# Symmetries of the q-deformed real projective line

Perrine Jouteur

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Perrine Jouteur. Symmetries of the q-deformed real projective line. 2025. hal-05048093

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# SYMMETRIES OF THE $q$-DEFORMED REAL PROJECTIVE LINE

## PERRINE JOUTEUR

ABSTRACT. We generalize in two steps the quantized action of the modular group on $q$-deformed real numbers introduced by Morier-Genoud and Ovsienko in [13]. First, we let the projective general linear group $PGL_2(\mathbb{Z})$ act on $q$-real numbers via a $q$-deformed action. The quantized matrices we get have combinatorial interpretations. Then we consider an extension of the group $PGL_2(\mathbb{Z})$ by the 2-elements cyclic group, and define a quantized action of this extension on $q$-real numbers. We deduce from these actions some underlying relations between $q$-real numbers, and between left and right versions of $q$-deformed rational numbers. In particular we investigate the case of some algebraic numbers of degree 4 and 6. We also prove that the way of quantizing real numbers defined in [13] is an injective process.

## CONTENTS

<table><tr><td>1. Introduction and main results</td><td>2</td></tr><tr><td>1.1. Background on $q$-deformed numbers</td><td>2</td></tr><tr><td>1.2. First extension of symmetries</td><td>4</td></tr><tr><td>1.3. Second extension of symmetries</td><td>5</td></tr><tr><td>1.4. Application to irrational algebraic numbers</td><td>6</td></tr><tr><td>2. Quantized action of the projective linear group</td><td>6</td></tr><tr><td>2.1. From $PSL_2(\mathbb{Z})$ to $PGL_2(\mathbb{Z})$</td><td>6</td></tr><tr><td>2.2. Quantized positive continued fractions</td><td>8</td></tr><tr><td>2.3. Palindromicity and positivity of $q$-traces in determinant $-1$</td><td>10</td></tr><tr><td>3. Duplication of $PGL_2(\mathbb{Z})$</td><td>12</td></tr><tr><td>3.1. Twisted matrices</td><td>12</td></tr><tr><td>3.2. Action on rational numbers</td><td>13</td></tr><tr><td>4. Focus on $q$-real numbers</td><td>14</td></tr><tr><td>4.1. Convergences</td><td>14</td></tr><tr><td>4.2. Injectivity of quantization</td><td>16</td></tr><tr><td>5. Application to $q$-deformed algebraic numbers</td><td>16</td></tr><tr><td>5.1. Degree 4</td><td>16</td></tr><tr><td>5.2. Degree 6</td><td>18</td></tr><tr><td>Appendix A. Fence poset combinatorics</td><td>20</td></tr><tr><td>Appendix B. Positivity of $q$-traces of determinant 1</td><td>21</td></tr><tr><td>Appendix C. Proof of Proposition 1.1</td><td>22</td></tr><tr><td>Acknowledgements</td><td>23</td></tr><tr><td>References</td><td>23</td></tr></table>
# 1. INTRODUCTION AND MAIN RESULTS

The $q$-deformed real numbers were defined by Morier-Genoud and Ovsienko in [13] and [15], using a quantization of the action of the modular group $PSL_2(\mathbb{Z})$ on the rational projective line. These quantized numbers happen to be related to many different areas, including Farey triangulations [20], friezes patterns [14], combinatorics of fence posets [11, 17], snake graphs [18, 2], cluster algebras and $F$-polynomials [13], Markov numbers and Diophantine approximation [6, 8, 5, 10], knot theory via the Kauffman bracket [21, 7, 23], and homological algebra [1, 3].

In this paper we extend the group of symmetries acting on the quantized real numbers. Given a real number $x$ and its $q$-deformation $[x]_q$, by Morier-Genoud and Ovsienko's construction one has the following relations:

$$ 
\begin{array}{l}
[x+1]_q = q[x]_q + 1, \\
(1) \qquad \text{and} \\
\left[\frac{-1}{x}\right]_q = -\frac{1}{q[x]_q}.
\end{array}
$$

We show that in addition the following holds

$$ (2) \qquad [-x]_q = \frac{-[x]_q + 1 - q^{-1}}{(q-1)[x]_q + 1} \quad \text{and equivalently,} \quad \left[\frac{1}{x}\right]_q = \frac{(q-1)[x]_q + 1}{q[x]_q + 1 - q}. $$

This induces a quantization of the action of $PGL_2(\mathbb{Z})$ on the real projective line.

As an application of these extended symmetries of quantized numbers, we get convenient formulas to work with $q$-deformed positive continued fractions (whereas in previous works the negative continued fractions were preferred, see [9]). In particular, we show the injectivity of the quantization map.

We also use the $q$-deformed action of $PGL_2(\mathbb{Z})$ to quantize some polynomial equations whose Galois groups act on roots by fractional-linear transformations, following the ideas of [19].

Moreover, we define a quantized action of $PGL_2(\mathbb{Z}) \times \mathbb{Z}_2$, where $\mathbb{Z} := \{1, -1\}$ is the group with two elements, on the rational projective line. This action exchanges the left and the right versions of $q$-rational numbers, with the vocabulary of [1]. For every rational number $x \in \mathbb{Q}$, we deduce this relationship between left and right $q$-rational numbers, which was first observed by Thomas in [22].

$$ (3) \qquad [x]_q^{\sharp} = \frac{[x]_{q^{-1}}^{\flat} + q - 1}{(1-q)[x]_{q^{-1}}^{\flat} + q} \quad \text{and} \quad [x]_q^{\flat} = \frac{[x]_{q^{-1}}^{\sharp} + q - 1}{(1-q)[x]_{q^{-1}}^{\sharp} + q}. $$

This extended quantized action provides a coherent framework explaining the relations, for $x \in \mathbb{Q}$,

$$ (4) \qquad [-x]_q = -q^{-1}[x]_{q^{-1}} \quad \text{and} \quad \left[\frac{1}{x}\right]_q = \frac{1}{[x]_{q^{-1}}}. $$

## 1.1. Background on $q$-deformed numbers.
Let $q$ be a formal parameter. The classical $q$-deformed integers are polynomials in $q$,

$$ [n]_q = \frac{1 - q^n}{1 - q} = 1 + q + \cdots + q^{n-1} \in \mathbb{Z}[q]. $$

More generally, $q$-deformed rational numbers are rational fractions in $q$, belonging to $\mathbb{Z}(q)$, and $q$-deformed irrational real numbers are formal Laurent series in $q$, that is they belong to $\mathbb{Z}[[q;][q^{-1}]]$:
$$ \mathbb{Z}[[q]][q^{-1}] := \left\{ \sum_{i \ge \nu} a_i q^i \mid a_i \in \mathbb{Z}, \nu \in \mathbb{Z} \right\}. $$

First, let us focus on q-deformed rational numbers. We will consider the extended rational line, $\mathbb{P}^1(\mathbb{Q}) = \mathbb{Q} \cup \{\infty\}$. The modular group $\mathrm{PSL}_2(\mathbb{Z})$ acts on $\mathbb{P}^1(\mathbb{Q})$ by fractional-linear transformations,

$$ \mathrm{PSL}_2(\mathbb{Z}) \times \mathbb{P}^1(\mathbb{Q}) \longrightarrow \mathbb{P}^1(\mathbb{Q}) \qquad \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) \cdot \frac{r}{s} = \frac{ar+bs}{cr+ds}. $$

Provided the q-deformed recursive relation $[x+1]_q = q[x]_q + 1$ corresponding to the quantization of the transformation $R: x \mapsto x+1$, there is only one way of quantizing the whole action of $\mathrm{PSL}_2(\mathbb{Z})$. This only way is the following,

$$ (5) \qquad R = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \quad \text{and} \quad S = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} $$

become

$$ (6) \qquad R_q = \begin{pmatrix} q & 1 \\ 0 & 1 \end{pmatrix} \quad \text{and} \quad S_q = \begin{pmatrix} 0 & -q^{-1} \\ 1 & 0 \end{pmatrix}. $$

In $\mathrm{PSL}_2(\mathbb{Z})$, the defining relations between $R$ and $S$ are $S^2 = (RS)^3 = \mathrm{Id}$. After quantization, the relations are unchanged (up to powers of $q$ and a sign) : $S_q^2 = \mathrm{Id}$ and $(R_q S_q)^3 = \mathrm{Id}$. The subgroup of $\mathrm{PGL}_2(\mathbb{Z}[q, q^{-1}])$ generated by $R_q$ and $S_q$ is thus isomorphic to the modular group [9], so this gives a well defined q-deformation of the action of $\mathrm{PSL}_2(\mathbb{Z})$ on $\mathbb{P}^1(\mathbb{Q})$. We will denote by $\mathrm{PSL}_{2,q}(\mathbb{Z})$ the image of $\mathrm{PSL}_2(\mathbb{Z})$ in $\mathrm{PGL}_2(\mathbb{Z}[q, q^{-1}])$ by the group homomorphism sending $R, S$ on $R_q, S_q$,

$$ \begin{array}{ccc} \mathrm{PSL}_2(\mathbb{Z}) & \xrightarrow{\approx} & \mathrm{PSL}_{2,q}(\mathbb{Z}) \subset \mathrm{PGL}_2(\mathbb{Z}[q, q^{-1}]) \\ M & \mapsto & M_q \end{array}. $$

The group $\mathrm{PSL}_{2,q}(\mathbb{Z})$ can also be understood via the reduced Burau representation of the braid group on three strands $B_3$. Let $\sigma_1$ and $\sigma_2$ be the two classical generators of $B_3$. The Burau representation of $B_3$ is

$$ \rho_3 : B_3 \longrightarrow \mathrm{SL}_2(\mathbb{Z}(q)), \qquad \sigma_1 \longmapsto \begin{pmatrix} q & 1 \\ 0 & 1 \end{pmatrix} \quad \text{and} \quad \sigma_2 \longmapsto \begin{pmatrix} 1 & 0 \\ -q & q \end{pmatrix}. $$

The quantized modular group is then precisely the projective version of the image of $B_3$ via this representation. Indeed, the centre $Z(B_3)$ of $B_3$ is the kernel of $\rho_3$, and the induced quotient homomorphism is an isomorphism of groups such that

$$ \begin{array}{ccc} B_3/Z(B_3) & \xrightarrow{\approx} & \mathrm{PSL}_{2,q}(\mathbb{Z}) \\ \sigma_1 & \mapsto & R_q \\ \sigma_1\sigma_2\sigma_1 & \mapsto & S_q \end{array}. $$

This point of view was used in [16] to give a partial answer to the question of faithfulness of specialization of the Burau representation. It was also used in [4] to quantize the rational projective plane.

As the action by fractional-linear transformations on $\mathbb{P}^1(\mathbb{Q})$ is transitive, the q-deformation of one point, say $\infty = \frac{1}{0}$, is enough to have the q-deformation of every point of the projective rational line. Yet, there is a subtlety : for an orbit of $\mathbb{P}^1(\mathbb{Z}(q))$ to be an actual q-deformation of $\mathbb{P}^1(\mathbb{Q})$, there must be in the orbit exactly one q-deformation for each rational number.
**Proposition 1.1.** There are only two $q$-deformations of $\infty$ in $\mathbb{P}^1(\mathbb{Z}(q))$ whose orbits under the action of $\text{PSL}_{2,q}(\mathbb{Z})$ satisfy the condition :

"For every $x \in \mathbb{P}^1(\mathbb{Q})$, there is exactly one element $f_x(q)$ of the orbit such that $f_x(1) = x$".

These two $q$-deformations of $\infty$ are

$$[\infty]_q^{\sharp} = \frac{1}{0} \quad \text{and} \quad [\infty]_q^{\flat} = \frac{1}{1-q}.$$

This property, which will be proved in Appendix C, leads to the definition of left and right versions of $q$-rational numbers.

**Definition 1.2.** Let $x \in \mathbb{P}^1(\mathbb{Q})$ and let $M \in \text{PSL}_2(\mathbb{Z})$ be such that $M \cdot \infty = x$. The two quantizations of $x$ are elements of $\mathbb{P}^1(\mathbb{Z}(q))$, given by

* Right version : $[x]_q^{\sharp} = M_q \cdot \frac{1}{0}$,

* Left version : $[x]_q^{\flat} = M_q \cdot \frac{1}{1-q}$.

In the rest of the article, except when necessary, we will skip the subscript "q" in the notation of $q$-rational numbers, writing $[x]^{\sharp}$ instead of $[x]_q^{\sharp}$ and $[x]^{\flat}$ instead of $[x]_q^{\flat}$.

Note that the right version is the original one, defined by Morier-Genoud and Ovsienko [13]. In particular, it recovers the usual $q$-integers of Gauss : for $n \in \mathbb{N}$,

$$[n]^{\sharp} = \frac{q^n - 1}{q - 1} = 1 + q + q^2 + \cdots + q^{n-1}.$$

The left version was defined in [1] by Bapat, Becker and Licata, introducing this terminology of left and right $q$-numbers. The left $q$-integers are :

$$[n]^{\flat} = \frac{q^{n+1} - q^n + q^{n-1} - 1}{q-1} = 1 + q + q^2 + \cdots + q^{n-2} + q^n.$$

The $q$-deformation of a real irrational number $x$ relies on the approximation of $x$ by any sequence of rational numbers converging to $x$. The Taylor series coming from the $q$-deformations of the rational numbers approaching $x$ stabilize and converge toward a formal Laurent series denoted by $[x]_q$ and being the $q$-deformation of $x$ by definition [15]. For more precise statements, see section 4.

We get two quantization maps, depending on which version one choose for rational numbers.

$$Q^{\square} : \begin{array}{rcl} \mathbb{R} & \to & \mathbb{Z}[[q))[q^{-1}] \\ x & \mapsto & \begin{cases} [x]^{\square} & \text{if } x \in \mathbb{Q} \\ [x]_q & \text{else} \end{cases} \end{array}, \text{ with } \square \in \{\sharp, \flat\}.$$

**Proposition 1.3.** The two quantizations maps $Q^{\sharp}$ and $Q^{\flat}$ are injective.

This proposition will be proved in section 4.2.

**1.2. First extension of symmetries.** The quantization of the action of $\text{PGL}_2(\mathbb{Z})$ on $\mathbb{P}^1(\mathbb{R})$ relies on the $q$-deformation of the transformation $x \mapsto -x$, corresponding to the matrix $N = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$, by

$$(7) \qquad N_q = \begin{pmatrix} -1 & 1 - q^{-1} \\ q - 1 & 1 \end{pmatrix}.$$

Note that $N_q$ has determinant $-q^{-1}(q^2 - q + 1)$. Let us then set

$$t = q^2 - q + 1.$$
We will work up to powers of $t$, so we denote by $\Lambda$ the localization of $\mathbb{Z}[q, q^{-1}]$ by $t$, and $\bar{\Lambda}$ its field of fractions.

$$ \Lambda := \mathbb{Z}[q, q^{-1}]_t, \quad \bar{\Lambda} = \mathbb{Z}(q). $$

The relations between $R, S, N$ in $\mathrm{PGL}_2(\mathbb{Z})$ are $S^2 = (RS)^3 = 1$ and $N^2 = (NR)^2 = (NS)^2 = 1$. We prove that they remain unchanged if we replace $R, S, N$ by $R_q, S_q, N_q$ in $\mathrm{PGL}_2(\Lambda)$. The subgroup of $\mathrm{PGL}_2(\Lambda)$ generated by $R_q, S_q$ and $N_q$ will be denoted by $\mathrm{PGL}_{2,q}(\mathbb{Z})$, it is isomorphic to $\mathrm{PGL}_2(\mathbb{Z})$.

**Theorem 1.4.** *There is a well-defined group homomorphism*

$$ (8) \qquad \begin{array}{ccc} \mathrm{PGL}_2(\mathbb{Z}) & \to & \mathrm{PGL}_{2,q}(\mathbb{Z}) \subset \mathrm{PGL}_2(\Lambda) \\ M & \mapsto & M_q \end{array}, $$

*sending $R, S, N$ to $R_q, S_q, N_q$, and it is an isomorphism of groups.*

This theorem will be proved in section 2.1. We will also describe the traces of these $q$-matrices $M_q$, showing they are palindromic polynomials (section 2.3). In Appendix A, we also prove that these $q$-traces are generating functions of some fence posets.

Under the larger action of $\mathrm{PGL}_2(\mathbb{Z})$ on $q$-deformed rational numbers, the left and right versions form a single orbit in $\mathbb{P}^1(\mathbb{Z}(q))$. On irrational real numbers, the action of $\mathrm{PGL}_2(\mathbb{Z})$ commutes with quantization. The first part of the following theorem dealing with rational numbers will be proved in section 2.1, and the second part on real numbers will be proved in section 4.1.

**Theorem 1.5.** Let $M \in \mathrm{GL}_2(\mathbb{Z})$, with $\det(M) = -1$. Let $x \in \mathbb{P}^1(\mathbb{Q})$. Then in $\Lambda$,

$$ M_q \cdot [x]^\sharp = [M \cdot x]^\flat \quad \text{and} \quad M_q \cdot [x]^\flat = [M \cdot x]^\sharp. $$

*Furthermore, for $x \in \mathbb{R} \setminus \mathbb{Q}$,*

$$ M_q \cdot [x]_q = [M \cdot x]_q. $$

1.3. **Second extension of symmetries.** The next step is to extend again the group acting on $\mathbb{P}^1(\mathbb{Z}(q))$, from $\mathrm{PGL}_2(\mathbb{Z})$ to $\mathrm{PGL}_2(\mathbb{Z}) \times \mathbb{Z}_2$. To do this, we consider the operator

$$ \tau: \begin{array}{ccc} \mathbb{P}^1(\mathbb{Z}(q)) & \to & \mathbb{P}^1(\mathbb{Z}(q)) \\ f & \to & f(q^{-1}) \end{array}. $$

This is an involution, quantizing the identity transformation. This will twist the action of every matrix $M \in \mathrm{PGL}_2(\mathbb{Z})$. In the following, we will denote by $\cdot$ the composition of operators of $\mathbb{P}^1(\mathbb{Z}(q))$. We define

$$ (9) \qquad I_q = \begin{pmatrix} 1 & q-1 \\ 1-q & q \end{pmatrix}, \quad \text{and} \quad \bar{I}_q := I_q \cdot \tau. $$

This operator $\bar{I}_q$ is another quantization of the identity $\mathrm{Id}$ of $\mathrm{PGL}_2(\mathbb{Z})$, and is almost the same as the $q$-rational transition map of Thomas in [22], which makes transition between left and right $q$-rationals.

For any $M \in \mathrm{PGL}_2(\mathbb{Z})$, the twisted $q$-deformation of $M$ is the operator

$$ \bar{M}_q := M_q \cdot \bar{I}_q : f(q) \mapsto M_q I_q \cdot f(q^{-1}). $$

**Theorem 1.6.** *The following map is a well-defined group isomorphism.*

$$ (10) \qquad \begin{array}{ccc} \mathrm{PGL}_2(\mathbb{Z}) \times \mathbb{Z}_2 & \to & \mathrm{PGL}_{2,q}(\mathbb{Z}) \times \{\mathrm{Id}, \bar{I}_q\} \\ (M, \epsilon) & \mapsto & \left\{ \begin{array}{ll} (M_q, \mathrm{Id}) & \text{if } \epsilon = 1 \\ (M_q, \bar{I}_q) & \text{if } \epsilon = -1 \end{array} \right. \end{array}. $$

*It gives rise to an action*

$$ (11) \qquad \begin{array}{ccc} (\mathrm{PGL}_{2,q}(\mathbb{Z}) \times \{\mathrm{Id}, \bar{I}_q\}) \times \mathbb{P}^1(\mathbb{Z}(q)) & \to & \mathbb{P}^1(\mathbb{Z}(q)) \\ ((M_q, T), f) & \mapsto & M_q \cdot T \cdot f = \begin{cases} M_q \cdot f & \text{if } T = \mathrm{Id} \\ \bar{M}_q \cdot f & \text{if } T = \bar{I}_q \end{cases} \end{array}. $$
The action of twisted matrices of determinant $-1$ commutes with left and right quantizations whereas in determinant 1 they are exchanged.

**Theorem 1.7.** Let $M \in \mathrm{PGL}_2(\mathbb{Z})$. Let $x \in \mathbb{P}^1(\mathbb{Q})$. If $\det(M) = -1$, then

$$ \overline{M}_q \cdot [x]^\sharp = [M \cdot x]^\sharp \quad \text{and} \quad \overline{M}_q \cdot [x]^\flat = [M \cdot x]^\flat. $$

A contrario, if $\det(M) = 1$, then

$$ \overline{M}_q \cdot [x]^\sharp = [M \cdot x]^\flat \quad \text{and} \quad \overline{M}_q \cdot [x]^\flat = [M \cdot x]^\sharp. $$

Theorems 1.7 and 1.6 will be proved in section 3.

1.4. **Application to irrational algebraic numbers.** As an illustration of what can be done with the quantized $\mathrm{PGL}_2(\mathbb{Z})$, we study $q$-deformations of algebraic equations. Even if we know how to quantize irrational numbers, there is still many interesting questions unsolved for $q$-deformed algebraic numbers.

The case of algebraic numbers of degree 2 has already been fully understood by Leclere and Morier-Genoud [9]. In [19], Ovsienko and Ustinov partly covered the case of equations of degree 3. Here we focus on degree 4 and degree 6, considering equations of the form

$$ (12) \qquad x^4 - bx^2 + 1 = 0 \quad \text{and} \quad x^6 - 3x^5 - bx^4 + (2b+5)x^3 - bx^2 - 3x + 1 = 0. $$

Exploiting symmetries of these equations, we are able to quantize the Vieta's relations. It is the topic of Section 5.

## 2. QUANTIZED ACTION OF THE PROJECTIVE LINEAR GROUP

In this section, we give the details of the quantized action of $\mathrm{PGL}_2(\mathbb{Z})$ on $\mathbb{P}^1(\mathbb{Z}(q))$ that we have introduced in section 1.2 above.

2.1. **From $\mathrm{PSL}_2(\mathbb{Z})$ to $\mathrm{PGL}_2(\mathbb{Z})$.** Recall the usual presentation by generators and relations of $\mathrm{PGL}_2(\mathbb{Z})$.

**Lemma 2.1.** *The group $\mathrm{PGL}_2(\mathbb{Z})$ is isomorphic to*

$$ \langle R, S, N \mid S^2 = (RS)^3 = 1, N^2 = (NR)^2 = (NS)^2 = 1 \rangle, $$

with

$$ R = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \quad S = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \quad N = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}. $$

The first two generators and relations are those defining $\mathrm{PSL}_2(\mathbb{Z})$, which have already been quantized in [13] and [9]. Thus we just have to define a quantized version of $N$ and check the three last relations to have a $q$-deformed version of $\mathrm{PGL}_2(\mathbb{Z})$.

**Definition 2.2.** Put $N_q = \begin{pmatrix} -1 & 1 - q^{-1} \\ q - 1 & 1 \end{pmatrix}$.

We will work in the localization of $\mathbb{Z}[q, q^{-1}]$ by $t = 1 - q + q^2$.

$$ \Lambda := \mathbb{Z}[q, q^{-1}]_t, \quad \bar{\Lambda} = \mathbb{Z}(q). $$

**Proposition 2.3.** In $\mathrm{PGL}_2(\Lambda)$, the $q$-deformed matrices $R_q, S_q$ and $N_q$ satisfy

$$ N_q^2 = (N_q R_q)^2 = (N_q S_q)^2 = t \mathrm{Id}, $$

and there is no other relation between $R_q, S_q$ and $N_q$.
*Proof.* Easy computations ensure the relations to be true. If there was another relation in PGL₂(Λ) of the form ℜ(R<sub>q</sub>, S<sub>q</sub>, N<sub>q</sub>) = Id, then by evaluating q = 1 we would get a relation in PGL₂(Z) between R, S and N, so ℜ(R, S, N) would be in the ideal generated by the previous relations, and then ℜ(R<sub>q</sub>, S<sub>q</sub>, N<sub>q</sub>) was not new. □

Therefore over Λ, the relations of PGL₂(Z) and only them are satisfied by R<sub>q</sub>, S<sub>q</sub> and N<sub>q</sub>, and Theorem 1.4 is proven.

*Example 2.4.* Now we have a *q*-deformation of the operation $x \mapsto 1/x$ given by the matrix $J = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ of determinant $-1$.

$$J = NS \xrightarrow{q} J_q = N_q S_q = \begin{pmatrix} q-1 & 1 \\ q & 1-q \end{pmatrix}.$$

Note that we can choose R and J to be generators of PGL₂(Z), as we have the following relations

$$R^{-1}JRJR^{-1} = S \text{ and } JRJR^{-1}JR = N.$$

In the quantized setting, these relations become

$$R_q^{-1} J_q R_q R_q^{-1} = t S_q \text{ and } J_q R_q R_q^{-1} J_q R_q = t q N_q.$$

Let us consider the quantized action

$$\begin{array}{ccc} \mathrm{PGL}_{2,q}(\mathbb{Z}) \times \mathbb{P}^1(\mathbb{Z}(q)) & \longrightarrow & \mathbb{P}^1(\mathbb{Z}(q)) \\ (M_q, f) & \longmapsto & M_q \cdot f. \end{array}$$

First, recall that the action of PSL₂,q(Z) commutes with both left and right quantizations.

**Proposition 2.5** ([9, 1]). Let $x \in \mathbb{P}^1(\mathbb{Q})$ and let $M \in \mathrm{PSL}_2(\mathbb{Z})$. Then

$$M_q \cdot [x]^b = [M \cdot x]^b \text{ and } M_q \cdot [x]^\sharp = [M \cdot x]^\sharp.$$

We have a similar result for matrices of determinant $-1$. This is the first part of Theorem 1.5.

**Proposition 2.6.** Let $M \in \mathrm{PGL}_2(\mathbb{Z})$, with $\det(M) = -1$. Let $x \in \mathbb{P}^1(\mathbb{Q})$. Then in Λ,

$$M_q \cdot [x]^\sharp = [M \cdot x]^b \text{ and } M_q \cdot [x]^b = [M \cdot x]^\sharp.$$

*Proof.* We can write $M = M^+ N$ with $\det(M^+) = 1$, so that $M_q = M_q^+ N_q$. Let us consider first the case $x = \infty$. Then

$$M_q \cdot [\infty]^\sharp = M_q^+ N_q \cdot \frac{1}{0} = M_q^+ \cdot \frac{1}{1-q} = [M^+ \cdot \infty]^b.$$

Now for $x \in \mathbb{Q}$, there is a matrix $A \in \mathrm{PSL}_2(\mathbb{Z})$ such that $A \cdot \infty = x$, and then

$$M_q \cdot [x]^\sharp = M_q A_q \cdot [\infty]^\sharp = [M A \cdot \infty]^b = [M \cdot x]^b.$$

The equality $M_q \cdot [x]^b = [M \cdot x]^\sharp$ is symmetric. □

*Example 2.7.* Let us take the matrix $N = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$, and $x = n \in \mathbb{N}^*$. Then

$$\begin{align*} [N]_q \cdot [n]^\sharp &= \frac{[-n]^\sharp + 1 - q^{-1}}{(q-1)[n]^\sharp + 1} \\ &= q^{-n}(-[-n]^\sharp + 1 - q^{-1}) \\ &= -q^{-1} - q^{-2} - \dots - q^{-n} + q^{-n} - q^{-(n+1)} \\ &= -q^{-1} - q^{-2} - \dots - q^{-(n-1)} - q^{-(n+1)} \end{align*}$$
On the other hand,

$$[N \cdot n]^b = [-n]^b = -q^{-1} - q^{-2} - \dots - q^{-(n-1)} - q^{-(n+1)}.$$

So $[N]_q \cdot [n]^b = [N \cdot n]^b$. In the other way around,

$$ \begin{aligned} [N]_q \cdot [n]^b &= \frac{-[n]^b + 1 - q^{-1}}{(q-1)[n]^b + 1} = -q^{-1} \frac{1 + q^2 + q^3 + \dots + q^{n-1} + q^{n+1}}{q^{n+1} - q^n + q^{n-1}} \\ &= -q^{-1} \frac{1 - q^n + q^2[n]^b}{q^{n-1}t} \\ &= -q^{-1} \frac{[n]^b}{q^{n-1}} \\ &= -q^{-1}[n]_{q^{-1}}^b \\ &= [-n]^b. \end{aligned} $$

So $[N]_q \cdot [n]^b = [-n]^b$.

**Example 2.8.** Let us consider now $x = \frac{3}{2}$ and $M = \begin{pmatrix} 3 & 1 \\ 1 & 0 \end{pmatrix}$. Then $M$ decomposes as $M = R^3 J$, so

$$M_q = R_q^3 J_q = \begin{pmatrix} q^4 + q^2 + q & 1 \\ q & 1 - q \end{pmatrix}.$$

On one side, $[M \cdot \frac{3}{2}]^b = [\frac{11}{3}]^b = \frac{q^5+2q^4+2q^3+3q^2+2q+1}{q^2+q+1}$.

On the other side, one can compute $[\frac{3}{2}]^b = \frac{1+q^2+q^3}{1+q^2}$. Then applying the quantized matrix $M_q$ and cancelling a factor $t$ above and below the fraction,

$$M_q \cdot \left[ \frac{3}{2} \right]^b = M_q \cdot \frac{1 + q^2 + q^3}{1 + q^2} = \frac{q^5 + 2q^4 + 2q^3 + 3q^2 + 2q + 1}{q^2 + q + 1}.$$

2.2. **Quantized positive continued fractions.** The initial definition of $q$-rational numbers was based on continued fractions [13]. Each rational number $u/v$ can be written as

$$ \frac{u}{v} = a_1 + \frac{1}{a_2 + \frac{1}{\ddots + \frac{1}{a_n}}}, \text{ with } a_1 \in \mathbb{Z} \text{ and } a_i \in \mathbb{N}^*, i \ge 2, $$

which is also denoted by $u/v = [a_1, a_2, \dots, a_n]$. The sequence $(a_1, a_2, \dots, a_n)$ is unique when the parity of $n$ is fixed. This continued fraction expansion is translated in the matrix setting by

$$ (13) \qquad \begin{pmatrix} u & u' \\ v & v' \end{pmatrix} = \begin{pmatrix} a_1 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} a_2 & 1 \\ 1 & 0 \end{pmatrix} \cdots \begin{pmatrix} a_n & 1 \\ 1 & 0 \end{pmatrix}, $$

where $u'/v' = [a_1, a_2, \dots, a_{n-1}]$. To work with matrices of determinant 1, it was necessary to choose $n$ even and to consider pairs of elementary matrices, using the relation

$$R_q^{a_1} J_q R_q^{a_2} J_q = t R_q^{a_1} L_q^{a_2}, \text{ where } L_q = \begin{pmatrix} q & 0 \\ q & 1 \end{pmatrix}.$$

The following property is the $q$-deformation of equation (13).
**Proposition 2.9** (Proposition 4.3 of [13]). Let $u/v$ be a rational number and let $[a_1, a_2, \dots, a_{2m}]$ be its even positive continued fraction expansion. Let us set $u_{n-1}/v_{n-1} = [a_1, a_2, \dots, a_{2m-1}]$ and

$$ \left[ \frac{u}{v} \right]^\sharp = \frac{U^\sharp}{V^\sharp} \quad \text{and} \quad \left[ \frac{u_{2m-1}}{v_{2m-1}} \right]^\sharp = \frac{U_{2m-1}^\sharp}{V_{2m-1}^\sharp}. $$

Then it holds

$$ R_q^{a_1} J_q R_q^{a_2} J_q \cdots R_q^{a_{2m}} J_q = q^{\min(0, a_1)} t^m \begin{pmatrix} qU_q^\sharp & U_{2m-1}^\sharp \\ qV_q^\sharp & V_{2m-1}^\sharp \end{pmatrix}. $$

The elementary matrices involved in the decomposition 13 have determinant $-1$, and can be $q$-deformed directly as

$$ R^a J = \begin{pmatrix} a & 1 \\ 1 & 0 \end{pmatrix} \xrightarrow{q} R_q^a J_q = \begin{pmatrix} q[a]^b & 1 \\ q & 1-q \end{pmatrix}. $$

With these $q$-deformed matrices of determinant $-1$, we get the left version of Proposition 2.9.

**Proposition 2.10.** Let $u/v$ be a rational number and let $[a_1, a_2, \dots, a_{2m+1}]$ its odd positive continued fraction expansion, with $m \ge 1$. Let us set $u_{2m}/v_{2m} = [a_1, \dots, a_{2m}]$ and

$$ \left[ \frac{u}{v} \right]^b = \frac{U^b}{V^b} \quad \text{and} \quad \left[ \frac{u_{2m}}{v_{2m}} \right]^b = \frac{U_{2m}^b}{V_{2m}^b}. $$

Then it holds

$$ R_q^{a_1} J_q R_q^{a_2} J_q \cdots R_q^{a_{2m+1}} J_q = q^{\min(0, a_1)} t^m \begin{pmatrix} qU_q^b & U_{2m}^b \\ qV_q^b & V_{2m}^b \end{pmatrix}. $$

*Proof.* We consider $u_{2m-1}/v_{2m-1} = [a_1, \dots, a_{2m-1}]$, and its right quantization $[u_{2m-1}/v_{2m-1}]^\sharp = U_{2m-1}^\sharp/V_{2m-1}^\sharp$. According to the even case,

$$ R_q^{a_1} J_q R_q^{a_2} J_q \cdots R_q^{a_{2m}} J_q = q^{\min(0, a_1)} t^m \begin{pmatrix} qU_{2m}^\sharp & U_{2m-1}^\sharp \\ qV_{2m}^\sharp & V_{2m-1}^\sharp \end{pmatrix}. $$

Therefore

$$ R_q^{a_1} J_q R_q^{a_2} J_q \cdots R_q^{a_{2m}} J_q = q^{\min(0, a_1)} t^m \begin{pmatrix} q(q[a_{2m+1}]^b U_{2m}^\sharp + U_{2m-1}^\sharp) & qU_{2m}^\sharp + (1-q)U_{2m-1}^\sharp \\ q(q[a_{2m+1}]^b V_{2m}^\sharp + V_{2m-1}^\sharp) & qV_{2m}^\sharp + (1-q)V_{2m-1}^\sharp \end{pmatrix}. $$

An induction on $m$ shows that

$$ U^b = q[a_{2m+1}]^b U_{2m}^\sharp + U_{2m-1}^\sharp \quad \text{and} \quad V^b = q[a_{2m+1}]^b V_{2m}^\sharp + V_{2m-1}^\sharp. $$

Finally, as indicated in Appendix A of [1], $qU_{2m}^\sharp + (1-q)U_{2m-1}^\sharp = U_{2m}^b$ and same for denominators. $\square$

**Example 2.11.** Let us detail the case of $u/v = 7/5 = [1, 2, 1, 1] = [1, 2, 2]$. With the even continued fraction, we get

$$ R_q J_q R_q^2 J_q R_q J_q R_q J_q = \begin{pmatrix} q^5 + 2q^4 + 2q^3 + q^2 + q & q^3 + q^2 + q + 1 \\ q^4 + 2q^3 + q^2 + q & q^2 + q + 1 \end{pmatrix}, $$

so that $[7/5]^\sharp = \frac{1+q+2q^2+2q^3+q^4}{1+q+2q^2+q^3}$. For the odd continued fraction,

$$ \begin{aligned} R_q J_q R_q^2 J_q R_q^2 J_q &= \begin{pmatrix} q^8 + 2q^6 + 2q^4 + q^3 + q & q^5 + 2q^2 - q + 1 \\ q^7 + q^5 + q^4 + q^3 + q & q^4 - q^3 + 2q^2 - q + 1 \end{pmatrix} \\ &= t \begin{pmatrix} q^6 + q^5 + 2q^4 + q^3 + q^2 + q & q^3 + q^2 + 1 \\ q^5 + q^4 + q^3 + q^2 + q & q^2 + 1 \end{pmatrix}. \end{aligned} $$
thus $\left[\frac{7}{5}\right]^b = \frac{1+q+q^2+2q^3+q^4+q^5}{1+q+q^2+q^3+q^4}$.

**2.3. Palindromicity and positivity of q-traces in determinant** $-1$. One of the main properties of $q$-deformed matrices of PSL$_2(\mathbb{Z})$ is that their traces are palindromic polynomials [9]. This property still holds for the $q$-deformed matrices of determinant $-1$.

**Proposition 2.12.** Let $M \in \text{PGL}_2(\mathbb{Z})$, with $\det(M) = -1$. Then $\text{Tr}(M_q)$ is a palindromic polynomial with positive integer coefficients, up to a signed power of $q$.

Our proof of this proposition relies on negative continued fractions, also called Hirzebruch-Jung continued fractions. For a rational number $x \in \mathbb{Q}$, there are uniquely determined integers $c_1 \in \mathbb{Z}$, $c_2, \dots, c_k \ge 2$, such that

$$x = c_1 - \frac{1}{c_2 - \frac{1}{\ddots - \frac{1}{c_k}}}$$

This negative continued fraction expansion of $x$ is denoted by $x = \llbracket c_1, c_2, \dots, c_k \rrbracket$.

We will need the following lemma which follows directly from Proposition 2.9, and Theorem 1.5.

**Lemma 2.13.** Let $u/v$ be a rational number and let $\llbracket c_1, \dots, c_k \rrbracket$ its negative continued fraction expansion, with $c_1 \in \mathbb{Z}$ and $c_i \ge 2$ for all $i \ge 2$. Consider

$$NR^{c_1}SR^{c_2}S \cdots R^{c_k}S = \begin{pmatrix} -u & u' \\ v & -v' \end{pmatrix},$$

where $u'/v' = \llbracket c_1, \dots, c_{k-1} \rrbracket$. Then the $q$-deformation of this matrix is

$$N_q R_q^{c_1} S R_q^{c_2} S \cdots R_q^{c_k} S = q^{\min(0, c_1) - 1} \begin{pmatrix} U^b & -q^{c_k-1} (U')^b \\ V^b & -q^{c_k-1} (V')^b \end{pmatrix},$$

$$\text{where } \left[\frac{-u}{v}\right]^b = \frac{U^b}{V^b} \text{ and } \left[\frac{-u'}{v'}\right]^b = \frac{(U')^b}{(V')^b}.$$

*Proof of Proposition 2.12.* • Let us first show that $\text{Tr}(M_q)$ is a palindromic polynomial in $q$. We can write $M = NM(c_1, c_2, \dots, c_k)$, where

$$M(c_1, c_2, \dots, c_k) = R^{c_1} S R^{c_2} S \cdots R^{c_k} S.$$

Let us compute $^t M_{q^{-1}}$.

$$\begin{align*}
{}^t M_{q^{-1}} &= \begin{pmatrix} [c_k]^\sharp_{q^{-1}} & 1 \\ -q^{1-c_k} & 0 \end{pmatrix} \cdots \begin{pmatrix} [c_2]^\sharp_{q^{-1}} & 1 \\ -q^{1-c_2} & 0 \end{pmatrix} \begin{pmatrix} [c_1]^\sharp_{q^{-1}} & 1 \\ -q^{1-c_1} & 0 \end{pmatrix} {}^t N_{q^{-1}} \\
&= q^{-\Sigma(c_i-1)} \begin{pmatrix} [c_k]^\sharp_q & q^{c_k-1} \\ -1 & 0 \end{pmatrix} \cdots \begin{pmatrix} [c_2]^\sharp_q & q^{c_2-1} \\ -1 & 0 \end{pmatrix} \begin{pmatrix} [c_1]^\sharp_q & q^{c_1-1} \\ -1 & 0 \end{pmatrix} {}^t N_{q^{-1}} \\
&= q^{-\Sigma(c_i-1)} N M_q(c_k, \dots, c_2, c_1) N^t N_{q^{-1}},
\end{align*}$$

because the conjugation by $N$ turn the signs of the antidiagonale to the opposite. Therefore $N^t N_{q^{-1}} = N_q N$, so taking the trace we get:

$$\text{Tr}(M_{q^{-1}}) = q^{-\Sigma(c_i-1)} \text{Tr}(M_q(c_k, \dots, c_2, c_1) N_q).$$

It is thus enough to show that $\text{Tr}(M_q(c_k, \dots, c_2, c_1) N_q) = \text{Tr}(N_q M_q(c_1, c_2, \dots, c_k))$. Let us set

$$M_q(c_1, c_2, \dots, c_k) = \begin{pmatrix} A & B \\ C & D \end{pmatrix} \text{ and } M_q(c_k, \dots, c_2, c_1) = \begin{pmatrix} \tilde{A} & \tilde{B} \\ \tilde{C} & \tilde{D} \end{pmatrix}.$$
With these notations, $\mathrm{Tr}(N_q M_q(c_1, c_2, \dots, c_k)) = -A + (1 - q^{-1})C + (q - 1)B + D$, and same for $\mathrm{Tr}(N_q M_q(c_k, \dots, c_2, c_1))$ with $\tilde{A}, \tilde{B}, \tilde{C}, \tilde{D}$. In the proof of Lemma 3.8 of [9], the following relations are settled, using induction on the length of the continued fraction $[[c_1, \dots, c_k]]$ and on the integer $c_k$:

$$ \left\{ \begin{array}{ll} A + D = \tilde{A} + \tilde{D} & (1) \\ C - qB = \tilde{C} - q\tilde{B} & (2) \\ A + B - C = \tilde{A} + \tilde{B} - \tilde{C} & (3) \end{array} \right. . $$

Taking (2) + (3), we get

$A - (q-1)B = \tilde{A} - (q-1)\tilde{B}$, and taking $(1) - q^{-1}(2) - (3)$, we get $D + (1 - q^{-1})C = \tilde{D} + (1 - q^{-1})\tilde{C}$, and this shows exactly that

$$ \mathrm{Tr}(N_q M_q(c_1, c_2, \dots, c_k)) = \mathrm{Tr}(N_q M_q(c_k, \dots, c_2, c_1)). $$

Therefore $\mathrm{Tr}(M_{q^{-1}}) = q^{-\sum(c_i-1)} \mathrm{Tr}(M_q)$, meaning that $\mathrm{Tr}(M_q)$ is a palindromic polynomial.

• Now we are showing that the coefficients of $\mathrm{Tr}(M_q)$ are of the same sign. We keep the same notations as above, $M = NM(c_1, c_2, \dots, c_k)$. It was explained in [12] that we can choose the $c_i$'s to be such that $c_2, \dots, c_{k-2} \ge 2$, and either $c_{k-1} \ge 2$ and $c_k \ge 2$, either $c_{k-1} \ge 2$ and $c_k = 1$, either $c_{k-1} = c_k = 1$.

First, let us suppose that $c_i \ge 2$ for all $2 \le i \le k$. If $c_1 \ge 1$, the rational number given by $[[c_1, \dots, c_k]]$ is nonnegative, and the Lemma 2.13 ensures that $\mathrm{Tr}(M_q)$ has negative coefficients. If $c_1 \le -1$, then we can use the relation $N_q M_q(c) = -q^c M_q(-c)N_q$ to get

$$ \mathrm{Tr}(M_q) = \mathrm{Tr}(N_q M_q(c_1, c_2, \dots, c_k)) = -q^{c_1} \mathrm{Tr}(N_q M_q(c_2, \dots, c_k, -c_1)). $$

According to the previous case, here $\mathrm{Tr}(M_q)$ has positive coefficients.

Now let us suppose that $c_1 = 0$. Then using the relations $N_q S_q = -S_q N_q$ and $S_q^2 = -q^{-1} \mathrm{Id}$,

$$ \mathrm{Tr}(N_q M_q(0, c_2, \dots, c_k)) = - \mathrm{Tr}(S_q N_q M_q(c_2, \dots, c_k)) = q^{c_2-1} \mathrm{Tr}(N_q M_q(c_3, \dots, c_k - c_2)), $$

thus if $c_k - c_2 \ne 0$ the previous cases apply, and if $c_k - c_2 = 0$, we get the result by induction on the length $k$ of the continued fraction.

On a second hand, let us suppose that $c_i \ge 2$ for all $2 \le i \le k$ and $c_k = 1$.

Using the relations $R_q S_q R_q = -q^2 S_q R_q^{-1} S_q$ and $R_q^{-1} N_q = q^{-1} N_q R_q$, we get that

$$ \mathrm{Tr}(N_q M_q(c_1, \dots, c_{k-1}, 1)) = q \mathrm{Tr}(N_q M_q(c_1, \dots, c_{k-1} - 1) R_q^{-1}) = \mathrm{Tr}(N_q M_q(c_1 + 1, c_2, \dots, c_{k-1} - 1)). $$

Therefore if $c_{k-1} \ge 3$, the previous cases give the result. Else, $c_{k-1} - 1 = 1$ and we can conclude by induction on $k$.

Finally, the case where $c_i \ge 2$ for all $2 \le i \le k-2$ and $c_{k-1} = c_k = 1$ is very similar to the previous case. $\square$

**Example 2.14.** Let us consider the matrix $M = \begin{pmatrix} -7 & 3 \\ 5 & -2 \end{pmatrix}$, which is decomposed on $R, S, N$ as

$$ M = N R^2 S R^2 S R^3 S. $$

After quantization, we get

$$ \mathrm{Tr}(M_q) = \mathrm{Tr} \begin{pmatrix} -q^5 - q^4 - q^3 - 2q^2 - q - 1 & q^5 + q^3 + q^2 \\ q^6 + q^5 + q^4 + q^3 + q^2 & -q^6 - q^4 \end{pmatrix} = -(1 + q + 2q^2 + q^3 + 2q^4 + q^5 + q^6). $$

This is a palindromic polynomial, but notice that it is not unimodular.
### 3. DUPLICATION OF PGL₂(Z)

The aim is to define an action of PGL₂(Z) × Z₂ on P¹(Z(q)), containing the previous q-deformed action of PGL₂(Z).

**3.1. Twisted matrices.** Let us consider the operator

$$ \tau: \begin{array}{ccc} \mathbb{P}^1(\mathbb{Z}(q)) & \rightarrow & \mathbb{P}^1(\mathbb{Z}(q)) \\ f & \rightarrow & f(q^{-1}) \end{array}. $$

**Proposition 3.1.** Up to powers of $t = q^2 - q + 1$, there is only one matrix $I_q$ in PGL₂(Λ) such that the operator $I_q \cdot \tau$ commutes with the quantized action of PGL₂(Z) on P¹(Z(q)). This matrix is

$$ I_q = \begin{pmatrix} 1 & q-1 & 0 \\ 1-q & q & 0 \end{pmatrix}, $$

and it satisfies $(I_q \cdot \tau)^2 = q^{-1}t \operatorname{Id}$.

*Proof.* Let $\tilde{I}_q = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, be such that $\tilde{I}_q \cdot \tau$ commutes with every matrix $A_q$ for $A \in \mathrm{PGL}_2(\mathbb{Z})$. In particular,

$$ \tilde{I}_q \cdot \tau \cdot R_q = R_q \cdot \tilde{I}_q \cdot \tau \begin{pmatrix} aq^{-1} & a+b \\ cq^{-1} & c+d \end{pmatrix} \cdot \tau = \begin{pmatrix} aq+c & qb+d \\ c & d \end{pmatrix} \cdot \tau, $$

$$ \text{so } \begin{cases} a = aq + c \\ qa + qb = qb + d \\ qc + qd = d \end{cases} \text{, and then } c = (1-q)a, \text{ and } d = qa : $$

$$ \tilde{I}_q = \begin{pmatrix} a & b \\ (1-q)a & qa \end{pmatrix}. $$

Same with $S_q$:

$$ \tilde{I}_q \cdot \tau \cdot S_q = S_q \cdot \tilde{I}_q \cdot \tau \begin{pmatrix} b & -qa \\ qa & (q^2 - q)a \end{pmatrix} \cdot \tau = \begin{pmatrix} (1-q^{-1})a & -a \\ a & b \end{pmatrix} \cdot \tau, $$

so $b = (q-1)a$. Finally, we can check that $\tilde{I}_q \cdot \tau \cdot N_q = N_q \cdot \tilde{I}_q \cdot \tau$.

To have $\det(\tilde{I}_q)$ invertible in $\Lambda$, the coefficient $a$ must be invertible so $a$ is a power of $tq$. $\square$

**Definition 3.2.** We define $\tilde{I}_q := I_q \cdot \tau$ as the second quantization of the identity of PGL₂(Z) (the first quantization of identity being just identity of PGL₂(Λ)). Then, for every matrix $M \in \mathrm{PGL}_2(\mathbb{Z})$, we define the operator

$$ \bar{M}_q := M_q \cdot \tilde{I}_q \text{ acting on } \mathbb{P}^1(\mathbb{Z}(q)). $$

*Remark 3.3.* As $\tilde{I}_q^2 = \operatorname{Id}$ (up to $q$ and $t$), for all matrices $A, B$ in PGL₂(Z), we have

$$ \bar{A}_q \bar{B}_q = A_q B_q \text{ as operators of } \mathbb{P}^1(\mathbb{Z}(q)). $$

*Example 3.4.* For the generators $R, S$ and $N$, we get

$$ \bar{R}_q = \begin{pmatrix} 1 & q^2 \\ 1-q & q \end{pmatrix} \cdot \tau, \quad \bar{S}_q = \begin{pmatrix} q-1 & -q \\ q & q^2-q \end{pmatrix} \cdot \tau \quad \text{and} \quad \bar{N}_q = \begin{pmatrix} -t & 0 \\ 0 & qt \end{pmatrix} \cdot \tau = \begin{pmatrix} -1 & 0 \\ 0 & q \end{pmatrix} \cdot \tau. $$
Recall the other usual generator $L = RSR$ ($L$ together with $R$ generate $PSL_2(\mathbb{Z})$). In $PSL_2(\Lambda)$, it becomes

$$L_q = \begin{pmatrix} 1 & 0 \\ 1 & q^{-1} \end{pmatrix}.$$

We can check that

$$\overline{R}_q \overline{S}_q \overline{R}_q = q^{-1} t \begin{pmatrix} q & q^2 - q \\ 1 & q^2 \end{pmatrix} \cdot \tau = \overline{L}_q.$$

For the matrix $J = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, the second quantification is trivial :

$$\overline{J}_q = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \cdot \tau.$$

### 3.2. Action on rational numbers.

**Definition 3.5.** We define the quantized duplication of $PGL_2(\mathbb{Z})$ to be the action

$$\begin{array}{rcl} (\mathrm{PGL}_{2,q}(\mathbb{Z}) \times \mathbb{Z}_2) \times \mathbb{P}^1(\mathbb{Z}(q)) & \to & \mathbb{P}^1(\mathbb{Z}(q)) \\ ((M_q, \epsilon), f) & \mapsto & \left\{ \begin{array}{ll} M_q \cdot f & \text{if } \epsilon = 1 \\ \overline{M}_q \cdot f & \text{if } \epsilon = -1 \end{array} \right. . \end{array}$$

**Proposition 3.6.** Let $M \in \mathrm{PGL}_2(\mathbb{Z})$. Let $x \in \mathbb{P}^1(\mathbb{Q})$. Then in $\Lambda$, if $\det(M) = 1$,

$$\overline{M}_q \cdot [x]^\sharp = [M \cdot x]^b \quad \text{and} \quad \overline{M}_q \cdot [x]^b = [M \cdot x]^\sharp,$$

and if $\det(M) = -1$,

$$\overline{M}_q \cdot [x]^\sharp = [M \cdot x]^\sharp \quad \text{and} \quad \overline{M}_q \cdot [x]^b = [M \cdot x]^b.$$

*Proof.* These equalities are direct consequences of what is above, and the of behaviour of $\overline{I}_q$ :

$$\overline{I}_q \cdot [x]^\sharp = [x]^b \quad \text{and} \quad \overline{I}_q \cdot [x]^b = [x]^\sharp.$$

□

*Example 3.7.* Let us take $M = R$ and $x = n \in \mathbb{N}^*$. We have

$$\begin{align*} \overline{R}_q \cdot [n]^\sharp &= \begin{pmatrix} 1 & q^2 \\ 1-q & q \end{pmatrix} \cdot [n]_{q^{-1}}^\sharp = \frac{[n]_{q^{-1}}^\sharp + q^2}{(1-q)[n]_{q^{-1}}^\sharp + q} \\ &= q^{n-1}[n]_{q^{-1}}^\sharp + q^{n+1} \\ &= [n]_q^\sharp + q^{n+1} \\ &= [n+1]^b. \end{align*}$$

**Corollary 3.8.** Let $x \in \mathbb{P}^1(\mathbb{Q})$. Then

$$[-x]^\sharp = -q^{-1}[x]_{q^{-1}}^\sharp = \frac{-[x]^b_q + 1 - q^{-1}}{(q-1)[x]^b_q + 1}.$$

$$\left[ \frac{1}{x} \right]^\sharp = \frac{1}{[x]_{q^{-1}}^\sharp} = \frac{(q-1)[x]^b_q + 1}{q[x]^b_q + 1 - q}.$$
## 4. FOCUS ON q-REAL NUMBERS

**4.1. Convergences.** Recall the definition of q-deformed real numbers, in [15]. This takes place in the ring $\mathbb{Z}[[q]]$ of formal Laurent series in $q$,

$$ \mathbb{Z}[[q]][[q^{-1}]] = \left\{ \sum_{k \ge \nu} x_k q^k \mid x_k \in \mathbb{Z}, \nu \in \mathbb{Z} \right\}, $$

endowed with its usual metric relying on the valuation $\nu$ of formal Laurent series,

$$ d(f,g) = \frac{1}{2^{\nu(f-g)}}. $$

Thus saying that a sequence $(x_n = \sum_{k \ge \nu_n} x_{n,k} q^k)_n$ in $\mathbb{Z}[[q]]$ converges toward $y = \sum_{k \ge \nu} y_k q^k$ means exactly that the valuations $\nu_n$ stabilize to $\nu$ and the coefficients $x_{n,k}$ stabilize to $y_k$ as $n$ grows,

$$ \forall k \in \mathbb{Z}, \exists N \in \mathbb{N}, \forall n \ge N, x_{n,k} = y_k. $$

**Proposition 4.1** ([15]). Let $x \in \mathbb{R} \setminus \mathbb{Q}$ be an irrational number. Then for any sequence of rational numbers $(x_n)_n$ converging to $x$, the Taylor expansions of $[x_n]^\sharp$ converge toward a Laurent series that does not depend on the choice of the sequence $(x_n)_n$.

**Definition 4.2.** With the previous notations, the resulting series of the convergence of the Taylor expansions of $([x_n]^\sharp)_n$ is defined to be the quantization of $x$ and is denoted by $[x]_q$.

For irrational numbers, the left and right quantizations are the same.

**Proposition 4.3.** Let $x \in \mathbb{R} \setminus \mathbb{Q}$ be an irrational number. Let $(x_n)_n$ be a sequence of rational numbers converging to $x$. Then, the sequence $([x_n]^\sharp - [x_n]^\flat)_n$ converges to the zero series.

*Proof.* We can suppose that $(x_n)_n$ is actually the sequence of convergents of $x : x_n = [a_0, a_1, a_2, \dots, a_n]$, with $a_0 \in \mathbb{Z}$ and $a_i \in \mathbb{Z}_{\ge 1}$ for all $i \ge 1$.

Let us write, for all $n \in \mathbb{N}$, $[x_n]^\sharp = \pm q^d \frac{\mathcal{R}_n(q) + (1-q)\mathcal{R}_{n-1}(q)}{\mathcal{S}_n(q) + (1-q)\mathcal{S}_{n-1}(q)}$, with $\mathcal{R}_n$ and $\mathcal{S}_n$ two coprime monic polynomials in $q$ with constant coefficient 1. The integer $d$ depends on $x$ as

$$ d = \begin{cases} 0 & \text{if } x \ge 1 \\ \lfloor 1/x \rfloor & \text{if } 0 < x < 1 \\ \lfloor x \rfloor & \text{if } x < 0 \end{cases}. $$

The left version of $x_n$ is then

$$ [x_n]^\flat = \pm q^d \frac{q \mathcal{R}_n(q) + (1-q) \mathcal{R}_{n-1}(q)}{q \mathcal{S}_n(q) + (1-q) \mathcal{S}_{n-1}(q)}, $$

so

$$ [x_n]^\sharp - [x_n]^\flat = \pm q^d \frac{(1-q)(\mathcal{R}_n(q)\mathcal{S}_{n-1}(q) - \mathcal{R}_{n-1}(q)\mathcal{S}_n(q))}{q\mathcal{S}_n(q)^2 + (1-q)\mathcal{S}_n(q)\mathcal{S}_{n-1}(q)} = \frac{(1-q)q^{d+a_1+\cdots+a_{n-1}}}{q\mathcal{S}_n(q)(\mathcal{S}_n(q) - \mathcal{S}_{n-1}(q)) + \mathcal{S}_{n-1}(q)}. $$

And the series $1/(q\mathcal{S}_n(q)(\mathcal{S}_n(q) - \mathcal{S}_{n-1}(q)) + \mathcal{S}_{n-1}(q))$ starts with the coefficient 1 thanks to $\mathcal{S}_{n-1}$, so the series $[x_n]^\sharp - [x_n]^\flat$ starts with $q^{d+a_1+\cdots+a_{n-1}}$, and thus converges to the zero series when $n$ grows. $\square$

**Corollary 4.4.** Let $x \in \mathbb{R} \setminus \mathbb{Q}$ be an irrational number. Let $(x_n)_n$ be a sequence of rational numbers converging to $x$. Then the sequence of left q-deformed numbers $([x_n]^\flat)_n$ converges to $[x]_q$.

The following corollary is the second part of Theorem 1.5.
**Corollary 4.5.** Let $x \in \mathbb{R} \setminus \mathbb{Q}$ be an irrational number. Let $M \in \mathrm{PGL}_2(\mathbb{Z})$, with $\det(M) = -1$. Then

$$M_q \cdot [x]_q = [M \cdot x]_q.$$

*Proof.* Let $(x_n)_{n \ge 0}$ be any sequence of rational numbers converging to $x$. For all $n \ge 0$, by the first part of Theorem 1.5,

$$M_q \cdot [x_n]^\sharp = [M \cdot x_n]^\flat.$$

Then by Proposition 4.1 and continuity, $M_q \cdot [x_n]^\sharp$ converges toward $M_q \cdot [x]_q$ in the sense of Laurent series, and by Corollary 4.4 $[M \cdot x_n]^\flat$ converges toward $[M \cdot x]_q$. Therefore

$$M_q \cdot [x]_q = [M \cdot x]_q.$$

□

*Example 4.6.* For every irrational number $x$, we have the formulas

$$[-x]_q = [N]_q \cdot [x]_q \quad \text{and} \quad \left[\frac{1}{x}\right]_q = [NS]_q \cdot [x]_q.$$

More concretely,

$$[-x]_q = \frac{-[x]_q + 1 - q^{-1}}{(q-1)[x]_q + 1} \quad \text{and} \quad \left[\frac{1}{x}\right]_q = \frac{(q-1)[x]_q + 1}{q[x]_q + 1 - q}.$$

These formulas could not be obtained without a quantization of the $\mathrm{PGL}_2(\mathbb{Z})$-action. In [9], it has already been noticed that for $x \in \mathbb{Q}$,

$$\left[\frac{1}{x}\right]^\sharp = \frac{1}{[x]_{q^{-1}}^\sharp} = \overline{J_q} \cdot [x]^\sharp,$$

but this formula could not be generalized to the case of irrational numbers, as there is no clear way of defining the operator $\tau$ on formal Laurent series.

*Example 4.7.* The radius of convergence of the series $[\frac{1}{\pi}]_q$ is the same as the one of $[\pi]_q$, and one can compute the terms of the quantized series of $1/\pi$ with the formula

$$\left[\frac{1}{\pi}\right]_q = \frac{(q-1)[\pi]_q + 1}{q[\pi]_q + 1 - q}.$$

The first 30 terms of the series $[1/\pi]_q$ are

$$q^3 - q^5 - q^6 + q^7 + 2q^8 - 4q^{10} - q^{11} + 5q^{12} + 5q^{13} - 6q^{14} - 11q^{15} + 3q^{16} + 20q^{17} + 6q^{18} - 28q^{19} - 26q^{20} + 31q^{21} + 58q^{22} - 17q^{23} - 103q^{24} - 28q^{25} + 146q^{26} + 131q^{27} - 165q^{28} - 299q^{29} + 94q^{30}...$$

Finally, the terminology of left and right quantizations comes from the following.

**Proposition 4.8** ([15, 1]). Let $x \in \mathbb{Q}$ be a rational number.

(i) Let $(x_n)_n$ be a sequence of rational numbers converging to $x$ by the right. Then the sequences of right and left $q$-deformed numbers $([x_n]^\sharp)_n$ and $([x_n]^\flat)_n$ converge to $[x]^\sharp$.

(ii) Let $(x_n)_n$ be a sequence of rational numbers converging to $x$ by the left. Then the sequences of right and left $q$-deformed numbers $([x_n]^\sharp)_n$ and $([x_n]^\flat)_n$ converge to $[x]^\flat$.
4.2. **Injectivity of quantization.** We are proving Proposition 1.3 for the right quantization. The proof is the same in the left case. So let us consider the right quantization map

$$Q^{\sharp}: \mathbb{R} \to \mathbb{Z}[[q]]\lbrack q^{-1}\rbrack.$$

On the rational numbers, it is injective, because one can evaluate at $q = 1$ to recover the rational number from its quantization.

Let $x$ and $x'$ be two irrational numbers such that $[x]_q = [x']_q$. The continued fractions of $x$ and $x'$ are infinite :

$$x = [a_1, a_2, a_3, \dots], \quad x' = [a'_1, a'_2, a'_3, \dots].$$

By Theorem 2 of [15], we know that the integer part of a real number can be read directly on its $q$-deformation, so we get that $a_1 = a'_1$.

Then put

$$y = \frac{1}{x - a_1} = [a_2, a_3, \dots] \quad \text{and} \quad y' = \frac{1}{x' - a_1} = [a'_2, a'_3, \dots].$$

We have $y = JR^{-a_1} \cdot x$ and $y' = JR^{-a_1} \cdot x'$, so

$$[y]_q = [JR^{-a_1}]_q \cdot [x]_q = [JR^{-a_1}]_q \cdot [x']_q = [y']_q.$$

Again $a_2 = a'_2$, and by induction for all $i$, $a_i = a'_i$, so $x = x'$. Proposition 1.3 is proved.

5. APPLICATION TO $q$-DEFORMED ALGEBRAIC NUMBERS

As an illustration of the extended quantized actions on $\mathbb{P}^1(\mathbb{Z}(q))$, we quantize some algebraic equations, in a compatible way with the $q$-deformations of their roots (being thus algebraic real numbers).

5.1. **Degree 4.** We consider the following particular case of algebraic equation of degree 4, with $b$ an integer :

$$(14) \qquad x^4 - bx^2 + 1 = 0.$$

We want to study the $q$-deformations of the solutions and the relations they share. Assuming there are four distinct real solution $x_1, x_2, x_3$ and $x_4$, we denote by $\sigma_1, \sigma_2, \sigma_3$ and $\sigma_4$ the elementary symmetric polynomials :

* $\sigma_1 = x_1 + x_2 + x_3 + x_4 = 0$;

* $\sigma_2 = x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4 = -b$;

* $\sigma_3 = x_1x_2x_3 + x_1x_2x_4 + x_1x_3x_4 + x_2x_3x_4 = 0$;

* $\sigma_4 = x_1x_2x_3x_4 = 1$.

**Notation 5.1.** For $i \in \{1, 2, 3, 4\}$, the $q$-deformed number $[x_i]_q$ will be denoted $X_i$, and the elementary polynomials in the $X_i$'s will be denoted by $\Sigma_1, \Sigma_2, \Sigma_3, \Sigma_4$.

**Lemma 5.2.** The equation (14) has four distinct real solutions if and only if $b > 2$. Moreover, the polynomial $x^4 - bx^2 + 1$ is irreducible over $\mathbb{Q}$ except when $b = n^2 \pm 2$, for $n \in \mathbb{N}$.

**Proposition 5.3.** Let us suppose that $b > 2$ in equation (14). Then the $q$-deformed numbers $X_1, X_2, X_3$ and $X_4$ are solutions of the following quantized equation

$$X^4 - \Sigma_1 X^3 + \left( \frac{q + q^{-1} - 3}{q - 1} \Sigma_1 + 2q^{-1} \right) X^2 + q^{-1} \Sigma_1 X + q^{-2} = 0,$$

where $\Sigma_1 = X_1 + X_2 + X_3 + X_4$ is a Laurent series in $q$.

In other words, the Vieta's relations become

* $\Sigma_4 = q^{-2}$,

* $\Sigma_3 = -q^{-1}\Sigma_1$,

* $(q-1)\Sigma_2 = (q + q^{-1} - 3)\Sigma_1 + 2(1 - q^{-1})$.
Proof. As the equation (14) is invariant under the two transformations $x \mapsto -x$ and $x \mapsto 1/x$, the matrices $N$ and $J$ act on the roots.

<figure><img src="image_2.png" /><figcaption>FIGURE 1. Galois group action on the roots by linear fractional transformations</figcaption></figure>

After quantization, $[J]_q$ and $[N]_q$ also act on the $X_i$'s, and then

$$ \begin{pmatrix} -1 & 1 - q^{-1} \\ q - 1 & 1 \end{pmatrix} X_4 = X_1 \quad \text{so} \quad (q-1)X_1 X_4 = -X_1 - X_4 + 1 - q^{-1}, $$

and same for $X_2$ and $X_3$: $(q-1)X_2 X_3 = -X_2 - X_3 + 1 - q^{-1}$. Letting $[J]_q$ and $[S]_q$ act, we get four more equations:

$$ qX_1X_2 = (q-1)(X_1 + X_2) + 1, \quad qX_3X_4 = (q-1)(X_3 + X_4) + 1, \quad qX_1X_3 = -1 \text{ and } qX_2X_4 = -1. $$

Using these equations, we get the quantized Vieta's relations. For instance,

$$ \begin{aligned} (q-1)\Sigma_2 &= (q-1)(X_1X_2 + X_3X_4 + X_1X_4 + X_2X_3 + X_2X_4 + X_1X_3) \\ &= q^{-1}(q-1)^2(X_1 + X_2 + X_3 + X_4) + 2q^{-1}(q-1) + (-X_1 - X_2 - X_3 - X_4 + 2(1-q^{-1})) - 2q^{-1}(q-1) \\ &= (q-2+q^{-1})\Sigma_1 - \Sigma_1 + 2(1-q^{-1}) \\ &= (q-3+q^{-1})\Sigma_1 + 2(1-q^{-1}), \end{aligned} $$

and similarly for the other Vieta's relations. $\square$

**Remark 5.4.** The Galois group of the polynomial in equation (14) is isomorphic to the group $\{\text{Id}, J, S, N\}$ which is the Klein four group, acting on the roots $x_1, x_2, x_3, x_4$ exactly as shown in Figure 5.1. Indeed, if $\sigma \in \text{Gal}_{\mathbb{Q}}(x^4 - bx^2 + 1)$, then it is determined by the image of one root $x_1$ of $x^4 - bx^2 + 1$ because all the roots are algebraic expressions of $x_1$.

**Example 5.5.** Let us consider more specifically the case $b = n^2 - 2$, for $n \ge 3$. Then the polynomial splits in two quadratic polynomials

$$ x^4 - (n^2 - 2)x^2 + 1 = (x^2 - nx + 1)(x^2 + nx + 1). $$

The solutions $X_1, X_2, X_3, X_4$ are then quadratic irrational numbers. Following [9], these quantized numbers are solutions in radicals of quantized quadratic equations. For instance, let us detail the way $x^2 - nx + 1$ is quantized. The two roots $x_1, x_2$ of this polynomial satisfy $x_i^2 = nx_i - 1$, so they are fixed points of the matrix $M = \begin{pmatrix} n & -1 \\ 1 & 0 \end{pmatrix} \in \text{PSL}_2(\mathbb{Z})$. In terms of the two generators $R$ and $S$ of the modular group (see (5)), the matrix $M$ is $R^n S$. Then the $q$-deformed numbers $X_1, X_2$ are solutions of a quadratic equation given by the equality $M_q \cdot X_i = X_i$, where

$$ M_q = R_q^n S_q = \begin{pmatrix} [n]^\sharp & -q^{n-1} \\ 1 & 0 \end{pmatrix}. $$

That is, the equation $x^2 - nx + 1 = 0$ is quantized by

$$ X^2 - [n]^\sharp X + q^{n-1} = 0. $$
Likewise, the equation $x^2 + nx + 1$ is quantized by

$$X^2 - [-n]^\sharp X + q^{-n-1} = 0.$$

Therefore, the Vieta's formulas become:

$$\begin{align*}
\Sigma_4 &= q^{-2}, \quad \Sigma_3 = q^{n-1}[-n]^\sharp + q^{-n-1}[n]^\sharp, \\
\Sigma_2 &= [n]^\sharp[-n]^\sharp + q^{-n-1} + q^{n-1} \text{ and } \Sigma_1 = [n]^\sharp + [-n]^\sharp.
\end{align*}$$

The case $b = n^2 + 2$ is similar, except that here we will get a matrix with determinant $-1$. Indeed, the decomposition is

$$x^4 - (n^2 + 2)x^2 + 1 = (x^2 - nx - 1)(x^2 + nx - 1),$$

and the roots $x_1, x_2$ of $x^2 - nx - 1$ are fixed points of the matrix $\begin{pmatrix} n & 1 \\ 1 & 0 \end{pmatrix}$, which is $NR^{-n}S$. Its quantization is

$$\begin{pmatrix} [n]^\flat & q^{-1} \\ 1 & q^{-1} - 1 \end{pmatrix}.$$

Therefore the quantized equations are

$$X^2 + (q^{-1} - 1 - [n]^\flat)X - q^{-1} = 0 \text{ and } X^2 + (q^{-1} - 1 - [-n]^\flat)X - q^{-1} = 0,$$

and the Vieta's relations for the equation $x^4 - (n^2 + 2)x^2 + 1 = 0$ become

$$\begin{align*}
\Sigma_4 &= q^{-2}, \quad \Sigma_3 = 2(q^{-2} - q^{-1}) - q^{-1}([n]^\flat + [-n]^\flat), \\
\Sigma_2 &= [n]^\flat[-n]^\flat - (q^{-1} - 1)([n]^\flat + [-n]^\flat) + 1 - 4q^{-1} + q^{-2} \text{ and } \Sigma_1 = 2(1 - q^{-1}) + [n]^\flat + [-n]^\flat.
\end{align*}$$

**5.2. Degree 6.** We consider the following equation, depending in one integer parameter $b$:

$$(15) \qquad x^6 - 3x^5 - bx^4 + (2b+5)x^3 - bx^2 - 3x + 1 = 0.$$

We keep the same notations as in the case of degree 4, the quantized solutions are denoted $X_i$ ($1 \le i \le 6$) and the symmetric polynomials $\Sigma_i$'s.

**Lemma 5.6.** The equation (15) has six distinct real solutions if and only if $b \ge 1$. In this case, it is irreducible over $\mathbb{Q}$ if and only if $b \ne 2$ and $b \notin \{k^2 + k + 1 | k \ge 0\}$.

*Proof.* Set $f(x) = x^6 - 3x^5 - bx^4 + (2b+5)x^3 - bx^2 - 3x + 1$.

This polynomial is invariant under the action of $J$ and of $\Gamma := \begin{pmatrix} 1 & -1 \\ 1 & 0 \end{pmatrix}$. So the set of roots is also stable by these transformations, and the roots are $x_1, \Gamma(x_1), \Gamma^2(x_1), 1/x_1, 1/\Gamma(x_1), 1/\Gamma^2(x_1)$. Therefore either all the roots are real, either all the roots are nonreal complex numbers. Then it is straightforward to check that the roots are real if and only if $b \ge 1$.

Now let us suppose that $b \ge 1$. If the polynomial $f$ is not irreducible over $\mathbb{Q}$, it has either two factors of degree 3 or three factors of degree 2.

* Let us suppose that $f$ splits into two factors of degree 3. Then these two factors must be

$$(x - x_1)(x - \Gamma(x_1))(x - \Gamma^2(x_1)) \text{ and } \left(x - \frac{1}{x_1}\right)\left(x - \frac{1}{\Gamma(x_1)}\right)\left(x - \frac{1}{\Gamma^2(x_1)}\right).$$

So there is an integer $a$ such that $x_1\Gamma(x_1) + x_1\Gamma^2(x_1) + \Gamma(x_1)\Gamma^2(x_1) = a$, that is

$$x_1^3 - (3+a)x_1^2 + ax_1 + 1 = 0,$$

so the polynomial $f$ splits in two factors that are exactly the one considered in [19],

$$f(x) = (x^3 - (3+a)x^2 + ax + 1)(x^3 + ax^2 - (3+a)x + 1) \quad (\star).$$
Identifying coefficient by coefficient, we get that $a$ must satisfy $a^2 + 3a + 3 - b = 0$, therefore $4b - 3$ must be a square integer, that is $b = k^2 + k + 1$, for some $k \in \mathbb{N}$.

Conversely, if $b = k^2 + k + 1$ for a $k \in \mathbb{N}$, then the factorization $(\star)$ works, with $a = k - 1$.

• Let us suppose that $f$ splits into three factors of degree 2. Then one can check that the only possible configuration is when $b = 2$ and

$$f(x) = (x^2 - 3x + 1)(x^2 - x - 1)(x^2 + x - 1).$$

Thus $f$ is reducible over $\mathbb{Q}$ if and only if $b = 2$ or $b = k^2 + k + 1$ for some $k \in \mathbb{N}$.

<figure><img src="image_3.png" /></figure>

**Proposition 5.7.** Let us suppose that $b \ge 1$. After the $q$-deformation of the six roots of (15), the Vieta's relations are quantized as

• $\Sigma_6 = 1$,

• $\Sigma_5 = -\Sigma_1 + 6$,

• $\Sigma_4 = -5\Sigma_1 + \Sigma_2 + 15$,

• $\Sigma_3 = -5\Sigma_1 + 2\Sigma_2 + 10$,

• $(q-1)\Sigma_2 = (q^2 + q - 4 + q^{-1})\Sigma_1 + 3(-q^2 + q + 2 - q^{-1})$.

*Proof.* We use the symmetries of (15), which is invariant under the action of $J$ and $\Gamma$.

<figure><img src="image_4.png" /><figcaption>FIGURE 2. Galois group action on the roots by linear fractional transformations</figcaption></figure>

After quantization, these transformations give

$$ \begin{align*} X_1 X_2 &= q^{-1}(q-1)(X_1 + X_2) + q^{-1}, & X_1 X_5 &= X_1 - 1, & X_2 X_6 &= X_2 - 1, & X_1 X_4 &= q(X_1 + X_4) + 1 - q, \\ X_4 X_5 &= q^{-1}(q-1)(X_4 + X_5) + q^{-1}, & X_3 X_5 &= X_5 - 1, & X_4 X_6 &= X_6 - 1, & X_2 X_3 &= q(X_2 + X_3) + 1 - q, \\ X_3 X_6 &= q^{-1}(q-1)(X_3 + X_6) + q^{-1}, & X_1 X_3 &= X_3 - 1, & X_2 X_4 &= X_4 - 1, & X_5 X_6 &= q(X_5 + X_6) + 1 - q. \end{align*} $$

The $q$-deformed Vieta's relations follow.

<figure><img src="image_5.png" /></figure>

*Example 5.8.* Let us first take $b = 2$, so that the equation splits into $(x^2 - 3x + 1)(x^2 - x - 1)(x^2 + x - 1) = 0$. The quantization of this equation comes from the quantization of quadratic equations, and it is

$$(X^2 - [3]^\sharp X + q^2)(qX^2 - (q^2 + q - 1)X - 1)(q^2X^2 - (q^2 - q - 1)X - q) = 0.$$

Therefore in this case, $\Sigma_1 = q + 2q^2 + 3q^3 + 2q^4 + q^5$ is finite, and we can check all the Vieta's relations of Proposition 5.7.

*Example 5.9.* Now let us take $b = 3$ so that the equation splits into

$$(x^3 - 3x^2 + 1)(x^3 - 3x + 1) = 0.$$
According to [19], these two factors are q-deformed as

$$ (X^3 - B_1(q)X^2 + (B_1(q) - 3)X + 1)(X^3 - B_2(q)X^2 + (B_2(q) - 3)X + 1) = 0, $$

with $B_1(q) = X_1(q) + X_3(q) + X_5(q)$ (resp. $B_2(q) = X_2(q) + X_4(q) + X_6(q)$) the sum of odd (resp. even) roots.

We can check again that the relations of Proposition 5.7 are satisfied. Note that unlike the quadratic case, here it seems that the series $B_1(q)$ and $B_2(q)$ are actually infinite.

**Remark 5.10.** As all roots of the polynomial defining (15) are algebraic expressions of one (arbitrarily chosen) root $x_1$, the Galois group is isomorphic to the subgroup of $\mathfrak{S}_6$ generated by $(1 \ 2)(3 \ 6)(4 \ 5)$ and $(1 \ 3 \ 5)(2 \ 6 \ 4)$, and its action on the roots is realized by the linear fractional transformations generated by $J$ and $\Gamma$, as depicted in Figure 5.2.

## APPENDIX A. FENCE POSET COMBINATORICS

One of the key properties of traces of q-deformed matrices of PSL$_2,q(Z)$ is that they can be realized as generating functions, at least in the positive case. This property was the main tool used by Oğuz and Ravichandran in [17] to prove a conjecture of Morier-Genoud and Ovsienko about unimodality of q-traces (see also [11]). For matrices of determinant $-1$, we find the same kind of underlying combinatorics, with slight modifications.

**Definition A.1.** Let $(a_1, \dots, a_m)$ be a sequence of positive integers. We define the left circular fence poset associated to this sequence, denoted here by $\overline{F}^{\flat}(a_1, \dots, a_m)$, to be

<figure><img src="image_6.png" /></figure>

Let us also define an admissible ideal of this poset to be a subset $I \subset \overline{F}^{\flat}(a_1, \dots, a_m)$ such that

(i) for all $x \in I$, if $y \le x$ then $y \in I$;

(ii) if $0 \notin I$, then the two white vertices are together in $I$ or together in $I^c$.

**Example A.2.** The left circular fence poset associated with the sequence $(1, 2, 2)$ looks like

<figure><img src="image_7.png" /></figure>

The admissible ideals of this poset are

$$ \emptyset \{0\} \{0,3\} \{2,3\} \{0,2,3\} \{0,1,2,3\} \{0,3,4,5\} \{0,2,3,4,5\} \{0,1,2,3,4,5\}. $$

**Proposition A.3.** Let $(a_1, \dots, a_m)$ be a sequence of positive integers with m odd.

Let $M_q := R_q^{a_1} J_q R_q^{a_2} J_q \cdots R_q^{a_m} J_q$ be the associated q-deformed matrix.

Then $t^{-\frac{m-1}{2}} \mathrm{Tr}(M_q)$ is the generating function of admissible ideals of $\overline{F}^{\flat}(a_1, \dots, a_m)$.
Proof. According to the previous Lemma 2.10, and keeping the same notations, Tr($M_q$) = $t^{\frac{m-1}{2}}$($qU^b + (V')^b$). In Appendix A of [1], it is given a combinatorial interpretation of left q-rational numbers. In particular, $U^b$ is the generating function of lower ideals of the poset $F =$

<figure><img src="image_8.png" /></figure>

Similarly, $(V')^b$ is the generating function for the poset $F' =$

<figure><img src="image_9.png" /></figure>

Let us denote by $\bar{J}^b$ the poset of lower ideals of the left circular fence poset $\bar{F}^b(a_1, \dots, a_m)$ and by $J$ (resp. $J'$) the poset of lower ideals of $F$ (resp. $F'$). The poset $J$ is embedded in $\bar{J}^b$ by sending an ideal $I$ of $J$ on $I \cup \{0\}$ in $\bar{J}^b$ (as the image contains the vertex 0, the dashed edge is not working), and the poset $J'$ is embedded in $\bar{J}^b$ by sending an ideal $I'$ of $J'$ on the same ideal in $\bar{J}^b$, so that it does not contain 0 and the dashed edge is effective. This builds a bijection between $\bar{J}^b$ and $J \sqcup J'$, which is translated for the generating functions by the equality

$$qU^b + (V')^b = \sum_{\bar{I} \in \bar{J}^b} q^{|\bar{I}|}.$$

□

**Example A.4.** We consider again the sequence $(1, 2, 2)$. The associated matrix is

$$M_q = R_q J_q R_q^2 J_q R_q^2 J_q = t \begin{pmatrix} q^6 + q^5 + 2q^4 + q^3 + q^2 + q & q^3 + q^2 + 1 \\ q^5 + q^4 + q^3 + q^2 + q & q^2 + 1 \end{pmatrix},$$

so Tr($t^{-1}M_q$) = $1 + q + 2q^2 + q^3 + 2q^4 + q^5 + q^6$.

This corresponds to the generating function of admissible ideals of the poset described in exemple A.2 just above.

## APPENDIX B. POSITIVITY OF q-TRACES OF DETERMINANT 1

We would like to add here a result about positivity of q-traces for matrices of determinant 1. It was originally studied by Leclere and Morier-Genoud in [9], but it happens that their statement (Lemma 3.10) was slightly incorrect. We propose a little modification of this lemma to make it work, and give a proof following the proof in [9].

**Lemma B.1.** Let $c_1, \dots, c_k$ be integers greater or equal to 2. Let

$$M_q(c_1, \dots, c_k) = R^{c_1} S R^{c_2} S \cdots R^{c_k} S$$

be the associated matrix of PSL$_2(\mathbb{Z})$. Then the q-trace of this matrix is a polynomial in $q$ with positive coefficients.
*Proof.* We will write $P \ge 0$ to say that the polynomial $P$ has positive coefficients. We start an induction on $k$, aiming to prove that

$$ (H) \quad \forall k \ge 1, \quad \forall (c_1, \dots, c_k) \in \mathbb{Z}_{\ge 2}^k, \text{ if } M_q(c_1, \dots, c_k) = \begin{pmatrix} A & B \\ C & D \end{pmatrix}, \text{ we have} \\ A \ge 0, \quad -B \ge 0, \quad A+D \ge 0 \text{ and } A+B \ge 0. $$

For $k=1$,

$$ M_q(c_1) = \begin{pmatrix} [c_1]^\sharp & -q^{c_1-1} \\ 1 & 0 \end{pmatrix}. $$

And (H) is satisfied.

Let $k \ge 2$ and let us suppose that (H) is true for any $k_0 \le k$. Let $c_1, \dots, c_k, c$ be $k+1$ integers greater or equal to 2. We set

$$ M_q(c_1, \dots, c_k) = \begin{pmatrix} A & B \\ C & D \end{pmatrix}, $$

so that by hypothesis we have $A \ge 0, -B \ge 0, A+D \ge 0$ and $A+B \ge 0$.

Moreover, as the $c_i$'s are greater than 2, the rational number computed by their negative continued fraction is greater than 1, and by total positivity of $q$-rational numbers, we can deduce that $A-C \ge 0$.

Then one can compute

$$ M_q(c_1, \dots, c_k, c) = \begin{pmatrix} [c]^\sharp A + B & -q^{c-1}A \\ [c]^\sharp C + D & -q^{c-1}C \end{pmatrix} = \begin{pmatrix} A' & B' \\ C' & D' \end{pmatrix}. $$

Then $A' = q[c-1]^\sharp A + (A+B) \ge 0, -B' = q^{c-1}A \ge 0$, and

$$ \begin{aligned} A' + D' &= q[c-1]^\sharp A + A + B - q^{c-1}C \\ &= q[c-2]^\sharp A + q^{c-1}(A-C) + (A+B) \\ &\ge 0. \end{aligned} $$

Likewise,

$$ \begin{aligned} A' + B' &= [c]^\sharp A + B - q^{c-1}A \\ &= q[c-2]^\sharp A + (A+B) \\ &\ge 0. \end{aligned} $$

This concludes the induction and the proof. $\square$

<figure><img src="image_10.png" /></figure>

## APPENDIX C. PROOF OF PROPOSITION 1.1

We explain here why we consider only two $q$-deformations of the rational projective line, the left and the right versions, corresponding to the orbits in $\mathbb{P}^1(\mathbb{Q}(q))$ of $\frac{1}{0}$ and $\frac{1}{1-q}$ respectively.

Let $f(q) \in \mathbb{P}^1(\mathbb{Q}(q))$ be such that $f(1) = \frac{1}{0} \in \mathbb{P}^1(\mathbb{Q})$. Let $O_f$ be the orbit of $f(q)$ under the action of the projective modular space $\text{PSL}_{2,q}(\mathbb{Z})$. We are interested in the following proposition :

For all $x \in \mathbb{Q}$, there is exactly one element $g_x \in O_f$ such that $g_x(1) = x$.

This amounts to requiring that, via the isomorphism (8) between $\text{PSL}_2(\mathbb{Z})$ and $\text{PSL}_{2,q}(\mathbb{Z})$,

$$ (16) \qquad \text{Stab}_{\text{PSL}_{2,q}(\mathbb{Z})}(f(q)) \simeq \text{Stab}_{\text{PSL}_{2,q}(\mathbb{Z})} \left(\frac{1}{0}\right) = \langle R \rangle. $$
We need then to show that this equality (16) holds if and only if $f \in \{\frac{1}{0}, \frac{1}{1-q}\}$.

Let us take $f$ such that (16) holds. This means that $R_q \cdot f = f$, so $qf + 1 = f$. Let us write $f$ as a quotient of two coprime polynomials $f = \frac{h}{g}$.

We get $\frac{qh+g}{g} = \frac{h}{g}$, so $g(qh+g) = hg$. Then either $g=0$ meaning that $f = \frac{1}{0}$, either $g \neq 0$ and $(1-q)h = g$, meaning that $f = \frac{1}{1-q}$.

Conversely, the two solutions $f = \frac{1}{0}$ and $f = \frac{1}{1-q}$ satisfy (16).

## ACKNOWLEDGEMENTS

I would like to thank my advisor Sophie Morier-Genoud, and Valentin Ovsienko, who guided me all the way long.

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PERRINE JOUTEUR, LABORATOIRE DE MATHÉMATIQUES DE REIMS, UMR 9008 CNRS ET UNIVERSITÉ DE REIMS CHAMPAGNE-ARDENNE, U.F.R. SCIENCES EXACTES ET NATURELLES, MOULIN DE LA HOUSSE - BP 1039, 51687 REIMS CEDEX 2, FRANCE
*Email address: Perrine.jouteur@univ-reims.fr*