# Minimal algebraic solutions of the sixth equation of Painlevé

R. Conte

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R. Conte. Minimal algebraic solutions of the sixth equation of Painlevé. Theoretical and Mathematical Physics, 2025. hal-05033669

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# Minimal algebraic solutions of the sixth equation of Painlevé

Robert Conte¹,²

1. Université Paris-Saclay, ENS Paris-Saclay, CNRS, Centre Borelli, F-91190 Gif-sur-Yvette, France.

2. Department of Mathematics, The University of Hong Kong, Pokfulam, Hong Kong.

Theoretical and mathematical physics, accepted 07 April 2025

Keywords: sixth Painlevé equation, algebraic solutions.

PACS 02.30.Hq

AMS MSC 2000 33E17

## Abstract

For each of the forty-eight exceptional algebraic solutions $u(x)$ of the sixth equation of Painlevé, we build the algebraic curve $P(u, x) = 0$ of a degree conjectured to be minimal, then we give an optimal parametric representation of it. This degree is equal to the number of branches, except for fifteen solutions.

## Contents

<table><tr><td><b>1 Introduction and motivation</b></td><td><b>1</b></td></tr><tr><td><b>2 Statement of the problem</b></td><td><b>2</b></td></tr><tr><td><b>3 Results</b></td><td><b>4</b></td></tr><tr><td>3.1 Solutions of genus zero</td><td>5</td></tr><tr><td>3.1.1 Unfolded solutions of genus zero</td><td>5</td></tr><tr><td>3.1.2 Folded solutions of genus zero</td><td>7</td></tr><tr><td>3.2 Solutions of genus one</td><td>8</td></tr><tr><td>3.2.1 Unfolded solutions of genus one</td><td>9</td></tr><tr><td>3.2.2 Folded solutions of genus one</td><td>11</td></tr><tr><td>3.3 Folded solutions of genus higher than one</td><td>12</td></tr><tr><td><b>A Transformations conserving P<sub>VI</sub></b></td><td><b>13</b></td></tr><tr><td><b>B Optimal representations</b></td><td><b>14</b></td></tr><tr><td>B.1 Rational representations (genus zero)</td><td>14</td></tr><tr><td>B.2 Elliptic representations (genus one)</td><td>15</td></tr></table>

## 1 Introduction and motivation

Apart the transcendental general solution which characterizes it, the sixth equation $P_{VI}$ of Painlevé

$$ \frac{d^2 u}{dx^2} = \frac{1}{2} \left[ \frac{1}{u} + \frac{1}{u-1} + \frac{1}{u-x} \right] \left( \frac{du}{dx} \right)^2 - \left[ \frac{1}{x} + \frac{1}{x-1} + \frac{1}{u-x} \right] \frac{du}{dx} + \frac{u(u-1)(u-x)}{2x^2(x-1)^2} \left[ \theta_{\infty}^2 - \theta_0^2 \frac{x}{u^2} + \theta_1^2 \frac{x-1}{(u-1)^2} + (1-\theta_x^2) \frac{x(x-1)}{(u-x)^2} \right], $$
$$ (2\alpha, -2\beta, 2\gamma, 1 - 2\delta) = (\theta_{\infty}^2, \theta_0^2, \theta_1^2, \theta_x^2), \qquad (1) $$

admits exactly 48 exceptional algebraic solutions $u(x)$, in the sense that they are neither those mentioned by Picard [1, p. 298] when the four constant parameters $\theta_j$ vanish, nor some particular solutions of the Riccati equation admitting $P_{VI}$ as a differential consequence for some constraint between the $\theta_j$'s. They were discovered by various authors: Hitchin [2], Dubrovin [3], Dubrovin and Mazzocco [4], Kitaev [5] [6], Andreev and Kitaev [7], and all the others by Boalch [8] [9] [10] [11]. The proof that no other such solutions exist is due to Lisovyy and Tykhyy [12].

These 48 solutions are in fact equivalence classes of a group which leaves $P_{VI}$ form-invariant without constraining the $\theta_j$'s, group made of: the 4 sign changes of the $\theta_j$'s, the 4! homographies of $(u, x)$ and a unique birational transformation (see Appendix). Let us denote $P(u, x) = 0$ the algebraic curve representing an element of one of these equivalence classes. Its genus $g$ and number of branches $b$ (degree of the polynomial $P$ in $u$) are invariant under the group, but its degree $d$ (global degree of $P$ in $u$ and $x$) is not. In order to simplify the writing of the solutions, it is therefore convenient to lower this degree $d$, ideally to its minimal value $b$, by the repeated action of elements of the group.

*Remark.* The above notation $\theta = (\theta_{\infty}, \theta_0, \theta_1, \theta_x)$ for the sequence of the four monodromy exponents is the natural choice for at least three reasons. The first one is the writing of $P_{VI}$ in elliptic coordinates[13, 14, 15]

$$ \frac{d^2 U}{dX^2} = \frac{(2\omega)^3}{\pi^2} \sum_{j=\infty,0,1,x} \theta_j^2 \phi'(2\omega U + \omega_j, g_2, g_3), \qquad (2) $$

obviously form-invariant under any permutation of the four half-periods $\omega_j$ of the elliptic function $\phi$; defining $\theta_{\infty}$ with a shift of unity by $2\alpha = (\theta_{\infty} - 1)^2$ would break this invariance. The second one is the classical confluence [14] between the $P_n$'s, in which the four singularities $(\infty, 0, 1, x)$ of $u$ become successively $(\infty, 0, 1)$, $(\infty, 0)$, $(\infty)$, which defines an order in the sequence $(\theta_{\infty}, \theta_0, \theta_1, \theta_x)$; reordering this sequence into $(\theta_0, \theta_x, \theta_1, \theta_{\infty} - 1)$ [8, Eq. (4)] or into $(\theta_0, \theta_1, \theta_x, \theta_{\infty} - 1)$ [12, p. 125] would make one forget about this natural ordering. The third one is the choice of Painlevé [14] for the Greek alphabetical order $\alpha, \beta, \gamma, \delta$.

## 2 Statement of the problem

The goal of the present paper is therefore to solve the following problem.

**Problem.** For each of the 48 algebraic solutions, find a representative in its equivalence class whose degree $d$ equals the number of branches $b$.

As an example, consider one of the two solutions with $b = 5$ branches, the one found by Kitaev [5, Eq. (3.3)] which bears the number I21 in the tables de Boalch [8] and 2 in those of Lisovyy and Tykhyy [12]. Its genus is $g = 0$, its degree the nonoptimal value $d = b + 1$ and its number of terms is 12. An homography followed by the birational transformation simplifies it to $b = d = 5$ containing only 4 terms,

$$ \left\{ \begin{array}{l} \theta = \frac{(3, 1, 2, 1)}{5}, P = 9(2-x)u^5 + 15(x-4)u^4 + 5(10-x^2+9x)u^3 - 15(5x+3x^2)u^2 \\ \qquad + 45x^2u - x^2 - 8x^3, \\ \\ \theta = \frac{(1, 1, 2, 3)}{5}, P = (9x-8)u^5 - 5u^4x - 5x(x-3)u^3 - 5x(3x-1)u^2 + 5x^2u - 9x^2 + 8x^3, \\ \\ \theta = \frac{(0, 0, 1, 2)}{5}, P = u^5 - 5xu^3 + 5xu^2 - x^2. \end{array} \right. \qquad (3) $$

The sought simplification therefore consists in removing any fixed pole of $u$ (in the above example I21, the fixed pole $x = 2$), so that $u$ admits as only singularities the three critical fixed points $x = \infty, 0, 1$. For such removable fixed singularities, see for example the figures in [16].

There exists another transformation [17], called folding transformation, and it is unique [18], which leaves $P_{VI}$ form-invariant, but this is at the expense of two constraints between the $\theta_j$'s. A solution $\theta = (0, 0, 2a, 2b)$ (modulo permutations and sign changes) is folded into another solution $\theta = (a, a, b, b)$, an operation symbolized as "unfolded $\le$ folded" par Boalch.

Let us recall the partition of the 48 algebraic solutions in three disjoint types [19, p. 233] :
1. (genus zero) three solutions which depend on at most two arbitrary $\theta_j$'s;

2. (genus zero or one) thirty solutions having rational, non-arbitrary $\theta_j$'s, inequivalent under the folding transformation [17];

3. (genera 0, 1, 2, 3, 7) fifteen transformed of seven out of the thirty ones by folding(s).

The final result is the following.

**Conjecture.** Only 7 of the 30 unfolded solutions, and 8 of the 15 folded ones display a nonzero difference $d-b$, and this difference is minimal.

Before the present work, only 4 of the 30 unfolded solutions, and 3 of the 15 folded ones displayed a minimal $d-b$. The main reason is precisely the elegance of the method of Boalch. That method makes him find rational numbers $\theta_j$ whose denominators have a minimal gcd [8, Table 2], and it happens that the birational transformation doubles this gcd in most cases but that, surprisingly, it greatly simplifies (in the sense above defined) the representative. Table 1 displays numerous such examples : I22, I23, O09, I24, I25, etc.

Several facts support this conjecture.

1. Existence of several sets of siblings (in the sense defined by Boalch [8], i.e. whose all members have by definition the same values of $b$ and $x$), one member having an optimal degree $d=b$ and the others degrees $d \ge b+1$ impossible to lower by any transformation.

2. Probable nonexistence, for the solution I50 with $b=40$ branches, of a representative $d-b \le 5$ whose all $\theta_j$'s would be equal, so as to preserve its property to be doubly folded. Indeed, the representatives whose common value of the four $\theta_j$'s is respectively 3/20, 7/20, 13/20, 17/20 have degrees 46, 54, 78, 102 and a number of terms equal to 335, 663, 1647, 2631.

3. Failure to find a representative $d=b$ of the solution with the smallest $b$ having a degree $d \neq b$ (solution labeled K= Klein), after a triple loop on the $2^4$ sign changes, the 4! homographies and five birational transformations (three months of computation).
# 3 Results

They take two forms.

1. The solution of the above stated problem, summarized in Table 1: among the 30 + 15 solutions, the 7 + 8 without a representative $d = b$ have a minimal number $d - b$ of fixed poles equal to six for I50, I52, two for I38, 238, I46, O13, I51, I47, one for K, I34, I37, I43, I28, O12, I48. In particular, the associated representative possesses the maximal possible number of null $\theta_j$'s.

2. For a certain representative of each solution, a “simple” (in the senses of Appendix B) parametric representation listed in sections 3.1 and 3.2.

The representatives of these two forms may not be identical, but they only differ by a homography. Table 1 deserves some remarks.

1. All minimal solutions have all their $\theta_j$'s smaller than the unity.

2. The strategy of Dubrovin and Mazzocco [4] consisting in looking for solutions with three null $\theta_j$, which led them to the discovery of solutions (H3), (H3)', (H3)" (Table 1, last column), perfectly matches the property to be doubly foldable, a property which requires the presence of at least two null $\theta_j$'s.

3. Only four solutions, all doubly folded (O13, I50, I51, I52), are invariant under the 24 homographies, their $\theta_j$'s are all equal.

4. Only one solution (I33) is not invariant by any homography (apart the identity), this is the one called “generic” by Boalch. This is also the unique one whose four $\theta_j$'s are all different, whatever be the representative.

5. As noticed by Boalch [11], in each set of siblings (I22, I23), (I24, I25), (I26, I27), (I29, I30), (I34, I35), (I37, I38), (I39, I40), (237, 238, 239), (I42, I43), (I44, I45), (I47, I48), (I50,I51), all the members can be represented by the same value of $x$, at the possible expense of a greater degree or a higher number of terms. For instance, the minimal number of terms in the equivalence class of I24 is 11, not 17.

6. The siblings (I26, I27) ($b = d = 9$) seem to have not been noticed yet.

7. The solution I36 and the siblings (I42, I43) share the same elliptic curve, a fact yet unnoticed.

8. Two quadruplets of monodromy exponents, respectively $(0, 0, 1, 1)/6$ and $(1, 1, 1, 1)/12$, are common to more than one solution, respectively (T06, O11, I49) and (O12, I52).

9. The solution III has only one child, but this is the parent of O11 and T06.

10. All solutions with two null $\theta_j$'s are minimal.
## 3.1 Solutions of genus zero

The chosen representatives for the genus zero are those of Table 1, but other choices allow a better symmetry of the expressions, see Appendix B.1 for details.

### 3.1.1 Unfolded solutions of genus zero

* II, III, IV

$$ 
\begin{aligned}
II, b = 2, \theta &= (a, a, b, b), u = s, x = s^2, u^2 - x = 0, \\
III, b = 3, \theta &= (a, 2a, a, 1/3), u = s, x = \frac{s^3}{3s-2}, u^3 - 3xu + 2x = 0, \\
IV, b = 4, \theta &= (b, b, b, 1/2), u = s, x = -\frac{s^2(s^2-2s)}{2s-1}, u^4 - 2u^3 + 2xu - x = 0.
\end{aligned} 
\quad (5) $$

* I20=LT01.

$$ b = 5, \theta = \frac{(1, 3, 3, 4)}{15}, x = \frac{(2s-1)^2(2s+9)^3}{2(20s^2+27)^2}, u = \frac{(2s+9)(2s-1)(2s-3)}{2(20s^2+27)}. \quad (6) $$

* I21=LT02.

$$ b = 5, \theta = \frac{(0, 0, 1, 2)}{5}, x = \frac{(3s-1)^2(s+3)^3}{8s^3(s+5)^2}, u = \frac{(s+3)(3s-1)}{2s(s+5)}. \quad (7) $$

* O08=LT05.

$$ b = 6, \theta = \frac{(0, 0, 1, 5)}{12}, x = \frac{(3s-8)^4}{s^4(s-3)^2}, u = \frac{3s-8}{s(s-3)}. \quad (8) $$

* Siblings I23=LT06, I22=LT07.

$$ 
\begin{aligned}
b = 6, \theta_{I23} &= \frac{(1, 5, 7, 7)}{30}, \theta_{I22} = \frac{(1, 1, 5, 13)}{30}, x = -\frac{(2s-1)(5s-1)^2}{s^3(9s-2)(9s-5)^2}, \\
u_{I23} &= \frac{(5s-1)(2s-1)}{s^2(9s-5)}, u_{I22} = -\frac{5s-1}{s(9s-5)}.
\end{aligned} 
\quad (9) $$

* Klein=LT08.

$$ b = 7, \theta = \frac{(1, 1, 2, 1)}{7}, x = \frac{2(s-3)^3(s^2-6s+16)^2}{s^3(s^2-7s+14)^2}, u = \frac{(s^2-6s+16)(s-3)^2(s-4)}{s^2(s^2-7s+14)}. \quad (10) $$

* O09=LT10.

$$ b = 8, \theta = \frac{(1, 3, 3, 7)}{24}, x = -\frac{4(4+4s+3s^2)^2}{s^3(s^2+2s+4)^2(s+4)}, u = \frac{2(s+1)(4+4s+3s^2)}{s(s+4)(s^2+2s+4)}. \quad (11) $$

* Siblings I24=LT11, I25=LT12.

$$ 
\begin{aligned}
b = 8, \theta_{I24} &= \frac{(1, 5, 1, 7)}{20}, \theta_{I25} = \frac{(1, 3, 5, 3)}{20}, x = \frac{(s-3)^3(s+5)^5}{64s^3(s^2-6s+25)^2}, \\
u_{I24} &= \frac{(s+5)(s-3)(s^2-10s+5)}{8s(s^2-6s+25)}, u_{I25} = \frac{(s-5)(s+5)^2(s-3)^2}{16s^2(s^2-6s+25)}.
\end{aligned} 
\quad (12) $$

* I32=LT16.

$$ b = 10, \theta = \frac{(0, 0, 0, 1)}{5}, x = \frac{(s-1)^5(3s+1)^3(s^2+4s-1)}{256s^5(5s^2-1)}, u = -\frac{(3s+1)(s-1)^3}{16s^3}. \quad (13) $$
• I31=LT17.

$$b = 10, \theta = \frac{(0,0,0,3)}{5}, x = -\frac{(s^2 + s - 1)(2s + 1)^3}{s^5(s^2 - 1 - s)(s + 2)^3}, u = \frac{2s + 1}{s(s + 2)}. \quad (14)$$

• Siblings I29=LT18, I30=LT19.

$$b = 10, \theta_{I29} = \frac{(3,7,7,7)}{30}, \theta_{I30} = \frac{(1,1,9,1)}{30}, x = \frac{(2s^2 - s + 2)^2(s + 2)^5(3s - 2)}{8s^5(5s^2 + 12)^2}, \\ u_{I29} = \frac{(2s^2 - s + 2)(s + 2)^2(s^2 - s + 2)}{2s^3(5s^2 + 12)}, u_{I30} = \frac{(2s^2 - s + 2)(s + 2)^4}{4s^4(5s^2 + 12)}. \quad (15)$$

• I33=LT25.

$$b = 12, \theta = \frac{(1,7,11,17)}{60}, x = \frac{4(7s - 10)(s^2 + 20)^2(s^2 - 5s + 10)^3(s - 5)}{27s^5(s^2 - 4s + 20)^2(s - 4)^3}, \\ u = \frac{2(7s - 10)(s^2 + 20)(s^2 - 5s + 10)}{3s^2(s^2 - 4s + 20)(s - 4)^2}. \quad (16)$$
### 3.1.2 Folded solutions of genus zero

The correspondence between the $s$ of the folded solution and the $s$ of the unfolded one is not rational, but there exists a representation on $\mathbb{Q}$ for all the solutions but one (O13=LT30, which requires the extension $i$).

* O07=LT04.

$$b = 6, \theta = \frac{(1, 1, 5, 5)}{24}, x = \frac{(s+3)^3(3s+1)^3}{4s(9s^2+14s+9)^2}, u = \frac{(s+3)(3s+1)^2}{2(9s^2+14s+9)}. \quad (17)$$

* T06=LT03.

$$b = 6, \theta = \frac{(0, 0, 1, 1)}{6}, x = \frac{(s-1)^3(s-3)^3}{s^3(s-2)^3}, u = \frac{(s-3)(s-1)^2}{s(s-2)^2}. \quad (18)$$

* O10=LT09.

$$b = 8, \theta = \frac{(0, 0, 1, 1)}{4}, x = \frac{16(s-1)^3(s-3)^3}{s^3(s-2)^2(s-4)^3}, u = \frac{4(s-1)(s-3)^2}{s^2(s-2)(s-4)}. \quad (19)$$

* O13=LT30.

$$b = 16, \theta = \frac{(1, 1, 1, 1)}{8}, x = \frac{(s^2-1)^2(s^4+6s^2+1)^3}{32s^2(s^4+1)^3}, \\ u = \frac{(1+i)(s^2+(1-i)s+i)(s^2+2is+1)(s^2-1)(s^2-2is+1)^2}{8s(s^2-i)^2(s^2+i)(s^2+(1+i)s-i)}. \quad (20)$$

This representative is invariant under the $4!$ homographies. This is the only solution without any real branch [9, p 99]. Its representation depends on one algebraic number, normalized to $i$.

* I28=LT15.

$$b = 10, \theta = \frac{(1, 1, 2, 2)}{10}, x = \frac{(2s+3)^3(5s^2+4s+1)^2}{s^3(s^2+4s+5)^2(3s+2)^3}, u = \frac{(2s+3)(s-1)(5s^2+4s+1)}{s(s^2+4s+5)(3s+2)^2(s+1)}. \quad (21)$$

* O11=LT21.

$$b = 12, \theta = \frac{(0, 0, 1, 1)}{6}, x = \frac{(6s^2-8s+3)^2(2s-3)^4}{16s^4(2s^2-8s+9)^2(s-1)^4}, u = \frac{(2s-3)(6s^2-8s+3)}{4s(2s^2-8s+9)(s-1)^3}. \quad (22)$$
## 3.2 Solutions of genus one

The elliptic curve $t^2 = c_3s^3 + c_2s^2 + c_1s + c_0, c_3 \neq 0$, an affine transform of the curve of Weierstrass $\wp'(\lambda)^2 = 4\wp(\lambda)^3 - g_2\wp(\lambda) - g_3 = 4(\wp(\lambda) - e_1)(\wp(\lambda) - e_2)(\wp(\lambda) - e_3)$, is chosen so as to minimize the size of the integer numbers.

Each solution of genus $g$ higher than or equal to one admits a representative invariant under the involution $(u, x) \to (1-u, 1-x)$, this representative therefore admits for $g=1$ the representation [8],

$$x = \frac{1}{2} + R_1(s)t, u = \frac{1}{2} + R_2(s)t. \qquad (23)$$

This representative may not be the one in Table 1 (chosen to minimize the degree $d$ and the number of terms of the algebraic curve), but they differ by a homography.

This convention of simplicity is detailed in the Appendix B.2.
### 3.2.1 Unfolded solutions of genus one

All the representatives in this section are chosen invariant under the involution $(x, u, s, t) \to (1-x, 1-u, s, -t)$.

* Siblings I27=LT13, I26=LT14.

$$ 
\begin{aligned}
b &= 9, \theta_{I27} = \frac{(1, 2, 2, 2)}{15}, \theta_{I26} = \frac{(1, 1, 1, 8)}{15}, t^2 = \frac{s(2s+1)(5s+16)}{36}, s = \frac{12\varphi(\lambda) - 37}{30}, t = \frac{\varphi'(\lambda)}{15}, \\
x &= \frac{1}{2} - \frac{25s^4 - 40s^3 - 84s^2 - 136s - 8}{72s(2s+1)^2}t, u_{I27} = \frac{1}{2} + \frac{5s^2 + 2s + 2}{6s(2s+1)}t, u_{I26} = \frac{1}{2} - \frac{s-1}{s(2s+1)}t.
\end{aligned} 
\quad (24) $$

All the poles $s$ being also zeroes of $t$, the fields $x, u_{I27}, u_{I26}$ are polynomials of the elliptic functions of Jacobi.

* I36=LT22. The elliptic curve is identical to that of the siblings (I42, I43).

$$ 
\begin{aligned}
b &= 12, \theta = \frac{(1, 3, 3, 11)}{30}, t^2 = 3(5s - 2)(16s^2 - 25s + 10), \\
s &= \frac{\varphi(\lambda) + 157}{240}, t = \frac{\varphi'(\lambda)}{480}, g_2 = -7788, g_3 = 432856, \\
x &= \frac{1}{2} - \frac{160s^6 + 600s^5 - 2865s^4 + 4100s^3 - 2820s^2 + 960s - 128}{6s^2(16s^2 - 25s + 10)t}, u = \frac{1}{2} - \frac{31s^2 - 46s + 16}{2st}.
\end{aligned} 
\quad (25) $$

* Siblings I34=LT23, I35=LT24.

$$ 
\begin{aligned}
b &= 12, \theta_{I34} = \frac{(13, 5, 5, 23)}{60}, \theta_{I35} = \frac{(1, 5, 5, 11)}{60}, \\
t^2 &= s(32s^2 - 95s + 80), s = \frac{3\varphi(\lambda) + 95}{96}, t = \frac{\varphi'(\lambda)}{64}, \\
x &= \frac{1}{2} + \frac{2048s^6 - 15360s^5 + 48000s^4 - 77840s^3 + 65685s^2 - 20328s - 4000}{54(32s^2 - 95s + 80)t}, \\
u_{I34} &= \frac{1}{2} - \frac{64s^3 - 216s^2 + 249s - 200}{6(8s - 13)t}, u_{I35} = \frac{1}{2} + \frac{64s^3 - 288s^2 + 447s - 200}{18t}.
\end{aligned} 
\quad (26) $$

The fields $x$ and $u_{I35}$ are polynomials of the functions of Jacobi.

* Siblings I38=LT26, I37=LT27.

$$ 
\begin{aligned}
b &= 15, \theta_{I38} = \frac{(2, 2, 2, 3)}{15}, \theta_{I37} = \frac{(1, 1, 1, 6)}{15}, d_{I38} - b = 2, d_{I37} - b = 1, \\
t^2 &= 3(5s - 2)(4s^2 - 5s + 10), s = \frac{3\varphi(\lambda) + 11}{20}, t = \frac{9\varphi'(\lambda)}{40}, \\
x &= \frac{1}{2} - \frac{50s^7 - 140s^6 + 438s^5 - 490s^4 + 655s^3 + 1290s^2 - 640s + 1024}{486s^2(4s^2 - 5s + 10)^2}t, \\
u_{I38} &= \frac{1}{2} - \frac{10s^4 - 22s^3 + 51s^2 - 22s + 64}{54s(s + 2)(4s^2 - 5s + 10)}t, u_{I37} = \frac{1}{2} + \frac{10s^3 - 3s^2 + 30s - 64}{6st}.
\end{aligned} 
\quad (27) $$

* Siblings I40=LT28, I39=LT29. [8, p 208] [11, p 28].

$$ 
\begin{aligned}
b &= 15, \theta_{I39} = \frac{(2, 0, 0, 7)}{15}, \theta_{I40} = \frac{(1, 0, 0, 4)}{15}, d_{I39} - b = 2, d_{I40} - b = 1, \\
t^2 &= 3(s + 5)(4s^2 + 15s + 15), s = \frac{\varphi(\lambda) - 35}{12}, t = \frac{\varphi'(\lambda)}{24}, \\
x &= \frac{1}{2} - \frac{2s^7 + 10s^6 - 90s^4 - 135s^3 + 297s^2 + 945s + 675}{18(4s^2 + 15s + 15)^2(s^2 - 5)}t, \\
u_{I39} &= \frac{1}{2} - \frac{(2s^2 + 3s - 3)}{6(s + 1)(4s^2 + 15s + 15)}t, u_{I40} = \frac{1}{2} - \frac{2s^3 + 4s^2 - 9s - 15}{2t}.
\end{aligned} 
\quad (28) $$
• I41=LT31=(H3)”. [11, p 29].

$$
\begin{align*}
b &= 18, \theta = \frac{(0, 0, 0, 1)}{3}, t^2 = s(8s^2 - 11s + 8), s = \frac{12\wp(\lambda) + 11}{24}, t = \frac{\wp'(\lambda)}{2}, \\
x &= \frac{1}{2} + \frac{(s+1)(32s^8 - 320s^7 + 1112s^6 - 2420s^5 + 3167s^4 - 2420s^3 + 1112s^2 - 320s + 32)}{54s^2(s-1)(8s^2 - 11s + 8)t}, \\
u &= \frac{1}{2} - \frac{8s^3 - 12s^2 + 3s - 4}{6t}. \tag{29}
\end{align*}
$$

This representative is also invariant under the involution $(x, s, t) \rightarrow (x, 1/s, -t/s^2)$.

• Siblings 237=LT32, 238=LT33, 239=LT34.

$$
\begin{align*}
b &= 18, \theta_{LT32} = \frac{(1, 5, 5, 5)}{42}, \theta_{LT33} = \frac{(3, 3, 3, 7)}{21}, \theta_{LT34} = \frac{(1, 1, 1, 17)}{42}, \\
t^2 &= (2s-1)(4s^2-2s+7), s = \frac{\wp(\lambda)+6}{18}, t = \frac{\wp'(\lambda)}{54}, \\
x &= \frac{1}{2} - \frac{16s^9 - 72s^8 + 144s^7 - 336s^6 + 252s^5 - 504s^4 - 294s^3 + 225s^2 - 288s + 128}{54s^2(4s^2-2s+7)t}, \\
u_{LT32} &= \frac{1}{2} + \frac{8s^5 - 4s^4 + 20s^3 - 8s^2 - 5s + 16}{18st}, \\
u_{LT33} &= \frac{1}{2} + \frac{4s^4 - 4s^3 + 12s^2 - s + 16}{6s(2s+1)(4s^2-2s+7)t}, \\
u_{LT34} &= \frac{1}{2} - \frac{4s^3 + 3s - 16}{6st}. \tag{30}
\end{align*}
$$

• Siblings I43=LT37, I42=LT38.

The curve $(s, t)$ is identical to that of I36.

$$
\begin{align*}
b &= 20, \theta_{I43} = \frac{(7, 3, 3, 13)}{60}, \theta_{I42} = \frac{(1, 9, 9, 19)}{60}, \\
t^2 &= 3s(16s^2 - 61s + 64), s = \frac{\wp(\lambda) + 61}{48}, t = \frac{\wp'(\lambda)}{96}, g_2 = -7788, g_3 = 432856, \\
x &= \frac{1}{2} + \frac{P(s)}{6(16s^2 - 61s + 64)(2s^2 - 6s + 5)^2 t}, \\
P(s) &= 512s^{10} - 7680s^9 + 51840s^8 - 206560s^7 + 535380s^6 - 935448s^5 \\
&\quad + 1098280s^4 - 825660s^3 + 343875s^2 - 41120s - 13824, \\
u_{I43} &= \frac{1}{2} - \frac{32s^5 - 216s^4 + 590s^3 - 846s^2 + 689s - 288}{2(4s-7)(2s^2-6s+5)t}, \\
u_{I42} &= \frac{1}{2} + \frac{32s^5 - 256s^4 + 826s^3 - 1322s^2 + 1023s - 288}{2(2s^2-6s+5)t} \tag{31}
\end{align*}
$$

• I46=LT39. [8, p. 213].

$$
\begin{align*}
b &= 24, d = b + 2, \theta = \frac{(1, 1, 1, 3)}{12}, t^2 = (s+2)(8s^2 - 7s + 2), s = \frac{\wp(\lambda)}{2} - \frac{3}{8}, t = \frac{\wp'(\lambda)}{2}, \\
x &= \frac{1}{2} + \frac{(s^2 + 4s - 2)P(s)}{2(s+2)^2(3s^2 - 2s + 2)^2(8s^2 - 7s + 2)t}, \\
P(s) &= 8s^{10} + 16s^9 + 24s^8 - 84s^7 + 429s^6 - 312s^5 + 258s^4 - 288s^3 + 288s^2 - 128s + 32, \\
u &= \frac{1}{2} - \frac{4s^6 + 16s^5 + 9s^4 - 2s^3 - 34s^2 + 24s - 8}{2(3s^2 - 2s + 2)(s-2)t} \tag{32}
\end{align*}
$$
### 3.2.2 Folded solutions of genus one

* Siblings I44=LT36 et I45=LT35. Transport of $\theta_{I44} = (3,0,0,3)/10$ and of $\theta_{I45} = (1,0,0,1)/10$ [11, §4.1].

$$b = 20, \theta_{I44} = \frac{(0,3,3,0)}{10}, \theta_{I45} = \frac{(0,1,1,0)}{10}, t^2 = 3(s-1)(5s^2+5s-1), s = \frac{4}{15}\wp(\lambda), t = \frac{4}{15}\wp'(\lambda),$$

$$x = \frac{1}{2} - i \frac{P(s)}{576(5s^2 - 10s - 4)(5s^2 + 5s - 1)(s - 1)^2 t},$$

$$P(s) = 3125s^{10} - 12500s^9 + 48000s^7 - 35400s^6 - 117936s^5 + 191760s^4 + 27840s^3 - 58320s^2 + 10240s + 2240,$$

$$u_{I44} = \frac{1}{2} + i \frac{(5s^2 - 22s + 26)(5s^2 + 2s + 2)^2(5s - 2)(s + 2)}{144(5s^2 - 10s - 4)(5s^2 + 5s - 1)(s - 1)^2} + i \frac{(25s^3 - 30s^2 - 42s + 20)^2 + (54s)^2}{120(5s^2 - 10s - 4)(s - 1)t},$$

$$u_{I45} = \frac{1}{2} + i \frac{(5s^2 - 22s + 26)(5s^2 + 2s + 2)(5s - 2)(s + 2)}{24(s - 1)(5s^2 - 10s - 4)(5s^2 + 5s - 1)} + i \frac{5s^2 - 4s + 8}{2(5s^2 - 10s - 4)(5s^2 + 5s - 1)} t. \quad (33)$$

This representative is invariant under the involution $(x, u, s, i, t) \to (1-x, 1-u, s, -i, t)$.

*Remark.* This representation is obtained by folding the genus zero solutions I31 and I32 [11, §4.1], which defines the elliptic curve $T^2 = (9S^2 - 2S + 9)(S^2 - 2S + 17)$, then by the homography which sends to infinity one of the four zeroes of $T$ (for instance $1+4i$),

$$S = 1 + 4i + \frac{18(1 + 2i)}{X - \frac{3}{4}(11 - 3i)}, \quad T = \frac{48i(1 + 2i)Y}{[X - \frac{3}{4}(11 - 3i)]^2}, \quad Y^2 = \frac{4X - 15}{16X^2 + 60X - 45}. \quad (34)$$

The reason why the Cremona transformation [11, §4.1]

$$S = \frac{t - 9s - 81}{t + 3s - 9}, \quad T = 16 \frac{t^2 + 54t + 18s^2 + 540s + 405}{(t + 3s - 9)^2}, \quad t^2 = s^3 - 270s - 675, \quad (35)$$

does not display the invariance $(x, u) \to (1-x, 1-u)$ is its independence on the number $i$.

* O12=LT20. Transport of $\theta = (2,3,3,3)/6$ [9, p 99].

$$b = 28, d = b + 1, \theta = \frac{(1,1,1,1)}{12}, t^2 = (2s + 1)(9s^2 + 2s + 1), s = \frac{3\wp(\lambda) - 13}{54}, t = \frac{\wp'(\lambda)}{36},$$

$$x = \frac{1}{2} + \frac{27s^4 + 28s^3 + 26s^2 + 12s + 3}{(s+1)^3(9s^2 + 2s + 1)^2}s, \quad u = \frac{1}{2} + \frac{11s^3 + 5s + 1 + 7s^2}{2(s+1)^2t}. \quad (36)$$
### 3.3 Folded solutions of genus higher than one

* Siblings I47, I48 (genus two, hyperelliptic).

Transport of $\theta_{I47} = (7, 2, 2, 7)/30$ ($\theta_{I48} = (1, 4, 4, 1)/30$ is conserved) [11, §4.3].

$$b = 40, d_{I47} - b = 2, d_{I48} - b = 1, \theta_{I47} = \frac{(2, 7, 7, 2)}{30}, \theta_{I48} = \frac{(1, 4, 4, 1)}{30},$$

$$t^2 = (3s + 1)(s + 3)(s^2 + 1)(3s^2 + 4s + 3),$$

$$x = \frac{1}{2} + \frac{P(s)}{54}t,$$

$$P(s) = 81s^{14} + 270s^{13} + 567s^{12} + 540s^{11} + 621s^{10} + 1314s^9 + 2955s^8 + 3688s^7 + 2955s^6 + 1314s^5 + 621s^4 + 540s^3 + 567s^2 + 270s + 81,$$

$$u_{I47} = \frac{1}{2} - \frac{4s^2(3s^2 + 3s + 2)(2s^2 + 3s + 3)(9s^4 - 2s^2 + 9)}{9(s-1)(3s^2 + 4s + 3)(s^2 + 1)^2(s+1)^3(3s^2 + 2s + 3)} \\ + \frac{(3s^3 + 3s^2 + s - 3)(3s^3 - s^2 - 3s - 3)}{6(s+1)(3s^2 + 4s + 3)(s^2 + 1)^2(3s^2 + 2s + 3)}t,$$

$$u_{I48} = \frac{1}{2} + \frac{27s^9 + 63s^8 + 108s^7 + 36s^6 + 42s^5 + 130s^4 + 300s^3 + 228s^2 + 99s - 9}{18(s-1)(s+1)^4(s^2+1)^2(3s+1)(3s^2+4s+3)}t. \quad (37)$$

The involution $(s, t) \to (1/s, t/s^3)$ leaves invariant $x$, $u_{I48}$ and the third term of $u_{I47}$, it changes the sign of the second term of $u_{I47}$.

* I49 (genus three, hyperelliptic).

Transport of $\theta = (1, 0, 0, 1)/6$ [11, §4.4] [12].

$$b = 36, \theta = \frac{(0, 1, 1, 0)}{6}, \quad t^2 = \frac{1}{75}(s^2 + 2s + 5)(s^2 + 4s + 5)(3s^4 + 30s^3 + 110s^2 + 150s + 75),$$

$$x = \frac{1}{2} - \frac{5(s + 1)(5 + s)P(s)}{6(s^2 - 5)(3s^4 + 30s^3 + 110s^2 + 150s + 75)^2(s^2 + 2s + 5)^3(s^2 + 4s + 5)^3}t,$$

$$P(s) = 27s^{16} + 648s^{15} + 7452s^{14} + 53568s^{13} + 266292s^{12} + 968400s^{11} + 2714980s^{10} + 6371400s^9 \\ + 14138050s^8 + 31857000s^7 + 67874500s^6 + 121050000s^5 + 166432500s^4 + 167400000s^3 \\ + 116437500s^2 + 50625000s + 10546875,$$

$$u = \frac{1}{2} - \frac{2s(s^2 + 5s + 10)(3s^2 + 10s + 15)(2s^2 + 5s + 5)(3s + 5)^2(3 + s)^2}{3(s^2 - 5)(3s^4 + 30s^3 + 110s^2 + 150s + 75)(s^2 + 2s + 5)^2(s^2 + 4s + 5)} \\ - \frac{5(s^3 + 25s^2 + 75s + 75)(3s^3 + 15s^2 + 25s + 5)}{2(s^2 - 5)(3s^4 + 30s^3 + 110s^2 + 150s + 75)(s^2 + 2s + 5)^2}t. \quad (38)$$

This representative is invariant under the involution $(x, u, s\sqrt{5}, t) \to (1-x, 1-u, 1/(s\sqrt{5})), (t/25)(5/s)^4)$.

* Siblings I50, I51 (genus three, non-hyperelliptic).

The previously chosen representatives [11, §4.2] are minimal and invariant under the 4! homographies.

The folding of the siblings (I44, I45) generates a representation by two elliptic curves, then the homography $s \to 1 + 4/s$ simplifies the expressions of [11, §4.2],

$$b = 40, d_{I50} - b = 6, d_{I51} - b = 2, \theta_{I50} = \frac{(3, 3, 3, 3)}{20}, \theta_{I51} = \frac{(1, 1, 1, 1)}{20},$$

$$t_1^2 = -s(2s - 1)(s + 2), \quad t_2^2 = (2s - 1)(s^2 + 2s + 5),$$

$$x = \frac{1}{2} - \frac{s^{10} + 10s^9 + 45s^8 + 120s^7 + 190s^6 - 4s^5 - 410s^4 - 680s^3 + 25s^2 + 90s - 27}{16s^2(s^2 + 2s + 5)(s + 2)^3(2s - 1)^2}t_1,$$

$$u_{I50} = \frac{1}{2} - \frac{(s^2 + 4s - 1)(s^2 + 4s + 9)(s^2 + 1)^2}{4s(s^3 + 3s^2 + 15s + 1)(s + 2)t_2} - \frac{s^3 + 3s^2 + 3s - 3}{2(s^3 + 3s^2 + 15s + 1)s}t_1,$$

$$u_{I51} = \frac{1}{2} - \frac{(s^2 + 4s - 1)(s^2 + 4s + 9)(s^2 + 1)}{4s(s + 1)(s + 2)^2 t_2} - \frac{s - 3}{2s(s + 1)(s + 2)^2 t_1}. \quad (39)$$
The two elliptic curves admit a rational representation in terms of the genus three curve, whose minimal degree is four [11, §4.2],

$$ 
\begin{aligned}
& 5(p^4 + q^4) + 6(p^2q^2 + p^2 + q^2) + 1 = 0, \\
& s = -\frac{2p^2}{p^2 + q^2 + 1}, \quad t_1 = pt_2, \quad t_2 = -\frac{4(q^2 + 1)q}{(p^2 + q^2 + 1)^2} \cdot
\end{aligned} 
\quad (40) 
$$

## I52 (genus seven, non-hyperelliptic).

The folding of I49 generates a representation by two hyperelliptic curves, and the homography $s \to -1 + 3/s$ simplifies the representation of Ref. [11, §4.5],

$$ 
\begin{aligned}
b &= 72, d_{I52} - b = 6, \theta_{I52} = \frac{(1, 1, 1, 1)}{12}, \\
t_1^2 &= s(3s-2)(2s-3)(2s^2-s+2)(4s^2-7s+4), t_2^2 = -s^3(3s^2-4s+3)(4s^2-7s+4), \\
x &= \frac{1}{2} - \frac{(s^2-1)P(s)}{s(3s-2)(2s-3)(3s^2-4s+3)t_1^3}, \\
P(s) &= 864(s^{16} + 1) - 10368(s^{15} + s) + 59616(s^{14} + s^2) - 221184(s^{13} + s^3) + 599976(s^{12} + s^4) \\
&\quad - 1263960(s^{11} + s^5) + 2127908(s^{10} + s^6) - 2899008(s^9 + s^7) + 3212357s^8, \\
u_{I52} &= \frac{1}{2} + \frac{3(2s^2 - 2s + 1)^2(s^2 - 3s + 1)(s^2 - 2s + 2)(6s^4 - 6s^3 + s^2 - 6s + 6)s}{(3s-2)^2(2s-3)(2s^2 - s + 2)(2s^3 - 4s^2 + 6s - 3)t_2} \\
&\quad - \frac{s(6s^3 - 12s^2 + 8s + 1)}{2(3s-2)^2(2s^2 - s + 2)(2s^3 - 4s^2 + 6s - 3)t_1},
\end{aligned} 
\quad (41) 
$$

because it then gains the involution $(x, s, t_1) \to (x, 1/s, -t_1/s^4)$. This representative is invariant under the 4! homographies.

The two hyperelliptic curves admit a rational representation in terms of the genus seven curve, whose minimal degree is eight [11, §4.5],

$$ 
\begin{aligned}
& 9(q^6 p^2 + q^2 p^6) + 18q^4 p^4 + 4(q^6 + p^6) + 26(q^4 p^2 + q^2 p^4) + 8(q^4 + p^4) \\
&\quad + 57q^2 p^2 + 20(q^2 + p^2) + 16 = 0, \\
& p^2 + q^2 = S, p^2 - q^2 = D, (3S^2 + 5S + 8)^2 - (9S^2 + 14S + 41)D^2 = 0, \\
& s = \frac{3(S-1)}{4(S+1)} + \frac{(9S^2 + 14S + 41)D}{4(S+1)(3S^2 + 5S + 8)}, \\
& t_1 = \frac{(9S^2 + 14S + 41)pq}{8(S+1)^4} \left[ 3(S-17) + \frac{41S^2 + 46S + 329D}{3S^2 + 5S + 8} \right], \\
& t_2 = \frac{15S^3 - 43S^2 - 23S - 397}{8q(S+1)^3} + \frac{(5S^2 - 4S + 63)(9S^2 + 14S + 41)D}{8q(3S^2 + 5S + 8)(S+1)^3}.
\end{aligned} 
\quad (42) 
$$

# A Transformations conserving P<sub>VI</sub>

The Table 2 [15, p. 315] lists the 24 homographies.

The unique birational transformation between P<sub>VI</sub>(u, x, θ) and P<sub>VI</sub>(U, X, Θ) is the involution defined by [21, 22],

$$ 
\begin{pmatrix} \theta_{\infty} \\ \theta_0 \\ \theta_1 \\ \theta_x \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1 & -1 & -1 & -1 \\ -1 & 1 & -1 & -1 \\ -1 & -1 & 1 & -1 \\ -1 & -1 & -1 & 1 \end{pmatrix} \begin{pmatrix} \Theta_{\infty} \\ \Theta_0 \\ \Theta_1 \\ \Theta_x \end{pmatrix} + \frac{1}{2} \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}, 
\quad (43) 
$$

$$ \frac{N}{u - U} = \frac{x(x - 1)U'}{U(U - 1)(U - x)} + \frac{\Theta_0}{U} + \frac{\Theta_1}{U - 1} + \frac{\Theta_x - 1}{U - x} \quad (44) $$

$$ = \frac{x(x-1)u'}{u(u-1)(u-x)} + \frac{\theta_0}{u} + \frac{\theta_1}{u-1} + \frac{\theta_x - 1}{u-x}, \quad (45) $$

$$ N = 1 - \Theta_{\infty} - \Theta_{0} - \Theta_{1} - \Theta_{x} = (1/2) \sum (\theta_{j} - \Theta_{j}). \quad (46) $$
The unique folding transformation between $P_{VI}(u, x, \theta)$ et $P_{VI}(U, X, \Theta)$, found by Kitaev [17], has been interpreted by Manin [23] as a Landen transformation for the elliptic representation of $P_{VI}$. It can be written as [18, §3.2],

$$ \left\{ \begin{array}{l} x = \left( \frac{X^{-1/4} + X^{1/4}}{2} \right)^2, \quad u = \left( \frac{X^{-1/4}U^{1/2} + X^{1/4}U^{-1/2}}{2} \right)^2, \\ \forall (\lambda_1, \lambda_2) : \Theta = (\lambda_1, \lambda_1, \lambda_2, \lambda_2), \quad \theta = (2\lambda_1, 0, 0, 2\lambda_2). \end{array} \right. \qquad (47) $$

The quartic transformation [18, Eqs. (3.11)–(3.13)],

$$ \left\{ \begin{array}{l} x = X, \quad u = \frac{(U^2 - X)^2}{4U(U-1)(U-X)}, \\ \forall \lambda : \Theta = (\lambda, \lambda, \lambda, \lambda), \quad \theta = (4\lambda, 0, 0, 0), \end{array} \right. \qquad (48) $$

is essentially the square [18, Eq. (2.2)] of the transformation (47).

## B Optimal representations

Depending on its genus $g$, each solution (except the three non-hyperelliptic I50, I51, I52) can be represented by two rational functions $R_*$,

$$ (g=0) \quad x = R_1(s), \quad u = R_2(s), \qquad (49) $$

or by four rational functions $R_*$ and one polynomial $P$,

$$ (\text{elliptic or hyperelliptic}) \quad x = R_1(s) + R_2(s)t, \quad u = R_3(s) + R_4(s)t, \quad t^2 = P_{2g+1}(s), \qquad (50) $$

and an important practical question is to minimize the volume of these expressions.

Such a minimization has already been done mainly by Boalch with some improvements by Lisovyy and Tykhyy, but it dealt with representatives whose gap $d-b$ is sometimes high (see the case $d-b = 12, b = 18$ in the unique set “237, 238, 239” of three siblings elements, Table 1). We therefore put our effort on the minimal representatives.

A first lowering of this volume consists in choosing the arbitrary parameter $s$ so as to move the pole of $x(s)$ of maximal order to the origin.

Additional criteria allow one to obtain an even more compact representation [8]. If the equivalence class contains a representative whose curve $P(u, x) = 0$ is invariant under the involution $(x, u) \to (1-x, 1-u)$ resp. $(x, u) \to (1/x, 1/u)$ (respective numbers 3 and 8 in Table 2, see column “homographies” of Table 1), then there exists a choice of the parameter $s$ making $(x - 1/2, u - 1/2)$ odd in $s$ (resp. $(x, u)$. Let us make these criteria more precise for $g=0$ and $g=1$.

### B.1 Rational representations (genus zero)

The Klein solution, already representable by (10) (criterium of a minimal number of terms of $P(u, x)$), is equally representable by

$$ b = 7, \theta = \frac{(2, 1, 1, 1)}{7}, \quad u = \frac{1}{2} + \frac{3s^4 + 4s^2 + 9}{s(s^2 + 7)(s^2 + 3)}, \quad x = \frac{1}{2} + \frac{7s^6 + 14s^4 + 63s^2 + 108}{2s^3(s^2 + 7)^2}, \qquad (51) $$

(criterium of invariance under the involution $(x, u, s) \to (1-x, 1-u, -s)$) or by (cf. [10, p 171 Eq (7)]),

$$ b = 7, \theta = \frac{(1, 1, 2, 1)}{7}, \quad u = \frac{s(s^2 + s + 2)(2s + 1)^2}{(2s^2 + s + 1)(s + 2)^2}, \quad x = \frac{(2s + 1)^3(s^2 + s + 2)^2}{(2s^2 + s + 1)^2(s + 2)^3}, \qquad (52) $$

(criterium of invariance under the involution $(x, u, s) \to (1/x, 1/u, 1/s)$).

The degree being the same (here $d=8$), the representation (51) (which creates a parity in $s$ and therefore reduces the number of terms) is in principle twice less voluminous than (52) (which exchanges numerators and denominators without reducing the number of terms).
## B.2 Elliptic representations (genus one)

There exist three main representations of an elliptic function of $\lambda$.

1. Sum of derivatives of functions $\zeta(\lambda - \lambda_j)$ of Weierstrass (partial fraction decomposition ("décomposition en éléments simples") of Hermite [24]);

2. Product of integer powers (of both signs) of functions $\sigma(\lambda - \lambda_j)$ of Weierstrass;

3. Rational function of $\wp(\lambda)$ and $\wp'(\lambda)$.

The third one is practically the simplest one but it lacks unicity because of the addition formula of $\wp$ (the two others are insensitive to a translation of $\lambda$).

In order to minimize the size of the fractions (degrees and number of terms), it is necessary to perform a translation of $\lambda$ which moves to the origin the pole of $x$ of maximal order, like in the following example.

Consider an elliptic function with one pole of order one and one pole of order three, defined par the Hermite decomposition

$$ \left\{ \begin{array}{l} E(\lambda) = T_0'' + 5(T_0 - T_2), \quad T_0 = \zeta(\lambda), \quad T_1 = \zeta(\lambda + a) - \zeta(a), \quad T_2 = \zeta(\lambda + b) - \zeta(b), \\ g_2 = 2, \quad g_3 = 3, \quad \wp(a) = 1, \quad \wp(b) = 2, \quad \wp'(a) = i, \quad \wp'(b) = 5. \end{array} \right. \qquad (53) $$

The canonical representation $(\wp, \wp')$ of $E(\lambda)$

$$ E(\lambda) = \frac{25 + (1 + 2\wp(\lambda))\wp'(\lambda)}{2(\wp(\lambda) - 2)}, \qquad (54) $$

is optimal (triple pole at the origin), but its shifted by $a$ (triple pole at $a$),

$$ E(\lambda - a) = \frac{P_4(\wp(\lambda)) + P_2(\wp(\lambda))\wp'(\lambda)}{(\wp(\lambda) - (6 + 5i)/2)(\wp(\lambda) - 1)^3}, \qquad (55) $$

requires, in order to become optimal, a translation defined by the factor of the denominator having maximal multiplicity.

**Conflict of interest:** The author of this work declares that he has no conflicts of interest.

## Funding

This work was partially funded by the RGC (Research grants council), grant 106220136, during the author's visit to HKU in 2023-2024.

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<caption>Table 1: The 3+30+15 minimal exceptional solutions.</caption>

Theoretical and mathematical physics, accepted 07 April 2025
Columns: notation of Refs [8] and [12], genus *g* (preceded by H (hyperelliptic) or N (non-hyperelliptic) if *g* > 1) of the curve *P*(*u*, *x*) = 0, number *d* − *b* of fixed poles if nonzero, number of terms of *P*, sequence *θ* = (*θ*<sub>j</sub>), chain of foldings (filiation), set of siblings, list of homographies leaving invariant the curve *P* = 0 (their number is a divisor of 24 except 12, the identity is omitted), former best representative (*d* − *b*, number of terms, (*θ*<sub>j</sub>)), comment.

<table><thead><tr><th>B</th><th>LT</th><th>g</th><th>b</th><th>d-b</th><th>terms</th><th>(θ<sub>∞</sub>, θ<sub>1</sub>, θ<sub>x</sub>)</th><th>filiation</th><th>siblings</th><th>homographies</th><th>former d-b, terms, θ<sub>j</sub></th><th>comment</th></tr></thead><tbody><tr><td></td><td>II</td><td>0</td><td>2</td><td></td><td>2</td><td>(a, a, b, b)</td><td>II, II</td><td></td><td></td><td>2, 7, 8, 15, 16, 21, 22</td><td></td></tr><tr><td></td><td>III</td><td>0</td><td>3</td><td></td><td>3</td><td>(a, 2a, a, 1/3)</td><td>III, T06</td><td></td><td>18</td><td></td><td>cube</td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td>III, O11, O12</td><td></td><td></td><td></td><td>cube</td></tr><tr><td>IV</td><td>0</td><td>4</td><td></td><td></td><td>4</td><td>(b, b, b, 1/2)</td><td>IV, O10, O13</td><td></td><td>3, 8, 12, 14, 18</td><td></td><td>tetrahedron</td></tr><tr><td>120</td><td>1</td><td>0</td><td>5</td><td></td><td>7</td><td>(1, 3, 3, 4)/15</td><td></td><td></td><td>3</td><td>1, 13, (5, 6, 3, 5)/15</td><td></td></tr><tr><td>121</td><td>2</td><td>0</td><td>5</td><td></td><td>4</td><td>(0, 0, 1, 2)/5</td><td>121, 128</td><td></td><td>8</td><td>1, 12, (3, 1, 2, 1)/5</td><td></td></tr><tr><td>008</td><td>5</td><td>0</td><td>6</td><td></td><td>5</td><td>(0, 0, 1, 5)/12</td><td>008, O07</td><td></td><td>8</td><td>0, 14, (3, 4, 4, 9)/12</td><td></td></tr><tr><td>123</td><td>6</td><td>0</td><td>6</td><td></td><td>8</td><td>(1, 5, 7, 7)/30</td><td>122, 123</td><td></td><td>2</td><td>2, 16, (5, 6, 3, 6)/15</td><td></td></tr><tr><td>122</td><td>7</td><td>0</td><td>6</td><td></td><td>7</td><td>(1, 1, 5, 13)/30</td><td>122, 123</td><td></td><td>8</td><td>4, 23, (10, 3, 6, 3)/15</td><td></td></tr><tr><td>K</td><td>8</td><td>0</td><td>7</td><td>1</td><td>12</td><td>(1, 1, 2, 1)/7</td><td></td><td></td><td>6, 8, 10, 19, 23</td><td>3, 24, (3, 2, 2, 2)/7</td><td></td></tr><tr><td>009</td><td>10</td><td>0</td><td>8</td><td></td><td>13</td><td>(1, 3, 3, 7)/24</td><td></td><td></td><td></td><td>2, 24, (4, 4, 6, 3)/12</td><td></td></tr><tr><td>124</td><td>11</td><td>0</td><td>8</td><td></td><td>11</td><td>(1, 1, 7, 5)/20</td><td></td><td></td><td>8</td><td>1, 30, (2, 5, 2, 4)/10</td><td></td></tr><tr><td>125</td><td>12</td><td>0</td><td>8</td><td></td><td>12</td><td>(1, 3, 5, 3)/20</td><td>124, 125</td><td></td><td>6</td><td>2, 29, (2, 4, 5, 4)/10</td><td></td></tr><tr><td>127</td><td>13</td><td>1</td><td>9</td><td></td><td>15</td><td>(1, 2, 2, 2)/15</td><td>126, 127</td><td></td><td>2, 3, 4, 5, 6</td><td>5, 60, (9, 6, 10, 6)/15</td><td></td></tr><tr><td>126</td><td>14</td><td>1</td><td>9</td><td></td><td>12</td><td>(1, 1, 8, 1)/15</td><td>126, 127</td><td></td><td>3, 8, 12, 14, 18</td><td>6, 43, (10, 3, 3, 3)/15</td><td></td></tr><tr><td>132</td><td>16</td><td>0</td><td>10</td><td></td><td>15</td><td>(0, 0, 1, 5)</td><td>132, 145, 151</td><td></td><td>3, 8, 12, 14, 18</td><td>=</td><td>(H3) [11]</td></tr><tr><td>131</td><td>17</td><td>0</td><td>10</td><td></td><td>11</td><td>(0, 0, 3, 5)</td><td>131, 144, 150</td><td></td><td>3, 8, 12, 14, 18</td><td>=</td><td>(H3)'' [11]</td></tr><tr><td>129</td><td>18</td><td>0</td><td>10</td><td></td><td>18</td><td>(3, 7, 7, 7)/30</td><td></td><td></td><td></td><td>2, 43, (3, 5, 5, 5)/15</td><td>non-monic</td></tr><tr><td>130</td><td>19</td><td>0</td><td>10</td><td></td><td>15</td><td>(1, 9, 1)/30</td><td></td><td></td><td></td><td>6, 43, (9, 5, 5, 5)/15</td><td></td></tr><tr><td>136</td><td>22</td><td>1</td><td>12</td><td></td><td>23</td><td>(1, 3, 3, 11)/30</td><td></td><td></td><td>3</td><td>6, 82, (9, 5, 5, 3)/15</td><td></td></tr><tr><td>134</td><td>23</td><td>1</td><td>12</td><td>1</td><td>23</td><td>(5, 13, 5, 23)/60</td><td></td><td></td><td>18</td><td>6, 63, (15, 6, 10)/30</td><td></td></tr><tr><td>135</td><td>24</td><td>1</td><td>12</td><td></td><td>25</td><td>(1, 5, 11, 5)/60</td><td></td><td></td><td>6</td><td>6, 83, (15, 12, 10)/30</td><td>generic</td></tr><tr><td>133</td><td>25</td><td>0</td><td>12</td><td></td><td>24</td><td>(1, 7, 11, 17)/60</td><td></td><td></td><td>none</td><td>2, 52, (6, 12, 10, 15)/30</td><td>Valentiner</td></tr><tr><td>138</td><td>26</td><td>1</td><td>15</td><td>2</td><td>46</td><td>(2, 2, 2, 3)/15</td><td></td><td></td><td>137, 138</td><td>6, 104, (6, 5, 5, 5)/15</td><td>Valentiner</td></tr><tr><td>137</td><td>27</td><td>1</td><td>15</td><td>1</td><td>34</td><td>(1, 1, 6)/15</td><td></td><td></td><td>137, 138</td><td>3, 8, 12, 14, 18</td><td>12, 170, (12, 5, 5, 5)/15</td></tr><tr><td>140</td><td>28</td><td>1</td><td>15</td><td></td><td>26</td><td>(0, 0, 1, 4)/15</td><td>140, 148</td><td></td><td>139, 140</td><td>3, 8, 12, 14, 18</td><td>9, 176, (5, 9, 9, 10)/15</td></tr><tr><td>139</td><td>29</td><td>1</td><td>15</td><td></td><td>18</td><td>(0, 0, 7, 2)/15</td><td>139, 147</td><td></td><td>139, 140</td><td>8</td><td>2, 37, (2, 0, 0, 7)/15</td><td>[11, p28]</td></tr><tr><td>141</td><td>31</td><td>1</td><td>18</td><td></td><td>39</td><td>(0, 0, 0, 1)/3</td><td>141, 149, 152</td><td></td><td>3, 8, 12, 14, 18</td><td>6, 142, (2, 1, 1, 3)</td><td>(H3)''</td></tr><tr><td>237</td><td>32</td><td>1</td><td>18</td><td></td><td>54</td><td>(1, 5, 5, 5)/42</td><td></td><td></td><td>237, 238, 239</td><td>2, 3, 4, 5, 6</td><td>12, 264, (14, 12, 12)/21</td></tr><tr><td>238</td><td>33</td><td>1</td><td>18</td><td>2</td><td>58</td><td>(3, 3, 7, 3)/21</td><td></td><td></td><td>237, 238, 239</td><td>6, 8, 10, 19, 23</td><td>2, 96, (3, 7, 3, 3)/21</td></tr><tr><td>239</td><td>34</td><td>1</td><td>18</td><td></td><td>39</td><td>(1, 1, 1, 7)/42</td><td></td><td></td><td>237, 238, 239</td><td>3, 8, 12, 14, 18</td><td>12, 184, (14, 6, 6)/21</td></tr><tr><td>143</td><td>37</td><td>1</td><td>20</td><td>1</td><td>67</td><td>(3, 3, 7, 13)/60</td><td></td><td></td><td>142, 143</td><td>8</td><td>12, 251, (18, 10, 10, 15)/30</td><td>Kitaev</td></tr><tr><td>142</td><td>38</td><td>1</td><td>20</td><td></td><td>62</td><td>(1, 9, 9, 19)/60</td><td></td><td></td><td>142, 143</td><td>3</td><td>4, 136, (6, 10, 10, 15)/30</td><td>Kitaev</td></tr><tr><td>146</td><td>39</td><td>1</td><td>24</td><td>2</td><td>99</td><td>(1, 1, 1, 3)/12</td><td></td><td></td><td>3, 8, 12, 14, 18</td><td>12, 299, (3, 2, 2)/6</td><td>Valentiner</td></tr><tr><td>007</td><td>4</td><td>0</td><td>6</td><td></td><td>8</td><td>(1, 1, 5, 5)/24</td><td>O08, O07</td><td></td><td>7</td><td>2, 22, (4, 6, 3, 6)/12</td><td></td></tr><tr><td>T06</td><td>3</td><td>0</td><td>6</td><td></td><td>6</td><td>(0, 0, 1, 1)/6</td><td>III, T06</td><td></td><td>2, 7, 8</td><td>1, 13, (3, 3, 2, 2)/6</td><td></td></tr><tr><td>O10</td><td>9</td><td>0</td><td>8</td><td></td><td>9</td><td>(0, 0, 1, 1)/4</td><td>IV, O10, O13</td><td></td><td>2, 7, 8</td><td>2, 27, (1, 2, 2, 1)/4</td><td></td></tr><tr><td>O13</td><td>30</td><td>0</td><td>16</td><td>2</td><td>51</td><td>(1, 1, 1, 1)/8</td><td>IV, O10, O13</td><td></td><td>all</td><td>8, 153, (1, 2, 2, 2)/4</td><td></td></tr><tr><td>128</td><td>15</td><td>0</td><td>10</td><td>1</td><td>23</td><td>(1, 1, 2, 2)/10</td><td>121, 128</td><td></td><td>2, 7, 8</td><td>4, 57, (4, 5, 2, 5)/10</td><td></td></tr><tr><td>O11</td><td>21</td><td>0</td><td>12</td><td></td><td>21</td><td>(0, 0, 1, 1)/6</td><td>III, O11, O12</td><td></td><td>2, 7, 8</td><td>4, 73, (2, 2, 3, 3)/6</td><td></td></tr><tr><td>O12</td><td>20</td><td>1</td><td>12</td><td>1</td><td>28</td><td>(1, 1, 1, 1)/12</td><td>III, O11, O12</td><td></td><td>2, 7, 8, 15, 16, 21, 22</td><td>6, 93, (2, 3, 3, 3)/6</td><td></td></tr><tr><td>144</td><td>36</td><td>1</td><td>20</td><td></td><td>33</td><td>(0, 0, 3, 3)/10</td><td>131, 144, 150</td><td>144, 145</td><td>2, 7, 8</td><td>6, 90, (3, 0, 0, 3)/10</td><td>= [11, p28]</td></tr><tr><td>150</td><td>43</td><td>N3</td><td>40</td><td>6</td><td>335</td><td>(3, 3, 3, 3)/20</td><td>131, 144, 150</td><td>150, 151</td><td>all</td><td>=</td><td></td></tr><tr><td>145</td><td>35</td><td>1</td><td>20</td><td></td><td>59</td><td>(0, 0, 1, 1)/10</td><td>132, 145, 151</td><td>144, 145</td><td>2, 7, 8</td><td>2, 69, (1, 0, 1)/10</td><td>[11, p27]</td></tr><tr><td>151</td><td>44</td><td>N3</td><td>40</td><td>2</td><td>275</td><td>(1, 1, 1, 1)/20</td><td>132, 145, 151</td><td>150, 151</td><td>all</td><td>=</td><td></td></tr><tr><td>147</td><td>40</td><td>H2</td><td>30</td><td>2</td><td>134</td><td>(2, 2, 7, 7)/30</td><td>139, 147</td><td>147, 148</td><td>2, 7, 8</td><td>7, 195, (7, 2, 2, 7)/30</td><td></td></tr><tr><td>148</td><td>41</td><td>H2</td><td>30</td><td>1</td><td>145</td><td>(1, 1, 4, 4)/30</td><td>140, 148</td><td>147, 148</td><td>2, 7, 8</td><td>1, 170, (1, 4, 4, 1)/30</td><td></td></tr><tr><td>149</td><td>42</td><td>H3</td><td>36</td><td></td><td>153</td><td>(0, 0, 1, 1)/6</td><td>141, 149, 152</td><td></td><td>2, 7, 8</td><td>6, 246, (1, 0, 0, 1)/6</td><td>[11, p29]</td></tr><tr><td>152</td><td>45</td><td>N7</td><td>72</td><td>6</td><td>975</td><td>(1, 1, 1, 1)/12</td><td>141, 149, 152</td><td></td><td>all</td><td>=</td><td></td></tr></tbody></table>

(4)
<table><caption>Table 2: The 24 homographies $(u, x) \to (U, X)$ which conserve $P_{VI}$, ordered by increasing values of the order of the homography. The numbering in the first column is that of Ref. [20].</caption><thead><tr><th>num</th><th>order</th><th>&infin;01x</th><th>independent var</th><th>dependent var</th></tr></thead><tbody><tr><td>1</td><td>1</td><td>&infin;01x</td><td>x = X</td><td>u = U</td></tr><tr><td>7</td><td>2</td><td>0&infin;x1</td><td>x = X</td><td>u = X/U</td></tr><tr><td>15</td><td>2</td><td>1x&infin;0</td><td>x = X</td><td>u = (U &minus; X)/(U &minus; 1)</td></tr><tr><td>22</td><td>2</td><td>x10&infin;</td><td>x = X</td><td>u = X(U &minus; 1)/(U &minus; X)</td></tr><tr><td>6</td><td>2</td><td>&infin;x10</td><td>x = X/(X &minus; 1)</td><td>u = (U &minus; X)/(1 &minus; X)</td></tr><tr><td>11</td><td>2</td><td>01x&infin;</td><td>x = X/(X &minus; 1)</td><td>u = X(1 &minus; U)/((1 &minus; X)U)</td></tr><tr><td>18</td><td>2</td><td>10&infin;x</td><td>x = X/(X &minus; 1)</td><td>u = U/(U &minus; 1)</td></tr><tr><td>20</td><td>2</td><td>x&infin;01</td><td>x = X/(X &minus; 1)</td><td>u = &minus;X/(U &minus; X)</td></tr><tr><td>2</td><td>2</td><td>&infin;0x1</td><td>x = 1/X</td><td>u = U/X</td></tr><tr><td>8</td><td>2</td><td>0&infin;1x</td><td>x = 1/X</td><td>u = 1/U</td></tr><tr><td>3</td><td>2</td><td>&infin;10x</td><td>x = 1 &minus; X</td><td>u = 1 &minus; U</td></tr><tr><td>23</td><td>2</td><td>x01&infin;</td><td>x = 1 &minus; X</td><td>u = (1 &minus; X)U/(U &minus; X)</td></tr><tr><td>4</td><td>3</td><td>&infin;1x0</td><td>x = 1/(1 &minus; X)</td><td>u = (1 &minus; U)/(1 &minus; X)</td></tr><tr><td>10</td><td>3</td><td>0x1&infin;</td><td>x = 1/(1 &minus; X)</td><td>u = (U &minus; X)/((1 &minus; X)U)</td></tr><tr><td>14</td><td>3</td><td>1&infin;0x</td><td>x = 1/(1 &minus; X)</td><td>u = 1/(1 &minus; U)</td></tr><tr><td>24</td><td>3</td><td>x0&infin;1</td><td>x = 1/(1 &minus; X)</td><td>u = U/(U &minus; X)</td></tr><tr><td>5</td><td>3</td><td>&infin;x01</td><td>x = 1 &minus; 1/X</td><td>u = 1 &minus; U/X</td></tr><tr><td>12</td><td>3</td><td>01&infin;x</td><td>x = 1 &minus; 1/X</td><td>u = 1 &minus; 1/U</td></tr><tr><td>17</td><td>3</td><td>10x&infin;</td><td>x = 1 &minus; 1/X</td><td>u = (X &minus; 1)U/(X(U &minus; 1))</td></tr><tr><td>19</td><td>3</td><td>x&infin;10</td><td>x = 1 &minus; 1/X</td><td>u = (1 &minus; X)/(U &minus; X)</td></tr><tr><td>16</td><td>4</td><td>1x0&infin;</td><td>x = 1/X</td><td>u = (U &minus; X)/(X(U &minus; 1))</td></tr><tr><td>21</td><td>4</td><td>x1&infin;0</td><td>x = 1/X</td><td>u = (U &minus; 1)/(U &minus; X)</td></tr><tr><td>9</td><td>4</td><td>0x&infin;1</td><td>x = 1 &minus; X</td><td>u = 1 &minus; X/U</td></tr><tr><td>13</td><td>4</td><td>1&infin;x0</td><td>x = 1 &minus; X</td><td>u = (1 &minus; X)/(1 &minus; U)</td></tr></tbody></table>