J. Symbolic Computation (1997) 11, 1–000

Liouvillian and Algebraic Solutions of Second and Third Order Linear Differential Equations

MICHAEL F. SINGER^{†} and FELIX ULMER^{‡}

North Carolina State University, Department of Mathematics, Box 8205

Raleigh, N.C. 27695-8205

(Received in final form 12 August 1997)

In this paper we show that the index of a 1-reducible subgroup of the differential Galois group of an ordinary homogeneous linear differential equation L(y) = 0 yields the best possible bound for the degree of the minimal polynomial of an algebraic solution of the Riccati equation associated to L(y) = 0. For an irreducible third order equation we show that this degree belongs to {3, 6, 9, 21, 36}. When the Galois group is a finite primitive group, we reformulate and generalize work of L. Fuchs to show how to compute the minimal polynomial of a solution instead of the minimal polynomial of the logarithmic derivative of a solution. These results lead to an effective algorithm to compute Liouvillian solutions of second and third order linear differential equations.

0. Introduction

The computation of the algebraic solutions of a linear differential equation L(y) = 0 over the field of rational functions was a problem of great interest of the end of last century. P. Pepin, H. Schwarz, L. Fuchs, F. Klein, C. Jordan and others worked on this problem and gave a solution for second order equations (cf. (Baldassarri and Dwork (1979)), the intoduction of Boulanger (1898), and Gray (1986)). Many of the earliest contributions to the representation theory of finite groups have been made in connection with differential equation (e.g. Jordan’s Theorem) and it was the starting point for the classification of the finite primitive groups. In this paper we will focus on the ideas of Fuchs. In Fuchs (1878), Fuchs showed how the (then new) tools of invariant theory could be used to construct, in many cases, the minimal polynomial of an algebraic solution of a second order linear differential equation.

The more general question of finding the liouvillian solutions of a linear differential equation, in which case the differential Galois group can be infinite, leads to the theory of linear algebraic groups. But for a primitive unimodular Galois group, all liouvillian solutions are algebraic (cf. Ulmer (1992)) and in this case the approach of Fuchs can

† Partially supported by NSF Grant 90-24624

‡ Partially supported by Deutsche Forschungsgemeinschaft, while on leave from Universität Karlsruhe. The second author would like to thank North Carolina State University for its hospitality and partial support during the preparation of this paper.

0747–7171/90/000000 + 00 $03.00/0 \begin{matrix}c \\ ◯ \end{matrix} 1997 Academic Press Limited

be used. This leads to an effective method for computing the minimal polynomial of a solution in this case. This computation is much more linear than the computation of the minimal polynomial of the logarithmic derivative of a solution which is performed in the algorithm proposed by Kovacic for second order equations† and in the general algorithm proposed by the first author (cf. Kovacic (1986) and Singer (1981)). In the direct computation of a minimal polynomial of a solution, the knowledge of the finitely many possibilities for the differential Galois group can be used not only to bound the degree of the minimal polynomial, but also to compute the coefficients of this polynomial. In this paper we propose the following method for the computation of liouvillian solutions:

i) Case 1: If the differential Galois group is a reducible linear group, then a factorisation of the differential equation is used to reduce the problem to a linear differential equation of lower order. In this paper we show how this can be done for third order equations.

ii) Case 2: If the differential Galois group is an imprimitive linear group, then the algorithm proposed in Singer (1981) by the first author is used. For second (resp. third) order equations, this leads to the computation of a solution whose logarithmic derivative is algebraic of degree 2 (resp. 3), in which case this general algorithm is still practicable.

iii) Case 3: If the differential Galois group is a primitive finite linear group, then we show how the method of Fuchs can be extended to compute the minimal polynomial of an algebraic solution in a very efficient way.

In our approach, we assume that, over the differential field k of coefficients of L(y) = 0, algorithms computing a factorisation, a solution whose logarithmic derivative is in k (for case 2) and a solution which is in k (for case 3) of a of linear differential equation exist (see Section 1 for a discussion and references).

In this paper we discuss explicitely second and third order differential equation, but the extension of the method of Fuchs for case 3 to higher order equations is now straightfoward.

The paper is organized as follow: in the first section we derive some results from differential Galois theory. In the second section we show how, using factorisation, case 1 of a reducible third order linear differential equation can be be reduced to the problem of finding liouvillian solutions of a second order equation. In the next section we derive exact possible algebraic degrees of the logarithmic derivative of a second or third order equation. We then briefly discuss the algorithm given by the first author which is used in case 2, where the Galois group is an imprimitive linear group. In the last and main section we focus on differential equations with primitive differential Galois groups. We first compute a bound for the algebraic degree of a solution and then use the semiinvariants of the Galois group to compute the coefficients of the minimal polynomial of an algebraic solution. We also apply the method to a second and a third order linear differential equation with primitive Galois group and compute the minimal polynomial of a solution in both cases.

† In fact, an algorithm (with some mistakes) to find the minimal polynomial of the logarithmic derivatives of a solution of a second order linear differential equation was first given by Pepin one hundred years before Kovacic (1986) and Singer (1981) (cf. Pépin (1881)). Furthermore, in Pépin (1881), Pépin is able to use his method to verify the Schwarz list of hypergeometric equations with algebraic solutions (cf. Boulanger (1898))

1. Differential Galois Theory

In this section we first briefly review some facts about differential algebra and the existing algorithms for computing liouvillian solutions of linear differential equations. For a more complete exposition we refer to Kaplansky (1957), Kovacic (1986), Singer (1981) or Singer (1990). In the following we will use the same notation as in Ulmer (1992) or Singer and Ulmer (1992).

A differential field (k, δ) is a field k together with a derivation δ on k. A differential field extension of (k, δ) is a differential field (K, Δ) such that K is a field extension of k and Δ is an extension of the derivation δ to a derivation on K. In this paper we always assume that k is a field of characteristic 0 and that the field C = ker_{k}(δ) of constants of

δ in k is algebraically closed (e.g. (\overline{Q}(x), \frac{d}{dx})).

We also write y^{(n)} instead of δ^{n}(y) and y′, y′′, … for δ(y), δ^{2}(y), . . .. Unless otherwise stated, a differential equation L(y) = 0 over k always means an ordinary homogeneous linear differential equation

L(y) = y^{(n)} + a_{n−1}y^{(n−1)} + ⋯ + a_{1}y′ + a_{0}y = 0 (a_{i} ∈ k).

In the following we will have to compute rational solutions z of L(y) = 0 (i.e. z ∈ k), and solutions of L(y) = 0 whose logarithmic derivative is rational (i.e. z′/z ∈ k). Algorithms computing such solutions for various coefficient fields are described in Bronstein. (1992), Liouville (1833), Schlesinger (1895) (volume II, §177) and Singer (1991) (cf. Proposition (2.3)). In the following we always assume that k is a differential field

over which such solutions can be computed (e.g. C(x), \frac{d}{dx})). The computation of a solution whose logarithmic derivative is rational is usually much more difficult than the computation of a rational solution.

Definition 1.1. A differential field extension (K, Δ) of (k, δ) is a liouvillian extension if there is a tower of fields

k = K_{0} ⊂ K_{1} ⊂ ⋯ ⊂ K_{m} = K,

where K_{i+1} is a simple field extension K_{i}(η_{i}) of K_{i}, such that one of the following holds:

i) η_{i} is algebraic over K_{i}, or ii) δ (η_{i}) ∈ K_{i} (extension by an integral), or iii) δ (η_{i}) /η_{i} ∈ K_{i} (extension by the exponential of an integral).

A function contained in a liouvillian extension of k is called a liouvillian function over k.

In Kovacic (1986) an algorithm is given to find a basis of the liouvillian solutions of a second order linear differential equation with coefficients in k_{0}(x), where k_{0} is a finite algebraic extension of Q. In Singer (1981) the first author gives a procedure to find a basis of the liouvillian solutions of a linear differential equation L(y) = 0 of arbitrary degree n with coefficients belonging to a finite algebraic extension of Q(x).

We refer to Kaplansky (1957), Kovacic (1986), Singer (1981), Singer (1990), Ulmer (1992) or Singer and Ulmer (1992) for the definition of a Picard Vessiot extension (PVE) K associated with L(y) = 0, which can be viewed as a splitting field of L(y) = 0, and of the differential Galois group G(L) of L(y) = 0, which consists of the automorphisms of a PVE K of k that commute with δ.

If we choose a fundamental set of solutions {y_{1}, y_{2}, …, y_{n}} of the equation L(y) = 0, then for each σ ∈ G(L) we get σ(y_{i}) = ∑_{j=1}^{n} c_{ij}y_{j} , where c_{ij} ∈ C. This gives a faithful representation of G(L) as a subgroup of GL(n, C). Different choices of bases {y_{1}, y_{2}, …, y_{n}} give equivalent representations. This equivalence class of representations is fundamental to our approach. In the sequel we always consider this representation as the representation of G(L).

Many properties of the equation L(y) = 0 and of its solutions are related to the structure of the group G(L):

Theorem 1.1. (see e.g. Kolchin (1948)) A differential equation L(y) = 0 with coefficients in k has

i) only solutions which are algebraic over k if and only if G(L) is a finite group, ii) only liouvillian solutions over k if and only if the component of the identity G(L)^{∘}

of G(L) in the Zariski topology is solvable. In this case L(y) = 0 has a solution whose logarithmic derivative is algebraic over k.

The following theorem will enable us to always assume that the differential Galois group G(L) ⊆ GL(n, C) of a differential equation L(y) = 0 of degree n is unimodular.

Theorem 1.2. (Kaplansky (1957), p. 41) The differential Galois group of a differential equation of the form

L(y) = y^{(n)} + a_{n−1}y^{(n−1)} + ⋯ + a_{1}y′ + a_{0}y = 0  (a_{i} ∈ k)  (1.1)

is a unimodular group (i.e. G(L) ⊆ SL(n, C)) if and only if ∃W ∈ k, such that W′/W = a_{n−1}.

Using the variable transformation y = z · exp (−\frac{∫ a_{n−1}}{n} ) it is always possible to trans-

form a given differential equation L(y) into an equation L_{SL}(y) of the form:

L(y) = y^{(n)} + a_{n−2}y^{(n−2)} + ⋯ + a_{1}y′ + a_{0}y = 0  (a_{i} ∈ k).  (1.2)

For L(y) = y′′′ + a_{2}y′′ + a_{1}y′ + a_{0}y we get:

L_{SL}(y) = y′′′ + (a_{1} − \frac{a_{2}^{2}}{3} − a′_{2}) y′ + (a_{0} − \frac{a_{1}a_{2}}{3} − \frac{a′′_{2}}{3} + \frac{2a_{2}^{3}}{27} ) y.

Lemma 1.3. Let k ⊂ K be differential fields of characteristic zero with the same field of constants and y ∈ K with y′/y ∈ k. If y is algebraic over k, then the minimal polynomial of y over k is of the form y^{n} − a = 0 for some a ∈ k,

Proof. Let y′/y = u ∈ k and

y^{n} + a_{n−1}y^{n−1} + … + a_{0} = 0

be the minimal polynomial of y over k. Differentiating we have:

nuy^{n} + (a′_{n−1} + (n − 1)ua_{n−1})y^{n−1} + … + a′_{0} = 0.

Comparing coefficients we have

nua_{i} = a′_{i} + iua_{i}  (i = 1, …, n − 1).

If for some i, 0 < i < n, a_{i} ≠ 0, we have that

(n − i)u = \frac{a′_{i}}{a_{i}} ,

we then have

\frac{(y^{n−i}a_{i}^{−1})′}{y^{n−i}a_{i}^{−1}} = 0.

Therefore y^{n−i}a_{i}^{−1} is a constant (in k). This further implies that y would satisfy a polynomial of degree less than n, a contradiction. Therefore, for each i, 0 < i < n, we have a_{i} = 0. □

Corollary 1.4. Let k ⊂ K be as above, where the field of constants is algebraically closed.

i) If for y ∈ K algebraic over k we have y′/y algebraic over k of degree m, then the minimal polynomial P (Y) = 0 of y over k is of the form

Y^{i·m} + a_{m−1}Y^{i·(m−1)} + … + a_{1}Y^{i} + a_{0} (a_{j} ∈ k, m = [k(y′/y)/k])

ii) The extension k(y)/k(y′/y) is a normal extension. If H is the maximal subgroup of the Galois group G of K/k with the property that ∀h ∈ H, h(y′/y) = y′/y, then there is a normal subgroup N of H such that H/N is a cyclic group of order i.

iii) If T is a set of left coset representatives of H in G, then P (Y) = 0 can be written in the following way:

P (Y) = ∏_{σ∈T} (Y^{i} − (σ(y))^{i})  (1.3)

Proof. i) By the previous theorem, the minimal polynomial of y over k(y′/y) is of the form y^{i} − a = 0 for some a ∈ k(y′/y). Let m = [k(y′/y) : k], then y is a solution of a polynomial of the form:

a_{m}Y^{i·m} + a_{m−1}Y^{i·(m−1)} + … + a_{1}Y^{i} + a_{0} (a_{j} ∈ k)

Since [k(y) : k] = [k(y) : k(y′/y)] · [k(y′/y) : k] = i · m, y cannot be a solution of a polynomial of lower degree over k. Thus the above polynomial is the minimal polynomial of y over k.

ii) To the tower of fields k ⊆ k(y′/y) ⊆ k(y) ⊆ K corresponds the tower of groups G ⊇ H ⊇ N ⊇ {id}. Since k contains all the i-th roots of unity, the polynomial y^{i} − a = 0 splits over k(y′/y) and thus k(y) is a normal extension of k(y′/y). Thus N is a normal subgroup of H and the Galois group of k(y)/k(y′/y) is isomorphic to H/N and is a cyclic group.

iii) Since y^{i} is left fixed by the elements of H, we can use a set of left coset representatives T of H in G to write the minimal polynomial of y^{i} is the following way:

∏_{σ∈T} (Y − σ(y^{i})) .

This gives the following polynomial for y:

P (Y) = ∏_{σ∈T} (Y^{i} − (σ(y))^{i}) .

Comparing degrees as above, we get that P (Y) is the minimal polynomial of y.

□

In the following we will need some differential equations associated to L(y) = 0:

Theorem 1.5. (cf. Singer (1980)) Let L_{1}(y) = 0 and L_{2}(y) = 0 be linear differential equations of degrees respectively n_{1} and n_{2} and fundamental systems respectively S_{1} = {u_{1}, ⋯, u_{n_{1}}} und S_{2} = {v_{1}, ⋯, v_{n_{2}}}. Then one can construct a differential equation:

i) L(y) = L_{1}(y)◯ s L_{2}(y) = 0 of degree n_{3} ≤ n_{1}n_{2}, whose solution space is spanned by S = {u_{1}v_{1}, ⋯, u_{n_{1}}v_{1}, ⋯, u_{n_{1}}v_{n_{2}}}.

ii) L_{δ}(y) = 0 of degree n ≤ n_{1}, whose solution space is spanned by the set S_{δ} = {δ (u_{1}) , ⋯, δ (u_{n_{1}})}.

In Singer (1980) and Singer and Ulmer (1992) algorithms to construct the above equations are given. The equation

L^{◯ s m}(y) = \overbrace{L(y)◯ s ⋯ ◯ s L(y)}^{m} = 0

is called the symmetric power of order m of L(y) = 0 and is of order at most \binom{n+m−1}{n−1}.

Let L(y) have order n and let L(y) = 0 have solution space V in some Picard-Vessiot extension K of k. There is a natural G(L) morphism Φ_{m} of G(L) modules from the m^{th} symmetric power S^{m}(V) (c.f., Lang (1984), p. 586) into K given by sending z_{1}◯ s … ◯ s z_{m} to z_{1} · . . . · z_{m}. The image of this map is the solution space of L^{◯ s m}(y) = 0. If all representations of G(L) are completely reducible (e.g. if G(L) is finite), then the solution space of L^{◯ s m}(y) = 0 is G(L)-isomorphic to a direct summand of S^{m}(V) (cf. Singer and Ulmer (1992), Lemma 3.5). If I is an (semi-) invariant of degree m of the representation of G(L), then by the computation of the (semi-) invariant I we will always mean the computation of the image Φ_{m}(I) up to a constant multiple. For an invariant I of degree m, Φ_{m}(I) is a rational solution of L^{◯ s m}(y) = 0. If I is a semi-invariant, then there exists a one dimensional character χ of the group G(L) such that

∀σ ∈ G(L), σ(I) = χ(σ) · I.

In particular, if j is the smallest integer j such that χ^{j} is the trivial character, then Φ_{m}(I^{j}) is a rational solution of L^{◯ s (m·j)}(y) = 0. A bound for j follows from the character table of the group G(L). The one dimensional characters χ corresponding to the semi-invariants and the number of linear independent semi-invariants corresponding to a given character of degree one can be found by decomposing the character of the representation of G(L) on the m^{th} symmetric power S^{m}(V) (cf. Singer and Ulmer (1992), section 2.3). For later reference, we summarize those simple facts:

Lemma 1.6. Let L(y) = 0 be a linear differential equation with coefficients in k whose differential Galois group G(L) ⊂ GL(n, C) is finite. If I is a semi-invariant of degree m of G(L) and Φ_{m}(I) ≠ 0, then Φ_{m}(I) is a non trivial rational solution of L^{◯ s (m·i)}(y) = 0, where i divides the order of a one dimensional character χ of G(L). If I is an invariant, then i = 1. The possible characters χ corresponding to the semi-invariants can be found by decomposing the character of the representation of G(L) on the m^{th} symmetric power S^{m}(V).

2. Case 1: a reducible differential Galois group

The equation L(y) factors as a linear differential operator, if and only if G(L) is a reducible linear group (see e.g. Kolchin (1948)). The factorisation of a differential operator is not unique (see e.g. Singer and Ulmer (1992), section 3.2.1), but an algorithm for computing a factorisation of a differential operator with coefficients in \overline{Q}(x) is well known (see e.g. Grigor’ev (1990) and Schlesinger (1895)). For third order equations a factorisation can be found by computing the rational solutions of the Riccati equation of both L(y) = 0 and of its adjoint. In this section we show how, for a third order differential equation, one can use only one factorisation of L(y) in order to find all liouvillian solutions of L(y) = 0.

We will use the well known reduction method of d’Alembert, which allows one to reduce the order of a linear differential equation L(y) = ∑_{i=0}^{n} a_{i}y^{(i)} = 0 using a non trivial solution y_{1}. The problem of finding further solutions of L(y) = 0 reduces to finding the solutions of

L̃(y) = ∑_{i=0}^{n} a_{i}(y_{1} ∫ y)^{(i)} = 0,

since from a fundamental set of solutions y_{1}^{∗}, ⋯, y_{n−1}^{∗} of L̃(y) = 0 we get a fundamental system of solutions

y_{1}, y_{1} · ∫ (y_{1}^{∗}) , ⋯, y_{1} · ∫ (y_{n−1}^{∗})

of L(y) = 0.

If a second order equation is reducible, then after computing a solution whose logarithmic derivative is rational, one gets a second linearly independent liouvillian solution using the above. Thus, for second order equations, either none or all solutions are liouvillian. This is no longer true for higher order reducible equations:

Lemma 2.1. Let L(y) = y′′′+Ay′+By = 0 be a reducible third order differential equation with A, B ∈ \overline{Q}(x).

i) If L(y) = 0 has a solution z such that z′/z = u ∈ \overline{Q}(x), then the reduction method of d’Alembert gives the equation

L̃(y) = y′′ + 3u′y′ + (3u′′ + 3(u′)^{2} + A) y.

If L̃(y) = 0 has no non zero liouvillian solutions, then z is, up to a constant multiple, the only liouvillian solution of L(y) = 0. If L̃(y) = 0 has a non zero liouvillian solution, then applying again the method of d’Alembert gives 3 linear independent liouvillian solutions.

ii) If L(y) = 0 has no solution z such that z′/z = u ∈ \overline{Q}(x), then any factorisation algorithm will give a factorisation L(y) = L_{1}(L_{2}(y)), where L_{2}(y) is of order 2.

Either L_{2}(y) = 0 has only liouvillian solutions, in which case the procedure of d’Alembert will produce a third liouvillian solution of L(y) = 0 which is not a solution of L_{2}(y) = 0, or L(y) = 0 has no liouvillian solution.

Furthermore, one can determine algorithmically which of these cases hold.

Proof. If L(y) = 0 has a solution z such that z′/z = u ∈ \overline{Q}(x), then the reduction method of d’Alembert always gives an equation

L̃(y) = y′′ + 3u′y′ + (3u′′ + 3(u′)^{2} + A) y.

whose coefficients belong to \overline{Q}(x). The equation L̃(y) = 0 is a second order equation which can be solved using the Kovacic algorithm or the algorithm presented in this paper. Since a second order linear differential equation has either only liouvillian solution or no liouvillian solutions, we get the result.

If a third order differential equation L(y) = 0 has no solution z such that z′/z = u ∈ \overline{Q}(x), then any factorisation of L(y) will be of the form L_{1}(L_{2}(y)), where L_{2}(y) is a second order linear differential equation. We now apply the Kovacic algorithm to L_{2}(y) = 0. If L_{2}(y) = 0 has a liouvillian solution, then L_{2}(y) = 0 and thus L(y) = 0 will have two linear independent liouvillian solutions. Using the reduction method of d’Alembert we will get a third solution of L(y) = 0 which is not a solution of L_{2}(y) = 0. If V is the subspace of liouvillian solutions of L(y) = 0, then L_{2}(y) maps V into the solution space of L_{1}(y) = 0. If L_{2}(y) = 0 has no liouvillian solutions, then L_{2}(y) cannot vanish on V. So V has dimension at most 1. Since V is a G(L) invariant subspace of the solution space of L(y) = 0, there is a non zero z ∈ V so that z′/z ∈ \overline{Q}(x). Since we assume that there are no such solutions, V has dimension zero. □

3. Optimal bounds for the logarithmic derivative and case 2: an imprimitive differential Galois group

It is well known, that for a differential equation L(y) = 0, one can construct a non linear differential equation R(u) = 0, called the Riccati equation associated to L(y), such that the logarithmic derivative u = z′/z of any solution of L(y) = 0 is a solution of R(u) = 0. The Riccati equation associated to L(y) = y′′′ + a_{2}y′′ + a_{1}y′ + a_{0}y is R(u) = u′′ + 3uu′ + a_{2}u′ + u^{3} + a_{2}u^{2} + a_{1}u + a_{0}.

The known algorithms computing liouvillian solutions of a linear differential equation L(y) = 0 use the fact that if L(y) = 0 has a liouvillian solution, then L(y) = 0 has a solution z such that z′/z is algebraic of bounded degree. In Ulmer (1992) a sharp bound for the degree of the minimal polynomial of an algebraic solution of R(u) = 0 is derived.

In this section we will first derive the exact degrees of the minimal polynomial P (u) of an algebraic solution of R(u) = 0 for a third order differential equation and then present the general method given in Singer (1981) to compute the coefficients of P (u). If L(y) = 0 has a liouvillian solution, this, of course, allows us to find a liouvillian solution of the form y = e^{∫ u}. When G(L) is an imprimitive linear group, we show that the minimal degree of P (u) is 3 and we offer no alternative to the general method of Singer (1981). On the other hand, when G(L) is a finite primitive linear group (in which case the minimal degree of P (u) is much larger), we shall show in the next section how to determine directly the minimal polynomial of a solution of L(y) = 0. Nonetheless, we shall need the information found in this section.

3.1. The degree of an algebraic logarithmic derivative of a solution

In this section we assume the reader familiar with the notion of a reducible, imprimitive or primitive linear group and with the notion of a projective representation (see e.g. Huppert (1983), (Curtis and Reiner, I. (1962)), Issacs (1976) or Ulmer (1992)).

Since a normal abelian subgroup of a primitive group G is contained in the center Z(G) of G, we get from Jordan’s Theorem (see Jordan (1878) and (Curtis and Reiner (1962))) that for a finite primitive group G, there are only finitely many possible groups G/Z(G). If a group G̃ ⊆ P GL(n, C) is the image (under the canonical map) of a primitive subgroup of GL(n, C), we call G̃ a primitive subgroup of P GL(n, C).

Definition 3.1. A group G ⊆ GL(n, C) whose elements have a common eigenvector is called 1-reducible.

In Ulmer (1992) Theorem 3.4 it is proven that, if an irreducible differential equation L(y) = 0 has a liouvillian solution, then G(L) ⊆ GL(n, C) has a 1-reducible subgroup H of finite index and that there is a solution z of L(y) = 0 such that the algebraic degree of u = z′/z over k is ≤ [G(L) : H]. In fact the minimal index of a 1-reducible subgroup of G(L) is the best possible bound for the degree of an algebraic solution of the Riccati equation of L(y) = 0:

Lemma 3.1. If a differential equation L(y) = 0 of degree n has a solution z such that u = z′/z is algebraic of degree m, then G(L) ⊆ GL(n, C) has a 1-reducible subgroup H of index m.

Proof. Let H be the subgroup of G(L) which keeps u = z′/z fixed. For any σ ∈ G we have

(\frac{σ(z)}{z})′ = \frac{σ(z′)z − z′σ(z)}{z^{2}}

= σ(z) \frac{\frac{σ(z′)}{σ(z)} − \frac{z′}{z}}{z}
= \frac{σ(z)}{z} (σ (\frac{z′}{z} ) − \frac{z′}{z} )
= 0.

Thus σ(z)/z = c_{σ} ∈ C or σ(z) = c_{σ}z. This shows that z is a common eigenvector of H and that H must be a 1-reducible subgroup of G(L). Since H is the stabiliser of u, the orbit of u under the action of G(L) is of length [G(L) : H]. Thus [k(u) : k] = [G(L) : H].

□

A Schur representation group (Γ, π) of a group G (see e.g. Huppert (1983) p. 630) is a central extension of G having the universal property that, if a projective representation P of G of degree n is given, there exists a representation D of Γ such that the following diagram commutes:

[figure: G; Γ; GL(n, C); P GL(n, C); π; P_{n}; P; D]

where P_{n} : GL(n, C) ⟼ P GL(n, C) = GL(n, C)/Z(GL(n, C)) denotes the canonical homomorphism.

A Schur represention group is usually not uniquely defined, but for our purposes, the knowledge of only one Schur representation group (which by a theorem of I. Schur exists for any finite group G) is necessary (see Ulmer (1992)). There is a routine to construct a Schur representation group of a finite group in the group theory system Cayley, see Cannon (1984). We make the following definition:

Definition 3.2. We denote by F a function whose value F(n) gives the minimal value, such that for each finite primitive subgroup G ⊆ P GL(n, C), any primitive representation of degree n of a Schur representation group of G has a 1-reducible subgroup of index ≤ F(n).

In Ulmer (1992) it is shown that the above function F(n) is well defined. The following result of Ulmer (1992) shows that the bound in the imprimitive case is always small compared to the bound in the primitive case:

Theorem 3.2. If an irreducible differential equation L(y) = 0 of degree n with coefficients in a differential field k, whose field of constants is algebraic closed, has a liouvillian solution over k, then L(y) = 0 has a solution z such that

i) if G(L) ⊆ GL(n, C) is an imprimitive group, then u = z′/z is algebraic over k of degree at most max_{d|n,d>1} {d! · F(n/d)}.

ii) if G(L) ⊆ GL(n, C) is a primitive group, then u = z′/z is algebraic over k of degree at most F(n).

We note that if n is prime, one can get a better bound in the imprimitive case (Ulmer (1992), Lemma 4.2). In this case u = z′/z can be chosen to be algebraic of degree n.

In order to compute the bound F(n) one needs a list of the finite primitive subgroups of P GL(n, C). For n = 3 such a list is given for example in Blichfeld (1917):

(i) A_{6}, the alternating permutation group of 6 letters. (ii) G_{168}, the simple group of order 168. (iii) A_{5}, the alternating permutation group of 5 letters. (iv) H_{216}, the Hessian group of order 216, which is isomorphic to the permutation group

of 9 letters generated by the permutations (4, 5, 6)(7, 9, 8) and (1, 2, 4)(5, 6, 8)(3, 9, 7).

(v) H_{72}, the normal subgroup of order 72 of the group H_{216}. (vi) F_{36}, a normal subgroup of order 36 of the group H_{72} (there are 3 such groups,

which are all isomorphic).

From such a (finite) list of the finite primitive subgroups of P GL(n, C) the bound F(n) can always be computed using the characters of the subgroups of corresponding Schur representation groups. Let G be a finite primitive subgroup of P GL(n, C), Γ_{G} a Schur representation group of G and ρ(Γ_{G}) an irreducible representation of degree n of Γ_{G} with character ζ. The restriction of ρ(Γ_{G}) to a subgroup H is 1-reducible if and only if there is a one dimensional character ψ of H such that the scalar product (ψ, ζ|H) ≠ 0, where ζ|H denotes the restriction of ζ to H (cf. Curtis and Reiner (1962), §38). Considering the finitely many primitive groups G and the finitely many subgroups H of Γ_{G} will give F(n).

For n = 3 the computation is simplified by the fact that a 1-reducible subgroup ρ(H) of a irreducible finite group ρ(Γ_{G}) ⊆ GL(3, C) is either abelian or C^{3} is a direct sum of an irreducible one dimensional representation of H and an irreducible two dimensional represention of H. Let ζ be the character of ρ(Γ_{G}). If ρ(H) is 1-reducible, then either ζ|H is the sum of 3 characters of degree 1, or ζ|H is the sum of two characters χ_{1} and χ_{2} of H, where χ_{1}(1) = 1 and χ_{2}(1) = 2. We note also that, if a subgroup H of Γ has no irreducible character of degree 3, then ρ(H) must be 1-reducible.

We now look at a Schur representation group Γ of the above groups, constructed using Cayley (for the non simple groups these groups are not all isomorphic, so we will only give the generators and relations of the groups which have been used). Using character tables (also computed in Cayley) we performed the following case-by-case study (In the appendix the character tables of the subgroups of index ≤ 6 of the Schur representation group of A_{5} are given):

(i) Γ/Z(Γ) ≅ A_{6}. All subgroups of index 36 have no irreducible character of degree 3 and thus are 1-reducible. In order to see that 36 is the smallest index of a 1-reducible subgroup of ρ(Γ) we need to look at all subgroups whose index is less than 36 (We note that the kernel of an irreducible representation of degree 3 of Γ is always of order 2 and thus elements of order 3 have trace ≠ 3):

(a) Any subgroup H of index 30 contains an element g of order 4. If ζ is an irreducible character of degree 3 of Γ, then ζ(g) = −1. But for any irreducible character χ_{1} of degree 1 of H we get χ_{1}(g) = 1, and for any irreducible character χ_{2} of degree 2 of H we get χ_{2}(g) = 0. Thus ζ|H can not be the sum of 3 character of degree 1 and since χ_{1}(g) + χ_{2}(g) = 1, we have χ_{1} + χ_{2} ≠ ζ|H. Thus ρ(H) can not be a 1-reducible subgroup of ρ(Γ).

(b) Any subgroup H of index 20 contains an element g of order 3 with the property that for any irreducible character χ_{1} of degree 1 of H we get χ_{1}(g) = 1, and for any irreducible character χ_{2} of degree 2 of H we get χ_{2}(g) = 2. But for any irreducible character ζ of Γ, we have ζ(g) ≠ 3. Since ζ|H can not be the sum of 3 character of degree 1 and χ_{1} + χ_{2} ≠ ζ|H, ρ(H) can not be a 1-reducible subgroup of ρ(Γ).

(c) Any subgroup H of index 18 contains an element g of order 4. If ζ is an irreducible character of degree 3 of Γ, then ζ(g) = −1. But for any irreducible character χ_{1} of degree 1 of H we get χ_{1}(g) = 1, and for any irreducible character χ_{2} of degree 2 of H we get χ_{2}(g) = 0. Since ζ|H can not be the sum of 3 character of degree 1 and χ_{1} + χ_{2} ≠ ζ|H, ρ(H) can not be a 1-reducible subgroup of ρ(Γ).

(d) Any subgroup H of index 15 contains an element g of order 4. If ζ is an irreducible character of degree 3 of Γ, then ζ(g) = −1. But for any irreducible character χ_{1} of degree 1 of H we get χ_{1}(g) = 1, and for any irreducible character χ_{2} of degree 2 of H we get χ_{2}(g) = 0 or χ_{2}(g) = 2. Since ζ|H can not be the sum of 3 character of degree 1 and χ_{1} + χ_{2} ≠ ζ|H, ρ(H) can not be a 1-reducible subgroup of ρ(Γ).

(e) Any subgroup H of index 10 has no irreducible character of degree 2. For an element of order 3 of H and any irreducible character χ_{1} of degree 1 of H we get χ_{1}(g) = 1 and ζ|H can not be the sum of 3 character of degree 1. Thus ρ(H) can not be a 1-reducible subgroup of ρ(Γ).

(f) Any subgroup H of index 6 contains an element g of order 4. If ζ is an irreducible character of degree 3 of Γ, then ζ(g) = −1. But for any irreducible character χ_{1} of degree 1 of H we get χ_{1}(g) = 1, and for any irreducible character χ_{2} of degree 2 of H we get χ_{2}(g) = 0. Since ζ|H can not be the sum of 3 character of degree 1 and χ_{1} + χ_{2} ≠ ζ|H, ρ(H) can not be a 1-reducible subgroup of ρ(Γ).

(ii) Γ/Z(Γ) ≅ G_{168}. All subgroups of index 21 of G have no irreducible character of degree 3, and thus are 1-reducible. In order to see that 21 is the smallest index of a 1-reducible subgroup of Γ we need to look at all subgroups whose index is less than 21 (We note that the kernel of an irreducible representation of degree 3 of Γ is always of order 2 and thus elements of order 7 have trace ≠ 3):

(a) The subgroups of index 16 are all conjugate and have no representation of degree 2. For an element g of order 7 of H and any irreducible character χ_{1} of degree 1 of H we get χ_{1}(g) = 1 and ζ|H can not be the sum of 3 character of degree 1. Thus no subgroup of index 16 can be a 1-reducible subgroup of ρ(Γ).

(b) Any subgroup H of index 14 contains an element g of order 4. If ζ is an irreducible character of degree 3 of Γ, then ζ(g) = −1. But for any irreducible character χ_{1} of degree 1 of H we get χ_{1}(g) = 1, and for any irreducible character χ_{2} of degree 2 of H we get χ_{2}(g) = 0. Since ζ|H can not be the sum of 3 character of degree 1 and χ_{1} + χ_{2} ≠ ζ|H, ρ(H) can not be a 1-reducible subgroup of ρ(Γ).

(c) The subgroups of index 8 are all conjugate and have no irreducible character of degree 2. For an element of order 7 of H and any irreducible character χ_{1} of degree 1 of H we get χ_{1}(g) = 1 and ζ|H can not be the sum of 3 character of degree 1. Thus no subgroup of index 8 can be a 1-reducible subgroup of ρ(Γ).

(d) Any subgroup H of index 7 contains an element g of order 4 whose conjugacy class contains 6 elements. If ζ is an irreducible character of degree 3 of Γ, then ζ(g) = −1. But for any irreducible character χ_{1} of degree 1 of H we get χ_{1}(g) = 1, and for any irreducible character χ_{2} of degree 2 of H we get χ_{2}(g) = 0 or χ_{2}(g) = 2. Since ζ|H can not be the sum of 3 character of degree 1 and χ_{1} + χ_{2} ≠ ζ|H, ρ(H) can not be a 1-reducible subgroup of ρ(Γ).

(iii) Γ/Z(Γ) ≅ A_{5}. All subgroups of index 6 of Γ have no irreducible character of degree 3, and thus are 1-reducible. In order to see that 6 is the smallest index of a 1reducible subgroup of Γ we need to look at all subgroups whose index is less than 6. These non abelian groups are all conjugate and of index 5. A subgroup H of index 5 contains an element g of order 4. If ζ is an irreducible character of degree 3 of Γ, then ζ(g) = −1. But for any irreducible character χ_{1} of degree 1 of H we get χ_{1}(g) = 1, and for any irreducible character χ_{2} of degree 2 of H we get χ_{2}(g) = 0. Since ζ|H can not be the sum of 3 character of degree 1 and χ_{1} + χ_{2} ≠ ζ|H, ρ(H) can not be a 1-reducible subgroup of ρ(Γ).

(iv) From the presentation {a, b | a^{3} = b^{3} = (ab)^{4} = [(aba)^{2}, a] = id} of H_{216}, Cayley computes the following Schur representation group

{a, b, c, d | a^{3}c = b^{3} = (ab)^{4}c = a^{−1}b^{−1}a^{−1}(a^{−1}b^{−1})^{2}(aba)^{2}bd^{−1}
= [a, c] = [b, c] = [a, d] = [b, d] = [c, d] = id}.

We will only considere the faithfull representation of degree 3, since the non faithfull represenation of degree 3 of Γ has non central elements in its kernel. Any subgroup H of index 9 has an element g of order 4. If ζ is a faithfull irreducible character of degree 3 of Γ, then ζ(g) = 1. But for any irreducible character χ′ of degree 3 of H we must have χ′(g) = −1. Thus χ′ ≠ ζ|H and H must be a 1-reducible subgroup of Γ. In order to see that 9 is the smallest index of a 1-reducible subgroup of Γ we need to look at all subgroups whose index is less than 9. Since those groups are all non abelian and the irreducible representation ρ(Γ) of degree 3 is assumed faithfull, ζ|H can not be the sum of 3 character of degree 1.

(a) The subgroups of index 8 are all conjugate and have no irreducible character of degree 2. Thus no subgroup of index 8 can be a 1-reducible subgroup of ρ(Γ).

(b) The subgroups of index 6 are all conjugate and have no irreducible character of degree 2. Thus no subgroup of index 6 can be a 1-reducible subgroup of ρ(Γ).

(c) Any subgroup H of index 4 contains an element g of order 3 whose conjugacy class in H contains 1 element. If ζ is an irreducible character of degree 3 of Γ, then ζ(g) ≠ 3. But for any irreducible character χ_{1} of degree 1 of H we get χ_{1}(g) = 1, and for any irreducible character χ_{2} of degree 2 of H we get χ_{2}(g) = 2. Since χ_{1} + χ_{2} ≠ ζ|H, ρ(H) can not be a 1-reducible subgroup of ρ(Γ).

(d) Any subgroup H of index 3 contains an element g of order 3 whose conjugacy class in H contains 24 elements. If ζ is an irreducible character of degree 3 of Γ, then ζ(g) ≠ 3. But for any irreducible character χ_{1} of degree 1 of H we get χ_{1}(g) = 1, and for any irreducible character χ_{2} of degree 2 of H we get χ_{2}(g) = 2. Since χ_{1} + χ_{2} ≠ ζ|H, ρ(H) can not be a 1-reducible subgroup of ρ(Γ).

(v) From the presentation {a, b, c | a^{2}b^{−2} = aba^{−1}b = c^{4} = acb^{−1}c^{−2} = id} of H_{72}, Cayley computes the following Schur representation group Γ:

{a, b, c, d | a^{2}b^{−2} = aba^{−1}b = c^{4}d = acb^{−1}c^{−2}
= [a, d] = [b, d] = [c, d] = id}.

All subgroups of index 9 of Γ have no irreducible character of degree 3, and thus are 1-reducible. In order to see that 9 is the smallest index of a 1-reducible subgroup of Γ we need to look at all subgroups whose index is less than 9. Since those groups are all non abelian and all irreducible representations ρ(Γ) of degree 3 are faithfull, ζ|H can not be the sum of 3 character of degree 1.

(a) The subgroups of index 8 are all conjugate and have no irreducible character of degree 2. Thus no subgroup of index 8 can be a 1-reducible subgroup of ρ(Γ).

(b) Any subgroup H of index 4 contains an element g of order 3 whose conjugacy class contains one element. If ζ is an irreducible character of degree 3 of Γ, then ζ(g) ≠ 3. But for any irreducible character χ_{1} of degree 1 of H we get χ_{1}(g) = 1, and for any irreducible character χ_{2} of degree 2 of H we get χ_{2}(g) = 2. Thus χ_{1} + χ_{2} ≠ χ and ρ(H) can not be a 1-reducible subgroup of ρ(Γ).

(c) Any subgroup of index 2 has no irreducible character of degree 2. Thus no subgroup of index 2 can be a 1-reducible subgroup of ρ(Γ).

(vi) From the presentation {a, b | a^{4} = (ab^{−1})^{2} = b^{4} = (ab)^{3} = id} of F_{36}, Cayley computes the following Schur representation group Γ:

{a, b, c, d | a^{4}c^{−1}d^{2} = (ab^{−1})^{2}d = b^{4} = (ab)^{3}d
= [a, c] = [b, c] = [a, d] = [b, d] = [c, d] = id}.

All subgroups of index 6 of Γ have no irreducible character of degree 3, and thus are 1-reducible. In order to see that 6 is the smallest index of a 1-reducible subgroup of Γ we need to look at all subgroups whose index is less than 9. Since those groups are all non abelian and all irreducible representations ρ(Γ) of degree 3 are faithfull, ζ|H can not be the sum of 3 character of degree 1.

(a) The subgroups of index 4 are all conjugate and have no irreducible character of degree 2. Thus no subgroup of index 4 can be a 1-reducible subgroup of ρ(Γ).

(b) Any subgroup H of index 2 contains an element g of order 3 whose conjugacy class in H contains 1 element. If ζ is an irreducible character of degree 3 of Γ, then ζ(g) ≠ 3. But for any irreducible character χ_{1} of degree 1 of H we get χ_{1}(g) = 1, and for any irreducible character χ_{2} of degree 2 of H we get χ_{2}(g) = 2. Thus χ_{1} + χ_{2} ≠ ζ|H and ρ(H) can not be a 1-reducible subgroup of ρ(Γ).

Since, as noted above, any imprimitive subgroup of GL(3, C) has a 1-reducible subgroup of index 3 (Ulmer (1992), Lemma 4.2) we get:

Theorem 3.3. If an irreducible third order linear differential equation L(y) = 0 with coefficients in k has a liouvillian solution, then L(y) = 0 has a solution z, such that for the logarithmic derivative u = z′/z of z one of the following holds:

i) u is algebraic of degree 36 over k and G(L)/Z(G(L)) ≅ A_{6}. ii) u is algebraic of degree 21 over k and G(L))/Z(G(L) ≅ G_{168}. iii) u algebraic of degree 9 over k and G(L)/Z(G(L)) is isomorphic to H_{72} or H_{216}. iv) u is algebraic of degree 6 over k and G(L)/Z(G(L)) is isomorphic to F_{36} or A_{5}.

v) u is algebraic of degree 3 over k and G(L) ⊆ GL(3, C) is an imprimitive group.

For each group G(L) ⊆ GL(3, C) the numbers given above are best possible.

Proof. That there exists a solution whose logarithmic derivative is of the given degree follows from Theorem 3.4 of Ulmer (1992) and the previous discussion. From Lemma 3.1 we know that the degree [k(u) : k] is precisely the index of a 1-reducible subgoup of G(L). Since the above numbers are the minimal index of a 1-reducible subgroup of G(L), they are best possible. □

A similar calculation can be done to show that for second order equations, the best possible degrees are: 4 if G(L)/Z(G(L)) ≅ A_{4}, 6 if G(L)/Z(G(L)) ≅ S_{4} and 12 if G(L)/Z(G(L)) ≅ A_{5} (cf. Ulmer (1992)). This gives an alternative proof of theorem 1 of Kovacic (1986).

We point out that the above result is derived without explicitely determining the possible finite primitive unimodular Galois groups of L(y) = 0 but follows just from the knowledge of the list of the finite primitive subgroups of P GL(3, C).

3.2. Computing the coefficients of a minimal polynomial of an algebraic solution of known degree of the riccatti

Since we have just produced the exact minimal degrees of an algebraic solution u of the Riccati equation, we now briefly review the method given in Singer (1981) to computes the coefficients of the minimal polynomial of u. This method is the only known method which can be used in the case of an imprimitive differential Galois group of order n ≥ 3.

We start by describing an algorithm for finding all solutions y of L(y) = 0 such that δ(y)/y ∈ C(x). Let S be the set of singular points of L(y). At each point c ∈ S one can determine a finite set P_{c} of elements of C(x) of the form

f_{c} = \frac{α_{1c}}{x − c} + \frac{α_{2c}}{(x − c)^{2}} + … + \frac{α_{nc}}{(x − c)^{n}}

or f_{c} = α_{1c}x + … + α_{nc}x^{n} if c = ∞, such that if y is a solution of L(y) = 0 such that y′/y ∈ C(x) then

y = P (x)e^{(∫ ∑_{c∈S} f_{c})}

= P (x) ∏_{c∈S}(x − c)^{α_{1c}}e^{(\frac{α_{2c}}{x − c} + \frac{α_{3c}}{(x − c)^{2}} + … + \frac{α_{nc}}{(x − c)^{n}} )}

for some choice of f_{c} ∈ P and P (x) ∈ C[x]. Futhermore, the degree of any possible P (x) can be bounded in terms of the α_{1c}. A method for determining the sets P_{c} is given in (Schlesinger (1895), Vol. II.1 Section 177). A modern presentation using Newton polygons and emphasizing computational aspects and an implementation in DESIRE is given in Grigor’ev (1990) and Tournier (1987). This reduces the problem of finding such solutions to the problem of determining the coefficients of the possible P (x), a problem in linear algebra. A related method is given in Singer (1991).

Let L(y) = 0 be an irreducible differential equation and P (u) the minimal polynomial of an algebraic logarithmic derivative δ(y_{1})/y_{1} of a solution y_{1} of L(y) = 0. Any solution of P (u) = 0 is then a logarithmic derivative of a solution of L(y) = 0 and we get:

P (u) = u^{m} + c_{m−1}u^{m−1} + … + c_{0}
= (u − \frac{y′_{1}}{y_{1}} ) (u − \frac{y′_{2}}{y_{2}} ) … (u − \frac{y′_{m}}{y_{m}}) ,

where the y_{i} are solutions of L(y) = 0. The coefficients c_{i} ∈ k are homogeneous forms of degree m − i in the m logarithmic derivatives {y′_{1}/y_{1}, ⋯, y′_{m}/y_{m}}.

We have

c_{m−1} = \frac{y′_{1}}{y_{1}} + \frac{y′_{2}}{y_{2}} + ⋯ + \frac{y′_{m}}{y_{m}}
= \frac{(y_{1}y_{2} ⋯ y_{m})′}{y_{1}y_{2} ⋯ y_{m}} .

The product y_{1}y_{2} ⋯ y_{m} is a solution of L^{◯ s m}(y) = 0. Thus c_{m−1} is a rational logarithmic derivative of a solution of L^{◯ s}^{m}(y) = 0.

Using c_{m−1} we can now compute the other coefficients c_{i}. The coefficient c_{i} can be written as:

c_{i} = ∑_{1≤k_{1}<···<k_{i}≤m} (\frac{y′_{k_{1}}}{y_{k_{1}}} ⋯ \frac{y′_{k_{i}}}{y_{k_{i}}} ) .

Let

v_{i} = c_{i}y_{1}y_{2} … y_{n}
= ∑_{1≤k_{1}<···<k_{i}≤m} y′_{k_{1}}y′_{k_{2}} … y′_{k_{i}} ∏_{j≠k_{1},···,k_{i}} y_{j}.

Note that v_{i} is a solution of (L_{δ})^{◯ s i} (y)◯ s L^{◯ s (m−i)}(y), where L_{δ}(y) is a differential equation which is satisfied by the derivatives of solutions L(y) = 0. We then have:

\frac{v′_{i}}{v_{i}} = \frac{c′_{i}}{c_{i}} + \frac{(y_{1}y_{2} … y_{n})′}{y_{1}y_{2} … y_{n}}
= \frac{c′_{i}}{c_{i}} + c_{m−1}.

Therefore, v_{i} is the solution of a linear differential equation (that we can construct) and the logarithmic derivative of v_{i} is rational. We can describe all such solutions. Simillarly, y_{1}y_{2} … y_{n} is also a solution of a linear differential equation and its logarithmic derivative is also rational. Since c_{i} = v_{i}/(y_{1}y_{2} … y_{n}), we can determine the degrees of the numerator and denominator of c_{i} using the algorithm described at the beginning of this section to determine the possible candidates for v_{i} and (y_{1}y_{2} … y_{n}).

We therefore are able to determine bounds on the degrees of the numerators and denominators of the coefficients of a minimal polynomial. To determine the actual numbers that can appear as coefficients of these numerators and denominators, one must differentiate P (u) = 0 repeatedly, solve for the higher derivatives of u, and reduce the Riccati equation R(u) = 0 mod P (u). This will give algebraic conditions on the numbers appearing in the coefficients of P (u) (a similar method is used in the last section).

We note that for second order linear differential equations, R(u) has order 1. In this case , it is showed in Kovacic (1986) that for 0 ≤ i ≤ n − 2 a simple recursion gives each c_{i} in terms of the c_{j} with j > i. Therefore in the second order case it suffices to just find the possible c_{m−1}. We do not know a similar statement for higher order equations.

4. Case 3: a primitive unimodular Galois group

In this section we show that in this case, where the bound for the algebraic degree of an algebraic solution of the Riccati is large compared to the imprimitive case (cf.

Theorem 3.2 and 3.3), the difficult computation of a rational solution of some Riccati can be avoided. We will reduce the problem to the computation of a rational solution of some symmetric power and a Gröbner basis computation. In contrast to the previous section where only a projective representation of G(L) was used, a list of the possible Galois groups will be needed. We start by giving this list (taken from Blichfeld (1917) and (Miller, Blichfeld and Dickson (1938)) ) for second and third order equations.

4.1. The primitive unimodular groups of degree 2 and 3

4.1.1. The primitive unimodular groups of degree 2

Up to isomorphism, there are 3 primitive unimodular groups of degree 2. According to Miller, Blichfeld and Dickson (1938) we define the following matrices:

T = ( \begin{matrix}i & 0 \\ 0 & −i \end{matrix} ) S = ( \begin{matrix}\frac{i − 1}{2} & \frac{i − 1}{2} \\ \frac{i + 1}{2} & \frac{−i − 1}{2} \end{matrix} ) U = ( \begin{matrix}\frac{1 + i}{\sqrt{2}} & 0 \\ 0 & \frac{1 − i}{\sqrt{2}} \end{matrix} )

S′ = ( \begin{matrix}ξ^{3} & 0 \\ 0 & ξ^{2} \end{matrix} ) U′ = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) T′ = ( \begin{matrix}α & β \\ β & −α \end{matrix} ) ,

where ξ^{5} = 1, α = \frac{ξ^{4} − ξ}{\sqrt{5}} and β = \frac{ξ^{2} − ξ^{3}}{\sqrt{5}} .

Then the groups are the following:

(i) The icosahedral group A_{5}^{SL_{2}} of order 120 is generated by S′, U′ and T′. We have A_{5}^{SL_{2}}/{±1} is isomorphic to the alternating group A_{5} of 5 letters.

(ii) The octahedral group S_{4}^{SL_{2}} of order 48 is generated by S and U. We have S_{4}^{SL_{2}}/{±1} is isomorphic to the symmetric group S_{4} of 4 letters.

(iii) The Tetrahedral group of A_{4}^{SL_{2}} of order 24 is generated by S and T. We have A_{4}^{SL_{2}}/{±1} is isomorphic to the alternating group A_{4} of 4 letters.

When one looks at a character table of A_{4}^{SL_{2}}, one sees 3 irreducible representations of degree 3, but since in SL(2, C) the trace of an element of order 3 is −1, only one of these is in SL(2, C). The groups A_{5}^{SL_{2}} and S_{4}^{SL_{2}} have two non conjugate representation in SL(2, C). But the non equivalent representations can be obtained from each other using the Galois group of Q(\sqrt{2}, ξ) over Q. This follows from the fact that under the automorphism sending \sqrt{2} to −\sqrt{2} the trace of U will change and that under the automorphism sending \sqrt{5} ∈ Q(ξ) to −\sqrt{5} the trace of an element of order 10 will change. This will allow us to work with only one representation and to get the complete result by applying the corresponding automorphism.

4.1.2. The primitive unimodular groups of degree 3

Up to isomorphism, there are 8 primitive unimodular groups of degree 3. According to Miller, Blichfeld and Dickson (1938)† we define the following matrices:

E_{1} = ( \begin{matrix}1 & 0 & 0 \\ 0 & ξ^{4} & 0 \\ 0 & 0 & ξ \end{matrix} ) E_{2} = ( \begin{matrix}−1 & 0 & 0 \\ 0 & 0 & −1 \\ 0 & −1 & 0 \end{matrix} ) E_{3} = \frac{1}{\sqrt{5}} ( \begin{matrix}1 & 2 & 2 \\ 1 & s & t \\ 1 & t & s \end{matrix} )

E_{4} = \frac{1}{\sqrt{5}} ( \begin{matrix}1 & 2λ_{2} & 2λ_{2} \\ λ_{1} & s & t \\ λ_{1} & t & s \end{matrix} ) S = ( \begin{matrix}β & 0 & 0 \\ 0 & β^{2} & 0 \\ 0 & 0 & β^{4} \end{matrix} ) R = \frac{1}{\sqrt{−7}} ( \begin{matrix}a & b & c \\ b & c & a \\ c & a & b \end{matrix} )

S_{1} = ( \begin{matrix}1 & 0 & 0 \\ 0 & ω & 0 \\ 0 & 0 & ω^{2} \end{matrix} ) T = ( \begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix} ) U = ( \begin{matrix}ε & 0 & 0 \\ 0 & ε & 0 \\ 0 & 0 & εω \end{matrix} )

V = ρ ( \begin{matrix}1 & 1 & 1 \\ 1 & ω & ω^{2} \\ 1 & ω^{2} & ω \end{matrix} )  Z = ( \begin{matrix}ω & 0 & 0 \\ 0 & ω & 0 \\ 0 & 0 & ω \end{matrix} )

where ξ^{5} = 1, s = ξ^{2}+ξ^{3}, t = ξ+ξ^{4}, \sqrt{5} = t−s, ε^{6}+ε^{3}+1 = 0 (ε^{9} = 1), ω = −ε^{3}−1 (ω^{3} =

1), β^{7} = 1, a = β^{4} − β^{3}, b = β^{2} − β^{5}, c = β − β^{6}, \frac{1}{\sqrt{−7}} = \frac{β + β^{2} + β^{4} − β^{6} − β^{5} − β^{3}}{7} ,

λ_{1} = \frac{−1 ± \sqrt{−15}}{4}, λ_{2} = \frac{−1 ∓ \sqrt{−15}}{4} and ρ = \frac{1}{ω − ω^{2}} .

Then the groups are the following:

(i) The Valentiner group A_{6}^{SL_{3}} of order 1080 is generated by E_{1}, E_{2}, E_{3} and E_{4}. We have A_{6}^{SL_{3}}/ < Z > is isomorphic to the alternating group A_{6} of 6 letters.

(ii) The alternating group A_{5} of five letters generated by E_{1}, E_{2} and E_{3}. (iii) The direct product A_{5} × C_{3}, of A_{5} and the cyclic group C_{3} of three elements,

generated by E_{1}, E_{2}, E_{3} and Z. (iv) The simple group G_{168} of order 168 generated by S, T and R.

(v) The direct product G_{168} × C_{3}, of G_{168} and the cyclic group C_{3} of three elements, generated by S, T, R and Z.

(vi) The group H_{216}^{SL_{3}} of order 648 generated by S_{1}, T, V and U, whose projective

representation is the Hessian group of order 216. (vii) The group H_{72}^{SL_{3}} of order 216 generated by S_{1}, T, V and UV U^{−1}. (viii) The group F_{36}^{SL_{3}} of order 108 generated by S_{1}, T and V.

As in the previous case, all non equivalent representations of these groups in SL(3, C) can be obtained using the Galois group of the field to which the coefficients of the matrices belong. The group H_{72}^{SL_{3}} has two faithfull non conjugate representations in SL(3, C) which

† The matrix T used here corresponds to the inverse of the matrix T used in Miller, Blichfeld and Dickson (1938) and the definition of A_{6}^{SL_{3}} and A_{5} correspond to the definitions given in exercise 3 and 4 p. 252 of Miller, Blichfeld and Dickson (1938)

are sent to each other by the automorphism σ : ε ↦ ε^{2} of the Galois group of Q(ε)/Q, since the trace of M = UV U^{−1}V T^{−1}UV U^{−1} is ω, while the trace of σ(M) is ω^{2}. The automorphism σ also sends the two faithfull non conjugate representations of F_{36}^{SL_{3}} in SL(3, C) to each other, since the trace of M = V S^{2}T ST^{−1} is ω, while the trace of σ(M) is ω^{2}. This will allow us to work with only one representation and to get the complete result by applying the corresponding automorphism. We also note that for H_{72}^{SL_{3}} and F_{36}^{SL_{3}} only the representations where the elements of order 4 all have trace 1 belong to SL(3, C).

We point out that the above groups do not in general correspond to the Schur representation groups used in the previous section. For example, the group A_{6}^{SL_{3}} is of order 1080 while the Schur representation group of A_{6} is of order 2160.

4.2. Algebraic degree of a solution

From Theorem 3.3 and Theorem 3.8 of Ulmer (1992) we see that for the computation of an algebraic logarithmic derivative u = y′/y of a solution y, the most difficult cases, where the algebraic degree of u is large, are those of a finite primitive unimodular differential Galois group. In this case all solutions will be algebraic and we shall show how to compute the minimal polynomial of such a solution. From Corollary 1.4 we get that the number of coefficients of the minimal polynomials of y and y′/y are the same. In this section we will derive a bound for the algebraic degree of a solution y of L(y) = 0. We note that from k(y′/y) ⊆ k(y) the index of a 1 reducible subgroup of G(L) is a lower bound for the degree of a solution (cf. Lemma 3.1).

For a second order differential equation L(y) = 0 an old result of P. Pepin and L. Fuchs (see Fuchs (1875) and the introduction in Boulanger (1898)) shows that the degree of P (y) is always the largest possible degree, which is the order of G(L). This can be seen in the following way: If L(y) = y′′ − ry (r ∈ k), then the Wronskian y′_{1}y_{2} − y_{1}y′_{2} of two solutions y_{1} and y_{2} of L(y) = 0 is a constant c (see e.g. Kaplansky (1957), p. 40). Thus

\frac{y_{2}}{y_{1}} = c ∫ \frac{1}{y_{1}^{2}}

If y_{1} and y_{2} are algebraic over k, then the integral on the right hand side must be algebraic and thus a rational function in y_{1} over k. Since any solution can be used as y_{1}, we get that for any solution y_{1} of L(y) = 0 the field k(y_{1}) is the full Picard-Vessiot extension K associated to L(y) = 0 . Thus any solution is a primitive element of K and must be of degree G(L).

The following Theorem shows that the above result of Pepin and Fuchs no longer holds for third order differential equations:

Theorem 4.1. Let L(y) be an irreducible third order linear differential equation with Galois group a primitive group G(L) ⊂ SL(3, C). If L(y) = 0 has a liouvillian solution then all solution are algebraic and there is a solution z whose minimal polynomial P (Y) is of the form

Y^{d·m} + a_{m−1}Y^{d·(m−1)} + … + a_{1}Y^{d} + a_{0} (a_{i} ∈ k)

such that one of the following holds:

(i) If G(L) ≅ A_{6}^{SL_{3}}, m = 36 and d = 6.

(ii) If G(L) ≅ A_{5}, m = 6 and d = 2. (iii) If G(L) ≅ A_{5} × C_{3}, m = 6 and d = 6. (iv) If G(L) ≅ G_{168}, , m = 21 and d = 2.

(v) If G(L) ≅ G_{168} × C_{3}, , m = 21 and d = 6. (vi) If G(L) ≅ H_{216}^{SL_{3}}, m = 9 and d = 9. (vii) If G(L) ≅ H_{72}^{SL_{3}}, m = 9 and d = 3. (viii) If G(L) ≅ F_{36}^{SL_{3}}, m = 6 and d = 6.

The above numbers are also the minimal degree of an algebraic solution, except for the group F_{36}^{SL_{3}}, where there also exists a solution whose minimal polynomial is of degree 27 instead of 36.

In order to prove the above result we need a result linking the permutation representation of G(L) on the solutions of P (Y) = 0 and the linear representation of G(L) on the solutions of L(y) = 0.

Lemma 4.2. Let G be a finite group and V a finite irreducible G-module over a field of characteristic 0 with character χ. Let {v_{1}, …, v_{m}} be a G-invariant subset of V and V_{m} be the associated permutation G-module. Then V is a direct summand of V_{m} and

∑_{g∈G} χ(g) · fix(g) = t · |G| > 0,

where fix(g) is the number of vectors in {v_{1}, …, v_{m}} left fixed by g and t is the multiplicity of V in V_{m}.

Proof. By the definition of V_{m} we can identify a basis {z_{1}, …, z_{m}} of V_{m} with the set {v_{1}, …, v_{m}} in such a way that the action of G on these two sets are the same. We now define a map φ : V_{m} ⟼ V by φ(z_{i}) = v_{i} (i ∈ {1, … m}). This clearly defines a G-morphism. According to Maschke’s Theorem, V_{m} is completely reducible and thus the direct sum of the image φ(V_{m}) and the kernel of φ. Since V is an irreducible G-module, φ(V_{m}) = V. This shows that V is a direct summand of V_{m}. If we denote χ_{m} the character of V_{m}, then the orthogonality relations give us

∑_{g∈G} χ(g) · \overline{χ_{m}(g)} = t · |G| > 0.

With respect to the basis {z_{1}, …, z_{m}}, an element g ∈ G has a 1 in the (i, i) place if z_{i} is left fixed and a 0 otherwise. Therefore the trace of this matrix is fix(g) and the formula now follows. □

Proof of Theorem 4.1 For each possible group G(L) we know the index m of a one reducible subgroup H of smallest index (Theorem 3.3). If G(L) ≅ F̸_{36}^{SL_{3}} and G(L) ≅ G_{168} × C_{3}, then using Cayley one can show that all groups H of index m are conjugate so that one can choose any of these groups to perform the computations in the following. If G(L) ≅ F_{36}^{SL_{3}}, then two non conjugate groups H have to be considered. If G(L) ≅ G_{168} × C_{3}, then m = 21, but the groups of index 21 which have an irreducible representation of degree 3 cannot be 1-reducible, since from the character tables we get that in such a representation there is an element of order 2 whose trace is −1, but in any 1-dimensional representation of G_{168} × C_{3} this element has a trace 1 and in any 2-dimensional representation this element has trace 2. Those subgroups of G_{168} × C_{3} of index 21 which are always 1-reducible are all congugate and we can choose any of them.

From Theorem 3.3 we get that L(y) = 0 has a solution y such that [k(y′/y) : k] = m. From Corollary 1.4 we get that the differential Galois group of k(y)/k(y′/y) is cyclic. If K denotes the PVE of L(y) = 0, then the extension k(y) is an intermediate field of K/k(y′/y). The differential Galois group of K/k(y′/y) is isomorphic to H and the Galois group of k(y)/k(y′/y) is a cyclic factor of H (cf. Corollary 1.4). Since we know the possible groups H, using Cayley we can compute the order d of all possible cyclic factor groups of H. We get:

i) If G(L) ≅ A_{6}^{SL_{3}}, A_{5} × C_{3}, G_{168} × C_{3}, H_{72}^{SL_{3}} or F_{36}^{SL_{3}}, then d belongs to {1, 2, 3, 6}. ii) If G(L) ≅ A_{5} or G_{168} then d belongs to {1, 2}. iii) If G(L) ≅ H_{216}^{SL_{3}}, then d belongs to {1, 3, 9}.

From Corollary 1.4 we get that the minimal polynomial of y must be of the form:

P (Y) = Y^{d·m} + a_{m−1}Y^{d(m−1)} + … + a_{1}Y^{d} + a_{0} (a_{j} ∈ k)

Since G(L) ⊆ SL(3, C) is irreducible, the splitting field of P (Y) = 0 is a Picard Vessiot extension for L(y) = 0 (cf. Ulmer (1992), Corollary 2.4). Thus G(L) is the (classical) Galois group of P (Y) = 0 (cf. Ulmer (1992), Lemma 1.1) and must have a faithfull representation as a transitive permutation group of degree d · m. Using Cayley one can see that for a faithfull transitive representation of

(i) A_{6}^{SL_{3}} of degree 36 ≤ j ≤ (6 · 36), j must be 45, 90, 108 135, 180 or 216. (ii) A_{5} of degree 6 ≤ j ≤ (2 · 6), j must be 6, 10 or 12. (iii) A_{5} × C_{3} of degree 6 ≤ j ≤ (6 · 6), j must be 15, 18, 30 or 36. (iv) G_{168} of degree 21 ≤ j ≤ (2 · 21), j must be 21, 24, 28 or 42.

(v) G_{168} × C_{3} of degree 21 ≤ j ≤ (6 · 21), j must be 21, 24, 42, 63, 72, 84 or 126. (vi) H_{216}^{SL_{3}} of degree 9 ≤ j ≤ (9 · 9), j must be 81. (vii) H_{72}^{SL_{3}} of degree 9 ≤ j ≤ (6 · 9), j must be 27, 36 or 54. (viii) F_{36}^{SL_{3}} of degree 6 ≤ j ≤ (6 · 6), j must be 18, 27 or 36.

For each possible j above, we construct all transitive permutation representations of the corresponding differential Galois group G(L) (note that it is enougth to let G(L) act on the cosets of one representant of each set of conjugate subgroups) and compute the corresponding permutation representation P with character χ_{P} of degree j. If for all irreducible characters χ of degree 3 of G, the scalar product χ and χ_{P} is 0, then from Lemma 4.2 we get that G(L) cannot permute the j solutions according to P. If this is the case for all transitive permutation respresentations of degree j, then we can exclude the possibility j.

For the groups A_{6}^{SL_{3}}, A_{5}, G_{168}, G_{168} × C_{3} and H_{216}^{SL_{3}} only the numbers j = d · m given in the Theorem are still possible and must thus be minimal.

For the groups H_{72}^{SL_{3}} the values 3 · 9 and 6 · 9 and for F_{36}^{SL_{3}} the values 27 and 6 · 6 are still possible. We will show by examining the matrices defining the three dimensional representations that in both cases there is always a solution of degree 27 which will then be the minimal degree.

A representation G in SL(3, C) of the group H_{72}^{SL_{3}} is generated by the matrices S_{1}, T, V and UV U^{−1} given in section 4.1.2. A 1-reducible subgroup H of index 9 is generated by S_{1}T UV U^{−1}, S_{1}T UV U^{−1}V and S_{1}T^{−1}S_{1}^{−1}T. The solution y = (0, −1, 1) is left invariant by the first two matrices and sent to ωy by the last matrix. This shows that y^{3} is left invariant by H and, since by the above no solution of lower degree is possible, that for this representation of G(L) ≅ H_{72}^{SL_{3}} the equation L(y) = 0 has a solution of the form stated. Since we can map this representation of H_{72}^{SL_{3}} to a non equivalent one by the automorphism σ : ε ↦ ε^{2} of the Galois group of Q(ε)/Q and that the given eigenvector and eigenvalues of the elements of H belong to Q(ε), it follows that for the representation σ(G) the group σ(H) will have the above properties of H. This shows that a differential equation whose Galois group is isomorphic to H_{72}^{SL_{3}} will always have a solution of degree 3 · 9 and of the form stated.

A representation G in SL(3, C) of the group F_{36}^{SL_{3}} is generated by S_{1}, T and V of section 4.1.2. The cyclic subgroup F generated by S_{1}^{−1}T^{−1}S_{1}V^{−1} is of order 4 and index 27. The solution y = (−ε^{3}, 1, 0) is left invariant by F. Since from the above we know that no solution of lower degree is possible, the degree of y must be 27. This shows that for this representation of F_{36}^{SL_{3}}, a solution of degree 27 always exists. Since we can map this representation of H_{72}^{SL_{3}} to a non equivalent one by the automorphism σ : ε ↦ ε^{2} of the Galois group of Q(ε)/Q and that the given eigenvector and eigenvalues of the elements of H belong to Q(ε), it follows that for the representation σ(G) the group σ(F) will have the above properties of F. This shows that a differential equation whose Galois group is isomorphic to F_{36}^{SL_{3}} will always have a solution of degree 27. On the other hand, since 6 · 6 is the only possibility left of the form 6 · i, there will also always exists a solution of degree 6 · 6. This completes the proof of Theorem 4.1.

We also note that the result of Fuchs and Pepin, which states that for second order equations the degree of an algebraic solution always corresponds to the order of the primitive unimodular group G(L) (i.e. any solution is a primitive element of the PVE) can also be proven using the method above.

4.3. Decomposition of the coefficients in terms of (semi-)invariants

In this section as in the previous one we deal with the case of a differential equation L(y) = 0 whose Galois group G(L) is a finite primitive unimodular group. We show how the coefficients of the minimal polynomial of a solution of L(y) = 0 can be computed using a basis of the ring of invariants of G(L) (see e.g. Cox, Little and O’Shea (1992), Chapter 7). This approach is not new and has been successfully used in Fuchs (1875) for the case of second order differential equation. In this section we will describe this procedure and show how it can be generalized to higher order equations.

Let L(y) = 0 be a differential equation of degree n with finite primitive differential Galois group G(L) ⊆ SL(n, C). Let {y_{1}, y_{2}, …, y_{n}} be a basis of the solution space of L(y) = 0 corresponding to the representation G(L), H a 1-reducible subgroup of G(L) of minimal index m and T a set of left coset representatives of H in G(L). Let y be a common eigenvector of H, then by Corollary 1.4 we get that the minimal polynomial of y is the form

P (Y) = ∏_{σ∈T} (Y^{d} − (σ(y))^{d})
= Y^{d·m} + α_{d(m−1)}Y^{d(m−1)} + … + α_{d}Y^{d} + α_{0} (m = |G(L)/H|),

where any coefficient α_{i} is a polynomial of degree d·m−i in {y_{1}, y_{2}, …, y_{n}}. By construction these polynomials are invariant under the action of G(L) and thus can be expressed in terms of the elements of a basis of the ring of invariants (or semi-invariants) of G(L). This can be done in the following way:

i) Choose a representation G(L) ⊆ SL(n, C) of the differential Galois group of L(y) = 0 (i.e. fix a basis of the solution space) and compute a basis {b_{1}(y_{1}, …, y_{n}), …, b_{j}(y_{1}, …, y_{n})} of the ring of invariants of G(L).

ii) Compute a 1-reducible subgroup H of minimal index of G(L) and a common eigen-

vector y for the matices of H. iii) Compute a set of left coset representatives T of H in G(L). iv) Compute the polynomial

P (Y) = ∏_{σ∈T} (Y^{d} − σ(z)^{d})
= Y^{d·m} + γ_{m−1}(y_{1}, …, y_{n})Y^{d·(m−1)} + … + γ_{0}(y_{1}, …, y_{n}),

where m = |G(L)/H| and d is the index of a normal subgroup F of H such that H/F is a cyclic group.

v) Using the Gröbner basis algorithm, express γ_{i}(y_{1}, …, y_{n}) in terms of polynomials in the invariants {b_{1}(y_{1}, …, y_{n}), … b_{j}(y_{1}, …, y_{n})} (cf. Cox, Little and O’Shea (1992), Chapter 7, §3, Prop. 7. In practice an Ansatz turned out to be more effective).

The above computation has to be done once for the finitely many primitive finite subgroups of SL(n, C).

In the following we will use semi-invariants of G(L) to represent the coefficients of P (Y), since they are usually of lower degree.

4.3.1. Second order equations

For second order equations the decomposition of the coefficients α_{i}(y_{1}, …, y_{n}) in terms of {b_{1}(y_{1}, …, y_{n}), … b_{j}(y_{1}, …, y_{n})} has been computed by Fuchs in Fuchs (1875) for A_{4}^{SL_{2}} and S_{4}^{SL_{2}}. In the following we show that the result for second order equations can be obtained by our approach and restate Fuchs’ results.

The second order case is simplified by the following facts:

i) According to the result of Pepin and Fuchs, we must have d · m = |G(L)| (i.e. any solution is a primitive element of the PVE associated to L(y) = 0).

ii) Any one reducible subgroup is abelian and (assuming G(L) unimodular) is a cyclic group, so that a common eigenvector is just an eigenvector of a generator.

We have to deal with each group separetely.

The tetrahedral group A_{4}^{SL_{2}}:

We consider the algebraic extension Q(ω) of the rational numbers, where ω is a root of ω^{4} − 2ω^{3} + 5ω^{2} − 4ω + 1. We have i = \sqrt{−1} = −2ω^{3} + 3ω^{2} − 9ω + 4 and \sqrt{−3} = 4ω^{3} − 6ω^{2} + 16ω − 7. The group A_{4}^{SL_{2}} is generated by the matrices S and T of section 4.1.1 which are defined in Q(ω).

We denote {y_{1}, y_{2}} the basis corresponding to the above representation. In this representation, the ring of semi-invariants of A_{4}^{SL_{2}} is generated by (see Miller, Blichfeld and Dickson (1938), p. 224):

I_{1} = y_{1}^{4} + 2\sqrt{−3}y_{1}^{2}y_{2}^{2} + y_{2}^{4},
I_{2} = y_{1}y_{2}(y_{1}^{4} − y_{2}^{4}),
I_{3} = y_{1}^{4} − 2\sqrt{−3}y_{1}^{2}y_{2}^{2} + y_{2}^{4},

together with the relation 12\sqrt{−3}I_{2}^{2}−I_{1}^{3}+I_{3}^{3} = 0. We will only need I_{1} and I_{2} to represent the coefficients of P (Y).

A maximal 1-reducible subgroup of A_{4}^{SL_{2}} is the cyclic group of order 6 generated by the matrix T S^{−1} which has an eigenvector z = (ω^{3} − ω^{2} + 3ω − 1)y_{1} + y_{2}. A set of left coset representatives T of < T S^{−1} > in A_{4}^{SL_{2}} is

{id, S, S^{−1}, T S}

and the minimal polynomial of z is given by:

P (Y) = ∏_{σ∈T} (Y^{6} − σ(z)^{6}),

which is:

Y^{24} + 48ω^{3}(y_{2}y_{1}^{5} − y_{2}^{5}y_{1}) Y^{18} + ((−780ω^{2} + 780ω − 210)y_{1}^{12} + (−1824ω^{3} + 2736ω^{2} − 1368ω + 228)y_{2}^{2}y_{1}^{10}

+(25740ω^{2} − 25740ω + 6930)y_{2}^{4}y_{1}^{8} + (3648ω^{3} − 5472ω^{2} + 2736ω − 456)y_{2}^{6}y_{1}^{6}

\begin{matrix}+(25740ω^{2} − 25740ω + 6930)y_{2}^{8}y_{1}^{4} \\ +(−1824ω^{3} + 2736ω^{2} − 1368ω + 228)y_{2}^{10}y_{1}^{2} + (−780ω^{2} + 780ω − 210)y_{2}^{12}) Y^{12} \end{matrix}

\begin{matrix}+ ((6816ω^{3} + 2496ω^{2} − 7488ω + 2496)y_{2}y_{1}^{17} \\ +(97088ω^{3} − 326784ω^{2} + 255744ω − 61568)y_{2}^{3}y_{1}^{15} \end{matrix}

+(−231744ω^{3} − 84864ω^{2} + 254592ω − 84864)y_{2}^{5}y_{1}^{13} +(−291264ω^{3} + 980352ω^{2} − 767232ω + 184704)y_{2}^{7}y_{1}^{11} +(291264ω^{3} − 980352ω^{2} + 767232ω − 184704)y_{2}^{11}y_{1}^{7} +(231744ω^{3} + 84864ω^{2} − 254592ω + 84864)y_{2}^{13}y_{1}^{5} +(−97088ω^{3} + 326784ω^{2} − 255744ω + 61568)y_{2}^{15}y_{1}^{3} +(−6816ω^{3} − 2496ω^{2} + 7488ω − 2496)y_{2}^{17}y_{1}) Y^{6}

+ ((780ω^{2} − 780ω + 209)y_{1}^{24} + (−8688ω^{3} + 13032ω^{2} − 6672ω + 1164)y_{2}^{2}y_{1}^{22} +(−135720ω^{2} + 135720ω − 36366)y_{2}^{4}y_{1}^{20} +(304080ω^{3} − 456120ω^{2} + 233520ω − 40740)y_{2}^{6}y_{1}^{18} +(1134900ω^{2} − 1134900ω + 304095)y_{2}^{8}y_{1}^{16} +(−295392ω^{3} + 443088ω^{2} − 226848ω + 39576)y_{2}^{10}y_{1}^{14} +(1194960ω^{2} − 1194960ω + 320188)y_{2}^{12}y_{1}^{12} +(−295392ω^{3} + 443088ω^{2} − 226848ω + 39576)y_{2}^{14}y_{1}^{10} +(1134900ω^{2} − 1134900ω + 304095)y_{2}^{16}y_{1}^{8} +(304080ω^{3} − 456120ω^{2} + 233520ω − 40740)y_{2}^{18}y_{1}^{6} +(−135720ω^{2} + 135720ω − 36366)y_{2}^{20}y_{1}^{4} +(−8688ω^{3} + 13032ω^{2} − 6672ω + 1164)y_{2}^{22}y_{1}^{2} +(780ω^{2} − 780ω + 209)y_{2}^{24})

Using the Gröbner basis algorithm in Axiom (cf. Jenks and Sutor (1992)) we can represent the coefficients of P (Y) as polynomials in the invariants I_{1} and I_{2}:

Y^{24} + (48ω^{3}I_{2}) Y^{18}

+ ((−780ω^{2} + 780ω − 210)I_{1}^{3} + (−6144ω^{3} + 9216ω^{2} − 4608ω + 768)I_{2}^{2}) Y^{12}

+ ((6816ω^{3} + 2496ω^{2} − 7488ω + 2496)I_{2}I_{1}^{3}

+ (167936ω^{3} − 565248ω^{2} + 442368ω − 106496)I_{2}^{3}) Y^{6}

+(780ω^{2} − 780ω + 209)I_{1}^{6}

The above representation shows that the semi-invariant I_{2} is a rational function and thus an invariant, and that I_{1} is the cube root of a rational function. This last fact can also be derived using the one dimensional characters of A_{4}^{SL_{2}} (cf. Lemma 1.6), since in the decomposition of the character of the sixth symmetric product of a two dimensional character of A_{4}^{SL_{2}} there is exactly 1 one dimensional character ϕ which is of order 3 (i.e. ϕ^{3} = 1, ϕ ≠ 1). This also shows that, up to a constant, there is exactly one solution of L^{◯ s 6}(y) = 0 which is the cube root of a rational function.

We note that there are other minimal polynomials of solutions that can be derived using either another representation of the group, another eigenvector of a cyclic subgroup of order 4 or another cyclic subgroup of order 4. In Fuchs (1878) p. 21 the following decomposition of a polynomial P (Y) is given:

Y^{24} − 3φ Y^{18} + (−3χ_{1}^{3} − 78χ^{3}) Y^{12} + (−χ_{1}^{3}φ + 10χ^{3}φ) Y^{6} − 27χ^{6},

where χ is an invariant of degree 4 of A_{4}^{SL_{2}}, χ_{1} (the Hessian of χ) is another invariant of degree 4 and φ (the jacobian of χ_{1} and χ) is an invariant of degree 6. We note that our invariants differ from those of Fuchs because we select a different basis for our representation.

The octahedral group S_{4}^{SL_{2}}:

We consider the algebraic extension Q(i, \sqrt{2}) of the rational numbers. The group S_{4}^{SL_{2}} is generated by the matrices S and U of section 4.1.1.

We denote {y_{1}, y_{2}} the basis corresponding to the above representation. In this representation, the ring of semi-invariants of S_{4}^{SL_{2}} is generated by (see Miller, Blichfeld and Dickson (1938), p. 224):

I_{1} = y_{1}y_{2}(y_{1}^{4} − y_{2}^{4}),
I_{2} = y_{1}^{8} + 14y_{1}^{4}y_{2}^{4} + y_{2}^{8},
I_{3} = y_{1}^{12} − 33y_{1}^{8}y_{2}^{4} − 33y_{1}^{4}y_{2}^{8} + y_{2}^{12},

Together with the relation 108I_{1}^{4} − I_{2}^{3} + I_{3}^{2} = 0. We will only need I_{1} and I_{2} to represent the coefficients.

A maximal 1-reducible subgroup of S_{4}^{SL_{2}} is the cyclic group of order 8 generated by the above matrix U and y_{1} is an eigenvector of U. A set of left coset representatives T of < U > in S_{4}^{SL_{2}} is

{id, S, US, U^{2}S, S^{−1}U, SUS}

and the minimal polynomial of z is given by:

P (Y) = ∏_{σ∈T} (Y^{8} − σ(z)^{8})

Using the Gröbner basis algorithm in Axiom II we can represent the coefficients of P (Y) as polynomials in I_{1} and I_{2}. We get the following representation for P (Y):

Y^{48} − \frac{5}{4}I_{2} Y^{40} + \frac{35}{128}I_{2}^{2} Y^{32} + (\frac{1351}{128}I_{1}^{4} − \frac{25}{1024}I_{2}^{3}) Y^{24}

+ (\frac{265}{1024}I_{2}I_{1}^{4} + \frac{65}{65536}I_{2}^{4}) Y^{16} + (\frac{39}{32768}I_{2}^{2}I_{1}^{4} − \frac{1}{65536}I_{2}^{5}) Y^{8} + \frac{1}{65536}I_{1}^{8}

The above representation shows that I_{2} is a rational function and that I_{1} is the fourth root of a rational function. Since in the decomposition of the character of the sixth symmetric product of a faithfull two dimensional character of S_{4}^{SL_{3}} there is exactly 1 one dimensional character ϕ which is of order 2 (i.e. ϕ^{2} = 1, ϕ ≠ 1), we get that I_{1} is the square root of a rational function (cf. Lemma 1.6). This is also derived by L. Fuchs (cf. Fuchs (1878) p. 13). The decomposition of the characters also shows that there will be, up to a constant, exactly one rational solution of L^{◯ s 6}(y) = 0 and exactly one solution of L^{◯ s 6}(y) = 0 which is the square root of a rational function, and thus, up to a constant, exactly one choice for I_{1} and I_{2}.

The above polynomial was obtained using only one representation of S_{4}^{SL_{2}}, which has in fact two faithfull non equivalent representations in SL(2, C). Since there is an automorphism σ of the Galois group of Q(i, \sqrt{2})/Q sending this representation to a non equivalent representation, σ(P (Y)) would be the minimal polynomial of a solution for this representation. Since the above decomposition of P (Y) contains only rational coefficients, any representation of S_{4}^{SL_{2}} in SL(2, C) will lead to a solution whose minimal polynomial is of the above form.

We again note that there are other minimal polynomials of solutions that can be derived using either another representation of the group, another eigenvector of a cyclic subgroup of order 6 or another cyclic subgroup of order 6. In Fuchs (1878) p. 21 the following decomposition of a polynomial P (Y) is given:

Y^{48} − 20χ_{1} Y^{40} + 70χ_{1}^{2} Y^{32} + (−100χ_{1}^{3} − 14 · 3088χ^{4}) Y^{24}
+(65χ_{1}^{4} + 40 · 424χ_{1}χ^{4}) Y^{16} + (−16χ_{1}^{5} − 1248χ_{1}^{2}χ^{4}) Y^{8} − 16^{2}χ^{8},

where χ is an invariant of degree 6 of S_{4}^{SL_{2}} and χ_{1} (the Hessian of χ) is another invariant of degree 8.

The icosahedral group A_{5}^{SL_{2}}:

We consider the algebraic extension Q(ξ) of the rational numbers, where ξ^{5} = 1. The group A_{5}^{SL_{2}} is generated by the matrices S′, U′ and T′ of section 4.1.1.

We denote {y_{1}, y_{2}} the basis corresponding to the above representation. In this representation, the ring of invariants of S_{4}^{SL_{2}} is generated by (see Miller, Blichfeld and Dickson (1938), p. 224):

I_{1} = y_{1}y_{2}(y_{1}^{10} + 11y_{1}^{5}y_{2}^{5} − y_{2}^{10}),
I_{2} = −y_{1}^{20} − y_{2}^{20} + 228(y_{1}^{15}y_{2}^{5} − y_{1}^{5}y_{2}^{15}) − 494y_{1}^{10}y_{2}^{10},
I_{3} = y_{1}^{30} + y_{2}^{30} + 522(y_{1}^{25}y_{2}^{5} − y_{1}^{5}y_{2}^{25}) − 10005(y_{1}^{20}y_{2}^{10} + y_{1}^{10}y_{2}^{20}),

Together with the relation I_{3}^{2} + I_{2}^{3} − 1728I_{1}^{5} = 0.

A maximal 1-reducible subgroup of A_{5}^{SL_{2}} is the cyclic group of order 10. Such a group H is generated by the matrix U^{2}S, which has an eigenvector z = y_{1}. A set of left coset representatives T of H in A_{5}^{SL_{2}} is

{id, U, T^{−1}, UT^{−1}, ST^{−1}, T ST, S^{−2}T, T S^{−2}T, UST S^{−2}, US^{−1}T, S^{2}T^{−1}, US^{−2}T^{−1}}

and the minimal polynomial of z is given by:

P (Y) = ∏_{σ∈T} (Y^{10} − σ(z)^{10})

Using an Ansatz we can represent the coefficients of P (Y) as polynomials in the invariants I_{1}, I_{2} and I_{3} we get the following representation for P (Y):

Y^{120} + \frac{374}{625}I_{2} Y^{100} − \frac{1001}{3125}I_{3} Y^{90} − \frac{142373}{1953125}I_{2}^{2} Y^{80}

+\frac{78254}{9765625}I_{2}I_{3}Y^{70} + (\frac{832814147}{5273437500000}I_{2}^{3} − \frac{8910209}{42187500000}I_{3}^{2}) Y^{60}

−\frac{81631}{30517578125}I_{3}I_{2}^{2} Y^{50} + (\frac{−39788034}{152587890625}I_{1}^{5}I_{2} − \frac{158499}{19073486328125}I_{2}^{4}) Y^{40}

+ (\frac{−611864}{762939453125}I_{1}^{5}I_{3} + \frac{1254}{95367431640625}I_{2}^{3}I_{3}) Y^{30}

+ (\frac{103862}{476837158203125}I_{1}^{5}I_{2}^{2} + \frac{3124}{298023223876953125}I_{2}^{5}) Y^{20}

+ (\frac{4}{2384185791015625}I_{1}^{5}I_{2}I_{3} − \frac{1}{298023223876953125}I_{2}^{4}I_{3}) Y^{10}

+\frac{1}{298023223876953125}I_{1}^{10}

Since A_{5}^{SL_{2}} has only one irreducible character of degree 1, any invariant will be rational and thus not hard to compute. Since there is no polynomial relation between I_{1}, I_{2} and I_{3}, we get that, up to a constant multiple, there will be one polynomial solution of L^{◯ s 12}(y) = 0, L^{◯ s 20}(y) = 0 and L^{◯ s 30}(y) = 0. This can also be derived from the decomposition of the characters of the 12-th, 20-th and 30-th symmetric product of the irreducible characters of degree 3 of A_{5}^{SL_{2}}.

As for S_{4}^{SL_{2}} there is an element σ of the Galois group of Q(ξ)/Q sending the above representation in a non equivalent representation. Thus σ(P (Y)) would be the minimal polynomial of a solution for this representation. Since the above decomposition of P (Y) contains only rational coefficients, any representation of A_{5}^{SL_{2}} in SL(2, C) will lead to a solution whose minimal polynomial is of the above form.

In Fuchs (1878) (cf. p. 16) no explicit decomposition of a polynomial P (Y) is given.

4.3.2. Third order equations

For third order equations it is not the case that any solution of L(y) = 0 is a primitive element of the PVE extension associated to L(y) = 0. For G(L) ≅ H_{216}^{SL_{3}} the order of the minimal polynomial of a primitive element (which always exists) is 648, while from Corollary 4.1 there is a solution whose monic minimal polynomial is of order 81 where at most 9 non zero coefficients have to be computed.

In this section we present results involving groups, so that only the decomposition of the coefficients in terms of the fundamental invariants remains to be done. To illustrate the procedure for third order differential equations, we perform the decomposition of the minimal polynomial in the case G(L) ≅ A_{5}.

We consider each group separetely:

The Valentiner group A_{6}^{SL_{3}}:

The group A_{6}^{SL_{3}} is generated by the matrices E_{1}, E_{2}, E_{3} and E_{4} given in section 4.1.2. The 1-reducible subgroups of index 36 are all conjugate. Such a 1-reducible group H is generated by E_{1}, (E_{3}E_{1}^{2}E_{4}E_{1}^{−1})^{2} and E_{2}. If we denote {y_{1}, y_{2}, y_{3}} the basis of the solution space corresponding to the above representation, then the solution z = y_{1} spanns a one dimensional invariant subspace of H. A set of left coset representatives T of H in A_{6}^{SL_{3}} is

{id, E_{3}, E_{4}, E_{3}E_{4}, E_{1}E_{3}, E_{3}E_{1}E_{3}, E_{1}E_{4}, E_{4}E_{1}E_{4}, E_{1}^{−2}E_{3}, E_{1}^{−2}E_{4}, E_{4}E_{1}^{−1}E_{3},
E_{1}E_{4}E_{1}^{−1}E_{3}, E_{1}^{2}E_{4}E_{1}^{−1}E_{3}, E_{1}^{2}E_{4}E_{1}^{−1}, E_{3}E_{1}^{2}E_{4}E_{1}^{−1}, E_{1}^{−1}E_{3}E_{1}^{2}E_{4}E_{1}^{−1},
E_{1}^{−1}E_{3}E_{4}, E_{4}E_{1}^{−1}E_{3}E_{4}, E_{1}E_{4}E_{1}^{−1}E_{3}E_{4}, E_{2}E_{1}^{2}E_{4}E_{1}^{−1}E_{3}, E_{1}^{2}E_{3}, E_{1}E_{4}E_{3},
E_{4}E_{1}E_{4}E_{3}, E_{1}^{−2}E_{4}E_{3}, E_{4}E_{1}^{−2}E_{4}E_{3}, E_{2}E_{1}E_{4}E_{1}^{−2}E_{3}, E_{4}E_{1}^{2}E_{3}E_{4}E_{1}^{−1},
E_{3}E_{1}^{−2}E_{3}E_{4}, E_{1}E_{3}E_{1}^{−2}E_{4}, E_{1}^{2}E_{3}E_{4}, E_{3}E_{1}^{−1}E_{3}E_{4}, E_{2}E_{4}E_{1}^{−1}E_{3},
E_{1}^{−1}E_{4}E_{1}E_{3}, E_{3}E_{1}^{2}E_{3}E_{4}E_{1}^{−1}, E_{1}E_{4}E_{1}^{−1}E_{3}E_{4}E_{1}E_{3}, E_{1}E_{4}E_{1}^{2}E_{4}E_{3}E_{1}^{−1}}

and the minimal polynomial of z is given by:

P (Y) = ∏_{σ∈T} (Y^{6} − σ(z)^{6})

The simple group A_{5}:

The group A_{5} is generated by the matrices E_{1}, E_{2} and E_{3} given in section 4.1.2. The 1reducible subgroups of index 6 are all conjugate. Such a 1-reducible group H is generated by E_{1} and E_{2}. If we denote {y_{1}, y_{2}, y_{3}} the basis of the solution space corresponding to the above representation, then the solution z = y_{1} spanns a one dimensional invariant subspace of H. A set of left coset representatives T of H in A_{5} is

{id, E_{3}, E_{1}E_{3}, E_{3}E_{1}E_{3}, E_{1}^{−2}E_{3}, E_{1}^{2}E_{3}}

and the minimal polynomial of z is given by:

P (Y) = ∏_{σ∈T} (Y^{2} − σ(z)^{2})

In the given representation (cf. reference to F. Klein in Miller, Blichfeld and Dickson (1938), p. 254), the ring of invariants of A_{5} ⊂ SL(3, C) is generated by:

I_{1} = y_{1}^{2} + y_{2}y_{3},
I_{2} = 8y_{1}^{4}y_{2}y_{3} − 2y_{1}^{2}y_{2}^{2}y_{3}^{2} + y_{2}^{3}y_{3}^{3} − y_{1}(y_{2}^{5} + y_{3}^{5})
I_{3} = 320y_{1}^{6}y_{2}^{2}y_{3}^{2} − 160y_{1}^{4}y_{2}^{3}y_{3}^{3} + 20y_{1}^{2}y_{2}^{4}y_{3}^{4} + 6y_{2}^{5}y_{3}^{5}
−4y_{1}(y_{2}^{5} + y_{3}^{5})(32y_{1}^{4} − 20y_{1}^{2}y_{2}y_{3} + 5y_{2}^{2}y_{3}^{2}) + y_{2}^{10} + y_{3}^{10},

and an invariant I_{4} of degree 15. Using an Ansatz we can represent the coefficients of P (Y) as polynomials in the invariants I_{1}, I_{2} and I_{3}. We get the following representation for P (Y):

Y^{12} − 2I_{1} Y^{10} + \frac{7}{5}I_{1}^{2} Y^{8} + (−\frac{12}{25}I_{1}^{3} + \frac{2}{25}I_{2}) Y^{6} + (\frac{11}{125}I_{1}^{4} − \frac{6}{125}I_{1}I_{2}) Y^{4}

+ (−\frac{26}{3125}I_{1}^{5} + \frac{6}{625}I_{1}^{2}I_{2} − \frac{64}{625 · 320}I_{3}) Y^{2} + (\frac{1}{3125}I_{1}^{6} + \frac{1}{3125}I_{2}^{2} − \frac{2}{3125}I_{2}I_{1}^{3})

The group A_{5} × C_{3}:

This group is the direct product of the previous group with the center of SL(3, C) generated by Z. The 1-reducible subgroups of index 6 are all conjugate. From the previous case we get a 1-reducible group H × C_{3} generated by E_{1}, E_{2} and Z which has the same set of left coset representatives T in A_{5} × C_{3} has H in A_{5}. Since C_{3} consists of scalar multiplications, the common eigenvector z = y_{1} of H given in the previous case will be a common eigenvector for H × C_{3}. The function z^{3} of the PVE is left invariant by the normal subgroup C_{3}, which shows that k(z^{3})/k has Galois group A_{5}. Since z′/z is left fixed by H, the same holds for (z^{3})′/z^{3}. From Corollary 1.4 and the bound for d computed in Theorem 4.1 we get that the minimal polynomial P_{3}(Y) of z^{3} over k is given by:

∏_{σ∈T} (Y^{2} − σ(z^{3})^{2})

The following polynomial P (Y) has z as a solution:

P (Y) = ∏_{σ∈T} (Y^{6} − σ(z)^{6})

and comparing degrees (cf. Theorem 4.1), we get that P (Y) is the minimal polynomial of z.

The simple group G_{168}:

The group G_{168} is generated by the matrices S, T and R given in section 4.1.2. The 1-reducible subgroups of index 21 are all conjugate. Such a 1-reducible group H is generated by S^{−2}RS and RS^{−1}RT S. If we denote {y_{1}, y_{2}, y_{3}} the basis of the solution space corresponding to the above representation, then the solution

z = (β^{5} + β^{4} + β^{2} + 1) y_{1} + (β^{5} + β) y_{2} + y_{3}

spanns a one dimensional invariant subspace of H. A set of left coset representatives T of H in G_{168} is

{id, T, T^{−1}, S^{−1}T^{−1}, S^{−2}, S^{−1}, SR, RSR, RS^{−1}RS^{−1}, RS^{−3}, S, T S^{−1}, S^{−1}T,
SRT, RSRT, RS^{−1}RS^{−1}T, ST SR, RST SR, RT S^{2}, T S^{2}, RSRT^{−1}}

and the minimal polynomial of z is given by:

P (Y) = ∏_{σ∈T} (Y^{2} − σ(z)^{2})

The group G_{168} × C_{3}:

This group is the direct product of the previous group with the center of SL(3, C) generated by Z. The 1-reducible subgroups of index 21 are not all conjugate. From the previous case we get a 1-reducible group H × C_{3} generated by S^{−2}RS, RS^{−1}RT S and Z which has the same set of left coset representatives T has H in G_{168}. Since C_{3} consists of scalar multiplications, the common eigenvector z of H given in the previous case will be a common eigenvector for H × C_{3}. As in the case A_{5} × C_{3} we get the following minimal polynomial for z over k.

P (Y) = ∏_{σ∈T} (Y^{6} − σ(z)^{6}).

The group H_{216}^{SL_{3}}:

The group H_{216}^{SL_{3}} is generated by the matrices S_{1}, T, V and UV given in section 4.1.2. The 1-reducible subgroups of index 9 are all conjugate. Such a 1-reducible group H is generated by U^{2}V^{−1}S_{1} and V^{−1}U^{2}S_{1}. If we denote {y_{1}, y_{2}, y_{3}} the basis of the solution space corresponding to the above representation, then the solution −ε^{3}y_{1} + y_{2} spanns a one dimensional invariant subspace of H. A set of left coset representatives T of H in H_{216}^{SL_{3}} is

{id, V, V^{2}, V^{−1}, T^{−1}, S^{−1}, V^{−1}T^{−1}, V S^{−1}, V^{2}UT^{−1}S}

and the minimal polynomial of z is given by:

P (Y) = ∏_{σ∈T} (Y^{9} − σ(z)^{9})

The group H_{72}^{SL_{3}}:

The group H_{72}^{SL_{3}} is generated by the matrices S_{1}, T, V and UV U^{−1} given in section 4.1.2. The 1-reducible subgroups of index 9 are all conjugate. Such a 1-reducible group H is generated by S_{1}T UV U^{−1}, S_{1}T UV U^{−1}V and S_{1}T^{−1}S_{1}^{−1}T. If we denote {y_{1}, y_{2}, y_{3}} the basis of the solution space corresponding to the above representation, then the solution z = −y_{2} + y_{3} spanns a one dimensional invariant subspace of H. A set of left coset representatives T of H in H_{72}^{SL_{3}} is

{id, T, T^{−1}, S, S^{−1}, T S, U^{−2}V U^{−1}, ST S, U^{−1}V^{−1}U} ,

and the minimal polynomial of z is given by:

P (Y) = ∏_{σ∈T} (Y^{3} − σ(z)^{3})

The group F_{36}^{SL_{3}}:

The group F_{36}^{SL_{3}} is generated by S_{1}, T and V of section 4.1.2. A 1-reducible group H is generated by S^{−1}T, ST S and V^{2}T. If we denote {y_{1}, y_{2}, y_{3}} the basis of the solution space corresponding to the above representation, then the solution y_{1} + y_{2} + ωy_{3} spanns a one dimensional invariant subspace of H. A set of left coset representatives T of H in F_{36}^{SL_{3}} is

{id, V, V^{2}, V^{−1}, T^{−1}, V^{−1}T^{−1}} ,

and the minimal polynomial of z is given by:

P (Y) = ∏_{σ∈T} (Y^{6} − σ(z)^{6})

5. Computing a solution

Let L(y) = 0 be an equation of degree n with coefficients in k and finite primitive Galois group G(L) ⊂ SL(n, C). Then the decomposition of the coefficients of the minimal polynomial P (Y) of a solution given in the previous section reduces the computation of the coefficients of P (Y) to the computation of finitely many semi-invariants of G(L) and of the constants of the semi-invariants, which can be done using a Gröbner basis (for second order equation only gcd computation are needed). In fact, since the coefficients are invariants of G(L), only powers of the semi-invariants which are rational functions have to be computed. This reduces the computations to the computation of rational solutions of some symmetric power of L(y) = 0.

To compute the coefficients of P (Y) one can proceed in the following way:

From the a representation of the minimal polynomial P (Y) of a solution y of L(y) = 0 in terms of the semi-invariants of G one can compute the coefficients of P (Y) in the following way:

i) Compute the set L of i-th symmetric powers L^{◯ s i}(y) of L(y) = 0, where i belongs to the set of orders of the semi-invariants appearing in the decomposition of the coefficients of P (Y).

ii) for each equation in L, compute a solution f_{i} such that f_{i}^{j} ∈ k, where j belongs to the set of orders of the one dimensional characters of G(L) (cf. Lemma 1.6). In general the possible j’s can be further restricted by looking at the decomposition of P (Y) in terms of the semi-invariants of G(L) and noting that the coefficients of P (Y) must be rational.

iii) For each possible value f_{i} (defined up to a constant c_{i}) of the invariants I_{i} obtained, replace I_{i} by c_{i} · f_{i} in P (Y) = 0. This gives a new polynomial Q(Y) = 0 whose coefficients are polynomials in the variables c_{i} over k.

iv) For s ≤ n, using Q(Y) = 0, express the derivatives Y^{(s)} of Y as a polynomial in Y and replace Y^{(s)} by this value in L(y) = 0. This gives a polynomial \overline{P} (Y) = 0 whose coefficients are polynomials in the variables c_{i} over k.

v) compute the rest R(Y) of the division of \overline{P} (Y) by Q(Y) and determine the constants c_{i} by setting all coefficients of R(Y) equal to 0. For i ≥ 2 this can be done using a Gröbner basis.

vi) If a non trivial solution set {c_{i}} is found, then replacing c_{i} by these values in Q(Y) gives the minimal polynomial of a solution of L(y) = 0.

The above method is based on the fact that the polynomial Q(Y) is at least square free. If the cases are considered without knowing the group G(L), then one has to start with the cases where the polynomial to be constructed is of smallest degree (e.g. first the case A_{4}^{SL_{2}}, then S_{4}^{SL_{2}} and then A_{5}^{SL_{2}}) and test if the resuling polynomial is square free. Another possibility would be to use the method given in Singer and Ulmer (1992) to determine the differential Galois group of L(y) = 0 and make sure that the assumption on G(L) is correct.

Example. We now apply the above method to compute the solutions of the differential equation

L(y) = \frac{d^{2} y}{dx^{2}} + (\frac{3}{16x^{2}} + \frac{2}{9(x − 1)^{2}} − \frac{3}{16x(x − 1)}) y = 0

which is also studied in Kovacic (1986), p. 23 and Ulmer (1991), p 452. From the result of Kovacic (1986) we know that G(L) ≅ A_{4}^{SL_{2}}. This could also be computed using the result of Singer and Ulmer (1992) by showing that L^{◯ s 2}(y) is irreducible and L^{◯ s 3}(y) is reducible.

In Section 4.3.1 we decomposed the coefficients of P (Y) and noted that the only semiinvariants present are I_{1}, a semi-invariant of degree 4, and I_{2}, a semi-invariant of degree 6. Furthermore, one can see from the form of the coefficients of P (Y) (or from the orders of the associated characters) that I_{1}^{3} and I_{2} are rational. To compute I_{1} we compute the 4-th symmetric power of L(y) = 0:

L^{◯ s 4}(y) = \frac{d^{5} y}{dx^{5}}

+\frac{5(32x^{2} − 27x + 27)}{36x^{2}(x − 1)^{2}} \frac{d^{3} y}{dx^{3}}

−\frac{5(64x^{3} − 81x^{2} + 135x − 54)}{24x^{3}(x − 1)^{3}} \frac{d^{2} y}{dx^{2}}

+\frac{5(1760x^{4} − 2970x^{3} + 6615x^{2} − 5103x + 1458)}{324x^{4}(x − 1)^{4}} \frac{d y}{dx}

−\frac{5(1792x^{5} − 3780x^{4} + 10395x^{3} − 11718x^{2} + 6561x − 1458)}{324x^{5}(x − 1)^{5}} y

= 0 .

We must find a solution y of this equation such that y^{3} ∈ C(x). Since only I_{1}^{3} is needed, it is enough to compute the rational solution y^{3} of L^{◯ s 12}(y) = 0 (cf. Lemma 1.6). To compute y one can either use the algorithm described at the begining of Section 3.2 or more simply proceed as follows: Let

y = (P (x) ∏_{i}(x − α_{i})^{n_{i}})^{1/3} ,

where P (x) is a polynomial, {α_{i}} are the singular points of L^{◯ s 4}(y) and n_{i} are nonnegative integers. This implies that for each i, n_{1}/3 is an exponent at α_{i} and that the exponent at infinity is \frac{−1}{3} (deg(P) + ∑ n_{i}). Checking the possibilities shows that P (x) must be constant and that I_{2} must be a constant multiple of x(x − 1)^{4/3} or x(x − 1)^{5/3}.

From a similar computation we get the solution x^{2}(x − 1)^{2} for L^{◯ s 6}(y) = 0 (see e.g. Ulmer (1991)).

We set I_{1} = c_{1}x(x − 1)^{4/3} and I_{2} = c_{2}x^{2}(x − 1)^{2} in P (Y) and get a polynomial Q(Y) whose coefficients are polynomials in c_{1} and c_{2}. Using Q(Y) = 0 we write Y′ and Y′′ as a fraction of polynomials in Y and substitute those in L(y) = 0. The numerator of the rational function in Y obtained in this way is a polynomial \overline{P} (Y) having a common solution with Q(Y) = 0. The pseudo remainder R(Y) of \overline{P} (Y) and Q(Y) = 0 is a polynomial in Y of degree at most 23 which must be zero. Thus all coefficients of R(Y) must be zero, which gives a set of polynomials in c_{1}, c_{2} and x. Equating coefficients of powers of x to zero, gives a set of polynomials in c_{1} and c_{2}. We now can compute a Gröbner basis to find the constants c_{1} and c_{2} and get:

[432c_{1}^{2}c_{2}^{4} + c_{1}^{4}, (−48ω^{3} + 72ω^{2} − 192ω + 84)c_{1}^{2}c_{2}^{4} + c_{1}^{3}c_{2}^{2}]

In order for Q(Y) = 0 to be an irreducible polynomial we must have c_{1} ≠ 0 and c_{2} ≠ 0. Setting c_{2} = 1 we get c_{1} = 48ω^{3} − 72ω^{2} + 192ω − 84, which we also write \sqrt{−432}. This gives the following minimal polynomial of a solution of L(y) = 0:

Y^{24} + (48ω^{3}x^{2}(x − 1)^{2}) Y^{18}

+ ((−780ω^{2} + 780ω − 210)(\sqrt{−432})^{3}x^{3}(x − 1)^{4}

+(−6144ω^{3} + 9216ω^{2} − 4608ω + 768)x^{4}(x − 1)^{4}) Y^{12}

+ ((6816ω^{3} + 2496ω^{2} − 7488ω + 2496)(\sqrt{−432})^{3}x^{5}(x − 1)^{6}

+ (167936ω^{3} − 565248ω^{2} + 442368ω − 106496)x^{6}(x − 1)^{6}) Y^{6}

+(780ω^{2} − 780ω + 209)(\sqrt{−432})^{6}x^{6}(x − 1)^{8} □

The above shows that for the Tetrahedral group (as for any second order equation with primitive unimodular Galois group) the computation of the minimal polynomial of a solution is reduced to the computation of two semi-invariants (i.e. solutions of symmetric powers whose power is rational) and two constants. In fact one constant can be choosen arbitrary so that only one constant remains to be computed. This shows that for second order equations the constant can be computed using only gcd computations. In the paper of Fuchs it is shown that using the Hessian H(I_{1}) and Jacobian J(I_{1}, H(I_{1})) of an invariant I_{1} of lowers degree, one gets the other invariants. For the tetrahedral group A_{4}^{SL_{2}} one has (Miller, Blichfeld and Dickson (1938), p. 226)

I_{3} = \frac{H(I_{1})}{48\sqrt{−3}} = \frac{1}{48\sqrt{−3}} | \begin{matrix}\frac{∂^{2}I_{1}}{∂y_{1}∂y_{1}} & \frac{∂^{2}I_{1}}{∂y_{1}∂y_{2}} \\ \frac{∂^{2}I_{1}}{∂y_{2}∂y_{1}} & \frac{∂^{2}I_{1}}{∂y_{2}∂y_{2}} \end{matrix} |

I_{2} = \frac{J(I_{1}, I_{3})}{−32\sqrt{−3}} = \frac{1}{−32\sqrt{−3}} | \begin{matrix}\frac{∂I_{1}}{∂y_{1}} & \frac{∂I_{1}}{∂y_{2}} \\ \frac{∂I_{3}}{∂y_{1}} & \frac{∂I_{3}}{∂y_{2}} \end{matrix} |

Fuchs then shows how, as a function in x, for a given differential equation L(y) = y′′ − r(x)y = 0 the Hessian χ_{1} and Jacobian φ of an invariant χ of minimal degree can be written as a polynomial in χ(x) and derivatives of χ(x). He proves the following relations (Fuchs (1878), pp. 21-22):

χ_{1}(x) = c [(\frac{d log χ(x)}{dx})^{2} + 4\frac{d^{2} log χ(x)}{dx^{2}} − 16r] χ(x)^{2}
φ(x) = \sqrt{−χ_{1}(x)^{3} + 64χ(x)^{3}},

where c is a constant that can be computed (Fuchs (1878), p. 22). If we let χ be the invariant I_{1} used in Miller, Blichfeld and Dickson (1938), these formulas also yield expressions for I_{2} and I_{3}.

This reduces the above to the computation of one semi-invariant of lowest degree and one constant.

We now compute an example of a third order differential equation:

Example. We now apply the above method to compute the solutions of the differential equation

L(y) = \frac{d^{3}y}{dx^{3}} + \frac{21(x^{2} − x + 1)}{25x^{2}(x − 1)^{2}} \frac{dy}{dx} + \frac{21(−2x^{3} + 3x^{2} − 5x + 2)}{50x^{3}(x − 1)^{3}} y ,

which is irreducible and has Galois group A_{5} (cf. Singer and Ulmer (1992), section 5).

Since the invariants of A_{5} ⊂ SL(3, C) are of order 2, 6 and 10, we have to compute rational solutions of the second, 6-th and 10-th symmetric powers of L(y) = 0. The equation L^{◯ s 2}(y) = 0 has no non trivial rational solution. The subspaces of rational solutions of L^{◯ s 6}(y) = 0 and L^{◯ s 10}(y) = 0 each are one dimensional and generated by x^{4}(x − 1)^{4} and x^{6}(x − 1)^{6}(x^{2} − x + 1) respectively.

We set I_{1} = 0, I_{2} = c_{2}x^{4}(x − 1)^{4} and I_{3} = c_{3}x^{6}(x − 1)^{6}(x^{2} − x + 1) in P (Y) and get a polynomial Q(Y) whose coefficients are polynomials in c_{2} and c_{3}. Using Q(Y) = 0 we write Y′ and Y′′ as a fraction of polynomials in Y and substitute those in L(y) = 0. The numerator of the rational function in Y obtained this way is a polynomial \overline{P} (Y) having a common solution with Q(Y) = 0. The pseudo remainder R(Y) of \overline{P} (Y) and Q(Y) = 0 is a polynomial in Y of degree at most 11 which must be zero. Thus all coefficients of R(Y) must be zero, which gives a set of polynomials in c_{2} and c_{3}. We now can compute a Gröbner basis to find the constants c_{2} and c_{3} and get:

[−\frac{1}{256}c_{2}^{3}c_{3}^{3} + c_{2}^{8}, −\frac{1}{256}c_{2}c_{3}^{4} + c_{2}^{6}c_{3}, −\frac{1}{256}c_{3}^{5} + c_{2}^{5}c_{3}^{2}]

Since I_{1} = 0, we must have c_{2} ≠ 0 and c_{3} ≠ 0 in order for Q(Y) = 0 to be an irreducible polynomial with Galois group A_{5}. Setting c_{3} = 1 we get c_{2}^{5} = 1/256. This gives the following minimal polynomial of a solution of L(y) = 0:

Y^{12} + \frac{2}{25} \frac{x^{4}(x − 1)^{4}}{2 ^{5} \sqrt{8}} Y^{6} − \frac{64x^{6}(x − 1)^{6}(x^{2} − x + 1)}{625 · 320} Y^{2} + \frac{1}{3125}(\frac{x^{4}(x − 1)^{4}}{2 ^{5} \sqrt{8}} )^{2}

□

The above method turned out to be very efficient. The computation of a minimal polynomial (assuming known the decomposition of the minimal polynomial in term of the invariants) could be done in both cases in much less than 1 CPU hour on an IBM RISC 6000 with 64MB of main memory. An attempt to compute the minimal polynomial of the logarithmic derivative of a solution of the above second order equation (which is used as an example in Kovacic (1986)) using the implementation in MAPLE of the Kovacic algorithm on a SUN 4 with 20MB of main memory did not give any result within 10 CPU hours.

We note, that the above computations can always be done in an algebraic extension Q(α) of Q containing the finite singular points of L(y) = 0, the entries of the matrices in the transversal T , the coordinates of the common eigenvector of the choosen 1-reducible subgroup H and the coefficients of the invariants of G(L). Thus no additional algebraic extension is needed at runtime. However, as show in both examples, the coefficients of P (Y) = 0 (i.e. the result of the Gröbner basis computation) do not in general belong to Q(α)(x). Note that the final Gröbner basis computation will yield polynomials whose roots generate an extension F of Q(α) such that the coefficients belong to F (x).

A. The character tables of the subgroups of index ≤ 6 of the Schur representation group of A_{5}

We denote G the Schur representation group of A_{5} wihch is of order 120. The table of the irreducible characters of degree ≤ 3 of G produced by Cayley is:

\begin{matrix}class & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ conj & 1 & 1 & 20 & 30 & 12 & 12 & 20 & 12 & 12 \\ order & 1 & 2 & 3 & 4 & 5 & 5 & 6 & 10 & 10 \\ χ_{1} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ χ_{2} & 2 & −2 & −1 & 0 & z_{1} & −1 − z_{1} & 1 & −z_{1} & 1 + z_{1} \\ χ_{3} & 2 & −2 & −1 & 0 & −1 − z_{1} & z_{1} & 1 & 1 + z_{1} & −z_{1} \\ χ_{4} & 3 & 3 & 0 & −1 & −z_{1} & 1 + z_{1} & 0 & −z_{1} & 1 + z_{1} \\ χ_{5} & 3 & 3 & 0 & −1 & 1 + z_{1} & −z_{1} & 0 & 1 + z_{1} & −z_{1} \end{matrix}

where z_{1} = −1 − ω^{2} − ω^{3} and ω = e^{2πi/5}.

The subgroups of G of index ≤ 6 are of index 5 or 6.

The subgroups of index 5 are all congugate and the table of the irreducible characters of degree ≤ 3 of of such a group produced by Cayley is:

\begin{matrix}class & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ conj & 1 & 1 & 4 & 4 & 6 & 4 & 4 \\ order & 1 & 2 & 3 & 3 & 4 & 6 & 6 \\ χ_{1} & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ χ_{2} & 1 & 1 & −1 − J & J & 1 & −1 − J & J \\ χ_{3} & 1 & 1 & J & −1 − J & 1 & J & −1 − J \\ χ_{4} & 2 & −2 & −1 & −1 & 0 & 1 & 1 \\ χ_{5} & 2 & −2 & −J & 1 + J & 0 & J & −1 − J \\ χ_{6} & 2 & −2 & 1 + J & −J & 0 & −1 − J & J \\ χ_{7} & 3 & 3 & 0 & 0 & −1 & 0 & 0 \end{matrix}

where J = e^{2πi/3}.

The subgroups of index 6 are all congugate and the table of the irreducible characters of degree ≤ 3 of of such a group produced by Cayley is:

\begin{matrix}class & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ conj & 1 & 1 & 5 & 5 & 2 & 2 & 2 & 2 \\ order & 1 & 2 & 4 & 4 & 5 & 5 & 10 & 10 \\ χ_{1} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ χ_{2} & 1 & −1 & −I & I & 1 & 1 & −1 & −1 \\ χ_{3} & 1 & 1 & −1 & −1 & 1 & 1 & 1 & 1 \\ χ_{4} & 1 & −1 & I & −I & 1 & 1 & −1 & −1 \\ χ_{5} & 2 & −2 & 0 & 0 & z_{1} & −1 − z_{1} & −z_{1} & 1 + z_{1} \\ χ_{6} & 2 & −2 & 0 & 0 & −1 − z_{1} & z_{1} & 1 + z_{1} & −z_{1} \\ χ_{7} & 2 & 2 & 0 & 0 & z_{1} & −1 − z_{1} & z_{1} & −1 − z_{1} \\ χ_{8} & 2 & 2 & 0 & 0 & −1 − z_{1} & z_{1} & −1 − z_{1} & z_{1} \end{matrix}

where I = e^{2πi/4}, z_{1} = −1 − ω^{2} − ω^{3} and ω = e^{2πi/5}.

Acknowledgment: We would like to thank J. Cannon for his support regarding the group theory system Cayley and M. Bronstein for his help with Axiom.

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