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?

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About Edray Herber Goins, Ph.D.

Edray Herber Goins grew up in South Los Angeles, California. A product of the Los Angeles Unified (LAUSD) public school system, Dr. Goins attended the California Institute of Technology, where he majored in mathematics and physics, and earned his doctorate in mathematics from Stanford University. Dr. Goins is currently an Associate Professor of Mathematics at Purdue University in West Lafayette, Indiana. He works in the field of number theory, as it pertains to the intersection of representation theory and algebraic geometry.
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