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

Galois Groups 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 12 August 1997)

Using the representation theory of groups, we are able to give simple necessary and sufficient conditions regarding the structure of the galois groups of second and third order linear differential equations. These allow us to give simple necessary and sufficient conditions for a second order linear differential equation to have liouvillian solutions and for a third order linear differential equation to have liouvillian solutions or be solvable in terms of second order equations. In many cases these conditions also allow us to determine the group.

1. Introduction

Let k be a differential field†† with algebraically closed field of constants C and L(y) = 0 a linear differential equation‡‡ with coefficients in this field. One can form the m^{th} symmetric power L^{◯ s m}(y) of L(y) which is the smallest order nonzero linear differential equation satisfied by the m^{th} power of any solution of L(y) = 0. In this paper we show how factorization properties of these symmetric powers can be used to determine structural properties of the galois groups of second and third order linear differential equation. This in turn will allow us to give necessary and sufficient conditions for these linear differential equations to have liouvillian solutions. For example we show (Corollary 4.4):

Let L(y) = y′′ + ry = 0 be a second order linear differential equation with r ∈ k. L(y) = 0 has liouvillian solutions if and only if L^{◯ s 6}(y) is reducible.

For third order equations we have similar conditions and are also able to characterize those equations that are solvable in terms of lower order equations (Corollary 4.8):

† 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. †† of characteristic zero as are all the fields in this paper. ‡‡ All linear differential equations in this paper are homogeneous

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

Let L(y) = y′′′ + ry′ + sy = 0 be a third order linear differential equation with r, s ∈ k. L(y) = 0 is solvable in terms of lower order linear differential equations if and only if L^{◯ s 4}(y) has order less than 15 or is reducible.

Factorization properties can also be used to determine galois groups in many cases. For example (see Section 2.2 for the definition of the Tetrahedral Group and Theorem 4.3):

Let L(y) = y′′ + ry = 0 be a second order linear differential equation with r ∈ k. The galois group of L(y) = 0 is the Tetrahedral Group A_{4}^{SL_{2}} if and only if L^{◯ s 2}(y) is irreducible and L^{◯ s 3}(y) is reducible.

Our results show that one can reduce many questions concerning the galois groups of linear differential equations to factoring associated differential equations. This underscores the importance of finding efficient factorization algorithms, (c.f., [0, 0]).

The main tool of this paper is representation theory and the results spring from the following facts. The first fact (due to Chevalley) is that if one is given an algebraic subgroup H of GL(n, C) then there is a faithful representation Φ : GL(n, C) → GL(m, C) for some m such that Φ(H) is uniquely determined by its set of invariant subspaces in C^{m} (c.f., Theorem 11.2 of [0]). The second fact is that given a faithful representation of an algebraic group, any other representation can be constructed from this representation using the tools of linear algebra, i.e., tensor product, duals, direct sums and subspaces. Furthermore, if the group is the galois group of a linear differential equation and the representation is the representation on the solution space of the linear differential equation, then one can mimic this construction at the level of the equation to produce an equation whose solution space corresponds to the other representation (c.f., [0, 0, 0, 0, 0]). The final fact that we use is that the solution space of a linear differential equation has a subspace of dimension m invariant under the action of the galois group if and only if the equation has a factor of order m, [0]. Combining these facts one sees that one should be able to determine the galois group of a linear differential equation by considering the factorization properties of certain associated operators. This philosophy has been successfully used in [0, 0, 0, 0, 0]. In this paper, we apply this philosophy to the study of second and third order linear differential equations. Except for the last fact we do not use the full theoretical power of the above facts, but rather calculate directly for the groups involved. In particular we show that in this case it is enough to consider just symmetric powers of small order.

The paper is organized in the following manner. Section 2 contains a description of the groups that can appear as galois groups of second and third order linear differential equations as well as facts about their representation theory. Section 3 reviews facts from the formal theory of linear differential equations and galois theory. In section 4, we present the main results. Section 5 is devoted to examples.

We wish to thank John Cannon for making available to us a copy of Cayley and the the Institute für Algorithmen und Kognitive Systeme of the University of Karlsruhe for allowing us to use their computer algebra systems.

2. Group Theory

To any homogeneous linear differential equation L(y) = 0 of order n with coefficients in a differential field k (with algebraically closed field of constants C), one can associate a group of n × n matrices G(L) ⊆ GL(n, C) called the differential galois group of L(y) = 0 (see [0] or [0] for an exposition of this theory; for concreteness, one may let k = \overline{Q}(x) and C = \overline{Q}) . Differential and algebraic properties of the equation are mirrored by group theoretic properties of this group. This section is devoted to studying properties of subgroups of SL(n, C). The reader interested only in the form of the algorithms and willing to accept certain group theoretic facts, may proceed to the next section.

Let V denote a finite dimensional vector space of dimension n over an algebraically closed field C.

Definition 2.1. A subgroup G of GL(V) is said to act irreducibly if the only G-invariant subspaces of V are {0} and V. The group G is completely reducible if there are minimal G-invariant subspaces V_{1}, ⋯, V_{k} such that V = V_{1} ⊕ ⋯ ⊕ V_{k}.

According to Maschke’s theorem, every finite subgroup of GL(V) is completely reducible. We shall see (in sections 3 and 4) that L(y) = 0 has a galois group that acts reducibly if and only if L(y) is reducible and that one can test directly if this occurs (A differential equation L(y) with coefficients in k is called reducible if L(y) can be written as L_{1}(L_{2}(y)), where L_{1}(y) and L_{2}(y) are differential equations with coefficients in k of order > 0. L(y) = 0 is irreducible if it is not reducible). We shall need finer group theoretic information when the equation (and therefore the galois group) is irreducible. The following definitions are crucial to studying this situation.

Definition 2.2. Let G be a subgroup of GL(n, C) acting irreducibly, i.e. G is a linear group acting irreducibly on the vector space V of dimension n over C. Then G is called imprimitive if, for k > 1, there exist subspaces V_{1}, ⋯, V_{k} such that V = V_{1}⊕⋯⊕V_{k} and, for each g ∈ G, the mapping V_{i} → g(V_{i}) is a permutation of the set S = {V_{1}, …, V_{k}}. The set S is called a system of imprimitivity of G. If all the subspaces V_{i} are one dimensional, then G is called monomial. An irreducible group G ⊆ GL(n, C) which is not imprimitive is called primitive.

We note that since an imprimitive group G is assumed to act irreducibly on V, we have that G acts transitively on the V_{i}. In particular, all the V_{i} have the same dimension.

Definition 2.3. A group G ⊆ GL(n, C) whose elements have a common eigenvector is called 1-reducible.

In [0] 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 this section we will analyse imprimitive and primitive subgroups of G ⊆ GL(n, C) and also see what consequences we can draw from assuming that such a group has a 1-reducible subgroup of finite index.

2.1. Imprimitive Groups

When n is prime any system of imprimitivity for an imprimitive subgroup G ⊆ GL(n, C) contains only subspaces of dimension one (and therefore G must be a monomial group). The subgroup leaving one of these subspaces fixed will be a 1-reducible subgroup of index n. We therefore have the following:

Proposition 2.1. ([0]) Let n be a prime number and let G ⊆ GL(n, C) be an imprimitive group. Then G is a monomial group and contains a 1-reducible subgroup of index n.

2.2. Primitive Groups

The differential galois group G(L) of a linear differential equation is a linear algebraic group which, after a suitable change of variables (cf. Theorem 3.3), can be assumed to be unimodular, i.e. G(L) ⊆ SL(n, C). We thus restrict ourselves to linear algebraic subgroups of SL(n, C) (see [0], [0] or [0] for the appropriate definitions). We have the following general result:

Lemma 2.2. Let G ⊆ GL(n, C) be a primitive group. If H is a normal 1-reducible subgroup of G, then H is a subgroup of the group of scalar matrices.

Proof. We say a subspace W ⊆ C^{n} is a maximal eigenspace of H if each element of H acts by scalar multiplication on W and W is maximal with respect to this property. Let W be the set of maximal eigenspaces of H. By hypothesis, this set is non-empty. If W_{1}, …, W_{m+1} are elements of W such that W_{m+1} ∩ (W_{1} + … + W_{m}) ≠ {0}, then one can easily show that m = 1 and W_{1} = W_{2}. This implies that W is finite and that the sum V′ of the elements of W is a direct sum. Note that H is normal in G so G permutes the elements of W and so leaves V′ invariant. Since G is irreducible, we have V′ = C^{n} and so W is a system of imprimitivity of G, unless W contains just one element. Therefore, we can conclude that the elements of H are all scalar matrices. □

Proposition 2.3. Let G ⊆ SL(n, C) be a primitive linear algebraic group. Then:

1 either G is finite or G^{o}, the connected component of the identity of G, is a semisim-

ple subgroup of GL(n, C), 2 if G also contains a 1-reducible subgroup of finite index, G must be finite, 3 if n = 2 or 3, and G^{o} is semisimple, then G^{o} acts irreducibly on C^{n}.

Proof. (c.f., [0] p. 301 for a similar result) Assume that G is primitive and not finite. Let R(G^{o}) be the radical of G^{o}. Note that R(G^{o}) is normal in G. Since R(G^{o}) is connected and solvable, the Lie-Kolchin Theorem ([0], p. 113) implies that the elements of R(G^{o}) have a common eigenvector, i.e. R(G^{o}) is 1-reducible. Therefore, we can conclude from Lemma 2.2 that the elements of R(G^{o}) are all scalar matrices. Since there are only a finite number of such matrices in SL(n, C), we must have that R(G^{o}) is trivial and so G^{o} is semisimple. This proves 1.

If G also contains a 1-reducible subgroup of finite index, then it contains a 1-reducible normal subgroup of finite index. Lemma 2.2 implies that this latter group consists only of scalar matrices and, since G ⊆ SL(n, C), must be finite. Therefore, G is finite. This proves 2.

If G^{o} is semisimple, then any invariant subspace has a complementary invariant subspace. If n = 2 or 3, and G^{o} has a non-trivial invariant subspace, then G^{o} must have an invariant subspace of dimension 1. This means that G^{o} is 1-reducible and so by 2., G is finite, a contradiction. Therefore G^{o} acts irreducibly. □

Proposition 2.3 reduces the question of finding the primitive subgroups of SL(n, C) to the question of finding the finite primitive subgroups and the semisimple subgroups. We begin with the latter. A connected semisimple group is a quotient (by a finite group) of a direct product of simple groups ([0], p. 167). The simple algebraic groups and their representations are well understood. In particular, by comparing dimensions one can see that the only semisimple subgroup of SL(2, C) is SL(2, C). Therefore any primitive proper subgroup of SL(2, C) is finite. For n = 3, it is shown in ([0], p. 674) that the only connected proper semisimple subgroup of SL(3, C) that acts irreducibly on C^{3} is conjugate to the representation of SL(2, C) given by

ρ_{3} ([ \begin{matrix}a & b \\ c & d \end{matrix} ]) = [ \begin{matrix}a^{2} & 2ab & b^{2} \\ ac & ad + bc & bd \\ c^{2} & 2cd & d^{2} \end{matrix} ]

This is just the irreducible three dimensional representation of SL(2, C) (see Proposition 2.4 and the discussion after it). ρ_{3}(SL(2, C)) ≅ SL(2, C)/{±1} and we shall refer to this group as P SL_{2}. The normalizer of P SL_{2} in SL(3, C) is P SL_{2} × C_{3} where C_{3} is the three element subgroup of scalar matrices ([0], p. 674). Therefore any non-finite proper primitive subgroup of SL(3, C) is conjugate to either P SL_{2} or P SL_{2} × C_{3}.

We now turn to the finite primitive subgroups of SL(2, C) and SL(3, C).

One knows the finite primitive subgroups of P GL(3, C) (c.f., [0]). From this list, one can derive the primitive subgroups of SL(3, C) (c.f., [0]). Any finite primitive group of SL(3, C) is isomorphic to one of the following groups:

1 The Valentiner Group A_{6}^{SL_{3}} of order 1080 generated as a transitive permutation group of 18 letters by:

(1,2,4)(3,8,13)(5,7,9)(6,10,12)(11,15,14),
(1,3)(2,6)(4,5)(7,12)(8,9)(10,13),
(1,4)(3,8)(5,9)(6,11)(10,14)(12,15),
(1,4,8,3,5,9)(2,7,13)(6,12,10)(11,16,14,17,15,18),
(1,5,8)(2,7,13)(3,4,9)(6,12,10)(11,15,14)(16,18,17).

We have A_{6}^{SL_{3}}/Z(A_{6}^{SL_{3}}) ≅ A_{6}. 2 The simple group G_{168} of order 168 defined by:

{X,Y |X^{7}=(X^{4}Y)^{4}=(XY)^{3}=Y^{2}=id}.

3 G_{168} × C_{3}, the direct product of G_{168} with the cyclic group C_{3} of order 3.

4 A_{5}, the alternating group of five letters.

5 A_{5} × C_{3}, the direct product of A_{5} with a cyclic group C_{3} of order 3.

6 The group H_{216}^{SL_{3}} of order 648 defined by:

{U,V,S,T | U^{9}=V^{4}=T^{3}=S^{3}=(UV)^{3}=id,V S=T V
V T =S^{2}V,[U^{6},V]=[U^{6},T]=[U,S]=id,[U,V^{2}]=S}.

The group H_{216}^{SL_{3}}/Z(H_{216}^{SL_{3}}) is the hessian group of order 216.

7 The group H_{72}^{SL_{3}} of order 216 generated by the elements S, T, V and UV U^{−1} of

H_{216}^{SL_{3}}. 8 The group F_{36}^{SL_{3}} of order 108 generated by the elements S, T and V of H_{216}^{SL_{3}}.

If G/Z(G) is a simple group, then G also is a perfect group (i.e. equals its commutator group). In this case any representation of G belongs to SL(n, C). But for the groups H_{72}^{SL_{3}} and F_{36}^{SL_{3}} there exist irreducible representations in GL(3, C) which do not belong to SL(3, C). Note that if g ∈ SL(3, C) has order 2, then the trace of g is −1, and if g has order 4, then the trace cannot be I, −I or −1. Considering the character tables of these groups, one sees that this restriction implies that there are only two irreducible characters of degree 3 left for H_{72}^{SL_{3}} and F_{36}^{SL_{3}}. For the computations in the rest of the paper we will only have to consider these characters.

The finite primitive subgroups of SL(2, C) are isomorphic to one of the following groups (c.f., [0, 0]):

1 The icosahedral group A_{5}^{SL_{2}} of order 120 generated as a transitive permutation group of 24 letters by:

(1,4,2)(3,20,5)(6,13,14)(7,22,8)(9,24,10)(11,21,12)(15,23,16)(17,19,18),
(1,6,5,3,15,2)(4,12,21,20,22,7)(8,19,18,11,10,9)(13,17,16,23,24,14),
(1,3)(2,5)(4,20)(6,15)(7,21)(8,11)(9,18)(10,19)(12,22)(13,23)(14,16)(17,24),
(1,3)(2,5)(4,20)(6,15)(7,21)(8,11)(9,18)(10,19)(12,22)(13,23)(14,16)(17,24).

2 The octahedral S_{4}^{SL_{2}} of order 48 given by:

{X,Y |X^{3}=Y^{8}=id,Y X^{2}=XY^{3}}.

3 The tetrahedral group A_{4}^{SL_{2}} of order 24 generated by X and Y^{2}.

We note that, the tetrahedral group has two faithfull irreducible representations in GL(2, C) which do not belong to SL(2, C).

2.3. The characters of symmetric products

The main idea of this paper is that for n = 2 or 3, one can distinguish between the various primitive groups by decomposing small symmetric powers of the original representation. We do this calculation in this section.

The first step is to calculate the characters χ_{m} of the m^{th} symmetric powers. We follow the presentation in ([0], p. 181). Let z be a variable and define the functions on GL(n, C), q_{0}, …, q_{n}, via the formula:

det(I − zg) = q_{0} − q_{1}z + q_{2}z^{2} − … ± q_{n}z^{n}

for g ∈ G. Note that q_{0} = 1 and q_{n} = det(g). The characters χ_{m} of the symmetric power S^{m}(C^{n}), then satisfy the following recursion:

χ_{0} = 1
χ_{l} − q_{1}χ_{l−1} + q_{2}χ_{l−2} − … ± q_{n}χ_{l−n} = 0

for l = 1, 2, … and χ_{−1}, χ_{−2}, … are set equal to zero. If G ⊆ SL(2, C) we have that q_{0} = q_{2} = 1 and q_{1} = χ, the character of the representation of G on C^{2}. We get the following:

χ_{2} = χ^{2} − 1
χ_{3} = χ^{3} − 2χ
χ_{4} = χ^{4} − 3χ^{2} + 1
χ_{5} = χ^{5} − 4χ^{3} + 3χ
χ_{6} = χ^{6} − 5χ^{4} + 6χ^{2} − 1

For G ⊆ SL(3, C), we have that q_{0} = q_{3} = 1, q_{1} = χ, the character of the representation of G on C^{3} and q_{2} = χ̄, where χ̄(g) = χ(g^{−1}). We get the following:

χ_{2} = χ^{2} − χ̄
χ_{3} = χ^{3} − 2χχ̄ + 1
χ_{4} = χ^{4} − 3χ̄χ^{2} + 2χ + χ̄^{2}
χ_{5} = χ^{5} − 4χ̄χ^{3} + 3χ^{2} + 3χ̄^{2}χ − 2χ̄

For higher dimensional representations of G the power maps (cf. [0]) are needed in the formula for χ_{i} (see e.g. [0] for details and further references).

Proposition 2.4. (See [0], pages 150, 180) SL(n, C) acts irreducibly on C^{n} and all symmetric powers S^{m}(C^{n}) of C^{n}.

In fact, any irreducible representation of SL(2, C) is of the form S^{m}(C^{2}) for n = 0, 1, 2, … (S^{0}(C^{2}) is the trivial one dimensional representation). Note that P SL_{2} is just the image of SL(2, C) in SL(S^{2}(C^{2})) ≅ SL(3, C). For m odd, these representations are faithful representations of SL(2, C) and for m even, these have kernel {±1} (and so factor through P SL_{2}). Therefore the irreducible representations of P SL_{2} are S^{m}(C^{2}), m = 0, 2, 4, . . .. A character of a representation of SL(2, C) is determined by its behavior on the diagonalizable elements of SL(2, C), since these are Zariski dense in this group. If

ϕ_{m} is the character associated with S^{m}(C^{2}) and g = ( \begin{matrix}a & 0 \\ 0 & a^{−1} \end{matrix} ) is a diagonal element,

then ϕ_{m}(g) = a^{m} + a^{m−2} + … + a^{−m}. We use this notation in the following:

Proposition 2.5. Let χ be the character of the irreducible representation of P SL_{2} on C^{3} and χ_{m} the character of the representation on S^{m}(C^{3}). We then have the following decompositions:

1 χ_{2} = ϕ_{4} + ϕ_{0}

2 χ_{3} = ϕ_{6} + ϕ_{2}

3 χ_{4} = ϕ_{8} + ϕ_{4} + ϕ_{0} 4 χ_{5} = ϕ_{10} + ϕ_{6} + ϕ_{2}

Proof. To verify these formulas, it is enough to evaluate the χ_{m} on the diagonal el-

ements of P SL_{2}. Such a diagonal element is of the form ( \begin{matrix}a^{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & a^{−2} \end{matrix} ). Therefore, χ(g) = a^{2} + 1 + a^{−2} = χ̄(g). Using the formulas χ_{2} = χ^{2} − χ̄, χ_{3} = χ^{3} − 2χχ̄ + 1, etc., we are able to verify the above formulas. □

For each finite primitive subgroup of SL(2, C) and SL(3, C), using its character table (computed using the group theory system Cayley [0]) and the orthogonality relations of characters, we can decompose the characters of the symmetric product (computed using the computer algebra system AXIOM, cf. [0]). The result is summarised in the following two tables, where the numbers 4, 3^{2} in the column A_{5} and the row 3 of Table 2 means that the 3-th symmetric product of the character of any faithfull irreducible representation of A_{5} in SL(3, C) has an irreducible summand of degree 4 and two irreducible summands of degree 3.

A_{5}^{SL_{2}}  S_{4}^{SL_{2}}  A_{4}^{SL_{2}}
2  3  3  3

3  4  4  2^{2}

4  5  3, 2  3, 1^{2}

5  6  4, 2  2^{3}

6  4, 3  3^{2}, 1  3^{2}, 1

Table 1

P SL_{2}  A_{5}  G_{168}
P SL_{2} × C_{3} A_{5} × C_{3} F_{36}^{SL_{3}} G_{168} × C_{3} A_{6}^{SL_{3}} H_{72}^{SL_{3}} H_{216}^{SL_{3}}

2  5, 1  5, 1  3, 3  6  6  6  6

3  7, 3  4, 3^{2}  1^{2}, 4^{2}  7, 3  10  8, 2  8, 2

4  9, 5, 1  5^{2}, 4, 1  3^{5}  8, 6, 1  9, 6 6^{2}, 3 6^{2}, 3

5  11, 7, 3  5, 4, 3^{4}  3^{7}  8, 7, 3^{2}  15, 3^{2} 6, 3^{5} 9, 6, 3^{2}

Table 2

For later use we note that in the decomposition of the third symmetric product of F_{36}^{SL_{3}}, the two one dimensional characters ψ_{i} are of order 4, i.e. ψ_{i}^{4} = 1 but ψ_{i}^{2} ≠ 1.

3. Differential Equations

3.1. 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 [0, 0, 0, 0].

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).

Definition 3.1. 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:

η_{i} is algebraic over K_{i}, or

δ (η_{i}) ∈ K_{i} (extension by an integral), or

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 [0] J. Kovacic gives an algorithm 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 [0] 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).

Definition 3.2. Let K_{1} and K_{2} be two differential extensions of k. A differential kisomorphism between K_{1} and K_{2} is a field isomorphism that leaves k fixed and commutes with δ. The differential galois group G(K/k) of a differential field extension K of k is the set of all differential k-automorphisms of K.

Definition 3.3. Let L(y) = 0 be a homogeneous linear differential equation of degree n with coefficients in a differential field k. A differential field extension K of k is called a Picard-Vessiot extension (PVE) of k for L(y) = 0 if the following holds:

K = k < y_{1}, y_{2}, …, y_{n} >, the differential field generated by k and y_{1}, y_{2}, …, y_{n}, where {y_{1}, y_{2}, …, y_{n}} is a fundamental set of solutions of L(y) = 0. K and k have the same field of constants.

We denote G(L) the differential galois group of a PVE associated to L(y) = 0.

A PVE of k associated with L(y) = 0 is well defined and unique up to differential k-isomorphisms if ker_{k}(δ) is an algebraically closed field of characteristic 0. It may be viewed as the splitting field for the equation L(y) = 0. The differential galois group G(L) of L(y) = 0 is a linear algebraic group, and there is a galois correspondence between differential subfields of K/k and linear algebraic subgroups of G(L) (see, e.g. [0, 0]). 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 basis {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).

The following known results show that many properties of the equation L(y) = 0 and of its solutions are related to the structure of the group G(L):

Theorem 3.1. (see e.g. [0], §22, [0], §33) The differential equation L(y) = 0 of degree n over k factors as a differential operator over k if and only if G(L) ∈ GL(n, C) is a reducible linear group. Let V be the solution space of L(y) = 0. G(L) leaves an m dimensional subspace of V invariant if and only if L(y) = L_{n−m}(L_{m}(y)) where L_{n−m}(y) and L_{m}(y) have coefficients in k and are of order m − n and m.

Theorem 3.2. (see e.g. [0]) A differential equation L(y) = 0 with coefficients in k has

only solutions which are algebraic over k if and only if G(L) is a finite group, 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 3.3. ([0], p. 41) The differential galois group of an differential equation of the form

L(y) = y^{(n)} + a_{n−2}y^{(n−2)} + ⋯ + a_{1}y′ + a_{0}y = 0  (a_{i} ∈ k)  (3.1)

is a unimodular group (i.e. G(L) ⊆ SL(n, C)).

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 (3.1). 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.

The above form is a sufficient but not necessary condition for G(L) to be unimodular.

3.2. Linear Operators

We first collect some basic facts on linear differential operators. Linear differential operators can be seen as skew polynomials which can be manipulated almost in the same way as ordinary polynomials.

3.2.1. Factorization of linear differential equations

Let k be a field and δ be a derivation on k. In order to define the notions of irreducibility and factorization for a linear differential equation

L(y) = a_{n}δ^{n}(y) + a_{n−1}δ^{(n−1)}(y) + ⋯ + a_{0}y = 0

of degree n and coefficients in k we look at the associated differential operator:

p(δ) = a_{n}δ^{n} + a_{n−1}δ^{(n−1)} + ⋯ + a_{0}

We now replace D^{n} by δ^{n} in p(δ) and consider

p(D) = a_{n}D^{n} + a_{n−1}D^{(n−1)} + ⋯ + a_{0}

as a skew polynomial in D. From δ(ay) = δ(a)y + aδ(y) one gets the rule Da = aD + a^{δ}. We denote k[D, δ] the set of all such skew polynomials. This is an example of what is called an Ore ring in the litterature.

In [0] an algebraic theory of k[D, δ] is given. It is shown there that the usual polynomial addition and a multiplication defined by Da = aD + a^{δ} and distributivity makes k[D, δ] into a (non commutative) ring which has a left and right euclidean algorithm. The degree of p(D) is defined to be the usual polynomial degree of p(D) in D. Since k is a field, the degree of a product is the sum of the degrees.

Definition 3.4. A linear differential operator p(D) ∈ k[D, δ] is reducible, if ∃q_{1}(D), q_{2}(D) ∈ k[D, δ] of degrees > 0 such that p(D) = q_{1}(D) q_{2}(D). If p(D) is not reducible, then p(D) is called irreducible. A linear differential equation is called reducible (resp. irreducible) if the associated differential operator is reducible (resp. irreducible).

If L(y) is reducible, then L(y) can be writen as L_{1}(L_{2}(y)), where L_{1}(y) and L_{2}(y) are differential equations of degree > 0. We point out that a factorization of differential equation is usually not unique: Example: We give two irreducible decompositions of a third order differential operator:

\frac{d^{3}}{dx^{3}} − \frac{8x^{2} − 2x − 3}{2x(4x + 1)} \frac{d^{2}}{dx^{2}} − \frac{12x + 11}{4x(4x + 1)} \frac{d}{dx} + \frac{4x + 5}{4x(4x + 1)}

= (\frac{d^{2}}{dx^{2}} + \frac{4x + 3}{2x(4x + 1)} \frac{d}{dx} − \frac{4x + 5}{4x(4x + 1)}) (\frac{d}{dx} − 1)

= (\frac{d}{dx} − \frac{4x^{2} + x − 1}{x(4x + 1)}) (\frac{d^{2}}{dx^{2}} + \frac{1}{2x} \frac{d}{dx} − \frac{1}{4x})

This shows that different irreducible decompositions, where the degrees are permuted, are possible.

In general, although the irreducible factors are not unique, their degrees are unique (up to permutation) (see [0]). By Theorem 3.1, any irreducible subspace of the solution space of L(y) = 0 corresponds to a right factor of L(y). Therefore for a completely reducible galois group (for example, a primitive galois group) the irreducible factors correspond to irreducible invariant subspaces. In this case, any permutation of the degrees gives a (possibly different) factorization. Further properties of factorizations and an algorithm computing a factorization of a reducible differential equation L(y) with coefficients in C(x) can be found in [0, 0, 0].

An Eisenstein criterium for linear differential equation is given in [0]. Another condition for irreducibility is given in [0], p. 293.

3.2.2. Symmetric power of a differential equation

In this section we show how, given some linear differential equations, one can construct the following equations:

Theorem 3.4. (cf. [0]) Let L_{1}(y) = 0 and L_{2}(y) = 0 be linear differential equations of degree respectively n_{1} and n_{2} and fundamental system respectively S_{1} = {u_{1}, ⋯, u_{n_{1}}} und S_{2} = {v_{1}, ⋯, v_{n_{2}}}. Then one can construct a differential equation:

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}}}.

L^{(1)}(y) = 0 of degree n ≤ n_{1}, whose solution space is spanned by the set S^{(1)} =

{δ (u_{1}) , ⋯, δ (u_{n_{1}})}.

Proof. In order to construct L_{1}(y)◯ s L_{2}(y) = 0 we take two “arbitrary” solutions u and v of L_{1}(y) = 0 resp. L_{2}(y) = 0 and differentiate their product:

Y = uv
δ(Y) = δ(u)v + uδ(v)
⋯
δ^{m}(Y) = ∑_{j=0}^{m} \binom{m}{j}δ^{j}(u)δ^{m−j}(v).

On the right side we can always replace terms δ^{n_{1}}(u) and δ^{n_{2}}(v) by derivatives of lower order using L_{1}(u) = 0 and L_{2}(v) = 0. On the right side there are then at most n_{1}n_{2} different terms δ^{i}(u)δ^{j}(v) where i < n_{1} and j < n_{2}. This shows that for some m ≤ n_{1}n_{2} the set {Y, δ(Y), ⋯, δ^{m}(Y)} is linear dependent over k, which gives a differential equation for Y = uv.

For L_{1} = ∑_{i}^{n_{1}} a_{i}δ^{i}(y) the differential equation L^{(1)}(y) is given by:

If a_{0} = 0, then L^{(1)}(y) = ∑_{i=1}^{n_{1}} a_{i}δ^{i−1}(y).

If a_{0} ≠ 0 then

L^{(1)}(y) = a_{n}δ^{n}(y) + ∑_{i=0}^{n−1}(δ(a_{i+1}) + a_{i})δ^{i}(y) − \frac{δ(a_{0})}{a_{0}} (∑_{i=1}^{n} a_{i}δ^{i−1}(y)) .

In [0] it is shown that S (resp. S^{(1)}) spans the solution space of L_{1}(y)◯ s L_{2}(y) = 0 (resp. L^{(1)}(y) = 0). □

An important special case of the above construction is:

Definition 3.5. The linear differential equation

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

is called symmetric power of order m of L(y) = 0.

In order to compute L^{◯ s m}(y), one can also differentiate u^{m}, where u is an “arbitrary” solution of L(y) = 0. It can be shown that the differential equation of lowest degree for u^{m} is just L^{◯ s m}(y) (note that u^{m} has to be a solution of L^{◯ s m}(y)). The order of L^{◯ s m} is at most \binom{n+m−1}{n−1}, where n is the degree of L(y) (cf. [0]).

Example: For the the Airy equation L(y) = \frac{d^{2}y}{dx^{2}} − xy = 0 we get

L^{◯ s 2}(y) = \frac{d^{3}y}{dx^{3}} − 4x\frac{dy}{dx} − 2y

L^{◯ s 6}(y) = \frac{d^{7}y}{dx^{7}} − 56x\frac{d^{5}y}{dx^{5}} − 140\frac{d^{4}y}{dx^{4}} + 784x^{2} \frac{d^{3}y}{dx^{3}} + 2352x\frac{d^{2}y}{dx^{2}}

−4(576x^{3} − 295)\frac{dy}{dx} − 3456x^{2}y.

Note that (L^{◯ s 2})^{◯ s 3} = L^{◯ s 6}(y).

Let L(y) have order n and let L(y) = 0 have solution space V in some Picard-Vessiot K extension of k. There is a natural map Φ_{m} of the m^{th} symmetric power S^{m}(V) (c.f., [0], 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. The following lemma summarizes the properties of Φ_{m} needed later:

Lemma 3.5. 1 Φ_{m} is a G(L) morphism of G(L) modules.

2 If all representations of G(L) are completely reducible, then the solution space of

L^{◯ s m}(y) = 0 is G(L)-isomorphic to a direct summand of S^{m}(V). 3 If n = 2, then Φ_{m} is a bijection for all m. In particular, L^{◯ s m}(y) = 0 has order

m + 1. 4 If n = 3 and Φ_{i} is a bijection for i < m, then the dimension of the kernel of Φ_{m}

is at most 1. In this case, the order of L^{◯ s m}(y) = 0 is either \frac{1}{2}(m + 2)(m + 1) or \frac{1}{2}(m + 2)(m + 1) − 1.

5 If n = 3 and Φ_{i} is a bijection for i < m − 1, then the dimension of the kernel of Φ_{m} is at most 3. In this case, the order of L^{◯ s m}(y) = 0 is at least \frac{1}{2}(m + 2)(m + 1) − 3.

Proof. 1. is obvious and 2. follows from 1. and complete reducibility. Now assume n = 2. If Φ_{m} is not injective, then there is a homogeneous polynomial F of degree m with coefficients in C such that F (y_{1}, y_{2}) = 0 for some linearly independent solutions y_{1} and y_{2} of L(y) = 0. Since C is algebraically closed, F may be written as a product of linear polynomials, so F (y_{1}, y_{2}) = 0 would imply that y_{1} and y_{2} are linearly dependent, a contradiction. This proves 3.

Assume n = 3 and assume Φ_{i} is a bijection for i < m. Let {y_{1}, y_{2}, y_{3}} be a basis of V. We then have that if P ≠ 0 is a homogeneous polynomial of degree i, i < m, then P (y_{1}, y_{2}, y_{3}) ≠ 0. Let W = {F | F is a homogeneous polynomial of degree m with coefficients in C such that F (y_{1}, y_{2}, y_{3}) = 0}. Each non-zero F in W must be irreducible, since otherwise we would have P (y_{1}, y_{2}, y_{3}) = 0 for some homogeneous P of degree less than m. If the kernel of Φ_{m} has dimension at least 2, then there would be two relatively prime irreducible homogeneous polynomials F_{1} and F_{2} such that F_{1}(y_{1}, y_{2}, y_{3}) = F_{2}(y_{1}, y_{2}, y_{3}) = 0. The resultant Res_{y_{3}}(F_{1}, F_{2}), of F_{1} and F_{2} with respect to y_{3}, is an homogeneous polynomial F (y_{1}, y_{2}) of degree m^{2} in y_{1} and y_{2} which must be zero. As in the previous case, a factorization of F (y_{1}, y_{2}) yields a contradiction. This proves 4.

Again assume n = 3 and assume Φ_{i} is a bijection for i < m − 1. Let {y_{1}, y_{2}, y_{3}} be a basis of V. If Φ_{m−1} is a bijection then by what we have just shown, the kernel of Φ_{m} has dimension at most 1. Assume Φ_{m−1} is not a bijection. This means that the kernel of Φ_{m−1} has dimension 1. Identifying the symmetric powers with spaces of homogeneous polynomials, we let P be a homogeneous polynomial of degree m − 1 that spans this kernel. We see, as above, that P must be irreducible. Let W = {F | F is a homogeneous polynomial of degree m with coefficients in C such that F (y_{1}, y_{2}, y_{3}) = 0}. If F ∈ W and P does not divide F, then arguing with resultants as in the previous case, we would have a contradiction. Therefore, P divides all the elements of W. This means that W is a subspace of the space of homogeneous polynomials of degree m spanned by Y_{1}P, Y_{2}P and Y_{3}P. Therefore the dimension of W is at most 3. □

The following will give us a criterium to test if the galois group is monomial. As we have noted at the beginning of section 2.1, if n is prime, then a subgroup of GL(n, C) is imprimitive if and only if it is monomial. If n is not prime, there are always non-monomial imprimitive subgroups of GL(n, C).

Proposition 3.6. If an irreducible linear differential equation L(y) = 0 of order n with coefficients in k has a monomial differential galois group G(L) ⊆ SL(n, C), then the n-th symmetric power L^{◯ s n}(y) = 0 of L(y) = 0 has a solution which is the square root of an element of k.

Proof. If G(L) ⊆ SL(n, C) is a monomial group, then there is a basis {y_{1}, ⋯, y_{n}} of the solution space of L(y) = 0 such that all matrices σ ∈ G(L) contain only one non zero element in any row and any column. Such a matrix σ has n non zero entries a_{1}, ⋯, a_{n}, and since it is an element of a unimodular group, its determinant ±a_{1}a_{2} ⋯ a_{n} is 1. For any σ ∈ G(L) we get

σ(y_{1}y_{2} ⋯ y_{n}) = ±(a_{1}a_{2} ⋯ a_{n})(y_{1}y_{2} ⋯ y_{n}) = ±det(σ) · (y_{1}y_{2} ⋯ y_{n})
= ±y_{1}y_{2} ⋯ y_{n}.

This shows that (y_{1}y_{2} ⋯ y_{n})^{2} is invariant under G(L) and thus belongs to k. Since y_{1}y_{2} ⋯ y_{n} is a solution of L^{◯ s n}(y) = 0, we get that L^{◯ s n}(y) = 0 has a solution which is the square root of an element of k. □

4. Main results

4.1. Galois groups and symmetric powers

In this section we describe the behavior of the galois group in terms of properties of various symmetric powers of the differential equation. This will give necessary and sufficient conditions for a second or third order linear differential equation to have a liouvillian solution. We start with second order equations. In what follows k will always be a differential field with algebraically closed field of constants C.

Theorem 4.1. Let L(y) = 0 be a second order homogeneous linear differential equation with coefficients in k and unimodular differential galois group.

1 L(y) is reducible if and only if L(y) = 0 has a solution y ≠ 0 such that y′/y ∈ k.

In this case G(L) ⊆ SL(n, C) is reducible. 2 Assume L(y) is irreducible. Then G(L) is imprimitive if and only if L^{◯ s 2}(y) = 0

is reducible. In this case L^{◯ s 2}(y) = 0 has a solution y ≠ 0 such that y^{2} ∈ k. and G(L) ≅ C^{∗} ⋊ Z/2Z or the dihedral group D_{2n}.

3 Assume G(L) is primitive. Then L^{◯ s 6}(y) = 0 is reducible if and only if G(L) is a

finite group. 4 G(L) ≅ SL(2, C) if none of the above hold.

Proof. Theorem 3.1 handles case 1. Therefore assume that L(y) is irreducible. In this case the galois group is either primitive or imprimitive. As we noted in the discussion following Proposition 2.3, the only primitive subgroups of SL(2, C) are either finite or all of SL(2, C).

Assume L^{◯ s 2}(y) is reducible. Lemma 3.5 implies that the solution space of L^{◯ s 2}(y) is G(L)-isomorphic to the second symmetric power of the solution space of L(y) = 0. Therefore this symmetric power must be reducible. Table 1 shows that G(L) cannot be a finite primitive group. Proposition 2.4 shows that G(L) cannot be SL(2, C). Therefore G(L) must be imprimitive. Proposition 3.6 implies that L^{◯ s 2}(y) has a solution y ≠ 0 such that y^{2} ∈ k. Furthermore G(L) must be a monomial group which in this case means that it is a subgroup of C^{∗} ⋊ Z/2Z. Either it is the full group or it must be a proper subgroup, in which case it is finite and must be a dihedral group. Conversely, if G(L) is imprimitive, proposition 3.6 implies that L^{◯ s 2}(y) is reducible.

Now assume that G(L) is primitive. We then have that G(L) is one of the finite primitive groups or all of SL(2, C). Table 1 implies that L^{◯ s 6}(y) is reducible. Conversely, Proposition 2.4 implies that L^{◯ s 6}(y) is irreducible if G(L) ≅ SL(2, C).

Finally, any proper subgroup of SL(2, C) is either reducible, imprimitive or a finite primitive group so the final statement above is true. □

In cases 1, 3, and 4 of the above, one can give simple criteria to determine the galois group. We do this in the next three propositions. Case 2 is more problematic and we discuss this following these three results.

Proposition 4.2. Let L(y) = 0 be a second order homogeneous linear differential equation with coefficients in k and unimodular differential galois group. Assume L(y) is reducible (and so has a solution y ≠ 0 such that y′/y ∈ k).

1 If L(y) = 0 has a unique (up to constant multiple) solution y ≠ 0 such that y′/y ∈ k, then G(L) is conjugate to a subgroup of

T = \{( \begin{matrix}a & b \\ 0 & a^{−1} \end{matrix} ) | a, b ∈ C, a ≠ 0\}

Furthermore, G(L) is a proper subgroup of T if and only if y^{m} ∈ k for some positive integer m. In this case, G(L) is conjugate to

T_{m} = \{( \begin{matrix}a & b \\ 0 & a^{−1} \end{matrix} ) | a, b ∈ C, a^{m} = 1\}

where m is the smallest positive integer such that y^{m} = 1. 2 If L(y) = 0 has two linearly independent solutions y_{1} and y_{2} such that y′_{i}/y_{i} ∈ k, i =

1, 2, then G(L) is conjugate to a subgroup of

D = \{( \begin{matrix}a & 0 \\ 0 & a^{−1} \end{matrix} ) | a ∈ C, a ≠ 0\}

In this case, y_{1}y_{2} ∈ k. Furthermore, G(L) is conjugate to a proper subgroup of D if and only if y_{1}^{m} ∈ k for some positive integer m. In this case G(L) is a cyclic group of order m where m is the smallest positive integer such that y_{1}^{m} ∈ k.

Proof. If L(y) = 0 has a solution y ≠ 0 such that y′/y ∈ k then y is an eigenvector for all the elements of G(L) so G(L) is conjugate to a subgroup of T. If L(y) = 0 has two linearly independent solutions y ≠ 0 such that y′/y ∈ k then the elements of G(L) have two independent common eigenvectors so G(L) is conjugate to a subgroup of D.

Assume case 1 holds and select a basis of the solution space such that G(L) ⊆ T. The

map sending σ = ( \begin{matrix}a & b \\ 0 & a^{−1} \end{matrix} ) to a is a homomorphism of G(L) to C^{∗}. If the image

of G(L) is a proper subgroup of C^{∗}, then this image is a finite cyclic group. Therefore, G(L) = T_{m}. Let y be a common eigenvector of G(L). We then have σ(y^{m}) = (ay)^{m} = y^{m} for any σ ∈ G(L) since a is an m^{th} root of 1. Clearly, m is the smallest positive integer such that y^{m} ∈ k. This proves 1.

Assume L(y) = 0 has two linearly independent solutions y_{1} and y_{2} such that y′_{i}/y_{i} ∈ k, i = 1, 2. With respect to y_{1} and y_{2} G(L) may be identified with a subgroup of D. D is isomorphic to C^{∗} so any proper subgroup is a finite cyclic group. Furthermore, for any σ ∈ G(L), σ(y_{1}y_{2}) = ay_{1}a^{−1}y_{2} = y_{1}y_{2} so y_{1}y_{2} ∈ k. The remaining claim follows as above.

□

When k = C(x), x′ = 1, one has algorithms to decide this question (see the factorization algorithms mentioned above and also [0, 0]). [0] also mentions the idea of seeing how many solutions one has of a specified form to determine the galois group.

Proposition 4.3. Let L(y) = 0 be a second order homogeneous linear differential equation with coefficients in k and unimodular differential galois group. Assume G(L) is primitive.

1 L^{◯ s 3}(y) factors over k if and only if G(L) ≅ A_{4}^{SL_{2}}. In this case, L^{◯ s 3}(y) = L_{1}(L_{2}(y)) where L_{1}(y) and L_{2}(y) have order 2.

2 Assuming L^{◯ s 3}(y) irreducible, then L^{◯ s 4}(y) factors over k if and only if G(L) ≅ S_{4}^{SL_{2}}. In this case, L^{◯ s 4}(y) = L_{1}(L_{2}(y)) where L_{1}(y) and L_{2}(y) have orders 3 and 2.

3 Assuming L^{◯ s 4}(y) irreducible, then L^{◯ s 6}(y) factors over k if and only if G(L) ≅ A_{5}^{SL_{2}}. In this case, L^{◯ s 6}(y) = L_{1}(L_{2}(y)) where L_{1}(y) and L_{2}(y) have orders 4 and 3.

4 G(L) ≅ SL(2, C) if and only if L^{◯ s 6}(y) is irreducible over k.

Proof. Lemma 3.5 states that the solution space of L^{◯ s m}(y) = 0 is isomorphic to the m^{th} symmetric power of V, the solution space of L(y) = 0. Table 1 gives the decompositions of these spaces for small m and Proposition 3.1 gives the first three results. Proposition 2.4 gives the final result. □

Proposition 4.4. Let L(y) = 0 be a second order homogeneous linear differential equation with coefficients in k and unimodular differential galois group. G(L) ≅ SL(2, C) if and only if L^{◯ s 6}(y) is irreducible over k. Therefore, L(y) = 0 has liouvillian solutions if and only if L^{◯ s 6}(y) is reducible.

Proof. If G(L) ≅ SL(2, C), then Proposition 2.4 implies L^{◯ s 6}(y) is irreducible over k. Conversely, if G(L) is a proper subgroup of SL(2, C), then Theorem 4.1 implies that L(y), L^{◯ s 2}(y), or L^{◯ s 6}(y) are reducible. If L(y) is reducible then the solution space V of L(y) = 0 has a G(L) invariant subspace W. S^{6}(W) will be a G(L) invariant subspace of S^{6}(V), so L^{◯ s 6}(y) will be reducible. Similarly, if W is a proper G(L) invariant subspace of S^{2}(V), then S^{3}(W) will be a proper G(L) invariant subspace of S^{6}(V). The final statement follows from the fact that any proper subgroup of SL(2, C) has a component of the identity that has dimension less than 3. □

Proposition 4.4 can be improved if one knows a priori that L(y) = 0 has no algebraic solutions (for example, if k = C(x) and L(y) = 0 is not fuchsian). In this case, the proof shows that G(L) ≅ SL(2, C) if and only if L^{◯ s 2}(y) is irreducible over k (in fact, if and only if L^{◯ s 2}(y) = 0 has a solution y ≠ 0 such that y′/y ∈ k). This fact is the basis of the necessary conditions for liouvillian solutions developed by Kaplansky [0]. The above shows that they are also sufficient in this case.

We note that it is not so simple to distinguish between the cases of a finite and an infinite group when G(L) is imprimitive. This question is discussed in [0] and depends on being able to decide : given an element u algebraic over k, determine if there is a non-zero integer n such that y′/y = nu has a solution y algebraic over k. This question is decidable when k = C(x), x′ = 1.

We now discuss the relationship between our ideas and Kovacic’s algorithm and we assume the reader is familliar with [0]. Kovacic’s algorithm deals with four cases. His first case corresponds to the case when the equation (and therefore also the group) is reducible. The second case corresponds to the galois group being an imprimitive group. Kovacic shows that in this case the linear differential equation L(y) = 0 has a solution y ≠ 0 such that u = y′/y satisfies an irreducible polynomial equations u^{2} + a_{1}u + a_{0} = 0. Kovacic’s algorithm attempts to find this equation. The coefficient a_{1} is of the form a_{1} = \frac{y′_{1}}{y_{1}} + \frac{y′_{2}}{y_{2}} = \frac{(y_{1}y_{2})′}{y_{1}y_{2}} for some solutions y_{1}, y_{2} of L(y) = 0. In particular, this implies that L^{◯ s 2}(z) = 0 has a nonzero solution z such that z′/z is rational. By considering the structure of the galois group, Kovacic can in fact show that z^{2} is rational. Kovacic further shows that the coefficient a_{0} is completely determined once a_{1} is known. Therefore, a careful reading of Kovacic’s proof shows that he has proven parts 1. and 2. of Theorem 4.1. Case 3 of Kovacic’s algorithm corresponds to the galois group being a primitive proper subgroup of SL(2, C) and therefore being finite. Kovacic shows that in this case, L(y) = 0 has a solution y ≠ 0 such that u = y′/y satisfies an irreducible polynomial equation u^{m} + a_{m−1}u^{m−1} + … + a_{0} = 0 for m = 4, 6, or 12. The form of a_{m−1} shows that for m = 4, 6, or 12 L^{◯ s m}(z) = 0 has a nonzero solution z such that z′/z is rational. In particular, L^{◯ s m}(z) = 0 will have a factor of order 1 in one of these cases. Kovacic again shows that the other coefficients a_{i} are completely determined by a_{n−1} and so, he shows that (assuming cases 1 and 2 do not hold) a necessary and sufficient condition that L(y) = 0 has an algebraic solution is that for m = 4, 6, or 12 the equation L^{◯ s m}(z) = 0 has a nonzero solution z such that z′/z is rational. Kovacic shows this using the internal structure (e.g., existence of ”large” abelian subgroups) of the finite primitive subgroups of SL(2, C). This could also be shown just from the representation theory in the spirit of Theorem 4.1. When one uses the representation theory as we did, one is naturally led to consider higher order factors of the symmetric powers. One can then find necessary and sufficient conditions in terms of the factorization of symmetric powers of relatively small order. This leads to our distinction of the cases of the algorithm in [0] using L^{◯ s m}(z) = 0 for m ∈ {2, 3, 4, 6} instead of m ∈ {2, 4, 6, 12} (see [0]: p. 5 and p. 32):

Corollary 4.5. Let L(y) = 0 be a second order homogeneous linear differential equation with coefficients in k and unimodular differential galois group. Then one of the following holds:

1 The equation L(y) = 0 has a solution of the form e^{∫ ω} where ω ∈ k if and only if

L(y) = 0 is reducible. 2 Assume that the above does not hold. The equation L(y) = 0 has a solution of the

form e^{∫ ω} where ω is algebraic over k of degree 2 if and only if L^{◯ s 2}(y) = 0 is reducible. In this case G(L) is an imprimitive subgroup of SL(2, C).

3 Assume that the above does not hold, then

(a) The equation L(y) = 0 has an algebraic solution of the form e^{∫ ω} where ω is algebraic over k of degree 4 if and only if L^{◯ s 3}(y) = 0 is reducible. In this case G(L) ≅ A_{4}^{SL_{2}}.

(b) The equation L(y) = 0 has an algebraic solution of the form e^{∫ ω} where ω is algebraic over k of degree 6 if and only if L^{◯ s 4}(y) = 0 is reducible and L^{◯ s 3}(y) = 0 is irreducible. In this case G(L) ≅ S_{4}^{SL_{2}}

(c) The equation L(y) = 0 has an algebraic solution of the form e^{∫ ω} where ω is algebraic over k of degree 12 if and only if L^{◯ s 6}(y) = 0 is reducible and L^{◯ s 4}(y) = 0 is irreducible. In this case G(L) ≅ A_{5}^{SL_{2}}.

4 The differential equation has no liouvillian solutions. In this case G(L) ≅ SL(2, C).

Proof. The first case is trivial, so assume G(L) acts irreducibly.

If G(L) ⊆ SL(2, C) is an irreducible algebraic group which is not imprimitive and not finite, then the last case holds.

If G(L) ⊆ SL(2, C) is a finite primitive group, then G(L) ≅ A_{4}^{SL_{2}}, S_{4}^{SL_{2}} or A_{5}^{SL_{2}}. Table 1 proves the facts about the decomposition of the symmetric powers. That the algebraic degree 4, 6, 12 for ω is best possible follows from [0] p. 32 (or [0, 0]).

Since G(L) acts irreducibly (i.e. case 1 does not hold), we get from Theorem 4.1 and Table 1 that G(L) is imprimitive if and only if L^{◯ s 2}(y) = 0 is reducible. The fact that the algebraic degree 2 for ω is best possible follows from [0] p. 32 (or [0, 0]). □

We do not make any claims that the above conditions yield, at present, an algorithm that is better than Kovacic’s. Kovacic analyses the situation much further and gives more information than we do above (see [0] for improvements of Kovacic’s algorithm and applications and [0] for generalizations to higher order equations of some of Kovacic’s other ideas and necessary conditions). We do claim that our results show the importance of factorization algorithms and the need for finding more efficient ways to factor linear operators. They are also readily generalized to higher order equations and can be used over any differential field in which there exists an algorithm to factor differential operators.

We now turn to third order equations. We state our results first for groups that are not primitive and then for primitive groups.

Theorem 4.6. Let L(y) = 0 be a third order linear differential equation with coefficients in a differential field k with algebraically closed field of constants whose differential galois G(L) group is unimodular.

1 L(y) = 0 is reducible if and only if L(y) = 0 has a solution y ≠ 0 such that y′/y ∈ k or L^{∗}(y) = 0, the adjoint of L(y) = 0, has a solution y ≠ 0 such that y′/y ∈ k (if L(y) = y′′′ + py′ + qy = 0, then L^{∗}(y) = y′′′ + py′ − (q − p′)y = 0) .

2 Assume L(y) is irreducible. Then G(L) is imprimitive if and only if L^{◯ s 3}(y) = 0 has a solution y ≠ 0 such that y^{2} ∈ k. In this case G(L) is isomorphic to a subgroup of C^{∗} ⋊ S_{3}, where S_{3} is the symmetric group on three letters. If G(L) is isomorphic to a subgroup of C^{∗} ⋊ A_{3}, where A_{3} is the alternating group on three letters, then the above solution y is already in k.

3 Assume L(y) is irreducible and 2. does not hold, then G(L) is a primitive group.

Proof. L(y) = 0 is reducible if and only if L(y) = L_{2}(L_{1}(y)) or L(y) = L_{1}(L_{2}(y)) where L_{1}(y) and L_{2}(y) are of order 1 and 2 respectively with coefficients in k. If L(y) = L_{1}(L_{2}(y)) then taking adjoints, we have L^{∗}(y) = L_{2}^{∗}(L_{1}^{∗}(y)). Therefore (1) holds.

Now assume that G(L) is irreducible. If G(L) is imprimitive, then Proposition 2.1 implies that L^{◯ s 3}(y) = 0 has a solution y ≠ 0 such that y^{2} ∈ k. Now assume the conditions of 2. hold. Table 2 implies that these conditions cannot hold if G(L) is primitive, unless G(L) ≅ F_{36}^{SL_{3}}. Since y^{2} ∈ k, we must have that χ^{2} = 1 where χ is the associated character of the one dimensional invariant subspace generated by y of the third symmetric power of the solution space S of L(y) = 0. If G(L) ≅ F_{36}^{SL_{3}}, then all one dimensional characters in the decomposition of the third symmetric power of S have degree 4, i.e. χ^{2} ≠ 1, χ^{4} = 1 (see remark after Table 2). We cannot be in this case. □

Theorem 4.7. Let L(y) = 0 be a third order linear differential equation with coefficients in a differential field k with algebraically closed field of constants, whose differential galois group G(L) is unimodular. Assume that G(L) is primitive.

1 If L^{◯ s 2}(y) has order 5 or factors then G(L) is isomorphic to P SL_{2}, P SL_{2} × C_{3}, A_{5}, A_{5} × C_{3} or F_{36}^{SL_{3}}. In this case one of the following holds G(L) ≅ F_{36}^{SL_{3}} if and only if L^{◯ s 2}(y) has a factor of order 3, or G(L) ≅ A_{5} or A_{5} × C_{3} if and only if L^{◯ s 3}(y) has a factor of order 3 and a factor of order 4, or G(L) ≅ P SL_{2} or G(L) ≅ P SL_{2} × C_{3} if and only if the previous two cases do not hold.

2 If L^{◯ s 2}(y) has order 6 and is irreducible, then one of the following holds

G(L) ≅ G_{168} or G_{168} × C_{3} if and only if L^{◯ s 3}(y) has a factor of order 3. G(L) ≅ A_{6}^{SL_{3}} if and only if L^{◯ s 4}(y) is reducible and L^{◯ s 3}(y) is irreducible. G(L) ≅ H_{72}^{SL_{3}} if and only if L^{◯ s 5}(y) has more than 2 factors of order 3. G(L) ≅ H_{216}^{SL_{3}} if and only if L^{◯ s 5}(y) has exactly 2 factors of order 3 and L^{◯ s 2}(y) has a factor of degree 2. The galois group is SL(3, C) if and only if none of the above happen.

Proof. The proof proceeds by examining Table 2. Let V be the solution space of L(y) = 0. If L^{◯ s 2}(y) has order 5 or factors, then S^{2}(V) must have an invariant subspace. This can only happen if G(L) is one of the groups mentioned. Lemma 3.5.3 implies that the order of L^{◯ s 2}(y) is at least 5, so G(L) ≅ F_{36}^{SL_{3}} if and only if L^{◯ s 2}(y),has a factor of order 3. Lemma 3.5.4 implies that L^{◯ s 3}(y) has order at least 7. Therefore if G(L) ≅ A_{5} or A_{5} × C_{3} then L^{◯ s 3}(y) has a factor of order 3. This does not happen in the other cases considered, therefore 1. holds.

Assume L^{◯ s 2}(y) has order 6 and is irreducible. Table 2 implies that G(L) ≅ G_{168} or G_{168} × C_{3} or A_{6}^{SL_{3}} or H_{72}^{SL_{3}} or H_{216}^{SL_{3}} or SL(3, C). From Table 2 we get that G(L) ≅ G_{168} or G_{168}×C_{3} if and only if L^{◯ s 3}(y) has a factor of order 3. If G(L) ≅ G̸_{168} or G_{168}×C_{3}, then none of the m^{th} symmetric powers, m = 2, 3, 4, 5 of V have a 1 dimensional invariant subspace for these groups, so the m^{th} symmetric powers of L(y) have order exactly \frac{1}{2}(m+2)(m+1). Table 2 describes how these symmetric powers factor and 2. summarizes the distinguishing cases.

If the galois group is SL(3, C) then all symmetric powers are irreducible, so the theorem follows. □

One can use Table 2 to state other necessary and sufficient conditions for the primitive groups. For example, if L^{◯ s 2}(y) has order 6 and is irreducible then G(L) ≅ A_{6}^{SL_{3}} if and only if L^{◯ s 3}(y) is irreducible and L^{◯ s 4}(y) factors. Since it is not clear which criteria will be most usefull, we have just stated one set of criteria to give a taste of what can be done. The above theorems allows us to give criteria for a third order linear differential equation to be solvable in terms of lower order linear differential equations (c.f., [0], for third order equations this concept coincides with the concept of “solving in terms of second order equations” or “eulerian” [0, 0]).

Corollary 4.8. A third order linear differential equation L(y) = 0 with coefficients in a differential field k with algebraically closed field of constants and whose differential galois group G(L) is unimodular, is solvable in terms of lower order linear differential equations if and only if L^{◯ s 4}(y) has order less than 15 or factors.

Proof. We first note that L(y) = 0 is solvable in terms of lower order linear differential equations if and only if G(L) is a proper subgroup of SL(3, C). To see this we use the criterion of [0], Theorem 1 (c.f., [0] Theorem 5.1, p. 48): L(y) = 0 cannot be solved in terms of lower order linear differential equations if and only if G(L) has a lie algebra g that is simple and such that if ρ : g → gl(m, C) with m < n, then ρ ≡ 0. The simple lie subalgebras of sl(3,C) are sl(3,C) and sl(2,C) (c.f., [0]). Since sl(2,C) has a representation of smaller degree, the only simple lie algebra having no nontrivial representations of lower order is sl(3,C). Therefore L(y) = 0 cannot be solved in terms of lower order linear differential equations if and only if G(L) = SL(3, C). Therefore if L^{◯ s 4}(y) has order less than 15 or factors, then G(L) ≠ SL(3, C), so L(y) = 0 is solvable in terms of lower order linear differential equations.

Now we note that G(L) is a proper subgroup of SL(3, C) if and only if G(L) acts reducibly or is imprimitive or is a primitive proper subgroup of SL(3, C). If G(L) acts reducibly, then the solution space V of L(y) = 0 has an invariant one or two dimensional subspace W. S^{4}(W) will be a proper invariant subspace of S^{4}(V), so L^{◯ s 4}(y) has order less than 15 or factors. If G(L) is imprimitive, then Theorem 4.6 implies that L^{◯ s 3}(y) has a factor of order 1. Therefore the solution space Z of L^{◯ s 3}(y) = 0 has a one dimensional invariant subspace. Let p span this space and y_{1}, y_{2}, y_{3} be a basis of V. Then y_{1}p, y_{2}p, y_{3}p spans an invariant subspace of the solution space of L^{◯ s 4}(y) = 0 and so must factor or have order at most 3. Finally, if G(L) is a primitive proper subgroup of SL(3, C), then Table 2 shows that L^{◯ s 4}(y) has order less than 15 or factors. □

One also can give necessary and sufficient conditions for the existence of liouvillian solutions:

Corollary 4.9. Let L(y) = 0 be an irreducible third order linear differential equation with coefficients in a differential field k with algebraically closed field of constants whose differential galois G(L) group is unimodular. L(y) = 0 has a liouvillian solution if and only if

1 L^{◯ s 4}(y) has order less than 15 or factors, and

2 one of the following holds:

L^{◯ s 2}(y) has order 6 and is irreducible, or L^{◯ s 3}(y) has a factor of order 4.

Proof. L(y) = 0 has a liouvillian solution if and only if it is solvable in terms of lower order linear differential equations and its galois group is not P SL_{2} or P SL_{2} × C_{3}. The result now follows from Theorem 4.6 and Table 2. □

We now show how our apporach can be used to distinguish the different cases for the algebraic degree of the logarithmic derivative of a liouvillian solution in the algorithm given in [0] using the bounds given in [0] Theorem 5.2 and the improvement of this bounds given in [0].

Corollary 4.10. Let L(y) = 0 be an irreducible third order linear differential equation with coefficients in a differential field k with algebraically closed field of constants whose differential galois G(L) group is unimodular.

1 L(y) has a solution whose logarithmic derivative is algebraic of degree 3 if and only if L^{◯ s 3}(y) has a solution y ≠ 0 such that y^{2} ∈ k. In this case G(L) is an imprimitive subgroup of SL(3, C).

2 If the above does not hold, then

(a) L(y) has an algebraic solution whose logarithmic derivative is algebraic of degree 6 if and only if L^{◯ s 3}(y) has an irreducible factor of order 4. In this case G(L) ≅ A_{5}, A_{5} × C_{3} or F_{36}^{SL_{3}}.

(b) L(y) has an algebraic solution whose logarithmic derivative is algebraic of degree 9 if and only if L^{◯ s 3}(y) has an irreducible factor of order 2. In this case G(L) ≅ H_{216}^{SL_{3}} or H_{72}^{SL_{3}}.

(c) L(y) has an algebraic solution whose logarithmic derivative is algebraic of degree 21 if and only if L^{◯ s 3}(y) has an irreducible factor of order 3 and L^{◯ s 2}(y) is irreducible of degree 6. In this case G(L) ≅ G_{168} or G_{168} × C_{3}.

(d) L(y) has an algebraic solution whose logarithmic derivative is algebraic of degree 36 if and only if L^{◯ s 3}(y) is irreducible of degree 10 and L^{◯ s 4}(y) is reducible. In this case G(L) ≅ A_{6}^{SL_{3}}.

3 If none of the above holds, then L(y) = 0 has no liouvillian solutions.

Proof. If L^{◯ s 3}(y) has a solution y ≠ 0 such that y^{2} ∈ k, then G(L) is an imprimitive subgroup of SL(3, C) (Theorem 4.6). In this case L(y) = 0 has a solution whose logarithmic derivative is algebraic of degree 3 (Theorem 5.2 of [0]). The only if part follows from the fact that the finite primitive group the bounds given in the Theorem are best possible (cf. [0]), Theorem 4.4). If the first case does not hold, then G(L) is a primitive subgroup of SL(3, C). If L^{◯ s 3}(y) has a factor of order 4, or L^{◯ s 2}(y) is irreducible of degree 6 or L^{◯ s 3}(y) is irreducible of degree 10, then G(L) ≅ P̸ SL_{2} or P SL_{2} × C_{3}. If L^{◯ s 3}(y) or L^{◯ s 4}(y) is reducible, then G(L) ≅ S̸L(3, C) (cf. Proposition 2.4). Thus G(L) is a finite primitive subgroup of SL(3, C). The conditions of the symmetric powers of L(y) = 0 now follow from Table 2 and the algebraic degree of the logarithmic derivative from Theorem 4.4 of [0]. This proves 2.

If G(L) is not an imprimitive group and not a finite primitive subgroup of SL(3, C), then the irreducible equation L(y) = 0 has no liouvillian solutions (cf. [0]), Corollary 3.7). □

The above result shows that with the necessary and sufficient conditions given in this paper, one has to look for at most one possible degree of logarithmic derivative of the solution of L(y) = 0. This gives a substantial simplification of the algorithm given in [0] for third order differential equations.

5. Examples

Using our results, we want to decide if the differential equation

L(y) = \frac{d^{3} y}{dx^{3}} + \frac{32x^{2} − 27x + 27}{36x^{2}(x − 1)^{2}} \frac{d y}{dx} − \frac{64x^{3} − 81x^{2} + 135x − 54}{72x^{3}(x − 1)^{3}} y = 0

has a liouvillian solution.

The equation L(y) = 0 is reducible if and only if L(y) = 0 or its adjoint L^{∗}(y) = 0 has a right factor of order 1. This is equivalent to saying that either L(y) = 0 or L^{∗}(y) = 0 has a solution whose logarithmic derivative is rational (cf. [0]). Since no such solution exists (this could be computed for example using an algorithm implemented by Manuel Bronstein in the computer algebra system AXIOM, cf. [0]), we get that L(y) = 0 is irreducible.

We now test if the differential galois group is an imprimitive subgroup of SL(3, C). This is the case (cf. Theorem 4.6) if and only if L^{◯ s 3}(y) =

\frac{d^{7}y}{dx^{7}} + \frac{224x^{2} − 189x + 189}{18x^{2}(x − 1)^{2}} \frac{d^{5}y}{dx^{5}} + \frac{−2240x^{3} + 2835x^{2} − 4725x + 1890}{36x^{3}(x − 1)^{3}} \frac{d^{4}y}{dx^{4}}

+\frac{340480x^{4} − 574560x^{3} + 1263465x^{2} − 969570x + 280665}{1296x^{4}(x − 1)^{4}} \frac{d^{3}y}{dx^{3}}

+\frac{−358400x^{5} + 756000x^{4} − 2036475x^{3} + 2275560x^{2} − 1284255x + 289170}{432x^{5}(x − 1)^{5}} \frac{d^{2}y}{dx^{2}}

+ (\frac{1003520x^{6} − 2540160x^{5} + 8042895x^{4} − 11711070x^{3}}{576x^{6}(x − 1)^{6}}

+\frac{9723735x^{2} − 4309200x + 793800}{576x^{6}(x − 1)^{6}}) \frac{d y}{dx}

+ (\frac{−1576960x^{7} + 4656960x^{6} − 16875810x^{5} + 30150225x^{4}}{864x^{7}(x − 1)^{7}}

+\frac{−32863320x^{3} + 21565845x^{2} − 7858620x + 1224720}{864x^{7}(x − 1)^{7}}) y

has a solution y such that y^{2} ∈ \overline{Q}(x).

Since x^{2}(x − 1)^{2} is a solution of L^{◯ s 3}(y) = 0, we get that G(L) is an imprimitive subgroup of SL(3, C). Thus L(y) = 0 has a solution whose logarithmic derivative is algebraic of degree 3 (cf., Corollary 4.10).

We note that our approach does not determine the group G(L) in the imprimitive case. But since in this example L(y) = 0 is the second symmetric power of the equation

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

whose differential galois group is A_{4}^{SL_{2}} ([0], p. 23), we get by construction that G(L) ≅ A_{4} (from Table 1 it now also follow that L(y) = 0 is irreducible).

For third order differential equations very few examples can be found in the literature. We shall show how one can construct such examples for the primitive groups. Assume, we are given a finite group G and a differential equation of arbitrary order with G as its galois group. Let us also assume we know that G has an irreducible representation of degree n. We shall show how to construct a differential equation of order n having the image of G in GL(n) as its galois group. The idea behind this construction is that such a differential equation will occur as a factor of some other equation that we can construct. This will also allow us to construct a differential equation for a group G from the knowledge of an irreducible polynomial P (Y) ∈ \overline{Q}(x)[Y] whose Galois group is G.

The validity of our construction depends on the following result of Burnside which shows that if V is a faithful G-module, then any irreducible G-module is a G-summand

of V^{⊗n} = \underbrace{V ⊗ … ⊗ V}_{n times} for some n ≥ 1:

Theorem 5.1. ([0],[0] p. 25) Let V be a finite dimensional vector space and G ⊂ GL(V) a finite group. If W is a finite dimensional vector space on which G acts irreducibly, then for some n ≥ 1, W appears as a direct summand of V^{⊗n}.

We shall also need the following result which shows that one can construct a linear differential equation whose solution space is isomorphic to the tensor product of the solution spaces of two given linear differential equations.

Proposition 5.2. Let L_{1}(y) = 0 and L_{2}(y) = 0 be linear differential equations with coefficients in \overline{Q}(x) of orders n and m respectively. One can effectively construct a linear differential equation L_{1} ⊗ L_{2}(y) = 0 with coefficients in \overline{Q}(x) having the following property: if K is a Picard-Vessiot extension of \overline{Q}(x) such that L_{1}(y) = 0 (resp. L_{2}(y) = 0) has n (resp. m) linearly independent solutions in K, then the solution space of L_{1}⊗L_{2}(y) = 0 is G(K/\overline{Q}(x)) isomorphic to V_{1}⊗V_{2} where V_{1} (resp. V_{2}) is the solution space of L_{1}(y) = 0 (resp. L_{2}(y) = 0) in K.

Proof. Let L_{2}^{(i)}(y) = 0 be the differential equation whose solution space is {y^{(i)} | y ∈ V_{2}} and let α be a nonsingular point of L_{1}◯ s L_{2}^{(i)}(y) for i = 0, …, m − 1. Such a point exists and can be effectively found since each L_{2}^{(i)}(y) = 0 and so, each L_{1}◯ s L_{2}^{(i)}(y) can be effectively constructed. For notational convenience, we assume α = 0. Let t be any integer greater than or equal to the orders of the L_{1}◯ s L_{2}^{(i)}(y). Note that if z is a solution of L_{1}◯ s L_{2}^{(i)}(y) = 0 and z(0) = z′(0) = … = z^{(t)}(0) = 0, then z = 0. Let K be a Picard-Vessiot extension of \overline{Q}(x) and let w_{1}, …, w_{m} be a basis of the solution space V_{2} of L_{2}(y) = 0 in K. Let u_{i} = ∑_{j=1}^{m−1} x^{j·t}w_{i}^{(j)}. One can show that the u_{i} are linearly

independent (since det(w_{i}^{(j)}) ≠ 0) and that they form a basis of a G(K/\overline{Q}(x)) module isomorphic to V_{2}. Let y_{1}, …, y_{n} be a basis of V_{1} in K and consider the elements {y_{i}u_{j}} in K. We claim that these are linearly independent over the constants. To see this, let ∑_{i,j} c_{ij}y_{i}u_{j} = 0 for some constants c_{ij}. This implies that ∑_{k=0}^{m−1} x^{k·t} ∑_{i,j} c_{ij}y_{i}w_{j}^{(k)} = 0.

Note that each z_{k} = ∑_{i,j} c_{ij}y_{i}w_{j}^{(k)} is a solution of L_{1}◯ s L_{2}^{(k)}(y) = 0 and so has a zero of

order at most t − 1 at x = 0 if z_{k} ≠ 0. Therefore each term x^{k·t} ∑_{i,j} c_{ij}y_{i}w_{j}^{(k)} is either zero or has a zero of order between k · t and k · t + t − 1. Since these terms sum to zero, each of them must equal zero. Therefore, for each k, 0 ≤ k ≤ m − 1, ∑_{i,j} c_{ij}y_{i}w_{j}^{(k)} = 0.

Since det(w_{j}^{(k)}) ≠ 0, we have for each j, 1 ≤ j ≤ m, ∑_{i=1}^{n} c_{ij}y_{i} = 0. Since the y_{i} are linearly independent, we have all c_{ij} = 0. Therefore the elements {y_{i}u_{j}} are linearly independent over the constants. One now sees that they form the basis of a G(K/\overline{Q}(x)) module isomorphic to V_{1} ⊗ V_{2}. Proceeding as in Theorem 3.4, one sees that these form the basis of the solution space of a linear differential equation L_{1} ⊗ L_{2}(y) = 0 with coefficients in \overline{Q}(x) and that this operator can be constructed just knowing L_{1}(y), L_{2}(y) and the integer t. □

Corollary 5.3. Let L(y) = 0 be a linear differential equation of order n with coefficients in \overline{Q}(x). For any m, one can effectively construct a linear differential equation L_{1}^{⊗m}(y) = 0 with coefficients in \overline{Q}(x) having the following property: if K is a Picard-Vessiot extension of \overline{Q}(x) such that L_{1}(y) = 0 has n linearly independent solutions in K, then the solution space of L_{1}^{⊗m}(y) = 0 is G(K/\overline{Q}(x)) isomorphic to V^{⊗m} where V is the solution space of L(y) = 0.

Combining Theorem 5.1 and Corollary 5.3, we see that the desired differential equation will eventually occur as a factor of some L^{⊗m}(y) = 0. The orders of these equations grow very quickly. Sometimes one can find the desired differential equation by looking at only symmetric powers: Example: The differential equation

\frac{d^{2}y}{dx^{2}} + \frac{21}{100} \frac{x^{2} − x + 1}{x^{2}(x − 1)^{2}} y = 0

is irreducible and G(L) ≅ A_{5}^{SL_{2}} (see [0], p. 342). According to Table 1, the third symmetric power of this differential equation

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

is irreducible and has galois group A_{5}. In order to prove that G(L) ≅ A_{5} using factorization of differential operators over over \overline{Q}(x), it is enough (cf., Theorem 4.6 and 4.7) to show that:

1 L(y) is irreducible.

2 L^{◯ s 3} has no solution y such that y^{2} ∈ \overline{Q}(x).

3 L^{◯ s 2} has order 5 or factors.

4 L^{◯ s 3} has a factor of order 3.

We note that, since L^{◯ s 2} is the fourth symmetric power of the above second order equation, L^{◯ s 2} will be of order 5 in this case (cf., Lemma 3.5). In this case, the fact that L^{◯ s 3} has no solution y such that y^{2} ∈ \overline{Q}(x) will follow from a factorization of L^{◯ s 3}, which (if G(L) ≅ A_{5}) will have no factor of order 1.

Assume we are given an irreducible polynomial P (Y) ∈ \overline{Q}(x) such that the galois group of P (Y) = 0 is G and an irreducible representation of G ∈ GL(V). Assume P (Y) has degree n. Differentiating P (Y) = 0, and successively solving for the derivatives of Y and reducing mod P (Y). we get for i = 0, …, n expressions of the form Y^{(i)} = a_{i,o} + a_{i,1}Y + … + a_{i,n−1}Y^{n−1}, with the a_{i,j} ∈ \overline{Q}(x). These n + 1 expressions in the n terms Y^{j} must be linearly dependent over \overline{Q}(x), so we can find a linear differential equation L(y) = y^{(n)} + … + a_{0}y = 0 with n′ ≤ n and coefficients in \overline{Q}(x) whose solution space is spanned by the roots of P (Y) = 0. Note that G(L) is G. For some value of m, L^{⊗m}(y) = 0 has a solution space having a subspace G(L) isomorphic to W. Therefore some factor of L^{⊗m}(y) = 0 will have a solution space G(L) isomorphic to W.

There are two problems in using the above method. The first is that one needs to determine the representation (or at least its character) of G(L) on the solution space of L(y) = 0 in order to be able to predict for which value of m L^{⊗m}(y) = 0 has a solution space having a subspace isomorphic to W. The second problem is that L^{⊗m}(y) = 0 may have many factors of the same order whose solution spaces are different G-modules. One is faced with the problem of determining which factor gives the desired representation. Nonetheless, the above argument shows that such an operator always exists. We now give an example, where some of these problems can be avoided.

Example: We will show how a third order differential equation with differential galois group G_{168} can be constructed using the polynomial

Y^{7} − 56Y^{6} + 609Y^{5} + 1190Y^{4} + 6356Y^{3} + 4536Y^{2} − 6804Y − xY^{3}(Y + 1) − 5832,

which is irreducible over \overline{Q}(x) and has galois group G_{168} ≅ P SL_{2}(7) (se e.g. [0], p. 188). Using the variable transformation Z = Y − \frac{1}{7} we get the irreducible polynomial

Y^{7} − 735 Y^{5} − (x + 10290) Y^{4} − (33x − 4116) Y^{3} − (408x − 979608) Y^{2}
−(2240x − 7020524) Y − (4608x − 15731352),

which we denote P (Y), whose galois group over \overline{Q}(x) is also G_{168}. We denote y_{1}, …, y_{7} the solutions of P (Y) = 0 and K the splitting field of P (Y) = 0 over \overline{Q}(x). The functions y_{1}, …, y_{7}, whose derivatives all belong to K, will satisfy a differential equation L(y) = 0 of order at most 7 with coefficients in \overline{Q}(x). If L(y) = 0 is of degree 7, then G(L) is equivalent to a permutation representation of degree 7 of G_{168}. Such a permutation representation has an invariant subspace generated by y_{1} +…+y_{7}. Since y_{1} +…+y_{7} = 0 by construction, the functions y_{1}, …, y_{7} must satisfy a differential equation L(y) = 0 of degree at most 6, and computation shows that L(y) = 0 is in fact of degree 6. This differential equation is not of the form given in Theorem 3.3, but since G_{168} is a perfect group, the corresponding differential galois group will be unimodular. The group G_{168} has 6 irreducible characters: the trivial charcater χ_{1}, two characters χ_{3,1} and χ_{3,2} of degree 3 and the characters χ_{6}, χ_{7}, χ_{8} of degree 6, 7 and 8. According to these characters, if L(y) = 0 is reducible, then it either has an irreducible factor of order 3 or only irreducible factors of order 1. The last case clearly can not happen. Thus, if a factorization of L(y) = 0 does not produce an irreducible third order equation (one can show that this is not possible), then L(y) = 0 is an irreducible equation of order 6 with G(L) ≅ G_{168}. Using the above Corrolary 5.3 we can construct an equation L^{⊗m}(y) = 0 whose solution space is isomorphic to V^{⊗m} where V is the solution space of L(y) = 0. From Theorem 5.1 we get that for some m the character of G(L^{⊗m}) contains a character of degree 3 of G_{168}. Decomposing powers of χ_{6} we get:

(χ_{6})^{2} = χ_{1} + 2χ_{6} + χ_{7} + 2χ_{8}
(χ_{6})^{3} = 2χ_{1} + 3χ_{3,1} + 3χ_{3,2} + 10χ_{6} + 8χ_{7} + 10χ_{8}

Thus by factoring L^{⊗3}(y) = 0 one will get an irreducible third order differential equation with galois group G_{168}.

The above example shows that working with tensor products instead of symmetric products leads to differential equations of very large order containing a large amount of redundancy. This is the reason why we have stated our main results using symmetric powers instead of tensor powers.

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