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Algebraic aspects of linear differential and difference equationsHendriks, Peter Anne

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Chapter 2

Galois Action on Solutions of a

Di�erential Equation

2.1 Introduction

Consider a second order di�erential equation of the form y00 + ay0 + by = 0 with

a; b 2 Q(x). If one works over the algebraic closure Q of Q then di�erential

Galois theory is available. This theory tells us that there exists a Picard{Vessiot

�eld F � Q(x) satisfying

1. F � Q(x) is an extension of di�erential �elds.

2. The �eld of constants of F is Q.

3. The equation has two Q-linear independent solutions in F .

4. F is minimal with respect to the conditions (2) and (3).

The theory continues as follows:

1. F is unique up to d-isomorphism.

2. Let G be the group of d-isomorphism of F over Q(x). Then the set of

G-invariant elements of F is Q(x).

3. Let V � F be the set of solutions of the di�erential equation. Then V is

a two-dimensional vector space over Q. The group G acts faithfully on V

and the image of G in Aut(V ) is a linear algebraic subgroup.

A solution of the Riccati equation is an u 2 F of the form v0

vwith v 2 V and

v 6= 0. Kovacic's algorithm tries to compute the solutions of the Riccati equation.

The �rst part of this algorithm tries to �nd solutions u = v0

v2 Q(x). This means

that Qv is a G-invariant line in V .

7

The origin of this paper is the question to �nd out for which �eld C � Q the

solution u lies in C(x).

Suppose that the group G has the following form: V has a basis v1; v2 such

that G consists of all g 2 Aut(V ) with g(v1) = cv1 and g(v2) = c�1v2 with

c 2 Q�. Then V has only two G-invariant lines and ui =

v0ivi; i = 1; 2; are the

only solutions of the Riccati which lie in Q(x).

The intuitive reasoning is now the following: A Galois automorphism of Q

can at most permute fu1; u2g. Hence u1; u2 lie in C(x) where [C : Q] � 2. An

explicit example of this situation is the equation y00 = 2x�4y. The �eld C is here

Q(p2).

In the next two sections we will develop the theory justifying this intuitive

idea. We note that there is no suitable di�erential Galois theory which works

over a constant �eld of characteristic 0 which is not algebraically closed. The

next sections provides some theory in a special case. In later sections we will

summarize and correct the results of [HP93] and we will give some more appli-

cations. Further a theory of symmetries and forms of a di�erential operator is

developed in order to construct di�erential operators with special features with

respect to the �eld of constants which is needed for the Riccati equation.

We would like to thank M. F. Singer and F. Ulmer for pointing out a mistake

in [HP93]. Unfortunately, it was too late to correct the paper before publication.

2.2 The universal �eld of k((x))

Let k be a �eld of characteristic 0 with algebraic closure �k. One considers dif-

ferential equations over the di�erential �eld K = k((x)) with the di�erentiation0 = x d

dxgiven by (

Panx

n)0 =Pnanx

n. We want to make a universal �eld exten-

sion � K for which every di�erential equation over K has a full set of solutions.

More precisely one de�nes as follows:

1. � K is an extension of d-�elds.

2. The �eld of constants of is �k.

3. For every (homogeneous linear) di�erential equation

y(d) + a1y(d�1) + :::+ ad�1y

(1) + a0y = 0

with coe�cients in a �nite extension of K the space of solutions in is a

vector space over �k of dimension d.

8

4. is minimal with respect to the conditions (2) and (3).

As is well known condition (3) is equivalent to (3�) Every matrix di�erential

equation v0 = Av, where the matrix A has coe�cients in a �nite extension K has

a fundamental matrix with coe�cients in .

Theorem 2.2.1 with the above properties exists and is unique up to isomor-

phism.

Proof. The �eld must certainly contain the following things:

1. �k.

2. For every m � 1 an element x1

m .

3. A solution of y0 = 1, since has two linearly independent solutions of

y00 = 0.

4. A non zero solution of y0 = qy for every q 2 [m�1x�1=m�k[x�1=m].

One would like to see the solutions of the equations above as certain functions.

This is however a priori not possible, since we would have to explain then what

the product of a formal Laurent series with a solution of say y0 = 1 means. There-

fore we will build up in a very algebraic way.

Let Q denote the additive group (or �k-vector space) [m�1�k[x�1=m]. Further,

�K denotes the algebraic closure of K. We note that �K is generated over K

by �k and the elements x1

m ; m � 1. We remark that, in general, �k((x)) is not

algebraic over K, but every element of �K can be represented by a Laurent series

in �k((x1

m )) for some m � 1. We start with a polynomial ring in many variables

R0 := �K[L;E(q)]q2Q and we de�ne a di�erentiation on R0 by:

1. For f =P

�2Q c�x� 2 �K one has f 0 =

P�2Q �c�x

� 2 �K. This is the unique

extension of the di�erentiation 0 = x d

dxto �K.

2. L0 = 1.

3. E(q)0 = qE(q).

It is easily veri�ed that the above de�nes a unique di�erentiation on R0. Let

I � R0 denote the ideal generated by E(q1 + q2) � E(q1)E(q2) (all q1; q2 2 Q)and E(�)�x� (all � 2 Q). The ideal I is invariant under di�erentiation. Indeed,

(E(q1 + q2)� E(q1)E(q2))0 = (q1 + q2)(E(q1 + q2)� E(q1)E(q2)), and

9

(E(�)� x�)0 = �(E(�)� x�)

Let R := R0=I and let l and e(q) denote the images of L and E(q) in R. Since

the ideal I is invariant under di�erentiation one �nds a di�erentiation on R and

one has the following properties:

1. l0 = 1.

2. e(q)0 = qe(q) for all q 2 Q.

3. e(q1 + q2) = e(q1)e(q2) for all q1; q2 2 Q.

4. e(�) = x� for all � 2 Q.

2

Lemma 2.2.2 The ring R has no zero divisors and there is no prime ideal (6=0; R) in R which is invariant under di�erentiation.

Proof. Choose q1; :::; qs 2 Q such that 1; q1; :::; qs are linearly independent over

Q. Consider the subring

A := �K[l; e(q1); e(�q1); :::; e(qs); e(�qs)] of R

Every �nite subset of R lies in such a subring of A. One can see that there are only

the trivial relations e(q1)e(�q1) = 1; :::; e(qs)e(�qs) = 1 among the generators of

A over �K. That means that A is isomorphic to the ring �K[X; T1; T�11 ; :::; Ts; T

�1s ].

This ring is a localization of the polynomial ring �K[X; T1; :::; Ts]. In particular,

A and R have no zero divisors. It is easily seen that A can have no non-trivial

ideal that is invariant under di�erentiation. This is implies the same statement

for R. 2

Lemma 2.2.3 Let denote the �eld of fractions of R, then is a d-�eld con-

taining �K and the �eld of constants of is �k.

Proof. Let u 2 satisfy u0 = 0 and u 6= 0. For suitable q1; :::; qs 2 Q such

that 1; q1; :::; qs are linearly independent over Q one can write u = f

gwhere

f; g 2 A := �K[l; e(q1); :::; e(qs)]. The generators of A over �K have no relations

and so A is an ordinary \free" polynomial ring over �K in s + 1 generators. One

may assume that g.c.d.(f; g) = 1. Now u0 = 0 implies that f 0g = fg0 and in

particular f divides f 0 and g divides g0.Write f as a �nite sum

Pq2Nq1+:::+Nqs aqe(q) with each aq 2 �K[l] and where

N denotes the set of non-negative integers. Then f 0 =P(a0q + qaq)e(q) has a

total degree which is less or equal to the total degree of f . Hence f 0 = Bf for

some B 2 �K and also g0 = Bg. For each q one �nds the relation a0q + qaq = Baq.

We will need the following result.

10

Lemma 2.2.4 If the equation a0 + ha = 0 with h 2 �K has a non-zero solution

in �K[l], then h 2 Q+ [n�1x1=n�k[[x1=n]].

Proof. Write a = a0 + a1l + ::: + adld with all ai 2 �K and with ad 6= 0. Using

l0 = 1 and looking at the coe�cient of ld, one �nds a0d + had = 0. Write ad =

cx�(1 +P

�>0 c�x�) with c 2 �k c 6= 0 and � 2 Q. Then it is clear that h has the

form speci�ed in the lemma. 2

We continue the proof of Lemma 2.2.3. If aq 6= 0 then B must have the

form B = q + h where h 2 Q + [n�1x1=n�k[[x1=n]]. This can happen for only

one q since 1; q1; :::; qs are supposed to be linearly independent over Q. Hence

f = aqe(q) and similarly g = bre(r) for some r 2 Nq1 + ::: + Nqs. However

q � r 2 Q+ [n�1x1=n�k[[x1=n]] implies that q = r. Since g.c.d.(f; g) = 1 it follows

that q = r = 0 and f; g 2 �K[l] and B 2 Q + [n�1x1=n�k[[x1=n]]. We may assume

that g is monic as polynomial in l. Then g divides g0 implies that g = 1 since the

degree of g0 is less then the degree of g. Hence u = f 2 �K[l] and f 0 = 0. Write

f = f0 + f1l+ :::+ fdld with f0; :::; fd 2 �K and fd 6= 0. Looking at the coe�cient

of ld in f 0 = 0 one �nds that fd = c 2 �k and c 6= 0. If d � 1 then the coe�cient

of ld�1 gives the equation f 0d�1 = �dc. It is easily seen that no element in �K

satis�es this relation. Hence d = 0 and u 2 �k. This �nishes the proof of Lemma

2.2.3. 2

Lemma 2.2.5 Every di�erential equation over a �nite extension of K has a basis

of solutions in .

Proof. We will use here the language of di�erential modules over K. Such a

module (M; �) is a �nite dimensional vector space M over K with an operator �

on it, satisfying: � is additive and �(km) = k0m+ k�(m) for k 2 K and m 2M .

It is well known how to translate a homogeneous linear di�erential equation (or a

di�erential equation in matrix form) into a di�erential module. In the following

we will replaceK by �nite extensions ofK without changing the notation. We use

now the formal classi�cation of di�erential equations (cf. [Lev75]). After a �nite

�eld extension of K, the di�erential moduleM is a direct sum of submodules N .

For each N there is a q 2 Q such that the operator � � q is nilpotent. Then one

easily sees that the �eld �K(l; e(q)) � contains d linearly independent solutions

of N , where d denotes the dimension of N . Hence has the desired property. 2

Lemma 2.2.6 The �eld is the universal extension of K.

Proof. The minimality of has already been discussed. The other properties are

contained in Lemmas 2.2.3 and 2.2.5.

Thus we have shown the existence of a universal �eld. Let 0 be another

universal �eld. Then there is a di�erential homomorphism � : R! 0. Accordingto lemma 2.2.2, the kernel of � is 0 and � extends to a di�erential homomorphism

� of to 0. The minimality of 0 implies that � is an isomorphism. 2

11

Corollary 2.2.7 Let G+ be the group of all automorphisms � of the �eld such

that � is the identity on K and � commutes with the di�erentiation on . Let G

denote the normal subgroup of G+ consisting of the � which are the identity on�k. Then the following sequence is split exact.

0! G! G+ ! Gal(�k=k)! 0

Proof. Let � 2 Gal(�k=k). Extend � �rst to an automorphism � 0 of �K by

� 0(P

�2Q a�x�) =

P�(�a�)x

�. One de�nes an automorphism �+ 2 G+ extending

� 0 by �+l = l and �+e(q) = e(� 0q). It is easily seen that �+ 2 G+ and that

� 7! �+ is a homomorphism of groups. This is the required splitting. 2

2.3 The k-structure on the space of solutions

A k-structure on a vector space V over �k is a k-linear subspace W of V such

that V = �k k W . Such a k-structure induces a Gal(�k=k)-action on V by

�(P

i fiwi) =P

i(�fi)wi for fi 2 �k and wi 2 W .

Consider a di�erential equationM of order d over k(x). Fix an element a 2 k.This gives an embedding k(x) � k((x � a)) � with the universal �eld of

k((x � a)). The space of solutions V = V (M) � of M is a vector space

over �k with dimension d. Let G and G+ be as in Corollary 2.2.7. The space

V is invariant under G+ and the splitting of Corollary 2.2.7 gives an action of

Gal(�k=k) on V . Let Va � V denote the set of Gal(�k=k)-invariant elements.

Lemma 2.3.1 V = �k k Va and the action of Gal(�k=k) on V is induced by this

k-structure on V .

Proof. We note that any element ! 2 needs only �nitely many elements of �k

for its de�nition. Hence the subgroup of Gal(�k=k) �xing ! is open. Some open

normal subgroup N of Gal(�k=k) �xes a basis fv1; :::; vdg of V . Let W denote the

set of N -invariant elements of V and let k0 be the �eld of N -invariant elements of�k. On W = k0v1 + :::+ k0vd acts the Galois group Gal(k0=k) = Gal(�k=k)=N . For

� 2 Gal(k0=k) one denotes by A(�) 2 Gl(d; k0) the matrix which expresses the

vectors �(v1); : : : ; �(vd) in the basis fv1; : : : ; vdg. For �; � 2 Gal(k0=k) one �ndsthe formula A(��) = A(�) �(A(�)). In other words, � 7! A(�) is a 1-cocycle.

Its image in H1(Gal(k0=k); Gl(d; k0)) is trivial since proposition 3 on page 159

of [Ser68] states that the set H1(Gal(k0=k); Gl(d; k0)) is trivial. This means that

W also has a basis w1; : : : ; wd over k0 such that �(wi) = wi for every i. One

�nds that W = k0 k W0 with W 0 := kw1 + : : : + kwd. It follows at once that

W 0 is equal to Va, the set of Gal(�k=k)-invariant elements of V . This proves the

lemma. 2 A Picard{Vessiot �eld of the equation M over k(x) is the �eld F �

generated over �k(x) by a basis of V . The di�erential Galois group G(M) of

12

M over �k(x) is the group of the automorphisms of F=�k(x) commuting with the

di�erentiation on F . Fix a basis fv1; :::; vdg of Va. Then G(M) can be identi�ed

with an algebraic subgroup of Gl(d; �k). This algebraic subgroup is unique up

to conjugation with an element in Gl(d; �k). The coordinate ring of Gl(d; �k) is�k[Xi;j;

1D], where D = det(Xi;j). Let I � �k[Xi;j;

1D] denote the (radical) ideal

which de�nes G(M).

Corollary 2.3.2 G(M) is de�ned over the �eld k.

Proof. For any � 2 Gal(�k=k) and g 2 G(M) one has �g��1 2 G(M). This

implies that the ideal I is invariant under � . A standard argument implies that I

is generated by J := I\k[Xi;j;1D]. Let G(M)a denote the algebraic group de�ned

by J . Then G(M) = G(M)a k�k. 2

2.3.1 Remarks

1. For two elements a; b 2 k, the k-structures Va and Vb of V seem to be un-

related. It is not clear how the two groups G(M)a and G(M)b are related.

2. The map ! : M 7! V (M)a from di�erential equations over k(x) to vector

spaces over k commutes with \all constructions of linear algebra". In more

sophisticated terms (see [DM82]) it is a faithful, exact, k-linear, -functorof the rigid abelian category of left k(x)[@]-modules of �nite dimension over

k(x) to the category of �nite dimensional vector spaces over k.

If we restrict ! to the subcategory generated by a �xed M , then it can

be seen that the a�ne group scheme Aut(!) over k coincides with the

algebraic group G(M)a over k de�ned above.

3. Let Z � V = V (M) be an algebraic subset that is invariant under the

di�erential Galois group G(M) of M . For any � 2 Gal(�k=k) the set �Z

is again an algebraic subset of V that is invariant under G(M). A case of

special interest is Z is a (�nite union of) G(M)-invariant subspace.

2.4 Rational solutions of Riccati

One considers a linear homogeneous di�erential equation of order n over k(x) or

k((x)). In the �rst case we �x an a 2 k and an embedding k(x) � k((x�a)) � .

This induces the k-structure on the solution space V of the di�erential equation.

In the second case the k-structure on V is given in Section 2.3. The di�erential

13

Galois group of the equation is denoted by G. A rational solutions u of the Ric-

cati equation determines a line �ky � V , where y 2 V satis�es u = y0

y, which is

invariant under G. Conversely, if �ky � V is a line, invariant under G, then u = y0

y

is a rational solution of the Riccati equation. In this section we give estimates for

the degree of the �eld extension of k over which a solution of the Riccati equation

is de�ned.

Lemma 2.4.1 Suppose that there is a rational solution of the Riccati equation,

i.e. a solution in �k(x) or �k k k((x)).

1. If there are only �nitely many solutions, then each one of them is de�ned

over a �eld k0 � k with degree � n.

2. If there are in�nitely many solutions of the Riccati equation, then there is

a solution which is de�ned over a �eld k0 with degree � n

2.

Proof. For every algebraic character � : G! �k� one de�nes the vector spaceV� := fv 2 V j g(v) = �(g)v for all g 2 Gg. For the � with V� 6= 0 one hasPV� � V . Any � 2 Gal(�k=k) acts on the algebraic characters by � 7! ��

where ��(g) = �(�g��1). For every non zero V� one considers the stabilizer

H � Gal(�k=k) of V�. The index of this stabilizer is � n

dim V�. Hence V� is de�ned

over a �eld k0 � k of degree � n

dim V�. Inside V� one can take a line which is also

de�ned over k0. The solution y corresponding to this line gives a solution u = y0

y,

which lies in k0(x) or k0((x)), of the Riccati equation. If there are only �nitely

many solutions of the Riccati equation, then all the non-zero V� have dimension

1. Hence any choice of of an G-invariant line de�nes a �eld extension of degree

� n over k. In the second case, some V� has dimension � 2. A suitable line in

V� is therefore de�ned over a �eld k0 with [k0 : k] � n

2. Furthermore, V� contains

lines which are de�ned over arbitrary big extensions of k. 2

2.5 Algebraic solutions of the Riccati equation

and examples

One considers as before a di�erential equation over the �eld K which is either

k(x) or k((x)). The vector space V over �k of solutions is given a k-structure as

in section 2.3. The di�erential Galois group is called G.

A solution u = y0

yof the Riccati equation, algebraic of degree m, corresponds

to a line L = �ky � V such that its stabilizer H � G is a subgroup of index m.

The element u satis�es a monic irreducible equation P of degree m over �kK. The

collection of all zeroes of P corresponds to the orbit GL of the line L. We want

to �nd the smallest �eld extension k0 of k such that the coe�cients of P are in

14

k0K. This is equivalent to �nding the smallest �eld extension k0 � k such that

the orbit GL is de�ned over k.

Let us de�ne an m-line to be a G-invariant subset of V consisting of m lines

through 0, such that G acts transitively on the m lines. The collection of all

m-lines is denoted by Lm. This set can be in�nite.

Let us de�ne an m-group to be the conjugacy class [H] of an algebraic sub-

group H of index m such that H is the stabilizer of at least one line in V . The

collection of all m-groups is denoted by Hm. This set is �nite.

The map � : Lm ! Hm is de�ned as follows. The image of an m-line L1 [::: [ Lm is the conjugacy class of the stabilizer of L1 (or any Li). The map � is

by de�nition surjective. More precisely, let H be an algebraic subgroup of index

m such that the collection

X := fL j L is a line with stabilizer gHg�1 for some g 2 Gg

is not empty. Then ��1([H]) is the set of G-orbits of X.

On Lm the Galois group Gal(�k=k) acts in the obvious way. On Hm we let any

� 2 Gal(�k=k) act by �([H]) = [��1H� ]. The map � is equivariant for the de�ned

actions.

Proposition 2.5.1 Let u be an algebraic solution of the Riccati equation and let

k0 be the minimal extension of k such that the minimal monic equation of u over�kK is de�ned over k0K. Let L be the line in V corresponding to u and let H � G

denote the stabilizer of L. Then

[k0 : k] � (#Gal(�k=k)[H])(#��1([H]))

Proof. Since k0 is also the minimal extension for which the m-line [L] := GL is

de�ned over k0, the inequality follows at once from the obvious inequality

(#Gal(�k=k)[L]) � (#Gal(�k=k)[H])(#��1([H])):

2

Remark 2.5.2 For m = 1 the last proposition gives the inequality

[k0 : k] � the number of lines �xed by G:

This is in accordance with lemma 2.4.1.

15

2.5.1 Equations of order two

One considers an equation of order two over the �eld k(x). This equation can be

normalized and has then the form y00 = ry, with r 2 k(x). The corresponding

Riccati equation is u0 + u2 = r. The di�erential Galois group G is an algebraic

subgroup of Sl(2; �k) and is unique up to conjugation. According to [Kov86] one

has the following result:

Theorem 2.5.3 The following statements hold.

1. If there is no algebraic solution over �k(x) of the Riccati equation, then

G = Sl(2; �k).

2. If there is an algebraic solution of the Riccati equation, then the minimal

degree m of such an equation can be 1; 2; 4; 6; 12 and

(a) If m = 1 then G � f c d

0 c�1

!j c 2 �k�; d 2 �kg.

(b) If m = 2 then G = D1 or G = Dn with n � 3. (The dihedral groups).

(c) If m = 4 then G is the tetrahedral group.

(d) If m = 6 then G is the octahedral group.

(e) If m = 12 then G is the icosahedral group.

Using the inequality in Proposition 2.5.1 and Lemma 2.4.1 one �nds

Theorem 2.5.4 Assume that the Riccati equation u0 + u2 = r 2 k(x) has a

solution which is algebraic over �k(x). Let m be the minimal degree, as in Theorem

2.5.3. Then the coe�cients of the monic minimal polynomial of u over �k(x) lie

in a �eld k0(x) with [k0 : k] � 3. More precisely:

1. If m = 1 and G is the multiplicative group Gm or a �nite cyclic group of

order > 2 then [k0 : k] � 2.

2. If m = 2 and G = D4, then [k0 : k] = 3 or k0 = k.

3. If m = 4 and G is the tetrahedral group, then [k0 : k] � 2.

4. In all other cases k0 = k.

Proof. We will give the proof case by case.

1. If m = 1 and there are in�nitely many solutions or precisely one solution

of Riccati in �k(x), then by Lemma 2.4.1 one has k0 = k. In the remaining

case [k0 : k] � 2.

2. If m = 2 and G 6= D4, then G has precisely one algebraic subgroup H of

index 2. Hence #Gal(�k=k)[H] = 1. Further #��1([H]) = 1.

16

3. If m = 2 and G = D4, then H2 has three elements. For each such [H] 2 H2

one has #��1([H]) = 1.

4. For the tetrahedral group one has #H4 = 1 and #��1([H]) = 2.

5. For the octahedral group one has #H6 = 1 and #��1([H]) = 1.

6. For the icosahedral group one has #H12 = 1 and #��1([H]) = 1.

2

2.5.2 Equations of order three

A third-order equation can be normalized in the form y000 + py0 + qy = 0 with

p; q 2 k(x). The di�erential Galois group G is an algebraic subgroup of Sl(3; �k).

The Riccati equation reads u00 + 3uu0 + u3 + pu+ q = 0. The analogue Theorem

2.5.5 of Theorem 2.5.3 is proved in [SU93]. We will use their terminology and

description of �nite primitive groups. In particular, a subgroup H of Sl(3; �k) is

called 1-reducible if the elements of H have a common eigenvector.

Theorem 2.5.5 The following statements hold.

1. If there is no algebraic solution over �k(x) of the Riccati equation then G

satis�es one of the following properties:

(a) G = Sl(3; �k);

(b) G=Z(G) = PSL(2; �k) where Z(G) denotes the center of G;

(c) G is reducible but not 1-reducible.

2. If there is an algebraic solution of the Riccati equation then the minimal

degree m of such an equation can be 1; 3; 6; 9; 21; 36 and

(a) If m = 1 then G is a 1-reducible group.

(b) If m = 3 then G is an imprimitive group.

(c) If m = 6 then G=Z(G) is isomorphic to F36 or A5.

(d) If m = 9 then G=Z(G) is isomorphic to H72 or H216.

(e) If m = 21 then G=Z(G) = G168.

(f) If m = 36 then G=Z(G) = A6.

Using Lemma 2.4.1 and Proposition 2.5.1 one �nds

Theorem 2.5.6 Suppose that the Riccati equation has an algebraic solution and

let m be as in Theorem 2.5.5. There exists an algebraic solution of the Riccati

equation of degree m such that the coe�cients of the monic minimal equation

over �k(x) lie in k0(x) with

17

1. [k0 : k] � 3 if the equation is 1-reducible.

2. [k0 : k] � 4 if G is imprimitive.

3. [k0 : k] � 2 in case G=Z(G) = F36.

4. k0 = k in all other cases.

Proof.

1. If G is 1-reducible one applies lemma 2.4.1 and thus part (1) of the theorem

is proved.

2. Let G be imprimitive. A system of imprimitivity consists of three lines�kei; i = 1; 2; 3 such that fe1; e2; e3g is a basis of the solution space V and

such that G permutes the three lines. In particular, this is a 3-line for G.

If G has only one system of imprimitivity then this system is de�ned over

k and k0 = k holds.

If there is more than one system of imprimitivity then one easily sees that

Go = 1 and the group G is �nite. In particular, G is completely reducible.

For any 3-line fL1; L2; L3g for G the space L1+L2+L3 � V is G-invariant.

If this space has dimension 2 then G has also a one-dimensional invariant

subspace. This contradicts our assumption. Hence L3 is the set of all

systems of imprimitivity of G. According to [Bli17] there are at most four

systems of imprimitivity or in our language #L3 � 4. This proves part

(2) of the theorem. We remark that, according to Blichtfeldt (1917), there

are only two imprimitive groups with four systems of imprimitivity. The

groups have orders 27 and 54 respectively. The corresponding subgroups of

PGl(3) have orders 9 and 18. They are sometimes called H9 and H18.

3. Suppose now that G is a �nite primitive group and m the number de�ned

in theorem 2.5.5. Using the description of such groups given in [SU93], one

can verify that any 1-reducible subgroup H of index m in G is non-abelian.

Therefore H can only �x one line L in the solution space V . It follows that

[H] (the set of conjugates of H in G) consists of m elements. Moreover,

for any [H] 2 Hm the set ��1([H]) consists of one element. Therefore

[k0 : k] � #Hm = 1m#T , where T denotes the set of 1-reducible subgroups

of index m in G. Counting gives #T = m except in one case. This is the

case G=Z(G) = F36. In this case, m = 6 and #T = 12. This proves parts

(3) and (4) of the theorem.

2

18

2.5.3 Remarks

1. Consider a di�erential equation of order n over the �eld K = k((x)). Ac-

cording to [Lev75], there is a �eld extension K 0 � K of degree � n such

that the equation over K 0 has a one-dimensional factor. Let e denote the

rami�cation index of K 0 over K and f the degree of the residue �eld exten-

sion. Then ef = [K 0 : K] � n. The extension K 0 � K 0(x1

e ) is unrami�ed

and has degree � n. This can be reformulated as follows:

The minimal number m such that the Riccati equation has a solution u

which is algebraic over �kK is � n. The minimal monic equation of u over�kK has its coe�cients in a �eld k0K with [k0 : k] � n.

2. Let k be a sub�eld of the �eld of complex numbers. Then we replace

the �eld of formal Laurent series, used in remark (1) above, by the �eld

K of convergent Laurent series. Let G be the di�erential Galois group

of the equation of order n and let Go be the component of 1 in G. One

knows that the group G=Go does not change if one makes the step from

convergent Laurent series to formal Laurent series. This implies that this

group is cyclic. We note that there are in general no algebraic solutions of

the Riccati equation!

Any algebraic solution u of the Riccati equation corresponds to a line in

the solution space with stabilizer H � G0.

The group H is uniquely determined by its index in G and in particular

invariant under conjugation with any element in Gal(�k=k). Using Lemma

2.4.1 one �nds the following:

If the Riccati equation has an algebraic solution then there is an algebraic

solution u such that its monic minimal equation over the �eld �kK has co-

e�cients in a �eld k0K with [k0 : k] � n.

2.6 Symmetries

The aim of this section is to construct di�erential equations over the �eld k(x)

such that the algebraic solutions of degree m of the corresponding Riccati equa-

tion are de�ned over a proper extension of the base �eld k. There are two

ingredients in the construction:

1. The group of symmetries of the equation.

2. Forms of a di�erential equation over k(x).

19

Let k be a �eld of characteristic 0 with algebraic closure �k. The ring of

di�erential operators k(x)[@x] is the skew polynomial ring de�ned by the relation

@xx = x@x + 1. We note that k(x) is the unique maximal sub�eld of k(x)[@x].

Lemma 2.6.1 The k-algebra homomorphisms � : k(x)[@x] ! k(t)[@t] are given

by

�(x) = a and �(@x) =1

a0@t + b

with a 2 k(t) n k; a0 := d

dta and b 2 k(t).

Proof. � is determined by �(x) and �(@x). These two elements of k(t)[@t] satisfy

the relation �(@x)�(x)� �(x)�(@x) = 1. A straightforward calculation with this

relation yields the lemma. 2

Corollary 2.6.2 Let Aut(k(x)[@x]) denote the group of the k-algebra automor-

phisms of k(x)[@x]. There is a split exact sequence of groups

0! k(x)! Aut(k(x)[@x])! PGl(2; k)! 1

Proof. The element a = �(x) must satisfy k(x) = k(a). Therefore, any au-

tomorphism � induces an automorphism of k(x). This explains the map of

Aut(k(x)[@x])! PGl(2; k). Clearly this map is surjective and the kernel consists

of the automorphisms � with �(x) = x and �(@x) = @x + b where b 2 k(x) is

arbitrary. The splitting s is given by: if � 2 PGl(2; k) then s(�)(x) = �(x) and

s(�)(@x) =1

�(x)0@x. This proves the corollary. 2

2.6.1 Symmetries of L

Let L 2 k(x)[@x] n k(x). The group Symk(L) of the symmetries of L consists

of the automorphisms � of k(x)[@x] satisfying �(L) = L. The map Symk(L) !PGl(2; k) is injective as one easily sees.

In some interesting cases L itself has very few symmetries but there is a

non-trivial �nite group H of automorphisms � of k(x)[@x] satisfying �(L) = fL

for some f 2 k(x)�. In this situation one can change L into gL for a certain

g 2 k(x)� such that H is contained in the group of symmetries of gL. We will

call this normalizing L. That normalization is always possible can be seen as

follows:

Write f(�) for the element of k(x)� satisfying �(L) = f(�)L. Then � 2H 7! f(�) 2 k(x)� is a 1-cocycle. The corresponding cohomology group is

H1(H; k(x)�). We will show that this group is trivial. Indeed, let K � k(x)

denote the sub�eld of the H-invariant elements of k(x). Then K � k(x) is a Ga-

lois extension with Galois group H. By Hilbert 90 the group H1(H; k(x)�) = 1.

Therefore the 1-cocycle � 7! f(�) is trivial. This implies that we can choose a

20

g 2 k(x)� such that �(gL) = gL for every � 2 H.

In general, the group of symmetries of an operator L is trivial. In many cases

a study of the singular points of the operator shows that this group is a �nite

subgroup of PGl(2; k).

We will explain this and introduce some terminology. Let the operator L have

the form f(@nx+an�1@n�1x +: : :+a0). Then p 2 �k is called a singularity of L if some

ai has a pole at p. The point p = 1 is called a singular point if L expressed in

the local parameter t = x�1 is singular for t = 0. We note that an automorphism

� of k(x)[@x] not only changes the position of the singular points but singularities

can disappear and new ones can appear. We introduce the notion of essential

singularity in order to distinguish among the singular points.

A singular point p 2 �k [ f1g of L is called essential if there is no automor-

phism �, with �(x) = x, such that p is a regular point of �(L).

If p is an essential singular point of L then �(p) is an essential singular point of

�(L). Indeed, it su�ces to verify this statement for three types of automorphisms:

1. �(x) = x. By de�nition, this � preserves the essential singular points.

2. �(x) = x + � with � 2 k and �(@x) = @x. This case is trivial.

3. �(x) = x�1 and �(@x) = � 1x2@x. One can verify the statement by hand.

One can explain the notion of essential singularity as follows. Let L = f(@nx +

an�1@n�1x + : : : + a0) be a di�erential operator and let p 2 �k. We perform on

L the following operation: �rst, delete the term f ; second, make the unique

transformation x 7! x and @x 7! @x � an�1

nwhich kills the coe�cient of @n�1 of

L. The result is the di�erential operator M = @n + bn�2@n�2x + : : :+ b0. It is not

di�cult to see that L has an essential singularity at p if and only if p is a pole of

some of the bi.

Any L has only �nitely many essential singular points. If this number is � 3

then Symk(L) is obviously a �nite group.

Let H be a �nite group of automorphisms of k(x)[@x]. The set of H-invariant

elements of k(x)[@x] can be identi�ed with k(t)[@t]. Indeed, choose t such that

k(t) is the sub�eld of k(x) consisting of the H-invariant elements. One can

choose b 2 k(x) such that @x := 1t0@x + b is H-invariant. This follows from

H1(H; k(x)) = 0. Let � : k(t)[@t]! k(x)[@x] denote the inclusion. A di�erential

operator L 2 k(x)[@x] has H as group of symmetries, if and only if L = �(M) for

some M 2 k(t)[@t].This shows that any �nite subgroup of PGl(2; k) occurs as group of symme-

tries of some di�erential operator.

21

2.6.2 Transforming algebraic solutions of Riccati

Let L 2 k(x)[@x] n k(x) and a k-algebra homomorphism � : k(x)[@x] ! k(t)[@t]

be given. We extend � to a morphism of k(x)[@x] ! k(t)[@t] by extending the

homomorphism of �elds k(x) ! k(t) to the algebraic closures k(x) and k(t) of

those �elds. An algebraic solution u 2 k(x) of the Riccati equation means that

L factors as M(@x � u) in k(x)[@x]. Then �(L) factors as �(M)(�(@x) � �(u)).

Write �(@x) =1

�(x)0@t+b, then �(x)

0(�(u)�b) is a solution of the Riccati equationof �(L).

Let P (x; T ) 2 �k(x)[T ] be the monic irreducible polynomial of degree m satis-

�ed by u then �(x)0(�(u)�b) satis�es the monic polynomial (�(x)0)mP (�(x); 1�(x)0

T+

b) over �k(t). We will denote this polynomial by ��P .Let G be a group of symmetries of L. Suppose that the collection Rm(L)

of the algebraic solutions of degree m over �k(x) of the Riccati equation is �nite.

Then there are distinct monic irreducible polynomials Pi(x; T ); 1 � i � s of

degree m over �k(x) such that Rm(L) is equal to the set of solutions of P1:::Ps.

For every g 2 G and i the polynomial g�Pi belongs to fP1; :::; Psg. Hence G acts

as a group of permutations on fP1; :::; Psg.

2.6.3 Forms of a di�erential operator

Let L 2 k(x)[@x] n k(x) be a di�erential operator. A di�erential operator M 2k(x)[@x] of the formM = �(L) with � 2 Aut(�k(x)[@x]) is called a form of L. Two

formsM1;M2 are equivalent if there exists a 2 Aut(k(x)[@x]) with (M1) =M2.

We let the group Gal(�k=k) act on �k(x)[@x] by its natural action on �k and as

the identity on k(x)[@x]. Further Gal(�k=k) acts on Sym�k(L) by �(g) = �g��1

(with � 2 Gal(�k=k) and g 2 Sym�k(L)). In the following we will construct an

injective map � : Forms(L)! H1(Gal(�k=k); Sym�k(L)) , where Forms(L) is the

set of equivalence classes of forms of L.

For a form M = �(L) of L and a � 2 Gal(�k=k) one has that a(�) :=

��1����1 2 Sym�k(L). Moreover � 7! a(�) is a 1-cocycle, i.e. a(�1�2) =

a(�1)�1(a(�2)) for all �1; �2 2 Gal(�k=k). The class of � 7! a(�) in

H1(Gal(�k=k); Sym�k(L)) does not depend on the choice of �. Indeed, any other

choice for � has the form �g with g 2 Sym�k(L). The new 1-cocycle has the form

� 7! g�1��1��g��1 = g�1(��1����1)�g��1. By de�nition this is an equivalent 1-

cocycle. We will verify that the resulting map � : Forms(L)! H1(Gal(�k=k); Sym�k(L))

is injective. In general � will not be surjective. Consider the following two maps:

� : H1(Gal(�k=k); Sym�k(L))! H1(Gal(�k=k); Aut(�k(x)[@x]))

: H1(Gal(�k=k); Aut(�k(x)[@x])! H1(Gal(�k=k); PGl(2; �k))

22

The trivial element of those sets (i.e. the image of the trivial 1-cocycle) is denoted

by 1.

Lemma 2.6.3 The following statements hold.

1. � is injective.

2. The image of � is ��1(f1g).3. is surjective.

4. �1(f1g) = f1g.5. � is bijective if either H1(Gal(�k=k); PGl(2; �k)) is trivial or if the subgroup

Sym�k(L) of PGl(2; �k) lifts to a subgroup of Gl(2; �k).

Proof. We will prove this theorem case by case.

1. If theMi = �i(L) for i = 1; 2 induce the same element ofH1(Gal(�k=k); Sym�k(L)),

then there is a g 2 Sym�k(L) with

��12 ��2�

�1 = g�1(��11 ��1�

�1)�g��1 ;

for all � 2 Gal(�k=k). This implies that �1g��12 � = ��1g�

�12 for all � and so

�1g��12 2 Aut(k(x)[@x]). Then �1g�

�12 (M2) = M1 proves that M1 and M2

are equivalent.

2. Clearly, the image of �� is f1g. On the other hand, let a be a 1-cocycle

for Sym�k(L) with �(a) = 1. Then there exists a � 2 Aut(�k(x)[@x]) with

a(�) = ��1�(�) = ��1����1 for all � 2 Gal(�k=k). Put M := �(L). Then

�(M) = ����1(L) = �a(�)(L) = �(L) =M . Hence M 2 k(x)[@x] is a form

of L and M induces the 1-cocycle a.

3. This follows from the splitting of the group homomorphism

Aut(�k(x)[@x])! PGl(2; �k).

4. Let a be a 1-cocycle for Sym�k(L) which has trivial image inH1(Gal(�k=k); PGl(2; �k)).

Then a is equivalent to a 1-cocycle for the normal subgroup �k(x) ofAut(�k(x)[@x]).

Since �k(x) = �k k k(x) one has that H1(Gal(�k=k); �k(x)) = 0. Hence a is a

trivial 1-cocycle.

5. This follows from statements (1)-(4) and the well known fact that

H1(Gal(�k=k); Gl(2; �k)) = 1.

2

23

2.6.4 Construction of special di�erential equations

One �xes a di�erential equation L 2 k(x)[@x] and a number m � 1. One assumes

that the set Rm(L), as de�ned in Section 2.6.2, is �nite and that the monic

polynomials P1; :::; Ps actually lie in k(x)[T ].

Let G � Symk(L) be some �nite group which lifts to a subgroup of Gl(2; k).

This group acts as a permutation group on fP1; :::; Psg.Let h : Gal(�k=k)! G be a continuous homomorphism. Then one can see h as

an element of H1(Gal(�k=k); Sym�k(L)). There is an automorphism � of �k(x)[@x]

such that M := �(L) 2 k(x)[@x] such that �� = �h(�)� for all � 2 Gal(�k=k).The set f��P1; :::; ��Psg are the monic irreducible polynomials of degree m

over �k(x) with as set of zeroes all the solutions of the Riccati equation ofM which

are algebraic over �k(x) of degree m. Since �� = �h(�)� and �Pi = Pi for all � ,

one �nds that � 2 Gal(�k=k) permutes the set f��P1; :::; ��Psg in the same way

as h(�) permutes the fP1; :::; Psg. We can formulate the results above as follows.

Theorem 2.6.4 Let L 2 k(x)[@x] be a di�erential equation with a �nite group of

symmetries G � Symk(L) which lifts to a subgroup of Gl(2; k). Let m � 1 and

suppose that the set of algebraic solutions of degree m over �k(x) of the Riccati

equation of L is �nite and given as the set of zeroes of P1:::Ps where the Pi 2k(x)[T ] are distinct monic polynomials of degree m and irreducible as elements

of �k(x)[T ]. Let a continuous homomorphism h : Gal(�k=k)! G be given.

Then there is a form M = �(L) 2 k(x)[@x] of L with �� = �h(�)� for all

� 2 Gal(�k=k). The polynomials Qi := ��Pi 2 �k(x)[T ]; 1 � i � s are monic

and irreducible. The set of zeroes of Q1:::Qs coincides with the set of algebraic

solutions of degree m over �k(x) of the Riccati equation of M . The action of

Gal(�k=k) on fQ1; :::; Qsg coincides with the homomorphism Gal(�k=k)h! G !

Perm(fP1; :::; Psg), where Perm(fP1; :::; Psg) denotes the group of permutations

of fP1; :::; Psg.

2.6.5 Examples

We will give here examples to show that the inequalities in Theorem 2.5.4 and

Theorem 2.5.6, part 1 are sharp.

Cyclic groups

Consider the di�erential operator L = (x@x)2 � a with a 2 k�. There are two

singular points 0 and 1. They are essential singular points. The symmetry

group of L is the group of all the � 2 PGl(2; �k) for which f0;1g is invariant.

In particular, � given by �(x) = x�1 and �(x@x) = �x@x, is a symmetry. The

two solutions of the Riccati equation (with respect to 0 = x d

dx) are �pa. The

symmetry � permutes these two solutions.

24

1. Ifpa 62 Q then the di�erential Galois group is Gm. If, moreover,

pa 62 k

then the Riccati equation de�nes a degree two extension of k.

2. If ( tn)2 = a with t; n 2 Z; t; n � 1 and relative prime, then the di�erential

Galois group is cyclic of order n. The two solutions of the Riccati equation

are in k(x) and since they are permuted by �, one can apply Theorem 2.6.4.

Hence there is for every quadratic extension k0 � k a form M of L such

that the solutions of the Riccati equation of M are in k0(x) n k(x). In the

next subsection, we will give an explicit example M .

Cyclic groups again

Let k0 = k(�) with �2 2 k and � 62 k. Take u0; u1 2 k(x) and u1 6= 0 and write

u = u0 + �u1. The equation r = u0+ u2 2 k(x) is equivalent to u0 = �1=2u01u1

and

r = u20+u00+�

2u21. Clearly u0+�u1 and u0��u1 are two solutions of the Riccatiequation of y00 = ry. The equation y0 = (�1=2u01

u1+�u1)y has a di�erential Galois

group (over k0(x) or �k(x)) which can be �nite or the multiplicative group Gm.

For a general choice of u1 2 k(x) the di�erential Galois group will be Gm.

If one chooses u1 =a

b(x2��2) with a; b 2 Z n 0; b > 1 and g.c.d.(a; b)=1 then

y = (x � �)(1

2+ a

2b)(x + �)(

1

2� a

2b) satis�es the equation y0 = (�1=2u01

u1+ �u1)y.

Therefore the di�erential Galois group has order b if a; b are odd and has order

2b if a or b is even.

The group D4

The standard di�erential operator for this group is (see [BD79] with in their

notation the case e1 = e2 = e3 = 2)

L = @2x � (3

16x(x� 1)� 3

16x2� 3

16(x� 1)2)

The singular points are f0; 1;1g. They are essential singular points. The sym-

metry group of L is therefore contained in the subgroup of PGl(2;Q) which

leaves f0; 1;1g invariant. This group is isomorphic to S3 and lifts to a subgroup

of Gl(2;Q).

The operator L is actually not invariant under S3. One has to normalize L

into x2(1 � x2)L. The latter has S3 as group of symmetries. One can calcu-

late that u = 3x�1+px

4x(x�1)is a solution of the Riccati equation. The corresponding

�eld extension of Q(x) is Q(px). The other solutions of the Riccati equation

give the �eld extensions Q(px� 1) and Q(

qx(x� 1)). It follows that the six

25

quadratic solutions of the Riccati are the roots of three irreducible monic poly-

nomials P1; P2; P3 2 Q(x)[T ] and that the symmetry group of L acts as S3 on

fP1; P2; P3g. Hence we can apply Theorem 2.6.4. Consider L as given over some

�eld k � Q and let k0 � k be a Galois extension with Galois group isomorphic

to S3. Then there is a form M of L with coe�cients in k(x) such that the Galois

group Gal(k0=k) acts as S3 on fQ1; Q2; Q3g, where the Qi are the monic irre-

ducible polynomials of degree 2 over �k(x) which have as zeroes the six solutions

of the Riccati equation of M . The minimal �eld ki � k such that Qi 2 ki(x)[T ]

has degree 3 over k.

One can make this example more explicit as follows. Write k0 = k(a1; a2; a3)

such that the Galois group of k0=k acts as S3 on fa1; a2; a3g. Let the equation of

the ai over k be X3 � e1X2 + e2X � e3 = 0. De�ne � : �k(x)[@x] ! �k(x)[@x] by

�(x) = a2�a3a2�a1

x�a1x�a3 and a suitable �(@x) such that �(L) takes the form f(@2x � r)

for some f; r 2 �k(x). The term f disappears after a normalization of L. A long

calculation yields the expression for r:

3

16

(3e1e3 � e22) + (e1e2 � 9e3)x + (3e2 � e21)x2

(x3 � e1x2 + e2x� e3)2

The special choice: a1; a2; a3 are the three roots of X3�2, gives the nice equa-

tion @2x +27x

8(x3�2)2. The Galois group of F := Q( 3

p2; e

2�i3 ) acts as the group S3

on the three quadratic irreducible minimal polynomials fQ1; Q2; Q3g � F (x)[T ]

corresponding to six quadratic solutions of the Riccati equation.

The tetrahedral group

The standard example for the tetrahedral group is the di�erential operator (See

[BD79] with in their notation e1 = 2; e2 = e3 = 3).

L := @2x � (3

16x(x� 1)� 3

16x2� 2

9(x� 1)2)

The three singular points f0; 1;1g are essential singular points. Any symmetry

of L has to permute those three points. Further the singularities at 0 and 1 are

isomorphic and the singularity at1 is not isomorphic to the one at 0. Therefore

the only symmetry, 6= 1, is g given by g(x) = x

x�1and g(@x) = �(x�1)2@x+(x�1).

Actually, one has to normalize L into x4L and then g(x4L) = x4L. We note that

[BD79] contains a mistake at this point. There are eight algebraic solutions of

degree 4 of the Riccati equation and there are two irreducible monic polynomials

P1; P2 2 Q(x)[T ] of degree 4 such that the set of zeros of P1P2 is the set of those

eight solutions. Actually, one of the polynomials, say P1, is computed in [Kov86].

This polynomial turns out to lie in Q(x)[T ]. A calculation shows that g�P1 6= P1

26

and so g�P1 = P2. Therefore P1; P2 2 Q(x)[T ] and g permutes them. It seems

possible to make a proof of this without the explicit knowledge of P1.

Applying Theorem 2.6.4, one �nds a form �(M) of L over k and a quadratic

extension k0 � k such that the Galois group of k0=k permutes the corresponding

polynomials f��P1; ��P2g of M . Let k0 = k(�) with �2 2 k and � 62 k. Then one

can take for � the automorphism given by �(x) = 21+2�x

and suitable �(@x). A

calculation shows that M has the form

@2x +32�2

9(1� 4�2x2)2� 3�2

4(1� 4�2x2)

Order three with a 1-reducible Galois group

Let T 3 + pT + q 2 k[T ] denote an irreducible polynomial with Galois group S3.

Then the di�erential equation y000 + py0 + qy = 0 has as basis for the solutions

yi := e�ix where �1; �2; �3 are the three roots of the polynomial T 3+pT + q. The

only relation over the �eld �k(x) satis�ed by the yi is y1y2y3 = 1. The di�erential

Galois group is therefore a maximal torus in Sl(3; �k) and there are precisely

three solutions of the Riccati equation, namely �1; �2; �3. This shows that the

inequality in Theorem 2.5.6 part 1 is sharp.

Remarks

In the two remaining cases part 2 and 3 of Theorem 2.5.6 we are unable to give

explicit examples to show that the inequalities are sharp.

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