Computing the differential Galois group of a one-parameter
family of second order linear differential equations
Carlos E. Arreche

arXiv:1208.2226v1 [] 10 Aug 2012

A BSTRACT
We develop algorithms to compute the differential Galois group corresponding to a one-parameter
family of second order homogeneous ordinary linear differential equations with rational function
coefficients. More precisely, we consider equations of the form
∂2Y
∂Y
+ r1
+ r2Y = 0,
∂x2
∂x
where r1 , r2 ∈ C(x,t) and C is an algebraically closed field of characteristic zero.
We work in the setting of parameterized Picard-Vessiot theory, which attaches a linear differential
algebraic group to such an equation, that is, a group of invertible matrices whose entries satisfy a
system of polynomial differential equations, with respect to the derivation in the parameter-space. We
will compute the ∂t∂ -differential-polynomial equations that define the corresponding parameterized
Picard-Vessiot group as a differential algebraic subgroup of GL2 .
Contents
0

Introduction
0.1 Background . . . . . . . . . . . . . . . . . . .
0.2 This work . . . . . . . . . . . . . . . . . . . .
0.3 Organization . . . . . . . . . . . . . . . . . . .
1 Preliminaries
1.1 Differential algebra . . . . . . . . . . . . . . .
1.2 Linear Differential Algebraic Groups . . . . . .
1.3 Picard-Vessiot theory . . . . . . . . . . . . . .
1.4 Parameterized Picard-Vessiot theory . . . . . .
2 Algorithms for first order equations
2.1 First case: η ∈ K . . . . . . . . . . . . . . . .
2.2 Second case: η ∈
/K . . . . . . . . . . . . . . .
2.3 The PPV group of an intersection of PPV fields
3 Algorithms for second order equations
3.1 Kovacic’s algorithm . . . . . . . . . . . . . . .
3.2 Upper triangular case . . . . . . . . . . . . . .
3.3 Infinite dihedral case . . . . . . . . . . . . . .
3.4 Recovering the original group . . . . . . . . .
4 Concluding remarks
4.1 Consequences . . . . . . . . . . . . . . . . . .
4.2 Improvements . . . . . . . . . . . . . . . . . .
4.3 Future directions . . . . . . . . . . . . . . . .
References

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2010 Mathematics Subject Classification Primary 34M15; Secondary 12H20, 34M03, 20H20, 13N10, 37K20
Keywords: Differential Galois group, differential equations with parameters, parameterized Picard-Vessiot theory, linear differential
algebraic groups, Kovacic’s algorithm.
The author was partially supported by a Ford Foundation Predoctoral Fellowship and by NSF grant CCF-0952591.

C ARLOS E. A RRECHE
0. Introduction
0.1 Background
In differential Galois theory, one studies a differential equation with coefficients in a given differential field K
(§1.1), by investigating the differential field extension of K generated by the solutions for the equation, together
with their derivatives. This is in analogy with the classical Galois theory of polynomial equations, where one
studies such equations by studying the splitting field of the polynomial. In both cases, the algebraic relations
amongst the solutions for the equation are reflected in the algebraic structure of the group of automorphisms
of the field extension over the base field. The differential Galois theory of linear differential equations was
developed by Kolchin [18, 20], putting earlier work of Picard and Vessiot on a firm modern footing. The differential Galois group corresponding to a linear differential equation is a linear algebraic group over the subfield of
constants of K (§1.1).
In [3], a parameterized Picard-Vessiot theory for linear differential equations with parameters is developed in
close analogy with Kolchin’s differential Galois theory. This theory is a special case of an earlier generalization
of Kolchin’s theory, developed in [23]. In the parameterized theory, the base field K is a differential field, where
there are now two kinds of derivation: principal derivations and parametric derivations, so that the subfield of
constants, with respect to the principal derivations, is a differential field with respect to the parametric derivations. The PPV field (Definition 1.8) of a system of linear differential equations, with respect to the principal
derivations, is generated over the base field by the solutions for the system, together with their derivatives with
respect to all (principal, as well as parametric) derivations. The PPV groups (Definition 1.9) which arise in
this theory are linear differential algebraic groups (Definition 1.2), and their structure reflects the differentialalgebraic relations amongst the solutions, with respect to the parametric derivations. Linear differential algebraic
groups are the differential-algebraic analogues of linear algebraic groups—that is, they are subgroups of GLn
which are defined by the vanishing of systems of polynomial differential equations in the matrix entries.
The parametrized Picard-Vessiot theory has been finding a diverse number of applications in other areas. We
shall list a few examples, although we will not attempt to be exhaustive. In [12, §5], the theory is related to GaussManin connections [17, 24], leading to many potential applications in algebraic geometry. In [3, 11, 12, 25] the
theory is applied to the study of isomonodromy, and in [26] it is applied to monodromy-evolving deformations,
and to the study of some equations in mathematical physics; see [25, §1] for more references on this. In [13],
the authors apply a generalization of this theory to the study of difference equations, which has numerous
applications. See also [11], where the authors recast the theory in a new light and suggest several applications.
0.2 This work
We develop algorithms to compute the PPV group (Definition 1.9) associated to a second order homogeneous
linear differential equation defined over C(x,t); that is, an equation of the form
∂2Y
∂Y
+ r1
+ r2Y = 0,
2
∂x
∂x

(3.1)

∂
where r1 , r2 ∈ C(x,t), and the principal and parametric derivations are ∂x
and ∂t∂ , respectively. The key ingredient
in our approach is the Reduction Lemma 2.1, which reduces the determination of the PPV group to solving
some systems of linear equations. Our algorithms are Maple-ready, and we are currently working on writing a
computer implementation.
For technical reasons,1 it has become a tradition in the theory to assume that the field of constants, with
respect to the principal derivations, is differentially closed (§1.1). Therefore, we work over the field K := K0 (x),
where K0 is a ∂t∂ -differentially closed field extension of C(t). Working with differentially closed fields tends to be
an obstacle in practical applications of the theory, since such fields are usually unnaturally large (see [35, §1]).
1 In

[3], the authors impose this condition in order to ensure the existence of the PPV groups and extensions (see §1.4).

2

C OMPUTING G ALOIS GROUPS OF PARAMETERIZED SECOND ORDER EQUATIONS
The groups produced by our algorithm are defined over C(t) (Theorem 4.2), and we only perform computations
over finite algebraic extensions of C(t)—the differential closure plays a marginal role in our arguments. This is
in agreement with the stronger and more general results of [11, Theorem 2.8], which imply in particular that the
groups we wish to compute are actually defined over C(t), while it is shown in [35], by different methods, that
they are defined over C(t).
In [22], Kovacic developed an algorithm which computes all the “elementary" solutions to a second order
order equation of the form (3.2), if they exist. From these data, together with the classification of the algebraic
subgroups of SL2 , Kovacic’s algorithm decides which algebraic subgroup of SL2 is the (non-parameterized)
PV group (Definiton 1.7) of (3.2). The differential algebraic subgroups of SL2 (U ), where U is a universal
differential field,2 were classified in [32]. In [3, p. 137], the authors ask whether one can use this classification to
develop a parameterized analogue of Kovacic’s algorithm. Although we do not rely on the classification itself,
the methods in [32] were a great source of inspiration in developing the algorithms presented here, and the
outputs of our Algorithms 4 and 5 have the form of one of the groups in the classification in [32] (Theorem 4.1).
An algorithm to compute the PPV group of (3.2), provided that it is a Zariski-dense subgroup of SL2 , is
given in [8]. We develop algorithms to compute the PPV group of (3.1) in the remaining cases. Some of the
arguments we give in the other cases are similar to those in [8], but we work in different settings: in [8], the field
of principal constants is assumed to be universal, and a theorem of Seidenberg is applied in order to obtain a
concrete analytic interpretation for elements of this abstract universal differential field, as meromorphic functions
on some complex polydisk. Aided by this concrete interpretation, the author obtains his results. We will reprove
some of these results algebraically in the course of the proofs of Propositions 3.5 and 3.7, in order to be able to
apply them in our setting.
Lastly, we wish to point to a subtlety which is new to the parametrized setting. In Kovacic’s algorithm, one
performs a change of variables on (3.1) in order to obtain a new equation (3.2), whose PV group is a subgroup
of SL2 . Then, it is a simple matter to reconstruct the PV group of the original equation from these data, which
is always an almost direct product3 of the PV group of (3.2) and a change-of-variables group (§3.4). This is
no longer true4 for parameterized equations, where the richer differential-algebraic structure of the differential
Galois groups allows them to be related to each other in more complicated ways. We have solved the problem of
determining precisely how the PPV group of (3.1) is related to that of (3.2) in all cases, and we give an algorithm
to compute this relationship in §3.4.
0.3 Organization
We will now describe the contents of the paper in some more detail. In §1, we will briefly recall some basic
definitions and results from differential algebra, the theory of linear differential algebraic groups, Picard-Vessiot
theory, and parameterized Picard-Vessiot theory. In this section, we will also define most of the notation and
concepts which will be used later in the paper. In §2, we will describe three auxiliary algorithms which will form
the building blocks of our algorithms to compute the PPV group (Definition 1.9). Each algorithm in this section
is first followed by a concrete example of its implementation, and then by a proof that it gives the right answer.
In §3, we will describe algorithms to compute the PPV group of (3.1). We will first apply Algorithms 1 and 2
of §2 to compute the PPV group of an associated second order (3.2), which is a differential algebraic subgroup
of SL2 . Then, we will indicate how to recover the differential algebraic subgroup of GL2 corresponding to the
original (3.1), as an application of Algorithm 3 from §2. In §4, we state some consequences of the algorithms
and comment on improvements which will be relevant for practical applications.
2 The

notion of universal differential field [20] is stronger than that of being differentially closed.
is a quotient of the direct product of two groups, with finite kernel.
4 This answers a question of Ovchinnikov [28], who suggested to the author to treat the general case where the PPV group is not assumed
to be unimodular. He has pointed out to us that the parameterized and non-parameterized situations are still analogous: the PPV group
of (3.1) is a quotient of the direct product of the PPV group of (3.2) and the change-of-variables group, with zero-dimensional kernel.
3 This

3

C ARLOS E. A RRECHE
1. Preliminaries
In this section, we will briefly recall without proof some standard definitions and results. We will also set the
notation which we shall use for the rest of the paper. The algebraic closure of a field K will be denoted by K̄.
References will be given in each section.
Every field considered in this work will be of characteristic zero,
and C will always denote an algebraically closed field.
1.1 Differential algebra
We will briefly recall some definitions from differential algebra. We refer to [16, 20, 29] for what follows.
A differential field is a pair (K, ∆), where K is a field, and ∆ is a finite set of pairwise commuting derivations.
We abbreviate this by saying that K is a ∆-field. If ∆ = {δ} is a singleton, we write δ instead of {δ}, e.g. we say
K is a δ-field. We denote the n-fold composition of δ with itself by δn , with the convention that δ0 := idK .
We say that a field automorphism σ : K → K is a ∆-automorphism if σ ◦ δ = δ ◦ σ for every δ ∈ ∆. For any
subset Π ⊆ ∆, we denote by K Π the subset of elements c ∈ K such that δc = 0 for all δ ∈ Π. One can show that
K Π is a field, and we call it the subfield of Π-constants of K.
If L is a ∆-field, and K ⊆ L is a subfield such that δ(K) ⊂ K for each δ ∈ ∆, we say that L is a ∆-field extension
of K, and that K is a ∆-subfield of L. If y1 , . . . , ym ∈ L are such that
{δn yi | 1 6 i 6 m, δ ∈ ∆, n ∈ Z>0 }
is a set of generators for L as an overfield of K, we write L = Khy1 , . . . , ym i∆ . In this case, we also say that L is
differentially generated by {y1 , . . . , ym } over K, and that this is a set of differential generators for L over K.
Now let K be a δ-field. The ring of differential operators over K is denoted by K[δ] and consists of elements
n
of the form ∑m
n=0 rn δ with rn ∈ K. The multiplication in this ring is given by composition, and is determined by
n
the rule δ ◦ f = f ◦ δ + δ( f ) for f ∈ K. The integer m is the order of the differential operator D := ∑m
n=0 rn δ ,
and we denote it by ord(D ). We say D is monic if rm = 1.
The ring of differential polynomials over K (in m differential indeterminates) is
( j)

K {Y1 , . . . ,Ym }δ := K[Yi ],
( j)

the free (commutative) K-algebra on the infinite set of variables {Yi | 1 6 i 6 m, j ∈ Z>0 }. We give it a
( j)
( j+1)
(0)
δ-ring structure by setting δYi = Yi
. We often omit the zero superscript, and write Yi = Yi . The ring
K{Y1 , . . . ,Ym }δ is a free object in the category of δ-K-algebras. In particular, any ordered m-tuple (y1 , . . . ym ) of
( j)
elements of K defines a unique δ-homomorphism K{Y1 , . . . ,Ym }δ → K, determined by Yi 7→ δ j yi . We say that
K is differentially closed if every consistent system of differential polynomial equations with coefficients in K
has a solution in K.
1.2 Linear Differential Algebraic Groups
The theory of differential algebraic groups was initiated by Cassidy in [1], and we refer to that paper for more
information and complete proofs of the statements given here (see also [21]). For the rest of this section, we
assume that K is a differentially closed ∂-field, with subfield of constants K ∂ = C.
D EFINITION 1.1. Let V ⊆ K m . We say V is Kolchin-closed if there are differential polynomials p1 . . . pr in
K{Y1 , . . . ,Ym }∂ such that
V = {v ∈ K m | pi (v) = 0, ∀ 1 6 i 6 r}.
D EFINITION 1.2 (Cassidy [1, II, §5]). A linear differential algebraic group is a Kolchin-closed subgroup of
GLn (K) ⊂ K n×n .
4

C OMPUTING G ALOIS GROUPS OF PARAMETERIZED SECOND ORDER EQUATIONS
T HEOREM 1.3 (Cassidy [1, Prop. 11]). Let B be a proper differential algebraic subgroup of Ga (K). Then, there
exists a unique nonzero monic operator L ∈ K[∂] such that
Ga (K; L ) := {b ∈ Ga (K) | L b = 0} = B.
T HEOREM 1.4 (Cassidy [1, Prop. 31 and its Corollary]). Let A be a proper differential algebraic subgroup of
Gm (K). Then, either A = µn ⊂ C× , the group of nth roots of unity, for some n ∈ N, or else there exists a unique
nonzero monic operator L ∈ K[∂] such that
n
 
o
Gm (K; L ) := a ∈ Gm (K) L ∂a
=
0
= A.
a
T HEOREM 1.5 (Cassidy [1, Prop. 42]). Let H be a Zariski-dense differential algebraic subgroup of SL2 (K).
Then, either H = SL2 (K), or else H is conjugate to SL2 (C).
1.3 Picard-Vessiot theory
We refer to [29, Ch. 1] for what follows. Let K be a δ-field such that K δ is algebraically closed and let D be a
differential operator over K of order m.
D EFINITION 1.6. A Picard-Vessiot extension (or PV-extension) for the differential equation DY = 0 is a δ-field
extension of K, which we denote by PV(D /K), such that
(i) There are m distinct, K δ -linearly independent elements y1 , . . . , ym ∈ PV(D /K) such that D yi = 0 for all
1 6 i 6 m.
(ii) PV(D /K) = Khy1 , . . . , ym iδ .
(iii) PV(D /K)δ = K δ .
A PV extension for the operator D exists and is unique up to (non-unique) differential isomorphism, thus
justifying the notation PV(D /K).
D EFINITION 1.7. The Picard-Vessiot group of D over K (or PV-group) is the set of δ-field automorphisms of
PV(D ) which leave K point-wise fixed, and is denoted by Galδ (D /K).
The solution space of D is the K δ -vector subspace of PV(D /K) consisting of all y ∈ PV(D /K) such that
D y = 0, and we denote it by Sol(D ).
The K δ vector space Sol(D ) is m-dimensional, and it is stable
 under the action of Galδ (D /K). This action
defines an injective homomorphism Galδ (D /K) ,→ GL Sol(D ) , and one can show that its image is a linear
algebraic group. There is a Galois correspondence between Zariski-closed subgroups of Galδ (D /K) and intermediate differential field extensions between K and PV(D /K) (cf. Theorem 1.10).
1.4 Parameterized Picard-Vessiot theory
We are now ready to state the main results of the parameterized Picard-Vessiot theory. We will follow the
presentation in [3]. Let K be a ∆ := {δ, ∂}-field, such that K0 := K δ is a differentially closed ∂-field, and let
D ∈ K[δ] be a differential operator over K, of order m.
D EFINITION 1.8. A parameterized Picard-Vessiot extension (or PPVextension) for the differential equation

DY = 0 is a ∆-field extension of K, which we denote PPV(D /K), such that
(i) There are m distinct, K0 -linearly independent elements y1 , . . . , ym ∈ PPV(D /K) such that D yi = 0 for all
1 6 i 6 m.
(ii) PPV(D /K) = Khy1 , . . . , ym i∆ .
(iii) PPV(D /K)δ = K0 .
5

C ARLOS E. A RRECHE
A PPV extension for the operator D exists and is unique up to (non-unique) differential isomorphism, thus
justifying the notation PPV(D /K).
D EFINITION 1.9. The parameterized Picard-Vessiot group of D over K (or PPV group) is the set of ∆-field
automorphisms of PPV(D /K) which leave K point-wise fixed, and is denoted by Gal∆ (D /K).
The solution space of D is the K0 -vector subspace of PPV(D /K) consisting of all y ∈ PPV(D /K) such that
D y = 0, and we denote it by Sol(D ).
One can show that Sol(D ) is m-dimensional as a K0 -vector space, and the action of Gal∆ (D /K) on Sol(D )
identifies Gal∆ (D /K) with a subgroup of GL Sol(D ) . It is shown in [3] that the image of Gal∆ (D /K) under
this monomorphism is a linear differential algebraic group.
T HEOREM 1.10 (Parameterized Galois correspondence [3, Thm. 3.5]). Let D ∈ K[δ]. Let P be the set of ∆-field
extensions L of K contained in PPV(D /K), and let S be the set of Kolchin-closed subgroups H of Gal∆ (D /K).
Consider the maps ψ : S → P and ϕ : P → S defined by ψ(H) := {a ∈ PPV(D /K) | σa = a ∀σ ∈ H} and
ϕ(L) := Gal∆ (D /L). Then:
(i) ϕ defines an inclusion-reversing bijection P ↔ S, with inverse given by ψ.
(ii) L ∈ P is a PPV extension of K (for some operator D 0 ) if and only if ϕ(L) is a normal subgroup of
Gal∆ (D /K). In this case, σ(L) = L for all σ ∈ Gal∆ (D /K), and the restriction homomorphism σ 7→ σ|L is
a surjection Gal∆ (D /K)  Gal∆ (D 0 /K).
Remark 1.11. The δ-subfield Khy1 , . . . , ym iδ ⊆ PPV(D /K) defines a PV extension of (K, δ) for D (where
{y1 , . . . , ym } is any fixed K0 -basis of Sol(D ) in PPV(D /K)). The δ-field PV(D /K) is stabilized by Gal∆ (D /K),
and the resulting restriction homomorphism is in fact an injection
σ 7→ σ|PV(D /K) : Gal∆ (D /K) ,→ Galδ (D /K).
T HEOREM 1.12 (Cassidy-Singer [3, Prop. 3.6 (2)]). For D ∈ K[δx ], Gal∆ (D /K) is Zariski-dense in Galδx(D /K).
2. Algorithms for first order equations
In this section, we give three auxiliary algorithms which will be applied in the next section to compute the PPV
group of a second order equation. The main idea for Algorithm 1 is well-known (see [3, Example 7.1]); we
have merely modified it in order to avoid factoring polynomials into irreducible factors. Algorithm 1 also helps
illustrate Algorithms 2 and 3. Algorithm 2 solves an analogous problem to that solved by Algorithm 1, but in a
slightly different context. The main idea for Algorithm 2 comes from an argument given by Singer in the course
of the proof of [31, Prop. 4.1 (4)]. Algorithm 3 computes the PPV group corresponding to the intersection of the
PPV fields for two first order inhomogeneous equations.
S ETTING . For the rest of this paper we shall use the following notation:

(i) K0 is a differentially closed ∂t -field extension of C(t), ∂t , where ∂t denotes the derivation

d
dt

on C(t).

(ii) K := K0 (x), endowed with the structure of ∆ := {δx , ∂t }-field determined by setting δx x = 1, ∂t x = 0, and
δx a = 0 for all a ∈ K0 .
(iii) We consider the elements of K as rational functions in x with coefficients in K0 , and we denote the degree
of a polynomial in x and greatest common divisor of polynomials in x by degx and gcdx , respectively. We
adopt the convention that degx (0) = 0.
Consider the ∆-field extension F of K of given by F := K(η), where δx η := pη, ∂t η := qη, and p, q ∈ C(t)(x)
satisfy the integrability condition δx q = ∂t p. We do not exclude the possibility that η ∈ K or, equivalently,
6

C OMPUTING G ALOIS GROUPS OF PARAMETERIZED SECOND ORDER EQUATIONS
that F = K. We wish to compute the PPV group corresponding to the first order (inhomogeneous) differential
equation δxY = η over F.
We claim this is a differential algebraic subgroup of Ga (K0 ). To see this, consider the associated homogeneous equation [29, Ex. 1.18]: let P := δ2x − pδx . One can show that, if {1, θ} is a K0 -basis of Sol(P ),
then PPV(P /F) = Fhθi∆ . We may assume without loss of generality that δx θ = η, because δx (δx θ) = pδx θ
and δx θ 6= 0 together imply that δx θ = aη for some a ∈ K0× . It follows from this that δx (σθ − θ) = 0 for all
σ ∈ Gal∆ (P /F). The map σ 7→ bσ := (σθ − θ) defines an isomorphism of Gal∆ (P /F) onto a differential algebraic subgroup of Ga (K0 ). From now on, we will identify Gal∆ (P /F) with its image in Ga (K0 ) via this
isomorphism.
Simply put, each algorithm will use the input data to construct a system of homogeneous linear equations,
and then construct the output from a solution for the system which satisfies a certain minimality condition. The
construction of these systems of equations will rely on the following Lemma.
L EMMA 2.1 (Reduction Lemma). Let p, q ∈ K, η ∈ F, and P ∈ K[δx ] be defined as in the preceding discussion,
and let 0 6= L ∈ K0 [∂t ]. The following are equivalent:
(i) Gal∆ (P /F) ⊆ Ga (K0 ; L ).
(ii) L (η) = δx f for some f ∈ F.
(iii) L (η) = (δx h + ph)η for some h ∈ K.
Proof. Assume that Gal∆ (P /F) ⊆ Ga (K0 ; L ), or, equivalently, that L (bσ ) = 0 for all σ ∈ Gal∆ (P /F). Since
σ (L (θ)) = L (θ) + L (bσ ) = L (θ),
we have that L (θ) ∈ F by the Galois correspondence. Since δx L (θ) = L (η), setting f := L (θ) gives the implication (i) ⇒ (ii). Conversely, assume that there exists f ∈ F such that δx f = L (η). Then δx (L (θ) − f ) = 0, so
L (θ) = f + a for some a ∈ K0 . This implies that L (bσ ) = 0 for all σ, and we have shown that (ii) ⇒ (i).
To see that (iii) ⇒ (ii), let h ∈ K be as in part (iii) of the Lemma, and set f := hη.
n

Now we will show that (ii) ⇒ (iii). Since ∂ηt η ∈ K, we have that ∂ηt η ∈ K for any n, so L (η)
η ∈ K. If η ∈ K,
setting h := ηf establishes our claim. If η ∈
/ K, we have two cases to consider: either η is algebraic over K or η
is transcendental over K.
If η is algebraic over K of degree m, then {1, η, . . . , ηm−1 } is a basis for F as a K-vector space, so we may
L (η)
δx f
i
i
write f = ∑m−1
i=0 ai η with ai ∈ K in a unique way. Then δx f = ∑(δx ai + ipai )η and, since η = η ∈ K, we
have that
δx f m−1
= ∑ (δx ai + ipai )ηi−1 = δx a1 + pa1 ,
η
i=0
because the set {η−1 , 1, . . . , ηm−2 } is also a basis for F as a K-vector space. Therefore, δx f = δx (a1 η), and
setting h := a1 establishes our claim in this case.
Now assume that η is transcendental over K. Then f has a unique partial fraction decomposition:
c j,k
f = ∑ ai ηi + ∑
,
j
i
j,k (η − ek )
where ai ∈ K and c j,k , ek ∈ K̄, because the set
o
[n
1
B := {ηi | i > 0} ∪
j
>
1
j
(η−e)
e∈K̄

is a basis for K̄(η) as a K̄-vector space. Since the image of B under the invertible K̄-linear map given by
7

C ARLOS E. A RRECHE
multiplication by η−1 is still a basis, and L (η)
η =

δx f
η

∈ K, we see that

δx c j,k (η − ek ) − jc j,k (pη − δx ek )
δx f
= ∑(δx ai + ipai )ηi−1 + ∑
η
η(η − ek ) j+1
i
j,k
= ∑(δx ai + ipai )ηi−1 + ∑
i

j,k

δx ck, j − j pc j,k jc j,k (δx ek − pek )
+
η(η − ek ) j
η(η − ek ) j+1

= δx a1 + pa1 .
Therefore, δx f = δx (a1 η), and setting h := a1 establishes the Lemma.
Remark 2.2. The equivalence of (i) and (ii) in the Lemma is well-known (see [8, 31]). The fact that (iii) ⇒ (ii)
was used by Singer in [31]. We do not know of a proof of the implication (ii) ⇒ (iii) in the literature.
The following two algorithms compute the operator L ∈ K0 [∂t ] such that Gal∆ (P /F) ' Ga (K0 ; L ). Since
Ga (K0 ; L ) = Sol(L ), its dimension as a C-vector space is equal to ord(L ), and L is the operator of smallest
order such that the equivalent conditions of Lemma 2.1 are satisfied. Algorithm 1 applies only when η ∈ K,
while Algorithm 2 applies only when η ∈
/ K.
2.1 First case: η ∈ K
By part (ii) of Lemma 2.1, the operator L ∈ K0 [∂t ] such that Gal∆ (P /K) ' Ga (K0 ; L ) is the operator of smallest
order such that L η = δx f , for some f ∈ K = F. We will write L and f with undetermined coefficients, and use
this relation to obtain a system of linear equations in the coefficients of L and f .
A LGORITHM 1 (Primitive of a rational).
Input: η ∈ C(t)(x).
Output: L η := ∑`i=0 ai ∂ti ∈ C(t)[∂t ] such that Gal∆ (P /K) ' Ga (K0 ; L η ).
Step 1: Write
η = η0 +

η1
,
η2

with ηi ∈ C(t)[x] for 0 6 i 6 2, such that
gcdx (η1 , η2 ) = 1

degx (η1 ) < degx (η2 ).

and

Step 2: Let

η2
,
gcdx (η2 , δx η2 )
and let n be the smallest integer such that d n η is a polynomial in x. We remark that d is the product of the
irreducible factors of η2 , and n is the highest multiplicity of an irreducible factor of η2 .
Step 3: Let M := degx (d) and s := degx (η0 ). For each 0 6 N 6 M, write
!
(n+N−1)
M−1
l
s+1
N
ξ
x
∑
k,l
l=0
LN := ∑ αi ∂ti
and
fN := ∑ β j x j + ∑
,
dk
i=0
j=0
k=1
d :=

where the αi , β j , and ξk,l are undetermined coefficients.
Step 4: Treating each β j and ξk,l as a δx -constant, set
δx fN = LN (η),

(HN )

and then multiply each side of this equation by d n+N . The result will be an equality of polynomials in x, whose
coefficients are homogeneous linear forms in the αi , β j , and ξk,l . Equating coefficients of like-powers of x, we
obtain a system of
8

C OMPUTING G ALOIS GROUPS OF PARAMETERIZED SECOND ORDER EQUATIONS
M(n + N) + s
homogeneous linear equations with coefficients in C(t), in the
(N + 1) + (s + 2) + M(n + N − 1)
variables αi , β j and ξk,l .
Step 5: If N = M, the system of linear equations defined by (HN ) has a solution with not all αi being equal
to zero. Find the smallest nonnegative integer ` 6 M such that the system of linear equations defined by (HN ),
with N = `, has a solution with not all αi being zero.
Step 6: Find a solution
αi = ai

for 0 6 i 6 `,

βj = bj

for 0 6 j 6 s + 1,
for 1 6 k 6 n + N − 1 and 0 6 l 6 M − 1;

ξk,l = ck,l

for this system of homogeneous linear equations, such that each ai , b j , ck,l ∈ C(t) and a` = 1. This condition
determines a0 , . . . , a` uniquely.
Step 7: Set L η := ∑`i=0 ai ∂ti , and go to Output.
Remark 2.3. Upon inspection, we see that the system of linear equations of Step 4 has coefficients in the smallest
algebraic extension L of C(t) such that η ∈ L(x). Therefore, L η ∈ L[∂t ].
Remark 2.4. In practice, it is often possible to calculate L η more directly, as follows: let {d1 , . . . , dM } be the set
of poles of η, and let n be the maximum order of η at any of these poles. If
n

η=

n

M

ek,l

∑ b j x j + ∑ ∑ (x − dl )k

j=0

k=1 l=1

is the partial fraction decomposition of η, it is shown in [3, Example 7.1] that Ga (K0 ; L η ) is the C-span of
{e1,l }M
i=1 in K0 . Incidentally, this proves the claim made in Step 5, that the system obtained from (HN ) with
N = M has a solution with not all αi being 0, because

M > dimC Ga (K0 ; L η ) = ord(L η ).
If {ẽk }`k=1 is a basis of Ga (K0 ; L η ) considered as a C-vector space, then L η is given by ∑`i=0 ai ∂ti , where
 (0)

Y
Y (1) · · · Y (`)
 ẽ1 ∂t ẽ1 · · · ∂` ẽ1 
`
t 

det  .
= ∑ aiY (i) .

.
.
.
..
..
..  i=0
 ..
ẽ`

∂t ẽ` · · ·

∂t` ẽ`

Algorithm 1 circumvents the need to factorize the denominator of η, because performing such a factorization
could be computationally infeasible in practice. However, when such a factorization is available, it should be
more efficient to carry out this simpler algorithm, instead of Algorithm 1.
E XAMPLE 1. We will now apply Algorithm 1 with η :=

2t
,
x2 +t

to compute the PPV group corresponding to

∂Y /∂x = 2t/(x2 + t).

(2.1)

In this case, d = x2 + t, n = 1, M = 2, and s = 0. We write the operator LN and the rational function fN with
undetermined coefficients, as in Step 3, with N = 1:
ξ1,0 + ξ1,1 x
L1 := α1 ∂t∂ + α0
and
f1 := β0 + β1 x +
.
x2 + t
Step 4 then requires us to substitute these expressions in (HN ), with N = 1, to obtain
ξ1,1
−2ξ1,0 x − 2ξ1,1 x2 2tα0 + 2α1
∂ f1
−2tα1
= β1 + 2
+
=
+ 2
= L1 (η).
2
2
2
∂x
x +t
(x + t)
x +t
(x + t)2
9

(2.2)

C ARLOS E. A RRECHE
After multiplying by d 2 = (x2 +t)2 on both sides, we have that (2.2) holds if and only if the following polynomial


in x is zero
(−β1 )x4 + 2tα0 + 2α1 − 2tβ1 + ξ1,1 x2 + (2ξ1,0 )x + 2t 2 α0 − t 2 β1 − tξ1,1 .
Thus we obtain a system of linear equations by setting each coefficient equal to zero. We then find the solution
α0 = − 2t1 , α1 = 1, β1 = β0 = ξ1,0 = 0, ξ1,1 = −1, and we thus obtain an operator L ∈ C(t)[ ∂t∂ ] and a rational
function f ∈ C(x,t), defined as L := ∂t∂ − 2t1 and f := x−x
2 +t , which satisfy condition (ii) of Lemma 2.1. One can
check that there is no solution with α1 = 0, and therefore the PPV group of (2.1) is Ga (K0 ; ∂t∂ − 2t1 ).
P ROPOSITION 2.5 (Algorithm 1 is correct). Suppose that η ∈ K. Then, the Output of Algorithm 1 is the operator
which defines Gal∆ (P /K) as a subgroup of Ga (K0 ). In other words, Gal∆ (P /K) ' Ga (K0 ; L η ).
Proof. The solution found in Step 6, for the system of linear equations of Step 4, also defines a rational function
!
(n+`−1)
l
s+1
∑M−1
j
l=0 ck,l x
f := ∑ b j x + ∑
.
dk
j=0
k=1
By construction, L η ∈ K0 [∂t ] and f ∈ K satisfy part (ii) of Lemma 2.1. Therefore, Gal∆ (P /K) ⊆ Ga (K0 ; L η ).
We need to rule out the possibility that there exist L̃ ∈ K0 [∂t ] and f˜ ∈ K satisfying part (ii) of Lemma 2.1
with `˜ := ord(L̃ ) < ord(L η ) = `. We proceed by contradiction: assume that such L̃ and f˜ do exist. Write
!
M−1
l
`˜
s+1
T
c̃
x
∑
k,l
i
j
l=0
L̃ := ∑ ãi ∂t ,
and
f˜ := ∑ b̃ j x + ∑
,
dk
i=0
j=0
k=1
˜
˜
for some T ∈ N, with ãi , b̃ j , c̃k,l ∈ K0 . A calculation shows that d n+` L̃ (η) ∈ K0 [x]. Therefore, d n+` δx f˜ ∈ K0 [x],
and we may take T = n + `˜ − 1. But this means that αi = ãi , β j = b̃ j , ξk,l = c̃k,l is a solution for the system of
equations defined by (HN ) with N = `,˜ which contradicts the choice of `.

2.2 Second case: η ∈
/K
By part (iii) of Lemma 2.1, the desired L ∈ K0 [∂t ] is the operator of smallest order such that L (η) = (δx h + ph)η
for some h ∈ K. We will write L and h with undetermined coefficients, and use this relation to obtain a system
of linear equations in the unknown coefficients of L and h.
A LGORITHM 2 (Primitive of an exponential).
Input: p, q ∈ C(t)(x) such that ∂t p = δx q.
Output: L p,q := ∑`i=0 ai ∂ti ∈ C(t)[∂t ] such that Gal∆ (P /F) ' Ga (K0 ; L p,q ).
Step 1: Write

p = p0 +

p1
p2

and

q = q0 +

q1
,
q2

with pk , qk ∈ C(t)[x] for 0 6 k 6 2, such that
gcdx (p1 , p2 ) = 1 = gcdx (q1 , q2 ),
degx (p1 ) < degx (p2 ) and
Step 2: Let

degx (q1 ) < degx (q2 ).

p2 q2
,
gcdx p2 q2 , δx (p2 q2 )
let ν be the smallest integer such that d ν p and d ν q are both polynomials in x, and define
d :=

m := degx (d),

n := max{degx (p0 ), degx (q0 ), ν},

and

M := mn + n + 1.

We remark that d is the product of the irreducible factors of p2 q2 , and ν is the highest multiplicity of an irreducible factor of p2 or q2 .
10

C OMPUTING G ALOIS GROUPS OF PARAMETERIZED SECOND ORDER EQUATIONS
Step 3: Define a sequence {Ri } ⊂ K recursively: R0 := 1 and Ri := ∂t Ri−1 + qRi−1 for i > 1.
Step 4: For each nonnegative integer N 6 M, consider the inhomogeneous first order differential equation
(with undetermined coefficients α0 , . . . , αN ):
N

δxY + pY = ∑ αi Ri .

(2.3)

i=0

Find bounds5 SN , TN ∈ N, depending only on p, q, and N, such that the following condition holds: if a0 , . . . , aN ∈
K0 is any (N + 1)-tuple of elements, and hN ∈ K satisfies the equation obtained from (2.3) after replacing αi by
ai for each 0 6 i 6 N, then hN must be of the form
!
m−1
SN
TN
l
ξ
x
∑
k,l
l=0
,
(2.4)
hN = ∑ β j x j + ∑
dk
j=0
k=1
for some β0 , . . . , βSN , ξ1,0 , . . . , ξTN ,m−1 ∈ K0 . That is, the degree of the polynomial part of hN is bounded by SN ,
and the order of any pole of hN is bounded by TN , regardless of the values of α0 , . . . , αN in K0 .
Step 5: Replace Y with this expression for hN in (2.3), to obtain
N

δx hN + phN = ∑ αi Ri ,

(IN )

i=0

where the β j and ξk,l are to be treated as undetermined δx -constants. Multiplying by d TN +n on each side, we
obtain an equality of polynomials in x, whose coefficients are homogeneous linear forms in the αi , β j , and ξk,l .
Equating like-powers of x on both sides, we obtain a system of
m(TN + n) + SN + n
homogeneous linear equations with coefficients in C(t), in the
(N + 1) + (SN + 1) + mTN
variables αi , β j , and ξk,l .
Step 6: If N = M, the system of linear equations defined by (IN ) has a solution. Find the smallest nonnegative integer ` 6 M such that the system of linear equations defined by (IN ), with N = `, has a solution.
Step 7: Find a solution
αi = ai

for 0 6 i 6 `,

βj = bj

for 0 6 j 6 S` ,

ξk,l = ck,l

for 1 6 k 6 TN and 0 6 l 6 m − 1;

for this system of homogeneous linear equations, such that each ai , b j , ck,l ∈ C(t) and a` = 1. This condition
determines a0 , . . . , a` uniquely.
Step 8: Set L p,q := ∑`i=0 ai ∂ti , and go to Output.
Remark 2.6. Upon inspection, we see that the system of linear equations of Step 5 has coefficients in the smallest
algebraic extension L of C(t) such that p, q ∈ L(x). Therefore, L p,q ∈ L[∂t ].
Remark 2.7. As in [31, Prop. 4.1(4)], the operator L p,q produced by Algorithm 2 must be nonzero. Otherwise,
the element h ∈ C(t)(x) obtained from (2.4) by substituting β j = b j and ξk,l = ck,l would have to satisfy the (IN )
with N = 0 and α0 = 0, which is δx h = ph. This implies that h = aη for some a ∈ K0 , which would contradict
the assumption that η ∈
/ K.
5 The

computation of these bounds is elementary, but somewhat lengthy. We will indicate how to compute these bounds in Remark 2.9.

11

C ARLOS E. A RRECHE
E XAMPLE 2 (The Picard-Fuchs equation for the Legendre family of elliptic curves). Consider the first order
− 1
inhomogeneous equation
∂Y /∂x = x(x − 1)(x − t) 2 .
(2.5)
− 21
, and apply Algorithm 2 with inputs
We will take η := x(x − 1)(x − t)




1
∂η/∂x
1 1
1
1
∂η/∂t
1
p :=
=−
+
+
and
q :=
=
,
η
2 x x−1 x−t
η
2 x−t
to derive the Picard-Fuchs equation for the Legendre family of elliptic curves Et : y2 = x(x−1)(x−t) (cf. [24, pp.
77-78] and [12, Example 6.9]).
In Step 2, we compute d = x(x − 1)(x − t), ν = 1, m = 3, n = 1, and M = 5. Now let




1
∂q
1
1
3
2
, R2 =
, R3 = · · ·
R0 = 1, R1 = q =
+q =
2 x−t
∂t
4 (x − t)2
be the sequence defined in Step 3 (we will have no need to compute Ri for i > 2), and consider the inhomogeneous
equation with undetermined coefficients of Step 4, with N = 2






∂Y 1 1
1
1
α1
1
3α2
1
,
−
+
+
Y = α0 +
+
∂x 2 x x − 1 x − t
2 x−t
4
(x − t)2
A computation shows that we may take T2 = 1 and S2 = 1 (see Remark 2.9, or the proof of Proposition 3.5
ξ0,1
below). Now we write the rational function h2 = β0 + β1 x + x−t
with undetermined coefficients β0 , β1 , ξ0,1 as in
Step 4, to obtain (IN ) with N = 2 as in Step 5:




 3α  1 
ξ1,0
ξ1,0
α1
∂
1 1
1
1
1
2
β
+
β
x
+
−
+
+
β
+
β
x
+
=
α
+
.
(2.6)
0
1
0
1
0
x−t
2 x
x−1
x−t
x−t
2 x−t + 4
∂x
(x−t)2
After multiplying by x(x − 1)(x − t)2 to clear denominators,6 and then subtracting the left-hand side from the
right-hand side, we obtain that (2.6) holds if and only if the following polynomial in x is zero:

(α0 + 12 β1 )x4 + (−2t − 1)α0 − 21 α1 + 32 β0 − 12 tβ1 x3

+ (t 2 + 2t)α0 + ( 12 t + 12 )α1 − 43 α2 − ( 52 t + 1)β0 − 12 tβ1 + 52 ξ1,0 x2

+ −t 2 α0 − 21 tα1 + 34 α2 + (t 2 + 32 t)β0 + 12 t 2 β1 − (t + 2)ξ1,0 x + (− 21 t 2 β0 + 12 tξ1,0 )x0 .
Setting each coefficient equal to zero yields the system of linear equations defined in Step 5. One can check that
α0 = − 41 , α1 = 2t − 1, α2 = t(t − 1), β0 = 12 (t − 1), β1 = 12 , and ξ1,0 = 21 t(t − 1) is a solution,7 and that indeed
! !
∂2 η
∂η 1
∂
1
1 2 12 t(t − 1)
t(t − 1) 2 + (2t − 1) − η =
(t − 1) + x +
η .
∂t
∂t 4
∂x
2
2
x−t
After checking that the system of linear equations does not have a solution with α2 = 0, we conclude that the

2
PPV group corresponding to (2.5) is Ga K0 ; t(t − 1) ∂t∂ 2 + (2t − 1) ∂t∂ − 14 .
P ROPOSITION 2.8 (Algorithm 2 is correct). Suppose that η ∈
/ K. Then, the Output of Algorithm 2 is the operator
which defines Gal∆ (P /F) as a subgroup of Ga (K0 ). In other words, Gal∆ (P /F) ' Ga (K0 ; L p,q ).
Proof. Consider the sequence {Ri } ⊂ K defined in Step 3. Since ∂t η = qη, we have that ∂ti η = Ri η for each
!
i > 0, and therefore
N
N

∑ αi ∂ti η = ∑ αi Ri

i=0

η,

i=0

2
that here the algorithm calls for multiplying by d 2 = x(x − 1)(x − t) , but in this concrete example we see directly that this is
unnecessary. One can use square-free factorizations to do this in general (see Section 4).
7 We have taken α = t(t − 1) instead of α = 1, because this is how the Picard-Fuchs equation usually appears in the literature.
2
2
6 Note

12

C OMPUTING G ALOIS GROUPS OF PARAMETERIZED SECOND ORDER EQUATIONS
for each 0 6 N 6 M. Note also that the solution found in Step 7, for the system of polynomial equations defined
in Step 5, also defines a rational function h ∈ C(t)[x], obtained from the rational function (with undetermined
coefficients) h` of (2.4) by setting β j = b j and ξk,l = ck,l for each j, k, l:
!
m−1
S`
T`
l
c
x
∑
k,l
l=0
h := ∑ b j x j + ∑
∈ C(t)(x).
dk
j=0
k=1
By construction, L p,q ∈ K0 [∂t ] and h` ∈ K satisfy part (iii) of Lemma 2.1. Therefore, L p,q (bσ ) = 0 for every σ ∈
Gal∆ (P /F), where bσ is defined as in the discussion preceding Lemma 2.1. We need to rule out the possibility
that there exist L̃ ∈ K0 [∂t ] and h̃ ∈ K satisfying part (iii) of Lemma 2.1 with
`˜ := ord(L̃ ) < ord(L p,q ) = `.
˜

We proceed by contradiction: assume such L̃ and h̃ do exist. Let L̃ := ∑`i=0 ãi ∂ti , with ãi ∈ K0 . By Lemma 2.1,
`˜
h̃ satisfies the inhomogeneous equation
(2.7)
δx h̃ + ph̃ = ∑ ãi Ri .
i=0

If the bounds S`˜ and T`˜ of Step 4 exist, then h̃ must have the following form

S`˜
T`˜  m
∑l=0 c̃k,l xl
j
h̃ = ∑ b̃ j x + ∑
.
dk
j=0
k=1

(2.8)

This implies that setting ãi = αi , b̃ j = β j , and c̃k,l = ξk,l for each i, j, k, l is a solution for the system of homogeneous linear equations defined by (IN ) with N = `.˜ This contradicts the choice of `.
It remains to be shown that the bounds S`˜ and T`˜ of Step 4 exist. We will compute these bounds directly,
using a modification of the argument given in [30, Lemma 3.1], and we will see that they depend only on p, q,
and `.˜ Let e ∈ K0 be a pole of h̃ ∈ K0 (x). We claim that e must be a pole of either p or q. To see this, suppose
on the contrary that e is a pole of h̃, but not of p or q. If −T is the order of h̃ at e, with T > 1, then the order of
˜
δx h̃ + ph̃ is −T − 1 (because e is not a pole of p), while the order of ∑`i=0 ãi Ri is nonnegative (because e is not a
pole of q); this gives a contradiction. Now let e be a pole of p or q, and let
w2
+ (higher order terms)
p=
`˜
(x − e)τ2
w1
∑ ãi Ri = (x − e)τ1 + (higher order terms)
w3
i=0
h̃ =
+ (higher order terms).
(x − e)T
×

be the (x − e)-adic expansions of these elements, with w2 ∈ C(t) , and w1 , w3 ∈ K0× . If we substitute these
expressions in (2.7), we obtain

 

−Tw3
w2 w3
w1
+(higher order terms) +
+ (higher order terms) =
+(higher order terms).
(x − e)T +1
(x − e)T +τ2
(x − e)τ1
(2.9)
We see that either max{T + 1, T + τ2 } = τ1 , or else the lowest order terms in the left-hand side of (2.9) must
cancel. In order for this cancellation to take place, it is necessary that T + 1 = τ2 + 1 and Tw3 = w2 w3 , or,
equivalently, τ2 = 1 and T = w2 . In any case, we have that T 6 max{w2 , τ1 − 1} (if w2 is not an integer, we
deduce that T 6 τ1 − 1). Evidently, w2 depends only on p. Although τ1 depends on the ãi , as well as on q and `,˜
one can easily find a (generic) upper bound for τ1 which depends only on q and `,˜ and not on the specific values
˜ Now we may take T`˜ to be greater than any of the finitely
of the ãi ; namely, take τ1 := mini {orde (Ri ) | 0 6 i 6 `}.
many bounds so obtained for each pole of p and q.
The bound S`˜ is computed as follows: let S̃ be an upper bound for the degree of the polynomial part of each
Ri , for 0 6 i 6 `.˜ It follows from (2.7) that max{S`˜ − 1, degx (p0 ) + S`˜} 6 S̃, where p0 is defined as in Step 1.
This concludes the proof that the bounds T`˜ and S`˜ of Step 4 exist.
13

C ARLOS E. A RRECHE
Remark 2.9. In the preceding proof, we showed how to compute the bounds SN and TN of Step 4 by working
with the (x − e)-adic expansions of p and q, for each linear factor (x − e) of d over C(t)[x]. In practice, it may not
be feasible to compute a factorization of d into irreducible factors over C(t), let alone linear factors over C(t).
We will now indicate how to obtain the bounds SN and TN without factorizing d.
Let (a0 , . . . , aN ) be any (N + 1)-tuple of elements of K0 , and let h ∈ K be a solution for the equation obtained
from (2.3) by replacing αi with ai for each i:
N

δxY + pY = ∑ ai Ri .

(2.10)

i=0

If (x − e) is a factor of d in C(t)[x], the argument given in the proof of Proposition 2.8 shows that h can have a
pole at e of order at most τ1 − 1, where τ1 is defined as in the proof of Proposition 2.8, unless p has a pole at e
of order exactly 1. Given this a priori bound, it suffices to bound the order of a pole of h at the simple poles of p.
We now compute the divisor of simple poles of p:
d1 :=

gcdx (d, p2 )
.
gcdx (d, p2 , δx p2 )

This is the product of all linear factors of p2 which have multiplicity one in p2 . We will mimic the argument
given at the end of Proposition 2.8, using powers of d1 as denominators, instead of powers of (x − e). Write
N

w1

∑ ai Ri = d τ + (irrelevant terms),

i=0

1

p=

w2
+ (irrelevant terms),
d1

and

h=

w3
+ (irrelevant terms),
d1T

where w2 ∈ C(t)[x], w1 , w3 ∈ K0 [x], and degx (w j ) < degx (d1 ) for each 1 6 j 6 3. A term in ∑i a1 Ri , p, or h is
deemed “irrelevant” if its order at every linear factor of d1 is higher than −τ, −1, or −T , respectively. In order
to bound T , we proceed as before: substituting these expressions in (2.10), we obtain
!
!
−Tw3 δx d1
w2 w3
w1
(2.11)
+ (irrelevant terms) +
+ (irrelevant terms) = τ + (irrelevant terms).
T +1
T +1
d1
d1
d1
It follows that T + 1 > τ. If this inequality is strict, then the lowest order terms in the left-hand side of (2.11)
must cancel. For such cancellation to take place, it is necessary for (w2 w3 − Tw3 δx d1 ) to be divisible by d1 in
K0 [x]. This is almost what we want, except that this condition on T depends on w3 , which in turn depends on the
ai . However, since degx (w3 ) < degx (d1 ), we have the implication


d1 w3 (w2 − T δx d1 ) =⇒ gcdx d1 , (w2 − T δx d1 ) 6= 1.
If we now perform the Euclidean algorithm to compute this gcdx , we will obtain a series of residues. These
residues will be polynomials in x, whose coefficients are polynomials in C(t)[T ]. At least one of these residues
must be zero in order for the gcdx to be different from 1. Therefore, T must satisfy at least one of a finite
number of explicitly constructed polynomial equations with coefficients in C(t), and we may obtain a bound
on the size of any integer solution for a given polynomial equation, depending only on the coefficients of the
given polynomial. Having thus obtained the bound TN , we obtain the bound SN by comparing the degrees of the
polynomial parts of the right- and left-hand sides of (2.10), after replacing Y with h and clearing denominators.
2.3 The PPV group of an intersection of PPV fields
In this section we give an algorithm to compute the PPV group corresponding to the intersection of two PPV
fields of the form considered at the beginning of this section. We will apply this algorithm in §3.4 to recover the
PPV group corresponding to a second order equation from a pair of auxiliary differential algebraic groups.
Let η1 , η2 ∈ C(t)(x), and consider the first order inhomogeneous equation δxY = ηr , for r = 1, 2. If we let
pr := δηx ηr r , the discussion at the beginning of this section shows that the corresponding homogeneous equations
are given by the operators
14

C OMPUTING G ALOIS GROUPS OF PARAMETERIZED SECOND ORDER EQUATIONS

P1 := δ2x − p1 δx and P2 := δ2x − p2 δx ,
and that Gal∆ (Pr /K) is a differential algebraic subgroup of Ga (K0 ). Let E denote the PPV extension of PPV(P1 /K)
corresponding to the operator P2 , and let θ1 , θ2 ∈ E such that δx θr = ηr . Then, we have that
Khθr i∆ = PPV(Pr /K) ⊆ E = Khθ1 , θ2 i∆ , and we may define L := PPV(P1 /K) ∩ PPV(P2 /K) ⊆ E.
The group Gal∆ (Pr /L) is normal in Gal∆ (Pr /K), since the latter is abelian. It follows from the Galois correspondence (Theorem 1.10) that L is a PPV extension of K (for some operator). If we denote the corresponding PPV
group by Λ, the Galois correspondence implies that we have surjections πr : Gal∆ (Pr /K)  Λ, and that
Gal∆ (E/K) ' Gal∆ (P1 /K) ×Λ Gal∆ (P2 /K),
where the fibered product8 is taken with respect to the differential-algebraic homomorphisms πr .
The following algorithm computes Λ, as well as the maps π1 and π2 .
A LGORITHM 3 (Intersection).
Input: η1 , η2 ∈ C(t)(x).
Output: A differential algebraic group Λη1 ,η2 , and differential algebraic homomorphisms
πrη1 ,η2 : Gal∆ (Pr )  Λη1 ,η2 , for r = 1, 2, defined by
Λη1 ,η2 := Ga (K0 ; L 00 ) and

πηr 1 ,η2 : σ 7→ Lr0 (σ),

such that the PPV group of PPV(P1 /K) ∩ PPV(P2 /K) is isomorphic to Λη1 ,η2 and πηr 1 ,η2
is the corresponding surjection of PPV groups (cf. Theorem 1.10).
Step 1: Apply Algorithm 1 with inputs η1 and η2 , and let L1 := L η1 and L2 := L η2 be the Outputs. If
ord(L1 ) > ord(L2 ), proceed to Step 2.
Otherwise, reverse the roles of η1 and η2 , so that ord(L1 ) > ord(L2 ), and then proceed to Step 2.
(1)
(1)
Step 2: Write
η1
η2
(0)
(0)
η1 = η1 + (2)
and η2 = η2 + (2)
,
η1
η2
(i)

(i)

with9 η1 , η2 ∈ C(t)[x] for 0 6 i 6 2, such that
(1)

(2)

(1)

(2)

gcdx (η1 , η1 ) = 1 = gcdx (η2 , η2 ),
(1)

(2)

degx (η1 ) < degx (η1 ) and

(1)

(2)

degx (η2 ) < degx (η2 ).

(2) (2)

Step 3: Let

d :=

η1 η2
,

(2) (2)
(2) (2) 
gcdx η1 η2 , δx η1 η2

an let n ∈ N be the smallest nonnegative integer such that d n ηr is a polynomial in x, for each r = 1, 2. We remark
(2)
(2)
that d is the product of the irreducible factors of η1 and η2 , and n is the highest multiplicity of an irreducible
(2)
(2)
factor of η1 or η2 .
Step 4: Set
n
o
(0)
(0)
`r := ord(Lr ),
ν := `1 − `2 ,
s := max degx (η1 ), degx (η2 ) ,
and
M := degx (d).
For each ν 6 N 6 `1 , write

LN(1)

N

:= ∑ αi ∂ti ,
i=0

LN(2)

N−ν

:=

∑

j
β j ∂t ,

s+1

and

j=0

∑ γk x k + ∑
k=0

8 Recall
9 The

fN :=

(n+N−1)
l=1

m
∑M−1
m=0 ξl,m x
dl

that this is the subgroup of elements (σ1 , σ2 ) ∈ Gal∆ (P1 /K) × Gal∆ (P2 /K) such that π1 (σ1 ) = π2 (σ2 ).
(i)
parenthetical superscript in η j is an index, and does not denote an ith derivative.

15

!
,

C ARLOS E. A RRECHE
where the αi , β j , γk , and ξl,m are undetermined coefficients.
Step 5: Treating each γk and ξl,m as a δx -constant, set

LN(1) (η1 ) − LN(2) (η2 ) = δx fN ,

(JN )

and then multiply each side of this equation by d n+N . The result will be an equality of polynomials in x, whose
coefficients are linear forms in the αi , β j , γk , and ξl,m . Equating coefficients of like-powers of x, we obtain a
system of
M(n + N) + s
homogeneous linear equations with coefficients in C(t), in the
(N + 1) + (N − ν + 1) + (s + 2) + M(n + N − 1)
variables αi , β j , γk , ξl,m .
Step 6: If N = `1 , the system of linear equations defined by (JN ) has a solution with not all αi being zero
and not all β j being zero. Find the smallest nonnegative integer ω 6 `1 such that the system of linear equations
defined by (JN ), with N = ω, has a solution with not all αi being zero.
Step 7: If ω = `1 , set L10 := L1 , L20 := L2 , and L 00 := 1, and then go to Output.
Otherwise, find a solution
αi = ai

for 0 6 i 6 ω,

βj = bj

for 0 6 j 6 ω − ν,

γk = ck

for 0 6 k 6 s + 1
for 1 6 l 6 n + N − 1 and 0 6 m 6 M − 1;

ξl,m = el,m

such that each ai , b j , ck , el,m ∈ C(t) and aω = 1. This condition determines a0 , . . . , aω and b0 , . . . , bω−ν uniquely.
j

ω−ν
00
10
i
0
Step 8: Set L10 := ∑ω
i=0 ai ∂t and L2 := ∑ j=0 b j ∂t , let L ∈ C(t)[∂t ] be the unique monic operator such that
00
0
L1 = L L1 , and go to Output.
2

2

x+t
and η2 := x22t+t ; note that η1 = 12 tη2 + 1x .
E XAMPLE 3. We will now apply Algorithm 3 with inputs η1 := x x+t
3 +tx
2

The first step is to apply Algorithm 1 with inputs η1 and η2 , to obtain the operators L1 := ∂t∂ 2 − 2t1 ∂t∂ and L2 :=
∂
1
3
∂t − 2t (cf. Example 1). The polynomial d of Step 3 is given by d = x + tx. We then compute the integers
n = 1, `1 = 2, `2 = 1, ν = 1, M = 3, s = 0, and write down the pair of operators and the rational function with
undetermined coefficients of Step 4, with N = 1:
ξ1,0 + ξ1,1 x + ξ1,2 x2
L1(1) := α1 ∂t + α0 ,
L1(2) := β0 ,
and
f1 := γ0 + γ1 x +
.
x3 + tx
Then, we substitute these expressions in equation (JN ) as in Step 5, to obtain
 2 2 


ξ1,0 +ξ1,1 x+ξ1,2 x2
x+t
x2 +t 2 x+t
2t
∂
α1 ∂t∂ x x+t
+
α
·
−
β
·
=
γ
+
γ
x
+
.
(2.12)
0
0 x2 +t
0
1
3 +tx
∂x
x3 +tx
x3 +tx
After expanding this out and doing some simplification, we obtain
−α1 t 2 x2
(x3 +tx)2

+ α0 x

2 +(α t 2 +2α t)x+α t
0
1
0
x3 +tx

0
− x2tβ
2 +t = γ1 +

ξ1,1 +2ξ1,2 x
x3 +tx

−

(ξ1,0 +ξ1,1 x+ξ1,2 x2 )(3x2 +t)
.
(x3 +tx)2

After multiplying by d 2 = (x3 + tx)2 on both sides, and then subtracting the left-hand side from the right-hand
side, we obtain that (2.12) holds if and only if the following polynomial in x is zero:


γ1 x6 − α0 x5 + −t 2 α0 − 2tα1 − 2tβ0 + 2tγ1 − ξ1,2 x4 + −2tα0 − 2ξ1,1 x3

+ −t 3 α0 − t 2 α1 − 2t 2 β0 + t 2 γ1 − 3ξ1,0 + tξ1,2 x2 − t 2 α0 x − tξ1,0 x0 .
10 The

ring C(t)[∂t ] is a left and right Euclidean domain, and there are algorithms to compute this L 00 (see [29, §2.1]).

16

C OMPUTING G ALOIS GROUPS OF PARAMETERIZED SECOND ORDER EQUATIONS
Setting each coefficient equal to zero yields the system of homogeneous linear equations defined in Step 5.
We verify that α0 = 0, α1 = 1, β0 = 43 , γ1 = 0, ξ1,0 = 0, ξ1,1 = 0, ξ1,2 = − 12 t is a solution, and that indeed
3
∂
−tx2
∂
∂t η1 − 4 η2 = ∂x 2(x3 +tx) . The smallest value for N allowed by Step 4 is N = 1, since we assume N > ν = 1.
Therefore, ω = 1, and we have L10 = ∂t∂ , L20 = 34 , and L 00 = ∂t∂ − 2t1 .
Our proof that Algorithm 3 gives the right answer will rely on the following well-known result.
T HEOREM 2.10 (Kolchin-Ostrowski [19]). Let E be a δx -field extension of K such that E δx = K0 , and let
e1 , . . . , em , f1 , . . . , fn ∈ E such that δexiei ∈ K for each 1 6 i 6 m and δx f j ∈ K for each 1 6 j 6 n.
Then, there exists a nonzero polynomial G ∈ K[X1 , . . . , Xm ,Y1 , . . . ,Yn ] such that G(e1 , . . . , em , f1 , . . . , fn ) = 0 if
and only if at least one of the following holds:
mi
(i) There exist integers mi ∈ Z, not all zero, such that ∏m
i=1 ei ∈ K.
(ii) There exist elements c j ∈ K δx , not all zero, such that ∑nj=1 c j f j ∈ K.

P ROPOSITION 2.11 (Algorithm 3 is correct). Let η1 , η2 ∈ C(t)(x), and let Λη1 ,η2 and πηr 1 ,η2 be the Outputs of
Algorithm 3. Define the operators Pr ∈ K[δx ] for r = 1, 2, as well as the fields E and L as above, that is

E := PPV P2 /PPV(P1 /K) , and L := PPV(P1 /K) ∩ PPV(P2 /K) ⊆ E.
Then, πrη1 ,η2 : Gal∆ (Pr /K)  Λη1 ,η2 is a surjection, and L is the PPV extension corresponding to ker(πηr 1 ,η2 ) by
Theorem 1.10 applied to PPV(Pr /K), for each r = 1, 2.
Proof. Let θr ∈ E such that δx θr = ηr . Then, PPV(Pr /K) = Khθr i∆ , and since δx ∂tn θr = ∂tn ηr ∈ K, it is enough
to take derivatives with respect to ∂t only; that is, PPV(Pr /K) = Khθr i∂t . If we define `r as in Step 4, Lemma 2.1
implies that
PPV(Pr /K) = Khθr i∂t = K(θr , ∂t θr , . . . , ∂t`r −1 θr ),
and the transcendence degree of PPV(Pr /K) over K is `r .
The map σ 7→ (σθr − θr ) identifies Gal∆ (Pr /K) with Ga (K0 ; Lr ), for each r = 1, 2. By Theorem 1.10, there
is a unique ∂t -algebraic subgroup Hr 6 Gal∆ (Pr /K) such that L is the fixed field of Hr ; that is, L is precisely the
set of elements a ∈ PPV(Pr /K) such that σa = a for every σ ∈ Hr . On the other hand, if L , L 0 ∈ K0 [∂t ] are such
that Ga (K0 ; L 0 ) ⊆ Ga (K0 ; L ), then there exists L 00 ∈ K0 [∂t ] such that L = L 00 L 0 , and Ga (K0 ; L )/Ga (K0 ; L 0 ) '
Ga (K0 ; L 00 ) (see [29, p.45]). If Lr0 ∈ K0 [∂t ] is a right-hand factor of Lr , the fixed field of Ga (K0 ; Lr0 ) is precisely
KhLr0 (θr )i∆ . Therefore, there is a right-hand factor Lr0 of Lr such that L = KhLr0 (θr )i∆ .
We remark that every element a ∈ L\K defines an algebraic relation, as follows: since a ∈ PPV(P1 /K), there
is a rational function Φ ∈ K(Y1 , . . . ,Y`1 ) such that a = Φ(θ1 , . . . , ∂t`1 −1 θt ). Similarly, there is a rational function
Ψ ∈ K(Y1 , . . . ,Y`2 ) such that a = Ψ(θ2 , . . . , ∂t`2 −1 θ2 ). Setting these two expressions for a equal to each other and
clearing denominators, we obtain that L 6= K if and only if there is a nontrivial K-algebraic relation amongst the
elements θ1 , θ2 , ∂t θ1 , ∂t θ2 , . . . , ∂t`1 −1 θ1 , ∂t`2 −1 θ2 ∈ E.
By Definition 1.8, we have that E δx = K0 = K δx , and we may apply Theorem 2.10 with
( j
∂t θ1
for 0 6 j 6 `1 − 1, and
f j :=
j−`1
∂t θ2 for `1 6 j 6 `1 + `2 − 1.
Therefore, if L 6= K, we have that there exist elements ã0 , . . . ã`1 −1 , b̃0 , . . . , b̃`2 −1 ∈ K0 , not all zero, such that
`1 −1

∑

i=0

ãi ∂ti θ1 −

`2 −1

∑ b̃ j ∂tj θ2 ∈ K.

(2.13)

j=0

There is a nonnegative integer ω̃ ∈ N which is minimal with respect to the property that (2.13) holds for some
ãi , b̃ j ∈ K0 , with ãi = 0 for all i > ω̃, and ãω̃ 6= 0, and if we impose the further condition that ãω̃ = 1, this determines ã0 , . . . , ãω̃ , b̃0 , . . . , b̃`2 −1 uniquely. The minimality of ω̃ implies that L is ∆-generated over K by ∑i ãi ∂ti θ1
j
(and therefore also by ∑ j b̃ j ∂t θ2 ).
17

C ARLOS E. A RRECHE
We claim that ω = ω̃, where ω is defined as in Step 6. Note that this holds if and only if ai = ãi , and b j = b̃ j ,
where ai , b j ∈ C(t) are the elements found in Step 7. For definiteness, let us set
ω̃

∑ ãi ∂ti θ1 −

i=0

`2 −1

∑ b̃ j ∂tj θ2 = f˜ ∈ K,

(2.14)

j=0

`2 −1
i
0
where aω̃ = 1. If we define L̃10 := ∑ω̃
i=0 ãi ∂t , L̃2 := ∑ j=0 b̃ j ∂t , and
j

πr : Gal∆ (Pr /K) −→ Ga (K0 )
σ 7 −−−−→ L̃r0 (σθr − θr )


for r = 1, 2, we have that π1 Gal∆ (P1 /K) = π2 Gal∆ (P2 /K) , as differential algebraic
subgroups of Ga (K0 ).

If we let L̃ 00 ∈ K0 [∂t ] be the unique monic operator such that π1 Gal∆ (P1 /K) = Ga (K0 ; L̃ 00 ), it follows that
L̃ 00 L̃10 = L1 , and therefore `1 − ω̃ = ord(L̃ 00 ) = `2 − ord(L̃20 ), or in other words
ord(L̃20 ) = ω̃ − (`1 − `2 ).
Therefore, b̃ j = 0 for all j > ω̃ − ν, and b̃ω̃−ν 6= 0, where ν := `1 − `2 , as defined in Step 4. This implies that,
if we apply δx on both sides of the equality in (2.14), we obtain that ã0 , . . . , ãω̃ , b̃0 . . . , b̃ω̃−ν , together with the
coefficients of a base-d expansion of f˜, define a solution for the system of equations defined by (JN ), with ω̃ = N.
Therefore, ω = ω̃ and we obtain our result.
3. Algorithms for second order equations
Let K = K0 (x) be the ∆-field defined at the beginning of §2. In this section, we will describe an algorithm to
compute the PPV group corresponding to a parameterized differential equation of the form
δ2x Y + r1 δxY + r2Y = 0,

(3.1)

with ri ∈ C(x,t) for 0 6 i 6 2. After performing a change of variables (see [29, Ex. 1.35 (5)] or [22, p. 5]), we
may put this equation in the following form
δ2x Y − rY = 0,
(3.2)
1 2
1
where r := 4 r1 + 2 δx r1 − r2 .
The advantage of performing this change of variables is that, if D := δ2x − r, then Galδx(D /K) is an algebraic
subgroup of SL2 (K0 ) (and therefore Gal∆ (D /K) is a differential algebraic subgroup of SL2 (K0 )), so that there
are less possible candidates for (parameterized) Picard-Vessiot groups to consider. For the time being, we shall
limit our attention to equations of the form (3.2). In §3.4, we will show how to express the PPV group of the
original equation in terms of the PPV group of (3.2).
3.1 Kovacic’s algorithm
In [22], Kovacic describes an algorithm which: (i) finds all Liouvillian solutions of (3.2), whenever such solutions exist; and (ii) computes Galδx(D /K) from the data of these solutions (or their nonexistence). Kovacic’s
algorithm proceeds by first determining whether the Riccati equation [29, §4.1] for the operator D , which is
given by
RD (U) := δxU +U 2 − r = 0,
has a solution u ∈ K̄. If such a u exists, a computation shows that D factors
 over K(u)[δx ] into first order factors,
as D = (δx + u) ◦ (δx − u) [29, Ex. 1.36 (1b)]. Therefore, Galδx D /K(u) is reducible [7, Prop. 2.1].
D EFINITION 3.1. We denote the upper triangular and infinite dihedral subgroups of SL2 (K0 ) by UT(K0 ) and
D∞ (K0 ), respectively. They are defined as




 


a b
a 0
0
a
×
×
UT(K0 ) :=
a ∈ K0 , b ∈ K0
and
D∞ (K0 ) :=
,
a ∈ K0 .
0 a−1
0 a−1
−a−1 0
18

C OMPUTING G ALOIS GROUPS OF PARAMETERIZED SECOND ORDER EQUATIONS
For a proof of a significantly sharper version of the following result, see [29, §4.3.4] or Kovacic’s original
paper [22, §1.2]. We have only included those parts which are relevant for our purposes.
T HEOREM 3.2 (Cases of Kovacic’s algorithm). Kovacic’s algorithm falls into four main cases:
(i) If RD (u) = 0 for some u ∈ K, then Galδx(D /K) is conjugate to a subgroup of UT(K0 ).
(ii) If RD (u) = 0 for some u ∈ F\K, with F a quadratic extension of K, then Galδx(D /K) is conjugate to a
subgroup of D∞ (K0 ).
(iii) If Cases (1) and (2) do not hold, and RD (u) = 0 for some u ∈ K̄, then Galδx(D /K) is finite.
(iv) If Cases (1), (2) and (3) do not hold, then Galδx(D /K) = SL2 (K0 ).
T HEOREM 3.3 (Dreyfus [8, Prop. 9]). Suppose that Galδx(D /K) = SL2 (K0 ). Then, Gal∆ (D /K) is conjugate to
SL2 (C) if and only if the following inhomogeneous equation has a solution in K:
δ3x Y − 4rδxY − 2(δx r)Y = −2∂t r.

(3.3)

The following result is a corollary of Theorem 1.12.
T HEOREM 3.4. Galδx(D /K) is finite if and only if Gal∆ (D /K) is finite. In this case, Gal∆ (D /K) = Galδx(D /K).
One can show that (3.3) has a solution in K if and only if it has a solution in C(t)(x), and there is an algorithm
to decide whether or not this holds [29, Prop. 4.1 and Rmk. 4.5 (2)]. Therefore, Theorems 1.5, 3.3 and 3.4
completely solve the problem of determining Gal∆ (D /K) if Galδx(D /K) is finite or if Galδx(D /K) = SL2 (K0 ).
We shall now restrict our attention to cases (1) and (2) of Kovacic’s algorithm. From now on, we will
assume that Gal∆ (D /K) is infinite, unless explicitly stated otherwise. Our strategy is similar to the one followed
in [8], and the arguments given there were of great help to us in developing ours. Since we do not have an
interpretation for the solutions for our differential equations in terms of path integrals, we cannot directly apply
these results in our context. We will give differential-algebraic versions of the arguments presented in [8, §4.1,
§4.2], in order to prove that our Algorithms 4 and 5 are correct. In cases (1) and (2) of Kovacic’s algorithm, we
use the solutions for (3.2) to define semi-invariants of the linear differential algebraic group Gal∆ (D /K), in the
sense of [1, II,§7]. In the non-parameterized setting, this approach has led to generalizations and simplifications
of Kovacic’s algorithm which work over more general fields;11 see for example [15, 34]. See also [14] for an
efficient algorithm to compute first order right-hand factors of D .
3.2 Upper triangular case
Suppose that RD (U) = 0 has a solution u ∈ K. One can find such a u ∈ C(t)(x) explicitly (see [14, 34] and [29,
Prop. 4.9]).
A LGORITHM 4 (Upper triangular).
Input: u ∈ C(t)(x).
Output: A differential algebraic group UT(K0 ; Au , Bu ), defined by



a b
u
u
u u
a ∈ A ,b ∈ B ,
UT(K0 ; A , B ) :=
0 a−1
and such that Gal∆ (D /K) ' UT(K0 ; Au , Bu ).
Step 1: Apply Kovacic’s algorithm to compute Galδx(D /K). If it is found that



a b
Galδx(D /K) '
a ∈ µn , b ∈ K0
0 a−1
11 Kovacic’s

original algorithm assumes that the ground field K is C(x).

19

C ARLOS E. A RRECHE
for some n ∈ N, set Au := µn (recall µn denotes the group of nth roots of unity in C× ) and go to Step 4.
Otherwise, go to Step 2.
Step 2: Apply Algorithm 1 with input η := ∂t u, let L η be its Output, and set
Au := Gm (K0 ; L η ).
Step 3: If L η 6= 1, let

(
0
if Galδx(D /K) ' Gm (K0 );
B :=
Ga (K0 ) otherwise,
u

and go to Output.
If L η = 1, proceed to Step 4.
Step 4: Find v ∈ C(t)(x) such that δx v = ∂t u. Such a v exists whenever Au ⊆ Gm (C).
Step 5: Apply Algorithm 2 with inputs p := −2u and q := −2v, and let L p,q be its Output. Set
Bu := Ga (K0 ; L p,q ),
and go to Output.
P ROPOSITION 3.5 (Algorithm 4 is correct). Let u ∈ C(t)(x) such that RD (u) = 0. Then,
Gal∆ (D /K) ' UT(K0 ; Au , Bu ).
Proof (cf. [8, §2.2]). There is a basis {y1 , y2 } of Sol(D ) ⊂ PPV(D /K) such that δx y1 = uy1 and δx ( yy21 ) =
y−2
1 [22, §1.3]. This choice of basis identifies Gal∆ (D /K) with a subgroup of UT(K0 ), as follows: for σ ∈
σy2
y2
×
1
Gal∆ (D /K), define aσ := σy
y1 , and bσ := y1 − σy1 . A computation shows that aσ ∈ K0 , bσ ∈ K0 , and


aσ bσ
σ 7→
0 a−1
σ
is the embedding Gal∆ (D /K) ,→ SL2 (K0 ) determined by the choice of basis {y1 , y2 } of Sol(D ).
Moreover, the map χ : Gal∆ (D /K) → Gm (K0 ) : σ 7→ aσ is a differential-algebraic homomorphism (so its
kernel is Kolchin-closed). The PPV extension corresponding to ker(χ) by Theorem 1.10 is F := Khy1 i∆ , which
is a PPV extension of K for the operator δx − u. The induced character χ : Gal∆ ((δx − u)/K) → Gm (K0 ) is now
injective, and its image is a differential algebraic subgroup A of Gm (K0 ). If σ ∈ ker(χ), then bσ = σ( yy12 ) − yy21 , so
the map σ 7→ bσ identifies Gal∆ (D /F) with a differential algebraic subgroup B of Ga (K0 ), and we have that



a b
Gal∆ (D /K) ' UT(K0 ; A, B) =
a ∈ A, b ∈ B .
0 a−1
We will show that A = Au and B = Bu .
First we calculate A. If Gal∆ (D /K) is infinite, Theorem 1.12 implies that A = µn if and only if



a b
Galδx(D /K) '
a ∈ µn , b ∈ K0 .
0 a−1


If A is infinite, a computation shows that σ ∂yt y11 = ∂yt y11 + ∂at aσσ and δx ∂yt y11 = ∂t u. Applying Proposition 2.5 with
η = ∂t u and θ =

∂t y1
y1 ,

we deduce that A = Au .

In order to calculate the operator defining B, we have to consider the possibilities A ⊆ Gm (C) and A * Gm (C)
separately. The following result is proved in [32] under different hypotheses, but the proof works in our setting
as well.
L EMMA 3.6. Let B be a nontrivial proper differential algebraic subgroup of Ga (K0 ), and let a ∈ Gm (K0 ). Then,
B = aB := {ab | b ∈ B} if and only if ∂t a = 0.
20

C OMPUTING G ALOIS GROUPS OF PARAMETERIZED SECOND ORDER EQUATIONS
Proof of Lemma. Let L ∈ K0 [∂t ] be the unique monic operator defining B. We have that L 6= 0 and ord(L ) > 1.
If a ∈ Gm (K0 ), the monic operator defining aB is L 0 := a−1 ◦ L ◦ a. If ∂t a 6= 0, then L 6= L 0 , so B 6= aB.
Case 1: A * Gm (C). The group A acts on B via


  −1  

a 0
1 b
a
0
1 a2 b
=
.
0 a−1
0 1
0 a
0 1
By Lemma 3.6, in this case B is either 0 or Ga (K0 ), (because a ∈ C ⇔ a2 ∈ C). By Theorem 1.12, B = 0 if and
only if Galδx(D /K) ' Gm (K0 ), and we have that B = Bu in this case.
Case 2: A ⊆ Gm (C). In this case,
in [3] that there exists v ∈ K such that ∂t y1 = vy1 and δx v = ∂t u.
 it is shown
−2
−2
Note that this implies that δx y1 = −2uy1 and ∂t y−2
= −2vy−2
1
1 . Since
PPV(D /F) = Fhy2 i∆ = F
and δx

y2 
−2
y1 = y 1 ,

we apply Proposition 2.8 with η = y−2
1 and θ =

y2
y1 ∆
y2
y1

to deduce that B = Bu in this case also.

3.3 Infinite dihedral case
Suppose that RD (U) = 0 does not have a solution in K, but RD (u) = 0 for some u in a quadratic extension of K.
In this case, one can find φ ∈ C(t)(x) (see [30, 34], [29, Prop. 4.24]) such that the minimal polynomial of u over

K is:
U 2 + φU + 12 δx φ + 12 φ2 − r = 0.
A LGORITHM 5 (Infinite dihedral).
Input: φ ∈ C(t)(x).
Output: A differential algebraic group D∞ (K0 ; Aφ ), defined by

 


a 0
0
a
φ
D∞ (K0 ; Aφ ) :=
,
a
∈
A
,
0 a−1
−a−1 0
and such that Gal∆ (D /K) ' D∞ (K0 ; Aφ ).
Step 1: Set

η := 21 ∂t

Step 2: Apply Algorithm 2 with inputs p :=

p
4r − 2δx φ − φ2 .

δx η
η

and q :=

∂t η
η ,

and let L p,q be the Output.

Step 3: Set Aφ := Gm (K0 ; L p,q ), and go to Output.
P ROPOSITION 3.7 (Algorithm 5 is correct). Let u ∈ C(x,t) and φ ∈ C(t)(x), such that RD (u) = 0 and the minimal
polynomial of u over K is given by

U 2 + φU + 21 δx φ + 12 φ2 − r .
Then, Gal∆ (D /K) ' D∞ (K0 ; Aφ ).
Proof. (cf. [8, §2.3]). If v is the other root of the minimal polynomial of u, one can show that there is a basis
{y1 , y2 } of Sol(D ) such that δx y1 =uy1 and δx y2 = vy2 . Since K(u) ⊂ PPV(D /K), Theorem 1.10 gives a surjection Gal∆ (D /K)  Gal K(u)/K . After possibly replacing y2 by ay2 for some a ∈ K0× , one can show that
τ(y1 ) := y2 , τ(y2 ) := −y1 , τ(u) := v defines a lift τ ∈ Gal∆ (D /K) of the unique non-trivial K-automorphism of
K(u) [22, §4.3]. For σ ∈ Gal∆ (D /K) we define π(σ) := 0 if σ(u) = u, and π(σ) := 1 if σ(u) = v. If we define
!
σ ◦ τπ(σ) (y1 )
,
aσ :=
y1
21

C ARLOS E. A RRECHE
a computation shows that aσ ∈ K0× . One can show that the embedding of Gal∆ (D /K) in SL2 (K0 ), determined
by the choice of basis {y1 , y2 } of Sol(D ), identifies Gal∆ (D /K) with a subgroup of D∞ (K0 ), as follows:


π(σ)
aσ 0
0 1
σ 7→
.
0 a−1
−1 0
σ

The group Gal∆ (D /K) has two connected components, and the normal subgroup Gal∆ D /K(u) is the
connected component of the identity. One can show that σ(y1 y2 ) = ±y1 y2 for every σ ∈ Gal∆ (D /K). Therefore,

PPV D /K(u) = K(u)hy1 i∆ ,


and Gal∆ D /K(u) is the PPV group for the operator δx − u over K(u). For σ ∈ Gal∆ D /K(u) , we have that
1
aσ = σy
y1 . If we let A ⊂ Gm (K0 ) be the (isomorphic) image of Gal∆ D /K(u) under the differential-algebraic
homomorphism σ 7→ aσ , we have that Gal∆ (D /K) ' D∞ (K0 ; A) (defined as in the Output of Algorithm 5). We
will show that A = Aφ .


As in the upper triangular case, a calculation shows that σ ∂yt y11 = ∂yt y11 + ∂at aσσ and δx ∂yt y11 = ∂t u. Note that we
cannot apply Algorithm 22 to compute the PPV group corresponding to the inhomogeneous equation δxY = ∂t u,
ci
because the conditions ∂∂tt uu ∈ K and δ∂x ∂t ut u ∈ K are not satisfied. However, one can show that φ = ∑i x−e
, for some
i
ci ∂t ei
1
ei ∈ C(t) and ci ∈ 2 Z (see [22, Step 3, §4.1]), and a computation shows that ω := ∑i x−ei ∈ K satisfies δx ω = ∂t φ.
By making the substitution g := f + 12 ω, we see that, for any L ∈ K0 [∂t ],

there exists f ∈ K(u) such that L (∂t u) = δx f ⇐⇒ there exists g ∈ K(u) such that L ∂t u + 21 ∂t φ = δx g.
Now we may apply Proposition 2.8 with η = ∂t u + 21 ∂t φ and θ =
This concludes the proof that A = Aφ .

∂t y1
y1

+ 21 ω, because

δx η
η

and

∂t η
η

are in C(t)(x).

3.4 Recovering the original group
At the beginning of this section, we performed a change of variables on (3.1) to put it in the form (3.2). We will
now indicate how to compute the PPV group corresponding to (3.1). The corresponding problem in classical
Picard-Vessiot theory could be regarded as an exercise; indeed, Kovacic [22, §1.1], as well as numerous other
authors, consider the problem solved after computing the group corresponding to (3.2) only. In the parameterized
situation there is a new subtlety, stemming from the fact that Gm (K0 ) has many infinite differential algebraic
subgroups (cf. Theorem 1.4); while any proper algebraic subgroup of Gm (K0 ) must be finite cyclic. In general,
the differential-algebraic relations amongst the solutions for (3.2) and the solution for the change-of-variables
operator (i.e., the operator E defined below), form an obstruction to expressing the PPV group of (3.1) as an
almost direct product (that is, a finite-index quotient of the direct product) of the PPV groups corresponding
to these two operators. This is drastically different from the classical situation, where the PV-group of (3.1) is
always an almost-direct product of the PV-group of (3.2) and the change-of-variables group.
To fix notation, consider the differential operators

D := δ2x − r,

and
F := δ2x + r1 δx + r2 ,

where r := 14 r12 + 21 δx r1 − r2 . Let E := PPV E /PPV(D /K) , and let w ∈ E × such that E w = 0. In other words,
Khwi∆ = PPV(E /K) ⊂ E. Let y1 , y2 ∈ PPV(D /K) be a K0 -basis of Sol(D ). If we let zi := wyi ∈ E, one can
show that F zi = 0, and since z1 and z2 are K0 -linearly independent, they form a basis of Sol(F ) in E. Therefore,

E := δx + 21 r1 ,

Khz1 , z2 i∆ = PPV(F /K) ⊂ E = Khy1 , y2 , wi∆ .
Moreover, a computation shows that z1 δx z2 − z2 δx z1 = aw2 for some a ∈ K0× [29, Ex. 1.14 (5)], so either
E = Gal∆ (F /K), or else E is a quadratic extension of PPV(F /K), generated by w. For later use, we define
ν := [E : PPV(F /K)], the algebraic degree of this field extension. One can show that ν = 1 if and only if
Gal∆ (E /K) is finite of odd order. Now consider the ∆-subfield of E defined by L := PPV(D /K) ∩ PPV(E /K).
22

C OMPUTING G ALOIS GROUPS OF PARAMETERIZED SECOND ORDER EQUATIONS
Since Gal∆ (E /K) is a differential algebraic subgroup of Gm (K0 ), it is abelian, and the Kolchin-closed subgroup
corresponding to L, as a ∆-subfield of PPV(E /K), is normal. Therefore, L is a PPV extension of K (for some
operator). If we let Λ := Gal∆ (L/K), we have a lattice of PPV extensions over K:
E

PPV(D /K)
Gal∆ (D /K)

(3.4)

µν

PPV(E /K)

L

Gal∆ (E /K)

PPV(F /K)
Gal∆ (F /K)

Λ

K
The Galois correspondence (Theorem 1.10) implies that
Gal∆ (E/K) ' Gal∆ (D /K) ×Λ Gal∆ (E /K),
where the fibered product is taken with respect to the surjections ϕ : Gal∆ (D /K)  Λ and ψ : Gal∆ (E /K)  Λ,
induced from the lattice (3.4). It only remains to compute the middle term in the exact sequence
1 −→ µν −→ Gal∆ (D /K) ×Λ Gal∆ (E /K) −→ Gal∆ (F /K) −→ 1.

(3.5)

The computation of Gal∆ (E /K) is analogous to the computation of the diagonal part of Gal∆ (D /K) in the
upper triangular case (cf. the proof of Proposition 3.5): we first compute the PV group Galδx(E /K). If it is finite,
then Gal∆ (E /K) = Galδx(E /K). If Galδx(E /K) is infinite, one shows that σ 7→ σw
w =: eσ induces
 an isomorphism
of Gal∆ (E /K) onto a differential algebraic subgroup of Gm (K0 ), and we have that σ ∂wt w = ∂wt w + ∂et eσσ and
δx ∂wt w = − 21 ∂t r1 . An application of Algorithm 1, with input η := − 12 ∂t r1 , will give L η ∈ C(t)[∂t ] such that
Gal∆ (E /K) ' Gm (K0 ; L η ). We will now compute Λ, as well as the maps ϕ and ψ. There are two cases to
consider, depending on whether Gal∆ (E /K) is finite or infinite. We will apply the following result in the case
that Gal∆ (E /K) is finite.
L EMMA 3.8. The groups Galδx(D /K) and Gal∆ (D /K) have the same finite quotients.
Proof. If G is a linear differential algebraic group, we denote the connected component of the identity of G by
G0 . If ϕ : G  H is a ∂t -algebraic surjection onto a finite group H, then ϕ factors through G  G/G0 [1, p. 906].
Theorem 1.12 implies that Gal∆ (D /K)/Gal∆ (D /K)0 ' Galδx(D /K)/Galδx(D /K)0 .
Case 1: Suppose that Galδx(E /K) = µm for some m ∈ N, so that Λ is also finite, the order of Λ divides m,
and wm ∈ K [20]. We shall list all the finite cyclic quotients of the algebraic subgroups of SL2 (K0 ), in increasing
order of the largest finite cyclic group which they admit as a quotient, and examine the possibilities for Λ in each
case.
The groups Ga (K0 ), Gm (K0 ), UT(K0 ), and SL2 (K0 ) are connected, and the alternating group A5 is simple,
so none of these groups admit a nontrivial finite cyclic quotient. Therefore, if Galδx(D /K) is isomorphic to any
of these groups, then Λ must be trivial, and we have that
Gal∆ (F /K) ' µm/ν · Gal∆ (D /K) ⊂ GL2 (K0 ),
where µm/ν · Gal∆ (D /K) := {ζ · σ | ζ ∈ µm/ν , σ ∈ Gal∆ (D /K)} and µm/ν ⊂ C× denotes the (m/ν)th roots of unity.
Each of the following groups admits precisely one nontrivial finite cyclic quotient, of order 2: the dihedral
groups D2n , the infinite dihedral group D∞ (K0 ), and the symmetric group S4 . Therefore, both ϕ and ψ are determined by Λ in any of these cases. If m is odd, we conclude that Λ is trivial, and Gal∆ (F /K) ' µm · Gal∆ (D /K).
23

C ARLOS E. A RRECHE
If m is even, one checks whether the (unique) quadratic extension of K which is contained in PV(D /K) is
m
(algebraically) isomorphic to K(w 2 ) using finite Galois theory. This holds if and only if Λ = µ2 .
Similarly, the group A4 has a unique cyclic quotient, of order 3. If m is not divisible by 3, then Λ = {1} and
we are done. Otherwise, one needs to decide whether the unique cubic extension of K which is contained in
m
PV(D /K) is isomorphic to K(w 3 ); this will hold precisely when Λ = µ3 .
Lastly, for n ∈ N, the groups of the form



a b
a ∈ µn , b ∈ B ,
0 a−1
whether
one quotient of order l, for each l dividing n, given by
 B =nε{0} or B = Ga (K0 ), each have precisely
a b
n
l
7→ a for some ε ∈ N relatively prime to l . To compute Λ, we proceed as before: one has to find the
0 a−1
largest l ∈ N such that: (i) l divides gcd(m, n); and (ii) the unique finite cyclic extensions of K of degree l, which
are contained respectively in PV(E /K) and PV(D /K), are algebraically isomorphic. In this case, we have that
Λ = µl . If there is no such l ∈ N, then Λ = {1}.
Case 2: Now suppose that Gal∆ (E /K) is infinite, and therefore Kolchin-connected. In this case, Λ must
either be trivial or infinite. The only differential algebraic subgroups of SL2 (K0 ) which admit an infinite abelian
quotient are the groups UT(K0 ; A, B) (defined as in the Output of Algorithm 4). Therefore, Λ = {1} whenever
Gal∆ (D /K) is not of this form. The following algorithm computes Λ and the maps ϕ and ψ. For simplicity,
Algorithm 6 assumes that Gal∆ (D /K) is upper-triangular, and that Gal∆ (E /K) is infinite and has already been
computed.
A LGORITHM 6 (Recovering the original group in the upper triangular case).
Inputs: D = δ2x − r and E = δx + 21 r1 , with r, r1 ∈ C(x,t).
Output:A linear differential algebraic group ΛD ,E , and differential-algebraic homomorphisms
ϕD ,E : Gal∆ (D /K)  ΛD ,E

and

ψD ,E : Gal∆ (E /K)  ΛD ,E

such that ΛD ,E is the PPV group of PPV(D /K) ∩ PPV(E /K), and ϕD ,E and ψD ,E are the
corresponding diferential-algebraic surjections of PPV groups.
Step 1: Compute a solution u ∈ C(t)(x) to the Riccati equation RD (U) = 0. Apply Algorithm 4 with input
u, and let Gal∆ (D /K) ' UT(K0 ; Au , Bu ) be the Output.
Step 2: If Au ⊆ µ2 , compute a solution y1 ∈ C(x,t) to the differential equation δxY = uY . Apply Algorithm 3
1
η1 ,η2 , πη1 ,η2 , and πη1 ,η2 be the Outputs. Set
with inputs η1 := y−2
1 , η2 := − 2 ∂t r1 , and let Λ
1
2


η1 ,η2
η1 ,η2 ∂t σ
D ,E
η1 ,η2
D ,E
D ,E
Λ
:= Λ
,
ϕ
:= π1 ,
and
ψ
: σ 7→ π2
,
σ
and then go to Output.
If Au 6⊂ µ2 , proceed to Step 3.
Step 3: If Au = µn for some n > 3, set ΛD ,E := {1}, let ϕD ,E , ψD ,E be the trivial maps, and go to Output.
If Au is infinite, proceed to Step 4.
Step 4: If there exist integers12 m1 , m2 ∈ Z, not both zero, and f ∈ C(t)(x), such that
m1 u −

m2
δx f
r1 =
,
2
f

(3.6)

there is a unique pair of integers m1 and m2 satisfying (3.6) with m1 positive and as small as possible. Set
ΛD ,E := {σm1 | σ ∈ Au } = {σm2 | σ ∈ Gal∆ (E /K)},
12 See

ϕD ,E : σ 7→ σm1 ,

[7, Prop. 2.4] for an algorithm to compute these integers, if they exist.

24

and

ψD ,E : σ 7→ σm2 ,

C OMPUTING G ALOIS GROUPS OF PARAMETERIZED SECOND ORDER EQUATIONS
and go to Output.
Otherwise, proceed to Step 5.
Step 5: Apply Algorithm 3 with inputs η1 := ∂t u, η2 := − 12 ∂t r1 , and let Λη1 ,η2 , πη1 1 ,η2 , and πη2 1 ,η2 be the
Outputs. Set




η1 ,η2 ∂t σ
η1 ,η2 ∂t σ
D ,E
D ,E
η1 ,η2
D ,E
,
and
ψ
: σ 7→ π2
,
Λ
:= Λ
,
ϕ
: σ 7→ π1
σ
σ
and then go to Output.
Our proof that Algorithm 6 gives the right answer will rely on the following result.
L EMMA 3.9. Let E := δx − g for some g ∈ K, and let 0 6= ρ ∈ PPV(E /K) be a solution (i.e., δx ρ = gρ). Let `
be the order of the differential operator L ∈ K0 [∂t ] such that Gal∆ (E /K) ' Gm (K0 ; L ). Then, PPV(E /K) has


transcendence degree ` + 1 over K, and it is algebraically generated by ρ and ∂ρt ρ , ∂t ∂ρt ρ , . . . , ∂t`−1 ∂ρt ρ .
Proof. Since δx ρ = gρ, we have that in fact Khρi∆ = PPV(E /K) is generated over K by {∂tn ρ}, for n ∈ Z>0 .
That is, PPV(E /K) = Khρi∂t . A computation shows that


n
∂tn−1 ∂ρt ρ − ∂ρt ρ ∈ K ρ, ∂t ρ, . . . , ∂tn−1 ρ ,

and therefore Khρi∂t = K(ρ) ∂ρt ρ ∂t . One can show that δx ∂tn ∂ρt ρ = ∂tn+1 g ∈ K for everyn > 0, and our result
now follows from Theorem 2.10, since L is the operator of smallest order such that L ∂ρt ρ ∈ K (cf. Lemma 2.1
and the proof of Proposition 3.5).
P ROPOSITION 3.10 (Algorithm 6 is correct). Suppose that Gal∆ (D /K) is isomorphic to UT(K0 ; A, B), and that
Gal∆ (E /K) is infinite. Define the fields L and E as above, that is

E := PPV E /PPV(D /K) ,
and
L := PPV(D /K) ∩ PPV(E /K) ⊆ E.
Then, the Output of Algorithm 6 is correct. That is, Λ ' ΛD ,E , ϕ = ϕD ,E , and ψ = ψD ,E .
Proof. Let y1 , y2 ∈ PPV(D/K) be the solutions computed in Proposition 3.5, that is, y1 and y2 satisfy the relations δx y1 = uy1 and δx yy21 = y−2
1 . We will divide the proof in two cases, depending on whether A is finite or
infinite.
Case 1: If A = µn for some n ∈ N, the Galois correspondence implies that yn1 ∈ K, and we showed in the
proof of Proposition 3.5 that PPV(D /K) = K(y1 ) yy12 ∆ . Suppose that B = Ga (K0 ; L1 ), where 0 6= L1 ∈ C(t)[∂t ],
and let `1 := ord(L1 ). If we let θ1 := yy21 , the argument given at the beginning of the proof of Proposition 2.11
shows that in fact
PPV(D /K) = K(y1 , θ1 , ∂t θ1 , . . . , ∂t`1 −1 θ1 ).
On the other hand, if we let L2 ∈ C(t)[∂t ] such that Gal∆ (E /K) ' Gm (K0 ; L2 ), and θ2 :=
Lemma 3.9 to PPV(E /K) to conclude that

∂t w
w ,

we may apply

PPV(E /K) = K(w, θ2 , ∂t θ2 , . . . , ∂t`2 −1 θ2 ),
where `2 := ord(L2 ). Theorem 2.10 shows that
K(y1 ) ∩ K(θs , ∂t θs , . . . , ∂t`s −1 θs ) = K = K(w) ∩ K(θs , ∂t θs , . . . , ∂t`s −1 θs ),
for each s = 1, 2. Moreover, since y1 is algebraic and w is transcendental (over K), another application of Theoδ (y−2 )

rem 2.10 gives that K(y1 ) ∩ K(w) = K. Therefore, if we let p1 := xy−21 , p2 := − δ∂x ∂t rt 1r1 , and Ps := δ2x − ps δx for
1
s = 1, 2 (we remark that these are the homogeneous operators corresponding to the first order inhomogeneous
1
equations δxY = y−2
1 and δxY = − 2 ∂t r1 , respectively), we have that
PPV(Ps /K) = Khθs i∂t ⊂ E,

L = PPV(P1 /K) ∩ PPV(P2 /K) ⊂ E.

and
25

C ARLOS E. A RRECHE
If n 6 2, then y−2
1 ∈ K, in which case Proposition 2.11 establishes our contention. We claim that if n > 3,
then L = K. To see this, we proceed by contradiction: suppose that L 6= K. By Theorem 2.10, the elements
θ1 , θ2 , ∂t θ1 , ∂t θ2 , . . . , ∂t`1 −1 θ1 , ∂t`2 −1 θ2 ∈ E must satisfy a relation of the form
`1 −1

∑

ai ∂ti θ1 −

i=0

`2 −1

∑ b j ∂tj θ2 = f ∈ K,

(3.7)

j=0

with a0 , . . . , a`1 −1 , b0 , . . . , b`2 −1 ∈ K0 not all zero. Applying δx on both sides of the equality in (3.7) gives a contra/ K, where the sequence {Ri } ⊂ K is defined as in Step 3 of Algorithm 2.
diction, since ∑i ai ∂ti η1 = (∑i ai Ri )y−2
1 ∈
Case 2: If A is infinite, then y1 is transcendental over K. Let L3 ∈ C(t)[∂t ] such that A = Gm (K0 ; L3 ), and let

θ3 := ∂yt y11 . If we apply Lemma 3.9 to the field Khy1 i∆ = PPV (δx − u)/K with ρ = y1 , we obtain
Khy1 i∆ = K(y1 , θ3 , ∂t θ3 , . . . , ∂t`3 −1 θ3 ),
where `3 := ord(L3 ). We claim that L = Khy1 i∆ ∩ Khwi∆ ⊂ E (in other words, we may ignore the second
solution y2 ∈ PPV(D /K) in this case). To see this, let θ1 := yy12 and θ2 := ∂wt w as above. Since every element of (Khy1 , y2 i∆ ∩ Khwi∆ )\Khy1 i∆ defines a nontrivial algebraic relation over Khy1 i∆ amongst the elements
j
θ1 , θ2 , ∂t θ1 , ∂t θ2 , . . . , ∂t`1 −1 θ1 , ∂t`2 −1 θ2 (cf. the proof of Proposition 2.11), and since δx (∂ti θ1 ), δx (∂t θ2 ) ∈ Khy1 i∆ ,
Theorem 2.10 implies that, if Khwi∆ ∩ Khy1 , y2 i∆ 6⊂ Khy1 i∆ , then there is a nontrivial linear combination
`1 −1

∑

i=0

ai ∂ti θ1 −

`2 −1

∑ b j ∂ j θ2 = f ∈ Khy1 i∆

(3.8)

j=0

with ai , b j ∈ K0 . Define the sequence {Ri } ⊂ K by R0 := 1, Ri := ∂t Ri−1 − 2θ3 Ri−1 for i > 1 as in Step 3 of
Algorithm 2. Applying δx on both sides of the equality in (3.8), we obtain
!
`1 −1
1 `2 −1
j+1
+
b j ∂t r1 = δx f .
(3.9)
∑ ai Ri y−2
∑
1
2 j=0
i=0
d

`3 −1
m,l
There is a unique partial fraction decomposition: f = ∑k ck yk1 + ∑m,l (y −e
θ3 ), and
l with ck ∈ K(θ3 , . . . , ∂t
m)
1
dl,m , el algebraic over this field. Therefore, if we let e0 := 0, (3.9) implies that

δx dm,l (y1 − em ) − ldm,l (uy1 − δx em )
(y1 − em )l+1
m,l

δx f = ∑(δx ck + ku)yk1 + ∑
k

δx dm,l − ludm,l ldm,l (δx em − uem )
+
(y1 − em )l
(y1 − em )l+1
m,l

= ∑(δx ck + ku)yk1 + ∑
k

= δx c0 + (δx d0,2 − 2ud0,2 )y−2
1 .
j

j

This implies that δx (∑ j b j ∂t θ2 + c0 ) = 0, whence ∑ j b j ∂t θ2 = g − c0 for some g ∈ K0 ; or, in other words,
j
j
j
∑ j b j ∂t θ2 ∈ Khy1 i∆ . We have shown that if ∑ j b j ∂t θ2 ∈ Khy1 , y2 i∆ , then ∑ j b j ∂t θ2 ∈ Khy1 i∆ . This establishes
our claim that L = Khy1 i∆ ∩ Khwi∆ .
j
Since δyx 1y1 , δwx w ∈ K, and δx (∂ti θ2 ), δx (∂t θ3 ) ∈ K, Theorem 2.10 implies that
K(y1 ) ∩ K(θs , ∂t θs , . . . , ∂t`s −1 θs ) = K = K(w) ∩ K(θs , ∂t θs , . . . , ∂t`s −1 θs ),
for each s = 2, 3. Therefore, if L 6= K, precisely one of the following statements is true:
(i) L = K(y1 ) ∩ K(w).
(ii) L = K(θ2 , ∂t θ2 , . . . , ∂t`2 −1 θ2 ) ∩ K(θ3 , ∂t θ3 , . . . , ∂t`3 −1 θ3 ).
1 m2
By Theorem 2.10, (i) holds if and only if there exist integers m1 and m2 , not both zero, such that ym
∈ K,
1 w
and this holds if and only if the condition of Step 4 holds. In case (ii), Proposition 2.11 establishes our result.

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C OMPUTING G ALOIS GROUPS OF PARAMETERIZED SECOND ORDER EQUATIONS
4. Concluding remarks
4.1 Consequences
We mention two consequences of the algorithms.
T HEOREM 4.1. Every unimodular PPV group of a second order homogeneous linear equation over K is one of
the groups in Sit’s classification.
T HEOREM 4.2. Let D ∈ C(t)(x)[δx ] be an operator of order 1 or 2. Then, Gal∆ (D /K) is defined over C(t).
4.2 Improvements
In computer algebra, one usually computes the partial-fraction decomposition of a rational function by computing the square-free factorization13 of its denominator [10]. In the algorithms of §2, we stopped short of
computing a complete square-free factorization of the denominators involved, because the polynomial d computed in each algorithm was sufficient to establish our results; namely, that the algorithms work, their outputs are
defined over C(t), and that no factorizations of polynomials are necessary (beyond those which can be carried
out using the Euclidean algorithm). The use of a single denominator d (and its powers) also allowed us to reduce
the number of indices involved in our presentation. We believe that this has improved the clarity of our presentation. In practice, one should reduce the size of the system of linear equations considered in each algorithm by
working with square-free factorizations of the denominators.
Each algorithm in §2 constructs a first order inhomogeneous differential equation with undetermined coefficients (these are the equations (HN ), (IN ), and (JN )), and one then finds a minimal set of values for the
coefficients such that the differential equation has a rational solution. One could also solve these differential
equations abstractly, using any of the usual methods in computer algebra (e.g., Hermite reduction, RothsteinTrager, Risch, . . . ) [10], and then find a minimal set of values for the undetermined coefficients such that the
solution is rational. Although any of these approaches would have to be abstractly equivalent to ours, they may
well be more efficient in some situations. Therefore, the eventual implementation of our algorithms should first
decide, perhaps heuristically, how to set up the systems of linear equations in order to minimize the amount of
computation.
4.3 Future directions
The differential operators produced by the algorithms presented in §2 arise as solutions to a creative telescoping
problem, namely: given η ∈ F, where F is defined as in the beginning of §2, find L ∈ C(t)[∂t ] and f ∈ F
such that L (η) = δx f (see [4, §1] for a more precise and general definition); such an operator L is called a
telescoper for η. In [4] and [5], the authors propose algorithms which, in particular, solve the problem of finding
a telescoper L ∈ C(t)[∂t ] for η ∈ C(x,t) arbitrary. Lemma 2.1 replaces a telescoping problem over F with a
“twisted” telescoping problem over the smaller field K (see part (iii) of Lemma 2.1), for the restrictive class of
algebraic functions η ∈ C(x,t) such that ηn ∈ C(t)(x) for some n ∈ N, and also in the case that η is transcendental
over K, provided that δηx η and ∂ηt η belong to C(t)(x). It would be interesting to see whether one can use creative
telescoping to compute PPV groups for higher-order systems, or over more general fields than we consider here.
It would also be interesting to see whether our approach, which handles a very restrictive class of telescoping
problems, can be generalized in order to “descend” some telescoping problems to simpler base fields, with
Lemma 2.1 as a prototype.
We mentioned in the introduction a result of [11], which says that the PPV groups we wish to compute are
actually defined over C(t), while our algorithms find differential polynomial equations which are defined over
F is a field of characteristic zero, and p ∈ F[x] is a polynomial, we say that p = ∏ni=1 (pi )i is a square-free factorization of p if
pi ∈ F[x] for each i, and gcdx (pi , p j ) = 1 for i 6= j. The square-free factorization is unique up to multiplication by constant polynomials,
and it can be computed using only the Euclidean algorithm and algebraic differentiation.
13 If

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C ARLOS E. A RRECHE
C(t). This is because we work with the solutions to 3.1 found by Kovacic’s algorithm, which are not (and cannot
be) defined over C(t) in general. It would be desirable to have algorithms to compute the defining equations for
the PPV group over the smaller field C(t), but we do not yet know whether it is possible to extend our methods
to produce such an algorithm.
It would be desirable to have algorithms in the case of several parametric derivations. We do not yet know
whether it is possible to extend our methods to this more general setting in a straightforward way. In [12],
the authors reduce the number of compatibility conditions that one has to check when working with several
parametric derivations. It would be interesting to see if the results of [12] can be used to extend our methods to
the setting of several parametric derivations, by working with one parametric derivation at a time.
ACKNOWLEDGEMENTS
Professor Alexey Ovchinnikov suggested this problem to me last summer, and has been an invaluable source of
advice, knowledge and insight ever since. I am very thankful to him for his patience, kindness and generosity,
without which this work would not have been possible. I have also benefited enormously from many conversations about this work with the following people: Phyllis Cassidy, Richard Churchill, Thomas Dreyfus, Guillaume
Duval, Henri Gillet, Raymond Hoobler, Bill Keigher, Michael Singer, and William Sit. My deep thanks go to
all of them for taking the time to help me improve this work. I presented the results in this paper at the Kolchin
Seminar in New York, the Joint Mathematics Meetings in Boston, and the Workshop on Differential Schemes
and Differential Cohomology in Banff and Calgary, Canada. I thank the audiences of those talks for their very
valuable comments, suggestions, and criticisms.
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Carlos E. Arreche carreche@gc.cuny.edu
Mathematics Department, The Graduate Center of the City University of New York, New York, NY 10016

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