Quadratic Differential Equations in Three Variables without Multivalued Solutions: Part I
For ordinary differential equations in the complex domain, a central problem is to understand, in a given equation or class of equations, those whose solutions do not present multivaluedness. We consider autonomous, first-order, quadratic homogeneous equations in three variables, and begin the class...
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| Опубліковано в: : | Symmetry, Integrability and Geometry: Methods and Applications |
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| Дата: | 2018 |
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Інститут математики НАН України
2018
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| Цитувати: | Quadratic Differential Equations in Three Variables without Multivalued Solutions: Part I / A. Guillot // Symmetry, Integrability and Geometry: Methods and Applications. — 2018. — Т. 14. — Бібліогр.: 47 назв. — англ. |
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Digital Library of Periodicals of National Academy of Sciences of Ukraine| _version_ | 1859735153648074752 |
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| author | Guillot, A. |
| author_facet | Guillot, A. |
| citation_txt | Quadratic Differential Equations in Three Variables without Multivalued Solutions: Part I / A. Guillot // Symmetry, Integrability and Geometry: Methods and Applications. — 2018. — Т. 14. — Бібліогр.: 47 назв. — англ. |
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| container_title | Symmetry, Integrability and Geometry: Methods and Applications |
| description | For ordinary differential equations in the complex domain, a central problem is to understand, in a given equation or class of equations, those whose solutions do not present multivaluedness. We consider autonomous, first-order, quadratic homogeneous equations in three variables, and begin the classification of those that do not have multivalued solutions.
|
| first_indexed | 2025-12-07T14:35:16Z |
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Symmetry, Integrability and Geometry: Methods and Applications SIGMA 14 (2018), 122, 46 pages
Quadratic Differential Equations in Three Variables
without Multivalued Solutions: Part I
Adolfo GUILLOT
Instituto de Matemáticas, Universidad Nacional Autónoma de México,
Ciudad Universitaria, Mexico City 04510, Mexico
E-mail: adolfo.guillot@im.unam.mx
URL: http://www.matem.unam.mx/~guillot/
Received May 01, 2018, in final form November 05, 2018; Published online November 11, 2018
https://doi.org/10.3842/SIGMA.2018.122
Abstract. For ordinary differential equations in the complex domain, a central problem is
to understand, in a given equation or class of equations, those whose solutions do not present
multivaluedness. We consider autonomous, first-order, quadratic homogeneous equations in
three variables, and begin the classification of those which do not have multivalued solutions.
Key words: Painlevé property; univalence; semicompleteness; Chazy equation; Riccati equa-
tion; Kowalevski exponents
2010 Mathematics Subject Classification: 34M55; 34M45; 34M35
Contents
1 Introduction 2
2 Preliminaries 8
2.1 Univalence, semicompleteness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2.2 Radial orbits and Kowalevski exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2.3 Vector fields of Halphen type . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.4 Chazy and the homogeneous equations of the third order . . . . . . . . . . . . . . . . . . 14
2.5 Lins Neto’s vector fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3 First family 16
3.1 Exponents (−1,−1): Halphen’s equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3.2 Exponents (1, 1): reduction to two-dimensional equations . . . . . . . . . . . . . . . . . . 16
4 Second family 18
4.1 The plane associated to one of the distinguished radial orbits does not contain the other
radial orbit and vice-versa . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4.2 The plane associated to one distinguished radial orbit contains the other one (but the two
planes are different) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
4.3 Same invariant plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
5 Some Abelian surfaces with automorphisms and the associated equations 37
5.1 The Jacobian of Burnside’s curve and its order eight automorphism . . . . . . . . . . . . 38
5.2 The anharmonic square with an automorphism of order twelve . . . . . . . . . . . . . . . 40
6 Some Riccati equations with rational coefficients 42
6.1 First equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
6.2 Second equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
References 45
mailto:adolfo.guillot@im.unam.mx
http://www.matem.unam.mx/~guillot/
https://doi.org/10.3842/SIGMA.2018.122
2 A. Guillot
1 Introduction
In the complex domain, the solutions of an ordinary differential equation may be multivalued.
At the beginning of the XXth century, partly motivated by the success of the theory of elliptic
functions and by the work of Fuchs and Poincaré on equations of the first order, Painlevé posed
the problem “to determine all the algebraic differential equations of the first order, then of the
second, then of the third order etc., whose general solution is uniform” [42]. We will consider au-
tonomous quadratic homogeneous ordinary differential equations in dimension three, equations
of the form z′i = Pi(z1, z2, z3), i = 1, . . . , 3, for Pi a quadratic homogeneous polynomial; follow-
ing Painlevé’s behest, we will address the problem of classifying those which have exclusively
single-valued solutions.
One motivation for this study comes from investigating the “Painlevé property” (absence
of movable critical points) for nonautonomous ordinary differential equations. In one variable,
after the works of Fuchs and Poincaré on first-order equations (see [44]), Painlevé’s achievement
for equations of the second order [42] (corrected and completed by Gambier [13]) was followed
by the works of Chazy [6] for third-order equations (completed by Cosgrove [8]) and by those of
Cosgrove [9, 10] for fourth-order ones. First-order equations in more than one variable have not
received the same attention. For two variables, we have Garnier’s work [14] as well as Kimura and
Matuda’s announcement [33]. For three variables, there is, among others, Exton’s work [11], but,
to our best knowledge, there has been no attempt to systematically study first-order equations
in three variables.
If we have an equation of the form z′i = fi(z1, . . . , zn; t), with fi a polynomial of degree d+ 1
in z1, . . . , zn whose coefficients are analytic functions of t, and fi =
d+1∑
j=0
Pi,j , for Pi,j ho-
mogeneous of degree j in z1, . . . , zn then, for α 6= 0, ζi = α−1zi and τ = αdt, dζi/dτ =
d+1∑
j=0
αd+1−jPi,j
(
ζ;αdτ
)
. In the limit, as α → 0, we obtain the homogeneous autonomous sys-
tem dζi/dτ = Pi,d+1(ζ; 0). If all the solutions to the original equation are single-valued or if it
has the Painlevé property, the solutions to the latter will be single-valued [41, Section 6]. Thus,
the homogeneous and autonomous cases provide obstructions for the presence of the Painlevé
property or the absence of multivalued solutions in the inhomogeneous and nonautonomous
ones. Although simpler, these may still exhibit some complex phenomena, like the presence
of essential movable singularities (whose existence is, according to Painlevé [42], one of the
major difficulties in the subject). In this way, the analysis of quadratic homogeneous systems
contributes to the fulfillment of Painlevé’s program for first-order equations in three variables.
Homogeneous autonomous equations without multivalued solutions are also important for un-
derstanding holomorphic flows and, more generally, holomorphic actions of complex Lie groups
on complex manifolds. For instance, their study is a central ingredient in the results of Ghys and
Rebelo [15], from which we can extract the following statement: if X is a non-identically zero
semicomplete holomorphic vector field on the surface M having an isolated singularity at p ∈M
and if, in some chart centered at p, we have that X = X2 + X3 + · · · for Xj homogeneous of
degree j, X is, in a neighborhood of p, conjugated to X2 up to a reparametrization of the solu-
tions. Halphen systems, which occupy a distinguished place among homogeneous autonomous
equations, are essential to the understanding of holomorphic actions of SL(2,C) on complex
threefolds [21].
Lastly, some quadratic homogeneous systems are historically relevant, like Euler’s equation of
the top [36, Chapter VI] or Halphen’s system [28]. In the late XIXth century, Hoyer investigated,
within a class containing Euler’s equations, the quadratic homogeneous equations that could be
solved by elliptic functions [31], and Kowalevski studied some Lotka–Volterra homogeneous
equations [1, Chapter VI], a subject that foreshadowed her renowned work on the top [34].
Quadratic Differential Equations without Multivalued Solutions 3
We will study the quadratic differential equations under their vector field form. We will
consider vector fields on C3 of the form
3∑
i=1
Pi(z1, z2, z3)∂/∂zi with Pi a quadratic homogeneous
polynomial. We will say that a vector field is univalent or semicomplete if it does not have
multivalued solutions (these notions will be made precise in Section 2.1).
We aspire to have a classification of the univalent quadratic homogeneous vector fields on C3.
There are two situations where such a classification exists. A classification of the univalent
quadratic homogeneous systems that are symmetric with respect to the full permutation group
in three variables follows essentially from Chazy’s work (see Section 2.4.1). We have given a clas-
sification of the univalent quadratic homogeneous vector fields which are divergence-free [20,
Theorem C].
In this article we present the first part of a potentially comprehensive classification. In order to
state our results, let us mention some facts that will be later explained with detail. To a univalent
quadratic homogeneous vector field on C3 having an isolated singularity at the origin we may
associate seven unordered pairs of integers called Kowalevski exponents (Section 2.2.1), which
may be used to group them in six families (Section 2.2.3). Our main result is a classification of
the univalent vector fields in the first family (those for which the product of the two exponents
in one of these pairs is equal to 1) and in the second one (those for which the product of the
two exponents in two of these pairs is equal to 2).
Main Theorem. The quadratic homogeneous vector fields on C3 which have an isolated sin-
gularity at 0, are univalent, and are part of either the first or of the second family belong, up
to a linear change of coordinates, to the following list (equations XVII and XIX–XXVII are two
inequivalent equations each, one for each determination of the square root).
Vector fields of Halphen type:
[
α1x
2 + (1− α1)(xy − yz + zx)
] ∂
∂x
+
[
α2y
2 + (1− α2)(xy + yz − zx)
] ∂
∂y
+
[
α3z
2 + (1− α3)(−xy + yz + zx)
] ∂
∂z
, mi =
α1 + α2 + α3 − 2
αi
, (I)
x(x− y)
∂
∂x
+ y[y + 3x+ (n+ 2)z]
∂
∂y
+
(
z2 + βxy + 2xz
) ∂
∂z
,
β =
8m2n
(mn+ 2n+ 2m)(mn− 2n+ 2m)
, (II)
x[2x− (k + 6)y]
∂
∂x
+ y[(k + 6)y − kz] ∂
∂y
+
[
kz2 + (k − 2)(3y − 2z)x
] ∂
∂z
, (III)[
(5m− 6)x2 + 2(m− 6)xy − (7m− 6)(x+ y)z
] ∂
∂x
+
[
(m− 6)y2 − (m+ 6)(x+ y)z
] ∂
∂y
+ z[4mz − (5m− 6)x+ 3(m− 6)y]
∂
∂z
. (IV)
Vector fields all of whose solutions are rational:
x2
∂
∂x
+ y[my − (m− 1)x]
∂
∂y
+
1
4
[(
1− p2
)
x2 −
[
r2 − p2 +m2
(
1− q2
)]
xy +m2
(
1− q2
)
y2
] ∂
∂z
, (V)
x[x− y + (1− n)z]
∂
∂x
+ y(y − x+ nz)
∂
∂y
+
(
z2 + β−1xy
) ∂
∂z
,
β =
1
8
(2n− 1 + q)(2n− 1262− q), (VI)
4 A. Guillot
x[x− y + (1− n)z]
∂
∂x
+ y[y − x+ (n+ 3)z]
∂
∂y
+
(
z2 + β−1xy
) ∂
∂z
,
β = −1
4
(n+ 1 + 2q)(n+ 1− 2q). (VII)
Vector fields with a first integral reducing to Riccati equations with elliptic coefficients:
x(x− 2y)
∂
∂x
+ y(y − 2x)
∂
∂y
+
1
4
[(
1− p2
)
x2 +
(
p2 + q2 − 1− r2
)
xy +
(
1− q2
)
y2
] ∂
∂z
, (VIII)(
2y2 − x2
) ∂
∂x
+ xy
∂
∂y
+
1
8
[
2
(
1− p2
)
x2 +
(
q2 − r2
)
xy +
(
2p2 − q2 − r2
)
y2
] ∂
∂z
, (IX)
x(2x− 5y)
∂
∂x
+ y(y − 4x)
∂
∂y
+
1
4
[
4
(
1− q2
)
x2 +
(
4 + 4q2 + r2 − 9p2
)
xy +
(
1− r2
)
y2
] ∂
∂z
, (X)
x[x− y + (1− n)z]
∂
∂x
+ y[y − x+ (n− 2)z]
∂
∂y
+
(
z2 + β−1xy
) ∂
∂z
,
β =
1
24
(2n− 3 + q)(2n− 3− q), (XI)
x[x− y + (1− n)z]
∂
∂x
+ y[y − x+ (n− 5)z]
∂
∂y
+
(
z2 + β−1xy
) ∂
∂z
,
β =
1
12
(n− 3 + 2q)(n− 3− 2q), (XII)
x[x− y + (1− n)z]
∂
∂x
+ y[y − x+ (n− 3)z]
∂
∂y
+
(
z2 + β−1xy
) ∂
∂z
,
β =
1
8
(n− 2 + q)(n− 2− q), (XIII)
x[x− (1− q)y]
∂
∂x
+ y[(1− q)y − 5x− z] ∂
∂y
+
[
z2 − (q + 5)(q − 3)xy + (7 + q)xz
] ∂
∂z
, (XIV)[
(k − 1)z(x+ y)− (k + 1)x2
] ∂
∂x
+
[
(k + 5)y2 − (k + 7)(x+ y)z
] ∂
∂y
+ z[4z + (k + 1)x− (k + 5)y]
∂
∂z
. (XV)
Vector fields with a first integral reducing to constant vector fields on Abelian surfaces:[(
3x2 − y2 − 2xz
)
+ 2βy(x− z)
] ∂
∂x
+
[
2y(3x− 2z) + β
(
y2 − 3x2
)] ∂
∂y
+ 2[(z − 3x)− βy]z
∂
∂z
, β ∈ C, (XVI)
x(x− y)
∂
∂x
+ y(y − 4x− z) ∂
∂y
+
[
z2 −
(
1 + 2
√
5
)
xy + 4xz
] ∂
∂z
, (XVII)
x(x− y)
∂
∂x
+ y(y − 9x− z) ∂
∂y
+
(
z2 − 6xy + 14xz
) ∂
∂z
, (XVIII)
x(x− y)
∂
∂x
+ y(y − 4x− z) ∂
∂y
+
[
z2 +
1
11
(
17 + 12
√
3
)
xy + 4xz
]
∂
∂z
, (XIX)
x(x− y)
∂
∂x
+ y(y − 3x− z) ∂
∂y
+
[
z2 − 3
11
(
9 + 7
√
3
)
xy + 2xz
]
∂
∂z
, (XX)
Quadratic Differential Equations without Multivalued Solutions 5
x(x− y)
∂
∂x
+ y(y − 11x− z) ∂
∂y
+
[
z2 −
(
1 + 3
√
3
)
xy + 18xz
] ∂
∂z
, (XXI)[
x2 +
(
1−
√
5
2
)
yz
]
∂
∂x
+ y(2x+ y − 2z)
∂
∂y
+ z(z − 2x− y)
∂
∂z
, (XXII)[
x2 +
(
1 +
3
√
3
4
)
yz
]
∂
∂x
+ y(2x+ y − 2z)
∂
∂y
+ z(z − 2x− y)
∂
∂z
, (XXIII)
1
2
[
2x2 −
(
3
√
3 + 9
)
yz +
(
3 + 3
√
3
)
zx
] ∂
∂x
+
1
2
[
2
(
4 +
√
3
)
y2 −
(
21 + 9
√
3
)
yz +
(
7
√
3 + 9
)
zx
] ∂
∂y
+
[(
3 +
√
3
)
z − x−
(
4 +
√
3
)
y
]
z
∂
∂z
, (XXIV)[
3
(
1 +
√
3
)
zx− 4yz
√
3− x2
] ∂
∂x
+
[
3
√
3xz +
(
2
√
3− 4
)
y2 + 6
(
1−
√
3
)
yz
] ∂
∂y
+
[
x+
(
4− 2
√
3
)
y +
(√
3− 3
)
z
]
z
∂
∂z
, (XXV)[
2x2 +
(√
3− 6
)
yz
] ∂
∂x
+
[(
5− 2
√
3
)
y +
(
3
√
3− 9
)
z
]
y
∂
∂y
+
[(
3−
√
3
)
z − 2x+
(
2
√
3− 5
)
y
]
z
∂
∂z
, (XXVI)[
2
(
2−
√
2
)
yz + 2
(√
2− 1
)
zx− x2
] ∂
∂x
+
[(
8− 5
√
2
)
yz +
(
4
√
2− 5
)
zx+
(√
2− 2
)
y2
] ∂
∂y
+
[
x+
(
2−
√
2
)
(y − z)
]
z
∂
∂z
, (XXVII)[
2
(
2 +
√
2
)
x2 −
(
6 + 5
√
2
)
zx−
(
2− 3
√
2
)
zy
] ∂
∂x
+
[
2
(
2−
√
2
)
y2 −
(
2 + 3
√
2
)
zx−
(
6− 5
√
2
)
zy
] ∂
∂y
+
[
4z2 − 2
(
2 +
√
2
)
zx− 2
(
2−
√
2
)
zy
] ∂
∂z
. (XXVIII)
The seven pairs of Kowalevski exponents for each one of these appear in Table 1. For the equa-
tions to be univalent, it is necessary that the Kowalevski exponents are integers. Equations XVI
to XXVIII are always univalent; for equations I to XV, sufficient conditions for univalence
appear in Table 2 (these give, in each case, a Zariski-dense subset of the space of parameters).
The sufficient conditions for univalence in Table 2 need not be necessary ones. We will give
other sufficient conditions and mention the cases when these are necessary as we integrate these
equations (we will not give necessary and sufficient conditions for univalence for all equations).
We have grouped the equations according to their main geometric feature (which also de-
termines its integration). In the above equations, there is no phenomenon that is not already
present in the integration of the Chazy homogeneous equations (Section 2.4).
Equations of Halphen type will be defined in Section 2.3; they include equation I, the family
introduced by Halphen in [29]. (In family III, the case k = 6 is not of Halphen type; it has
a first integral and reduces to a Riccati equation with elliptic coefficients.) In equations I–IV,
for most cases (the “hyperbolic” ones), the solutions have a natural boundary. By Corollary D
in [27], these do not have rational first integrals. Some of the equations VIII–XV (which have
a first integral and reduce to Riccati equations with elliptic coefficients) have a supplementary
first integral; for these, the solutions are given by elliptic functions. For the vector fields that
reduce to constant vector fields on Abelian surfaces (Section 5), there is an automorphism of the
6 A. Guillot
Equation Kowalevski exponents
I (−1,−1), (1,m1), (1,−m1), (1,m2), (1,−m2), (1,m3), (1,−m3)
II (−1,−2), (1, 2), (1, n), (1,−n− 1), (−n, n+ 1), (1,m), (1,−m)
III (1, 2), (1, 2), (−2,−3), (−2, 3), (1, k − 1), (k, 1− k), (1,−k)
IV (−1,−2), (1, 2), (1, 3), (1,−3), (1,−m), (1,m− 1), (1−m,m)
V (1, 1), (1, q), (1,−q), (m, p), (m,−p), (−m, r), (−m,−r)
VI (1, 2), (1, 2), (n, 1− n), (1,−n), (1, n− 1), (−2, q), (−2,−q)
VII (1, 2), (1, 2), (n,−2− n), (−2,−n), (−2, n+ 2), (1, q), (1,−q)
VIII (1, 1), (3, p), (3,−p), (3, q), (3,−q), (3, r), (3,−r)
IX (1, 1), (2, p), (2,−p), (4, q), (4,−q), (4, r), (4,−r)
X (1, 1), (2, p), (2,−p), (3, q), (3,−q), (6, r), (6,−r)
XI (1, 2), (1, 2), (3,−n), (n, 3− n), (3, n− 3), (6, q), (6,−q)
XII (1, 2), (1, 2), (6,−n), (n, 6− n), (6, n− 6), (3, q), (3,−q)
XIII (1, 2), (1, 2), (4,−n), (n, 4− n), (4, n− 4), (4, q), (4,−q)
XIV (1, 2), (1, 2), (−2, 3), (2, 3), (6, q), (6,−6− q), (6 + q,−q)
XV (1, 2), (1, 2), (−1, 3), (1, 3), (6, k), (6,−6− k), (6 + k,−k)
XVI (1, 2), (1, 2), (2, 3), (2, 3), (2, 3), (−1, 3), (−1, 6)
XVII (1, 2), (1, 2), (−2, 3), (−3, 5), (3, 2), (2, 5), (−3, 10)
XVIII (1, 2), (1, 2), (−2, 3), (−13, 10), (13,−3), (2, 5), (2, 5)
XIX (1, 2), (1, 2), (−2, 3), (−3, 5), (3, 2), (3, 4), (−5, 12)
XX (1, 2), (1, 2), (−2, 3), (−1, 4), (1, 3), (2, 5), (−5, 12)
XXI (1, 2), (1, 2), (−2, 3), (−17, 12), (17,−5), (2, 5), (3, 4)
XXII (1, 2), (1, 2), (−1, 3), (1, 3), (−3, 5), (2, 5), (−3, 10)
XXIII (1, 2), (1, 2), (−1, 3), (1, 3), (−3, 5), (3, 4), (−5, 12)
XXIV (1, 2), (1, 2), (−1, 3), (1, 4), (2, 3), (−2, 7), (−7, 12)
XXV (1, 2), (1, 2), (−1, 3), (1, 4), (2, 3), (−2, 7), (−7, 12)
XXVI (1, 2), (1, 2), (−1, 3), (1, 4), (2, 3), (−2, 7), (−7, 12)
XXVII (1, 2), (1, 2), (−1, 3), (1, 4), (2, 3), (−3, 8), (−3, 8)
XXVIII (1, 2), (1, 2), (−1, 3), (1, 4), (2, 3), (−3, 8), (−3, 8)
Table 1. The seven pairs of Kowalevski exponents for each one of the equations appearing in the Main
Theorem.
Equation Conditions
I mi ≥ 2
II n ≥ 2 and m ≥ 2
III k ≥ 2
IV m ≥ 2
V m - p, m - r, p+ r or p− r equals (q − 1− 2j)m for j ∈ {0, . . . , q − 1}
VI 2 - q and q < 2n
VII 2 - n and n < 2q
VIII 3 - p, 3 - q and 3 - r
IX 2 - p
X 6 - r, 3 - q and 2 - p
XI 6 - q and 3 - n
XII 6 - n and 3 - q
XIII 2 - n and 2 - q
XIV 6 - q
XV 6 - k
Table 2. Some sufficient conditions for univalence.
Quadratic Differential Equations without Multivalued Solutions 7
Abelian surface automorphism order d equations k comments
Jacobian of
ζ2 = ξ5 − 1
induced by
(ζ, ξ) 7→ (−ζ, ωξ)
ω5 = 1
10
XVII
XVIII
XXII
3
reduced
Chazy IX
Ei × Ei (z, w) 7→ (iw,−i(z + w)) 12
XIX
XX
XXI
XXIII
5
reduced
Chazy X
Eρ × Eρ (z, w) 7→ (−ρz,−ρw) 6 XVI 5
Lins Neto’s
equations
Eρ × Eρ (z, w) 7→
(
ρ2w,−ρ2z
)
12
XXIV
XXV
XXVI
7
Jacobian of
ζ2 = ξ
(
ξ4 − 1
) induced by
(ζ, ξ) 7→
(
λζ, λ2ξ
)
λ8 = 1
8
XXVII
XXVIII
3
particular case
of Cosgrove’s
reduced F-V
Table 3. The Abelian surfaces with automorphisms related to equations XVI–XXVIII. Here, Eµ =
C/〈1, µ〉, ρ3 = 1. At the fixed point, the eigenvalues are λ, a primitive dth root of unity and λk.
surface associated to the vector field. The essential information appears in Table 3. Although
the occurrence in Chazy’s classification of vector fields that reduce to constant vector fields on
Abelian surfaces (Section 2.4) may be seen as anecdotal, the present classification exhibits the
ubiquity of this phenomenon in dimension three.
Halphen’s equations have, both in the “hyperbolic” and “parabolic” cases, movable singular-
ities. For all other equations, the solutions are given by meromorphic functions defined in the
whole complex line. Only the equations of Halphen type and those having exclusively rational
solutions have one (and only one) radial orbit with two positive exponents. It corresponds to
an asymptotic direction of escape for solutions defined in a neighborhood of infinity.
In broad lines, the strategy of the proof is, first, to use some arithmetical information to find
equations that are potentially univalent and then to try to integrate them. In Section 2 we give
the details on the first part and establish the terminology, conventions and notations that will be
used throughout the article. At various points we will need to solve some systems of Diophantine
equations. The nature of these equations makes it unreasonable to do it by hand but suitable
to do it with the help of the computer. We have written some programs that solve them in the
SageMath software [46]. They are available as ancillary files at the arXiv page of this article.1
A complete classification of all the univalent quadratic homogeneous vector fields on C3
which have an isolated singularity at 0 can, in principle, be achieved along the lines of this
article (classify those in the third, fourth and fifth families, and then those in the seventh
one) although it seems that, for this, a better knowledge of the algebraic relations between the
Kowalevski exponents is needed. On the other hand, it is possible (and this would certainly
condition the success of such a classification) that there is some vector field that has no evident
obstruction for being univalent but whose univalence status might prove very difficult to establish
(see the discussion around the history of the integration of the Chazy IX and Chazy X equations
in Section 2.4). After all, we do not have an a priori knowledge of the ways in which a quadratic
homogeneous vector field on C3 may be univalent (except in the case where it has an algebraic
first integral, where our joint work with J. Rebelo [27, Theorem B] gives a geometric description
of each level surface). This said, we do not know examples of univalent polynomial vector fields
1https://arxiv.org/abs/1804.10664.
https://arxiv.org/abs/1804.10664
8 A. Guillot
on C3 without an algebraic first integral which are of not of Halphen type, and it may very well
happen that all of these are.
There are univalent quadratic vector fields which do not have an isolated singularity at 0, and
which do not fall within the scope of the Main Theorem. Among them we find historically rele-
vant ones, like the previously mentioned examples of Euler’s equations of the top, or Halphen’s
system [28]
(xy − yz + zx)
∂
∂x
+ (yz − zx+ xy)
∂
∂y
+ (zx− xy + yz)
∂
∂z
. (1.1)
Some of these appear as degenerations of some families (the above equation is the limit of
equation I as mi → ∞), but not all of them do. A full classification of the univalent quadratic
vector fields would have to include these cases, where the obstructions for univalence given by
the Kowalevski exponents will be less strong and which may require other tools and techniques.
2 Preliminaries
We begin by giving some background as well as outlining the strategy of the proof of the Main
Theorem. Further details appear in [20]. For standard facts on differential equations in the
complex domain, including those concerning Riccati equations, we refer the reader to Hille’s [30]
and Ince’s books [32]. (The integration of some of the Riccati equations with elliptic coefficients
that appear here has been carried out in the companion article [25].) The elementary facts on
elliptic functions that we will use may be found in [37].
2.1 Univalence, semicompleteness
We will say that the holomorphic vector field X on the manifold M is univalent (following
Palais [43, Definition VI, Chapter III]) or semicomplete (following Rebelo [45, Definition 2.3])
if for each p ∈M there is a domain Ωp ⊂ C, 0 ∈ Ωp and a solution φ : Ωp →M of X, φ(0) = p,
such that φ × I : Ωp → M × C, t 7→ (φ(t), t) is proper. (For autonomous systems of ordinary
differential equations, where all critical points are movable, most notions related to the absence
of multivaluedness of the solutions coincide.)
2.2 Radial orbits and Kowalevski exponents
Let Vn denote the space of quadratic homogeneous vector fields on Cn. For X ∈ Vn, a radial
orbit of X is a line through the origin of Cn that is invariant by X; equivalently a radial orbit
of X is an orbit of Euler’s vector field E =
∑
i zi∂/∂zi, different from the origin, that is invariant
by X. A radial orbit ρ of a quadratic homogeneous vector field X is said to be nondegenerate
if X does not vanish at any point of ρ other than the origin. A quadratic homogeneous vector
field on Cn is said to be nondegenerate if all its radial orbits are nondegenerate (equivalently, if
its singularity at 0 is an isolated one). A generic quadratic homogeneous vector field on Cn is
nondegenerate and has 2n − 1 radial orbits.
2.2.1 The Kowalevski exponents of a radial orbit
There are two complex numbers that one can associate to a nondegenerate radial orbit which
give a first-order approximation to the dynamics of X in the neighborhood of it and in which
we can read an arithmetical obstruction for semicompleteness. These numbers, central in the
“Painlevé analysis” of homogeneous equations, are referred to as Kowalevski’s exponents (a term
coined by Yoshida [47], motivated by the relevant role that they play in Kowalevski’s analysis of
Quadratic Differential Equations without Multivalued Solutions 9
the top [34]), Fuchs indices, resonances, or eigenvalues. They are defined in a number of ways
and there are several interpretations to them. We refer to [16] and [7] for an overview as well as
to our own account [20, Section 2.1].
Let X ∈ Vn, X =
∑
i Pi(z1, . . . , zn)∂/∂zi. Let ρ be a nondegenerate radial orbit of X.
Let a = (a1, . . . , an) in ρ \ {0} be such that Pi(a) = ai (it is unique for, since X is quadratic,
Pi(ca) = c2ai). A parametrization of ρ as a solution of X is t 7→ (−a1/t, . . . ,−an/t). The
equation of variations associated to this solution reads
t
dvi
dt
= −
n∑
j=1
∂Pi
∂zj
(a1, . . . , an)vj ,
which may be written, for Mij = ∂Pi/∂zj(a1, . . . , an), as tξ′ = −Mξ. If λ is an eigenvalue
of −M and v ∈ Cn is an eigenvector associated to it, tλv is a solution to this equation. (By
Euler’s relation,
n∑
j=1
aj∂Pi/∂zj(a1, . . . , an) = 2Pi(a1, . . . , an) = 2ai, (I−M)a = −a, and thus −1
is always an eigenvalue of I−M .)
Definition 2.1. With the preceding notations, the Kowalevski exponents of the nondegenerate
radial orbit ρ of the vector field X are the roots of det([I−M ]− λI)/(λ+ 1).2
The Kowalevski exponents (or exponents, for short) of a nondegenerate radial orbit of a semi-
complete vector field are integers (for otherwise the equation of variations would have a mul-
tivalued solution). This result also follows from the fact that, for E =
∑
i zi∂/∂zi, if X is
semicomplete then for every λ ∈ C the vector field E − λX is semicomplete and all of its solu-
tions are, like those of E, 2iπ-periodic [20, Section 2.1]: if a is such that Pi(a) = ai then E −X
vanishes at a, where its linear part is given by I−M and where the periodicity of its solutions
implies that the eigenvalues of this linear part are integers.
As an example, consider the vector field x21∂/∂x1 +
n∑
i=2
λix1xi∂/∂xi together with its non-
degenerate radial orbit [1 : 0 : · · · : 0]. It has exponents −λ2, . . . , −λn and the solu-
tion
(
−t−1, t−λ2 , . . . , t−λn
)
, which is univalent only when λi ∈ Z for all i.
In a nondegenerate semicomplete vector field in Vn all the radial orbits are simple [20, Corol-
lary 2.7] and, in particular, for each radial orbit, the Kowalevski exponents are both nonzero (for
otherwise the radial orbit would not be a simple one). A nondegenerate semicomplete vector
field on C3 has seven couples of nonzero integers for exponents.
2.2.2 The associated foliations on projective spaces
A quadratic homogeneous vector field X on Cn induces a foliation F on CPn−1. Radial orbits
for X give singular points for F . For a nondegenerate radial orbit ρ of X, the characteristic num-
bers of the singularity of F induced by ρ are the exponents of ρ (these characteristic numbers are
defined only up to multiplication by a constant). Some obstructions for X to be semicomplete
may be read directly from F . For example, if X is a semicomplete vector field and ρ a nondegen-
erate radial orbit of X, F is linearizable and has rational characteristic numbers at the singularity
induced by ρ [20, Corollary 2.7]. (Lemma 4.1 will be a very useful consequence of this.)
For E =
∑
i zi∂/∂zi and ` a linear form in C3, the vector fields X and X+`E induce the same
foliation on CP2. In dimension three, for a radial orbit ρ which is nondegenerate for both X
and X + `E, the exponents with respect to X and with respect to X + `E coincide if and only
if `(ρ) ≡ 0 (see [19, Lemma 5] and [20, Corollary 2.7]).
2Warning! The eigenvalues of a radial orbit, as defined in [19] and used throughout [20], differ from the
Kowalevski exponents by a sign.
10 A. Guillot
For X ∈ V3 inducing the foliation F on CP2, if ρ is a nondegenerate radial orbit of X with
exponents (u, v), the number (u+ v)2/(uv) is called the Baum–Bott index of the singular point
of F associated to ρ. See [3, Chapter 3] for definitions and details.
2.2.3 The Kowalevski exponents as functions on the space of vector fields
In dimension three, a generic quadratic homogeneous vector field is locally characterized by its
Kowalevski exponents:
Theorem 2.2. Let Ψ: V3 99K Sym7
(
Sym2(C)
)
be the map that associates to a generic vector
field on C3 the two exponents of its seven radial orbits. The map Ψ has finite fibers. In other
words, for a generic vector field in V3 there are, up to linear equivalence, only finitely many
vector fields having the same set of exponents.
A proof of this theorem was allegedly given in [20, Proposition 3.2]. However, as J.V. Pereira
brought to our attention, the proof given there is incomplete. Let us complete the proof here.
It is sufficient to exhibit a vector field such that any deformation preserving the exponents
of its radial orbits comes from a linear change of coordinates. Consider the family
Xα = X0 +
(∑
αjzj
)∑
zi
∂
∂zi
,
for
X0 = z1(2z2 − z3)
∂
∂z1
+ z2(2z3 − z1)
∂
∂z2
+ z3(2z1 − z2)
∂
∂z3
,
αi ∈ C. Let X ′ be given by αi = 1 for all i. It is a nondegenerate vector field. Let us prove that
any deformation of X ′ within V3 that preserves its exponents is made of vector fields linearly
equivalent to X ′. All the vector fields Xα induce the same foliation F on CP2. The arguments
given in the proof of Proposition 3.2 in [20] do prove that any deformation of F that preserves
its Baum–Bott indexes (like one induced by a deformation of X ′ that preserves the exponents) is
given by linear changes of coordinates. This implies that ifX ′ admits a deformation preserving its
exponents then this deformation belongs, up to a linear change of coordinates, to the family Xα.
For the radial orbits ρ1 = [1 : 0 : 0], ρ2 = [0 : 1 : 0] and ρ3 = [0 : 0 : 1] of Xα, the sum of the
exponents of ρi is −1/αi. This completes the proof of Theorem 2.2.
Remark 2.3. Not all fibers of Ψ are finite: equation XVI gives a one-parameter family of
linearly inequivalent and nondegenerate vector fields having the same exponents.
Question 2.4. What is the degree of Ψ? Given a generic set of seven complex numbers whose
sum is one, Lins Neto proved that there are 7!/21 = 240 different linear equivalent classes of
foliations on CP2 of degree two having them as Baum–Bott indexes [39]. (The condition on the
sum is necessary by the theorem of Baum and Bott.) Since two vector fields in V3 which induce
the same foliation and which have the same eigenvalues are linearly equivalent [19, Lemma 5],
[20, Corollary 2.7], the degree of Ψ is at most 240.
A quadratic homogeneous vector field on C3 depends upon 18 parameters. The general
linear group of C3 has nine parameters and acts on this space. Since a generic vector field
has no linear symmetries, a quadratic homogeneous vector field depends essentially upon nine
parameters. Since each one of the seven radial orbits has two exponents, there are at least
five independent relations between them (actually, by Theorem 2.2, exactly five). We can give
explicitly some of these relations. Let X ∈ V3 have an isolated singularity at 0 and simple radial
Quadratic Differential Equations without Multivalued Solutions 11
orbits ρ1, . . . , ρ7. The radial orbit ρi has two exponents in C, that will be denoted by ui and vi.
They are bound by the relations R0, R1 and R2 given respectively by
7∑
i=1
1
uivi
= 1,
7∑
i=1
ui + vi
uivi
= 4,
7∑
i=1
(ui + vi)
2
uivi
= 16. (2.1)
Relation R2 is the Baum–Bott relation for the foliation on CP2 that a vector field in V3 naturally
induces. As explained in [19, Example 14], the other two relations may be obtained through
some Camacho–Sad and Baum–Bott relations of the foliation that a vector field in V3 naturally
induces on CP3. (See also [19, Corollary 12] for a unified proof of these relations.) These are all
the relations that we know of: we are missing two independent ones. This is one of the main
difficulties in the classification of the univalent vector fields on C3 with an isolated singularity.
Recent results by Kudryashov and Ramı́rez on a closely related subject [35] encourage a moderate
optimism towards the possibility of giving these missing relations explicitly (see also [26]).
A strategy for classifying nondegenerate semicomplete vector fields in V3 would be to classify
the vector fields having integral exponents and then to try to integrate each one of these vector
fields. For the first part, we may start by describing the sets of integers that may be exponents of
a nondegenerate semicomplete vector field. The first problem we face is that we do not know the
complete system of Diophantine equations we must solve, that we only know the relations (2.1),
and that a set of seven couples of integers satisfying this system needs not be the set of exponents
of a vector field. Despite this, we may still try to start by solving system (2.1). Let ξi = uivi so
that R0 reads
7∑
i=1
1
ξi
= 1. (2.2)
A nondegenerate semicomplete quadratic homogeneous vector field on C3 gives an integer so-
lution to this “Egyptian fractions” problem, and in order to classify nondegenerate quadratic
semicomplete vector fields on C3, one may start by solving this Diophantine equation. There are
infinitely many integer solutions to equation (2.2), which may be grouped in families. Families
give subvarieties of V3 where all nondegenerate semicomplete vector fields are to be found. As
in [20, Section 3.3], a nondegenerate quadratic homogeneous vector field is said to belong to the
first family if there is a radial orbit ρi such that ξi = 1; it is said to belong to the second family
if there exist i and j, i 6= j such that ξi = 2 and ξj = 2, (so that 1/ξ1 + 1/ξ2 = 1) and, more
generally, a solution ξ1, . . . , ξ7 to (2.2) is said to belong to the nth family if n is the smallest
cardinality of the subsets J of {1, . . . , 7} such that
∑
i∈J
1/ξi = 1. There is no sixth family. The
seventh family contains only finitely many solutions of equation (2.2).
2.2.4 Invariant planes
A vector field X in V3 may leave invariant (be tangent to) a plane Π in C3. Three (out of the
seven) radial orbits of X will be within this invariant plane. (We will represent invariant planes
schematically by diagrams like the one appearing in Fig. 1.) For each one of these radial orbits,
one of the two exponents corresponds to the exponent of the radial orbit of the restriction of X
to Π (we will denote it by ui). Denoting the other exponent by vi, we have the relations
1
u1
+
1
u2
+
1
u3
= 1, (2.3)
v1
u1
+
v2
u2
+
v3
u3
= 1. (2.4)
12 A. Guillot
u1 u2 u3
v1 v2 v3
Figure 1. Diagram representing an invariant plane. The invariant plane contains three radial orbits.
For each one, the exponent tangent to the plane is ui and the other one vi.
u1 u2 u3
v1
v2
v3
a1
a2
b1
b2
Figure 2. Diagrams for vector fields having two invariant planes.
The first one is the relation for quadratic homogeneous vector fields on C2. The second one is
the Camacho–Sad relation [5] for the foliation induced on CP2 with respect to the invariant line
induced by Π. Notice that the exponents of two of the radial orbits within Π determine the
exponents of the third.
In order to classify the nondegenerate quadratic homogeneous vector fields on C2 we may list
the solutions to the Diophantine equation (2.3). They are
(2, 3, 6), (2, 4, 4), (3, 3, 3), (1,m,−m), m ∈ Z \ {0}. (2.5)
Each one of these triples can be realized by a unique (up to linear changes of coordinates) vector
field which is, moreover, semicomplete, and which may be explicitly solved (by elliptic functions
in the first three cases, by rational ones in the other); see [2, 14, 15], and [20, Example 2.12].
We will come back to this integration in Section 3.2.
Some of the equations that we will consider have two invariant planes. We will represent
them by diagrams like the one in Fig. 2. The union of the two invariant planes will contain five
radial orbits. In the diagram, the five pairs of circles joined by a dotted line give the Kowalevski
exponents of each one of these five radial orbits. The two continuous lines represent the two
invariant lines, and the three circles on top of each one of these lines, the three Kowalevski
exponents tangent to each one of them. For example, for the diagram in Fig. 2 the first invariant
plane has the diagram in Fig. 1, the second one the diagram in Fig. 3; these intersect at the
radial orbit with exponents (u2, v2).
Within the space of quadratic homogeneous vector fields on C3, those having an invariant
plane form a codimension one subvariety. From the discussion in Section 2.2.3 we know that, in
restriction to it, the fourteen Kowalevski exponents depend upon eight parameters: there must
be at least six independent relations. We have the three relations (2.1) plus the relations (2.3)
and (2.4), that hold in the presence of an invariant line. Thus, even in this restricted case, there
are still missing relations. In collaboration with V. Ramı́rez, we have recently investigated the
missing relations in this setting [26].
Quadratic Differential Equations without Multivalued Solutions 13
a1 v2 a2
b1 u2 b2
Figure 3. The second invariant plane.
2.3 Vector fields of Halphen type
The only known examples of semicomplete quadratic homogeneous vector fields on C3 that do
not have a first integral belong to this class. We have studied their geometry and dynamics in
see [21, Section 3], where details on what follows may be found. Let E =
∑
i zi∂/∂zi.
Definition 2.5. A quadratic homogeneous vector field X on C3 is said to be of Halphen type
if there exists a rational vector field C on C3, homogeneous of degree 0, such that [C,X] = 2E.
The homogeneity of the vector fields involved in this definition implies, by Euler’s relation,
that [E,X] = X and [E,C] = −C. The vector fields give thus a representation of the Lie
algebra sl(2,C) into the Lie algebra of rational vector fields on C3. The prototypical example is
given by Halphen’s equation (1.1), introduced in [28], which is of Halphen type with respect to
the vector field
∑
i ∂/∂zi, this is,
[∑
i ∂/∂zi, X
]
= 2E.
For a vector field X of Halphen type with respect to a vector field C, an adapted function
is a primitive common first integral of E and C. Recall that the Schwarzian derivative of f(t)
with respect to t, {f, t} is given by
{f, t} =
f ′′′
f ′
− 3
2
(
f ′′
f ′
)2
.
If X is a vector field of Halphen type, ξ is an adapted function and ξ(t) the evaluation of ξ on
a solution of X, we may define the projective invariant
{t, ξ} = − 1
(ξ′)2
{ξ, t}, (2.6)
which is a function of ξ(t) and is, moreover, independent of the chosen solution of X. For
example, for equation I we have the adapted function ξ = (x− y)/(y − z), and
− 1
(ξ′(t))2
{ξ(t), t} =
1
2
(
1− 1
m2
1
)
(ξ − 1)ξ −
(
1− 1
m2
2
)
(ξ − 1) +
(
1− 1
m2
3
)
ξ
ξ2(ξ − 1)2
, (2.7)
for mi = (α1 + α2 + α3 − 2)/αi. We have given a criterion for univalence of vector fields of
Halphen type in terms of the projective invariant (2.6); see [21, Proposition 4]. For systems
where it is given by (2.7), a sufficient condition for semicompleteness is that mi ∈ Z, m2
i 6= 1. In
this case, a key fact is that, when restricted to the upper half plane, t(ξ) uniformizes a triangle
bounded by circular arcs having internal angles π/mi [30, Section 10.2].
In the hyperbolic case (
∑
1/mi < 1), undoubtedly the most interesting one, the triangle can
be seen as a geodesic triangle in the hyperbolic plane which, moreover, tessellates it. The asso-
ciated equation of Halphen type is related to the group of direct symmetries of this tessellation,
the triangle group T (m1,m2,m3). The domains where the solutions of the equation are defined
have a circular natural boundary and the equations do not have a meromorphic first integral
(see [21, Theorem A] or [27, Corollary D] for different proofs of this last fact).
There are semicomplete quasihomogeneous vector fields of Halphen type in C3 which are not
associated to hyperbolic geometry and whose solutions are defined in domains with a fractal
14 A. Guillot
boundary [22] or whose complement is a Cantor set [24, Section 4.11]. We do not know of any
algebraic autonomous differential equation in dimension three having a univalent solution that
has a natural boundary and which is not of Halphen type.
2.4 Chazy and the homogeneous equations of the third order
A classification closely related to the one that concerns us lies at the beginning of Chazy’s effort
to parallel Painlevé’s studies on second-order equations for equations of the third order [6]. As
one of the first steps in this program, Chazy sought to establish the equations of the form
φ′′′ = aφ4 + bφ2φ′ + c(φ′)2 + dφφ′′, (2.8)
with a, b, c, d ∈ C, which have only single-valued solutions. The vector field
X = y
∂
∂x
+ z
∂
∂y
+
(
ax4 + bx2y + cy2 + dxz
) ∂
∂z
,
equivalent to (2.8), is quasihomogeneous when x, y and z are given, respectively, the weights 1, 2
and 3: for L = x∂/∂x+2y∂/∂y+3z∂/∂z, [L,X] = X. Our problem is thus analogous to Chazy’s.
Equation (2.8) has three similarity solutions (the natural analogue of radial orbits, orbits
where X and L are collinear). From each one of these solutions may extract one pair of expo-
nents, which must satisfy the relations
3∑
i=1
1
uivi
=
1
6
,
3∑
i=1
ui + vi
uivi
= −7
6
,
3∑
i=1
(ui + vi)
2
uivi
=
49
6
.
A differential equation (2.8) depends essentially upon three parameters and thus, in this case, we
know all the relations between the exponents. In order to classify the semicomplete equations in
family (2.8), we may first solve this system of Diophantine equations, and then, for each solution,
study each one of the differential equations it determines, in order to assess if all of its solutions
are single-valued. It is essentially in this way that Chazy obtained some families of equations
(some of them featuring a parameter) usually bearing the names Chazy I through Chazy XII.
Chazy integrated most of these equations except those labeled IX and X, for which we had to
wait almost 90 years until Cosgrove achieved their integration [8]. Let us briefly describe the
geometry of those which will be related to our classification (see [23] for details).
Chazy IX. The equation is
φ′′′ = 54φ4 + 72φ2φ′ + 12(φ′)2. (2.9)
When considered as a polynomial vector field on C3, it has a polynomial quasihomogeneous
first integral of degree 10 and commutes with a quasihomogeneous polynomial vector field of
degree 4 having the same first integral. A generic level surface S of this first integral is an affine
surface endowed with two commuting vector fields. Furthermore, by the quasihomogeneity of
the first integral, S has a natural action of Z/10Z preserving the restriction of the vector field
up to a constant factor. On the other hand, consider the genus-two curve x2− y5− 1. It has an
automorphism of order 10 given by (x, y) 7→ (−x, ωy) for ω a fifth root of unity. This induces
an automorphism of order 10 on the Jacobian J of the curve. This automorphism preserves two
constant vector fields up to a constant factor. The generic level surface S embeds into J , by
mapping the vector field of the Chazy IX equation to one of these two constant vector fields
(and the one that commutes with it to the other one). This mapping is equivariant with respect
to the corresponding actions of Z/10Z.
Quadratic Differential Equations without Multivalued Solutions 15
Chazy X. Given by the two equations
φ′′′ = 6φ2φ′ +
3
11
(
9 + 7
√
3
)(
φ2 + φ′
)2
, (2.10)
one for each determination of
√
3. It was integrated by Cosgrove through an “unexpectedly
complicated” formula [8]. We may describe the accompanying geometry as follows. When seen
as a quasihomogeneous polynomial vector field on C3, it has a quasihomogeneous polynomial
first integral of degree 12 and commutes with a quasihomogeneous polynomial vector field of
degree 6 having the same first integral, so that the generic level surface S is endowed with two
commuting vector fields and an action of Z/12Z that preserves each one of them up to a constant
factor. On the other hand, consider the elliptic curve Ei = C/Λ, for Λ = 〈1, i〉. It has an order
four automorphism induced by multiplication by i. The square Ei×Ei has an automorphism of
order 12, given by the order four automorphism of Ei acting diagonally and by the order-three
map (z, w) 7→ (w,−z −w), which commutes with the previous one. The cyclic group generated
by these symmetries preserves two vector fields up to a constant factor each. The generic level
surface S embeds equivariantly into Ei×Ei by mapping the vector field of the Chazy X equation
into one of these two vector fields.
Chazy XI. The one-parameter family of equations
φ′′′ =
1− p2
2
φφ′′ +
(
1− p2
2
+ 6
)
(φ′)2 − 3
(
1− p2
)
φ2φ′ +
3
(
1− p2
)2
8
φ4.
These can be reduced to Riccati equations with elliptic coefficients. We have the first inte-
gral g3 = 4z3 − (z′)2 for z = φ′ − 1
4
(
1 − p2
)
φ2 and thus 1
4
(
1 − p2
)
φ is a solution to the Riccati
equation f ′ = f2+ 1
4
(
1−p2
)
℘, the Riccati form of Lamé’s equation, for the Weierstrass function ℘
such that (℘′)2 = 4℘3 − g3. The equation is univalent if p is not a multiple of 6.
Chazy XII. The one-parameter family
φ′′′ = 2φφ′′ − 3(φ′)2 +
4
36− k2
(
6φ′ − φ2
)2
,
probably the most famous of Chazy’s equations; it is of Halphen type. It corresponds to the
vector field
X = y
∂
∂x
+ z
∂
∂y
+
[
2xz − 3y2 +
4
36− k2
(
6y − x2
)2] ∂
∂z
,
which is quasihomogeneous with respect to the vector field L = x∂/∂x + 2y∂/∂y + 3z∂/∂z,
[L,X] = −X. For C = 3∂/∂x+ x∂/∂y + 3y∂/∂z, [L,C] = −C and [C,X] = L, so X is indeed
of Halphen type. For the adapted function
ξ =
(
36− k2
k2
) (
x3 − 9xy + 9z
)2(
6y − x2
)3 ,
we have (2.7) for m1 = 3, m2 = 2 and m3 = k. The equation is semicomplete if k ∈ Z, k ≥ 2.
2.4.1 The Chazy equations as quotients of the quadratic symmetric ones
If X is a quadratic homogeneous vector field on C3 which is invariant under the group S3 of
permutations of the three variables, for a solution (z1(t), z2(t), z3(t)) of X, φ(t) = z1(t) + z2(t) +
z3(t) is a solution to an equation of the form (2.8) which depends only on X. For instance, the
equations in the Chazy XII family are the quotients of the symmetric instances of equations I
(Halphen’s equations). The quotient of a univalent quadratic vector field will be a univalent
equation of the form (2.8).
16 A. Guillot
2.5 Lins Neto’s vector fields
They are those of equation XVI. We introduced them in [18], where their integration was studied.
The foliations that these vector fields induce on CP2 give one of the remarkable families of
foliations that Lins Neto constructed in relation to the Poincaré problem [38].
For ρ3 = 1, let Λ = 〈1, ρ〉 and let Eρ be the elliptic curve C/Λ. It has an automorphism
of order six induced by multiplication by −ρ on C. The Abelian variety Eρ × Eρ has a cyclic
automorphism of order six given by the diagonal action of the latter. Every constant vector
field on C2 induces a vector field on Eρ×Eρ which is preserved, up to a constant factor, by this
cyclic automorphism. All the vector fields of equations XVI commute. They have a common
polynomial homogeneous first integral of degree six. The generic level surface of this first integral
can be compactified into Eρ × Eρ by an embedding that maps every vector field in the family
into a constant one. See [18] for details.
3 First family
We now begin the classification of the nondegenerate vector fields belonging to the first fam-
ily, which is characterized by the existence of a radial orbit whose exponents are either (1, 1)
or (−1,−1). A rough classification of the nondegenerate semicomplete vector fields within the
first family was carried out in [20, Proposition 3.8].
3.1 Exponents (−1,−1): Halphen’s equations
If a nondegenerate vector field in V3 has a radial orbit at [1 : 1 : 1] with exponents (−1,−1) and
radial orbits at [1 : 0 : 0], [0 : 1 : 0] and [0 : 0 : 1], it belongs to the family I (briefly discussed in
Section 2.3); see [20, Proposition 3.8].
3.2 Exponents (1, 1): reduction to two-dimensional equations
A nondegenerate vector field in V3 with a radial orbit at [0 : 0 : 1] having exponents (1, 1) is of
the form
P (x, y)
∂
∂x
+Q(x, y)
∂
∂y
+
[
z2 + 2z`(x, y) +R(x, y)
] ∂
∂z
, (3.1)
for ` a linear homogeneous function and R a quadratic homogeneous one (upon replacing z
by z + `(x, y) we may suppose that ` ≡ 0). The linear projection onto (x, y) gives a quadratic
homogeneous vector field, which must be semicomplete and nondegenerate. According to Sec-
tion 2 in [15] (see also [2]), this reduced equation is, up to a linear change of coordinates, one of
the following ones:
1. x(x− 2y)∂/∂x+ y(y − 2x)∂/∂y,
2. x(x− 3y)∂/∂x+ y(y − 3x)∂/∂y,
3. x(2x− 5y)∂/∂x+ y(y − 4x)∂/∂y,
4. x2∂/∂x+ y[my − (m− 1)x]∂/∂y, m ∈ Z \ {0}.
We must, in each case, determine the polynomials R that make the vector field (3.1) semicom-
plete.
Quadratic Differential Equations without Multivalued Solutions 17
3.2.1 Case (1): equation VIII
By setting
R =
1
4
(
1− p2
)
x2 +
1
4
(
p2 + q2 − 1− r2
)
xy +
1
4
(
1− q2
)
y2,
with p, q, r ∈ C, we get the most general quadratic polynomial of the form under consideration.
The exponents of the radial orbits are (1, 1), (3, p), (3,−p), (3, q), (3,−q), (3, r), (3,−r), so
we must have that p, q, r ∈ Z. The involution (x, y, z) 7→ (y, x, z) exchanges the roles of p
and q, while the involution (x, y, z) 7→ (x − y,−y, z) exchanges those of q and r. (The roles
played by p, q and r are thus completely symmetric.) The reduced equation (1) has the first
integral c = xy(x−y), which is also a first integral of the full system. By the homogeneity of the
equation we need only consider one nonzero level curve. A solution to the reduced equation is
x =
√
−g3
℘
, y =
1
2
(
℘′
℘
+
√
−g3
℘
)
,
for the Weierstrass elliptic function ℘ such that (℘′)2 = 4℘3− g3 (see [37, Chapter 6] for general
facts about these). Substituting in equation (3.1), we obtain the Riccati equation with elliptic
coefficients
z′ = z2 − g3
16
(
3− 2p2 − 2r2 + q2
) 1
℘2
+
√
−g3
8
(
p2 − r2
) ℘′
℘2
+
1
16
(
1− q2
)(℘′
℘
)2
.
This is equation II in [25], where we have established that the equation is semicomplete whenever
• neither p nor q nor r are multiples of 3, or when
• 3 | p but 6 - p, q2 = r2, 3 - q, 3 - r (and the analogous conditions obtained by permuting p, q
and r).
3.2.2 Case (2): equation IX
The change of coordinates (u, v) = (x+y, x−y) maps the reduced vector field to
(
2v2−u2
)
∂/∂u+
uv∂/∂v, which has the first integral c = v2
(
v2 − u2
)
. Let
R =
1
4
(
1− p2
)
u2 +
1
8
(
q2 − r2
)
uv +
1
8
(
2p2 − q2 − r2
)
v2.
The exponents of the radial orbits are (1, 1), (2, p), (2,−p), (4, q), (4,−q), (4, r), and (4,−r).
The parameters q and r play symmetric roles, for the involution (u, v, z) 7→ (u,−v, z) exchanges
them while preserving p. A solution to the reduced equation for c = 1 is given by v = i sn(t),
u = v′/v, where sn(t) is Jacobi’s sinus amplitudinis elliptic function of modulus k2 = −1, the
one such that (sn′(t))2 = 1 − sn4(t) (see [37, Chapter 2] for details about these functions).
Substituting, we obtain the Riccati equation with elliptic coefficients
w′ = w2 +
1
8
(
q2 + r2 − 2
)
sn2(t) +
i
8
(
q2 − r2
)
sn′(t) +
1
4
(
1− p2
)
sn−2(t).
This is equation V in [25], where we have established that the equation is semicomplete when-
ever q and r are odd and either
• p is odd or
• p is even but not a multiple of 4 and q2 = r2.
18 A. Guillot
3.2.3 Case (3): equation X
Set
R =
(
1− q2
)
x2 +
1
4
(
4 + 4q2 + r2 − 9p2
)
xy +
1
4
(
1− r2
)
y2.
The exponents of the radial orbits are (1, 1), (2, p), (2,−p), (3, q), (3,−q), (6, r) and (6,−r).
The reduced equation has the first integral g3 = −4xy2(x− y)3 and the solutions
x =
1
2
g3
℘℘′
, y = 2
℘2
℘′
,
for the Weierstrass elliptic function ℘ such that (℘′)2 = 4℘3 − g3. Substituting, we obtain the
Riccati equation
z′ = z2 +
1
(℘′)2
[
g23
4
(
1− q2
)
℘−2 +
g3
4
(
4 + 4q2 + r2 − 9p2
)
℘+
(
1− r2
)
℘4
]
.
This is equation III in [25]. It is semicomplete whenever 6 - r, 3 - q and 2 - p.
3.2.4 Case (4): equation V
Setting
R =
1
4
[(
1− p2
)
x2 −
[
r2 − p2 +m2
(
1− q2
)]
xy +m2
(
1− q2
)
y2
]
,
the exponents of the radial orbits are found to be (1, 1), (1, q), (1,−q), (m, p), (m,−p), (−m, r),
and (−m,−r). The reduced equation has the first integral xmy/(x− y) and the solution
x = −1
t
, y = − tm−1
tm − 1
,
corresponding to a nonradial level curve. Substituting these in the original equation, we obtain
the Riccati equation with rational coefficients
z′ = z2 +
1
4t2
[
m2
(
1− q2
)( tm
tm − 1
)2
−
[
r2 − p2 +m2
(
1− q2
)]( tm
tm − 1
)
+
(
1− p2
)]
.
It will be integrated in Section 6.1. The results there obtained show that this Riccati equation
will have exclusively single-valued solutions if m - p, m - r and if, for some j ∈ {0, . . . , q − 1},
either p− r = (q− 1− 2j)m or p+ r = (q− 1− 2j)m. Moreover, in these cases, all the solutions
will be rational ones.
4 Second family
This second family is characterized by the existence of two radial orbits such that the exponents
of each one of them are either (1, 2) or (−1,−2). We will refer to these two radial orbits as the
distinguished ones. We begin by establishing two lemmas.
Lemma 4.1. In a degree two foliation on CP2 a singularity with exponents [1 : 2] is linearizable
if and only if the line tangent to the exponent 1 is invariant.
Quadratic Differential Equations without Multivalued Solutions 19
Proof. The most general foliation of degree two on CP2 having a nondegenerate singularity at
(0, 0), whose linearization has exponents [1 : 2] and such that the coordinate axes are tangent
to the eigenspaces of the linear part is given by(
2y + a1x
2 + a2xy + a3y
2 + y
[
b1x
2 + b2xy + b3y
2
])
dx
−
(
x+ a4x
2 + a5xy + a6y
2 + x
[
b1x
2 + b2xy + b3y
2
])
dy.
The term a1x
2 is a resonant one and we must have a1 = 0 if the singularity is linearizable. In
this case y = 0 is an invariant line. �
Lemma 4.2. In a nondegenerate semicomplete vector field in V3, if the exponents of two ra-
dial orbits are both (−1,−2), there exists a third radial orbit whose exponents are either (1, 2)
or (1, 1).
Proof. Under the hypothesis, there are seven nondegenerate radial orbits, whose exponents,
(ui, vi), are nonzero integers. Relation R1 reads
−1
1
− 1
2
− 1
1
− 1
2
+
7∑
i=3
(
1
ui
+
1
vi
)
= 4,
and thus
7∑
i=3
(1/ui+1/vi) = 7. If among the ten numbers u3, . . . , u7, v3, . . . , v7 we have n1 times 1
and n2 times 2 then, since for the remaining ones ui ≥ 3 if ui > 0 and since the sum of the
positive terms in
7∑
i=3
(1/ui + 1/vi) is at least 7,
n1
1
+
n2
2
+
10− (n1 + n2)
3
≥ 7,
and thus 4n1 + n2 ≥ 22. Since n1 + n2 ≤ 10, n1 ≥ 4. Suppose that (ui, vi) 6= (1, 1) for all i.
We have that n1 < 6 (for otherwise one radial orbit will have its two exponents equal to 1).
If n1 = 5 then n2 ≥ 2, and the n1 exponents 1 must belong to different radial orbits: one of
them must pair with one of the 2’s. If n1 = 4 then n2 = 6 and a similar reasoning applies. �
Let us now begin the classification in the second family. Cases appear according to the signs
of the exponents of the two distinguished radial orbits and to the configuration of the invariant
planes guaranteed by Lemma 4.1.
4.1 The plane associated to one of the distinguished radial orbits
does not contain the other radial orbit and vice-versa
We will now study the cases when, for one of the two distinguished radial orbits, the invariant
plane given by Lemma 4.1 does not contain the other radial orbit and vice-versa (in particular,
these invariant planes are different).
4.1.1 Positive exponents for both distinguished radial orbits
We have the diagram in Fig. 4 with m, n ∈ Z. By taking {x = 0} and {y = 0} as invariant
planes and fixing the two radial orbits with ξi = 2 at [0 : 1 : 0] and [1 : 0 : 0], we find the vector
field
x (x− y + [1− n]z)
∂
∂x
+ y (y − x+ [1−m]z)
∂
∂y
+
(
z2 + αxy
) ∂
∂z
, (4.1)
20 A. Guillot
1 m −m
2
n
m+ n
1
−n
2
m+ n
Figure 4. The configuration of Section 4.1.1.
for some α ∈ C. There are two further radial orbits whose exponents are, respectively, (p, q)
and (p,−q), for p and q given by
1
κ
+
1
p
=
1
2
, q2 =
α(m− n)2 − 8κ
α(2− κ)2
,
for κ = m+ n. In particular,
(κ, p) ∈ {(1,−2), (−2, 1), (4, 4), (3, 6), (6, 3)}. (4.2)
These will give, respectively, equations VI, VII, XIII, XI and XII. The image of (4.1) un-
der (x, y, z) 7→ (xy, z) = (w, z) is
(2− κ)wz
∂
∂w
+
(
z2 + αw
) ∂
∂z
.
In particular, we have the second-order equation
z′′ = (4− κ)zz′ − (2− κ)z3, (4.3)
which is independent of α. From the solutions to these equations, the solutions to the original
ones may be obtained by solving the (independent) Riccati equations
x′ = x2 + (1− n)z(t)x− w(t),
y′ = y2 + (1−m)z(t)y − w(t). (4.4)
For the first one, setting v = x+ 1
2(1− n)z(t) we have v′ = v2 +A, for
A =
1
4
(
1− n2
)
z2 +
(
1
2
α[1− n]− 1
)
w; (4.5)
there is an analogous equation for (4.4). Let us integrate the equations corresponding to the
values of κ in (4.2).
4.1.1.1 Equation XI, κ = 3. A solution to (4.3) is z = −℘′/℘ for ℘ the Weierstrass elliptic
function ℘ such that (℘′)2 = 4℘3 − g3. For (4.5),
A =
1
4
(
1− q2
)
℘− 1
4
(
1− n2
) g3
℘2
.
The corresponding Riccati equation is equation I in [25]. It is semicomplete if 6 - q and 3 - n.
Since n+m = 3, condition 3 - n implies that 3 - m and thus equation (4.4) is also semicomplete
under these conditions.
Quadratic Differential Equations without Multivalued Solutions 21
4.1.1.2 Equation XII, κ = 6. A solution to (4.3) is z = 1
2℘
′/℘, (℘′)2 = 4℘3−g3, and thus,
for (4.5),
A =
1
4
(
1− n2
)
℘− 1
4
(
1− q2
) g3
℘2
.
The corresponding Riccati equation is again equation I in [25]. It is semicomplete if 6 - n
and 3 - q. Since n + m = 6, condition 6 - n implies that 6 - m and thus equation (4.4) is also
semicomplete under these conditions.
4.1.1.3 Equation XIII, κ = 4. A solution to (4.3) is z = isn(t), for the Jacobi elliptic
function sn(t) of modulus k2 = −1, the one such that (sn′(t))2 = 1 − sn4(t). Accordingly,
in (4.5),
A =
1
8
[(
n2 + q2 − 2
)
sn2(t) +
(
q2 − n2
)
i sn′(t)
]
=
1
16
(
1− n2
)
(cn(t) + i dn(t))2 +
1
16
(
1− q2
)
(cn(t)− i dn(t))2,
for the other Jacobi elliptic functions cn(t) and dn(t) of the same modulus. This is equation IV
in [25]. It is semicomplete in all the cases where both n and q are odd and in the case where 4 - q,
4 - n and q2 = n2. Likewise, equation (4.4) will be semicomplete in all the cases where both m
and q are odd and in the case where 4 - q, 4 - m and q2 = m2. Thus, equation XIII is
semicomplete in the cases where n, m and q are all odd and in the special case m = 2, n = 2
and q = ±2.
4.1.1.4 Equation VI, κ = 1. For any quadratic polynomial u, z = −u′/u is a solution
to (4.3) and hence, for (4.5),
A =
1
4
(
1− n2
) [(u′)2 − 2uu′′
u2
]
+
1
8
(
1− q2
)u′′
u
. (4.6)
If n = −1 then (−n,m + n) = (1, 1), bringing us to back to Section 3.2; if n = 1 then m = 0,
which does not correspond to a nondegenerate vector field. We may thus suppose that n2 6= 1.
This equation will be studied in Section 6.2, where we will see that it has exclusively single-valued
solutions (actually, rational ones) if q is odd and q < 2n.
4.1.1.5 Equation VII, κ = −2. For a quadratic polynomial u, z = −1
2u
′/u is a solution
to (4.3). For (4.5),
A =
1
4
(
1− q2
) [(u′)2 − 2uu′′
u2
]
+
1
8
(
1− n2
)u′′
u
.
Since p = 1, q2 = 1 would imply the existence of a radial orbit with exponents (1, 1), placing us in
the setting of Section 3.2, so we may suppose that q2 6= 1. The Riccati equation we obtain is the
very same one associated to (4.6), that will be analyzed in Section 6.2. From the analysis there
done, the Riccati equation has only single-valued solutions (actually, rational ones) whenever n
is odd and n < 2q.
4.1.2 The exponents of the two distinguished radial orbits have different signs
We have, for some n ∈ Z, the diagram in Fig. 5. If n = −1 the vector field is degenerate
(for n+ 1 = 0) and if n = 1 it corresponds to one of the vector fields studied in Section 3.2; we
22 A. Guillot
1 −n− 1 n+ 1
2
1
−n
−1
1
−2
n
Figure 5. The configuration of Section 4.1.2.
may thus suppose that n2 6= 1. Placing the horizontal line at the plane {x = 0}, the vertical at
{y = 0}, the radial orbit of exponents (1, 2) at [0 : 1 : 0] and the one with exponents (−2,−1)
at [1 : 0 : 0], we find the vector field Y of equation II for some γ ∈ C. Let
γ =
8m2n
(mn+ 2n+ 2m)(mn− 2n+ 2m)
.
The other two radial orbits have exponents (−1,m) and (−1,−m); we may suppose that m2 6= 1
for otherwise we would be in the setting of Section 3.2. These vector fields are of Halphen type
with respect to C = ∂/∂x − y/x∂/∂y, for C is homogeneous of degree 0 and [C, Y ] = 2E. For
the adapted function ξ = (2γ/n)xy/z2,
− 1
ξ2
{ξ, t} =
1
2
(
1− 1
4
)
(ξ − 1)ξ −
(
1− 1
n2
)
(ξ − 1) +
(
1− 1
m2
)
ξ
ξ2(ξ − 1)2
.
Following the discussion in Section 2.3, equation II is associated to the triangle group T (2,m, n).
It is univalent when n,m ∈ Z, |n| ≥ 2 and |m| ≥ 2 (recall that we have dismissed the vector
fields where n2 = 1 or m2 = 1). These vector fields had been previously spotted in [20], where
it was shown that they are quotients of Halphen’s systems I having two coincident parameters.3
By Lemma 4.2, there is no need to consider the case where the exponents of both distinguished
radial orbits are positive.
4.2 The plane associated to one distinguished radial orbit
contains the other one (but the two planes are different)
4.2.1 One radial orbit with positive exponents and one with negative ones
Two out of the three eigenvalues of the restriction of the vector field to the invariant plane must
be either (−1, 2) or (1,−2). In the first case the restriction of the vector field to this plane
cannot be, by the enumeration in (2.5), univalent. In the second one we have the diagram in
Fig. 6. Since the plane tangent to the exponent −1 in the radial orbit with exponents (−2,−1)
is invariant, the above diagram extends into a diagram of the form portrayed in Fig. 7 for
some n ∈ Z. We may assume that n ≥ 2 since n − 1 and −n play symmetric roles. If n = 2,
(1, n − 1) = (1, 1), and we would be back to Section 3.2; if n = 3, (1, n − 1) = (1, 2) and we
would be in the setting of Section 4.1. We will thus suppose that n > 3. Relation R0 reads
Quadratic Differential Equations without Multivalued Solutions 23
1 −2 2
2 −1 −3
Figure 6. The invariant line for Section 4.2.1.
1 −2 2
2
−1
−3
1
1
n− 1
−n
Figure 7. The configuration in Section 4.2.1.
1
ξ6
+
1
ξ7
+
1
n(n− 1)
=
1
6
. (4.7)
Several cases arise when considering the solutions to this equation.
4.2.1.1 First case, ξ6 = 6, ξ7 = −n(n− 1). We have the three equations Ri for the three
unknowns v6, v7, n. From the solutions to these,
1. for (u6, v6) = (6, 1), u7 = 1
2
(
−7 +
√
97
)
and v7 = 1
2
(
−7−
√
97
)
, which are not integers;
2. for (u6, v6) = (3, 2), n = 2 (but we had previously supposed that n > 3);
3. for (u6, v6) = (−2,−3), n = 3/2 or n = −1/2 (which is not an integer);
4. for (u6, v6) = (−1,−6), n = 5/3 or n = −2/3.
We can thus discard all these cases.
4.2.1.2 Second case, ξ6 6= 6, ξ7 6= 6. Since n > 3, there are only finitely many solutions
to (4.7). We have used the computer to find them.4 Let us list these solutions, in order to
illustrate the nature of the problems that we are dealing with. The solutions are: for n = 4,
(3,−4), (4,−6) (8,−24), (9,−36), (10,−60), (11,−132), (13, 156), (14, 84), (15, 60), (16, 48),
(18, 36), (20, 30), (21, 28), (24, 24); for n = 5, (5,−12), (8,−120), (9, 180), (10, 60), (12, 30),
(15, 20); for n = 6, (3,−5), (5,−15), (7,−105), (8, 120), (9, 45), (10, 30), (12, 20), (15, 15);
for n = 7, (8, 56), (14, 14); for n = 8, (7, 168), (8, 42); for n = 9, (8, 36), (9, 24); for n = 10,
(10, 18); for n = 11, (11, 15); for n = 12, (4,−11); for n = 13, (12, 13); for n = 16, (10, 16);
for n = 19, (9, 19); for n = 21, (5,−28); for n = 25, (8, 25) ; for n = 30, (5,−29); for n = 43,
(7, 43). In particular, there is no solution for n > 43.
3By setting m = (α+2β−2)/β, n = −2(α+2β−2)/α (γ = α(α+2β−2)/(2β−1)) we find, in the coordinates
(ζ1, ζ2, ζ3) =
(
x+ α−1z, 2x, 1
2
x+ (2[2β − 1])−1y
)
, the vector field H2(α, β) in formula (3.1) of [20].
4See the ancillary files p1a.sage and p1b.sage.
24 A. Guillot
1 2 −2
2 1 3
Figure 8. The invariant line in Section 4.2.2.
1 2 −2
2
1
3
−m
m
−n
−m− n
Figure 9. The master configuration of Section 4.2.2.
Given n, for each one of these solutions (ξ6, ξ7) we have finitely many choices for u6, v6,
u7 and v7, which can be again obtained with the help of the computer. With these values we
would have all the potential values of ui and vi for all i, and we can check if they satisfy the
equations Ri. We have done this and have found that in there is no set of ui’s and vi’s that is
compatible with these relations in the above cases.
4.2.2 Positive exponents for the two radial orbits
These are the vector fields having a diagram of the form portrayed in Fig. 8. By Lemma 4.1,
another invariant plane passes through one of the radial orbit having exponents (2, 1), and we
have a diagram like the one in Fig. 9 for n,m ∈ Z. We may suppose that m > 0. Let k =
n(m+ n). Equation R0 reads
1
ξ6
+
1
ξ7
+
1
k
=
1
6
. (4.8)
We have infinitely many solutions for it if one of the involved integers is 6, finitely many other-
wise. The most general vector field having these data is
x(x− y)
∂
∂x
+ y(y + [n+ 1]x− z) ∂
∂y
+
(
z2 + αxy + [m+ 1]xz
) ∂
∂z
(4.9)
for some α ∈ C. If we set α = −
(
6σ2 + 2mσ + 5nσ + mn + n2
)
/σ, σ ∈ C, there is a radial
orbit ρ6 at [1 : σ : 2σ + n] and another one, ρ7, at [6σ : n(n + m) : 2n(n + m + 3σ)]. They are
exchanged by the involution σ 7→ 1
6n(n+m)/σ, so they are equivalent for most purposes. The
exponents of ρ6 are the roots of
ζ2 −
(
7 +
7 +m+ 2n
σ − 1
)
ζ +
6σ2 − n2 −mn
(σ − 1)2
, (4.10)
Quadratic Differential Equations without Multivalued Solutions 25
while those of ρ7 are the roots of
ζ2 −
(
7 +
6σ[7 +m+ 2n]
n2 +mn− 6σ
)
ζ +
6n(n+m)
(
n2 +mn− 6σ2
)(
n2 +mn− 6σ
)2 . (4.11)
Notice that the coefficient of the linear term is −7 in the first if and only if it is −7 in the
second (σ ∈ {0,∞} corresponds to degenerate vector fields) and that this happens if and only
if 7 +m+ 2n = 0. We have three subcases according to the nature of the solutions of (4.8).
4.2.2.1 First subcase: in (4.8), k = 6. Since n divides 6, the values that n may take are
only finitely many; in turn, the value of n determines that of m, for m = k/n− n. Among the
divisors of 6, if n ∈ {6, 3,−1,−2}, m < 0. We have the following remaining cases.
4.2.2.1.1 When (n,m) = (2, 1). Since there is a radial orbit with exponents (−n,−m)
= (−2,−1), we are in the setting of Section 4.1.2.
4.2.2.1.2 When (n,m)=(−3, 1). Since there is a radial orbit with exponents (m,−m−
n) = (1, 2), we are in the setting of Section 4.1.1.
4.2.2.1.3 When (n,m) = (1, 5). For the partial sum of R1 we have
5∑
i=1
(1/ui + 1/vi) =
5/3 and thus 1/u6 + 1/v6 + 1/u7 + 1/v7 = 7/3. If the number 1 appeared only once among u6,
v6, u7, v7, say, u6 = 1, we would have 1/v6 +1/u7 +1/v7 = 4/3. Up to ordering, the only way to
express 4/3 as a sum of inverses of integers without the summand 1 is 4/3 = 1/2+1/2+1/3 and
we would either have a radial orbit with exponents (1, 2), going back to Section 4.1.1, or one with
exponents (2, 2). In this last case, through this radial orbit with exponents (2, 2) we would have
three invariant planes [20, Lemma 3.6]. One of them, P , intersects the invariant plane associated
with the vertical line in the diagram of Fig. 9 at the radial orbit with exponents (−m,−n). Two
of the three exponents of the restriction to P would be 2 and −n = −1, but this is impossible by
the enumeration in (2.5). Hence, the number 1 appears at least twice as an exponent among u6,
v6, u7, v7 (but not within the same radial orbit since that would take us back to Section 3.2),
say u6 = 1 and u7 = 1. From R0, v6 = −v7, but from R1, 1/v6+1/v7 = 1/3, which is impossible.
(There is no vector field having these data.)
4.2.2.1.4 When (n,m) = (−6, 5). From R0, ξ7 = −ξ6. Sincem+2n = −7, from (4.10),
u6 and v6 are the roots of x2 − 7x + d6 (whose discriminant is 49 − 4d26), while u7 and v7 are
the roots of x2 − 7x− d6 (whose discriminant is 49 + 4d26). Both discriminants must be squares
of integers. Their sum is 98, but 98 cannot be written as the sum of two different squares of
integers.
4.2.2.2 Second subcase: in (4.8), ξ6 = 6, ξ7 = −k. Given u6 and v6 (integers such
that u6v6 = 6), we have four unknowns (m,n, u7, v7) and the three equations Ri, which we can
actually solve. The solutions, factoring out the symmetries of the problem, are the following.
4.2.2.2.1 When (u6, v6) = (−2,−3). We have n = −6, (u7, v7) = (6, q), m = q + 6,
σ = 1
2(1−q). From the vector field (4.9), after rescaling y by t 7→ (1−q)y, we find the family XIV.
The image of the vector field under
(x, y, z) 7→
(
x[x− 2y], 2x
[
x2 + (q + 3)xy + yz
])
= (u, v)
26 A. Guillot
is v∂/∂u + 6u2∂/∂v, which has the first integral g3 = 4u3 − v2 (it induces a homogeneous first
integral of degree 6 for the original vector field). The reduced equation has the solution (u, v) =
(℘, ℘′), for the Weierstrass elliptic function ℘ such that (℘′)2 = 4℘3 − g3. Solving for y and z
we find
y =
x2 − ℘
2x
, z =
℘′ + (q + 3)x℘− (5 + q)x3
x2 − ℘
.
Substituting in (4.9) reduces the system of equations to x′ = 1
2(q+ 1)x2 + 1
2(1− q)℘. If q = −1,
x is the Weierstrass ζ function, up to the addition of a constant. If q 6= −1, for w = 1
2(q + 1)x,
w′ = w2 + 1
4
(
1− q2
)
℘: the vector field is semicomplete whenever q ∈ Z \ 6Z (see [6, 25]). These
vector fields appeared in [17] as the family T3.
4.2.2.2.2 When (u6, v6) = (2, 3). We have n = −1, m = 1− k, (u7, v7) = (1,−k), σ =
1
12(k+6). From the vector field (4.9), upon rescaling the variables by (x, y, z) 7→ (2x, [k+6]y, kz),
we find the family of vector fields Y (k) of equation III. (The data for k = −6 is not realizable.)
When k 6= 6 these vector fields are of Halphen type: for
C = 3
∂
∂x
+
(
1− 3
y
x
) ∂
∂y
+
(
3− z
y
)
∂
∂z
,
we have that [C, Y ] = (6− k)E. They all have the common adapted function
ξ =
x(x− 6y)3(
x2 − 9xy + 9yz
)2 ,
for which
− 1
ξ2
{ξ, t} =
1
2
(
1− 1
4
)
(ξ − 1)ξ −
(
1− 1
9
)
(ξ − 1) +
(
1− 1
k2
)
ξ
ξ2(ξ − 1)2
,
corresponding, as described in Section 2.3, to the triangle group T (2, 3, k). Thus, when k 6= 6,
the vector field is semicomplete whenever k ∈ Z \ {−1, 0, 1}. It is, after exchanging y and z,
the vector field kH3a(1/k) in formula (3.5a) of [20] and it is thus a quotient of the symmetric
case of the Halphen system I when the three parameters coincide. These vector fields appeared
previously in [17] as the family T4.
When k = 6 the vector field commutes with C. The projection onto the orbit space of C
(x, y, z) 7→
(
4x[x− 6y], 16x
[
x2 − 9xy + 9yz
])
= (u, v)
maps the vector field to v∂/∂u + 6u2∂/∂v; the solution of this reduced equation is given
by (u, v) = (℘, ℘′), for the Weierstrass elliptic function ℘ such that (℘′)2 = 4℘3 − g3. Sol-
ving for y and z we obtain
y(t) =
4x2 − ℘
24x
, z(t) =
8x3 − 6x℘+ ℘′
6
(
4x2 − ℘
) ,
and substituting in (4.9), we have x′(t) = 1
2℘(t), and thus x(t) = −1
2ζ(t)+c, for the Weierstrass ζ
function. The vector field is also semicomplete for k = 6.
4.2.2.2.3 When (u6, v6) = (−1,−6). The equations impose the condition 1/n+1/(m+
n) = −7/6, and thus either (m + n = −6 and n = −1) or (m + n = −1 and n = −6). In the
first case we have m = −5 (but we supposed m > 0); in the second, since k = 6, we are back to
the first subcase.
Quadratic Differential Equations without Multivalued Solutions 27
4.2.2.2.4 When (u6, v6) = (1, 6). Since u6+v6 = 7, from (4.10) and (4.11), u7+v7 = 7
and m + 2n = −7 (we may set v7 = 7 − u7 and m = −7 − 2n). The relation R0 implies
that n2 + 7n − 7u7 + u27 = 0 and thus u7 = 7
2 ±
1
2
√
49− 28n− 4n2, which is an integer only
if n = 0 or n = −7 (which implies −m− n = 0). These correspond to degenerate vector fields.
4.2.2.3 Third subcase: in (4.8), κ 6= 6, ξ6 6= 6, ξ7 6= 6. There are 140 unordered triples
of integers that solve the equation 1/n1 + 1/n2 + 1/n3 = 6 under the condition that ni 6= 6 for
all i. From the list of such triples we can deduce all the possible values of ξ6, ξ7 and k. From
the values of the first two, we may deduce all potential values of (u6, v6) and (u7, v7), as well as
all factorizations of ξ6 and ξ7, and, from the latter, the putative values of n (among the factors
of k) and m (under the assumption that m > 0). We thus get possible values for ui, vi, which
can then be tested for compliance with the relations Ri. We obtain in this way the following
five solutions:5
1. k = 10, n = −5, m = 3, (u6, v6) = (−3, 10), (u7, v7) = (2, 5),
2. k = −30, n = −10, m = 13, (u6, v6) = (2, 5), (u7, v7) = (2, 5),
3. k = 10, n = −5, m = 3, (u6, v6) = (−5, 12), (u7, v7) = (3, 4),
4. k = 12, n = −4, m = 1, (u6, v6) = (−5, 12), (u7, v7) = (2, 5),
5. k = −60, n = −12, m = 17, (u6, v6) = (3, 4), (u7, v7) = (2, 5).
In all of them, m+ 2n+ 7 = 0. Substituting these values of (u6, v6) as roots of (4.10), we obtain
a quadratic equation for σ; after substituting one of the roots in (4.9), we obtain, respectively,
the equations XVII, XVIII, XIX, XX and XXI. The equation corresponding to the other value
of σ is obtained by changing the determination of the square root (except in equation XVIII,
where the two values of σ give linearly equivalent equations).
Let us integrate them by showing that each one of them has a polynomial first integral and
that its restriction to a particular level set of it is birationally equivalent to the restriction of
either the Chazy IX or the Chazy X equation to a level set of the corresponding first integral.
4.2.2.3.1 Equation XVII. It is birationally equivalent to the Chazy IX equation (2.9).
If φ is a solution to the latter, a solution to this equation is given by
x = −3
2
(√
5− 1
)
φ, y = −1
4
(√
5− 1
)6φ2 +
(√
5 + 1
)
φ′
φ
,
z =
1
2
(√
5− 1
)54φ3 −
(
3 +
√
5
)
φ′′ + 3
(√
5 + 1
)
φφ′
6φ2 +
(√
5 + 1
)
φ′
.
Reciprocally, from a solution to XVII, a solution to Chazy IX is given by the birational inverse
of the above,
φ = −1
6
(
1 +
√
5
)
x, φ′ = −1
6
(
1 +
√
5
)
x(x− y),
φ′′ = −1
6
(
1 +
√
5
)
x
(
2x2 + xy + yz
)
.
This proves that the vector field is semicomplete.6 It has a homogeneous polynomial first integral
of degree 10 and commutes with a homogeneous polynomial vector field of degree 4 having the
same first integral.
5See the files p2a.sage and p2b.sage.
6In order to integrate equation XVII, we started by looking (with the help of the computer) for homogeneous
polynomial first integrals and homogeneous polynomial vector fields that commuted with the vector field we
wished to integrate. Having established that the vector field had a homogeneous polynomial first integral of
28 A. Guillot
4.2.2.3.2 Equation XVIII. It is also birationally equivalent to Chazy IX. If φ is a so-
lution to the Chazy IX equation, a solution to this equation is given by
x = −φ, y = −φ
2 + φ′
φ
, z =
8φ3 + 6φφ′ − φ′′
φ2 + φ′
.
Reciprocally, from a solution (x, y, z) to this equation we may build a solution
φ = −x, φ′ = x(y − x), φ′′ = −x
(
2x2 + 6yx+ yz
)
,
to Chazy IX (this is the birational inverse of the previous map). Equation XVIII has a homo-
geneous polynomial first integral of degree 10 and commutes with a homogeneous polynomial
vector field of degree 14 (which, after dividing by the first integral, becomes a rational homo-
geneous vector field of degree 4 which commutes with the original one) having the same first
integral.
4.2.2.3.3 Equation XIX. It is birationally equivalent to the Chazy X equation (2.10).
If φ is a solution to the latter, a solution to this equation is given by
x =
1
2
(
3 +
√
3
)
φ, y =
1
6
(
3 +
√
3
)(−3 +
√
3
)
φ′ + 3φ2
φ
,
z = −1
2
(
3 +
√
3
)(−4 + 2
√
3
)
φ′′ + 9φ3 +
(
−3 +
√
3
)
φφ′(
−3 +
√
3
)
φ′ + 3φ2
.
Reciprocally, from a solution to XIX a solution to Chazy X is given by
φ =
1
3
(
3−
√
3
)
x, φ′ =
1
3
(
3−
√
3
)
x(x− y), φ′′ =
1
3
(
3−
√
3
)
x
(
2x2 + xy + yz
)
.
These two are birational inverses of one another. Equation XIX has a homogeneous polynomial
first integral of degree 12 and commutes with a homogeneous polynomial vector field of degree 6
having the same first integral.
4.2.2.3.4 Equation XX. It is birationally equivalent to Chazy X. If φ is a solution to
the Chazy X equation, a solution to this equation is given by
x = −φ, y = −φ
2 + φ′
φ
, z =
2φ3 − φ′′
φ2 + φ′
.
Reciprocally, from a solution to this equation, a solution to Chazy X may be given by the
birational inverse
φ = −x, φ′ = x(y − x), φ′′ = −x
(
2x2 + yz
)
of the above map. Equation XX has a homogeneous polynomial first integral of degree 12 and
commutes with a homogeneous polynomial vector field of degree 6 having the same first integral.
degree 10 and commuted with a homogeneous polynomial vector field of degree 4 having the same first integral,
just like the Chazy IX equation, it began to seem likely for both equations to be birationally equivalent (by some
birational mapping that preserved the homogeneity). We then looked for linear functions of x, y and z such that,
when evaluated at a solution to equation XVII, solved the Chazy IX equation, and obtained the present result.
Equations XVIII–XXI were integrated in the same way (taking into account that Chazy X has a polynomial first
integral of degree 12 and commutes with a homogeneous polynomial vector field of degree 6 having the same
first integral), considering the possibility that the homogeneous vector fields that commute with them could be
rational, and not necessarily polynomial.
Quadratic Differential Equations without Multivalued Solutions 29
(A) 1 −1 1
2 −2 −3
(B) 1 1 −1
2 2 3
Figure 10. The two possible configurations for Section 4.3.
4.2.2.3.5 Equation XXI. It is birationally equivalent to Chazy X. If φ is a solution to
the Chazy X equation, a solution to this equation is given by
x = − 1
11
(
1 + 2
√
3
)
φ, y =
1
11
(
1 + 2
√
3
)(1− 2
√
3
)
φ′ − φ2
φ
,
z =
1
11
(
1 + 2
√
3
)(13− 4
√
3
)
φ′′ − 10φ3 + 8
(
1− 2
√
3
)
φφ′(
1− 2
√
3
)
φ′ − φ2
.
Further, from a solution to this equation a solution to the Chazy X equation may be given by
φ =
(
1− 2
√
3
)
x, φ′ =
(
1− 2
√
3
)
x(x− y), φ′′ =
(
1− 2
√
3
)
x
(
2x2 + 8xy + yz
)
,
which is the birational inverse of the previous map. It has a homogeneous polynomial first
integral of degree 12 and commutes with a homogeneous polynomial vector field of degree 18
(which becomes rational of degree 6 after dividing by the first integral) having the same first
integral.
By Lemma 4.2, there is no need to consider the case where the exponents of both radial orbits
are positive.
4.3 Same invariant plane
We have one of the diagrams, (A) or (B), appearing in Fig. 10. In both cases, for R0 to hold
one must have
1
ξ4
+
1
ξ5
+
1
ξ6
+
1
ξ7
=
1
3
. (4.12)
As before, special situations arise whenever the equality is already realized by a proper subset
of the summands.
4.3.1 In (4.12), ξ4 = 3
This is the case where, in equation (4.12), we have ξ4 = 3. We either have (u4, v4) = (1, 3)
or (u4, v4) = (−1,−3). In both cases, the exponents are resonant, and there is a further ob-
struction on the vector fields for semicompleteness. This obstruction can be easily read in the
foliation induced on CP2 at the corresponding point p: the singularity of the foliation, which is
a resonant node, must be linearizable. This is equivalent to having, in the induced foliation, an
invariant curve tangent to the exponent ±1. One way to guarantee the existence of this curve is
to impose the invariance of the plane that passes through p tangent to the exponent ±1. This
will be the assumption in Sections 4.3.1.1 through 4.3.1.3. The invariance of this plane is not the
only way to overcome the linearizability obstruction, as we will show in Section 4.3.1.4. The re-
sults there obtained will be used in Sections 4.3.1.5 through 4.3.1.7 to produce the classification
in these cases.
30 A. Guillot
−1 1 1
−2
2
−3
−2
1
5
3
Figure 11. The first configuration of Section 4.3.1.1.
1 −1 1
2
−2
−3
2
1
−5
3
Figure 12. The second configuration of Section 4.3.1.1.
4.3.1.1 Case (A), (u4, v4) = (1, 3), with invariant plane. If the plane tangent to the
exponent 1 is invariant, we have a diagram of one of the forms portrayed in Figs. 11, 12 or 13.
For the first two, for the partial sum of R0 we have that
5∑
i=1
1/ξi = 9/10 6= 1 and for last
one
5∑
i=1
1/ξi = 14/15 6= 1. For each one of these cases we can (with the help of the computer)
find all the possible (finitely many) nonzero integers ξ6 and ξ7 that satisfy, together with the
data already obtained, equation R0. For each value of ξ6 and ξ7 we can find all possible values
of (u6, v6) and (v7, u7) and, with these, test for equations R1 and R2 (in order to find all seven
couples of integers containing the above set of five pairs and satisfying the three relations Ri).
We did not find any solutions in any of these three cases.7
4.3.1.2 Case (B), (u4, v4) = (1, 3), with invariant plane. We have a diagram of one
of the forms in Figs. 14 or 15. For the first,
5∑
i=1
1/ξi = 9/10 6= 1 and for the second one,
5∑
i=1
1/ξi = 14/15 6= 1. By performing the test just described, we find no solutions in the first
7See the file p3.sage.
Quadratic Differential Equations without Multivalued Solutions 31
1 1 −1
2
−3
−2
3
1
−5
3
Figure 13. The third configuration of Section 4.3.1.1.
1 1 −1
2
2
3
−2
1
5
3
Figure 14. One of the configurations of Section 4.3.1.2.
case. In the second case, we find that the solution can be completed by either ((u6, v6) = (2, 5),
(u7, v7) = (−3, 10)) or ((u6, v6) = (3, 4), (u7, v7) = (−5, 12)) – which will give, respectively,
equations XXII and XXIII – and that these are the only solutions.8
The vector fields(
x2 + [1− a]yz
) ∂
∂x
+ y(2x+ y − 2z)
∂
∂y
+ z(z − 2x− y)
∂
∂z
,
for a ∈ C, are the most general ones representing the configuration of the diagram in Fig. 15. Two
values of a differing by a sign give the same vector field: the involution (x, y, z) 7→ (x+ y,−y, z)
preserves the family and acts upon a by a 7→ −a. By setting a = −
(
σ2 + 15
)
/(8σ) we find
a radial orbit at [s+ 3 : −2s : 4] whose exponents are the roots of ζ2 − 7ζ + 15− σ2.
4.3.1.2.1 Equation XXII. The case (u6, v6) = (2, 5), (u7, v7) = (−3, 10) can be realized
by a = 1
2
√
5; in this way we find equation XXII. It has a homogeneous polynomial first integral
of degree 10 and commutes with a homogeneous polynomial vector field of degree 4 having the
same first integral (like Chazy IX). When restricted to a generic level surface of the first integral,
the vector field is birationally equivalent to the restriction of Chazy IX to a generic level surface
8For the analysis of both cases see the file p3.sage.
32 A. Guillot
1 −1 1
2
3
2
−3
1
5
3
Figure 15. The configuration of Section 4.3.1.2 leading to equations XXII and XXIII.
of its first integral: if φ is a solution to the Chazy IX equation, a solution to equation XXII is
given by
x =
3
2
(
1−
√
5
)
φ, y = −
3
(
1 +
√
5
)
(6φ2 +
[
1 +
√
5
]
φ′)2
2
([
1 +
√
5
]
φ′′ + 12φφ′
) ,
z = −
(
1 +
√
5
)
φ′′ + 12φφ′(
1 +
√
5
)
φ′ + 6φ2
;
reciprocally, from a solution (x, y, z) of equation XXII, we have the solution of Chazy IX
φ = −1
6
(
1 +
√
5
)
x, φ′ = − 1
12
(
1 +
√
5
)(
2x2 +
[
2−
√
5
]
zy
)
,
φ′′ = − 1
12
(
1 +
√
5
)(
4x3 +
[
4− 2
√
5
]
xyz −
[
2−
√
5
]
z2y
)
(these two transformations are birational inverses of one another).
4.3.1.2.2 Equation XXIII. The case (u6, v6) = (3, 4), (u7, v7) = (−5, 12) can be realized
by a = −3
4
√
3, obtaining equation XXIII. It has a homogeneous polynomial first integral of
degree 12 and commutes with a homogeneous sextic polynomial vector field having the same
first integral (like Chazy X). When restricted to a generic level surface of the first integral, it is
birationally equivalent to the restriction of Chazy X to a generic level surface: for a solution φ
of Chazy X, a solution to equation XXIII is given by
x =
1
2
(
3 +
√
3
)
φ, y =
2
11
(
1 + 2
√
3
)(3φ2 − [3−√3
]
φ′
)2
6φφ′ −
(
3−
√
3
)
φ′′
,
z =
(
3−
√
3
)
φ′′ − 6φφ′
3φ2 −
(
3−
√
3
)
φ′
;
reciprocally, for a solution (x, y, z) to XXIII, a solution to Chazy X is given by
φ =
1
3
(
3−
√
3
)
x, φ′ =
1
12
(
3−
√
3
)(
4x2 +
[
4 + 3
√
3
]
zy
)
,
φ′′ =
1
12
(
3−
√
3
)(
8x3 + 2
[
4 + 3
√
3
]
xzy −
[
4 + 3
√
3
]
z2y
)
(these two transformations are birational inverses of one another).
Quadratic Differential Equations without Multivalued Solutions 33
4.3.1.3 Cases (A) & (B), (u4, v4) = (−1,−3), with invariant plane. From the enu-
meration in (2.5), the exponents of the only two-dimensional nondegenerate quadratic homoge-
neous vector field with one radial direction with exponent −1 are (−1, 1, 1). The exponent −1
belongs to the fourth radial orbit and the exponents 1 to the other radial orbits in the invariant
plane. One of these must correspond to the exponent of a radial orbit in the original plane, but
all the exponents 1 are tangent to the original invariant plane. These data cannot be realized.
4.3.1.4 Case (B), (u4, v4) = (−1,−3) without invariant plane. Let us construct the
most general vector field having the required data and such that there is neither an obstruction
to semicompleteness given by the resonance at the fourth point nor an invariant plane making
this obstruction vanish. The most general vector field having z = 0 as an invariant plane, having
radial orbits with exponents (1, 2) at [1 : 0 : 0] and [0 : 1 : 0] (both with the exponent 1 along
the invariant plane) and a radial orbit at p = [0 : 0 : 1] with exponents whose sum is −4 is, up
to a rescaling of the variables,
(
x2 + αxz + βyz
) ∂
∂x
+
(
y2 + γxz − [α+ 2]yz
) ∂
∂y
+ z (z − x− y)
∂
∂z
, (4.13)
for α, β, γ ∈ C. If β and γ vanish simultaneously, there are two invariant lines through p, so we
may suppose that β 6= 0. The exponents at p are the roots of ξ2 − 4ξ + 3−
(
α2 + 2α+ βγ
)
and
we may thus set γ = −α(α+ 2)/β so that the exponents at p are (1, 3). At p, the exponent 1 is
tangent to the line ` given by αx+ βy = 0. Since this line is not invariant, α(α+ β) 6= 0.
The induced foliation on CP2 is, in the affine chart [u : v : 1], given by the kernel of the form(
u[α− 1] + βv + 2u2 + uv
)
dv −
(
−α[α+ 2]β−1u− [3 + α]v + uv + 2v2
)
du.
Doing a first blowup by v =
(
w − αβ−1
)
u we find the foliation
u
(
1 +
[
αβ−1 − 2
]
u− βw − uw
)
dw
−
(
2w − α
[
1 + αβ−1
]
β−1u+
[
1 + 2αβ−1
]
uw + βw2 − uw2
)
du.
A second blowup by w =
(
t+ α(β + α)β−2
)
u yields the foliation
(u+ · · · )dt−
(
t+ α(α+ β)
(
2α2 + α+ 2αβ + 3β
)
β−3u+ · · ·
)
du.
The condition for the original equation to induce a linearizable foliation at p is that
α(α+ β)
(
2α2 + α+ 2αβ + 3β
)
= 0.
Since ` is not invariant, we must have
β = −α(2α+ 1)
2α+ 3
. (4.14)
This gives the most general vector field having the sought conditions. Set α = 1
4(k−1) so that β =
−1
4(k+1)(k−1)/(k+5). After rescaling the variables by (x, y, z) 7→
(
−[k+1]−1x, [k+5]−1y, 14z
)
,
we obtain the vector field Y (k) in equation XV. This vector field is still the most general one
having the sought conditions, for it is nondegenerate whenever β is in C \ {0}. Let us integrate
equation XV. The image of Y (k) under
(x, y, z) 7→ (4z[x+ y],−16z[xy + xz + yz]) = (u, v)
34 A. Guillot
is v∂/∂u+ 6u2∂/∂v. The latter has the first integral g3 = 4u3 − v2, from which one can obtain
a first integral for the original equation. A solution of the reduced system is (u, v) = (℘(t), ℘′(t)),
for the Weierstrass function ℘ such that (℘′)2 = 4℘3 − g3. Solving for y and z,
y(t) = −℘
2 + x℘′
4x℘+ ℘′
, z(t) =
4x℘+ ℘′
4
(
4x2 − ℘
) ;
substituting into (4.13), the system reduces to x′(t) = −(k + 1)x2 + 1
4(k − 1)℘(t). For w =
−(k + 1)x, w′ = w2 + 1
4
(
1− k2
)
℘ and the equation is semicomplete if 6 - k. These vector fields
appeared in [17] as the family T2.
4.3.1.5 Case (A), (u4, v4) = (−1,−3) without invariant plane. We do not need to go
through a full analysis like the one of the preceding section for, as we previously mentioned, the
obstruction lies at the level of the foliation, and the vector fields of this section and of those of
the previous one induce the same foliations.
For X(m) = mY (6[1 − m]/m) + 2(m − 6)yE, for the vector field Y (·) of equation XV we
obtain equation IV. For β as in (4.14) we have β = 1
4(7m−6)(5m−6)m−1(m−6)−1: the vector
field is a nondegenerate one whenever β ∈ C \ {0}. This vector field has radial orbits with
exponents (1, 2) at [1 : 0 : 0], (−1,−2) at [0 : 1 : 0], and one with exponents (1, 3) at [0 : 0 : 1].
It induces a foliation on CP2 which is linearizable at this point (but such that the line tangent
to the exponent 1 is not invariant). These vector fields are of Halphen type. For
C = −x
2 − xz − yz
y2 − xz − yz
∂
∂x
+
∂
∂y
+
z(x− y)
y2 − xz − yz
∂
∂z
,
we have [C,X] = 2(m− 6)E. They have the adapted function
ξ =
z(x+ y)3
(xy + yz + zx)2
,
for which
− 1
ξ2
{ξ, t} =
1
2
(
1− 1
4
)
(ξ − 1)ξ −
(
1− 1
9
)
(ξ − 1) +
(
1− 1
m2
)
ξ
ξ2(ξ − 1)2
.
Thus, the vector field is semicomplete if m ∈ Z\{−1, 0, 1}. This family had previously appeared
in [17], as the family T1, and in [20], as the family H3c.
9
4.3.1.6 Case (A), (u4, v4) = (−1,−3), without invariant plane. By a discussion similar
to the one of the preceding section, these vector fields belong to the family X(m) − 8mzE
for m ∈ C, for X(m) in equation IV. The exponents at the radial orbits of the invariant line z = 0
are those of X(k) and those at [0 : 0 : 1] are (−1,−3). For this vector field, the exponents
at the radial orbit [2 : 2 : 1] are (3/(m + 3), 3(m − 1)/(m + 3)), and those at the radial
orbit
[
2m(m−6) : 2m(m+6) : (m−6)2
]
are (3m/(m−9),−3/(m−9)). If 3/(m+3) = n ∈ Z\{0},
−3/(m− 9) = n/(4n− 1) is not a nonzero integer.
9In the coordinates (ζ1, ζ2, ζ3) = (3z, y − 2z, x− 3y + 4z), we find the vector field −kH3c
(
k−1
)
, for
H3c(γ) = ζ1[8γζ1 + 12ζ2 − (6γ − 5)ζ3]
∂
∂ζ1
+
[
(6γ − 1)ζ22 + (2γ − 1)(8ζ2 + 3ζ3)ζ1
] ∂
∂ζ2
+
[
(6γ − 5)ζ23 + 8(3γ − 2)(3ζ2 + 2ζ3)ζ2
] ∂
∂ζ3
.
(This formula corrects formula (3.5c) in [20].)
Quadratic Differential Equations without Multivalued Solutions 35
4.3.1.7 Case (B), (u4, v4) = (−1,−3), without invariant plane. These vector fields
belong to the family Z(k) = Y (k)− 8zE for k ∈ C, for Y (k) of equation XV. The exponents at
the invariant line z = 0 are those of Y (k) and those at [0 : 0 : 1] are (−1,−3). The exponents
of Z(k) at the radial orbit ρ1 = [2 : 2 : 1] are (−k/3, 2 + k/3), and we thus need that k/3 ∈ Z.
The exponents of Z(k) at the radial orbit
[
2(1−k) : 2(1+k) : (1+k)2
]
are (6/(2k+3), k/(2k+3)).
Under the assumption that k/3 ∈ Z, the first one is an integer only if k = 0 or k = −3. In the
first case ρ1 has a zero exponent, which is incompatible with the regularity hypothesis; in the
second, the exponents of ρ1 are (1, 1), bringing us to the setting of Section 3.2.
4.3.2 In (4.12), 1/ξ4 + 1/ξ5 = 1/3.
Only three pairs of integers satisfy this equation: (4, 12), (6, 6), and (2,−6). Choosing one of
these fixes essentially five pairs of exponents.
4.3.2.1 If (ξ4, ξ5) = (2,−6). A radial orbit with product of exponents equal to 2 appears,
and places us in the setting of either Section 4.1 or 4.2.
4.3.2.2 If (ξ4, ξ5) = (4, 12) and either (u4, v4) = (2, 2) or (u4, v4) = (−2,−2). For
a radial orbit with exponents (2, 2), there exist three invariant planes containing this orbit
(Lemma 3.6 in [20]). Each one of these three invariant planes contains one further radial orbit
at its intersection with the original invariant plane. From (2.5), the sets of exponents of a non-
degenerate semicomplete quadratic homogeneous vector field on C2 containing the exponent 2
are (−2, 1, 2), (2, 3, 6) and (2, 4, 4). Thus, for each radial orbit in the invariant line, the exponent
that is not tangent to this line must belong to S = {−2, 1, 3, 4, 6}. This does not happen: both
in cases (A) and (B), one of these exponents is 2, which is not in S. Likewise, considering
a radial orbit with exponents (−2,−2), the only nondegenerate semicomplete quadratic homo-
geneous vector field on C2 having an exponent 2 has exponents (−2, 1, 2), but the exponents of
the original invariant plane are incompatible with this.
4.3.2.3 The “special” case. This is case (B) when (u4, v4) = (2, 3) and (u5, v5) = (2, 3).
The two radial orbits with exponents (2, 3) cannot be coplanar with one with exponents (1, 2)
in the invariant line, and we may thus assume that these four radial orbits are in general
position. The vector fields that have {z = 0} as an invariant line and radial orbits at [1 : q : 0]
for q2 = −3, each with exponents (1, 2), as in diagram (B), and that have radial orbits at [1 : 1 : 2]
and [1 : −1 : 2], with exponents (2, 3) each, form two one-parameter families. (If the position of
the other radial orbit within z = 0 is set to be at [1 : β : 0], β becomes a parameter.) One of
them is the family of Lins Neto’s vector fields (equation XVI, see Section 2.5). Vector fields in
the other family are linearly equivalent to vectors fields in this one.
4.3.2.4 The other cases. For all the remaining cases we can list all the possible values
of u4, v5, u5 and v5 and calculate the value of r5 =
5∑
i=1
(1/ui + 1/vi). It is equal to 4 in the
“special” case and is different from 4 in all the other cases. Leaving the special case aside, in
the solutions of the equation
1
u6
+
1
v6
+
1
u7
+
1
v7
= 4− r5,
one number (say u6) is determined up to a finite choice by the above equation: for example,
if r5 < 4, at least one of the summands must be positive and must belong to the interval [1, 4/(4−
36 A. Guillot
r5)], and so on. Once the value of u6 is determined, we may solve the three equations Ri in terms
of the three unknowns v6, u7 and v7. We obtain the following integer solutions.10 In case (B),
(u4, v4) = (2, 3), (u5, v5) = (−2,−3), (u6, v6) = (2, 3), (u7, v7) = (1,−6),
which, up to renumbering, falls within the setting defining the special case; in case (A),
(u4, v4) = (2, 3), (u5, v5) = (2, 3), (u6, v6) = (2, 3), (u7, v7) = (1,−6).
This last case is not realizable. If it were, following the discussion in Section 2.2.2, it would
differ from one of the vector fields of equation XVI by the addition of multiple of Euler’s vector
field
∑
i zi∂/∂zi by a homogeneous linear form, which must vanish at the three radial orbits
with exponents (2, 3) – which cannot be coplanar – but not at one of the radial orbits with
exponents (1, 2). This is impossible.
4.3.3 The remaining cases in (4.12)
We are in this case whenever ξ4, ξ5, ξ6 and ξ7 solve equation (4.12) but where the this equation
is not satisfied for any subset of the summands. The solutions to these equations are finite in
number. For each solution one may determine the finitely many values that (u4, v4), (u5, v5),
(u6, v6), and (u7, v7) may take and test, for each case, if the relations Ri are fulfilled. In this
way we obtain the following sets of four couples (u4, v4), (u5, v5), (u6, v6), and (u7, v7):
11
1. (2, 2), (1, 6), (1,−10), (6, 10),
2. (2, 2), (3, 5), (2, 12), (−2, 20),
3. (2, 2), (4, 4), (4, 8), (−4, 24),
4. (2, 2), (2, 4), (−6, 8), (−6, 8),
5. (1, 4), (1, 20), (−4,−7), (−4, 105),
6. (1, 4), (2, 3), (−3, 8), (−3, 8),
7. (1, 4), (2, 3), (−2, 7), (−7, 12).
The first one belongs to case (A), the rest to case (B). For each one of these cases, we must
first determine if the corresponding sets of exponents are realizable.
The argument given in Section 4.3.2.2 proves that cases (1) through (4) cannot be realized.
We are left with the exponents in (5), (6) and (7), all of which belong to case (B) and have a radial
orbit with eigenvalues (1, 4). Let us construct vector fields having these data. Choose z = 0
as the invariant line, and place the radial orbits ρ1 and ρ2 with exponents (1, 2) at [1 : 0 : 0]
and [0 : 1 : 0]. Let ρ4 and ρ5 be, respectively, radial orbits at [0 : 0 : 1] and [1 : 1 : 1]. The most
general vector field satisfying these conditions is
(
ax2 + dyz + ezx
) ∂
∂x
+
(
by2 + fyz + gzx
) ∂
∂y
+ z(cz − ax− by)
∂
∂z
,
with a = h − d − e, b = h − f − g, c = h − a − b. We would like ρ4 to be the radial orbit
with exponents (1, 4). The plane tangent to the exponent 4 cannot be invariant and contain
one of the radial orbits ρ1 or ρ2, for in that case the invariant plane would need to have a third
radial orbit with exponent 4, which cannot happen. In particular, the lines x = 0 and y = 0
cannot be simultaneously invariant. We impose the condition that u4 + v4 = 5 by setting
10See the file p4.sage.
11See the file p5.sage.
Quadratic Differential Equations without Multivalued Solutions 37
e = 1
2(9h− 3d− 2f − 3g). An explicit calculation shows that u5 + v5 = 5. In particular, case (5)
cannot be realized. The condition for ξ4 = 4 is that
2gd+ 3fg + 3fd− 9hf + 2f2 = 0. (4.15)
If d = −3
2f , the above equation reduces to f(5f+18h) = 0. If f = 0, the condition for ξ5 = 6
reads 13g2 − 42hg + 33h2 = 0; by solving it, we obtain equation XXVI up to a permutation
of x and y and a choice of the square root. If f = 18
5 h, the condition for ξ5 = 6 reads 325g2 −
204gh + 463h2 = 0; after solving it we find vector fields which have two nondegenerate radial
orbits whose exponents are not integers.
If 2d+3f 6= 0, we may set g = f(9h−3d−2f)/(2d+3f) in order to have (4.15). The resulting
vector fields form a variety that embeds into CP2, with homogeneous coordinates [d : f : h]. The
condition for the radial orbit ρ4 to be nondegenerate is that d2 − 3hd − f2 6= 0. We must now
impose two further conditions. The first: that the exponents at ρ5 be (2, 3). This can be done
since the radial orbits with exponents (1, 4) and (2, 3) cannot be coplanar with a radial orbit in
the invariant plane. The condition for this is that
36hd2f + 33h2d2 + 2f2d2 − 30f2dh− 90h2df − 2f3d− 2fd3
− 6hd3 + 54h2f2 + d4 + f4 = 0.
This is a rational quartic. Under the quadratic transformation
d = qr + 27pr, f = qr − 18pr, h = 5pr + pq, (4.16)
the equation becomes the quadratic one
4050rp− 350rq + 324pq − 3125r2 + q2 − 28431p2 = 0. (4.17)
The second: that the induced foliation on CP2 has a linearizable singularity at ρ4. The condition
for this is, after calculating it following a procedure analogous to the one done in Section 4.3.1.4,
(d− 9h− f)(3d+ 2f − 9h)
(
6hd+ df − f2
)(
d2 − f2 + 3hf − hd
)
×
(
f3 + 15f2h− df2 + 16dfh− 27fh2 − d2f − 6hd2 + d3 + 9h2d
)
×
(
d2 − 3hd+ 9hf − df
)(
27h2d− 16hd2 + d3 − 9dfh− d2f − df2 + f3
)
= 0.
We may parametrize rationally equation (4.17), substitute into (4.16) and then substitute into
this last equation which then becomes a polynomial equation in a single variable. Upon solving
this equation, we obtain the vector fields in equations XXIV through XXVIII, as well as other
vector fields obtained from these by either exchanging x and y or by replacing one of the square
roots in their expression with its Galois conjugate. A posteriori, equations XXIV through XXIV
correspond to case (7), equations XXVII through XXV to case (6). Their integration will be
carried out in the next section.
5 Some Abelian surfaces with automorphisms
and the associated equations
The Chazy IX and Chazy X equations and Lins Neto’s vector fields (equation XVI) give examples
of polynomial vector fields on affine surfaces (generic level surfaces of their homogeneous polyno-
mial first integrals) which can be embedded as Zariski-open subsets of Abelian surfaces and such
that, under this embedding, the vector fields become constant ones. The homogeneity of the
38 A. Guillot
first integral is associated to an automorphism of the Abelian surface, and this automorphism
preserves the vector field up to a constant factor.
Other special quasihomogenous vector fields are bound to arise from Abelian surfaces having
interesting cyclic automorphisms. Fujiki classified the finite automorphism groups of complex
tori of dimension two [12]. Table 6 in Fujiki’s work features the classification of nonspecial cyclic
automorphism groups of complex tori, which will be relevant to our results.
5.1 The Jacobian of Burnside’s curve and its order eight automorphism
Consider the genus two curve C of equation y2 = x
(
x4 − 1
)
. It is usually known as Burnside’s
curve, following Burnside’s work on its uniformization [4]. It has an automorphism group of
order 48; an order eight automorphism is given by (x, y) 7→
(
λ2x, λy
)
, for λ an eighth root of
unity. The Jacobian of C inherits an automorphism of order eight induced by this one. Let us
follow Mumford’s account on Jacobi’s work concerning the algebraic construction of the Jacobian
of C [40, Chapter IIIa]. Consider C4 with the coordinates (u1, u2, v1, v2). For
H = u2u
3
1 + u2v
2
1 − v22 − 2u22u1,
K = u41 − 2v1v2 + u22 − 3u21u2 + u1v
2
1,
an affine chart of the Jacobian J of C is the affine variety defined by H ≡ 0, K ≡ 1. These
relations ensure that, for the polynomials f = x
(
x4 − 1
)
, U = x2 + u1x+ u2 and V = v1x+ v2,
U divides f − V 2. The above order eight automorphism of C induces the order eight automor-
phism ` on J
(u1, u2, v1, v2) 7→
(
λ2u1, λ
4u2, λ
3v1, λ
5v2
)
, (5.1)
for which `∗H = λ2H, `∗K = K. The vector fields
X = v1
∂
∂u1
+ v2
∂
∂u2
+
1
2
(
2u2 − 3u21
) ∂
∂v1
+
1
2
(
u31 + v21 − 4u2u1
) ∂
∂v2
, (5.2)
Y = v2
∂
∂u1
+ (v2u1 − u2v1)
∂
∂u2
+
1
2
(
u31 − 4u2u1 + v21
) ∂
∂v1
+
1
2
(
u41 − 2u22 − u21u2 + u1v
2
1
) ∂
∂v2
of J are preserved by the cyclic automorphism (5.1) up a constant factor: λ for the first, λ3
for the second. When u1, u2, v1 and v2 are respectively given the weights 2, 4, 3 and 5, these
vector fields become quasihomogeneous of degrees two and four. Both H and K are common
first integrals of X and Y and are, respectively, quasihomogeneous of degrees 10 and 8. By
homogeneity considerations, all the varieties given by H ≡ 0 and K a nonzero constant, together
with their vector fields, are equivalent.
We will now integrate equations XXVII and XXVIII by showing that, when restricted to
a generic level surface of a first integral, they are birationally equivalent to J , and that this
birational equivalence maps the vector field of the equation to the vector field (5.2). These
equations have appeared in the analysis of other ones: if u1(t) is part of a solution of X,
y(t) = −1
4u1(t) solves Cosgrove’s reduced F-V equation y(iv) = 20yy′′ + 10(y′)2 − 40y3 (see [9,
Section 5]) in the particular case
(y′′′)2 − 24y′′′y′y − 192y5 + 80y′′y3 + 120(y′)2y2 + 4y′′(y′)2 − 8(y′′)2y = 0,
which has the first integral 20(y′)2y − 2y′′′y − 20y4 + (y′′)2.
Quadratic Differential Equations without Multivalued Solutions 39
5.1.1 Equation XXVII
It has the polynomial first integral of degree 8
Q = z3
[
zx−
(
1 +
√
2
)
y2 +
√
2yz
][
x3 +
(√
2− 1
)
y2z +
(
3
√
2− 4
)
yz2
+
(
3− 2
√
2
)
xz2 +
(
3
√
2− 5
)
x2z +
(√
2− 2
)
x2y +
(
8− 6
√
2
)
yzx
]
,
and commutes with a homogeneous polynomial vector field of degree 4 having the same first
integral. From a solution (x, y, z) to this equation, setting
u1 = 4
(
1−
√
2
)(
x+
√
2y
)
z,
v1 = 8
(√
2− 1
)[(
2− 2
√
2
)
yz +
(√
2− 2
)
zx− xy
]
z,
u2 = 16
(
7− 5
√
2
)[
zx−
(
1 +
√
2
)
y2 +
√
2yz
]
z2,
v2 = 32
(
7− 5
√
2
)[
zx−
(
1 +
√
2
)
y2 +
√
2yz
]
xz2
we obtain a solution to the vector fieldX in (5.2) taking values inH = 0, K =
(
4352−3072
√
2
)
Q.
An inverse to this map is given by
x =
u41 − 2u2u
2
1 + u1v
2
1 +
√
2v1v2
2
(
u1v2 +
√
2v1u2
) , y =
√
2v2 + v1u1
2
(
u21 − 2u2
) ,
z = −
(
1 +
√
2
)(
u1v2 +
√
2v1u2
)(
u21 − 2u2
)
2
(
2v22 − 4u31u2 + 4u22u1 + u51 + u21v
2
1 + 2
√
2u1v1v2
) .
To integrate this equation, based on the homogeneity properties of both equations, we looked
for solutions to (5.2) of the form u1 = P (x, y, z), with P a quadratic homogeneous polynomial.
(The vector field X determines v1 from the derivative of u1, u2 from the derivative of v1, v2
from the derivative of u2, and establishes an algebraic equation for P from the derivative of u2.)
5.1.2 Equation XXVIII
The vector field has the homogeneous first integral
Q =
[
x3z − 8
(
1 +
√
2
)
x2y2 −
(
2
√
2 + 3
)
y3z − 4yz3 +
(
12 + 8
√
2
)
xz3 −
(
6
√
2 + 10
)
x2z2
+
(
6 + 2
√
2
)
y2z2 +
(
12
√
2 + 11
)
zxy2 +
(
15 + 14
√
2
)
zx2y − 20
(
1 +
√
2
)
z2xy
]
×
[
x+
(
2
√
2− 3
)
y
]
z3
and commutes with a homogeneous polynomial vector field of degree 4. From a solution (x, y, z)
to the equation, setting
u1 = 16
(
17− 12
√
2
)[(
2
√
2− 2
)
y2 +
(
3− 2
√
2
)
yz − zx
]
,
u2 = 512
(√
2− 1
)[
x+
(
2
√
2− 3
)
y
]
(y − z)2z,
v1 =
64
7
(
6− 5
√
2
)[
7
(
3− 2
√
2
)
y2z − 7yzx+
(
5
√
2− 8
)
yz2
+
(
4 +
√
2
)
z2x+
(
8
√
2− 10
)
y3
]
,
v2 = 1024
(
3
√
2− 4
)(
2y2 − 3yz + zx
)[
x+
(
2
√
2− 3
)
y
]
(y − z)z,
we obtain a solution to (5.2) for H = 0 and K = 65536
(
17 − 12
√
2
)
Q. This solution has an
inverse: by setting
y =
(
2 +
√
2
)(
u1v1 −
√
2v2
)
8
(
u21 − 2u2
) ,
we can easily find x and z from u1 and v1. (The integration of this equation was obtained in
the same way as the previous one.)
40 A. Guillot
5.2 The anharmonic square with an automorphism of order twelve
For ρ3 = 1, let Λ = 〈1, ρ〉 and let Eρ be the elliptic curve C/Λ. It has an automorphism of order
three given by multiplication by ρ. The Abelian variety Eρ × Eρ has the cyclic automorphism
of order twelve induced by the linear mapping ` of C2, `(u, v) =
(
ρ2v,−ρ2u
)
, the composition
of the commuting mappings of orders three and four (u, v) 7→
(
ρ2u, ρ2v
)
and (u, v) 7→ (v,−u).
Let ℘ be the Weierstrass elliptic function of Eρ, and let g3 ∈ C∗ be such that (℘′)2 = 4℘3 − g3.
The field of meromorphic functions of Eρ × Eρ is generated by ℘(u), ℘′(u), ℘(v) and ℘′(v).
Since ℘(ρt) = ρ℘(t) and ℘′(ρt) = ℘′(t), the cyclic group generated by ` acts linearly upon the
vector space generated by these functions. A basis of eigenvectors is given by
a = ℘(u) + ℘(v), b = ℘(u)− ℘(v), c = ℘′(u) + i℘′(v), d = ℘′(u)− i℘′(v).
We have that `∗(a, b, c, d) =
(
λ8a, λ2b, λ9c, λ3d
)
for λ = −iρ.
We can embed a Zariski-open subset of Eρ × Eρ into C4 via (a, b, c, d). From the rela-
tion (℘′)2 = 4℘3 − g3, g3 = 1
2(a + b)3 − 1
4(c + d)2 and g3 = 1
2(a − b)3 + 1
4(c − d)2. From the
difference and sums of the right-hand sides we have the polynomials
∆ = 3a2b+ b3 − 1
2
(
c2 + d2
)
, 2G3 = a3 + 3ab2 − cd.
The image of the Zariski-open subset of Eρ × Eρ under the above-defined map is the variety A
given by ∆ ≡ 0 and G3 ≡ g3. The order 12 automorphism of Eρ × Eρ extends to C4 as
the linear automorphism (a, b, c, d) =
(
λ8a, λ2b, λ9c, λ3d
)
, which preserves A since `∗∆ = −∆
and `∗G3 = G3. The vector fields D± = ∂/∂u± i∂/∂v of C×C are preserved, up to a constant
multiplicative factor, by `: `∗D+ = λD+, `∗D− = λ7D−. Through the embedding, the vector
fields corresponding to D+ and D− are, respectively,
X+ = c
∂
∂a
+ d
∂
∂b
+ 6ab
∂
∂c
+ 3
(
a2 + b2
) ∂
∂d
,
X− = d
∂
∂a
+ c
∂
∂b
+ 3
(
a2 + b2
) ∂
∂c
+ 6ab
∂
∂d
.
Both ∆ and G3 are first integrals of these. If we give a and b the weight 2 and c and d the
weight 3, ∆ and G3 are quasihomogeneous of degree six, and D+ and D− are quasihomogeneous
of degree two. There is another interesting graduation for these objects taking values in Z/12Z.
If we give a, b, c and d the weights 8, 2, 9 and 3 in Z/12Z, respectively, X+ (resp. X−) is
quasihomogeneous of degree 2 (resp. 8), while ∆ and G3 are quasihomogeneous of degrees 6
and 0.
We will integrate equations XXIV, XXV and XXVI by showing that, when restricted to
a level set of a polynomial first integral, they are birationally equivalent to this algebraic model
of Eρ × Eρ with the vector field X+.
5.2.1 Equation XXIV
It has the homogeneous polynomial first integral of degree 12
Q =
[(
20 + 12
√
3
)
xz3 −
(
30 + 19
√
3
)
x2z2 +
(
6 + 6
√
3
)
x3z −
(
36 + 20
√
3
)
z3y
−
(
2 +
√
3
)
z2y2 +
(
66 + 36
√
3
)
xz2y −
(
14
√
3 + 26
)
x2zy + 2x4
]
×
[
zx−
√
3yz +
(√
3− 1
)
y2
]2
z4,
Quadratic Differential Equations without Multivalued Solutions 41
and commutes with a vector field of degree eight having the same first integral.12 From a solu-
tion (a, b, c, d) to X+,
x =
c
2a
, y = −
(
3 +
√
3
)
c4 + 48b2a4 − 6
(
4 +
√
3
)
a2bc2 + 2
√
3da3c
2a
[(
18 + 4
√
3
)
cba2 −
(
3 +
√
3
)
c3 − 6da3
] ,
z =
(
3−
√
3
)[(
18 + 4
√
3
)
cba2 −
(
3 +
√
3
)
c3 − 6da3
]
12
(
6ba2 − c2
)
a
gives a solution to equation XXIV. Reciprocally, from a solution to this equation in the level
set Q ≡ Q0, for θ2 =
(
26 + 15
√
3
)
/Q (θ is quasihomogeneous of degree −6 and is constant along
the solution),
a =
θ
2
(
5− 3
√
3
)[
6
(
11 + 6
√
3
)
z2xy − 2
(
13 + 7
√
3
)
x2zy + 6
(√
3 + 1
)
x3z
−
(
19
√
3 + 30
)
z2x2 + 2x4 −
(√
3 + 2
)
z2y2 + 4
(
3
√
3 + 5
)
xz3 − 4
(
9 + 5
√
3
)
z3y
]
×
[
zx−
√
3yz +
(√
3− 1
)
y2
]
z2,
b =
1
2
[
2x2 −
(√
3 + 3
)
yz +
(√
3 + 1
)
zx
]
,
c = 2xa,
d = 2x3 −
(
11 + 5
√
3
)
xyz + 3
(
1 +
√
3
)
x2z + 3
(
4 + 2
√
3
)
yz2 − 2
(
3 + 2
√
3
)
xz2,
we obtain a solution to X+ taking values in ∆ ≡ 0, G2
3 ≡
(
26 + 15
√
3
)
Q0.
13
5.2.2 Equation XXV
The integration is similar to that of the previous equation. It has the homogeneous polynomial
first integral of degree 12
Q =
[
4
√
3zx+ 8
(
1−
√
3
)
yz − x2
]
×
[
6zyx− 2
(
1−
√
3
)
yz2 +
(√
3− 2
)
xy2 − 2x2z −
√
3xz2 − 2
√
3y2z
]2
z4
and commutes with a vector field of degree 8 having the same first integral. For a solution (x, y, z)
to equation XXV and for θ2 = −1/Q, by setting
a = −θ
[
8
(
1−
√
3
)
yz + 4
√
3zx− x2
]
(x− z)
×
[
6zyx− 2
(
1−
√
3
)
yz2 − 2
√
3y2z −
√
3xz2 +
(√
3− 2
)
xy2 − 2x2z
]
z2,
b = x2 −
(
3 + 2
√
3
)
zx+ 2
(
1 +
√
3
)
yz,
c = −2θ
[(
3 +
√
3
)
zx−
(
2 +
√
3
)
yz − x2
][
8
(
1−
√
3
)
yz + 4
√
3zx− x2
]
×
[
2
(√
3− 1
)
yz2 − 2
√
3y2z −
√
3xz2 + 6zyx+
(√
3− 2
)
xy2 − 2x2z
]
z2,
d = 2
[
3
(
1 +
√
3
)
x2z − 3
(
1 +
√
3
)
z2x+ 4
√
3yz2 +
(
1− 4
√
3
)
zyx− x3
]
,
12While the degree of the first integral and the fact that the vector field commuted with a homogeneous
polynomial pointed towards a birational equivalence with the Chazy X equation, the degree of the commuting
polynomial ruled this out.
13In order to integrate this equation, from a solution (x, y, z) to equation XXIV, we looked for a solution of X+
where b was a quadratic polynomial P (x, y, z). Through X+, this determined d, which, in turn, determined a2.
The logarithmic derivative ξ = a′/a (which can be calculated from a2) satisfies the equation ξ′ = 6b − ξ2, and
this established an algebraic relation for P . From 2G3 = a
(
a2 + 3b2 + ξ
)
we determined a value of a, completing
the solution.
42 A. Guillot
we obtain a solution of X+. For the inverse, we can easily express y and z in terms of b, d and x.
Setting
x =
(
1 +
√
3
)(
bd−
√
3ca
)
3a2 + b2
,
completes the solution.
5.2.3 Equation XXVI
It has a homogeneous polynomial first integral of degree 12 and commutes with a homogeneous
polynomial vector field of degree 8 having the same first integral. Under the birational mapping
(homogeneous of degree one)
(x, y, z) 7→
(
x2z + (2−
√
3)(2z − 6x)yz + (3− 2
√
3)(xz − y2)z − (4
√
3− 7)xy2
[x+
(
1−
√
3
)
y − (2−
√
3)z](y − z)
,
(
√
3− 3)yz + xz + (2−
√
3)y2
y − z
,
z[x+
(
1−
√
3
)
y +
(√
3− 2
)
z]
y − z
)
the vector field of equation XXVI gets mapped to the one of equation XXIV. This proves that
its restriction to a level surface of its first integral is birationally equivalent to D+ on A.
6 Some Riccati equations with rational coefficients
We will now study the Riccati equations with rational coefficients appearing in our investigations.
They are of the form y′ = y2 + A(t) with A a rational function such that each one of its poles
is a double one. In [25, Section 3], we have stated conditions that, in some cases, guarantee
that all the solutions to such equations are meromorphic in the neighborhood of these double
poles. We will use these conditions to investigate the settings corresponding to equations having
exclusively single-valued solutions.
6.1 First equation
We begin with the equation z′ = z2 + t−2F (t) for
F (t) =
1
4
[
m2
(
1− q2
)( tm
tm − 1
)2
−
[
r2 − p2 +m2
(
1− q2
)]( tm
tm − 1
)
+
(
1− p2
)]
, (6.1)
which appeared in Section 3.2.4. We affirm that if m - p and for some j ∈ {0, . . . , q − 1},
either p− r = (q− 1− 2j)m or p+ r = (q− 1− 2j)m, all the solutions are meromorphic (notice
that if both conditions are satisfied we also have that m - r).
We begin by investigating the behavior of the solutions in the neighborhood of t = 0. Let ρ
be an mth root of unity. Since F (ρt) = F (t) and F (0) = 1
4
(
1− p2
)
, Lemma 3.1 in [25] allows us
to conclude that if m - p all the solutions are meromorphic in the neighborhood of t = 0.
We now study the solutions in the neighborhood of the values of t which are mth roots of
unity. The mapping (t, z) 7→
(
ρ−1t, ρz
)
is a symmetry of the equation. Since it permutes the mth
roots of unity, the behavior of the solutions at all of them is the same: if in the neighborhood
of t = 1 all the solutions to the equation are meromorphic then in the neighborhood of the
other mth roots of unity all the solutions to the equation will be meromorphic as well. In order
Quadratic Differential Equations without Multivalued Solutions 43
to consider the quotient by this symmetry, let w = tz, τ = tm (they are both invariant under
the above symmetry). Let G be the function such that G(tm) = F (t). We have
dw
dτ
=
1
mτ
(
w2 + w +G(τ)
)
.
Notice that since the mapping (t, z) 7→ (tm, tz) does not ramify along t = 1, the solutions of this
last equation in the neighborhood of τ = 1 correspond to those of (6.1) at t = 1. For s = 1− τ ,
q > 0 and
y =
1
m(s− 1)
(
w +
1
2
[1−m]
)
s+
1
2
(1− q),
s
dy
ds
= y2 + qy − 1
4m2
(
m2q2 − r2
1− s
+
p2 −m2
(1− s)2
)
s. (6.2)
In order to prove that in the neighborhood of s = 0 all the solutions to this last equation are
meromorphic, it suffices to show that this equation has a formal solution vanishing at 0 [25,
Section 3.1]. By setting r = mR, p = mP and y = −s(v′/v + 1
2(P + 1)/(s− 1)) for a solution y
to this last equation, we find that v is a solution of the hypergeometric equation
s(1− s)v′′ + [c− (a+ b+ 1)s]v′ − abv = 0,
for a = 1
2(P + R + 1 − q), b = 1
2(P − R + 1 − q) and c = 1 − q. If v =
∞∑
i=0
vis
i is a solution to
this equation then for every i
(i+ 1)(i+ c)vi+1 = (i+ a)(i+ b)vi.
The condition to have a non-identically zero formal solution is that for some j ∈ {0, . . . , q − 1},
either q− 1−P +R = 2j or q− 1−P −R = 2j. In such cases, which correspond to the second
one of the required conditions, we obtain a formal solution to (6.2) vanishing at 0.
Thus, when the stated conditions are satisfied, all the solutions to (6.1) will be meromorphic
in the neighborhood of the poles of the equation, and all the solutions will be single-valued.
In order to prove that the solutions are, moreover, rational, we must understand the behavior
of the solutions of (6.1) at t = ∞. By setting t = 1/τ and w = −t2z − t in (6.1), we have
that dw/dτ = w2 + τ−2F
(
τ−1
)
. Since F
(
τ−1
)
|τ=0 = 1
4
(
1 − r2
)
and F
(
(ρτ)−1
)
= F
(
τ−1
)
,
by Lemma 3.1 in [25], the solutions are meromorphic in the neighborhood of τ = 0 if m - r,
a condition implied by the two conditions that we already have. The solutions of (6.1) are
rational, for they are meromorphic at t =∞.
6.2 Second equation
We will now study the equations that appeared in Section 4.1.1 (cases κ = −1 and κ = 2), the
Riccati equations
y′ = y2 +
1
4
(
1− q2
) [(u′)2 − 2uu′′
u2
]
+
1
8
(
1− n2
)u′′
u
, (6.3)
where q, n ∈ Z, q2 6= 1, with u a quadratic polynomial in t. We claim that all the solutions of
these equations are meromorphic if n is odd and n < 2q.
The right-hand side the above equation has double poles at the zeros of u (even if they are
double ones). In order to prove that all the solutions to the above equation are single-valued, it
is sufficient to guarantee the existence of many local solutions defined in a neighborhood of the
poles.
44 A. Guillot
Multiplying u by a constant leaves the equation unchanged, so the parameters of the equation
are actually the roots of u. If y(t) is a solution to the above equation, ay(at+b) will be a solution
to the equation corresponding to replacing u by u(at+ b). There are two cases to consider, the
one where the roots of u are simple and the one where they are double.
If u has two different roots, we may suppose that u = t(t− 1), so that the equation becomes
y′ = y2 +
1
4
(
1− q2
) 1
t2(t− 1)2
+
1
4
(
1− n2
) 1
t(t− 1)
. (6.4)
The situations at both poles are equivalent (for we may replace t by 1− t following the previous
discussion), so we may concentrate exclusively in the one at t = 0. For q > 0, in the variable z =
ty + 1
2(1− q), we have the equation
tz′ = z2 + qz − 1
4
(
q2 − n2
1− t
+
q2 − 1
(1− t)2
)
t. (6.5)
Again, in order to prove that in the neighborhood of t = 0 all the solutions to equation (6.5) are
meromorphic, it suffices to show that there is a formal solution vanishing at 0 [25, Section 3.1].
But this equation is a special case of equation (6.2) for m = 1, p = q and r = n. Our previous
analysis proves that this equation has a formal solution if n is odd and n < 2q (supposing that
both of them are positive). In these situations, all the solutions to (6.4) are meromorphic in the
neighborhood of t = 0 and t = 1 and are thus single-valued.
In order to prove that the solutions are rational, we need to understand the behavior of the
solutions of (6.4) at t =∞. By setting t = 1
2(s− 1)/s and w = 1
2(1− 2t)2y + (2t− 1), we find
dw
ds
= w2 +
(
3− 4q2 + n2
)
s2 +
(
1− n2
)
4s2
(
1− s2
)2 .
The right-hand side is invariant under the involution s 7→ −s. Lemma 3.1 in [25] allows us to
conclude that this last equation has meromorphic solutions at s = 0 whenever n is odd. This
proves that, under the stated conditions, the solutions to the original equations are rational,
since they are meromorphic at infinity.
If in (6.3) u has a double root, the solutions will also be single-valued under the required con-
ditions, for the set of polynomials u giving exclusively single-valued solutions, which is a closed
one, contains, as we already proved, the polynomials with different roots. We can also establish
this directly: for u = t2, the equation becomes simply
y′ = y2 +
1− n2
4t2
,
which has, for c 6= 0, the solutions
y =
1− n− c(1 + n)tn
2t
(
ctn − 1
) ,
which are single-valued whenever n ∈ Z.
Acknowledgements
The author gratefully acknowledges support from grant PAPIIT IN102518 (UNAM, Mexico).
He thanks the referees for their thorough reading and helpful comments and suggestions.
Quadratic Differential Equations without Multivalued Solutions 45
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1 Introduction
2 Preliminaries
2.1 Univalence, semicompleteness
2.2 Radial orbits and Kowalevski exponents
2.3 Vector fields of Halphen type
2.4 Chazy and the homogeneous equations of the third order
2.5 Lins Neto's vector fields
3 First family
3.1 Exponents (-1,-1): Halphen's equations
3.2 Exponents (1,1): reduction to two-dimensional equations
4 Second family
4.1 The plane associated to one of the distinguished radial orbits does not contain the other radial orbit and vice-versa
4.2 The plane associated to one distinguished radial orbit contains the other one (but the two planes are different)
4.3 Same invariant plane
5 Some Abelian surfaces with automorphisms and the associated equations
5.1 The Jacobian of Burnside's curve and its order eight automorphism
5.2 The anharmonic square with an automorphism of order twelve
6 Some Riccati equations with rational coefficients
6.1 First equation
6.2 Second equation
References
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| id | nasplib_isofts_kiev_ua-123456789-209882 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1815-0659 |
| language | English |
| last_indexed | 2025-12-07T14:35:16Z |
| publishDate | 2018 |
| publisher | Інститут математики НАН України |
| record_format | dspace |
| spelling | Guillot, A. 2025-11-28T09:41:44Z 2018 Quadratic Differential Equations in Three Variables without Multivalued Solutions: Part I / A. Guillot // Symmetry, Integrability and Geometry: Methods and Applications. — 2018. — Т. 14. — Бібліогр.: 47 назв. — англ. 1815-0659 2010 Mathematics Subject Classification: 34M55; 34M45; 34M35 arXiv: 1804.10664 https://nasplib.isofts.kiev.ua/handle/123456789/209882 https://doi.org/10.3842/SIGMA.2018.122 For ordinary differential equations in the complex domain, a central problem is to understand, in a given equation or class of equations, those whose solutions do not present multivaluedness. We consider autonomous, first-order, quadratic homogeneous equations in three variables, and begin the classification of those that do not have multivalued solutions. The author gratefully acknowledges support from grant PAPIIT IN102518 (UNAM, Mexico). He thanks the referees for their thorough reading and helpful comments and suggestions. en Інститут математики НАН України Symmetry, Integrability and Geometry: Methods and Applications Quadratic Differential Equations in Three Variables without Multivalued Solutions: Part I Article published earlier |
| spellingShingle | Quadratic Differential Equations in Three Variables without Multivalued Solutions: Part I Guillot, A. |
| title | Quadratic Differential Equations in Three Variables without Multivalued Solutions: Part I |
| title_full | Quadratic Differential Equations in Three Variables without Multivalued Solutions: Part I |
| title_fullStr | Quadratic Differential Equations in Three Variables without Multivalued Solutions: Part I |
| title_full_unstemmed | Quadratic Differential Equations in Three Variables without Multivalued Solutions: Part I |
| title_short | Quadratic Differential Equations in Three Variables without Multivalued Solutions: Part I |
| title_sort | quadratic differential equations in three variables without multivalued solutions: part i |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/209882 |
| work_keys_str_mv | AT guillota quadraticdifferentialequationsinthreevariableswithoutmultivaluedsolutionsparti |