System of Equations for Computing Shabat Polynomials

A few weeks ago, Dong Quan Ngoc Nguyen (University of Notre Dame) came to visit here at Purdue. We spoke a little about the computer package Bertini (created by Daniel Bates, Jonathan Hauenstein, Andrew Sommese and Charles Wampler) and whether the homotopy continuation method can be used to compute Belyi maps and Shabat Polynomials. I’ve been working on setting up a system of polynomial equations whose solutions give the coefficients of the Belyi maps, so it really comes down to actually finding the solutions to these equations. The hope is that a polynomial homotopy continuation method will be much more efficient than say, using Groebner bases, to find all solutions!

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

Belyi Maps and Dessins d’Enfants

Say that we have a connected planar bipartite graph $\Gamma = \bigl( V, \, E \bigr)$ . The vertices $V = B \cup W$ consist of “black” vertices $B$ as well as “white” vertices $W$ , and the “midpoints” of the faces are $F$ . I am interested in the following

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

Its only critical values are $\{ 0, \, 1, \, \infty \} \subseteq \mathbb P^1(\mathbb C)$ .

$B = \beta^{-1}(0)$ are the “black” vertices.

$W = \beta^{-1}(1)$ are the “white” vertices.

$F = \beta^{-1}(\infty)$ are the “midpoints” of the faces.

$E = \beta^{-1} \bigl( [0,1] \bigr)$ as the edges is the inverse image of the interval from $0$ to $1$ .

We say $\beta: \mathbb P^1(\mathbb C) \to \mathbb P^1(\mathbb C)$ is a Belyi map, and $\Gamma$ 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 $b_i$ as the number of edges adjacent to each “black” vertex as indexed by $i \in I$ , $w_j$ as the number of edges adjacent to each “white” vertex as indexed by $j \in J$ , and $f_k$ as the number of edges adjacent to each face as indexed by $k \in K$ . Define the degree sequence as a valency list:

$\mathcal D = \biggl \{ \left \{ b_i \, \bigl| \, i \in I \right \}, \ \left \{ w_j \, \bigl| \, j \in J \right \}, \ \left \{ f_k \, \bigl| \, k \in K \right \} \biggr \};$

and denote the positive integer $n = |I| + |J| + |K| - 3$ . Define the $n$ polynomials $y_1, \, y_2, \, \dots, \, y_n$ in $\mathbb Z[x_1, x_2, \dots, x_n]$ in $n$ variables $x_1, \, x_2, \, \dots, \, x_n$ by the $3 \times 3$ determinants

$y_e = \left| \begin{matrix} B_0 & W_0 & F_0 \\[5pt] B_e & W_e & F_e \\[5pt] B_{n+1} & W_{n+1} & F_{n+1} \end{matrix} \right|$

for $e = 1, \, 2, \, \dots, \, n$ , each expressed in terms of the coefficients of the polynomials

$\begin{matrix} \displaystyle z^{b_{|I|}} \prod_{i = 1}^{|I|-1} \biggl( z - x_i \biggr)^{b_i} & = &\displaystyle \sum_{e = 0}^{n+1} B_e \, z^e \\[15pt] \displaystyle \bigl( z - 1 \bigr)^{w_{|J|}} \prod_{j = 1}^{|J|-1} \biggl( z - x_{|I|+j-1} \biggr)^{w_j} & = &\displaystyle \sum_{e = 0}^{n+1} W_e \, z^e \\[15pt] \displaystyle \prod_{k = 1}^{|K|-1} \biggl( z - x_{|I|+|J|+k-2} \biggr)^{f_k} & = &\displaystyle \sum_{e = 0}^{n+1} F_e \, z^e \end{matrix}$

each as elements of $\mathbb Z[x_1, x_2, \dots, x_n][z]$ . The following is relatively trivial to prove.

Proposition 1. Continue notation as above.

The Euler Characteristic and the Degree Sum Formula together imply
$n +1 = \displaystyle \sum_{i \in I} b_i = \sum_{j \in J} w_j = \sum_{k \in K} f_k = |I| + |J| + |K| - 2$ .

The following algebraic variety has dimension 0: $X = \left \{ P = (x_1, \, x_2, \, \dots, \, x_n) \in \mathbb A^n \ \biggl| \ y_1(P) = y_2(P) = \cdots = y_n(P) = 0 \right \}.$ In particular, Bezout’s theorem implies $\#X(\mathbb C) \leq \deg(y_1) \cdot \deg(y_2) \cdots \deg(y_n)$ .

Every point $P = (x_1, \, x_2, \, \dots, \, x_n)$ in $X(\mathbb C)$ gives rise to a Belyi map $\beta: \mathbb P^1(\mathbb C) \to \mathbb P^1(\mathbb C)$ such that $\Gamma$ is its Dessin d’Enfant, and we have the explicit points $\begin{matrix} B & = & \beta^{-1}(0) & = & \biggl \{ (0:1), \, (x_1:1), \, \dots, \, (x_{|I|-1}:1) \biggr \}, \\[15pt] W & = & \beta^{-1}(1) & = & \biggl \{ (1:1), \, (x_{|I|}:1), \, \dots, \, (x_{|I|+|J|-2}:1) \biggr \}, \\[15pt] F & = & \beta^{-1}(\infty) & =& \biggl \{ (1:0), \, (x_{|I|+|J|-1}:1), \, \dots, \, (x_{|I|+|J|+|K|-3}:1) \biggr \} \end{matrix}$ Rather explicitly, $\beta(z) = \dfrac {\left| \begin{matrix} F_0 & W_0 \\[5pt] F_{n+1} & W_{n+1} \end{matrix} \right|}{\left| \begin{matrix} B_0 & W_0 \\[5pt] B_{n+1} & W_{n+1} \end{matrix} \right|} \cdot \dfrac {\displaystyle z^{b_{|I|}} \prod_{i = 1}^{|I|-1} \biggl( z - x_i \biggr)^{b_i}}{\displaystyle \prod_{k = 1}^{|K|-1} \biggl( z - x_{|I|+|J|+k-2} \biggr)^{f_k}}.$

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

$\begin{aligned} \displaystyle \sum_{e = 1}^n y_e \, z^e & = \left| \begin{matrix} B_0 & W_0 & F_0 \\[15pt] \displaystyle \sum_{e = 0}^{n+1} B_e \, z^e & \displaystyle \sum_{e = 0}^{n+1} W_e \, z^e & \displaystyle \sum_{e = 0}^{n+1} F_e \, z^e \\[15pt] B_{n+1} & W_{n+1} & F_{n+1} \end{matrix} \right| \\[10pt] & = \left| \begin{matrix} F_0 & W_0 \\[5pt] F_{n+1} & W_{n+1} \end{matrix} \right| \cdot z^{b_{|I|}} \prod_{i = 1}^{|I|-1} \biggl( z - x_i \biggr)^{b_i} \\[10pt] & \quad - \left| \begin{matrix} F_0 & B_0 \\[5pt] F_{n+1} & B_{n+1} \end{matrix} \right| \cdot \bigl( z - 1 \bigr)^{w_{|J|}} \prod_{j = 1}^{|J|-1} \biggl( z - x_{|I|+j-1} \biggr)^{w_j} \\[10pt] & \qquad - \left| \begin{matrix} B_0 & W_0 \\[5pt] B_{n+1} & W_{n+1} \end{matrix} \right| \cdot \prod_{k = 1}^{|K|-1} \biggl( z - x_{|I|+|J|+k-2} \biggr)^{f_k}. \end{aligned}$

Finding All Points $P \in X(\mathbb C)$ using Bertini

Given a degree sequence $\mathcal D$ , one can list the polynomials $y_1, \, y_2, \, \dots, \, y_n$ in $\mathbb Z[x_1, x_2, \dots, x_n]$ very quickly. I would like to find the points $P$ in $X(\mathbb C)$ using the following

Algorithm. Denote $F$ as that extension of $\mathbb Q$ which is the field generated by the coordinates of the points $P \in X(\mathbb C)$ .

Using both the Homotopy Continuation and Newton’s Method, numerically find all of the points $P$ in $X(\mathbb C)$ .

Using LLL, find a monic polynomial $\phi(z) \in \mathbb Z[z]$ such that $F \simeq \mathbb Q[z] / \bigl( \phi(z) \bigr)$ . Note that the roots $z_P$ of $\phi(z)$ are in one-to-one correspondence with the points $P = (x_1, \, x_2, \, \dots, \, x_n)$ in $X(\mathbb C)$ .

Also using LLL, find polynomials $\phi_1, \, \phi_2, \, \dots, \, \phi_n$ in $\mathbb Q[z]$ such that $y_e \bigl( \phi_1(z), \, \phi_2(z), \, \dots, \, \phi_n(z) \bigr) \equiv 0 \quad \text{mod} \ \phi(z)$ for $e = 1, \, 2, \, \dots, \, n$ . Then $P = \bigl( \phi_1(z_P), \, \phi_2(z_P), \, \dots, \, \phi_n(z_P) \bigr)$ for each root $z_P$ of $\phi(z)$ .

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.

Shabat Polynomials and Trees

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.

From Wikipedia

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 $d+1$ vertices and hence $d$ edges, its degree sequence is in the form

$\mathcal D = \left \{ \{ m_0, \, m_1, \, \dots, \, m_d \},\ \{ 2, \, 2, \, \dots, \, 2 \}, \ \{ 2 \, d \} \right \}$

where we must have $m_0 + m_1 + m_2 + \cdots + m_d = 2 \, d$ . Here $|I| = d+1$ , $|J| = d$ , and $|K|= 1$ , so that $n = 2 \, d - 1$ 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 $n = 2 \, d - 1$ nontrivial complex numbers $x_1, \, x_2, \, \dots, \, x_n$ such that
$z^{m_d} \, \bigl( z-1 \bigr)^{m_0} \, \bigl( z - x_1 \bigr)^{m_1} \, \cdots \, \bigl( z - x_{d-1} \bigr)^{m_{d-1}} + x_d^2 - \left( z^d + \sum_{j=0}^{d-1} x_{j+d} \, z^j \right)^2 = \displaystyle \sum_{e = 1}^{n} y_e \, z^e$ is identically zero as a polynomial in $z$ , then the tree of interest is the Dessin d’Enfant of the Belyi map $\beta(z) = - \dfrac {1}{x_d^2} \cdot z^{m_d} \, \bigl( z-1 \bigr)^{m_0} \, \bigl( z - x_1 \bigr)^{m_1} \, \cdots \, \bigl( z - x_{d-1} \bigr)^{m_{d-1}}$ .

This Belyi map is not just a rational function of degree $n + 1 = 2 \, d$ ; it is a polynomial. We say that such a Belyi map is a Shabat Polynomial. Note that there are $n$ equations $y_1 = y_2 = \cdots = y_n = 0$ in $n$ variables $x_1, \, x_2, \, \dots, \, x_n$ , so again we can construct an algebraic variety $X$ of dimension 0 and ask for all points $P \in X(\mathbb C)$ .

Worked Examples of Shabat Polynomials

Given a collection of $d+1$ positive integers $m_0, \, m_1, \, \dots, \, m_d$ such that $m_0 + m_1 + m_2 + \cdots + m_d = 2 \, d$ , one can list the polynomials $y_1, \, y_2, \, \dots, \, y_n$ in $\mathbb Z[x_1, x_2, \dots, x_n]$ very quickly. Remember that the goal is to find the $n$ variables $x_1, \, x_2, \, \dots, \, x_n$ such that we have the $n$ equations $y_1 = y_2 = \cdots = y_n = 0$ .

Example 1

Say that $d = 1$ so that $n = 1$ . We have a degree sequence $\mathcal D = \bigl \{ \{ m_0, \, m_1\}, \ \{ 2 \}, \ \{ 2 \} \bigr \}$ such that $m_0 + m_1 = 2$ , so we must have $m_0 = m_1 = 1$ . We have the polynomial $z^{m_1} \, (z - 1)^{m_0} + x_1^2 - (z + x_1)^2 = y_1 \, z$ where $y_1 = - 1 - 2 \, x_1$ . Setting $y_1 = 0$ , we find $x_1 = -1/2$ . Hence the Shabat polynomial must be

$\beta(z) = - \dfrac {1}{x_1^2} \cdot z^{m_1} \, ( z-1)^{m_0} = 4 \, z \, (1 - z).$

Example 2

Say that $d = 2$ so that $n = 3$ . We have a degree sequence $\mathcal D = \bigl \{ \{ m_0, \, m_1, \, m_2 \}, \ \{ 2, \, 2 \}, \ \{ 4 \} \bigr \}$ such that $m_0 + m_1 + m_2 = 4$ , so up to permutation we must have $m_0 = m_1 = 1$ and $m_2 = 2$ . We have the polynomial

$z^{m_2} \, (z - 1)^{m_0} \, (z - x_1)^{m_1} + x_2^2 - (z^2 + x_3 \, z + x_2)^2 = y_1 \, z + y_2 \, z^2 + y_3 \, z^3$

where

$\begin{matrix} y_1 & = & - 2 \, x_2 \, x_3 \\[15pt] y_2 & = & - 2 \, x_2 - x_3^2 \\[15pt] y_3 & = & 1 - x_1 - 2 \, x_3 \end{matrix}$

Setting $y_1 = y_2 = y_3 = 0$ , we find two solutions: either $(x_1, \, x_2, \, x_3) = (1, \, 0, \, -1)$ or $(x_1, \, x_2, \, x_3) = ( -1, \, -1/2, \, 0)$ . We throw out the first solution because we will need to divide by $x_2$ , so from the second solution we find the Shabat polynomial

$\beta(z) = - \dfrac {1}{x_2^2} \cdot z^{m_2} \, ( z-1)^{m_0} \, ( z - x_1)^{m_1} = 4 \, z^2 \, (1 - z^2).$

Example 3

Consider the star graph $K_{1,d}$ on $d$ edges. It has degree sequence

$\mathcal D = \left \{ \{1, \, \dots, \, 1, \, d \},\ \{ 2, \, 2, \, \dots, \, 2 \}, \ \{ 2 \, d \} \right \}.$

We have already seen this when $d = 1, \, 2$ as in the two previous examples.

From Wikipedia

We are interested in an identity of the form $z^d \, ( z-1)^{1} \, ( z - x_1)^{1} \, \cdots \, \bigl( z - x_{d-1} \bigr)^{1} + x_d^2 - \left( z^d + \sum_{j=0}^{d-1} x_{j+d} \, z^j \right)^2 = 0$ for all $z$ . Let $\zeta = e^{2 \pi \sqrt{-1}/d}$ denote a primitive $d$ th root of unity. If we choose

$x_k = \left \{ \begin{matrix} \zeta^k & k = 1, \, 2, \, \dots, \, (d-1), \\[15pt] -1/2 &k = d, \\[15pt] 0 &k = (d+1), \, \dots, \, (2 \, d - 1); \end{matrix} \right.$

then we recover the identity $z^d \, (z^d - 1) + \left( \dfrac {1}{2} \right)^2 - \left( z^d - \dfrac 12 \right)^2 = 0$ . Hence the corresponding Shabat polynomial is $\beta(z) = 4 \, z^d \, (1 - z^d)$ .

Example 4

Now consider any tree with $d = 5$ edges having the degree sequence

$\mathcal D = \left \{ \{ 2, \, 2, \, 1, \, 1, \, 1, \, 3 \},\ 2, \, 2, \, 2, \, 2, \, 2 \}, \ \{ 10 \} \right \}.$

Since $n = 2 \, d - 1 = 9$ , we have the polynomial

$z^3 \, ( z-1)^2 \, ( z - x_1)^2 \, ( z - x_2)^1 \, ( z - x_3)^1 \, ( z - x_4)^1 + x_5^2 - \left( z^5 + x_9 \, z^4 + x_8 \, z^3 + x_7 \, z^2 + x_6 \, z + x_5 \right)^2 = y_1 \, z + y_2 \, z^2 + y_3 \, z^3 + y_4 \, z^4 + y_5 \, z^5 + y_6 \, z^6 + y_7 \, z^7 + y_8 \, z^8 + y_9 \, z^9$

in terms of

$\begin{matrix} y_1 & = & -2 \, x_5 \, x_6 \\[5pt] y_2 & = & -x_6^2 - 2 \, x_5 \, x_7 \\[10pt] y_3 & = & -x_1^2 \, x_2 \, x_3 \, x_4 - 2 \, x_6 \, x_7 - 2 \, x_5 \, x_8 \\[10pt] y_4 & = & x_1^2 \, x_2 \, x_3 + x_1^2 \, x_2 \, x_4 + x_1^2\, x_3 \, x_4 + 2 \, x_1 \, x_2 \, x_3 \, x_4 + 2 \, x_1^2 \, x_2 \, x_3 \, x_4 \\[5pt] & & \quad - x_7^2 - 2 \, x_6 \, x_8 - 2 \, x_5 \, x_9 \\[10pt] y_5 & = & - x_1^2 \, x_2 - x_1^2 \, x_3 - 2 \, x_1\, x_2 \, x_3 - 2 \, x_1^2 \, x_2 \, x_3 - x_1^2\, x_4 -2 \, x_1\, x_2 \, x_4 \\[5pt] & & \quad - 2 \, x_1^2 \, x_2 \, x_4 - 2 \, x_1 \, x_3 \, x_4 - 2 \, x_1^2 \, x_3 \, x_4 - x_2 \, x_3 \, x_4 - 4 \, x_1\, x_2 \, x_3 \, x_4 \\[5pt] & & \qquad - x_1^2 \, x_2 \, x_3 \, x_4 - 2 \, x_5 -2 \, x_7 \, x_8 - 2 \, x_6 \, x_9 \\[10pt] y_6 & = & 2 \, x_1^2 \, x_2 + x_1^2 \, x_2 \, x_3 + 2 \, x_1^2 \, x_3 + x_1^2 \, x_2 \, x_4 + x_1^2 \, x_3 \, x_4 + 2 \, x_1^2 \, x_4 + x_1^2 + 2 \, x_1 \, x_2 \\[5pt] & & \quad + 4 \, x_1 \, x_2 \, x_3 + 2 \, x_1 \, x_3 + 4 \, x_1 \, x_2 \, x_4 + 2 \, x_1 \, x_2 \, x_3 \, x_4 + 4 \, x_1 \, x_3 \, x_4 + 2 \, x_1 \, x_4 \\[5pt] & & \qquad - x_8^2 + x_2 \, x_3 + x_2 \, x_4 + 2 \, x_2 \, x_3 \, x_4 + x_3 \, x_4 - 2 \, x_6 - 2 \, x_7 \, x_9 \\[10pt] y_7 & = & -x_1^2 \, x_2 - x_1^2 \, x_3 - x_1^2 \, x_4 - 2 \, x_1^2 - 4 x_1 \, x_2 - 2 \, x_1 \, x_2 \, x_3 - 4 \, x_1 \, x_3 \\[5pt] & & \quad - 2 \, x_1 \, x_2 \, x_4 - 2 \, x_1 \, x_3 \, x_4 - 4 \, x_1 \, x_4 - 2 \, x_1 - x_2 - 2 \, x_2 \, x_3 - x_3 \\[5pt] & & \qquad - 2 \, x_2 \, x_4 - x_2 \, x_3 \, x_4 - 2 \, x_3 \, x_4 - x_4 - 2 \, x_7 - 2 \, x_8 \, x_9 \\[10pt] y_8 & = & x_1^2 + 2 \, x_1 \, x_2 + 2 \, x_1 \, x_3 + 2 x_1\, x_4 + 4 \, x_1 - x_9^2 + 2 \, x_2 + x_2 \, x_3 + 2 \, x_3 \\[5pt] & & \quad + x_2 \, x_4 + x_3 \, x_4 + 2 \, x_4 - 2 \, x_8 + 1 \\[10pt] y_9 & = & -2 - 2 \, x_1 - x_2 - x_3 - x_4 - 2 \, x_9 \end{matrix}$

Question. Can Bertini find all solutions to $y_1 = y_2 = \cdots = y_9 = 0$ ?

I only know that one solution corresponds to

$x_1 = \dfrac {5}{3}$ , $x_5 = - \dfrac {2}{9}$ , $x_6 = 0$ , $x_7 = 0$ , $x_8 = \dfrac {25}{9}$ , and $x_9 = - \dfrac {10}{3}$ . In particular, one set of identities which works is as follows:

$z^3 \, \bigl( z-1 \bigr)^2 \left( z - \dfrac {5}{3} \right)^2 \left( z^3 - \dfrac {4}{3} \, z^2 - \dfrac {8}{9} \, z - \dfrac {4}{9} \right)^1 + \left( \dfrac {2}{9} \right)^2 - \left( z^5 - \dfrac {10}{3} \, z^4 + \dfrac {25}{9} \, z^3 - \dfrac {2}{9} \right)^2 = 0.$

Using the substitution $z \mapsto (4-z)/3$ we find a corresponding Shabat polynomial

$\beta(z) = -\dfrac { (z^2 - 1)^2 \, (z^3 - 8 \, z^2 + 8 \, z + 44) \, (z-4)^3}{2916}.$

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

$\mathcal D = \left \{ \{ 1, \, 1, \, 1, \, 2, \, 2, \, 3 \},\ \{ 2, \, 2, \, 2, \, 2, \, 2 \}, \ \{ 10 \} \right \},$

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

$z^3 \, \bigl( z-1 \bigr)^1 \left( z^2 + \dfrac {5}{3} \, z + \dfrac {40}{9} \right)^1 \left( z^2 + \dfrac {4}{3} \, z + \dfrac {8}{3} \right)^2 + \left( \dfrac {32}{9} \right)^2 - \left( z^5 + \dfrac {5}{3} \, z^4 + \dfrac {40}{9} \, z^3 - \dfrac {32}{9} \right)^2 = 0.$

Using the substitution $z \mapsto (z-2)/3$ , we find that a completely different polynomial:

$\beta(z) = -\dfrac {(z^2 + 20)^2 \, (z-5) \, (z^2 + z + 34) \, (z-2)^3}{746496}.$

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