## Monodromy Groups and Compositions of Shabat Polynomials

Last week, the great Naiomi Cameron visited me for a few days to discuss some new directions about Shabat polynomials.  I’ve been horrible about posting on this blog, so now that I’ve been motivated to work on Shabat polynomials again, I figured it’s time for me to write!

As considered in the 1994 paper by Georgii Borisovich Shabat and Alexander Zvonkin entitled Plane trees and algebraic numbers, the rational function

$\displaystyle \beta(z) = - \dfrac {4}{531441} \, (z - 1) \, z^3 \, \bigl( 2 \, z^2 + 3 \, z + 9 \bigr)^3 \, \bigl( 8 \, z^4 + 28 \, z^3 + 126 \, z^2 + 189 \, z + 378 \bigr)$

is a Shabat polynomial which happens to be the composition $\beta = \phi \circ \Phi$ of two other Shabat polynomials.  The first has monodromy group $G_\phi = Z_2$ as a cyclic group, while the second has monodromy group $G_\phi = A_7$ as an alternating group.  The monodromy group of the composition has order $|G_\beta| = 12 \, 700 \, 800 = |G_\phi| \cdot |G_\Phi|^2$. Do we have $G_\beta = Z_2 \ltimes \bigl( A_7 \times A_7 \bigr)$ as the wreath product of $G_\Phi$ by $G_\phi$?

## Belyi Maps, Degree Sequences, and Passports

Let me begin by establishing some notation and making some remarks.  A lot of this to begin with will be background information, so I apologize to those who already know this stuff pretty well.

Let $X$ be a compact connected Riemann surface. Remember that the only case we care about is when $X$ is the Riemann sphere. Indeed, we have a bijection which comes from stereographic projection:

$\displaystyle \begin{matrix} X = \mathbb P^1(\mathbb C) = \mathbb C \cup \{ \infty \} & \longrightarrow & S^2(\mathbb R) = \left \{ (u,v,w) \in \mathbb A^3(\mathbb R) \, \biggl| \, u^2 + v^2 + w^2 = 1 \right \} \\[10pt] z = x + i \, y = \dfrac {u + i \, v}{1 - w} & \mapsto &(u,v,w) = \left( \dfrac {2 \, x}{x^2 + y^2 + 1}, \ \dfrac {2 \, y}{x^2 + y^2 + 1}, \ \dfrac {x^2 + y^2 - 1}{x^2 + y^2 + 1} \right) \end{matrix}$

A Belyi map $\beta: X \to \mathbb P^1(\mathbb C)$ is a morphism which is branched only above three points $\{ w_0, \, w_1, \, w_\infty \} \subseteq \mathbb P^1(\mathbb C)$. When $X = \mathbb P^1(\mathbb C)$ is a Riemann sphere, a Belyi map must be the ratio of two polynomials having complex coefficients. Rather implicitly:

$\displaystyle \dfrac {w_1 - w_\infty}{w_1 - w_0} \, \dfrac {\beta(z) - w_0}{\beta(z) - w_\infty} = \beta_0 \cdot \dfrac {\prod_{i = 1}^n \bigl(z - z_0^{(i)} \bigr)^{b_i}}{\prod_{k = 1}^p \bigl(z - z_\infty^{(k)} \bigr)^{f_k}} = 1 + \beta_1 \cdot \dfrac {\prod_{j = 1}^m \bigl(z - z_1^{(j)} \bigr)^{w_j}}{\prod_{k = 1}^p \bigl(z - z_\infty^{(k)} \bigr)^{f_k}}.$

Once we let $e_i$ be the number of times $b_i$ ($w_i$, or $f_i$, respectively) appears in the exponents $\{ b_1, \, \dots, \, b_n \}$ ($\{ w_1, \, \dots, \, w_m \}$, or $\{ f_1, \, \dots, \, f_p \}$, respectively), we define the degree sequence $\mathcal D$ of a Belyi map as a multiset of multisets:

$\displaystyle \mathcal D = \biggl \{ \bigl \{ b_1, \, \dots, \, b_n \bigr \}, \, \bigl \{ w_1, \, \dots, \, w_m \bigr \}, \, \bigl \{ f_1, \, \dots, \, f_p \bigr \} \biggr \}$

and the passport $\Pi$ of a Belyi map as shorthand notation for the degree sequence:

$\displaystyle \Pi = \left[ \prod_{i} {b_i}^{e_i}; \ \prod_j {w_j}^{e_j}; \ \prod_k {f_k}^{e_k} \right]$

where

$\displaystyle \deg(\beta) = \sum_{i = 1}^n b_i = \sum_{i} e_i \, b_i = \sum_{j = 1}^m w_j = \sum_{j} e_j \, w_j = \sum_{k = 1}^p f_k = \sum_{k} e_k \, f_k.$

These quantities give combinatorial information about the critical points of $\beta$, namely, the three sets $\beta^{-1}(w_0) = \bigl \{ z_0^{(1)}, \, \dots, \, z_0^{(n)} \bigr \}$, $\beta^{-1}(w_1) = \bigl \{ z_1^{(1)}, \, \dots, \, z_1^{(m)} \bigr \}$, and $\beta^{-1}(w_\infty) = \bigl \{ z_\infty^{(1)}, \, \dots, \, z_\infty^{(p)} \bigr \}$. For example, they correspond to partitions of the integer $N = \deg(\beta)$ into $n$, $m$, and $p$ parts, respectively.

## Shabat Polynomials

We say $\beta$ is a Shabat polynomial if $\beta$ is a Belyi map and $\beta^{-1}(w_\infty) = \{ z_\infty \}$ consists of just $p = 1$ point in the preimage. By pre- and post-composing with Moebius transformations, say

$\displaystyle \begin{matrix} \mathbb P^1(\mathbb C) & Z & 0 & 1 & \infty \\[10pt] \downarrow & \downarrow & \downarrow & \downarrow & \downarrow \\[10pt] \mathbb P^1(\mathbb C) &z =\dfrac {z_\infty \, (z_0 - z_1) \, Z + z_0 \, (z_1 - z_\infty)}{(z_0 - z_1) \, Z + (z_1 - z_\infty)} & z_0 & z_1 & z_\infty \\[10pt] \downarrow & \downarrow & \downarrow & \downarrow &\downarrow \\[10pt]\mathbb P^1(\mathbb C) & w = \beta(z) & w_0 & w_1 & w_\infty\\[10pt] \downarrow & \downarrow & \downarrow & \downarrow &\downarrow \\[10pt]\mathbb P^1(\mathbb C) &W = \dfrac {w_1 - w_\infty}{w_1 - w_0} \, \dfrac {w - w_0}{w - w_\infty} & 0 & 1 & \infty \end{matrix}$

we may as well assume that a Shabat polynomial $\beta: X \to \mathbb P^1(\mathbb C)$ is a Belyi map branched only above $\{ 0, \, 1, \, \infty \}$ such that $\beta \bigl( \{ 0, \, 1, \, \infty \} \bigr) \subseteq \{ 0, \, 1, \, \infty \}$ and $\beta^{-1}(\infty) = \{ \infty \}$. In particular, we can factor a Shabat polynomial as

$\displaystyle \beta(z) = \beta_0 \prod_{i = 1}^n \bigl(z - z_0^{(i)} \bigr)^{b_i} = 1 + \beta_1 \prod_{j = 1}^m \bigl(z - z_1^{(j)} \bigr)^{w_j}.$

We often omit the information about infinity and write the degree sequence/passport of a Shabat polynomial as

$\displaystyle \begin{matrix} \mathcal D & = & \biggl \{ \bigl \{ b_1,\, b_2, \, \dots, \, b_n \bigr \}, \, \bigl \{ w_1, \, w_2, \, \dots, \, w_m \bigr \} \biggr \} \\[5pt] \Pi & = &\displaystyle \left[ \prod_{i} {b_i}^{e_i}; \ \prod_j {w_j}^{e_j} \right] \end{matrix}$

where as before

$\displaystyle \deg(\beta) = \sum_{i = 1}^n b_i = \sum_{i} e_i \, b_i = \sum_{j = 1}^m w_j = \sum_{j} e_j \, w_j.$

There is a simple way to generate lots of examples of Shabat polynomials. Say that $\Phi: X \to \mathbb P^1(\mathbb C)$ is one example of a Shabat polynomial. It is well-known that we can generate more Shabat polynomials $\beta = \phi \circ \Phi : X \to \mathbb P^1(\mathbb C)$ as the composition of two maps $\Phi: X \to \mathbb P^1(\mathbb C)$ and $\phi: \mathbb P^1(\mathbb C) \to \mathbb P^1(\mathbb C)$ as long as $\phi$ is a Shabat polynomial which satisfies $\phi \bigl( \{ 0, \, 1, \, \infty \} \bigr) \subseteq \{ 0, \, 1, \, \infty \}$. One obvious such map is $\phi(z) = 4 \, z \, (1-z)$, although of course there are many others. I will return to this construction later.

## Dessins d’Enfants and Monodromy Groups

Given a Belyi map $\beta: X \to \mathbb P^1(\mathbb C)$, its Dessin d’Enfant $\Delta_\beta$ is a bipartite graph which can be embedded on the Riemann sphere $X = \mathbb P^1(\mathbb C)$. It is defined as follows.

• The “black” vertices are $B = \beta^{-1}(w_0) = \bigl \{ z_0^{(1)}, \, \dots, \, z_0^{(n)} \bigr \}$;
• The “white” vertices are $W = \beta^{-1}(w_1) = \bigl \{ z_1^{(1)}, \, \dots, \, z_1^{(m)} \bigr \}$;
• The “midpoints of the faces” are $F = \beta^{-1}(w_\infty) = \bigl \{ z_\infty^{(1)}, \, \dots, \, z_\infty^{(p)} \bigr \}$; and
• The “edges” are $E = \beta^{-1}\bigl( [0,1] \bigr)$ as the inverse image of the line segment $0 \leq w \leq 1$.

The Dessin d’Enfant will have $|B| + |W| = n + m$ vertices, $|E| = \deg(\beta) = N$ edges, and $|F| = p$ faces, so the Euler characteristic forces $N = m+n+p - 2$. When $\beta$ is a Shabat polynomial, there is only $p = 1$ face, so the corresponding Dessin d’Enfant will be a tree.  There is a nice post on MathOverflow which discusses this in a bit more detail.

In addition to Belyi maps/Shabat polynomials $\beta$ and degree sequences $\mathcal D$/passports $\Pi$, there is a third object of interest. The monodromy group $G_\beta$ is a transitive subgroup of $S_N$ which is constructed as follows.

1. Label the $N$ edges of $\Delta_\beta$ from 1 through $N$.
2. For each “black” vertex $z_0^{(i)} \in B$, read off the labels of the $b_i$ adjacent edges going counterclockwise, say as a $b_i$-cycle $\sigma_0^{(i)} = \bigl( B_1 \, B_2 \, \cdots \, B_{b_i} \bigr)$ in the symmetric group.
3. For each “white” vertex $z_1^{(i)} \in W$, read off the labels of the $w_j$ adjacent edges going counterclockwise, say as a $w_j$-cycle $\sigma_1^{(j)} = \bigl( W_1 \, W_2 \, \cdots \, W_{w_j} \bigr)$ in the symmetric group.
4. For each “midpoint of edge” $z_\infty^{(k)} \in F$, read off the labels of the $f_k$ adjacent edges going counterclockwise, say as a $f_k$-cycle $\sigma_\infty^{(k)} = \bigl( F_1 \, F_2 \, \cdots \, F_{f_k} \bigr)$ in the symmetric group.
5. Form that subgroup $G_\beta = \left \langle \sigma_0, \, \sigma_1, \, \sigma_\infty \right \rangle$ of $S_N$ which is generated by the products of disjoint cycles $\displaystyle \sigma_0 = \prod_{i = 1}^n \sigma_0^{(i)}$, $\displaystyle \sigma_1 = \prod_{j = 1}^m \sigma_1^{(j)}$ and $\displaystyle \sigma_\infty = \prod_{k = 1}^p \sigma_\infty^{(k)}$.

If $\sigma_0', \, \sigma_1', \, \sigma_\infty'$ is a different set of permutations arising from this construction but from a different labeling of the edges, then there is a permutation $\tau \in S_N$ so that these are simultaneously conjugate to $\sigma_0, \, \sigma_1, \, \sigma_\infty$, that is, $\sigma_0' = \tau \, \sigma_0 \, \tau^{-1}$, $\sigma_1' = \tau \, \sigma_1 \, \tau^{-1}$, and $\sigma_\infty' = \tau \, \sigma_\infty \, \tau^{-1}$. This permutation $\tau$ must send the edge labels $B_i \mapsto B'_i$, $W_j \mapsto W'_j$, and $F_k \mapsto F'_k$.

It is not too difficult to recover the Dessin d’enfant $\Delta_\beta$ from a triple $(\sigma_0, \, \sigma_1, \, \sigma_\infty)$. Indeed, the number $n$ ($m$, respectively) of disjoint cycles in the decomposition of $\sigma_0$ ($\sigma_1$, respectively) gives the number of “black” (“white”, respectively) vertices, and one can read off the cycle notation to find the labels of the edges which are adjacent to each “black” (“white”, respectively) vertex. One can connect the edges then slide around the vertices as necessary so that the faces match the disjoint cycle decomposition of $\sigma_\infty$.

## Relating Belyi Maps and Monodromy Groups

Fix a positive integer $N$, and let $S_N$ denote the symmetric group of degree $N$, and let $\mathcal D = \bigl \{ \{ b_1,\, b_2, \, \dots, \, b_n \}, \, \{ w_1, \, w_2, \, \dots, \, w_m \}, \, \{ f_1, \, f_2, \, \dots, \, f_p \} \bigr \}$ be a multiset of multisets such that $N = \sum_i b_i = \sum_j w_j = \sum_k f_k = m + n + p - 2$.

Proposition. The following are equivalent:

• $\mathcal D$ is the degree sequence of a Belyi map $\beta: X \to \mathbb P^1(\mathbb C)$ of degree $N$ on the Riemann sphere $X = \mathbb P^1(\mathbb C)$.
• There is a triple $(\sigma_0, \, \sigma_1, \, \sigma_\infty)$ of elements in $S_N$ which can be expressed as a disjoint product of cycles $\displaystyle \sigma_0 = \prod_{i = 1}^n \bigl( B_1 \, \cdots \, B_{b_i} \bigr)$, $\displaystyle \sigma_1 = \prod_{j = 1}^m \bigl( W_1 \, \cdots \, W_{w_j} \bigr)$, and $\displaystyle \sigma_\infty = \prod_{k = 1}^p \bigl( F_1 \, \cdots \, F_{f_k} \bigr)$ such that $\sigma_0 \, \sigma_1 \, \sigma_\infty = 1$ and $G = \left \langle \sigma_0, \, \sigma_1, \, \sigma_\infty \right \rangle$ is a transitive subgroup of $S_N$.

The result was established by Adolf Hurwitz in his 1891 paper Ueber Riemann’sche Flachen mit gegebenen Verzweigungspunkten. I’ve explained how to construct a degree sequence from a Belyi map, but of course the opposite direction is an active area of research.

Proposition. All permutations $\sigma_0, \, \sigma_0'$ of cycle type $(b_1, \, b_2, \, \dots, \, b_n)$ are conjugate, that is, $\sigma_0' = \tau \, \sigma_0 \, \tau^{-1}$ for some $\tau$. In particular, all triples $(\sigma_0, \, \sigma_1, \, \sigma_\infty), \, (\sigma_0', \, \sigma_1', \, \sigma_\infty')$ of “cycle type $\mathcal D$” are conjugate, that is, $\sigma_0' = \tau_0 \, \sigma_0 \, \tau_0^{-1}$, $\sigma_1' = \tau_1 \, \sigma_1 \, \tau_1^{-1}$, and $\sigma_\infty' = \tau_\infty \, \sigma_\infty \, \tau_\infty^{-1}$ for some $\tau_0, \, \tau_1, \, \tau_\infty \in S_N$.

This is an elementary fact about permutations.

Proposition. There is a one-to-one correspondence between:

• Belyi maps $\beta: X \to \mathbb P^1(\mathbb C)$ of degree $N$ on some compact connected Riemann surface $X$.
• Triples $(\sigma_0, \, \sigma_1, \, \sigma_\infty)$ of elements in $S_N$, modulo simultaneous conjugation, such that $\sigma_0 \, \sigma_1 \, \sigma_\infty = 1$ and $G = \left \langle \sigma_0, \, \sigma_1, \, \sigma_\infty \right \rangle$ is a transitive subgroup of $S_N$.

Proposition. Say that $(\sigma_0, \, \sigma_1, \, \sigma_\infty)$ is a triple of elements in $S_N$ such that $\sigma_0 \, \sigma_1 \, \sigma_\infty = 1$, and denote $G = \left \langle \sigma_0, \, \sigma_1, \, \sigma_\infty \right \rangle$ as the subgroup of $S_N$ generated by them. Denote $C_G(\sigma) = \bigl \{ \gamma \in F \, \bigl| \, \gamma \, \sigma \, \gamma^{-1} = \sigma \bigr \}$ as the centralizer of $\sigma$ in $G$. There is a one-to-one correspondence between:

• The set of all triples $(\sigma_0', \, \sigma_1', \, \sigma_\infty')$, modulo simultaneous conjugation, such that $\sigma_0' \, \sigma_1' \, \sigma_\infty' = 1$ and $\sigma_0' = \tau_0 \, \sigma_0 \, \tau_0^{-1}$, $\sigma_1' = \tau_1 \, \sigma_1 \, \tau_1^{-1}$ for some $\tau_0, \, \tau_1 \in S_N$
• Double cosets in $C_G(\sigma_0) \backslash G \slash C_G(\sigma_1)$.

These correspondences between Belyi maps, triples $(\sigma_0, \, \sigma_1, \, \sigma_\infty)$, and double cosets $C_G(\sigma_0) \backslash G \slash C_G(\sigma_1)$ were established by Michael Klug, Michael Musty, Sam Schiavone, and John Voight, in their 2013 paper Numerical calculation of three-point branched covers of the projective line. You can find the paper online at http://arxiv.org/abs/1311.2081v3. For the third correspondence, I’ve explained how to construct a triple of permutations from a Belyi map. Here’s a quick sketch of the fourth correspondence: say that $(\sigma_0', \, \sigma_1', \, \sigma_\infty')$ and $(\sigma_0'', \, \sigma_1'', \, \sigma_\infty'')$ are two triples such that

$\displaystyle \begin{matrix} \sigma_0' \, \sigma_1' \, \sigma_\infty' = 1 & \qquad \qquad & \sigma_0' = \tau_0 \, \sigma_0 \, \tau_0^{-1} & \qquad \qquad & \sigma_0'' = \theta_0 \, \sigma_0 \, \theta_0^{-1} \\[15pt] \sigma_0'' \, \sigma_1'' \, \sigma_\infty'' = 1 & & \sigma_1' = \tau_1 \, \sigma_1 \, \tau_1^{-1} & & \sigma_1'' = \theta_1 \, \sigma_1 \, \theta_1^{-1} \end{matrix}$

for some $\tau_0, \, \tau_1, \, \theta_0, \, \theta_1 \in S_N$, so denote $\eta' = \tau_0^{-1} \, \tau_1$ and $\eta'' = \theta_0^{-1} \, \theta_1$. Assume that they generate the same the double coset $C_G(\sigma_0) \, \eta \, C_G(\sigma_1)$, that is, there exist $\gamma_0 \in C_G(\sigma_0)$ and $\gamma_1 \in C_G(\sigma_1)$ such that $\eta'' = \gamma_0 \, \eta' \, \gamma_1$. We have the identity

$\displaystyle \begin{matrix} \sigma_0'' & = & \tau \, \sigma_0' \, \tau^{-1} \\[5pt] \sigma_1'' & = & \tau \, \sigma_1', \, \tau^{-1} \\[5pt] \sigma_\infty'' & = & \tau \, \sigma_\infty', \, \tau^{-1} \end{matrix}$

in terms of $\tau = \theta_0 \, \gamma_0 \, \tau_0^{-1} = \theta_1 \, \gamma_1^{-1} \, \tau_1^{-1}$ so that the two triples are simultaneously conjugate. Conversely, assume that the triples are simultaneously conjugate. Then the elements $\gamma_0 = \theta_0^{-1} \, \tau \, \tau_0 \in C_G(\sigma_0)$ and $\gamma_1 = \tau_1^{-1} \, \tau^{-1} \, \theta_1 \in C_G(\sigma_1)$, so that $\eta'$ and $\eta'' = \gamma_0 \, \eta' \, \gamma_1$ generate the same the double coset $C_G(\sigma_0) \, \eta \, C_G(\sigma_1)$.

## Algorithm: Counting Belyi Maps from Degree Sequences

The monodromy group $G_\beta$ helps to count the number of Belyi maps $\beta$ of degree $N$. Here’s an algorithm:

1. Fix a positive integer $N$.
2. Compute all multisets of multisets

$\displaystyle \mathcal D = \biggl \{ \bigl \{ b_1,\, b_2, \, \dots, \, b_n \bigr \}, \, \bigl \{ w_1, \, w_2, \, \dots, \, w_m \bigr \}, \, \bigl \{ f_1, \, f_2, \, \dots, \, f_p \bigl \} \biggr \}$

such that $N = \sum_i b_i = \sum_j w_j = \sum_k f_k = m + n + p - 2$.

3. For each $\mathcal D$, find a triple $(\sigma_0, \, \sigma_1, \, \sigma_\infty)$ of elements in $S_N$ which can be expressed as a disjoint product of cycles $\displaystyle \sigma_0 = \prod_{i = 1}^n \bigl( B_1 \, B_2 \, \cdots \, B_{b_i} \bigr)$, $\displaystyle \sigma_1 = \prod_{j = 1}^m \bigl( W_1 \, W_2 \, \cdots \, W_{w_j} \bigr)$, and $\displaystyle \sigma_\infty = \prod_{k = 1}^p \bigl( F_1 \, F_2 \, \cdots \, F_{f_k} \bigr)$ such that $\sigma_0 \, \sigma_1 \, \sigma_\infty = 1$ and $G = \left \langle \sigma_0, \, \sigma_1, \, \sigma_\infty \right \rangle$ is a transitive subgroup of $S_N$. If no such triple exists, there is no Belyi map for which $\mathcal D$ is a degree sequence.
4. Compute $C_G(\sigma_0)$ and $C_G(\sigma_1)$ as the centralizers of $\sigma_0$ and $\sigma_1$ in $G$. Given a double-coset representative $\eta$ in $C_G(\sigma_0) \backslash G \slash C_G(\sigma_1)$, the triple corresponding to $\sigma_0' = \sigma_0$ and $\sigma_1' = \eta \, \sigma_1 \, \eta^{-1}$ gives a unique Belyi map.

## Example #1

Let me give a relatively simple example. The rational $\beta(z) = z^N$ is a Shabat polynomial of degree $N$ whose passport is $\Pi = \bigl[ N^1; \, 1^N \bigr]$, Dessin d’Enfant is the Star Graph with $N$ spokes, and whose monodromy group is $G_\beta = Z_N$ as the cyclic group of order $N$. To see why the latter is true, observe that

• the “black” vertex is $z_0^{(1)} = 0$,
• the “white” vertices are $z_1^{(k)} = \cos (2 \pi k/N) + i \, \sin (2 \pi k / N)$ for $k = 1, \, 2, \, \dots, \, N$,
• the “midpoint of the face” is $z_\infty^{(1)} = \infty$, and
• the edges are the segments $e_k = \left \{ (1-t) \, z_0^{(1)} + t \, z_1^{(k)} \, \biggl| \, 0 \leq t \leq 1 \right \}$ for $k = 1, \, 2, \, \dots, \, N$.

If we label edge $e_k$ as the positive integer “$k$”, then $\sigma_0 = (1 \, 2 \, \cdots \, N)$ is an $N$-cycle while $\sigma_1 = 1$ is trivial. Hence the group generated by $\sigma_0$, $\sigma_1$, and $\sigma_\infty = (\sigma_0 \, \sigma_1)^{-1}$ is $G_\beta = \left \langle \sigma_0, \, \sigma_1, \, \sigma_\infty \right \rangle = Z_N$.

In fact, I can do even better: I claim this is the only Shabat polynomial associated with this passport! Indeed, if we have a triple $(\sigma_0, \, \sigma_1, \, \sigma_\infty)$ associated with the passport $\Pi = \bigl[ N^1; \, 1^N \bigr]$, then $\sigma_1$ must be the trivial permutation, and so the centralizer $C_G(\sigma_1) = G$. This means there are no nontrivial double cosets in $C_G(\sigma_0) \backslash G \slash C_G(\sigma_1)$, so there can only be one Shabat polynomial.

## Example #2

As considered in the 1994 paper by Georgii Borisovich Shabat and Alexander Zvonkin entitled Plane trees and algebraic numbers, the rational function

$\displaystyle \beta(z) = - \dfrac {4}{531441} \, (z - 1) \, z^3 \, \bigl( 2 \, z^2 + 3 \, z + 9 \bigr)^3 \, \bigl( 8 \, z^4 + 28 \, z^3 + 126 \, z^2 + 189 \, z + 378 \bigr)$

is a Shabat polynomial of degree $N = 14$ whose passport is $\Pi = \bigl[ 3^3 \, 1^5; \ 2^7 \bigr]$. Observe that $\beta = \phi \circ \Phi$ is the composition of the two rational functions $\phi(z) = 4 \, z \, (1-z)$ and $\displaystyle \Phi(z) = - \dfrac {1}{729} \, (z - 1) \, (2 \, z^2 + 3 \, z + 9)^3$ which are both Shabat polynomials, one of degree $\deg(\phi) = 2$ with passport $\Pi_\phi = \bigl[ 1^2; \ 2^1 \bigr]$, and the other of degree $\deg(\Phi) = 7$ with passport $\Pi_\Phi = \bigl[ 3^2 \, 1^1; \ 3^1 \, 1^4 \bigr]$. Since $\Phi(z) = 1 - \bigl( 2 \, z - 1 \bigr)^2$, we see that $\Delta_\phi$ is the Star Graph and $G_\phi = Z_2$ is the cyclic group of order 2. It is easy to see that the monodromy group of $\Phi$ has the generators

$\begin{matrix} \sigma_0 & = & \bigl( 1 \ 5 \ 3 \bigr) \, \bigl(2 \ 4 \ 6) \bigr) \\[5pt] \sigma_1 & = & \bigl(3 \ 7 \ 4 \bigr) \\[5pt] \sigma_\infty & = & \bigl( 1 \ 3 \ 2 \ 6 \ 4 \ 7 \ 5 \bigr) \end{matrix}$

so that $G_\phi = \left \langle \sigma_0, \, \sigma_1, \, \sigma_\infty \right \rangle = A_7$ is the alternating group.  Here’s the strange part. The monodromy group of the composition $\beta = \phi \circ \Phi$ has the generators

$\begin{matrix} \sigma_0 & = & \bigl( 3 \ 7 \ 5 \bigr) \, \bigl( 4 \ 6 \ 8 \bigr) \, \bigl( 11 \ 13 \ 12 \bigr) \\[5pt] \sigma_1 & = & \bigl( 1 \ 3 \bigr) \, \bigl( 2 \ 4 \bigr) \, \bigl( 5 \ 11 \bigr) \, \bigl( 6 \ 12 \bigr) \, \bigl( 7 \ 9 \bigr) \, \bigl( 8 \ 10 \bigr) \, \bigl( 13 \ 14 \bigr) \\[5pt] \sigma_\infty & = & \bigl( 1 \ 5 \ 12 \ 4 \ 2 \ 8 \ 10 \ 6 \ 13 \ 14 \ 11 \ 7 \ 9 \ 3 \bigr) \end{matrix}$

so that $G_\beta = \left \langle \sigma_0, \, \sigma_1, \, \sigma_\infty \right \rangle$ is a group of order $|G_\beta| = 12 \, 700 \, 800 = |G_\phi| \cdot |G_\Phi|^2$.

Question$\beta = \phi \circ \Phi$ is a composition of Shabat polynomials, where $G_\phi = Z_2$ and $G_\Phi = A_7$. Since $|G_\beta| = |G_\phi| \cdot |G_\Phi|^2$, do we have $G_\beta = Z_2 \ltimes \bigl( A_7 \times A_7 \bigr)$ as the wreath product of $G_\Phi$ by $G_\phi$?

I’m really confused by this example because it seems to be part of a general phenomenon…  I will post more on this later as I come up with more examples — and possibly a proof!