$(b_j)$$, $$(c_k)$$ become respectively $$\infty,0,1$$. So according to our future needs, if $$\rho$$ is one such root among the three, we may arrange that $$(X-\rho Y)$$ becomes $$X$$ if we want to set $$\rho=0$$, then $$(X-\rho Y)$$ becomes $$Y$$ if we move $$\rho$$ to $$\infty$$, and $$(X-\rho Y)$$ becomes $$(X-Y)$$, when $$\rho$$ is moved to one. For this point, we only arrange that $$b_1=\infty$$, and still have the chance to move two points... We dehomogenize, so that we work further with \begin{aligned} p(x) &=(x-a_1)(x-a_2)\dots(x-a_{16})\ ,\\ q(x) &=(x-b_2)\dots(x-b_{12})\ ,\qquad\text{(b_1 was moved to \infty)}\\ s(x) &=(x-b'_1)\dots(x-b'_6)\ ,\\ r(x) &=(x-c_1)(x-c_2)\dots(x-c_{16})\ . \end{aligned} All above polynomials are monic, and the degrees are now $$16$$ for $$p,r$$, $$12-1=11$$ for $$q$$, and $$6$$ for $$s$$. We may and do take now $$A=C=1$$. It remains to realize in the factorial / UFD ring $$\bar{\Bbb Q}[x]$$ - best even with $$p,q,r,s$$ having coefficients in $$\Bbb Q$$ or some simple extension of $$\Bbb Q$$- a relation of the shape: $$(p-r)(p-\omega r)(p-\omega^2 r)=p^3 - r^3=B\; q^3s^2\ .$$ We have denoted by $$\omega$$ a primitive thirs root of unity. Let $$F=\Bbb Q(\omega)=\Bbb Q(\sqrt{-3})$$ be the corresponding cyclotomic field.

Here, $$p,q,s,r$$ have different roots, so we use UFD to get the following arithmetic information: \left\{ \begin{aligned} p-r &= B\; q_1^3\; s_1^2\ ,\\ p-\omega r &= (1-\omega)\; q_2^3\; s_2^2\ ,\\ p-\omega^2 r &= (1-\omega^2)\; q_3^3\; s_3^2\ . \end{aligned} \right. The pieces $$q_1,q_2,q_3;s_1s_2,s_3$$ are monic in the further search.

One of the factors in $$(p-r)(\dots)=p^3-r^3$$ must go down in degree. We have chosen the first one to do this. So the degree for $$q_2^3\; s_2^2$$, $$q_3^3\; s_3^2$$ is $$16$$. In the product $$p^3-r^2=(p-r)(\dots)$$ the degree goes down by three, so this happens in $$(p-r)$$, whose degree is thus $$13$$. We may even search for $$q_2,q_3$$, respectively $$s_2,s_3$$ which are mapped in each other by the conjugation in $$F$$, $$\omega\to\omega^2$$, which is also the complex conjugation after some choice of an embedding $$F\subset \mathbb C$$, e.g. via $$\omega=(-1+i\sqrt 3)/2$$. (So no such polynomial has rational roots.)

We may now extract $$p,r$$ from the first two equations above, and plug in in the third one. We obtain an equation only in the $$q_1,q_2,q_3;s_1,s_2,s_3$$ polynomials. This equation may be also obtained by observing that $$(p-r)+\omega(p-\omega r)+\omega^2(p-\omega^2 r)=0$$. So we would like to realize: $$B\; q_1^3s_1^2 = \omega (1-\omega)\; q_2^3s_2^2 + \omega^2(1-\omega^2)\; q_3^3s_3^2 \ .$$ Note that $$\omega (1-\omega)=-\omega^2(1-\omega^2)$$, so we may indeed get cancellations with monic polynomials on the R.H.S. above. I have to make a decision, so i am taking $$\deg q_1=3$$, $$\deg q_2=\deg q_3=4$$, $$\deg s_1=\deg s_2=\deg s_2=2$$.

It is time to remember that we may further norm two roots now, so $$q_1$$ is a good candidate to be taken in the form $$q_1=x(x-1)(x-b_4)\ .$$

I have to stop here, but i hope it is clear how the search goes on. Some simple sage program did not found an explicit solution, but i will come back here in the eventuality that i manage to find an explicit solution.

• hi dan_fulea, thanks for your answer, I wasn't aware of your answer and the bounty expiring, I will read your answer closely and come back to you asap... Nov 9, 2021 at 10:06
• Hi, (i) is this a typo in the exponent of $s$: "In the product $p^3-r^{\color{red}2}=(p-r)(\dots)$ the degree goes down by three"? (ii) why does go down by 3? Dec 14, 2021 at 9:35