Calculating the Galois Group of L_{1}(L_{2}(y)) = 0, L_{1}, L_{2} Completely Reducible Operators

P. H. Berman M.F. Singer^{∗} Department of Mathematics, Box 8205, North Carolina State University, Raleigh, NC 27695-8205

    phberman@eos.ncsu.edu    singer@math.ncsu.edu
          http://www.math.ncsu.edu/∼singer

June 18, 1998

1 Introduction

In [3] a decision procedure is described that computes the Galois group of a completely reducible linear differential operator with rational or algebraic function coefficients (i.e., a linear differential operator that is the least common left multiple of irreducible operators or, equivalently, one whose Galois group is a reductive group). At present, it is unknown how to calculate the Galois group of a general operator. In this paper, we push beyond the completely reducible case by showing how to compute the Galois group (and Picard-Vessiot extension) of an operator of the form L_{1} ∘ L_{2} where L_{1} and L_{2} are completely reducible and have rational function coefficients.

The paper is organized as follows. In Section 2 we show how to compute the Galois group and Picard-Vessiot extension of an equation of the form L(y) = b with L completely reducible. This corresponds to the case of L_{1} ∘ L_{2} where L_{1} = D − b′/b. In Section 3, we show how one can reduce the general case to the above case and give several examples.

∗The preparation was partially supported by NSF Grant CCR-93222422

We would like to thank Mark van Hoeij for making available to us his wonderful improvements to the DEtools package in Maple.

2 Calculating the Galois Group of L(y) = b, L Completely Reducible

Let k be a differential field of characteristic zero with algebraically closed field of constants C and let D = k[D] be the ring of differential operators with coefficients in k. For any L ∈ D and b ∈ k, we shall use the phrase the Picard-Vessiot extension of L(y) = b to denote the Picard-Vessiot extension of k corresponding to (D − b′/b)L, that is, the smallest Picard-Vessiot extension of k containing a full set of solutions of L(y) = 0 as well as a specific solution of L(y) = b. We shall call the Galois group of this Picard-Vessiot extension the Galois group of L(y) = b. We will combine the techniques of [1],[2] with those of [3] to show how, given L ∈ k[D], L completely reducible, and b ∈ k, one can calculate the Galois group and Picard-Vessiot extension of L(y) = b. Note that an operator is said to be completely reducible if it is the least common left multiple of irreducible operators. An equivalent condition is that the Galois group G be reductive ([7], p.125), that is, the largest normal subgroup of G consisting of unipotent elements is trivial (for equivalent conditions see Lemma 2.13 of [13] and Proposition 2.2 of [3]). The key is the following proposition, which is a slight modification of Théorème 1 of [2]. We say that an algebraic group is a vector group if it is isomorphic to (C^{n}, +) for some n.

Proposition 2.1 Let k, b and L be as above. The Galois group of L(y) = b is isomorphic to the semidirect product W ⋊ G_{L}, of the Galois group G_{L} of L(y) = 0 and a vector group W. Furthermore, if L_{1} ∈ k[D] is a monic operator of maximal order satisfying

1. L_{1}(y) = b has a solution f_{1} ∈ k, and

2. L = L_{1}L_{0} for some L_{0} ∈ k[D]

then W is G-isomorphic to the solution space of L_{0}. In addition such an L_{1} is unique.

Proof: We first note that the operator L_{1} = 1 satisfies conditions 1. and 2. with L_{0} = L. Therefore there will exist an operator L_{1} of maximal order satisfying these conditions.

Let K be the Picard-Vessiot extension of k corresponding to (D−b′/b)L and K_{L} ⊂ K be the Picard-Vessiot extension corresponding to L. If L = L_{1}L_{0}, then K_{L} will contain fundamental sets of solutions of both L_{0}(y) = 0 and L_{1}(y) = 0 and so it makes sense to speak of the action of G_{L} on the solution spaces of these two equations.

Let f ∈ K be a solution of L(y) = b. For any σ ∈ G = Gal(K/k), σ(f)−f is in the solution space V_{L} of L(y) = 0. Let Φ : G → V_{L} be the map sending σ to σ(f) − f. Let H be the normal subgroup of G leaving K_{L} fixed.

Since K = K_{L} < f >, we see that Φ is injective on H. For any σ ∈ G, τ ∈ H, we have that

Φ(στσ^{−1}) = στσ^{−1}(f) − f
= σ[τ(σ^{−1}(f) − f) + τ(f)] − f
= σ[σ^{−1}(f) − f] + στ(f) − f since τ fixes the elements of V_{L} ⊂ K_{L}
= σ[τ(f) − f]
= σΦ(τ)

This calculation (from the proof of Théorèm 1 of [2]) shows that Φ is a G-morphism, where the action of G on H is given by conjugation. Therefore, Φ identifies H with a G-invariant subspace W of V_{L}. Since G/H is isomorphic to the reductive group G_{L} and H is unipotent, H is the unipotent radical of G. Any linear algebraic group may be written as a semidirect product G = H ⋊P where H is the unipotent radical of G and P is a reductive group (called a Levi subgroup of G, [11]). Clearly, P is isomorphic to G_{L}.

Let L̃_{0} be the monic operator in k[D] whose solution space is W and let L = L̃_{1}L̃_{0}. Since σ(f) − f ∈ W for all σ ∈ H, we have that L̃_{0}(σ(f)) = L̃_{0}(f) for all σ ∈ H. Therefore, L̃_{0}(f) ∈ K_{L}. Let W_{1} be the solution space of L̃_{1} in K_{L} and let W_{L̃_{0}(f)} be the space spanned by W_{1} and L̃_{0}(f). For any σ ∈ G_{L}, σ(L̃_{0}(f)) is again a solution of L̃_{1}(y) = b and so the space W_{L̃_{0}(f)} is left invariant by G_{L}. Furthermore W_{L̃_{0}(f)}/W_{1} is a trivial one-dimensional G_{L}- module. Since G_{L} is a reductive group, W_{1} has a G_{L}-complement in W_{L̃_{0}(f)}. This implies that there is an element f_{0} ∈ K_{L} such that f_{0} ≡ L̃_{0}(f) mod W_{1} and f_{0} is left fixed by G_{L}. We conclude from this that f_{0} ∈ k and L̃_{1}(f_{0}) = b.

Now let L_{1} satisfy 1. and 2. above. Since L_{1}(L_{0}(f)) = b, we have that L_{0}(f) − f_{1} ∈ W_{1}, where W_{1} is the solution space of L_{1}. In particular, L_{0}(f) ∈ K_{L}. Therefore, for any σ ∈ H, L_{0}(σ(f) − f) = 0. This implies that the image of Φ lies in the solution space of L_{0}. Therefore, L̃_{0} divides L_{0} on the right and so the order of L_{1} is at most the order of L̃_{1}. If these two orders are the same and L_{1} is monic, then L̃_{0} = L_{0} and so we must then have that L_{1} = L̃_{1}.

The following example illustrates this Proposition:

Example 2.2 Let k = C(x) and L = D^{2}−4xD+(4x^{2}−2) = (D−2x)∘(D−2x). A basis for the solution space of equation L(y) = 0 is {e^{x^{2}}, xe^{x^{2}}} so the Galois group of this homogeneous equation over k is C^{∗}.

For any (c, d) ∈ C^{2}, (c, d) ≠ (0, 0), we have that (c + dx)e^{x^{2}} is a solution of L(y) = 0 and so L has a right factor of the form D −(2x+ \frac{d}{c+dx}). Furthermore, all right factors of order one are of this form. Therefore the formula

L = (D − (2x − \frac{d}{c + dx})) ∘ (D − (2x + \frac{d}{c + dx}))

with (c, d) ≠ (0, 0) yields a parameterization of all irreducible factorizations of L.

We shall now compute the Galois groups of L(y) = b where b = 4x^{2} − 2, 1 and \frac{1}{x}.

(i) b = 4x^{2} − 2. In this case the equation L(y) = b has the rational solution y = 1. This implies that the W of Proposition 2.1 is trivial and so the Galois group of L(y) = b is C^{∗}.

(ii) b = 1. A partial fraction computation shows that L(y) = 1 has no rational solutions. Now let us search for first order left factors L_{1} of L such that L_{1}(y) = 1 has a rational solution. A calculation shows that the equation

y′ − (2x − \frac{d}{c + dx})y = 1  (1)

has a rational solution y = f if and only if z = (c + dx)f is a rational solution of

z′ − 2xz = c + dx  (2)

(c.f., Lemma 2.4). The rational solutions of (2) must be polynomials and one sees that this has a polynomial solution if and only if c = 0. Therefore the space W of Proposition 2.1 is the solution space of y′ − (2x + \frac{1}{x})y = 0, that is, the space spanned by xe^{x^{2}} in the solution space of L(y) = 0. Therefore the Galois group of L(y) = 1 is C ⋊ C^{∗}.

(iii) b = \frac{1}{x}. We shall show that for any (c, d) ≠ (0, 0), the equation

y′ − (2x − \frac{d}{c + dx})y = \frac{1}{x}  (3)

has no rational solution. This implies that L(y) = \frac{1}{x} also has no rational solution and so the W of Proposition 2.1 is the solution space of L(y) = 0. Therefore the Galois group of L(y) = \frac{1}{x} is C^{2} ⋊ C^{∗}. Equation (3) has a rational solution y = f if and only if z = (c + dx)f is a rational solution of

z′ − 2xz = \frac{c + dx}{x} .  (4)

If c ≠ 0 then any rational solution of (4) must have a pole at x = 0. Comparing orders of the left and right hand side of this equation yields a contradiction. Therefore c = 0. Similar considerations show that z′ − 2xz = d can never have a rational solution if d ≠ 0.

Proposition 2.1 allows us to give a detailed description of the Picard-Vessiot extension of k corresponding to L(y) = b.

Corollary 2.3 Let k, b, L be as above and let K_{L} be the Picard-Vessiot extension of k corresponding to L(y) = 0. Let L_{1} ∈ k[D] be the unique monic operator of maximal order satisfying

1. L_{1}(y) = b has a solution f_{1} ∈ k, and

2. L = L_{1}L_{0} for some L_{0} = D^{t} − b_{t−1}D^{t−1} − … − b_{o} ∈ k[D].

Then the Picard-Vessiot extension K of k corresponding to L(y) = b is the field K_{L}(z_{0}, z_{1}, …, z_{t−1}) where z_{0}, z_{1}, …, z_{t−1} are algebraically independent, z′_{i} = z_{i+1} for i = 0, …, t − 2 and z′_{t−1} = f_{1} + b_{0}z_{0} + … b_{t−1}z_{t−1}.

Proof: Let V be the solution space of (D − b′/b)L in K. The linear operator L_{0} maps V onto the solution space of (D − b′/b)L_{1}. Therefore, there exists a z ∈ V such that L_{0}(z) = f_{1}. Since L(z) = b, we have that K = K_{L} < z >. Since the Galois group of K over K_{L} is a vector group of dimension t, we have that K is a purely transcendental extension of K_{L} of transcendence degree t. Therefore K = K_{L}(z, z′, …, z^{(t−1)}). The elements z_{i} = z^{(i)} satisfy the conclusion of the Corollary.

Proposition 2.1 also implies that in order to find the Galois group of L(y) = b we must

1. Calculate the Galois group G_{L} of L, and 2. Find the monic operator L_{1} of maximal order satisfying 1. and 2. of

Proposition 2.1 and identify the action of G on the solution space of L_{0}.

The first task was dealt with in [3]. In this paper it is shown (Theorem 4.1^{1} and its proof) how for all points z_{0} ∈ ℂ outside some finite set (depending on L), one can calculate a matrix representation for the Galois group of L in the basis {y_{0}, …, y_{n−1}} of the solution space given by y_{i}^{(j)}(z_{0}) = δ_{i,j}.

Dealing with the second task will occupy the remainder of this section. We begin by recalling some basic definitions and facts concerning linear differential equations.

Two operators L_{2} and L_{1} are said to be equivalent if the D-modules D/DL_{2} and D/DL_{1} are D-isomorphic (see [13]). This is equivalent to the statement that the two operators have the same order m and that there exist operators R, S of orders at most m − 1 with GCRD(R, L_{1}) = 1 such that

L_{2}R = SL_{1} .  (5)

Note that such an operator R can be used to define a map 1 ↦ R which gives the isomorphism from D/DL_{2} to D/DL_{1}. We note that the ring D is a left and right euclidean domain. In particular given operators U, V ∈ D an extended euclidean algorithm yields operators A, B ∈ D, ord A < ord V, ord B < ord U, such that AU + BV = GCRD(U, V).

Lemma 2.4 Let L_{1}, L_{2} ∈ D be equivalent operators and S ∈ D as in equation (5). The equation L_{1}(y) = b, b ∈ k has a solution in k if and only if the equation L_{2}(y) = S(b) has a solution in k.

Proof: The extended euclidean algorithm yields R̃ and L̃_{1} in D such that R̃R + L̃_{1}L_{1} = 1 and ord R̃ < ord L_{1}. The map v ↦ R(v) is an isomorphism of V_{L_{1}} (the solution space of L_{1}) onto V_{L_{2}} and the map w ↦ R̃(w) is the inverse of this isomorphism ([13]). Since L_{1}R̃ and RR̃ − 1 vanish on V_{L_{2}}, we have that L_{2} divides both of these operators. Therefore there exist S̃ and L̃_{2} ∈ D such that L_{1}R̃ = S̃L_{2} and RR̃ + L̃_{2}L_{2} = 1.

We now claim that S̃S + L_{1}L̃_{1} = 1. We have that

(S̃S + L_{1}L̃_{1})L_{1} = S̃SL_{1} + L_{1}L̃_{1}L_{1}
= S̃L_{2}R + L_{1}(1 − R̃R)
= S̃L_{2}R + L_{1} − L_{1}R̃R
= S̃L_{2}R + L_{1} − S̃L_{2}R
= L_{1},

1This Theorem is stated in terms of matrix systems Y′ = AY but the translation to scalar equations L(y) = 0 is immediate.

and the equation follows after cancelling L_{1} on the right.

To prove one direction of the lemma, suppose L_{1}(f) = b for some f ∈ k. If h = R(f) ∈ k, then L_{2}(h) = SL_{1}(f) = S(b) as desired. To prove the other direction, suppose L_{2}(h) = S(b) for some h ∈ k. Let f = R̃(h) + L̃_{1}(b) ∈ k. Then

L_{1}(f) = L_{1}R̃(h) + L_{1}L̃_{1}(b)
= S̃L_{2}(h) + (1 − S̃S)(b)
= S̃S(b) + b − S̃S(b)
= b,

completing the proof.

Any operator can be written as a product of irreducible operators and for any other factorization, one has the same number of irreducible factors. Moreover, after a possible renumbering, the irreducible factors are equivalent. By definition any completely reducible operator L can be written as the least common left multiple of a finite set of irreducible operators. Any left or right factor will therefore be equivalent to the least common left multiple of some subset of these operators. When k is a finite algebraic extension of C(x), where C is a computable algebraically closed field of characteristic zero, one can effectively factor any differential operator into a product of irreducible differential operators, determine if an operator is completely reducible and, if so, effectively write it as a least common left multiple of irreducible operators [3].

We shall attack the second task above in the following way. We start by writing L as the least common left multiple of a set of irreducible operators T = {T_{1}, …, T_{s}}. Any monic operator L_{1} dividing L on the left is equivalent to a least common left multiple of elements from T . We fix a subset of T and let L_{2} be the least common left multiple of elements of this subset. We will show below that one can parameterize all pairs of elements (L_{1}, S), ord S < ord L_{1} = ord L_{2}, where S and L_{2} are as in equation (5) and L_{1} divides L on the left. We furthermore will show that one can decide if there are values of the parameters so that L_{1}(y) = b has a solution in k. Performing these tasks over all subsets of T , we will eventually find an operator L_{1} of maximal order satisfying conditions 1. and 2. of Proposition 2.1. We will then show how to describe the action of G_{L} on the solution space of L_{0} where L = L_{1}L_{0}.

The following three lemmas are used to describe the set of pairs (L_{1}, S) mentioned above. The fourth lemma will be used to decide if there are values of the parameters so that L_{1}(y) = b has a solution in k.

Let k = C(x). For a = \frac{p}{q} ∈ k, p, q ∈ C[x], (p, q) = 1, we define deg a = max(deg p, deg q). For L = a_{n}D^{n} + a_{n−1}D^{n−1} + … + a_{0} ∈ k[D], we define deg L = max_{1≤i≤n}(deg a_{i}) and ord L = n. Given operators L and L_{2}, we will want to parameterize all pairs of operators (L_{1}, S) satisfying:

1. L_{1} is a monic operator equivalent to L_{2} that divides L on the left, and

2. ord S ≤ ord L_{2} − 1 and L_{2}R = SL_{1} for some R ∈ D, GCRD(L_{1}, R) = 1.

Lemma 2.5 Let T_{1} and T_{2} be operators with coefficients in C(x) of orders n and m and degrees N and M respectively. If T_{3} is an operator with coefficients in C(x) such that T_{3}T_{2} = T_{1}, then deg T_{3} ≤ (n − m + 1)^{2}M + N.

Proof: Let T_{1} = ∑_{i=0}^{n} a_{i}D^{i}, T_{2} = ∑_{i=0}^{m} b_{i}D^{i} and T_{3} = ∑_{i=0}^{n−m} c_{i}D^{i}. The equation T_{3}T_{2} = T_{1} yields a system of (algebraic) linear equations

( a_{n} ; a_{n−1} ; ⋮ ; a_{0} ) = B ( c_{n−m} ; c_{n−m−1} ; ⋮ ; c_{0} ) ,

where B is a matrix whose entries are sums of terms of the form D^{j}(b_{i}), 0 ≤ j ≤ n−m. Therefore deg B ≤ (n−m+1)M. A solution of this system will be unique so the matrix B has rank n−m. Therefore there is an (n−m)×(n−m) invertible submatrix B̃ of B. For convenience of notation we shall assume this is formed by the first n − m rows of B. We then have

( c_{n−m} ; c_{n−m−1} ; ⋮ ; c_{0} ) = B̃^{−1} ( a_{n} ; a_{n−1} ; ⋮ ; a_{m+1} ) .

Since deg B̃^{−1} ≤ (n − m + 1) deg B̃ ≤ (n − m + 1)^{2}M, we have that

deg ( c_{n−m} ; c_{n−m−1} ; ⋮ ; c_{0} ) = deg(B̃^{−1} ( a_{n} ; a_{n−1} ; ⋮ ; a_{m+1} ))

≤ (n − m + 1)^{2}M + N .

Lemma 2.6 Let k = C(x) and L, L_{2} be monic operators in D of orders n and m respectively.

1. For any i, 0 ≤ i ≤ ord L one can effectively find an integer n_{i} such that if L = L_{1}L_{0} with monic L_{1}, L_{0} ∈ D and ord L_{1} = i then deg L_{0} ≤ n_{i}.

2. One can effectively find an integer N such that if L_{2}R̃ = S̃L for some R̃, S̃ ∈ D with ord R̃ < ord L and ord S̃ < ord L_{2}, then deg S̃ ≤ N.

3. One can effectively find an integer M such that if L_{1} is a monic operator equivalent to L_{2}, dividing L on the left, then there exist R and S in D such that L_{2}R = SL_{1}, ord R < ord L_{1}, ord S < ord L_{2} and deg R, deg S ≤ M.

Proof: 1. This fact is well known (c.f., Section 5.1 of [3] and [5]; the latter paper also contains explicit bounds) and so we only outline the proof. Let L = L_{1}L_{0} and ord L_{1} = n − i and let {y_{1}, …, y_{n−i}} be a fundamental set of solutions of L_{1}(y) = 0. The coefficients of L_{1} are quotients a/w where w is the wronskian determinant of {y_{1}, …, y_{n−i}} and a is the determinant of some (n − i) × (n − i) submatrix of W = (y_{l}^{(j)})_{l=1,...,n−i}^{j=0,...,n−i} (note that w is also the determinant of such a matrix). Since the logarithmic derivative of w is in k, we have that the logarithmic derivative of a is also in k. Furthermore, the determinants of (n − i) × (n − i) submatrices of W satisfy an equation L^{∧i}(y) = 0 where L^{∧i} is an operator that can be effectively constructed from L. Therefore the coefficients of L_{0} are quotients of two solutions of L^{∧i}(y) = 0, each of which has logarithmic derivative in k. One can effectively find sets of rational functions {g_{r}}, {f_{rs}} such that the elements e^{∫ g_{r}} are algebraically independent over k and if y satisfies L^{∧i}(y) = 0, y′/y ∈ k, then for some r there are constants {c_{s}} such that y = (∑ c_{s}f_{rs})e^{∫ g_{r}}. If a quotient of two such elements lies in k it will be of the form \frac{∑ d_{s}f_{rs}}{∑ c_{s}f_{rs}} for some constants c_{s}, d_{s}, and so deg L_{0} can be bounded in terms of the degrees of the f_{rs}.

2. We will consider the dual equation R̃^{∗}L_{2}^{∗} = L^{∗}S̃^{∗} formed by taking adjoints (the adjoint of an operator L = ∑_{i=0}^{n} a_{i}D^{i} is the operator L^{∗} = ∑_{i=0}^{n}(−1)^{i}D^{i}a_{i}, [12], Ch. 10). Ore showed (see [13] for a modern presentation and references) that there exists an (m × m) matrix A whose entries lie in D and can be calculated from the coefficients of L and L_{2} such that S̃^{∗} = s_{0} + s_{1}D + … s_{m−1}D^{m−1} satisfies such an equation if and only if A(s_{o}, … s_{m−1}) = 0. Using row and column operations one can find a set of linear scalar equations equivalent to the system A(s_{0}, …, s_{m−1}) = 0 (see [4]). Using standard algorithms to find rational solutions of scalar linear differential equations one can therefore find a bound on deg S̃^{∗} and therefore on deg S̃ = deg S̃^{∗∗}.

3. Let L = L_{1}L_{0}. From 1., one can effectively calculate an integer n_{i} such that deg L_{0} ≤ n_{i}. Lemma 2.5 then allows us to calculate an integer m_{i} such that deg L_{1} ≤ m_{i}. If L_{2}R = SL_{1} with ord R < ord L_{1} and ord S < L_{2} then L_{2}(RL_{0}) = SL_{1}L_{0} = SL. From 2., we have that there is a computable integer N such that deg S ≤ N. Let i = ord L_{2} = ord L_{1}. Taking adjoints in the equation L_{2}R = SL_{1} we have that R^{∗}L_{2}^{∗} = L_{1}^{∗}S^{∗}. One can bound deg L_{1}^{∗}S^{∗} in terms of deg L_{1}^{∗} and deg S^{∗}. Applying Lemma 2.5 to the equation R^{∗}L_{2}^{∗} = L_{1}^{∗}S^{∗} allows us to bound deg R^{∗} and therefore deg R.

Before stating the next lemma, we introduce the following notion. Let L = ∑_{i=0}^{n} a_{i}D^{i} ∈ C(x)[D] be an operator of order n and degree at most m where each

a_{i} = \frac{∑_{j=0}^{m} b_{i,j}x^{j}}{∑_{j=0}^{m} c_{i,j}x^{j}}

with b_{i,j}, c_{i,j} ∈ C. We may identify the operator L with the vector (b_{0,0}, b_{0,1}, …, c_{n−1,m}) ∈ C^{2(n+1)(m+1)}. We say that a set of operators L is constructible if, under this identification, they form a constructible subset of C^{2(n+1)(m+1)}.

Lemma 2.7 Let k, L and L_{2} be as in the hypotheses of Lemma 2.6.

1. The set of pairs of monic operators (L_{1}, L_{0}), ord L_{1} = m, ord L_{0} = n−m such that L = L_{1}L_{0} forms a constructible set whose defining equations can be explicitly computed.

2. Let n_{m} be as in Lemma 2.6.1 and M be as in Lemma 2.6.3. The set P_{L_{2}} of triples of operators (L_{1}, R, S) where

(a) ord L_{1} = m; deg L_{1} ≤ n_{m}; ord R, S ≤ m − 1; deg R, S ≤ M,

(b) L_{1} divides L on the the left,

(c) GCRD(L_{1}, R) = 1,

(d) L_{2}R = SL_{1} (and so L_{1} is equivalent to L_{2})

is constructible. Furthermore, one can effectively calculate the defining equations of P_{L_{2}}.

3. The set M_{L_{2}} of pairs (L_{1}, S) such that for some R ∈ D, (L_{1}, R, S) ∈ P_{L_{2}} is a constructible set. Furthermore, one can effectively calculate the defining equations of M_{L_{2}}.

Proof: 1. By Lemmas 2.5 and 2.6.1, we can effectively bound the degrees of L_{1} and L_{0}. Comparing the coefficients of D^{i} in the equation L = L_{1}L_{0} yields defining conditions for the constructible set.

2. Projecting the set defined in 1. yields the set L_{1} of L_{1} that divide L on the left. Therefore this is a constructible set. Lemma 2.6.3 yields a bound on the degrees of R and S. Therefore, the set of (L_{1}, R, S) such that L_{2}R = SL_{1}, L_{1} ∈ L_{1} is a constructible set. Imposing the condition that GCRD(L_{1}, R) = 1 yields a constructible subset and projecting (L_{1}, R, S) ↦ (L_{1}, R) yields P_{L_{2}}.

3. This follows from the fact that the projection of a constructible set is constructible.

Let L = y^{(n)} + a_{n−1}y^{(n−1)} + … + a_{0}y ∈ C(x)[D], let α ∈ C and let m_{α} = min_{0≤i≤n}{m_{i} − i} where m_{i} is the order of a_{i} at α. Let S_{L} be the set of pairs (m, α) where m is a negative integer, α ∈ C and L((x − α)^{m}) has order at α strictly larger than m + m_{α}. If (m, α) ∈ S_{L}, then α must be a finite singular point of L. Furthermore, for any finite singular point α, there is a nonzero polynomial equation (the indicial equation at α) such that the set of m with (m, α) ∈ S_{L} is precisely the set of roots of this equation. This equation can be effectively determined from the (x−α)-adic expansion of the a_{i}. In particular, the set S_{L} is a finite set that can be effectively determined. Similar calculations can be done at infinity (i.e., at 0 for the equation obtained after replacing x by 1/t and d/dx by −t^{2}d/dt) to determine a finite set S_{L}^{∞} of positive integers m having the property that L(x^{m}) has order at infinity strictly larger than −m + m_{∞}. Here m_{∞} = min_{0≤i≤n}{m̃_{i} + i} where m̃_{i} is the order of a_{i} at infinity.

Lemma 2.8 Let k = C(§), N an integer and L = y^{(n)} + … + a_{0}y ∈ D. The set V of (c_{0}, …, c_{N}, d_{0}, …, d_{N}) ∈ C^{2N+2} such that

L(y) = \frac{c_{N}x^{N} + … + c_{0}}{d_{N}x^{N} + … + d_{0}}  (6)

has a solution in k, is constructible. Furthermore, one can effectively find the defining equations of V.

Proof: Although this result can be deduced from the results of [5], we present a direct proof here. We shall show that there is an a priori bound on the degrees of the denominator and numerator of such a solution. Let f = \frac{p}{q} ∈ k, (p, q) = 1 be a solution of (6) for some fixed (c_{0}, …, c_{N}, d_{0}, …, d_{N}) ∈ C^{2N+2}. We first claim that q divides

Q = ∏_{(m,α)∈S_{L}}(x − α)^{−m_{α}} · (d_{N}x^{N} + … + d_{0}) .

Let

q = ∏(x − α)^{n_{α}}

be the factorization of q. If (−n_{α}, α) ∈ S_{L}, then (x − α)^{n_{α}} divides Q. If (−n_{α}, α) ∉ S_{L} then L(f) has a pole of order n_{α}+n at α. Therefore (x−α)^{n_{α}+n} divides d_{N}x^{N} + … + d_{0} and so our first claim is proved. In particular, we have a bound for the degree of the denominator of f. To bound the degree of the numerator, we expand at infinity. We have that f = x^{deg p−deg q} + smaller powers of x . If deg p−deg q > deg(c_{N}x^{N} +…+c_{0})−deg(d_{N}x^{N} +…+ d_{0}), then (deg p − deg q) ∈ S_{L}^{∞}. Therefore, deg p − deg q < max_{m∈S_{L}^{∞}}{m, N}, so we can bound the degree of p.

Let M be a bound for the degree of the denominators and numerators of possible f. The set of (h_{M}, … h_{0}, k_{M}, …, k_{0}, c_{N}, …, c_{0}, d_{N}, … d_{0}) ∈ C^{2M+2N+4} such that

L(\frac{h_{M}x^{M} + … + h_{0}}{k_{M}x^{M} + … + k_{0}} ) = \frac{c_{N}x^{N} + … + c_{0}}{d_{N}x^{N} + … + d_{0}}

is clearly a constructible set. Projection yields the set V.

We note that the previous lemma is true when k is replaced by an algebraic extension of C(x) and the parameterized rational function is replaced by a parameterized set of elements from k, but for simplicity we have only stated and proved it in the more restricted context. The fact that the set of factorizations of a linear differential operator constitutes a constructible set is well known (c.f., [5] and [14]). Since we need more information, we have presented our method above.

We are now in a position to deal with the second of the above tasks and present the complete algorithm to calculate the Galois group of L(y) = b.

Algorithm I

Input: A completely reducible n^{th} order operator L ∈ C(x)[D] and an element b ∈ C(x).

Output: A set of equations in n^{2} variables defining the Galois group G_{L} of L(y) = 0, an integer t and a rational homomorphism Φ : G_{L} → GL_{t}(C) such that the Galois group of L(y) = b is C^{t} ⋊ G_{L} where the action of G_{L} on C^{t} by conjugation is given by Φ.

1. Write L as a least common left multiple of a set T = {T_{1}, …, T_{s}} of irreducible operators (using, for example, the algorithms in the Appendix of [3]).

2. If L = L_{1}L_{0} then complete reducibility implies that L_{1} is equivalent to the least common left multiple of some subset of T . Fix some subset of T and let L_{2} be the least common left multiple of its elements. For this operator, apply Lemma 2.7.3 to construct the set M_{L_{2}}.

3. Let (L_{1}, S) ∈ M_{L_{2}}. Lemma 2.4 implies that L_{1}(y) = b has a solution in C(x) if and only if the equation

L_{2}(y) = S(b)  (7)

has a solution y ∈ C(x). Apply Lemma 2.8 to equation (7) to determine the set of (L_{1}, S) for which this equation has a rational solution.

4. Repeat steps 2. and 3. until one finds an L_{2} of maximal order so that the set R_{L_{2}} of (L_{1}, S) ∈ M_{L_{2}} for which the equation (7) has a rational solution is nonempty. In this case, Proposition 2.1 implies that there exists a unique L_{1} such that (L_{1}, S) ∈ R_{L_{2}} for some S.

5. We write L = L_{1}L_{0} and let t be the order of L_{0}. Find defining equations of the Galois group G_{L} of L with respect to a basis of the solution space that contains a basis of the solution space of L_{0}(y) = 0 (the results of [3] allow one to do this). In this basis the Galois group will be in block triangular form. Restriction to the space W ≃ C^{t} (i.e., selecting an appropriate block) yields the desired rational map Φ : G_{L} → GL_{t}(C) that gives the action of G_{L} on C^{t} in C^{t} ⋊ G_{L}.

Remarks: 1. One can easily interpret the calculations in Example 2.2 in terms of the method given above. From the factorization L = (D−2x)∘(D−2x) one sees that all factors of L are equivalent to D−2x. All left first order factors are given parametrically by D − (2x − \frac{d}{c+dx}), (c, d) ≠ (0, 0). Therefore to decide if L has a first order left factor L_{1} such that L_{1}(y) = b has a rational solution, we must decide if there exists (c, d) ≠ (0, 0) such that y′ − (2x − \frac{d}{c+dx})y = b has a rational solution. A calculation using Lemma 2.4 implies that this is equivalent to deciding if y′ − 2xy = (c + dx) · b has a rational solution. Lemma 2.8 gives a method to find all (c, d) such that this latter equation has a rational solution.

2. The results of [3] allow one to construct a presentation of the Picard-Vessiot extension associated to L(y) = 0. These can be combined with Corollary 2.3 to give a presentation of the Picard-Vessiot extension associated with L(y) = b.

3 Calculating the Galois Group of L_{1}(L_{2}(y)) = 0, L_{1}, L_{2} Completely Reducible.

We shall show how to reduce this question to the question dealt with in the previous section. This reduction was made by D. Bertrand in [1] in the context of D-modules and this section is devoted to making Bertrand’s reduction explicit and effective.

Despite the title of this section, we will find it useful to deal with first order systems Y′ + AY = 0. We begin by recalling certain facts concerning such systems (c.f., [3], [6]). Let k be a differential field and let A_{1}, A_{2} ∈ M_{n}(k), the ring of n × n matrices over k. We say that the systems Y′ + A_{1}Y = 0 and Y′ + A_{2}Y = 0 are equivalent if there exists a matrix B ∈ GL(n, k) such that A_{2} = −B′B^{−1} + BA_{1}B^{−1}. Let K be a Picard-Vessiot extension of k with Galois group G and assume that Y′ + A_{1}Y = 0, Y′ + A_{2}Y = 0 both have fundamental sets of solutions in K^{n}. One can show that these two systems are equivalent if and only if their solution spaces in K^{n} are isomorphic as Gmodules (in fact, if Y_{1} is a fundamental solution matrix of Y′ + A_{1}Y = 0 then Y_{2} = BY_{1} is a fundamental solution matrix of Y′ + A_{2}Y = 0). To each scalar equation L(y) = y^{(n)} + a_{n−1}y^{(n−1)} + … + a_{0}y_{0} = 0 one can associate a matrix equation Y′ + A_{L}Y = 0 where

A_{L} = ( \begin{matrix}0 & −1 & 0 & … & 0 \\ 0 & 0 & −1 & … & 0 \\ … & … & … & … & … \\ a_{0} & … & … & … & a_{n−1} \end{matrix} ) .

The equation L(y) = b can then be written as Y′ + A_{L}Y = B where B = (0, 0, ….b)^{T}. Conversely if k contains non-constants then any system is equivalent to one of the form Y′ +A_{L}Y = 0 [9]. Finally, it can be shown that if L = L_{1}L_{2} where L_{1} and L_{2} are completely reducible operators, then Y′ + A_{L}Y = 0 is equivalent to a system of the form

Y′ + ( \begin{matrix}A_{2} & C \\ 0 & A_{1} \end{matrix} ) Y = 0  (8)

where the Y′ + A_{1}Y = 0 and Y′ + A_{2}Y = 0 are completely reducible (c.f., [3], Section 2.3). In fact, A_{1} may be taken to be A_{L_{1}}, A_{2} may be taken to be A_{L_{2}}, and C may be taken to be the m×n matrix with entry −1 in the m, 1 position and 0’s everywhere else, where n is the order of L_{1} and m is the order of L_{2}. The following result is a concrete realization of Lemme 1 and the discussion in Section 2 of [1].

Lemma 3.1 If Y_{1} and Y_{2} are fundamental solution matrices of Y′ + A_{1}Y = 0 and Y′ + A_{2}Y = 0 respectively then

Y = ( \begin{matrix}Y_{2} & UY_{1} \\ 0 & Y_{1} \end{matrix} )  (9)

is a fundamental solution matrix of equation (8) if and only if U satisfies U′ + (A_{2}U − UA_{1}) = −C.

Proof: Substitute and calculate.

We note that the equation V′ + (A_{2}V − V A_{1}) = 0 arises naturally when studying systems of linear differential equations. If Y′ + A_{1}Y = 0 and Y′ + A_{2}Y = 0 are two such systems then the map Y ↦ V Y with V an m×n matrix with entries in k maps solutions of Y′ + A_{1}Y = 0 to solutions of Y′ + A_{2}Y = 0 if and only if V′ + (A_{2}V − V A_{1}) = 0. This allows one to identify solutions of this latter equation with elements of HOM(W_{1}, W_{2}) where each W_{i} is the solutions space of Y′ +A_{i}Y = 0. One can furthermore identify HOM(W_{1}, W_{2}) with W_{1}^{∗} ⊗ W_{2} and rewrite the differential equation V′ + (A_{2}V − V A_{1}) = 0 in terms of this identification. We shall now make this explicit^{2}.

Let V be an m × n matrix and let v_{i} be the i^{th} column of V. Let Ṽ be the column vector (v_{1}^{T} , …, v_{n}^{T})^{T}. A calculation shows that V satisfies V′ + A_{2}V − V A_{1} = 0 if and only if Ṽ satisfies Ṽ′ + (−A_{1}^{T} ⊗ I_{m} + I_{n} ⊗ A_{2})Ṽ = 0. Furthermore, if Y_{1} and Y_{2} are fundamental solution matrices of Y′ + A_{1}Y = 0 and Y′ + A_{2}Y = 0 respectively, then a calculation shows that (Y_{1}^{−1})^{T} ⊗ Y_{2} is a fundamental solution matrix of Ṽ′ + (−A_{1}^{T} ⊗ I_{m} + I_{n} ⊗ A_{2})Ṽ = 0. Denoting the columns of C by c_{i} we will let C̃ be the column vector (c_{1}^{T} , …, c_{n}^{T})^{T}.

Lemma 3.2 Assume that k contains a nonconstant. Let Y′ + A_{1}Y = 0 and Y′+A_{2}Y = 0 be completely reducible equations and let K be the Picard-Vessiot extension of k corresponding to equation (8). Let F ⊂ K be the Picard-Vessiot extension corresponding to

Y′ + ( \begin{matrix}A_{2} & 0 \\ 0 & A_{1} \end{matrix} ) Y = 0 .  (10)

Then 1. F contains the Picard-Vessiot extension E corresponding to

2One can formulate this in terms of D-modules (although we shall not need this). Let M_{1} and M_{2} be the D-modules associated to Y′ +A_{1}Y = 0 and Y′ +A_{2}Y = 0 (see [3]). One can put a natural D-module structure on HOM_{k}(M_{1}, M_{2}) such that V ∈ HOM_{k}(M_{1}, M_{2})

defines a D-module isomorphism if and only if V′ + (A_{2}V − V A_{1}) = 0. The D-modules HOM_{k}(M_{1}, M_{2}) and M_{1}^{∗} ⊗ M_{2} are isomorphic D-modules and the above identification of the two equations V′ + (A_{2}V − V A_{1}) = 0 and Ṽ′ + (−A_{1}^{T} ⊗ I_{m} + I_{n} ⊗ A_{2})Ṽ = 0 makes explicit what we need from this identification.

Ṽ′ + (−A_{1}^{T} ⊗ I_{m} + I_{n} ⊗ A_{2})Ṽ = 0  (11)

and so the Galois group G(E/k) is a quotient of G(F/k).

2. K = F (Ṽ) where Ṽ is a solution of

Ṽ′ + (−A_{1}^{T} ⊗ I_{m} + I_{n} ⊗ A_{2})Ṽ = −C̃.  (12)

3. The Galois group G(E(Ṽ)/k) is the semidirect product W ⋊ G(E/k) of the Galois group G(E/k) and a vector space W. Furthermore, the Galois group G(K/k) is the semidirect product W ⋊ G(F/k) where the action of G(F/k) on W is given by composing the quotient map G(F/k) → G(E/k) and the action of G(E/k) on W.

Proof: The first two statements follow from the discussion preceding the Lemma.

To prove the third statement we note that since Y′ + A_{1}Y = 0 and Y′ + A_{2}Y = 0 are completely reducible equations, the Galois group G(F/k) is a reductive group. Since the Galois group G(E/k) is a quotient of G(F/k) it is also reductive. If L̂ is a scalar equation such that Y′ + AL̂Y = 0 is equivalent to equation (11), then equation (12) is equivalent to L̂(y) = b̂ for some b̂ ∈ k. Since G(E/k) is reductive, L̂ is a completely reducible operator and we can apply Proposition 2.1 to conclude the first part of the third statement.

To prove the second part, we consider the following diagram:

[figure: F(Ṽ) =K; F; E(Ṽ); E; k]

We will first show that G(F (Ṽ)/F) ≃ G(E(Ṽ)/E). To do this it suffices to show that F ∩ E(Ṽ) = E (c.f., Lemma 5.10 of [8]). Since G(E(Ṽ)/E) is abelian (it is the vector group W), F ∩ E(Ṽ) is a Picard-Vessiot extension of k. The Galois group G(F ∩ E(Ṽ)/E) is a quotient of W and so is unipotent. Since G(F ∩ E(Ṽ)/E) is also a quotient of the reductive group G(F/k) it is also reductive and therefore must be trivial. Therefore F ∩ E(Ṽ) = E.

Let us denote by W̃ the Galois group G(K/F). As we have just shown this is isomorphic to W, the Galois group G(E(Ṽ)/E. Since G(K/F)/W̃ ≃ G(F/k) is reductive, we have that W̃ is the unipotent radical of G(K/k). Therefore G(K/k) is isomorphic to W̃ ⋊P where P is a Levi factor isomorphic to G(F/k) (via the map that takes a σ ∈ G(K/k) and restricts to F) and W̃ is isomorphic to W (via the map that takes σ ∈ G(K/k) and restricts to E(Ṽ)). We now consider the action of P on W̃ by conjugation. Let P_{E} be the subgroup of P that leaves E elementwise fixed. This is a normal subgroup of P and so is reductive. If σ ∈ P_{E}, then σ leaves E(Ṽ) invariant and so restriction gives a homomorphism of P_{E} to G(E(Ṽ)/E). Since this latter group is unipotent, the homomorphism must be trivial. Therefore any element of P_{E} leaves Ṽ fixed and so must commute with any element of W̃. Therefore, the action of P on W̃ factors through the action of P/P_{E} on W̃, and this is the same as the action of G(E/k) on W.

This last result and its proof tell us how to compute the Galois group of equation (8) when k = C(x).

Algorithm II

Input: A system of linear differential equations (8) where Y′ + A_{1}Y = 0 and Y′ + A_{2}Y = 0 are completely reducible with A_{1} ∈ M_{n}(C(x)), A_{2} ∈ M_{m}(C(x)).

Output: A system of equations in m + n variables defining the Galois group G(F/k) ⊂ GL_{n+m}(C) of the Picard-Vessiot extension corresponding to the system (10), an integer t and a rational homomorphism Φ : G(F/k) → GL_{t}(C) such that the Galois group of (8) is C^{t} ⋊ G(F/k) where the action of G(F/k) on C^{t} by conjugation is given by Φ.

1. One first calculates the Galois group G(F/k) of equation (10) using the results of [3]. This Galois group will be represented as matrices acting on diag(Y_{1}, Y_{2}) where Y_{1} is a fundamental solution matrix of Y′ + A_{1}Y = 0 and Y_{2} is a fundamental solution matrix of Y′ +A_{2}Y = 0. One can easily calculate the action of G(F/k) on (Y_{1}^{−1})^{T} ⊗Y_{2} and so calculate the Galois group G(E/k) of equation (11) as well as the map G(F/k) → G(E/k).

2. Find a scalar equation L̂(y) = 0 equivalent to the equation (11) as well as an element b̂ ∈ k so that equation (12) is equivalent to L̂(y) = b̂ (an algorithm to do this is presented in [9]; in the examples below ad hoc methods are used). Using the transformation of Y′ + AL̂Y = 0 to Ṽ′ + (−A_{1}^{T} ⊗ I_{n} + I_{m} ⊗ A_{2})Ṽ = 0 allows us to calculate the action of G(E/k) on the solution space of L̂(y) = 0.

3. Proposition 2.1 allows us to calculate a vector group W so that the Galois group of L̂(y) = b̂ (and so of equation (12)) is W ⋊ G(E/k).

4. Lemma 3.2 now tells us that the Galois group of equation (8) is the group W ⋊G(F/k) where the action of G(F/k) on W (i.e., the homomorphism Φ) can be calculated from the information we have.

Remark: As in the case of the equation L(y) = b, the algorithms of [3] can be combined with the above to give a presentation of the Picard-Vessiot extension corresponding to L_{1}(L_{2}(y)) = 0.

We will now give three examples of this method. In these examples we will start with an equation of the form L_{1}(L_{2}(y)) = 0 with coefficients in k = C(x). The Galois group G(F/k) that is the Galois group of equation (10) in Lemma 3.2, is the same as the Galois group of LCLM(L_{1}, L_{2}). In the examples we shall apply ad hoc methods to calculate this Galois group. We will then calculate a scalar equation equivalent to the system (11) as well as the matrix B defining this equivalence. This will allow us to find a scalar equation L̂(y) = b̂ equivalent to the system (12). We then apply the methods of Section 2 to calculate the vector space W.

Example 3.3 Consider the equation L(y) = 0, where L = L_{1} ∘ L_{2}, L_{1} = D^{2} − x, L_{2} = D^{2} + \frac{1}{x}D + 1.

The Galois group of this equation is an extension of the Galois group GL̃ of L̃ = LCLM(L_{1}, L_{2}). Since L_{1} and L_{2} are both known to have Galois group isomorphic to SL_{2}(C) (L_{1} is a form of Airy’s equation and L_{2} is a Bessel equation), GL̃ is a subgroup of SL_{2}(C) × SL_{2}(C).

According to ([10], p. 1158), if GL̃ is a proper subgroup of SL_{2}(C)×SL_{2}(C), then there exist a scalar matrix R = diag(α, α) with entries in a quadratic extension of C(x) and a matrix S ∈ GL_{2}(C(x)) such that

Wr(y_{1}, y_{2}) = R · S · Wr(z_{1}, z_{2}),

where {y_{1}, y_{2}} (resp., {z_{1}, z_{2}}) is a basis for a fundamental solution space of L_{1} (resp., L_{2}). Assuming this is the case, one can then show that

Wr(y_{1}^{2}, y_{1}y_{2}, y_{2}^{2}) = R̂ · Wr(z_{1}^{2}, z_{1}z_{2}, z_{2}^{2})

for some matrix R̂ ∈ GL_{2}(C(x)). Since {y_{1}^{2}, y_{1}y_{2}, y_{2}^{2}} (resp., {z_{1}^{2}, z_{1}z_{2}, z_{2}^{2}}) is a basis for the fundamental solution space of L_{1}^{Ⓢ2} (resp., L_{2}^{Ⓢ2}), we see that such an equation holds if and only if L_{1}^{Ⓢ2} and L_{2}^{Ⓢ2} are equivalent over C(x).

In our case, we claim that L_{1}^{Ⓢ2} and L_{2}^{Ⓢ2} are inequivalent (and therefore that GL̃ = SL_{2}(C) × SL_{2}(C)). The expanded version of DEtools developed by Mark van Hoeij for MapleV.5 allows one to calculate symmetric powers, LCLM’s and a basis of the ring of D-module endomorphisms of D/DL for an operator L ∈ D = C(x)[D]. Using this we proceed as follows. A calculation shows that

M = LCLM(L_{1}^{Ⓢ2}, L_{2}^{Ⓢ2})

is of order 4. If L_{1}^{Ⓢ2} and L_{2}^{Ⓢ2} were equivalent then D/DM would be the direct sum of two isomorphic D-modules. The endomorphism ring of D/DM would therefore have dimension 4. Using the eigenring command in DEtools one sees that this ring has dimension 2 and the desired result follows.

We now consider the equation

Ṽ′ + HṼ = −C̃,

where

H = −A_{1}^{T} ⊗ I_{2} + I_{2} ⊗ A_{2},
A_{1} = [ \begin{matrix}0 & −1 \\ −x & 0 \end{matrix} ] ,
A_{2} = [ \begin{matrix}0 & −1 \\ 1 & \frac{1}{x} \end{matrix} ] ,
C̃ = (0, −1, 0, 0)^{T} .

A cyclic-vector computation shows that the system Ṽ′ + HṼ = 0 is equivalent to Z′ + KZ = 0, where

K = [ \begin{matrix}0 & −1 & 0 & 0 \\ 0 & 0 & −1 & 0 \\ 0 & 0 & 0 & −1 \\ g_{1} & g_{2} & g_{3} & g_{4} \end{matrix} ] ,

g_{1} = \frac{x^{6} + 3x^{5} + 3x^{4} + 5x^{3} + 6x^{2} + 3x − 3}{x(x^{3} + x^{2} + 1)},
g_{2} = −\frac{x^{6} + 5x^{5} + 3x^{3} − 7x^{2} − 1}{x^{3}(x^{3} + x^{2} + 1)},
g_{3} = −\frac{2x^{3} − 2x^{2} + 1}{x^{2}},
g_{4} = −\frac{x^{3} − 2}{x(x^{3} + x^{2} + 1)}.

The equivalence is given by the equation Z = BṼ , where

B = [ \begin{matrix}1 & 0 & 0 & 0 \\ 0 & 1 & −x & 0 \\ −1 + x & −\frac{1}{x} & −1 & −2x \\ 2 + \frac{1}{x} & \frac{3x^{3}−x^{2}+2}{x^{2}} & 3x − x^{2} & 0 \end{matrix} ] .

Therefore, the equation Ṽ′ + HṼ = −C̃ is equivalent to

Z′ + KZ = −BC̃.  (13)

(The reader can verify that K = −B′B^{−1}+BHB^{−1}.) Since K is in companionmatrix form, it is easy to convert (13) into the inhomogeneous scalar equation L̂(y) = b̂, where

L̂ = D^{4} + g_{4}D^{3} + g_{3}D^{2} + g_{2}D + g_{1} (g_{i} as above),
b̂ = \frac{x^{4} + 2x^{3} + x^{2} + 4x + 3}{x^{3} + x^{2} + 1}.

Computations using the DFactor and ratsols commands in DEtools show that L̂ is irreducible over C(x) and that this equation admits no rational solutions. Thus, the vector space W referred to in Lemma 3.2(3) is all of C^{4}. We conclude that the Galois group G_{L} is (SL_{2}(C) × SL_{2}(C)) ⋉ C^{4}.

Example 3.4 Consider the equation L(y) = 0, where L = L_{1} ∘ L_{2}, L_{1} = D^{2} + \frac{1}{x}D + 1, L_{2} = D^{2} − D.

The Galois group G_{L} of L is once again an extension of GL̃, the group of L̃ = LCLM(L_{1}, L_{2}). To calculate GL̃ note that the Galois group G_{L_{1}} of L_{1} is SL_{2} and the Galois group G_{L_{2}} of L_{2} is the multiplicative group C^{∗}. The group GL̃ is a subgroup of G_{L_{1}} ×G_{L_{2}} that projects surjectively onto each factor. The Theorem of [10] implies that, in this case, GL̃ = G_{L_{1}} × G_{L_{2}}.

We now consider the equation

Ṽ′ + HṼ = −C̃,

where

H = −A_{1}^{T} ⊗ I_{2} + I_{2} ⊗ A_{2},
A_{1} = [ \begin{matrix}0 & −1 \\ 1 & \frac{1}{x} \end{matrix} ] ,
A_{2} = [ \begin{matrix}0 & −1 \\ 0 & −1 \end{matrix} ] ,
C̃ = (0, −1, 0, 0)^{T} .

A cyclic-vector computation shows that the system Ṽ′ + HṼ = 0 is equivalent to Z′ + KZ = 0, where

K = [ \begin{matrix}0 & −1 & 0 & 0 \\ 0 & 0 & −1 & 0 \\ 0 & 0 & 0 & −1 \\ g_{1} & g_{2} & g_{3} & g_{4} \end{matrix} ] ,

g_{1} = \frac{10x^{4} + 5x^{3} − 6x^{2} + 6x + 3}{x^{2}(5x^{2} − 3)},
g_{2} = −\frac{10x^{3} + 15x^{2} + 9x + 12}{x(5x^{2} − 3)},
g_{3} = 3\frac{5x^{2} + 5x + 2}{5x^{2} − 3},
g_{4} = −2\frac{5x^{2} + 5x − 3}{5x^{2} − 3}.

The equivalence is given by the equation Z = BṼ , where

B = [ \begin{matrix}1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ −1 & 1 & \frac{1}{x} & 2 \\ −\frac{1}{x} & −2 & −1 & 3 + \frac{3}{x} \end{matrix} ] .

Therefore, the system Ṽ′ + HṼ = −C̃ is equivalent to

Z′ + KZ = −BC̃.

(The reader can again verify that K = −B′B^{−1} + BHB^{−1}.) Conversion to an inhomogeneous scalar equation yields L̂(y) = b̂, where

L̂ = D^{4} + g_{4}D^{3} + g_{3}D^{2} + g_{2}D + g_{1} (g_{i} as above),
b̂ = −2 + \frac{5x^{2} + 5x + 12}{5x^{2} − 3}.

Using the eigenring command of DEtools, one sees that the dimension of the endomorphism ring of D/DL̂ is two. Since L̂ is completely reducible, this implies that D/DL̂ is the direct sum of two nonisomorphic irreducible D-modules. This furthermore implies that L̂ has exactly two nontrivial irreducible right (resp., left) factors. A computation using the command endomorphism charpoly yields two different right factors. From these one calculates the unique left factors and then one can show that for neither of these left factors L̄ does the equation L̄(y) = b̂ have a rational solution. Since L̂(y) = b̂ also has no rational solutions, we conclude that G_{L} is (SL_{2}(C) × C^{∗}) ⋉ C^{4}.

Example 3.5 Consider the equation L(y) = 0, where L = L_{1} ∘ L_{2}, L_{1} = LCLM(D − 2x, D), L_{2} = D^{2}.

Here it is clear that GL̃, the group of L̃ = LCLM(L_{1}, L_{2}), is C^{∗}.

We now consider the equation

Ṽ′ + HṼ = −C̃,

where

H = −A_{1}^{T} ⊗ I_{2} + I_{2} ⊗ A_{2},
A_{1} = [ \begin{matrix}0 & −1 \\ 0 & −2x − \frac{1}{x} \end{matrix} ] ,
A_{2} = [ \begin{matrix}0 & −1 \\ 0 & 0 \end{matrix} ] ,
C̃ = (0, −1, 0, 0)^{T} .

A cyclic-vector computation shows that the system Ṽ′ + HṼ = 0 is equivalent to Z′ + KZ = 0, where

K = [ \begin{matrix}0 & −1 & 0 & 0 \\ 0 & 0 & −1 & 0 \\ 0 & 0 & 0 & −1 \\ 0 & 0 & h_{1} & h_{2} \end{matrix} ] ,

h_{1} = 2\frac{8x^{6} − 12x^{4} + 18x^{2} + 9}{4x^{4} + 3},
h_{2} = 4\frac{x(4x^{4} − 4x^{2} + 3)}{4x^{4} + 3}.

The equivalence is given by the equation Z = BṼ , where

B = [ \begin{matrix}0 & −x & −x & 0 \\ x & −1 & 2x^{2} & −x \\ −2x^{2} + 1 & 2x & −2x(2x^{2} − 1) & 4x^{2} \\ 2x(2x^{2} − 3) & −6x^{2} + 3 & 4x^{2}(2x^{2} − 3) & −6x(2x^{2} − 1) \end{matrix} ] .

Therefore, the equation Ṽ′ + HṼ = −C̃ is equivalent to

Z′ + KZ = −BC̃.

(The reader can once again verify that K = −B′B^{−1} + BHB^{−1}.) In this example, the equivalent inhomogeneous scalar equation is L̂(y) = b̂, where

L̂ = D^{4} + h_{2}D^{3} + h_{1}D^{2},
b̂ = 3 − 6x^{2} + 6\frac{4x^{4} − 8x^{2} − 5}{4x^{4} + 3}.

The eigenring command shows that the corresponding endomorphism ring has dimension 10 and yields a basis of this ring. Applying the endomorphism charpoly to each of these will yield a list of right factors and a simple calculation yields their corresponding left factors. Despite the fact that in this case there is an infinite set of left factors, there is a third order operator

L_{0} = D^{3} + \frac{8x^{6} − 12x^{4} + 6x^{2} + 3}{x(4x^{4} + 3)}D^{2} − 2\frac{8x^{6} + 3}{x^{2}(4x^{4} + 3)}D + 2\frac{8x^{6} + 12x^{4} − 6x^{2} + 3}{x^{3}(4x^{4} + 3)}

on this list of left factors such that L_{0}(y) = b̂ admits the rational solution

y = −\frac{1}{4}x(6x^{2} + 5). Meanwhile, another computation shows L̂(y) = b̂ admits no rational solutions. We are therefore able to avoid a calculation involving parameterized operators. Thus, we have

G_{L} = C ⋊ C^{∗}.

References

[1] Bertrand, D. Extensions de D-modules et groupes de Galois différentiels. In P -adic analysis (Trento, 1989), F. B. et al., Ed., Lecture Notes in Mathematics, 1454. Springer-Verlag, Berlin, Heidelberg, New York, 1990, pp. 125–141.

[2] Bertrand, D. Un analogue différentiel de la théorie de Kummer. In Approximations Diophantiennes et Nombres Transcendents, Luminy 1990, P. Philippon, Ed. Walter de Gruyter and Co., Berlin, 1992, pp. 39–49.

[3] Compoint, E., and Singer, M. F. Calculating Galois groups of completely reducible linear operators. To appear in the Jounral of Symbolic Computation, 1997; a preprint is available at

http://www.math.ncsu.edu/~singer.

[4] Davenport, J., and Singer, M. F. Elementary and liouvillian solutions of linear differential equations. Journal of Symbolic Computation 2, 3 (1986), 237–260.

[5] Grigoriev, D. Y. Complexity of factoring and calculating the gcd of linear ordinary differential operators. Journal of Symbolic Computation 10, 1 (1990), 7–38.

[6] Haefliger, A. Local theory of meromorphic connections in dimension one (Fuchs theory). In Algebraic D-Modules, Borel et al. Academic Press, 1987, ch. III., pp. 129–149.

[7] Humphreys, J. Linear Algebraic Groups. Graduate Texts in Mathematics. Springer-Verlag, New York, 1975.

[8] Kaplansky, I. An Introduction to Differential Algebra, second ed. Hermann, Paris, 1976.

[9] Katz, N. A simple algorithm for cyclic vectors. American Journal of Mathematics 109 (1987), 65–70.

[10] Kolchin, E. Algebraic groups and algebraic dependence. American Journal of Mathematics 90 (1968), 1151–1164.

[11] Mostow, G. D. Fully reducible subgroups of algebraic groups. American Journal of Mathematics 78 (1956), 211–264.

[12] Poole, E. G. C. Introduction to the Theory of Linear Differential Equations. Dover Publications, Inc, New York, 1960.

[13] Singer, M. F. Testing reducibility of linear differential operators: a group theoretic perspective. Applicable Algebra in Engineering, Communication and Computing 7 (1996), 77–104.

[14] Tsarev, S. P. An algorithm for complete enumeration of all factorizations of a linear ordinary differential operator. In Proceedings of the 1996 International Symposium on Symbolic and Algebraic Computation (1996), L. Y. N., Ed., ACM Press, pp. 226–231.