2 Almost-Belyi maps

Given a (squarefree) Pellian polynomial $D(t)$ of degree $2d$ and a solution $(A, B)$ of degree n of the corresponding Pell–Abel equation $A^2-D(t)B^2=1$ , we consider the map $\phi _A:=A^2 : \mathbb {P}^1(\mathbb {C}) \rightarrow \mathbb {P}^1(\mathbb {C})$ of degree $2n$ .

It is easy to see that the Pell–Abel equation forces the branching of the map $\phi _A$ to be ‘concentrated’ in $ 0,\ 1$ and $\infty $ . First, since $\phi _A$ is a polynomial, there is total ramification above $\infty $ . Moreover, as $\phi _{A}$ is a square and $\phi _A-1=DB^2$ , we have that the ramification indices above $0$ are all even, while the ones above $1$ are all even with the exception of $2d$ points (the simple roots of $D(t)$ ). Hence, counting the branching of $\phi _A$ as the sum of $e-1$ over the ramification indices e, we have that above $0$ the branching is at least n, above $1$ at least $n-d$ and above $\infty $ it is exactly $2n-1$ . On the other hand, by the Riemann–Hurwitz formula (see [Reference Hindry and Silverman17, Theorem A.4.2.5., p. 72]), the total branching is equal to $4n-2$ . This means that the branching outside $0,1,\infty $ is at most $d-1$ , and thus there cannot be more than $d-1$ further branch points. Following the third author [Reference Zannier34], we call $\phi _A$ an almost-Belyi map as its branching is ‘concentrated’ above $0,1, \infty $ (in the sense that the number of branch points outside this set does not depend on the degree n of $\phi _A$ but only on the degree of D which is fixed)Footnote ¹ .

To each such map, we can associate a monodromy permutation representation in the following way (for references see [Reference Fried15], [Reference Miranda21] or [Reference Völklein32]).

Let us call $\mathcal {B}=\{b_1, \ldots , b_{h}\}$ the set of the branch points of $\phi _A$ (where $h\le d+2$ ), and let us choose a base point q different from the $b_i$ . Let us moreover call $V:=\mathbb {P}^1(\mathbb {C})\setminus \mathcal {B}$ . The fundamental group $\pi _1(V,q)$ of V is a free group on h generators $[\gamma _1],\ldots ,[\gamma _{h}]$ modulo the relation

$$\begin{align*}[\gamma_1]\cdots[\gamma_{h}]=1, \end{align*}$$

where each $\gamma _i:[0,1]\rightarrow V$ is a closed path which winds once around $b_i$ . Now, consider the fiber $\phi _A^{-1}(q)$ above q, and denote by $q_1, \ldots , q_{2n}$ the $2n$ distinct points in this fiber. Every loop $\gamma \subseteq V$ based at q and not passing through the $b_i$ can be lifted to $2n$ paths $\tilde {\gamma }_1, \ldots , \tilde {\gamma }_{2n}$ , where $\tilde {\gamma }_j$ is the unique lift of $\gamma $ starting at $q_j$ . Hence, $\tilde {\gamma }_j(0)=q_j$ for every j. Now, consider the endpoints $\tilde {\gamma }_j(1)$ ; these also lie in the fiber $\phi _A^{-1}(q)$ and indeed form the entire preimage set $\{q_1, \ldots , q_{2n}\}$ . We call $\sigma (j)$ the index in $\{1, \ldots , 2n\}$ such that $\tilde {\gamma }_j(1)=q_{\sigma (j)}$ . The function $\sigma $ is then a permutation of the set $\{1, \ldots , 2n\}$ which depends only on the homotopy class of $\gamma $ ; therefore, we have a group homomorphism $\rho :\pi _1(V,q) \longrightarrow S_{2n}$ called the monodromy representation of the covering map $\phi _A$ . This is clearly determined by the images $\sigma _i=\rho ([\gamma _i])$ of the generators of $\pi _1(V,q)$ , and the image of $\rho $ must be a transitive subgroup of $S_{2n}$ , as V is connected.

Note that these permutations must have a precise cycle structure depending on the branch points of the map. Following the notation of [Reference Bilu5], we say that a branch point c of a rational map f is of type $(m_1, \ldots , m_k)$ if, for $ f^{-1}(c)=\{c_1, \dots , c_k\}$ , the point $c_j$ has ramification index $m_j$ for every $j=1, \ldots , k$ . The type of a branch point corresponds to the cycle structure of the corresponding permutation, meaning that, going back to $\phi _A$ , if $b_i$ is of type $(m_1, \ldots , m_k)$ , then $\sigma _i$ is the product of k disjoint cycles $\tau _1,\dots , \tau _k$ with $\tau _j$ of length $m_j$ .

We just showed how to associate to a covering a certain tuple of permutations satisfying some specific decomposition properties; the Riemann existence theorem tells us that we can do the opposite, that is, to any tuple of permutations satisfying certain properties we can associate a covering. One can actually say more: There exists a 1-1 correspondence between these two sets if one mods out by the following equivalence relations on both sides.

We can finally state this consequence of the Riemann existence theorem.

Theorem 2.3 ([Reference Miranda21], Corollary 4.10).

Fix a finite set $\mathcal B=\{b_1, \ldots , b_h\}\subseteq \mathbb {P}^1(\mathbb {C})$ . Then there is a 1-1 correspondence between

$$\begin{align*}\begin{Bmatrix} \text{equivalence classes}\\ \text{ of rational maps} \\ F:\mathbb{P}^1(\mathbb{C}) \longrightarrow \mathbb{P}^1(\mathbb{C}) \\ \text{of degree }2n \\ \text{whose branch points lie in }\mathcal{B} \end{Bmatrix} \text{ and } \begin{Bmatrix} \text{conjugacy classes } (\sigma_1, \ldots, \sigma_{h})\in S_{2n}^h \\ \text{such that }\sigma_1\cdots \sigma_{h}=\mathrm{id},\\ \text{the subgroup generated by the } \sigma_i \\ \text{is transitive and } \sum_{i,j}(m_{ij}-1)=4n-2, \\ \text{where } (m_{i1},\dots, m_{ik_i}) \text{ is the cycle structure of } \sigma_i \end{Bmatrix}. \end{align*}$$

Moreover, given the tuple of permutations $(\sigma _1, \ldots , \sigma _{h})$ , for each $i=1,\dots , h$ there are $k_i$ preimages $b_{i1}, \ldots , b_{ik_i}$ of $b_i$ for the corresponding cover $F: \mathbb {P}^1(\mathbb {C}) \longrightarrow \mathbb {P}^1(\mathbb {C}) $ , with $e_{F}(b_{ij})=m_{ij}$ .

As seen before, given a Pell–Abel equation (1.2) with A of degree n and D of degree $2d$ , we get the map $\phi _A$ that is branched in $\{0,1,\infty \}$ and in further $k $ points, where $0\leq k \le d-1$ . Call $\mathcal {B}:=\{0,1,\infty ,b_1, \ldots , b_{k}\}$ and $V:=\mathbb {P}^1(\mathbb {C})\setminus \mathcal {B}$ . We fix a $q\in V$ , and we let $\rho :\pi _1(V,q) \longrightarrow S_{2n}$ be the associated monodromy representation (as explained above). We now let $\gamma _0,\gamma _{\infty },\gamma _{1}$ be closed paths winding once around $0,\infty ,1$ , respectively, and $\delta _i$ be a closed path winding once around $b_i$ for all $i=1, \ldots , k$ . Then $[\gamma _0] [\gamma _{\infty }] [\gamma _{1}] [\delta _1] \cdots [\delta _{k}]=1$ and all these classes generate $\pi _1(V,q)$ .

We let $\sigma _0=\rho (\gamma _0)$ , $\sigma _{\infty }=\rho (\gamma _{\infty })$ , $\sigma _1=\rho (\gamma _1)$ and $\tau _i=\rho (\delta _i)$ for $i=1, \ldots , k$ . Then we have that

1. 1 - $\sigma _0\sigma _{\infty }\sigma _1\tau _1\cdots \tau _{k}=\mathrm {id}$ ;

2. 2 - the subgroup of $S_{2n}$ generated by these permutations is transitive;

3. 3 - $\sigma _0$ and $\sigma _{\infty }$ do not fix any index;

4. 4 - $\sigma _1$ fixes exactly $2d$ indexes.

Moreover, concerning the decomposition of these permutations in disjoint cycles we have that

1. 5 - all nontrivial cycles in $\sigma _0$ and $\sigma _1$ have even length;

2. 6 - $\sigma _{\infty }$ must be a $2n$ -cycle;

3. 7 - the sum over all cycles of $\sigma _0, \dots , \tau _k$ of their lengths minus 1 must give $4n-2$ .

The permutations $\sigma _0, \dots , \tau _k$ depend on the chosen labelling of the preimages of q, so a Pell–Abel equation gives exactly one conjugacy class of $(k+3)$ -tuples and a $\mathcal {B}$ for some $k\in \{0,\dots , d-1\}$ .

On the other hand, fixing $k\in \{0,\dots , d-1\}$ and $\mathcal {B}$ , a conjugacy class of $(k+3)$ -tuples satisfying the above conditions gives a cover $\mathbb {P}^1(\mathbb {C})\rightarrow \mathbb {P}^1(\mathbb {C})$ that is a polynomial branched exactly in $\mathcal {B}$ . Moreover, the cycle structures of $\sigma _0$ and $\sigma _1$ make this polynomial satisfy a Pell–Abel equation.

We see an example.

3 Proof of Theorem 1.1

This section is devoted to proving Theorem 1.1. For the reader’s convenience, we recall the notations and the statement of the theorem. Given a point $\overline {s}=(s_1, \ldots , s_{2d})\in \mathcal {B}_{2d}$ , we denote by $H_{\overline s}$ a nonsingular model of the hyperelliptic curve defined by $y^2=t^{2d}+ \cdots +s_{2d}$ with two points at infinity that we call $\infty ^+$ and $\infty ^-$ , and by $\mathcal {J}_{\overline s}$ its Jacobian variety. We denote by $\mathcal {J} \rightarrow S=\mathcal {B}_{2d}$ the abelian scheme of relative dimension $d-1$ whose fibers are the $\mathcal {J}_{s}$ and by $\sigma :S \rightarrow \mathcal {J}$ the section corresponding to the point $[\infty ^+-\infty ^-]$ . We recall moreover that $\tilde {S}$ is the universal covering of $S(\mathbb {C})$ .

Proof. As the irreducible components of $\mathcal {P}_{2d}$ correspond to rational values of the Betti map, we clearly have that the number of components of $\mathcal {P}_{2d}$ is countable. Moreover, they are components of projections on $\mathcal {B}_{2d}$ of intersections of torsion subgroup schemes of $\mathcal {J}$ with the image of the section $\sigma $ . Therefore, they are algebraic subvarieties of $\mathcal {B}_{2d}$ .

Fix now a component U of $\mathcal {P}_{2d}$ of dimension $>d+1$ mapping to a rational point of denominator n. Then every point of U corresponds to a Pellian polynomial with a solution of degree n. We let W be the closure of

$$ \begin{align*} \{(a_0,\dots ,a_{2n})\in \mathbb{A}_{\mathbb{C}}^{2n+1}: a_{0}t^{2n}+a_{1}t^{2n-1}+\dots+ a_{2n} &=A^2 \mbox{ is a square} \\ &\qquad\text{and } A^2-D_{\overline{s}}B^2=1 \mbox{ for some }\overline{s}\in U \} \end{align*} $$

in

$$ \begin{align*} \mathcal{A}_{2n}:=\{(a_0,\dots ,a_{2n})\in \mathbb{A}_{\mathbb{C}}^{2n+1}: a_{0}\neq 0 \}. \end{align*} $$

Note that W is the closure of the projection on $\mathcal {A}_{2n}$ of

$$ \begin{align*} \{ (a_0,\dots ,a_{2n},&b_0,\dots, b_{2n-2d},\overline{s})\in \mathcal{A}_{2n}\times \mathbb{A}_{\mathbb{C}}^{2n-2d+1}\times U: \\ &a_{0}t^{2n}+\dots+ a_{2n} =A^2 \text{ and } b_{0}t^{2n-2d}+\dots+ b_{2n-2d} =B^2 \mbox{ are squares} \\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{and } A^2-D_{\overline{s}}B^2=1 \}. \end{align*} $$

Such projection has finite fibers, therefore W has dimension $>d+1$ .

Note that it is possible to compose any degree $2n$ complex polynomial f with a linear polynomial and obtain a monic polynomial with no term of degree $2n-1$ . Moreover, there are at most $2n$ possible polynomials of this special form that can be obtained in this way from a given f. We let

$$ \begin{align*} \mathcal{C}_{2n}:=\{(c_2,\dots ,c_{2n})\in \mathbb{A}_{\mathbb{C}}^{2n-1} \} \text{ and } \mathcal{L}:=\{(a,b)\in \mathbb{A}_{\mathbb{C}}^2: a\neq 0\} \end{align*} $$

and define the morphism $\varphi :\mathcal {C}_{2n}\times \mathcal {L}\rightarrow \mathcal {A}_{2n}$ ,

$$ \begin{align*} \varphi((c_2,\dots ,c_{2n}),(a,b)) &=(a_0,\dots ,a_{2n}) \iff \\ & a_0t^{2n}+\dots +a_{2n}=(at+b)^{2n}+c_{2}(at+b)^{2n-2}+\cdots+ c_{2n-1}(at+b)+ c_{2n}. \end{align*} $$

By the above considerations, $\varphi $ is surjective and the fibers are finite of cardinality at most $2n$ .

If we let $\pi $ be the projection $ \mathcal {C}_{2n}\times \mathcal {L}\to \mathcal {C}_{2n}$ and $Z=\pi (\varphi ^{-1}(W))$ , as the fibers of $\pi $ have dimension 2, Z must have dimension $>d-1$ .

Now, for $\overline {c}=(c_2,\dots ,c_{2n})\in Z$ , the corresponding polynomial $f_{\overline {c}}(t)=t^{2n}+c_{2}t^{2n-2}+\cdots + c_{2n}$ is a square and we have $f_{\overline {c}}(t)-D(t)B(t)^2=1$ for some polynomials $D(t),B(t)$ with D of degree $2d$ . We let

$$ \begin{align*} \tilde{Z}&= \{ (\overline{c},c^{\prime}_{2},\dots, c^{\prime}_{2n-1},r_1,\dots, r_{2n-1})\in Z \times \mathbb{A}_{\mathbb{C}}^{2n-2}\times \mathbb{A}_{\mathbb{C}}^{2n-1} : f^{\prime}_{\overline{c}}= 2n t^{2n-1}+ c^{\prime}_{2} t^{2n-3}+\dots+ c^{\prime}_{2n-1} \\ &\qquad\qquad\qquad\qquad\text{and } f^{\prime}_{\overline{c}} = 2n(t^{2n-1}+ \sigma_2(r_1,\dots, r_{2n-1})t^{2n-3}+\dots+ \sigma_{2n-1}(r_1,\dots, r_{2n-1})) \}, \end{align*} $$

where $\sigma _i(X_1,\dots , X_{2n-1})$ is the i-th elementary symmetric polynomial in $2n-1$ variables and $f^{\prime }_{\overline {c}}$ is the derivative of $f_{\overline {c}}$ . In other words, given $\overline c \in Z$ , $c_2',\ldots , c_{2n-1}'$ are the coefficients of the derivative of the polynomial $f_{\overline c}(t)$ and $r_1,\ldots , r_{2n-1}$ are the ramification points of the same polynomial (not necessarily distinct). Clearly, $\tilde {Z}$ must have dimension $>d-1$ .

Finally, if we consider the morphism $\psi :\tilde {Z} \to \mathbb {A}_{\mathbb {C}}^{2n-1} $ defined by

$$ \begin{align*}(\overline{c},c^{\prime}_{2},\dots, c^{\prime}_{2n-1},r_1,\dots, r_{2n-1})\mapsto (f_{\overline{c}}(r_1), \dots , f_{\overline{c}}( r_{2n-1})),\end{align*} $$

then $f_{\overline {c}}(r_1), \dots , f_{\overline {c}}( r_{2n-1})$ are the (not necessarily distinct) branch points of $f_{\overline {c}}$ . By the Riemann existence theorem and the above considerations, the fibers of $\psi $ are finite and have cardinality at most $2n$ , and therefore $\psi (\tilde {Z})$ has dimension $>d-1$ .

This gives a contradiction because the considerations of the previous section imply that the polynomial $f_{\overline {c}}$ , since it fits in a Pell–Abel equation, may have not more than $d+1$ finite branch points, two of which are 0 and 1. Therefore, not more than $d-1$ branch points are allowed to vary and $\psi (\tilde {Z})$ must have dimension $\leq d-1$ , as wanted.

4 Chebychev polynomials and powers of solutions of the Pell–Abel equation

This section is devoted to describing the group of solutions of a Pell–Abel equation. As seen in the introduction, if a Pell–Abel equation $A^2-DB^2=1$ has a nontrivial solution, then it has infinitely many ones, obtained by taking powers $A_m+\sqrt D B_m=(A+\sqrt D B)^m$ . Our goal for this section is to prove that $A_m^2=f_m(A^2)$ , for some polynomial $f_m$ related to the m-th Chebychev polynomial.

We recall that Chebychev polynomials of the first kind are defined recursively as

$$ \begin{align*} \begin{cases} T_0(t)=1, \\ T_1(t)=t, \\ T_{n+1}(t)=2tT_n(t)-T_{n-1}(t), \end{cases} \end{align*} $$

so $T_n$ is a polynomial of degree n, and, for every $n>1$ , the two terms of highest degree of $T_n$ are $2^{n-1}t^n$ and $-n2^{n-3}t^{n-2}$ . This polynomial satisfies many important properties; for example, for every $n\ge 0$ , we have that $T_n(t+t^{-1})=t^n+t^{-n}$ . Explicitly, we have that

(4.1) $$ \begin{align} T_{n}(t):=\sum _{h=0}^{[n/2]}\binom{n}{2h}t^{n-2h}(t^{2}-1)^{h}, \end{align} $$

where $[\cdot ]$ denotes the floor function. For a survey on these topics, see [Reference Lidl, Mullen and Turnwald19].

Suppose now that $(A,B)$ is a solution of the Pell equation $A^2-DB^2=1$ . Then $(A_m,B_m)$ defined by $A_m+\sqrt D B_m=(A+\sqrt D B)^m$ is another solution of the same Pell–Abel equation, and $A_m=\sum _{j=0}^{[m/2]} \binom {m}{2j} (DB^2)^j A^{m-2j}$ . Using that $DB^2=A^2-1$ , we have

(4.2) $$ \begin{align} A_m=\sum_{j=0}^{[m/2]} \binom{m}{2j} (A^2-1)^j A^{m-2j}=T_m(A). \end{align} $$

For the rest of the paper, we will denote by $\phi $ the square function $\phi (t)=t^2$ and by $\phi _g$ the composition $\phi \circ g=g^2$ .

For our purposes, it will be useful to express $\phi _{A_m}$ as a function of $\phi _A$ . By equation (4.2), one can see that $\phi _{A_m}=f_m(\phi _A)$ , where

(4.3) $$ \begin{align} f_{2k}(w):=\left ( \sum_{j=0}^{k} \binom{2k}{2j} (w-1)^j w^{k-j} \right )^2, \end{align} $$

and

(4.4) $$ \begin{align} f_{2k+1}(w):=w\left ( \sum_{j=0}^{k} \binom{2k+1}{2j} (w-1)^j w^{k-j} \right )^2. \end{align} $$

For every $m\ge 1$ we call $f_m$ the m-th power polynomial. Notice that $f_m \circ \phi =\phi \circ T_m$ .

For our purposes we will also need a description of the branch locus of the $f_m$ ; this can be easily done by looking at the branch locus of the Chebychev polynomials which is well-known in the literature. We recall that, given a polynomial f, we say that a branch point b of f is of type $(\mu _1, \ldots , \mu _k)$ if the $\mu _i$ are the ramification indexes of the points in the preimage of b. This is also equal to the array of the multiplicities of the roots of $f(t)-b$ .

Proof. For $m=2$ we have that $f_m(t)=(2t-1)^2$ , and the proposition holds trivially. We will then assume that $m\ge 3$ .

The Chebychev polynomials are strictly related to Dickson polynomials $D_m(x,a)$ (for a definition of these polynomials and their properties, see [Reference Bilu5, Section 3]). Indeed, we have that $T_m(t)=\frac {1}{2}D_m(2t,1)$ . Using this relation and [Reference Bilu5, Proposition 3.3, (b)] we have that, if $m\ge 3$ , the finite branching of $T_m$ happens only in $\pm 1$ . If m is odd, both $\pm 1$ are of type $(1,2, \ldots , 2)$ , while if m is even we have that $1$ is of type $(1,1,2, \ldots , 2)$ and $-1$ is of type $(2,\ldots ,2)$ . Using that $\phi \circ T_m= f_m \circ \phi $ , the finite branch points of $f_m$ are exactly the ones of $\phi \circ T_m$ ,that is, $0$ and $1$ . Moreover, since $0$ is not a branch point for $T_m$ , then the ramification points of $f_m\circ \phi $ have ramification index $2$ .

Assume first that m is even. By equation (4.3), since $f_m$ is a square, then $0$ is a branch point for $f_m$ . As $0$ is not contained $f_m^{-1}(0)$ , then it is of type $(2,2, \ldots , 2)$ . Let us now consider the branch point $1$ ; notice that $\{0,1\}\subseteq f_m^{-1}(1)$ . Since $\phi $ is ramified in $0$ , we have that $f_m$ is unramified in $0$ . Using that $f_m \circ (1-t)=f_m$ , this implies that $f_m$ is unramified also in $1$ . By comparing the other ramification indexes, it follows that $1$ is of type $(1,1,2, \ldots , 2)$ as wanted.

Assume now m odd. By equation (4.4), we have $f_m$ is unramified in $0$ , and all the other points in the preimage of $0$ have ramification index $2$ , so $0$ is of type $(1,2, \ldots , 2)$ . Using the relation

(4.5) $$ \begin{align} (1-t) \circ f_m \circ (1-t)= f_m, \end{align} $$

we have that the type of $0$ and the type of $1$ are the same, and $1$ is the unramified point over $1$ , concluding the proof.

5 Monodromy groups and polynomial decompositions

The problem of finding (functional) decompositions of polynomials is very classical and was studied first by Ritt in [Reference Ritt26] and then by several authors (see, for instance, [Reference Fried14]). In this section, we recall some basic facts that will be used in the paper.

Let f be a nonconstant polynomial in $\mathbb {C}[t]$ , let u be transcendental over $\mathbb {C}$ and L be the splitting field of $f(t)-u$ over $\mathbb {C}(u)$ . The monodromy group $\text {Mon}(f)$ of f is the Galois group of L over $\mathbb {C}(u)$ , viewed as a group of permutations of the roots of $f(t)-u$ . By Gauss’ lemma, it follows that $f(t) - u$ is irreducible over $\mathbb {C}(u)$ , so $\text {Mon}(f)$ is a transitive permutation group. If x is a root of $f(t)-u$ in L, then $u=f(x)$ and $\text {Mon}(f)=\text {Gal}(L/\mathbb {C}(f(x)))$ . We denote moreover by H the stabilizer of x, that is, $H=\text {Gal}(L/\mathbb {C}(x))$ . Note that, by Theorem 2.3, the polynomial f of degree n seen as a map $\mathbb {P}^1(\mathbb {C})\rightarrow \mathbb {P}^1(\mathbb {C})$ corresponds to a conjugacy class of h-tuples of permutations in $S_n$ , where h is the cardinality of the branch locus of f. Then, once we fix a labelling of the roots of $f(t)-u$ (which corresponds to fixing a representative $(\sigma _1, \ldots , \sigma _h)$ in the conjugacy class), the monodromy group of f is isomorphic to the subgroup of $S_n$ generated by $\sigma _1, \ldots , \sigma _h$ . Applying Lüroth’s theorem [Reference Schinzel28, Theorem 2] and [Reference Schinzel28, Theorem 4], we have the following proposition.

Using the Galois correspondence, this shows that the study of the polynomial decompositions of f reduces to the study of subgroups of the monodromy group of f containing H.

We give the following definition.

Notice that, if $f=g\circ h$ , we can always change the decomposition up to composing with linear polynomials, so it makes sense to study the polynomial decompositions up to linear equivalence. Moreover, notice that two linearly equivalent polynomials have the same monodromy group.

We are now interested in computing the monodromy group of the polynomials $f_m$ defined in the previous section.

6 Characterization of primitivity

In this section, we want to give a criterion to detect the primitivity of a solution of the Pell–Abel equation in terms of the associated conjugacy class of permutations. We start with a definition.

It is well-known (see [Reference Ritt26]) that, if $f= g \circ h$ is a polynomial of degree r which is the composition of two polynomials g and h, where $\deg h=s$ and $\deg g=r$ , then $\text {Mon}(f)$ is r-imprimitive. Indeed, if q is not a branch point for f, then the set of its preimages via f is partitioned in subsets whose elements map via h to the same preimage of q via g.

We now give a criterion for determining whether a solution $(A,B)$ of a Pell–Abel equation is primitive or an m-th power for some $m\geq 2$ , and this behavior is completely determined by the monodromy group $G_A$ of $\phi _A$ . Note that $G_A$ is always $2$ -imprimitive since $\phi _A$ is the square of a polynomial.

We will prove the above theorem in the next section. Before this, we show how Corollary 1.5 can be deduced from Theorem 6.2.

Proof of Corollary 1.5.

We prove that a solution $(A,B)$ associated to the conjugacy class of the tuple $(\sigma _0,\sigma _{\infty },\sigma _1,\tau _1, \ldots ,\tau _{d-1})$ given in Example 2.4 is primitive for every $n\ge d \ge 2$ .

We recall that $\sigma _{\infty }=(2n, \ldots , 1)$ , $\sigma _0=(1,2n) (2,2n-1)\cdots (n,n+1)$ , $\sigma _1=(1,2n-1) (2,2n-2)\cdots (n-d,n+d)$ and $\tau _i=(n-i,n+i)$ , for $i=1, \ldots , d-1$ .

Suppose by contradiction that $(A,B)$ is the m-th power of another solution $(A',B')$ for some $m \mid n$ , $1<m<n$ . By Theorem 6.2, $G_A$ has to preserve a $2m$ -partition of $\{1, \ldots , 2n\}$ . We see that the only $2m$ -partition preserved by $\sigma _{\infty }$ is $\mathcal F_{2m}:=\{F_1,\ldots , F_{2m}\}$ , where for every $1\le h\le 2m$ , we set

$$ \begin{align*}F_h= \{ a \in \{1,\dots , 2n \}: a \equiv h\ \ \mod 2m \}.\end{align*} $$

On the other hand, $\tau _{1}=(n-1, n+1)$ and $n-1,\ n+1$ are not in the same congruence class modulo $2m$ if $m\ge 2$ , so $\mathcal F_{2m}$ is not preserved by $G_A$ . This implies that $G_A$ does not preserve any $2m$ -partition, hence proving the theorem.

7 Proof of Theorem 6.2

In view of what we explained in the previous section, one direction is quite easy. Indeed, let $(A,B)$ be a solution of a Pell–Abel equation that is the m-th power of another solution $(A', B')$ . In Section 4, we showed that $\phi _{A}=f_m(\phi _{A'})=f_m \circ \phi \circ A'$ , where $\phi (t)=t^2$ . As $f_m$ is a polynomial of degree m, it follows immediately that $G_{A}$ is $2m$ -imprimitive.

The proof of the converse is much more involved.

Let us consider a solution $(A,B)$ with A of degree n and the associated almost-Belyi map $\phi _A$ . By Theorem 2.3, the map $\phi _A$ corresponds to the conjugacy class of a $(k+3)$ -tuple of permutations in $S_{2n}$ given by $\Sigma _A:=(\sigma _0,\sigma _{\infty },\sigma _1,\tau _1, \ldots ,\tau _{k})$ with $k \le d-1$ , where we recall that $\sigma _{\infty }$ is a $2n$ -cycle, $\sigma _0$ is the product of disjoint cycles of even length, $\sigma _1$ fixes exactly $2d$ indexes and is the product of disjoint cycles of even length; see properties 1 to 7 after Theorem 2.3. We call $b_1, \ldots , b_k$ the further k branch points different from $0,1, \infty $ .

Conjugating by a suitable permutation, we can and will always assume that $\sigma _{\infty }=(2n,2n-1,\ldots , 1)$ .

We recall that the sum over all cycles of $\sigma _0, \dots , \tau _k$ of their lengths minus 1 must give $4n-2$ . Looking at the cycle structure of $\sigma _0$ , $\sigma _\infty $ and $\sigma _1$ we have that $\sigma _1$ fixes $2d$ indexes and $\sum _{i,j}(m_{ij}-1)\leq d-1$ , where $(m_{i1},\dots , m_{ih_i})$ is the cycle structure of $\tau _i$ . Therefore, there are at least two indexes which are fixed by $\sigma _1$ and, at the same time, by every $\tau _i$ . Without loss of generality, we are going to assume that one of them is $2n$ . Moreover, each $\tau _i$ must fix at least $2n-2(d-1)$ indexes.

The monodromy group $G_A$ of $\phi _A$ is the subgroup of $S_{2n}$ generated by $\sigma _0,\sigma _{\infty },\sigma _1,\tau _1, \ldots ,\tau _{k}$ ; by assumption, $G_A$ preserves a $2m$ -partition of $\{1, \ldots ,2n\}$ . As mentioned before, the only $2m$ -partition preserved by $\sigma _{\infty }$ is $\mathcal F_{2m}=\{F_1,\ldots , F_{2m}\}$ , where, for every $h=1,\ldots , 2m$ , we defined $F_h=\{j\in \{1, \ldots , 2n\} : j\equiv h\ \ \mod 2m\}$ ; we can then assume that $G_A$ preserves this specific partition.

For $h=1, \ldots , m$ we let $E_h=\{ j\in \{1, \ldots , 2n\} : j\equiv h\ \ \mod m \}$ . Then $\mathcal E_m=\{E_1, \ldots , E_m\}$ gives another partition of $\{1,\ldots , 2n\}$ and we have that $E_h=F_h \cup F_{m+h}$ . We start by proving the following lemma.

Lemma 7.1. Suppose $G_A$ preserves the $2m$ -partition $\mathcal F_{2m}$ , then $G_A$ preserves the m-partition $\mathcal E_m$ . More precisely, the action of $\sigma _0,\sigma _{\infty },\sigma _1,\tau _1, \ldots ,\tau _{k}$ on $\mathcal F_{2m}$ and $\mathcal E_m$ is the following:

(7.1) $$ \begin{align} \begin{cases} \sigma_{\infty}(F_1)=F_{2m} \mbox{ and } \sigma_{\infty}(F_i)= F_{i-1} \ \forall{i=2,\ldots, 2m};\\ \sigma_1(F_{2m})=F_{2m} \mbox{ and } \sigma_1(F_i)=F_{2m-i}\ \forall i=1, \ldots, 2m-1; \\ \sigma_0(F_i)=F_{2m-i+1} \quad \forall i=1,\ldots, 2m; \\ \tau_j(F_i)=F_i \ \forall i=1, \ldots, 2m \mbox{ and } j=1,\ldots, k. \end{cases} \end{align} $$

and:

$$ \begin{align*} \begin{cases} \sigma_{\infty}(E_1)=E_{m} \mbox{ and } \sigma_{\infty}(E_i)= E_{i-1} \ \forall{i=2,\ldots, m};\\ \sigma_1(E_m)=E_{m} \mbox{ and } \sigma_1(E_i)=E_{m-i}\ \forall i=1, \ldots, m-1; \\ \sigma_0(E_i)=E_{m-i+1}\quad \forall i=1,\ldots, m; \\ \tau_j(E_i)=E_i \ \forall i=1, \ldots, m \mbox{ and } j=1,\ldots, k. \end{cases} \end{align*} $$

In particular, by equation (7.1), $\mathrm {Fix}(\sigma _1):=\{a\in \{1,\ldots , 2n\}\ |\ \sigma _1(a)=a\}\subseteq E_m$ .

Proof. By construction $\sigma _1(2n)=2n$ , thus we have $\sigma _1(F_{2m})=F_{2m}$ . Moreover, we have that $|F_h|=\frac {n}{m} \ge d$ for all h, and for all i at most $2d-2$ indexes are not fixed by $\tau _i$ . This implies that $\tau _i(F_h)=F_h$ for every $i=1, \ldots , k$ and $h=1, \ldots , 2m$ .

In what follows the indexes of the $F_h$ are considered modulo $2m$ . Using $\sigma _0\sigma _{\infty }\sigma _1\prod _{i=1}^{k}\tau _i=\mathrm {id}$ , we deduce that $\sigma _0\sigma _{\infty }\sigma _1(F_h)=F_h$ for every $h=1, \ldots , 2m$ . As we know that $\sigma _{\infty }( F_{h})=F_{h-1}$ , this gives that

(7.2) $$ \begin{align} \begin{cases} \sigma_0(F_{i_1})=F_{i_2} \ \Rightarrow\ \sigma_1(F_{i_2-1})=F_{i_1}; \\ \sigma_1(F_{j_1})=F_{j_2} \ \Rightarrow\ \sigma_0(F_{j_2})=F_{j_1+1}. \end{cases} \end{align} $$

Moreover, we claim that, if $\sigma _1(F_{k_1})=F_{k_2}$ for some indexes $k_1$ and $k_2$ , we must have that $\sigma _1(F_{k_2})=F_{k_1}$ . Indeed, suppose by contradiction that this is not the case; since the nontrivial cycles appearing in $\sigma _1$ have even length, there exist pairwise distinct $k_1, k_2,k_3, k_4\in \{1,\dots , 2m\}$ such that $\sigma _1(F_{k_1})=F_{k_2}$ , $\sigma _1(F_{k_2})=F_{k_3}$ and $\sigma _1(F_{k_3})=F_{k_4}$ . Now, each of the $n/m$ elements of $F_{k_1}$ gives a contribution of at least $3$ to the branching above 1, where we recall that we count the branching of $\phi _A$ above a point p as the sum of $e-1$ over the ramification indices e of the preimages of p.

Using moreover that $\sigma _1$ fixes $2d$ points, we have that the branching above $1$ must be at least $3\frac {n}{m}+(n-2\frac {n}{m}-d)\ge n$ since $\frac {n}{m}\ge d$ , and this is impossible as the branching above $1$ can be at most $n-1$ (recall properties 1–7 after Theorem 2.3).

Therefore, we can rewrite equation (7.2) as

(7.3) $$ \begin{align} \begin{cases} \sigma_0(F_{i_1})=F_{i_2} \ \Rightarrow\ \sigma_1(F_{i_1})=F_{{i_2}-1}; \\ \sigma_1(F_{j_1})=F_{j_2} \ \Rightarrow\ \sigma_0(F_{j_2})=F_{j_1+1}. \end{cases} \end{align} $$

Now, since $\sigma _1(F_{2m})=F_{2m}$ , using equation (7.3) we have that

$$ \begin{align*}\sigma_1(F_{2m})=F_{2m} \quad \mbox{ and } \quad \sigma_1(F_i)=F_{2m-i}\ \forall i=1, \ldots, 2m-1,\end{align*} $$

and

$$ \begin{align*}\sigma_0(F_i)=F_{2m-i+1} \quad \forall i=1,\ldots, 2m.\end{align*} $$

Now, for every $h=1,\ldots , m$ , the set $E_h$ is equal to $F_h \cup F_{m+h}$ . Using the previous relations, we have that $\sigma _1(E_h)=F_{2m-h}\cup F_{m-h}=E_{m-h}$ ; $\sigma _0(E_h)=F_{2m-h+1}\cup F_{m-h+1}=E_{m-h+1}$ , $\tau _i$ stabilizes $\{E_1, \ldots , E_m\}$ for every $i=1, \ldots , d-1$ and $\sigma _{\infty }(E_h)=F_{h-1}\cup F_{m+h-1}=E_{h-1}$ , which concludes the proof.

For $G < S_{2n}$ and $C_1,\ldots , C_{\ell }\subseteq \{1, \ldots , 2n\}$ , we define

$$ \begin{align*}\mathrm{Stab}_G(C_1, \ldots, C_{\ell}):=\bigcap_{i=1}^{\ell} \mathrm{Stab}_G(C_i),\end{align*} $$

where, for every $i=1, \ldots , \ell $ ,

$$ \begin{align*}\mathrm{Stab}_G(C_i)=\{\sigma\in G : \sigma(C_i)=C_i \}.\end{align*} $$

First, we prove this preliminary lemma.

Proof. As by assumption $G_A$ preserves $\mathcal {F}_{2m}$ , by Lemma 7.1 it preserves also $\mathcal {E}_m$ , hence it induces an action on $\{F_1, \ldots , F_{2m}\}$ and on $\{E_1, \ldots , E_m\}$ .

Let us define $H_1:=\mathrm {Stab}_{G_A}(F_{2m})$ and $H_2:=\mathrm {Stab}_{G_A}(E_m)$ . As $2n \in F_{2m} \subseteq E_m$ , by Lemma 7.1 it follows that $S\subseteq H_1 \subseteq H_2$ .

The group $G_A$ acts transitively on $\{1,\ldots , 2n\}$ ; therefore, the action on $\{F_1, \ldots , F_{2m}\}$ and on $\{E_1, \ldots , E_m\}$ is transitive as well, so the orbits of $F_i$ and $E_j$ have cardinality $2m$ and m, respectively. This implies that $|G_A|/|H_2|=|G_A\cdot E_m|=m$ and $|G_A|/|H_1|=|G_A\cdot F_{2m}|=2m$ , where $G_A \cdot C$ denotes the orbit of C with respect to the action of $G_A$ . This proves that $[G_A:H_2]=m$ and $[H_2:H_1]=2$ as wanted.

We now prove the second part of the statement.

For $m>2$ let us define $K:= \mathrm {Stab}_{G_A}(E_1, \ldots , E_m)$ . Note that $K \subseteq H_2$ and, if $m>2$ , then $K\neq H_2$ (as $\sigma _1 \in H_2$ and $\sigma _1 \not \in K$ ). To conclude the proof, we have to show that K is a normal subgroup of $G_A$ and $G_A/K\cong D_{2m}$ .

First, let us prove that $K\lhd G_A$ . Take $h\in K$ and $g\in G_A$ , then, if $g(E_i)=E_{j_i}$ for some $j_i$ , we have that

$$\begin{align*}g^{-1}hg(E_i)=g^{-1}h(E_{j_i})=g^{-1}(E_{j_i})=E_i \quad \mbox{for all } i=1,\ldots, m, \end{align*}$$

so $g^{-1}hg \in K$ as wanted.

Finally, we show that $G_A/K\cong D_{2m}$ . As $G_A$ induces an action on $\{E_1,\ldots , E_m\}$ , we can define a homomorphism $\varphi : G_A \rightarrow S_m$ given by $\varphi (\alpha )=\beta $ where $\beta \in S_m$ is defined by $\beta (i)=j$ if $\alpha (E_i)=E_j$ . Since $K=\mathrm {Stab}_{G_A}(E_1, \ldots , E_m)$ , then $\varphi $ induces an injective homomorphism $\tilde \varphi : G_A/K \rightarrow S_m$ . We want now to prove that $\tilde \varphi (G_A/K)\cong D_{2m}$ . By Lemma 7.1, $\tilde \varphi (\tau _i)=\mathrm {id}$ for every $i=1, \ldots , k$ , so $\tilde \varphi (G_A/K)$ is generated by the images of $\sigma _{\infty }$ and $\sigma _1$ , which we denote by r and s, respectively. We have to prove that $r^m=\mathrm {id}$ , $s^2=\mathrm {id}$ and $srsr=\mathrm {id}$ . Notice that, if $\sigma _{\infty }=(2n, 2n-1, \ldots , 1)$ , then $r=(m,m-1, \ldots , 1)$ , so $r^m=\mathrm {id}$ . Moreover, by Lemma 7.1, s is a product of transpositions, so it has order $2$ as wanted. Let us finally prove that $srsr=\mathrm {id}$ . By Lemma 7.1 we have that $\sigma _1(E_m)=E_m$ , $\sigma _1(E_i)=E_{m-i}$ for all $i=1,\dots , m-1$ and $\sigma _{\infty }(E_1)=E_{m}$ and $\sigma _{\infty }(E_j)=E_{j-1}$ for all $j=2, \ldots , m$ . So, for every $i=1,\ldots , m$ ,

$$\begin{align*}srsr(E_i)=rsr(E_{m-i})=sr(E_{m-i-1})=r(E_{i+1})=E_i. \end{align*}$$

This shows that $G_A/K\cong D_{2m}$ , concluding the proof.

Using the previous lemma, we can finally prove the following result about the polynomial decomposition of $\phi _A$ , which concludes the proof of Theorem 6.2.

Proof. By Lemma 7.2, there exists a chain of groups

$$ \begin{align*}\mathrm{Stab}_{G_A}(\{2n\}) < H_1 < H_2 < G_A,\end{align*} $$

where $[G_A:H_2]=m$ and $[H_2:H_1]=2$ , where we recall that $H_1=\mathrm {Stab}_{G_A}(F_{2m})$ and $H_2=\mathrm {Stab}_{G_A}(E_m)$ . By Galois theory, this implies that there exists a tower of subfields of $\mathbb {C}(t)$

$$ \begin{align*}T:=\mathbb{C}(\phi_A) \subset L_2 \subset L_1 \subset \mathbb{C}(t),\end{align*} $$

with $[L_2:\mathbb {C}(\phi _A)]=m$ and $[L_1:L_2]=2$ . Together with Proposition 5.1, this implies that

$$ \begin{align*}\phi_A=h_2 \circ h_1 \circ z,\end{align*} $$

where $h_1,h_2$ and z are polynomials with $\deg h_1=2$ and $\deg h_2=m$ .

Let us first prove that $h_2$ is linearly equivalent to $f_m$ . If $m=2$ , this is trivial since polynomials of degree $2$ are linearly equivalent to each other, so let us assume that $m>2$ . By Lemma 7.2, the subgroup $K=\mathrm {Stab}_{G_A}(E_1, \ldots , E_m)$ is a normal subgroup of $G_A$ contained in $H_2$ and such that $G_A/K\cong D_{2m}$ . Then K corresponds to a field $F\subseteq \mathbb {C}(t)$ containing $L_2$ and such that $F/T$ is a normal extension. We want to show that $F/T$ is the Galois closure of $L_2/T$ .

As the action of $G_A$ on $\{E_1, \ldots , E_m\}$ is transitive, the subgroups $\mathrm {Stab}_{G_A}(E_1), \ldots , \mathrm {Stab}_{G_A}(E_m)$ are exactly all the conjugates of $\mathrm {Stab}_{G_A}(E_m)=\text {Gal}(\mathbb {C}(t)/L_2)$ in $G_A$ . Therefore, the Galois closure of $L_2/T$ will be equal to the subfield of $\mathbb {C}(t)$ corresponding to $\mathrm {Stab}_{G_A}(E_1)\cap \ldots \cap \mathrm {Stab}_{G_A}(E_m)=K$ that is F as wanted.

As by construction $\text {Gal}(F/K)=\text {Mon}(h_2)\cong D_{2m}$ , by Proposition 5.3 we have that the polynomial $h_2$ is linearly equivalent to the Chebychev polynomial $T_m$ , and so to $f_m$ as proved in Section 5. This gives that there exist linear polynomials $\alpha _1$ and $\alpha _2$ such that

$$\begin{align*}\phi_A=\alpha_1\circ f_m \circ \alpha_2 \circ h_1 \circ z= \alpha_1\circ f_m \circ h_3 \circ z, \end{align*}$$

where $h_3=\alpha _2 \circ h_1$ is a polynomial of degree $2$ .

We want to prove that $\phi _A= f_m \circ \phi \circ A'$ where $ A'$ fits in the same Pell–Abel equation of A.

For the rest of the proof, as we are dealing with polynomials, we only consider branching at finite points.

By Section 4, $f_m$ has two branch points, namely $0$ and $1$ . As $\alpha _1$ is a linear map, $\alpha _1\circ f_m$ will be branched only at $\xi _0=\alpha _1(0)$ and $\xi _1=\alpha _1(1)$ with the same ramification indexes. As the branch points of $\alpha _1\circ f_m$ lie among the ones of $\phi _A$ , this means that $\xi _0, \xi _1 \in \{0,1,b_1, \ldots , b_{k}\}$ . We now consider two cases and write $\alpha _1(t)=pt+q$ . We recall that in Proposition 4.1 we gave a characterization of the ramification behavior of $f_m$ , and we are going to use this for the rest of the proof.

If m is even, as 0 is the only point with no simple preimage via $\phi _{A}$ , we necessarily have $\xi _0=0$ and thus $q=0$ . If $m>2$ , we have that the branching above $\xi _1$ is at least $(m-2)/2 \cdot (n/m)\geq (m/2-1)d>d-1$ while above any of $b_1,\dots , b_k$ is at most $d-1$ . This forces $\xi _1=1$ and $p=1$ . If $m=2$ , we have $\alpha _1\circ f_2=p (2t-1)^2=(2(\sqrt {p} t +(1-\sqrt {p})/2)-1)^2=f_2\circ (\sqrt {p} t +(1-\sqrt {p})/2)$ . In any case, we have or we may suppose that $\alpha _1=t$ .

Assume now m is odd; then the branching above $\xi _0$ is at least $(m-1)/2 \cdot (n/m)\geq d>d-1$ and the same for $\xi _1$ . As before, above $b_1,\dots , b_k$ the branching is at most $d-1$ . This implies that $\{\xi _0,\xi _1\} = \{0,1\}$ . In particular, if $\xi _0=0$ and $\xi _1=1$ , then $\alpha _1=t$ while if $\xi _0=1$ and $\xi _1=0$ we have $\alpha _1=1-t$ . We can reduce to the first case by noticing that $(1-t) \circ f_m = f_m \circ (1-t)$ .

Finally, we have

$$ \begin{align*} \phi_A= f_m \circ h_3 \circ z, \end{align*} $$

where $h_3$ is a polynomial of degree $2$ .

We write $w=h_3 \circ z$ and let r be the unique finite branch point of $h_3$ . We claim that $r\in \{0,1\}$ . Suppose not; the fact that $f_m(\{0,1\})\subseteq \{0,1\}$ and that $\phi _A$ fits in a Pell–Abel equation implies that $w^{-1}(\{0,1\}) $ contains exactly $2d$ simple preimages, and thus the sum of the branching above 0 and 1 of w is at least $2(n/m)-d$ . On the other hand, the branching of w above r is at least $n/m$ making the whole branching to be at least $3(n/m)-d\geq 2(n/m)=\deg w$ . This contradicts the Riemann–Hurwitz formula, and we have $r\in \{0,1\}$ .

If m is odd, then we must have $r=0$ because $1$ , being the only simple preimage of 1 via $f_m$ must have simple preimages via w.

If m is even and $r=1$ , we observe that $f_m=f_m\circ (1-t) $ . We then can simply replace $h_3$ by $(1-t)\circ h_3$ , and thus we may assume $r=0$ .

In any case, we have $h_3=st^2=(\sqrt {s}t)^2$ and then, possibly composing with a suitable linear polynomial on the right, we can assume that $h_3$ is exactly the square function $\phi $ , so

$$ \begin{align*}\phi_A= f_m \circ \phi \circ A'.\end{align*} $$

We are left with proving that $A'$ is another solution of the same Pell–Abel equation, that is, that $\phi _{A'}-1=D B^{\prime 2}$ for some polynomial $B'$ . Notice that, since A is a solution of the Pell–Abel equation, $\phi _A^{-1}(1)$ contains exactly $2d$ simple points $\{r_1, \ldots , r_{2d}\}$ (the zeros of D); as $1$ (and $0$ if m is even) is the only point in $f_m^{-1}(1)$ with ramification index $1$ and $\phi _{A'}=\phi \circ A'$ is a square, then $\{r_1, \ldots r_{2d}\} \subset \phi _{A'}^{-1}(1)$ . This shows that $(\phi _{A'}-1)/D={B'}^2$ for some polynomial $B'$ as wanted.

Corollary 7.4. Let m be a positive integer. Let $(A,B)$ be a solution of degree n of the Pell–Abel equation $A^2-DB^2=1$ , and, for $i=1, \ldots , 2m$ , let $F_i=\{j\in \{1, \ldots , 2n\}\ |\ j\equiv i\ \ \mod 2m\}$ , and, after conjugating, assume that $\sigma _{\infty }=(2n, 2n-1, \ldots , 1)$ and that one of the points which are fixed by $\sigma _1$ and by all the $\tau _i$ is $2n$ . Then $(A,B)$ is the $m^{th}$ -power of a solution $(A', B')$ of the same Pell–Abel equation if and only if the permutations $\sigma _{\infty }, \sigma _0, \sigma _1, \tau _1, \ldots , \tau _{k}$ associated to A satisfy the following conditions:

(7.4) $$ \begin{align} \begin{cases} \sigma_{\infty}(F_1)=F_{2m} \mbox{ and } \sigma_{\infty}(F_h)= F_{h-1} \quad \forall{h=2,\ldots, 2m};\\ \sigma_1(F_{2m})=F_{2m} \mbox{ and } \sigma_1(F_h)=F_{2m-h}\quad \forall h=1, \ldots, 2m-1; \\ \sigma_0(F_i)=F_{2m-i+1} \quad \forall i=1,\ldots, 2m; \\ \tau_j(F_i)=F_i \quad \forall i=1, \ldots, 2m \mbox{ and } j=1,\ldots, k. \end{cases} \end{align} $$

8 Counting the conjugacy classes of permutations

In this section, we want to show how to apply Theorem 2.3 and use combinatorial arguments to count equivalence classes of Pellian polynomials with a solution of degree n.

As discussed in the introduction, the advantage of the combinatorial argument is that it allows to compute the precise number instead of only an asymptotic formula, and this can be done algorithmically given n and d. On the other hand, the combinatorics becomes more difficult as soon as d grows due to the number of different configurations that the permutations can have. For this reason, in this section we are going to stick to the case $d=2$ and to maximal branching, that is when the map $\phi _A$ is branched over $0$ , $1$ , $\infty $ and over another point which we will denote by b. In this case, using the arguments of Section 2 and the Riemann–Hurwitz formula, if we count the branching of $\phi _A$ as the sum of $e-1$ over the ramification indices e, we have that, above $0$ the branching is exactly n, above $1$ is $n-1$ , above $\infty $ it is $2n-1$ and above b is $1$ . This implies that the cycle decomposition of $\sigma _0, \sigma _{\infty }, \sigma _1$ and $\tau _1=:\tau $ is fixed, and we have that:

1. 1 - $\sigma _0$ is the product of n disjoint transpositions;

2. 2 - $\sigma _{\infty }$ is a $2n$ -cycle;

3. 3 - $\sigma _1$ is the product of $n-1$ disjoint transpositions;

4. 4 - $\tau $ is a transposition.

Moreover, we know that $\sigma _0, \sigma _{\infty }, \sigma _1$ and $\tau $ satisfy $\sigma _0 \sigma _{\infty }\sigma _1\tau =$ id.

As said before each conjugacy class of $4$ -tuples $(\sigma _0,\sigma _{\infty }, \sigma _1, \tau )$ has at least one representative with $\sigma _{\infty }=(2n,2n-1, \ldots , 1)$ since $\sigma _{\infty }$ is a $2n$ -cycle and one can relabel the element using a suitable conjugation. Moreover, there are at least two indexes that are fixed by $\sigma _1$ and by $\tau $ ; without loss of generality we will always assume that one of the two indexes is $2n$ . We will call such a representative special.

Now, since $\sigma _1$ is the product of $n-1$ disjoint transpositions and $\tau $ is a transposition, we have three different cases:

• • $\sigma _1$ and $\tau $ are disjoint permutations;

• • $\sigma _1\tau $ is the product of $n-2$ transpositions and a $3$ -cycle;

• • $\sigma _1\tau $ is the product of $n-3$ transpositions and a $4$ -cycle.

In what follows, we are going to study the special $4$ -tuples, analysing in particular how many special $4$ -tuples we have in each conjugacy class. As the conjugation preserves the cycle decomposition, we have to analyse these three cases separately.

8.1 The disjoint case

Let $\Sigma $ be a special $4$ -tuple, that is, $\sigma _{\infty }=(2n, 2n-1, \ldots , 1)$ and $\sigma _1$ and $\tau $ both fix $2n$ . Suppose moreover that $\sigma _1$ and $\tau $ are disjoint permutations.

Since $2n$ is fixed by both $\sigma _1$ and $\tau $ and $\sigma _0\sigma _{\infty }\sigma _1\tau =$ id, then $2n$ is fixed by $\sigma _0\sigma _{\infty }$ ; moreover, as $\sigma _0\sigma _{\infty }\sigma _1\tau =$ id, we must have that $\sigma _0(2n)=1$ . Since $\sigma _0$ is composed only by transpositions, we have also that $\sigma _0(1)=2n$ , hence $\sigma _1\tau (2n-1)=1$ . Now as $\sigma _1\tau $ is composed only by transpositions, we have that $\sigma _1\tau (1)=2n-1$ , so $\sigma _0(2n-1)=2$ and so on. Going on with this argument, we have that $\sigma _0$ must have the form $(1,2n)(2,2n-1) \cdots (n,n+1)$ and $\sigma _0\sigma _{\infty }$ fixes also n. In this case, $\sigma _1 \tau =(1,2n-1)(2,2n-2) \cdots (n-1, n+1)$ and we may choose as $\tau $ any of the $n-1$ transpositions $(1,2n-1),(2,2n-2), \cdots , (n-1, n+1)$ .

Now, assume that $\Sigma '=(\sigma _0',\sigma _{\infty },\sigma _1',\tau ')$ is another special $4$ -tuple lying in the same conjugacy class of $\Sigma $ , that is, $\Sigma '=\gamma ^{-1}\Sigma \gamma $ for some $\gamma \in S_{2n}$ ; then, since by assumption $\sigma _{\infty }$ has to be fixed by the conjugation, we have that $\gamma $ will be a power of $\sigma _{\infty }$ . Furthermore, since we want $\sigma _1'$ and $\tau '$ both to fix $2n$ , then the conjugation has to permute $2n$ with n (which is the only other point fixed by $\sigma _1$ and $\tau $ ); consequently, $\gamma =\sigma _{\infty }^n$ . Notice finally that, given a special $4$ -tuple $\Sigma $ , we have $\sigma _{\infty }^{-n}\Sigma \sigma _{\infty }^n= \Sigma $ if and only if n is even and $\tau =\left ( \frac {n}{2}, 2n-\frac {n}{2} \right )$ .

This implies that:

• • if n is odd, then we have $\frac {1}{2}(n-1)$ conjugacy classes, because every conjugacy class contains two special $4$ -tuples;

• • if n is even, then we have $\frac {1}{2}n$ conjugacy classes because every conjugacy class contains exactly two special $4$ -tuples, except for the conjugacy class of the $4$ -tuple with $\tau =\left (\frac {n}{2}, 2n-\frac {n}{2}\right )$ which contains only one special $4$ -tuple.

In general we have

(8.1) $$ \begin{align} \#\ \textrm{conjugacy classes}= \left \lfloor \frac{n}{2} \right \rfloor. \end{align} $$

We point out that, in the disjoint case, Corollary 7.4 gives an easier way to detect whether the solution $(A,B)$ is primitive by looking only at the corresponding $\tau $ of a special representative.

Proposition 8.1. Consider the $4$ -tuple $\Sigma =(\sigma _0,\sigma _{\infty },\sigma _1,\tau )$ with $\sigma _{\infty }=(2n,2n-1,\ldots ,1)$ and $\sigma _0=(1,2n)(2,2n-1)\ldots (n,n+1)$ associated to a solution $(A,B)$ of a Pell–Abel equation. Then $(A,B)$ is the m-th power of another solution if and only if $\tau =(h,2n-h)$ with $m\mid h$ .

In particular, there are exactly $\varphi (n)/2$ equivalence classes corresponding to primitive solutions.

Proof. By Corollary 7.4, $\Sigma $ corresponds to an m-th power if and only if equation (7.4) holds. The conditions on $\sigma _0$ , $\sigma _\infty $ are always satisfied. The condition on $\tau $ is satisfied if and only if $\tau =(h,2n-h)$ with $h \equiv 2n-h\equiv -h\ \ \mod {2m}$ , that is, $m\mid h$ . Finally, if $m\mid h$ , also $\sigma _1$ satisfies the conditions of equation (7.4), proving the claim.

8.2 The 3-cycle case

Let $\Sigma =(\sigma _0,\sigma _{\infty },\sigma _1,\tau )$ be a special $4$ -tuple, that is, with $\sigma _{\infty }=(2n, \ldots , 1)$ and such that $\sigma _1$ and $\tau $ fix both $2n$ , and assume that $\sigma _1\tau $ decomposes as a product of $n-3$ transpositions and a 3-cycle.

First, we are going to prove the following result that describes the possible shapes of $\sigma _0$ and $\sigma _1\tau $ of such a $4$ -tuple.

Proposition 8.2. Let $\Sigma =(\sigma _0,\sigma _{\infty }, \sigma _1, \tau )$ be a special $4$ -tuple such that $\sigma _1\tau $ contains a $3$ -cycle. Then there exist $h,k$ with $1\le h\leq n-2$ , $h< k <2n-h$ and $h\equiv k\ \ \mod 2$ such that

(8.2) $$ \begin{align} \sigma_0 &=\prod_{i=1}^h(i,2n+1-i) \prod_{j=1}^{\frac{k-h}{2}}(h+j, k+1-j) \prod_{t=1}^{\frac{2n-h-k}{2}}(k+t, 2n-h+1-t);\\ \sigma_1\tau & =\prod_{i=1}^{h-1}(i,2n-i)\prod_{j=1}^{\frac{k-h}{2}-1} (h+j, k-j) \prod_{t=1}^{\frac{2n-h-k}{2}}(k+t,2n-h-t)\ (2n-h, h, k). \nonumber \end{align} $$

In particular, $\sigma _1\tau $ fixes 3 indexes: $2n, \frac {k+h}{2}$ and $\frac {2n-h+k}{2}.$ Moreover, there are three different $\Sigma $ satisfying equation (8.2) corresponding to different choices of $\tau $ , that is, $\tau \in \{(h,k), (h, 2n-h), (k, 2n-h)\}$ .

Proof. Let $\Sigma =(\sigma _0,\sigma _{\infty },\sigma _1,\tau )$ be a special $4$ -tuple and assume that $\sigma _1\tau $ is a product of $n-3$ disjoint transpositions and a disjoint 3-cycle. In this case, $\sigma _1\tau $ fixes $3$ indexes, and one of them is $2n$ . Then, since $\sigma _0\sigma _{\infty }\sigma _1\tau =\mathrm {id}$ , we must have that $\sigma _0(2n)=1$ . As $\sigma _0$ is the product of n disjoint transpositions, we have also that $\sigma _0(1)=2n$ , hence $\sigma _1\tau (2n-1)=1$ . Now, we have two choices: Either $\sigma _1\tau (1)=2n-1$ (so $(2n-1,1)$ is one of the disjoint transpositions in the decomposition of $\sigma _1\tau $ ), or $\sigma _1\tau (1)=k\neq 2n-1$ (giving rise to the 3-cycle in the product). We can go on as in the disjoint case described in the previous section until we reach the situation in which there exist $1\le h\le n-2$ and $h+1\le k\le 2n-h-1$ such that $\sigma _0$ contains the disjoint transpositions $(1, 2n),(2,2n-1)\cdots (h,2n-h+1)$ and $\sigma _1\tau $ contains the disjoint transpositions $(1,2n-1)\ldots , (2n-h+1,h-1)$ and the 3-cycle $(2n-h,h,k)$ .

Notice that $k\neq h+1, 2n-h-1$ , otherwise $\sigma _0\sigma _{\infty }\sigma _1\tau =\mathrm {id}$ would imply that $\sigma _0$ fixes $h+1$ (or in the second case $2n-h-1$ ), which is impossible because $\sigma _0$ has no fixed points. So, $h+1 <k< 2n-h-1$ . Now, if $\sigma _1\tau (h)=k$ , as $\sigma _0\sigma _{\infty }\sigma _1\tau =\mathrm {id}$ , we have $\sigma _0(k)=h+1$ and so $\sigma _0(h+1)=k$ . This means that $\sigma _1\tau (h+1)=k-1$ and $\sigma _1\tau (k-1)=h+1$ and so on. This process ends when one reaches the conditions $\sigma _0 \left( \frac {k+h}{2} \right)=\frac {k+h}{2}+1$ and vice versa, $\sigma _1\tau \left(\frac {k+h}{2}-1 \right)=\frac {k+h}{2}+1$ and vice versa and $\frac {k+h}{2}$ is fixed by $\sigma _1\tau $ . Notice that this gives the additional condition that $h\equiv k\ \ \mod 2$ .

On the other hand, if $\sigma _1\tau $ contains the 3-cycle $(2n-h,h,k)$ , we also have that $\sigma _1\tau (k)=2n-h$ , giving other conditions to satisfy. In fact, $\sigma _0\sigma _{\infty }\sigma _1\tau =\mathrm {id}$ implies $\sigma _0(2n-h)=k+1$ and $\sigma _0(k+1)=2n-h$ . This means, as before, that $\sigma _1\tau (2n-h-1)=k+1$ and vice versa, and so on. This process ends when one reaches the conditions $\sigma _0 \left( \frac {2n-h+k}{2}\right)=\frac {2n-h+k}{2}+1$ and viceversa, $\sigma _1 \tau \left( \frac {2n-h+k}{2}-1\right)=\frac {2n-h+k}{2}+1$ and $\frac {2n-h+k}{2}$ is fixed by $\sigma _1\tau $ , proving the first part of the assertion.

Finally, let us denote by

$$\begin{align*}\alpha_{hk}:= \sigma_1\tau (2n-h, h, k)^{-1}, \end{align*}$$

that is, the product of the $n-2$ disjoint transpositions that appear in the decomposition of $\sigma _1\tau $ .

Notice that, if $\sigma _1\tau $ contains the $3$ -cycle $(2n-h, h, k)$ , then we have three choices for $\sigma _1$ and $\tau $ , namely $\sigma _1=\alpha _{hk} (2n-h, h)$ and $\tau =(2n-h,k)$ or $\sigma _1=\alpha _{hk} (h,k)$ and $\tau =(h, 2n-h)$ or $\sigma _1=\alpha _{hk}(k, 2n-h)$ and $\tau =(k,h)$ , giving three different special $4$ -tuples as wanted.

We now want to count how many special $4$ -tuples are contained in each conjugacy class. First of all, notice that if we have two special $4$ -tuples $\Sigma =(\sigma _0,\sigma _{\infty },\sigma _1,\tau )$ and $\Sigma '=(\sigma _0',\sigma _{\infty }',\sigma _1',\tau ')$ with $\sigma _{\infty }=\sigma _{\infty }'=(2n, 2n-1, \ldots , 1)$ and such that $\Sigma '=\gamma ^{-1} \Sigma \gamma $ for some $\gamma \in S_{2n}$ , then $\gamma $ must be equal to a power of $\sigma _{\infty }$ . We will consider this conjugation from a geometric point of view; namely, if we consider the regular $2n$ -gon with vertices labelled by $1,2,\ldots , 2n$ , then conjugating by a permutation of the form $\sigma _{\infty }^\ell $ with $\ell \in \{1,\ldots ,2n\}$ corresponds to a rotation of angle $\frac {\pi }{n} \ell $ .

Let us denote by $p:=\frac {k+h}{2}$ and $q:=\frac {2n-h+k}{2}$ the two points different from $2n$ fixed by $\sigma _1\tau $ . Then, if we conjugate the set $\{2n, q, p\}$ by a power of $\sigma _{\infty }$ , it corresponds to a rotation of the triangle of vertices $\{2n, q, p\}$ . As we want $\Sigma '$ to be special, then $\sigma _1'\tau $ has to fix $2n$ , hence we are interested in the rotations which send one of the vertices to $2n$ . This means that we can conjugate only by $\sigma _{\infty }^{2n-p}$ or by $\sigma _{\infty }^{2n-q}$ . In the first case, the set of fixed points $\{2n, p, q\}$ is sent to $\{2n, 2n-p, q-p\}$ , while, in the second case, it is sent to $\{2n, 2n-(q-p), 2n-q\}$ .

We have the following result:

Proposition 8.3. Let $\Sigma $ be a special $4$ -tuple $(\sigma _{\infty }, \sigma _0,\sigma _1,\tau )$ such that $\sigma _1\tau $ contains the $3$ -cycle $(2n-h,h,k)$ . Then the conjugacy class of $\Sigma $ contains exactly three special $4$ -tuples.

Proof. Since the $4$ -tuple $\Sigma =(\sigma _{\infty }, \sigma _0,\sigma _1,\tau )$ is special, we have that $\sigma _{\infty }=(2n, 2n-1, \ldots , 1)$ and $\sigma _1\tau $ fixes $2n$ ; moreover, by assumption it contains the $3$ -cycle $(2n-h,h,k)$ . Then, as proved before, $\sigma _0$ and $\sigma _1\tau $ will have the shape (8.2). As discussed before, if $\Sigma '\neq \Sigma $ is a special $4$ -tuple conjugated to $\Sigma $ , then $\Sigma '=\gamma ^{-1}\Sigma \gamma $ with $\gamma \in \left \{ \sigma _{\infty }^{2n-\frac {k+h}{2}}, \sigma _{\infty }^{2n-\frac {2n-h+k}{2}} \right\}$ . To prove the assertion, we have to show that no $4$ -tuple is fixed by such a conjugation. To do this, we consider two cases.

Assume first that $3 \mid n$ and that the $3$ -cycle contained in $\sigma _1\tau $ is $ \left(\frac {n}{3}, \frac {5}{3}n, n \right)$ , that is, $(h,k)= (\frac {n}{3}, n)$ . By the description above, $\tau \in \left\{ \left(\frac {n}{3}, n \right), \left(n, \frac {5}{3}n \right), \left(\frac {n}{3}, \frac {5}{3} n \right) \right\}$ and $\sigma _1\tau $ fixes $ \left\{\frac {2n}{3}, \frac {4n}{3},2n \right\}$ (which are the vertices of an equilateral triangle). Let us assume without loss of generality that $\tau = \left(\frac {n}{3}, n \right)$ . If we conjugate $\Sigma $ by $\gamma \in \left\{\sigma _{\infty }^{\frac {2n}{3}},\sigma _{\infty }^{\frac {4n}{3}} \right\}$ , then $\gamma ^{-1}\sigma _0 \gamma =\sigma _0$ and $\gamma ^{-1}(\sigma _1\tau ) \gamma =\sigma _1\tau $ ; hence, $\Sigma $ will be conjugated to $\Sigma '= \left(\sigma _\infty , \sigma _0, \sigma _1', \left(\frac {5}{3}n, n \right) \right)$ and $\Sigma "= \left(\sigma _\infty , \sigma _0, \sigma _1", \left(\frac {n}{3}, \frac {5}{3} n \right)\right)$ .

Assume now $(h,k)\neq (\frac {n}{3}, n)$ . As the conjugation preserves the disjoint cycle structure, $\sigma _1\tau $ will contain a $3$ -cycle and by construction one of the indexes fixed by $\sigma _1\tau $ is $2n$ , so the conjugated $4$ -tuples will have the shape (8.2). We have only to compute the images of the $3$ -cycles. A direct computation shows that, if we conjugate by $\sigma _{\infty }^{2n-\frac {k+h}{2}}$ we have $\Sigma '$ with $(h',k')= \left( \frac {k-h}{2}, 2n- \frac {3h+k}{2}\right)$ while if we conjugate by $\sigma _{\infty }^{2n-\frac {2n-h-k}{2}}$ we have $\Sigma '$ with $(h',k')= \left( n- \frac {h+k}{2}, n+\frac {3h-k}{2}\right)$ , as wanted.

Using these two propositions we can count the different conjugacy classes of $4$ -tuples such that $\sigma _1\tau $ contains a $3$ -cycle.

Let us consider a special $4$ -tuple $\Sigma =(\sigma _0,\sigma _{\infty },\sigma _1,\tau _1)$ such that $\sigma _1\tau _1$ contains the $3$ -cycle $(2n-h,h,k)$ . By equation (8.2), we have $h\equiv k\ \ \mod 2$ ; moreover, every choice of the couple $(h,k)$ gives rise to three different choices of the couple $(\sigma _1,\tau )$ , and by the previous proposition, each conjugacy class contains exactly three special $4$ -tuples. Consequently, the number of conjugacy classes is equal to the number of different choices of the couple $(h,k)$ , that is, to the number of couples $\{h,k\}$ such that $1\le h\le n-2$ , $h+1<k<2n-h-1$ and $h\equiv k$ (mod $2$ ), hence

$$\begin{align*}\# \textrm{ conjugacy classes}= \sum_{h=1}^{n-2} \frac{2n-2(h+1)}{2}=\sum_{h=1}^{n-2} (n-h-1)= \sum_{j=1}^{n-2} j= \frac{(n-1)(n-2)}{2}. \end{align*}$$

8.3 The 4-cycle case

Let $\Sigma =(\sigma _0,\sigma _{\infty },\sigma _1,\tau )$ be a special $4$ -tuple, that is, with $\sigma _{\infty }=(2n, \ldots , 1)$ and such that $\sigma _1$ and $\tau $ fix both $2n$ , and assume that $\sigma _1\tau $ is a product of $n-4$ disjoint transpositions and a disjoint $4$ -cycle.

As in the previous section, our first result describes the shape of a special $4$ -tuple of this kind.

Proposition 8.4. Let $\Sigma =(\sigma _0,\sigma _{\infty }, \sigma _1, \tau )$ be a special $4$ -tuple such that $\sigma _1\tau $ contains a $4$ -cycle. Then there exist $1\le h< k_1 <k_2<2n-h$ with $h\equiv k_1\equiv k_2\ \ \mod 2$ such that

(8.3) $$ \begin{align} \sigma_0 &=\prod_{i=1}^h(i,2n+1-i) \prod_{j=1}^{\frac{k_1-h}{2}}(h+j, k_1+1-j) \prod_{t=1}^{\frac{k_2-k_1}{2}} (k_1+t, k_2+1-t) \nonumber \\[-2pt] &\prod_{v=1}^{\frac{2n-h-k_2}{2}}(k_2+v, 2n-h+1-v); \\[-2pt] \sigma_1\tau &=\prod_{i=1}^{h-1}(i,2n-i)\prod_{j=1}^{\frac{k_1-h}{2}-1} (h+j, k_1-j) \prod_{t=1}^{\frac{k_2-k_1}{2}-1} (k_1+t, k_2-t) \nonumber \\[-2pt] &\prod_{v=1}^{\frac{2n-h-k_2}{2}-1}(k_2+v,2n-h-v)\ (2n-h, h, k_1, k_2). \nonumber \end{align} $$

In particular $\sigma _1\tau $ fixes four indexes: $2n, \frac {k_1+h}{2}$ , $\frac {k_1+k_2}{2}$ and $\frac {2n-h+k_2}{2}$ . Moreover, there are two different $\Sigma $ satisfying equation (8.3) corresponding to different choices of $\tau $ , that is, $\tau \in \{(h,k_2), (k_1, 2n-h)\}$ .

Proof. Let $\Sigma =(\sigma _0,\sigma _{\infty },\sigma _1,\tau )$ be a special $4$ -tuple, and assume that $\sigma _1\tau $ is a product of $n-4$ disjoint transpositions and a disjoint $4$ -cycle. In this case, $\sigma _1\tau $ fixes $4$ indexes, and one of them is $2n$ . Then, as $\sigma _0\sigma _{\infty }\sigma _1\tau =\mathrm {id}$ , we have that $\sigma _0(2n)=1$ . As $\sigma _0$ is composed only by transpositions, we have also that $\sigma _0(1)=2n$ , hence $\sigma _1\tau (2n-1)=1$ . Now, we have two choices: Either $\sigma _1\tau (1)=2n-1$ (so $(2n-1,1)$ is one of the disjoint transpositions in the decomposition of $\sigma _1\tau $ ), or $\sigma _1\tau (1)=k\neq 2n-1$ (giving rise to the $4$ -cycle in the product). We go on as in the disjoint case described in Subsection 8.1 until we reach the situation in which there exist $1\le h\le n-2$ , $h+1\le k_1\le 2n-h-3$ , $k_1+1\le k_2\le 2n-h-1$ such that $\sigma _0$ contains the disjoint transpositions $(1, 2n),(2,2n-1)\cdots (h,2n-h+1)$ and $\sigma _1\tau $ contains the disjoint transpositions $(1,2n-1),\ldots , (2n-h+1,h-1)$ and the 4-cycle $(2n-h,h,k_1, k_2)$ . It is easy to see that $k_1\neq h+1, 2n-h-3$ and $k_2\neq k_1+1, 2n-h-1$ . For example, let us assume by contradiction that $k_1=h+1$ ; then, as we have that $\sigma _0\sigma _{\infty }\sigma _1\tau _1=\mathrm {id}$ , we would have that $\sigma _0$ fixes $h+1$ , which is impossible because $\sigma _0$ has no fixed points. For the other cases we can argue similarly. So, $h+2 \le k_1 \le 2n-h-4$ and $k_1+2\le k_2\le 2n-h-2$ , which implies that $1\le h\le n-3$ .

Now, if $\sigma _1\tau (h)=k_1$ , then by $\sigma _0\sigma _{\infty }\sigma _1\tau =\mathrm {id}$ we have $\sigma _0(k_1)=h+1$ and so $\sigma _0(h+1)=k_1$ . This means that $\sigma _1\tau (h+1)=k_1-1$ and $\sigma _1\tau (k_1-1)=h+1$ and so on. This process ends when one reaches the conditions $\sigma _0 \left( \frac {k_1+h}{2}\right)=\frac {k_1+h}{2}+1$ and vice versa, $\sigma _1\tau \left(\frac {k_1+h}{2}-1 \right)=\frac {k_1+h}{2}+1$ and vice versa and $\frac {k_1+h}{2}$ is fixed by $\sigma _1\tau $ . Notice that also in this case this gives the additional condition that $h\equiv k_1\ \ \mod 2$ . Here we also see that $k_2$ cannot lie between h and $k_1$ .

On the other hand, if $\sigma _1\tau $ contains the 4-cycle $(2n-h,h,k_1,k_2)$ , we also have that $\sigma _1\tau (k_1)=k_2$ and $\sigma _1\tau (k_2)=2n-h$ , giving other conditions to satisfy. Indeed, by $\sigma _0\sigma _{\infty }\sigma _1\tau =\mathrm {id}$ , we have $\sigma _0(2n-h)=k_2+1$ and $\sigma _0(k_2+1)=2n-h$ . This implies that $\sigma _1\tau (2n-h-1)=k_2+1$ and vice versa, and so on. This process ends when one reaches the conditions $\sigma _0 \left(\frac {2n-h+k_2}{2}\right)=\frac {2n-h+k_2}{2}+1$ and vice versa, $\sigma _1 \tau \left( \frac {2n-h+k_2}{2}-1\right)=\frac {2n-h+k_2}{2}+1$ and $\frac {2n-h+k_2}{2}$ is fixed by $\sigma _1\tau $ . With the same argument, as $\sigma _1\tau (k_1)=k_2$ , by $\sigma _0\sigma _{\infty }\sigma _1\tau =\mathrm {id}$ , we have $\sigma _0(k_1+1)=k_2$ and $\sigma _0(k_2)=k_1+1$ . This means that $\sigma _1\tau (k_1+1)=k_2-1$ and vice versa, and so on. This process ends when one reaches the conditions $\sigma _0 \left( \frac {k_1+k_2}{2} \right)=\frac {k_2+k_2}{2}+1$ and vice versa, $\sigma _1 \tau \left( \frac {k_1+k_2}{2}-1\right)=\frac {k_1+k_2}{2}+1$ and $\frac {k_1+k_2}{2}$ is fixed by $\sigma _1\tau $ . This gives, as before, the additional condition that $k_1\equiv k_2\ \ \mod 2$ . This proves that $\sigma _0$ and $\sigma _1\tau $ have the shape (8.3), proving the first part of the proposition.

In particular $\sigma _1\tau $ fixes 4 indexes, that is, $\frac {k_1+h}{2}$ , $\frac {k_1+k_2}{2}$ and $\frac {2n-h+k_2}{2}$ and $2n$ . Let us denote by

$$\begin{align*}\beta_{hk_1k_2}=: \sigma_1\tau (2n-h, h, k_1, k_2)^{-1}, \end{align*}$$

that is, the product of the disjoint transpositions appearing in the decomposition of $\sigma _1\tau $ . Then we have two different choices for the couple $(\sigma _1,\tau )$ giving the same product, namely $\sigma _1=\beta _{hk_1k_2} (2n-h, h)(k_1,k_2)$ and $\tau =(2n-h,k_1)$ or $\sigma _1=\beta _{hk_1k_2} (h,k_1)(2n-h, k_2)$ and $\tau =(h, k_2)$ . This concludes the proof.

As in the previous section, we want to count the number of special $4$ -tuples such that $\sigma _1\tau $ contains a $4$ -cycle in a given conjugacy class.

Notice that, if we have two special $4$ -tuples $\Sigma =(\sigma _0,\sigma _{\infty },\sigma _1,\tau _1)$ and $\Sigma '=(\sigma _0',\sigma _{\infty }',\sigma _1',\tau _1')$ with $\sigma _{\infty }\kern1.5pt{=}\kern1.5pt\sigma _{\infty }'\kern1.5pt{=}\kern1.5pt(2n, 2n-1, \ldots , 1)$ and such that $\Sigma '\kern1.5pt{=}\kern1.5pt\gamma ^{-1} \Sigma \gamma $ for some $\gamma \kern1.5pt{\in}\kern1.5pt S_{2n}$ , then $\gamma $ must be a power of $\sigma _{\infty }$ .

Let us call $p:= \frac {k_1+h}{2},q:=\frac {k_1+k_2}{2}$ and $r:=\frac {2n-h+k_2}{2}$ the three indexes other than $2n$ fixed by $\sigma _1\tau $ . Since we are interested in conjugations by permutations of the form $\sigma _{\infty }^\ell $ that send the set $\{p,q,r,2n\}$ into a set containing $\{2n\}$ , the only admissible $\gamma $ will be $\sigma _{\infty }^{2n-p}$ , $\sigma _{\infty }^{2n-q}$ or $\sigma _{\infty }^{2n-r}$ . In particular, in the first case the set $\{2n, p, q, r\}$ is sent to $\{2n, 2n-p, q-p, r-p\}$ ; in the second case to $\{2n, 2n-(q-p), r-q, 2n-q\}$ and, in the last case, to $\{2n-r, 2n-(r-p), 2n-(r-q), 2n\}$ .

The following proposition describes how many special $4$ -tuples we have in every conjugacy class of a $4$ -tuple; in this case, this depends on the configuration of $\{p,q,r,2n\}$ .

Proposition 8.5. Let $\Sigma =(\sigma _0, \sigma _{\infty },\sigma _1,\tau _1)$ be a special $4$ -tuple such that $\sigma _1\tau $ contains the $4$ -cycle $(h,k_1,k_2,2n-h)$ , then we have two possibilities.

• • If n is even and the $4$ -cycle is of the form $(h,n-h,n+h,2n-h)$ , then the conjugacy class of $\Sigma $ contains only two special $4$ -tuples;

• • If not, then the conjugacy class of $\Sigma $ contains exactly four special $4$ -tuples.

We point out that, looking at the symmetry of the problem, the number of special $4$ -tuples contained in the conjugacy class of $\Sigma $ depends on the configuration of the points fixed by $\sigma _1\tau $ . Indeed, in the first case, the set of fixed points is exactly $ \left\{\frac {n}{2}, n, \frac {3}{2}n, 2n \right\}$ , which are the vertices of a square.

Proof. Since $4$ -tuple $\Sigma =(\sigma _{\infty }, \sigma _0,\sigma _1,\tau )$ is special, we have that $\sigma _{\infty }=(2n, 2n-1, \ldots , 1)$ and $\sigma _1\tau $ fixes $2n$ ; moreover, by assumption it contains the $4$ -cycle $(h,k_1,k_2,2n-h)$ . Then, as proved before, $\sigma _0$ and $\sigma _1\tau $ will have the shape (8.3). As discussed before, if $\Sigma '\neq \Sigma $ is a special $4$ -tuple conjugated to $\Sigma $ then $\Sigma '=\gamma ^{-1}\Sigma \gamma $ with $\gamma \in \left\{ \sigma _{\infty }^{2n-\frac {h+k_1}{2}}, \sigma _{\infty }^{2n-\frac {k_1+k_2}{2}}, \sigma _{\infty }^{2n-\frac {2n-h+k_2}{2}} \right\}$ . We distinguish two cases.

Suppose that n is even and the $4$ -cycle is of the form $(h,n-h,n+h,2n-h)$ Footnote ² ; then the points fixed by $\sigma _1\tau $ are exactly $ \left\{\frac {n}{2}, n, \frac {3}{2}n, 2n \right\}$ . As discussed before, the other special $4$ -tuples contained in the conjugacy class of $\Sigma $ will be of the form $\gamma ^{-1}\Sigma \gamma $ , with $\gamma \in \left\{ \sigma _{\infty }^{\frac {n}{2}}, \sigma _{\infty }^ {n}, \sigma _{\infty }^{\frac {3}{2}n}\right\}$ . But for this special configuration, a direct computation shows that $\sigma _{\infty }^{-n}\Sigma \sigma _{\infty }^n=\Sigma $ , and $\sigma _{\infty }^{-n/2}\Sigma \sigma _{\infty }^{n/2}=\sigma _{\infty }^{n/2}\Sigma \sigma _{\infty }^{-n/2}\neq \Sigma $ , hence there are only two special $4$ -tuples contained in the conjugacy class of $\Sigma $ .

Suppose now that the $4$ -cycle contained in $\sigma _1\tau $ is not of the previous shape; we have that either $k_1\neq n-h$ or $k_2\neq n+h$ . In this case, a direct computation shows that if we conjugate $\Sigma $ by some $\gamma \in \left\{ \sigma _{\infty }^{2n-\frac {h+k_1}{2}}, \sigma _{\infty }^{2n-\frac {k_1+k_2}{2}}, \sigma _{\infty }^{2n-\frac {2n-h+k_2}{2}} \right\}$ , then $\sigma _0$ is not fixed, hence every conjugation gives rise to a special $4$ -tuple in the conjugacy class of $\Sigma $ , which concludes the proof.

Using the proposition, we can count the number of different conjugacy classes such that $\sigma _1\tau $ contains a $4$ -cycle; in this case, we have to distinguish the case n odd and n even.

Suppose first n odd; in this case, by Proposition 8.5, every conjugacy class contains exactly four special $4$ -tuples. Moreover, a special $4$ -tuple containing a $4$ -cycle $(h,k_1,k_2, 2n-h)$ with $h<k_1<k_2$ will have the shape (8.3), and there are two different choices of the couple $(\sigma _1, \tau )$ which gives the same permutation $\sigma _1\tau $ .

Putting all together, we have to count the number of ordered $3$ -tuples $h<k_1<k_2$ with $h\equiv k_1 \equiv k_2\ \ \mod 2$ and such that $1\le h\le n-3$ , $h+2\le k_1\le 2n-4-h$ and $k_1+2\le k_2\le 2n-2-h$ , which are equal to

$$\begin{align*}C_1= \sum_{h=1}^{n-3} \left [ \sum_{\substack{ k=h+2 \\ k \equiv h\ \ \mod 2 }}^{2n-4-h} \left ( n-1 -\frac{k+h}{2} \right ) \right ]. \end{align*}$$

Then we have that, if n is odd, the number of conjugacy classes of a $4$ -tuple such that $\sigma _1\tau $ contains a $4$ -cycle is

$$\begin{align*}\# \mbox{ conjugacy classes}= 2\cdot \frac{C_1}{4}= \frac{C_1}{2}. \end{align*}$$

Suppose now n even; in this case, we have to distinguish the cases in which the $4$ -cycle is of the special form $(h,n-h,n+h,2n)$ , which are $C_2=\frac {n}{2}-1$ . Recall that, for every configuration of the $4$ -cycle, we have two choices of the couple $(\sigma _1, \tau )$ giving rise to the same product. For this special configuration of the $4$ -cycle, the conjugacy class contains indeed only two special $4$ -tuples. The number of conjugacy classes in this case is given by

$$\begin{align*}\# \mbox{ conjugacy classes}= 2\cdot \frac{C_1-C_2}{4}+ 2\cdot \frac{C_2}{2}= \frac{C_1+C_2}{2}. \end{align*}$$