## Monodromy Groups and Compositions of Belyi Maps

A few weeks ago, I considered how the monodromy groups of Shabat polynomials change under composition by considering several examples.  I would like to explain a general phenomenon by considering the composition of Belyi maps on the sphere.

Say that we have two Belyi maps, namely $\phi, \, \beta: \mathbb P^1(\mathbb C) \to \mathbb P^1(\mathbb C)$ such that the composition $\Phi = \beta \circ \phi$ is also a Belyi map. (For example, a sufficient condition here is that $\beta \bigl( \{ 0, \, 1, \, \infty \} \bigr) \subseteq \{ 0, \, 1, \, \infty \}$.) I am interested in computing the monodromy group of the compositon $\Phi$. To this end, I will show the following.

Proposition. Say that $\text{Mon}(\beta) \subseteq S_N$ and $\text{Mon}(\phi) \subseteq S_M$ are the monodromy groups of $\beta$ and $\phi$, respectively, as subgroups of the symmetric groups $S_N$ and $S_M$, respectively. Then $\text{Mon}(\Phi) \subseteq S_M \wr S_N$ is a subgroup of the wreath product $S_M \wr S_N := {S_M}^N \rtimes S_N$ of the symmetric groups.

## Monodromy Groups of Belyi Maps

First, let me review how to compute the monodromy group $\text{Mon}(\beta)$ of a Belyi map $\beta$ of degree $N = \text{deg}(\beta)$.

Let $y_0$ be a complex number in the “thrice punctured sphere” $Y = \mathbb P^1(\mathbb C) - \{ 0, \, 1, \, \infty \}$. We know that the fundamental group $\pi_1(Y, \, y_0)$, the group of closed loops $\gamma: [0,1] \to Y$ satisfying $\gamma(0) = \gamma(1) = y_0$ modulo equivalence, is the free group on two generators. We will choose these generators to be the closed loops $\gamma_0, \, \gamma_1: [0,1] \to Y$ explicitly given by $\gamma_0(t) = y_0 \, e^{2 \pi i t}$ and $\gamma_1(t) = 1 + (y_0 - 1) \, e^{2 \pi i t}$ as loops around 0 and 1, respectively.

Now the Belyi map $\beta: \mathbb P^1(\mathbb C) \to \mathbb P^1(\mathbb C)$ restricts to an $N$-fold cover $\beta: X \to Y$ whose domain is the punctured sphere $X = \mathbb P^1(\mathbb C) - \beta^{-1} \bigl( \{ 0, \, 1, \, \infty \} \bigr)$. Let $\beta^{-1}(y_0) = \{x_1, \, x_2, \, \dots, \, x_N \}$ be those complex numbers which form the preimages of $y_0$. For each preimage $x_p$, there are unique paths $\widetilde{\gamma}_0^{(p)}, \, \widetilde{\gamma}_1^{(p)}: [0,1] \to X$ such that

$\left. \begin{matrix} \beta \circ \widetilde{\gamma}_0^{(p)} = \gamma_0 & \quad & \widetilde{\gamma}_0^{(p)}(0) = x_p \\[15pt] \beta \circ \widetilde{\gamma}_1^{(p)} = \gamma_1 & \quad & \widetilde{\gamma}_1^{(p)}(0) = x_p \end{matrix} \right \} \qquad \text{for} \qquad p = 1, \, 2, \, \dots, \, N.$

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

$\left.\begin{matrix} \dfrac {d \widetilde{\gamma}_0^{(p)}}{dt} = \left( 2 \, \pi \, i \, \dfrac {\beta}{\beta'} \right) \circ \widetilde{\gamma}_0^{(p)} & \quad & \widetilde{\gamma}_0^{(p)}(0) = x_p \\[15pt] \dfrac {d \widetilde{\gamma}_1^{(p)}}{dt} = \left( 2 \, \pi \, i \, \dfrac {\beta - 1}{\beta'} \right) \circ \widetilde{\gamma}_1^{(p)} & \quad & \widetilde{\gamma}_1^{(p)}(0) = x_p \end{matrix} \right \} \qquad\text{for} \qquad p = 1, \, 2, \, \dots, \, N.$

Since $\beta \bigl( \widetilde{\gamma}_0^{(p)}(1) \bigr) = \gamma_0(1) = y_0$, we see that $\widetilde{\gamma}_0^{(p)}(1) = x_{p'}$ for some $p'$. In particular, there exist permutations $\sigma_0, \, \sigma_1 \in S_N$ such that $\widetilde{\gamma}_0^{(p)}(1) = x_{\sigma_0(p)}$ and $\widetilde{\gamma}_1^{(p)}(1) = x_{\sigma_1(p)}$ for all $p = 1, \, 2, \, \dots, \, N$. Hence we have a group homomorphism

$\beta_\ast: \qquad \begin{matrix} \pi_1(Y, \, y_0) & \quad & \to & \quad & S_N \\[15pt] \gamma \sim {\gamma_0}^m \ast {\gamma_1}^n & & \mapsto & & \sigma = {\sigma_0}^m \circ {\sigma_1}^n \end{matrix}$

The set $\text{Mon}(\beta)$, called the monodromy group of $\beta$, is that subgroup of $S_N$ which is the image of this group homomorphism.

## Example #1

Consider the Belyi map $\beta: \mathbb P^1(\mathbb C) \to \mathbb P^1(\mathbb C)$ given by $\beta(z) = \dfrac {(z^2 + 1)^2}{4 \, z^2}$ of degree $N = 4$. We will show that $\text{Mon}(\beta) \simeq Z_2 \times Z_2$ is the Klein Vierergruppe.

Set $Y = \mathbb P^1(\mathbb C) - \{ 0, \, 1, \, \infty \}$ and $X = \mathbb P^1(\mathbb C) - \{0, \, \pm 1, \, \pm i, \, \infty \}$, so that $\beta: X \to Y$ is a 4-fold cover of punctured spheres. Choose $y_0 \in Y$, and write $y_0 = e^{2 \pi i p_0} = 1 + e^{2 \pi i q_0}$ for some complex numbers $p_0$ and $q_0$. Then $\beta^{-1}(y_0) = \{x_1, \, x_2, \, x_3, \, x_4 \}$ in terms of the complex numbers

$\begin{matrix} x_1 & = & + \sqrt{y_0} + \sqrt{y_0 - 1} & = & + e^{ \pi i p_0} + e^{ \pi i q_0} & & \\[5pt] x_2 & = & - \sqrt{y_0} - \sqrt{y_0 - 1} & = & - e^{ \pi i p_0} - e^{ \pi i q_0} & = & - x_1 \\[5pt] x_3 & = & + \sqrt{y_0} - \sqrt{y_0 - 1} & = & + e^{ \pi i p_0} - e^{ \pi i q_0} & = & + 1 / x_1 \\[5pt] x_4 & = & - \sqrt{y_0} + \sqrt{y_0 - 1} & = & - e^{ \pi i p_0} + e^{ \pi i q_0} & = & - 1/x_1 \end{matrix}$

It is easy to check that the paths $\widetilde{\gamma}_0^{(p)}, \, \widetilde{\gamma}_1^{(p)}: [0,1] \to X$ satisfy the differential equations

$\left. \begin{matrix} \dfrac {d \widetilde{\gamma}_0^{(p)}}{dt} = \pi i \, \dfrac {\bigl( \widetilde{\gamma}_0^{(p)} \bigr)^3 + \widetilde{\gamma}_0^{(p)}}{\bigl( \widetilde{\gamma}_0^{(p)} \bigr)^2 - 1} & \quad & \widetilde{\gamma}_0^{(p)}(0) = x_p \\[15pt] \dfrac {d \widetilde{\gamma}_1^{(p)}}{dt} = \pi i \, \dfrac {\bigl( \widetilde{\gamma}_0^{(p)} \bigr)^3 - \widetilde{\gamma}_0^{(p)}}{\bigl( \widetilde{\gamma}_0^{(p)} \bigr)^2 + 1} & \quad & \widetilde{\gamma}_1^{(p)}(0) = x_p \end{matrix} \right \} \qquad \text{for} \qquad p = 1, \, 2, \, 3, \, 4;$

which have the exact solutions

$\begin{matrix} \widetilde{\gamma}_0^{(p)}(t) = \dfrac {x_p^2 + 1}{2 \, x_p} \, e^{ \pi i t} + \sqrt{\dfrac {(x_p^2 + 1)^2}{4 \, x_p^2} \, e^{2 \pi i t} - 1} & \quad \implies \quad & \widetilde{\gamma}_0^{(p)}(1) = - \dfrac {1}{x_p}, \\[15pt] \widetilde{\gamma}_1^{(p)}(t) = \dfrac {x_p^2 - 1}{2 \, x_p} \, e^{ \pi i t} + \sqrt{\dfrac {(x_p^2 - 1)^2}{4 \, x_p^2} \, e^{2 \pi i t} + 1} & \quad \implies \quad & \widetilde{\gamma}_1^{(p)}(1) = + \dfrac {1}{x_p}. \\[15pt] \end{matrix}$

In particular,

$\begin{matrix} p = 1: & \qquad & \widetilde{\gamma}_0^{(p)}(0) = x_1 & & \widetilde{\gamma}_0^{(p)}(1) = x_4; & \qquad & \widetilde{\gamma}_1^{(p)}(0) = x_1 & & \widetilde{\gamma}_0^{(p)}(1) = x_3 \\[15pt] p = 2: & \qquad & \widetilde{\gamma}_0^{(p)}(0) = x_2 & & \widetilde{\gamma}_0^{(p)}(1) = x_3; & \qquad & \widetilde{\gamma}_1^{(p)}(0) = x_2 & & \widetilde{\gamma}_0^{(p)}(1) = x_4 \\[15pt] p = 3: & \qquad & \widetilde{\gamma}_0^{(p)}(0) = x_3 & & \widetilde{\gamma}_0^{(p)}(1) = x_2; & \qquad & \widetilde{\gamma}_1^{(p)}(0) = x_3 & & \widetilde{\gamma}_0^{(p)}(1) = x_1 \\[15pt] p = 4: & \qquad & \widetilde{\gamma}_0^{(p)}(0) = x_4 & & \widetilde{\gamma}_0^{(p)}(1) = x_1; & \qquad & \widetilde{\gamma}_1^{(p)}(0) = x_4 & & \widetilde{\gamma}_0^{(p)}(1) = x_2 \end{matrix}$

Hence $\sigma_0 = (1 \ 4) \ (2 \ 3)$ while $\sigma_1 = (1 \ 3) \ (2 \ 4)$, so that the monodromy group is given by

$\text{Mon}(\beta) = \left \langle \sigma_0, \, \sigma_1 \, \bigl| \, {\sigma_0}^2 = {\sigma_1}^2 = (\sigma_0 \circ \sigma_1)^2 = 1 \right \rangle \simeq Z_2 \times Z_2.$

## Monodromy of Compositions of Belyi Maps

Now let’s consider two Belyi maps $\phi, \, \beta: \mathbb P^1(\mathbb C) \to \mathbb P^1(\mathbb C)$ such that the composition $\Phi = \beta \circ \phi$ is also a Belyi map. Denote $N = \text{deg}(\beta)$ and $M = \text{deg}(\phi)$ so that $M \, N = \text{deg}(\Phi)$. Continuing notation as above, set

• $Y = \mathbb P^1(\mathbb C) - \{ 0, \, 1, \, \infty \}$ and choose $y_0 \in Y$;
• $X = \mathbb P^1(\mathbb C) - \beta^{-1} \bigl( \{ 0, \, 1, \, \infty \} \bigr)$ and $\beta^{-1}(y_0) = \{ \dots, \, x_p, \, \dots \}$ for $p = 1, \, 2, \, \dots, \, N$; and
• $Z = \mathbb P^1(\mathbb C) - \Phi^{-1} \bigl( \{ 0, \, 1, \, \infty \} \bigr)$ and $\Phi^{-1}(y_0) = \{ \dots, \, z_{pq}, \, \dots \}$ for $q = 1, \, 2, \, \dots, \, M$ such that $\phi(z_{pq}) = x_p$.

Hence $\Phi: Z \to X \to Y$ is an $M \, N$-fold cover.

As before, we will choose the generators of the fundamental group $\pi_1(Y, \, y_0)$ to be the closed loops $\gamma_0, \, \gamma_1: [0,1] \to Y$ explicitly given by $\gamma_0(t) = y_0 \, e^{2 \pi i t}$ and $\gamma_1(t) = 1 + (y_0 - 1) \, e^{2 \pi i t}$. For each preimage $z_{pq}$, there are unique paths $\widetilde{\gamma}_0^{(pq)}, \, \widetilde{\gamma}_1^{(pq)}: [0,1] \to X$ such that

$\left. \begin{matrix} \bigl( \beta \circ \phi \bigr) \circ \widetilde{\gamma}_0^{(pq)} = \gamma_0 & \quad & \phi \circ \widetilde{\gamma}_0^{(pq)} = \widetilde{\gamma}_0^{(p)} & \quad & \widetilde{\gamma}_0^{(pq)}(0) = z_{pq} \\[15pt] \bigl( \beta \circ \phi \bigr) \circ \widetilde{\gamma}_1^{(pq)} = \gamma_1 & \quad & \phi \circ \widetilde{\gamma}_1^{(pq)} = \widetilde{\gamma}_1^{(p)} & \quad & \widetilde{\gamma}_1^{(pq)}(0) = z_{pq} \end{matrix} \right \} \qquad \text{for} \qquad \left \{ \begin{matrix} p = 1, \, 2, \, \dots, \, N, \\[15pt] q = 1, \, 2, \, \dots,\, M. \end{matrix} \right.$

As we know how to compute $\text{Mon}(\beta)$, we already know there exist $\sigma_0, \, \sigma_1 \in S_N$ such that

$\left. \begin{matrix} \phi \bigl( \widetilde{\gamma}_0^{(pq)}(0) \bigr) = \widetilde{\gamma}_0^{(p)}(0) = x_{\sigma_0(p)} = \phi \bigl( z_{\sigma_0(p), q'} \bigr) \\[15pt] \phi \bigl( \widetilde{\gamma}_1^{(pq)}(0) \bigr) = \widetilde{\gamma}_1^{(p)}(0) = x_{\sigma_1(p)} = \phi \bigl( z_{\sigma_1(p), q''} \bigr) \end{matrix} \right \} \qquad \text{for} \qquad \left \{ \begin{matrix} p = 1, \, 2, \, \dots, \, N, \\[15pt] q, \, q', \, q'' = 1, \, 2, \, \dots,\, M. \end{matrix} \right.$

This means that we can find a $N$-tuples $\tau_0 = \left (\sigma_0^{(1)}, \, \sigma_0^{(2)}, \, \dots, \, \sigma_0^{(N)} \right)$ and $\tau_1 = \left (\sigma_1^{(1)}, \, \sigma_1^{(2)}, \, \dots, \, \sigma_1^{(N)} \right)$ in ${S_M}^N := S_M \times S_M \times \cdots \times S_M$ such that

$\left. \begin{matrix} \widetilde{\gamma}_0^{(pq)}(1) = z_{\sigma_0(p), q'} & \quad \text{where} \quad & q' = \sigma_0^{(p)}(q) \\[15pt] \widetilde{\gamma}_1^{(pq)}(1) = z_{\sigma_1(p), q''} & \quad \text{where} \quad & q'' = \sigma_1^{(p)}(q)\end{matrix} \right \} \qquad \text{for} \qquad \left \{ \begin{matrix} p = 1, \, 2, \, \dots, \, N, \\[15pt] q, \, q', \, q'' = 1, \, 2, \, \dots,\, M. \end{matrix} \right.$

Hence we have a group homomorphism

$\Phi_\ast: \qquad \begin{matrix} \pi_1(Y, \, y_0) & \quad & \to & \quad & S_M \wr S_N := {S_M}^N \rtimes S_N \\[15pt] \gamma_0 & & \mapsto & & \left ( \bigl(\sigma_0^{(1)}, \, \sigma_0^{(2)}, \, \dots, \, \sigma_0^{(N)} \bigr), \ \sigma_0 \right) \\[15pt] \gamma_1 & & \mapsto & & \left ( \bigl(\sigma_1^{(1)}, \, \sigma_1^{(2)}, \, \dots, \, \sigma_1^{(N)} \bigr), \ \sigma_1 \right) \end{matrix}$

The image $\text{Mon}(\Phi)$ of this group homomorphism is the monodromy group of $\Phi = \beta \circ \phi$. It is clear that there exists a subgroup $G \subseteq \text{Mon}(\Phi)$ such that we have a commutative diagram in the form

$\begin{matrix} 1 & \longrightarrow & {S_M}^N & \longrightarrow & S_M \wr S_N & \longrightarrow & S_N & \longrightarrow & 1 \\[15pt] & & \uparrow & & \uparrow & & \uparrow & & \\[15pt] 1 & \longrightarrow & G & \longrightarrow & \text{Mon}(\Phi) & \longrightarrow & \text{Mon}(\beta) & \longrightarrow & 1 \\[15pt] & & & & \left ( \bigl(\sigma^{(1)}, \, \sigma^{(2)}, \, \dots, \, \sigma^{(N)} \bigr), \ \sigma \right) & \mapsto & \sigma & & \end{matrix}$

## Example #2

For any positive integer $M$, consider the Belyi map $\Phi: \mathbb P^1(\mathbb C) \to \mathbb P^1(\mathbb C)$ given by $\Phi(z) = \dfrac {(z^M + 1)^2}{4 \, z^M} = \beta \bigl( \phi(z) \bigr)$ which is the composition of the two Belyi maps $\beta(z) = \dfrac {(z+1)^2}{4 \, z}$ and $\phi(z) = z^M$. It is easy to check that $\text{Mon}(\beta) \simeq Z_2$ and $\text{Mon}(\phi) \simeq Z_M$ are both cyclic groups, while $\text{Mon}(\Phi) \simeq D_M := Z_M \rtimes Z_2$ is a dihedral group. In particular, $\text{Mon}(\Phi)$ is a proper subgroup of the wreath product $\text{Mon}(\phi) \wr \text{Mon}(\beta) := (Z_M \times Z_M) \rtimes S_2$, also known as the generalized symmetric group.

## Example #3

The Belyi map $\Phi(z) = - \dfrac {4}{531441} \, (z - 1) \, z^3 \, \bigl( 2 \, z^2 + 3 \, z + 9 \bigr)^3 \, \bigl( 8 \, z^4 + 28 \, z^3 + 126 \, z^2 + 189 \, z + 378 \bigr) = \beta \bigl( \phi(z) \bigr)$ is the composition of the Belyi maps $\beta(z) = 4 \, z \, (1-z)$ and $\phi(z) = - \dfrac {1}{729} \, (z - 1) \, (2 \, z^2 + 3 \, z + 9)^3$. One checks that $\text{Mon}(\Phi) \simeq A_7 \wr Z_2$ is the wreath product of $\text{Mon}(\phi) \simeq A_7$ and $\text{Mon}(\beta) \simeq Z_2$.

(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!)

## Example #4

The Belyi map $\Phi(z) = 4 \, z^2 \, ( 1 - 2 \, z )^3 \, \bigl(1 + 3 \, z \bigr)^2 \, \bigl(15 - 10 \, z - 60 \, z^2 + 72 \, z^3 \bigr)$ is the composition of the Belyi maps $\beta(z) = 4 \, z \, (1-z)$ and $\phi(z) = ( 1- 2 \, z )^3 \, (1 + 3 \, z )^2$. It is easy to check that $\text{Mon}(\phi) \simeq S_5$ and $\text{Mon}(\beta) \simeq Z_2$. Curiously, the monodromy group $\text{Mon}(\Phi) \subseteq S_5 \wr S_2$ is a proper subgroup because it has order $\bigl| \text{Mon}(\Phi) \bigr| = 14 \, 400 = |S_5|^2$.  So what its the group $\text{Mon}(\Phi)$?

Question. Say that $\Phi = \beta \circ \phi$ is a Belyi map which is the composition of $\beta(z) = 4 \, z \, (1-z)$ and another Belyi map $\phi$ of degree $M = \text{deg}(\phi)$. Since $\text{Mon}(\beta) \simeq S_2$, we know that $\text{Mon}(\Phi) \subseteq (S_M \times S_M) \rtimes S_2$ is a subgroup of the wreath product of $S_M$ and $S_2$. Is it true that $\text{Mon}(\Phi) \subseteq \text{Mon}(\phi) \wr S_2$?