Liouvillian solutions of third order differential equations

Transcription

1 Article Submitted to Journal of Symbolic Computation Liouvillian solutions of third order differential equations Felix Ulmer IRMAR, Université de Rennes, 0 Rennes Cedex, France felix.ulmer@univ-rennes.fr Abstract The Kovacic algorithm and its improvements give explicit formulae for the Liouvillian solutions of second order linear differential equations. Algorithms for third order differential equations also exist, but the tools they use are more sophisticated and the computations more involved. In this paper we refine parts of the algorithm to find Liouvillian solutions of third order equations. We show that, except for finite groups and a reduction to the second order case, it is possible to give a formula in the imprimitive case. We also give necessary conditions and several simplifications for the computation of the minimal polynomial for the remaining finite set of finite groups (or any known finite group) by extracting ramification information from the character table. Several examples have been constructed, illustrating the possibilities and limitations. AMS classification: H0, 8Q0. Introduction and notation Classically the algorithm to compute Liouvillian solutions breaks into four cases. For third order equations L(y) = 0 over a differential field k with differential Galois group G we get:. G is a reducible linear group and there exists a decomposition L(y) = L (L (y)) into two operators of lower order.. G is an imprimitive linear group and there exits a solution z = e u of L(y) = 0 with [k(u) : k] =.. G belongs to a list of eight finite primitive linear groups. In particular all Liouvillian solutions are algebraic.. G is an infinite primitive linear group and L(y) = 0 has no Liouvillian solution.

2 Felix Ulmer: Liouvillian solutions of third order equations In this paper we improve the algorithms to compute Liouvillian solutions of third order linear differential equations. The paper is organised into four sections. Section gives a survey of the computation of Liouvillian solutions by providing a uniform approach for second and third order equations. In section we show that the imprimitive case can be solved using only linear algebra except if L(y) is the second symmetric power of some second order equation or if the differential Galois group is one of four finite imprimitive groups. The four exceptional imprimitive groups are characterised by the fact that the third symmetric power can be of order or that there are several minimal polynomials of degree of algebraic solutions of the Riccati equation associated to L(y). In the examples of section we show that all exceptional situations can indeed occur for those four groups. In section we consider the classical case k = C(x) and derive strong necessary conditions for G to be conjugate to a given finite group. We derive strong necessary conditions for the finite primitive group and for the four finite exceptional imprimitive groups. In particular the conditions will always allow to compute the exceptional imprimitive groups. The approach is based on the ramification data given by the local exponents. Using the character table of the groups we show how to restrict possible ramification data of factors in constructions and in particular of exponential solutions of symmetric powers. The examples in this section show that this approach substantially improves the previous algorithms (Hoeij et al., ; Singer and Ulmer, b, 7). Large parts of the approach generalise to higher order equations. In the paper k is a differential field whose field of constants C is algebraically closed of characteristic 0 (e.g. C(x) with the usual derivation d/dx). For the derivation δ of k and a k we write δ j (a) = a (j) and also a () = a, a () = a,.... Let L(y) = y (n) + a n y (n ) a y + a 0 y = 0, a i k () be a linear differential equation of order n over k. A solution of (). in k is a rational solution,. in an algebraic extension of k is an algebraic solution,. whose logarithmic derivative is in k is an exponential solution,. belonging to a field obtained from k by successive adjunction of exponentials, integrals and algebraic functions is a Liouvillian solution. ( ) an Using the variable transformation ỹ = y exp, it is always possible n to transform a given linear differential equation into the above form without losing the Liouvillian character of the solutions. We refer to (Magid, ; van der Put and Singer, 00) for an introduction to differential Galois theory and for definitions of a Picard-Vessiot extension K of k and differential Galois group G(L) for (). The group G(L) sends a solution of () to another solution of (). This action gives a faithful representation of G(L) as a subgroup of GL(n, C),

3 Felix Ulmer: Liouvillian solutions of third order equations also denoted G(L), whose character is denoted χ. The assumption that in () the coefficient a n = 0 implies that G(L) SL(n, C). A linear differential equation L(y) = 0 is reducible if there exists a decomposition L(y) = L (L (y)) into two operators of lower order. The equation () is reducible if and only if G SL(n, C) is a reducible linear group (cf. (Kolchin, 8) Section Theorem ). Definition.: Let G GL(V ) be a linear group acting irreducibly on the vector space V of dimension n over C. Then G is said to be imprimitive if there exist subspaces V,, V k with k > such that V = V V k and, for each σ G, the mapping V i σ(v i ) is a permutation of the set S = {V,..., 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(V ) which is not imprimitive is called primitive. Since an imprimitive group G is assumed to act irreducibly on V, G must permute the V i transitively. In particular, all the V i have the same dimension and if n is prime the group must be monomial. Our approach is based on a connection between Liouvillian solutions and polynomial semi-invariants of G. Definition.: Let V be a C-vector space, let Y,..., Y n be a basis for V, and let G GL(V ) be a linear group. Define a group action of σ G on I C[Y,..., Y n ] by σ (I(Y,..., Y n )) = I (σ(y ),..., σ(y n )). If a homogeneous I C[Y,..., Y n ] of degree m has the property that σ G, σ I(Y,..., Y n ) = ψ I (σ) I(Y,..., Y n ), with ψ I (σ) C then ψ I : G C is a one dimensional character of G and I is a semi-invariant of degree m of G with character ψ I. If ψ I is the trivial character l of G, then I is an invariant of degree m of G. If j N is minimal such as (ψ I ) j = l, then j is the order of the character ψ I and of the semi-invariant I. From (Singer and Ulmer, 7) Theorem we get that the existence of Liouvillian solutions is equivalent to the existence of semi-invariants that factor into linear forms. An algorithm to compute semi-invariants of G is given in (van Hoeij and Weil, 7). We will however, in general, use the concept of the value of a G semi-invariant. Definition.: Let L(y) be a n-th order linear differential equation with Galois group G GL(n, C). To a basis y,..., y n of the solution space of L(y) = 0 we associate the evaluation morphism Φ: C[Y,..., Y n ] K Y i y i The value of a semi-invariant of G GL(n, C) is its image in K under the above evaluation morphism.

4 Felix Ulmer: Liouvillian solutions of third order equations The morphism Φ has the following properties:. Φ restricts to a bijection between linear forms and solutions of L(y) = 0.. Semi-invariants of finite order j are sent to j-th roots of elements of k. In particular invariants are fixed under the action of G and are sent to elements of k. Definition.: Let L(y) = 0 be an n-th order homogeneous linear differential and let y,..., y n be a fundamental system of solutions. The differential equation L s m (y) whose solution space, denoted V m, is spanned by monomial of degree m in y,..., y n is called the m-th symmetric power of L(y) = 0. An algorithm to construct L s m (y) is given in (Singer, 80). The value of a semi-invariant of degree m and order j is a j-th root of a solution of L s m (y) = 0, cf. (Singer and Ulmer, b) Lemma... Algebraic solution of the Riccati equation The first three subsection introduce the basic idea and the notation and are well known (cf. (Bronstein, 00; Hoeij et al., )). If L(y) = 0 has a solution whose logarithmic derivative is algebraic with minimal polynomial Q k[u] of degree m, then all zeros u i of Q give a Liouvillian solution z i = e u i. Thus Q = ( U z z ) ( U z z )... ( U z m z m Multiplication of Q by m i= z i gives P K[U] of the form ( m ) z i U m (z z... z m z... z m z m ) U m +...+( ) m (z z... z m ). i= Note that the coefficients of P will be in k only after division by the leading coefficient m i= z i. Our goal in this section is to compute Q or equivalently to compute P. )... Computation of symmetric powers Let R m = k[z,0,..., Z m,n ] be a polynomial ring in nm variables. Using the differential equation () we define a derivation on R by. (f) = δ(f) for f k,. (Z i,s ) = Z i,s+ if 0 s < n,. (Z i,n ) = (a n Z i,n a Z i, + a 0 Z i,0 )

5 Felix Ulmer: Liouvillian solutions of third order equations For convenience we will use the notation Z i for Z i,0 and Z (j) i for Z i,j (0 < j < n). Note that the variables Z i are by construction solutions of (). For integers n > d d d m 0 we introduce the notation W d,...,d m for the monic polynomial corresponding to (d ) (d Z ) (d σ() Z m) σ(m) R m. σ S m Z σ() Example.: For n = the sum over the orbit under S of Z Z Z is Z Z Z + Z Z Z + Z Z Z. Thus W,0,0 is Z Z Z + Z Z Z + Z Z Z. If we start with the monomial S = Z Z Z m R m and take derivatives, then the resulting expressions will all be symmetric in the m indices and thus can be expressed as sums over k of the N = ( ) n+m n polynomials Wd,...,d m R m. For m we get S = Z Z Z m = W 0,0,...,0 S = Z () Z... Z m + Z Z ()... Z m Z... Z m Z m () = W,0,...,0 S = j m Z () j i Z i Z j + = W,0,...,0 + W,,0,...,0... =... Z () j Z () j j <j m i Z i Z j Z j Considering j derivatives of S we get a linear system of the form S W 0,...,0 S... = A W,0,...,0 j... S (j) W n,...,n If we take the first N = ( ) n+m n derivatives of S, then we will have more equations than unknowns on the right side, so the equations will be linearly dependent over k, giving a linear differential equation of degree N for S. We now consider the linear combination of the S (i) involving the lowest possible order of derivation, in other words where for the first time A j has a row rank less than j +. Then a unique, up to multiple, non zero element (b j, b j,..., b 0 ) k j+ exists such as S ( ) b0, b,..., b j S... = ( ) b 0, b,..., b j Aj = 0, S (j)

6 Felix Ulmer: Liouvillian solutions of third order equations i.e. for which j i=0 b is (i) = 0. For any m non zero solutions z,..., z m of L(y) = 0 we consider the differential evaluation morphism Ψ z,...,z m : R m K Z (j) i z (j) i ( j ) We have Ψ z,...,z m i=0 b is (i) = j i=0 b i ( m i= z i) (i) = 0, showing that L s m (y) = j i=0 b iy (i). Note that the above procedure is just the algorithm proposed in (Singer, 80, 8) to compute L s m (y) and that one could even start with S = Y m instead of Y Y Y m (cf. (Singer and Ulmer, a), Section..)... Computation of P (U) From (Singer and Ulmer, 7), Theorem and its proof we get that if g K is an exponential solution of L s m (y) and g = Ψ z,...,z m (S) for m N and non zero solutions z i of L(y) = 0, then for P K[U] given by ( m ) z i U m (z z... z m z... z m z m ) U m m ( ) m i= = Ψ z,...z m (W 0,...,0 )U m Ψ z,...z m (W,0,...,0 )U m ( ) m Ψ z,...z m (W,,..., ) we have that Q = P m i= z i k[u]. In this case all roots u i of Q give a Liouvillian solution e u i of L(y) = 0. If L s m (y) is of maximal order N = ( ) n+m n, then the matrix AN corresponding to Ψ z,...,z m (S) g Ψ z (Ψ z,...,z m (S)),...,z m (W 0,...,0 )... = g... = A Ψ z,...,z m (W,0,...,0 ) N... (Ψ z,...,z m (S)) (N ) g (N ) Ψ z,...,z m (W n,...,n ) will be invertible, allowing us to obtain the remaining coefficients of P or Q from g and its derivatives using linear algebra. We give an explicit example in the next section. Therefore the computation of Q k[u] is reduced to the computation of g and linear algebra if. g K is an exponential solution of L s m (y) such that g = Ψ zi,...,z m (S) for m non zero solutions z i of L(y) = 0. From (Bronstein, 00) Theorem.8 we get that for reductive groups it is enough to consider rational solutions, but to do this one must work with larger values of m. This has for example be proposed in (Ulmer and Weil, ). i= z i

7 Felix Ulmer: Liouvillian solutions of third order equations 7. L s m (y) is of maximal order N = ( ) n+m n. Since a solution z i is a linear combination over C of y,..., y n the first condition is equivalent to the fact that g K is the value of a semi-invariant of G that factors into linear forms ((Singer and Ulmer, 7), Theorem and its proof). Our goal is to look for conditions in order to be in the above situation... Second order Example We consider a second order example L(y) = y + a 0 y = 0 since for third order the results of the various steps are too long to be included. For S = Z Z R we have S = Z Z S = Z Z + Z Z S = a 0 Z Z + Z Z S = a 0 Z Z a 0 (Z Z + Z Z ) This gives the system: S 0 0 S S = 0 0 a 0 0 S a 0 a 0 0 W 0,0 W,0 W, = A W 0,0 W,0 W, From the matrix form we see that in this case the symmetric power L s (y) will always be of maximal order. To compute L s (y) we look for an element (b 0, b, b, b ) in the kernel of A t, ( a 0, a 0, 0, ) which gives us the following formula: a 0 0 a 0 a 0 0 L s (y) = a 0 y + a 0y + y = ( 0, 0, 0, 0 ), Since homogeneous forms in two variables, over an algebraic closed field, always factor into linear forms, we get that for second order equations L(y), any non zero exponential solution g of L s (y) = 0 is a product of solutions, i.e. g = z z, and will lead to a polynomial P (cf. (Ulmer and Weil, ) Theorem.). Thus for any exponential solution g of L s (y) by solving the linear system g g g = a 0 0 Ψ z,z (W 0,0 ) Ψ z,z (W,0 ) Ψ z,z (W, )

8 we get Felix Ulmer: Liouvillian solutions of third order equations 8 Q = U g g U + a 0g + g k[u] g Note that the formula is in fact (up to the computation of g and linear algebra) given by the above two matrices. Remark: The second order case is well known (cf. (Kovacic, 8; Ulmer and Weil, )). This case is characterised by the existence of various recursions (cf. (Bronstein et al., 7)) and simplified by the fact that forms in two variables over an algebraic closed field factor into linear forms. In particular L s m (y) is always of maximal order m + and a formula like the above can be given for the cases m {,,,, } of (Kovacic, 8). From (Kovacic, 8) we get that if no solution is found for m {,,,, }, then there are no Liouvillian solutions. For third order equations symmetric powers are not always of maximal order and simple recursions are no longer available but we will show that, up to a few exceptions, a formula similar to the above can be given.. Third order equations with an imprimitive group We now consider a third order equation L(y) = y +a y +a 0 y whose differential Galois group G is imprimitive. From (Singer and Ulmer, a) Theorem. we get that this case is characterised by the fact that L s (y) = 0 has an exponential solution whose square is rational, or equivalently that a (semi-)invariant of degree and order exists. The new difficulties, compared to second order equations, in the computation of the minimal polynomial Q of degree of an algebraic solution of the Riccati equation are:. L s (y) is not always of maximal order, in which case the matrix A is not invertible.. Not all exponential solutions of L s (y) = 0 are values of forms that factor into linear forms. Our goal is to characterise the above exceptional cases and to produce a formula in the general case where the order of L s (y) is 0 and where there is, up to multiples, a unique (semi-) invariant of degree and order (whose value, up to multiple, will be the unique solution of L s (y) whose square is in k)... The order of symmetric powers Lemma.: For a third order linear differential equation L(y) = y +a y +a 0 y over k the order of L s (y) is 7, or 0. The order of L s (y) is 7 if and only if a 0 = a. In this case L(y) = L s (y) where L(y) = y + a y. Proof: Let Φ m be the G-morphism obtained by restriction of the evaluation morphism Φ to the vector space of homogeneous forms of degree m of C[Y, Y, Y ] (cf. (Singer and Ulmer, a) Lemma.). Because Φ is a bijection, (Singer

9 Felix Ulmer: Liouvillian solutions of third order equations and Ulmer, a) Lemma. () implies that the kernel of Φ is at most of dimension, i.e. L s (y) is of order at least. If Φ is a bijection, then (Singer and Ulmer, a) Lemma. () implies that L s (y) is of order or 0. Suppose now that the kernel of Φ is of dimension and generated by F C[Y, Y, Y ]. Note that F must be irreducible since otherwise Φ is not a bijection. The homogeneous polynomials of degree divisible by F correspond to the C-Span of Y F, Y F, Y F and are in the kernel of Φ, showing that L s (y) is of order 7. If L s (y) is of order, there must exist an F C[Y, Y, Y ] in the kernel of Φ which is not a multiple of F. If F or F is of degree zero in Y, then we get a non constant homogeneous polynomial F C[Y, Y ] whose evaluation via Φ is zero. Since F is homogeneous and C algebraically closed, F factors as a product of linear forms i (α iy β i Y ). The evaluation via Φ gives a linear relation between the basis elements y, y and thus a contradiction. Denote F, C[Y, Y ] the resultant of F and F with respect to Y. Since F, F are relatively prime and are both of degree in Y, we get that F, C[Y, Y ] is homogeneous of degree. Arguing as above with F, we get a contradiction. Thus L s (y) is of order 7 in this case. This proves the first assertion. We want to characterise the fact that L s (y) is of order 7 or equivalently that L s (y) is of order. This has been done (Singer, 8) Lemma., but we will give a proof using the introduced formalism. According to section. we have that L s (y) is of order if and only if the matrix A is not invertible (i.e. if and only if (Z Z Z ),..., (Z Z Z ) () are linearly dependent). Performing a fraction free Gaussian elimination on the matrix (A ) t we get: 0 0 a 0 a 0 a a 0 a a a 0 a a 7a 0 + a a 0a 0a a (a a 0 ) Therefore L s (y) is of order 7 if and only if a = a 0. A direct computation shows that in this case L(y) = L s (y)... The four exceptional finite imprimitive groups We now want to characterise the cases where the order of L s (y) is or where there exists more than one, up to multiples, (semi-) invariants of degree and order. In order to avoid repetitions we shall consider the matrices: σ c = , σ t = and σ = , σ = ω ω ω ω ω. ω ω 0 0 0,

10 Felix Ulmer: Liouvillian solutions of third order equations 0 where ω + ω + = 0, and define the groups G 7 =< σ c, σ >, G =< σ c, σ, σ t >, G 8 =< σ c, σ > and G =< σ c, σ, σ t > whose respective orders correspond to the indices. We note that for each of the four groups all faithful irreducible unimodular characters of degree are conjugate. We consider an imprimitive group G SL(, C). This group will, in the basis Y, Y, Y corresponding to a system of imprimitivity, contain a diagonal element σ d = α β αβ and, since the group acts transitively on the system of imprimitivity, an element 0 0 µ ν µν 0 of order which we can assume to be of the conjugated form 0 0 σ c = We have a morphism ϕ: G S whose image Im(ϕ) is A or S (the transitive subgroups of S ). Since the Schur representation groups of A and S have no irreducible representation of degree, the irreducible group G must contain a non scalar diagonal element. Since conjugation by σ c permutes the entries of σ d cyclically, we may assume that in the above σ d we have α β. If Im(ϕ) = A then Y Y Y is an invariant and if Im(ϕ) = S then Y Y Y is a semi-invariant of order. The following shows that, with two exceptions, Y Y Y is, up to constant multiple, the only (semi-)invariant of degree and order. Theorem.: Let L(y) = y + a y + a 0 y be a third order linear differential equation with an unimodular imprimitive Galois group G. Then one of the following must be true. G has a two dimensional space of invariants of degree. In this case G = G 7.. G has no invariant of degree but a two dimensional space of semi-invariants of degree (for the same character) and order. In this case G = G.. G has up to multiples a unique (semi-)invariant of degree and order. This semi-invariant must factors into linear forms. If G has another semiinvariant of degree then one of the following holds (a) G also has (up to multiples) semi-invariants of degree and order corresponding to different characters. In this case G = G 8

11 Felix Ulmer: Liouvillian solutions of third order equations (b) G has also (up to multiples) semi-invariant of degree and order. In this case G = G Proof: We suppose G SL(, C), given in a basis Y, Y, Y corresponding to a system of imprimitivity, contains elements of the form σ c and σ d with α β. We want to find all possible semi-invariants of degree and order of G of the form: I = a Y Y Y Y + b Y Y Y + c Y + c Y + c Y Y Y Y which are not multiples of Y Y Y. If I is a semi-invariant of G of order, σ c (I) = ±I. By equating coefficients we find that a = a = a, b = b = b and c = c = c. We can therefore assume I to be of the form: Ĩ = ay + ay + ay + by Y + by Y + by Y + cy + cy + cy Y Y Y If Ĩ is a semi-invariant of G of order then σ d(ĩ) = λĩ where λ =. Equating coefficients and multiplying by powers of α and β which are not zero, we obtain a polynomial system in the variables a, b, c, d, α, β, λ. To this we add λ and the extra condition wαβ(α β) with a new variable w in order to guarantee that α, β are non zero and distinct. Computing a Gröbner basis for the corresponding system for a lex degree order in the variables w, a, b, c, d, α, β, λ we get the following equations not involving w: a(α + αβ + β ), a(β ), a(λ ), b, c, d(λ ), λ If Ĩ is not a multiple of Y Y Y then we must have a 0, from which we get β =. From α + αβ + β we get that α = Jβ or α = J β with J + J + = 0. Thus G contains at most six non scalar diagonal matrices. A group containing such a matrix, say σ and σ c will contain all other such matrices and must be conjugated to G 7. This is because all six matrices can be obtained from σ, σ and cyclic permutation of the entries via conjugation by σ c. Therefore if Im(ϕ) = A then G = G 7. Otherwise G 7 is a subgroup of index and thus a normal subgroup of G. Among the non isomorphic groups of order the group G is the only group having those properties and an irreducible unimodular representation of degree. By computing, one finds that G 7 has a two dimensional space of invariants of degree and that G has no invariants but a two dimensional space of semi-invariants of degree and order. This shows that only the three main cases stated can occur. Now we suppose that G is not isomorphic to G 7 or G. From the above we see that this implies that G contains a matrix σ d with β. If, besides multiples of Y Y Y, G has a semi-invariant I of degree, then σ c (I) λ I = 0 and σ d (I) λ I = 0 with λ, λ C. By equating coefficients and multiplying by powers

12 Felix Ulmer: Liouvillian solutions of third order equations of α and β which are not zero, we obtain a polynomial system in the variables a i, b i, c i, d, α, β, λ i. To this we add the extra condition wαβ(α β)λ λ (β ) with a new variable w in order to guarantee that α, β, λ, λ are non zero, α β and β. Computing a Gröbner basis for the corresponding system for a lex degree order in the variables w, λ, λ, d, a, a, a, b, b, b, c, c, c, α, β we get (among others) the following equations: a a a, a a a, a a a, a (α + αβ + β ), a (β + β + ), a (α + αβ + β ), a a, a (β + β + ), a (α + αβ + β ), a (β + β + ), b, b, b, c, c, c Therefore, if I is not a multiple of Y Y Y (i.e. some a i is non zero), then β is a primitive -th root of unity and α = Jβ or α = J β. There are at most twelve matrices of this form. A group containing a matrix of this form, say σ, and σ c will also contain {σ c (σ ) s σ c, σc (σ ) s σc s 8} and thus all twelve matrices and, by multiplication of those, all elements of G 7 (which we excluded via β ). Thus there is a unique such group which is conjugated to G 8. Therefore, if Im(ϕ) = A then G = G 8 otherwise G 8 is a subgroup of index and thus a normal subgroup of G which is of order. In this case the group of order is non abelian of exponent 8 and has a center of order. From the fifty-five non isomorphic groups of order only G satisfies those requirements and has a unimodular representation of degree. By computing all semi-invariants of degree of G 8 and G we get the result. Corollary.: Let L(y) = y + a y + a 0 y be a third order linear differential equation with unimodular imprimitive Galois group G. If L s (y) is not of order 7, then one of the following holds:. L s (y) is of order and G is isomorphic to G 7, G, G 8 or G.. L s (y) is of order 0 and one of the following holds: (a) L s (y) = 0 has up to multiple a unique solution whose square is rational. This must be the value of a semi-invariant that factor into linear forms. (b) L s (y) = 0 has a two dimensional space of rational solutions. In this case G = G 7. Within the two dimensional space of invariants of degree there are, up to multiples, four invariants that factor into linear forms. (c) L s (y) = 0 has a two dimensional space of non rational solutions whose square is rational. In this case G = G. Within the two dimensional space of semi-invariants of degree and order there are, up to multiples, four semi-invariants that factor into linear forms. Proof: The possible orders 7,, 0 for L s (y) follow from Lemma.. Let Φ be the G-morphism obtained by restriction of the evaluation morphism Φ to the

13 Felix Ulmer: Liouvillian solutions of third order equations vector space W of homogeneous forms of degree of C[Y, Y, Y ] ((Singer and Ulmer, a) Lemma.). Suppose that L s (y) is of order. The kernel of Φ is of dimension one generated by a homogeneous F C[Y, Y, Y ] of degree. Since, with the G- action defined in Definition., Φ is a G-morphism, we have σ G that Φ (σ F ) = σ(φ (F )) = 0 and thus that σ(f ) Ker(Φ ) is a multiple of F. Therefore F is a semi-invariant of G. Since the value of a semi-invariant which factors into linear forms (i.e. whose value is a product of non zero solutions) cannot be zero, we get that there must be at least semi-invariants of degree of G. By the previous result the only possibilities are the four given groups. If L s (y) is of order 0, then Φ is a bijection of W on the the solution space of L s (y) = 0. In this case there is a bijection between the semi-invariants of degree and order and the solutions of L s (y) = 0 whose square is rational. If Y, Y, Y is a basis corresponding to a system of imprimitivity, then Y Y Y is a semi-invariant that factors into linear forms and its value is a solution of L s (y) = 0 whose square is rational. From the previous result we get that only G 7 and G can, up to multiples, have more than one semi-invariant of degree and order and therefore more than a one dimensional space of solutions of L s (y) = 0 whose square is rational. The two exceptions now follow from the previous theorem. From (Hendriks and van der Put, ) or (Ulmer, ), Theorem. we get that G 7 and G both have, up to multiples, four semiinvariants of degree and order that factor into linear forms. In the next section we construct examples of equations whose differential Galois group corresponds to G 7, G, G 8 and G such that L s (y) is of order. For G 7 and G this shows that the semi-invariants that factor into linear forms may all have the same value. Therefore, in general, it is impossible to recover the minimal polynomial of an algebraic solution just from the value of a semi-invariant that factors into linear forms (cf. Example.). Corollary.: Let L(y) = y + a y + a 0 y be an irreducible third order linear with irreducible Galois group G SL(, C). If L s (y) is of order, then G is a finite group isomorphic to G 7, G, G 8, G or F SL, the primitive group of order 08. Proof: An irreducible group is either imprimitive or primitive. If G is imprimitive we get the result from the previous Corollary. If G is primitive then from (Singer and Ulmer, a) Table we get that only for F SL the order of L s (y) can be... The generic imprimitive case Consider L(y) = y + a y + a 0 y and suppose that G is an imprimitive group not conjugate to G 7, G and given in a monomial basis Y, Y, Y. If L s (y) is

14 Felix Ulmer: Liouvillian solutions of third order equations of maximal order 0, then a non zero element in the kernel of the transposed of S W 0,0,0 S... = A W,0, S (0) W,, gives the coefficients of L s (y) = 0 (cf. Section.). Since G is not conjugated to G 7, G, there exists up to multiples a unique non zero exponential solution g K of L s (y) = 0 with g k which must be the value of Y Y Y and therefore g = z z z for some solutions z i of L(y) = 0 (cf. previous section and (Singer and Ulmer, a), Theorem.). The square matrix A in g Ψ z,z,z (W 0,0,0 ) g... = A Ψ z,z,z (W,0,0 )... g () Ψ z,z,z m (W,, ) is invertible, allowing us to express the coefficients of Q k[u] using g and its derivatives (cf. Section.). The matrices A and A 0 have to be computed only once. The point in the above is that it is enough to know that g 0 is a product of solutions but that the actual decomposition is not needed. This follows also from ((Hoeij et al., ), Theorem.) and the method described there would also produce the above formula in this case. Example.: Consider the equation L(y) = 0 over C(x) given by d y dx + x 0x + 08 x (x x + ) dy dx 78x 7x 8 + 0x y = 0 8x (x x + x ) Factorisation shows that L is an irreducible operator. The equation L s (y) = 0 is of order 0 and has (up to multiples) a unique exponential solution g = x 7/ x / + x / with g C(x). Since the semi-invariant I whose value is g must factor into linear forms, our generic formula with m = can be applied. Solving the 0 0 system for g = x 7/ x / + x / we get Ψ z,z,z (W 0,0,0 ) = x 7/ x / + x / Ψ z,z,z (W,0,0 ) = 7 x/ x/ + x/ Ψ z,z,z (W,,0 ) = 87 x x + x / Ψ z,z,z (W,, ) = x 7 x + 7 x 08 8 x / (x )

15 Felix Ulmer: Liouvillian solutions of third order equations This gives us the following minimal polynomial Q k[u]: U 7 x x (x ) U + 87 x x + 8 x (x ) U x 7 x + 7 x 08 8 x (x ) Note that the above solutions z i are in fact e u i where the u i are the roots of Q, but that the computation of the z i is not needed in order to compute Q... Algorithm for third order equations: reducible and imprimitive case The following algorithm decides for an imprimitive group G if the formula can be used or if the group is among the four finite imprimitive groups. The steps must be performed in the given order:. The reducible case If L(y) = 0 is reducible, then a basis of Liouvillian solutions can be computed using a factorisation of L(y) and the reduction method of d Alembert (cf. (Singer and Ulmer, b) Section ).. If a 0 = a We then set L(y) = y + a y and look for Liouvillian solutions of L(y) = 0. If L(y) = 0 has two Liouvillian solutions z, z, then z, z z, z will be a basis of Liouvillian solutions of L(y) = 0. Otherwise L(y) = 0 does not have Liouvillian solutions but can be solved in terms of solutions of L(y) = 0 since y, y y, y will be a basis of solutions of L(y) = 0.. The imprimitive case Compute the exponential solutions of L s (y) = 0. (a) If L s (y) = 0 has a two dimensional space of rational solutions, then G = G 7. (b) If L s (y) = 0 has a two dimensional space of solutions which are square roots of rational functions, then G = G (c) If L s (y) is of order 0 and has an (up to multiple) unique non zero solution g whose square is rational, then G is an imprimitive group. There exists a Liouvillian solution and we are in the Generic case where the formula holds. Find the remaining coefficients of P (U) of degree by solving the linear system where a 0, a have been replaced by their values. (d) If L s (y) is of order and has an (up to multiple) unique non zero solution g whose square is rational, then G is isomorphic to G 7, G, G 8 or G. Otherwise G SL(, C) is a primitive linear group

16 Felix Ulmer: Liouvillian solutions of third order equations. The finite groups Throughout this section we will assume that k = C(x), δ = d and that the dx algorithm of the previous section showed that G is one of the four exceptional finite imprimitive groups or belongs to the finite list of primitive unimodular group ((Singer and Ulmer, a) Section.). In this case all Liouvillian solutions of L(y) = 0 are algebraic. Therefore L(y) has a Liouvillian solution if and only if G is one of the four exceptional imprimitive groups or one of the eight finite primitive subgroups of SL(, C) denoted A, A C, G SL 8, G SL 8 C, A SL, H SL, H SL 7 and F SL whose matrix generators can be found in (Singer and Ulmer, b) Section... Section. of (Singer and Ulmer, a) shows that it is always possible to compute G, if G belongs to a finite list of groups using the decompositions of symmetric powers. Therefore, after running the algorithm of the previous section, we will always be able to compute G. The determination of G is often simplified by the use of necessary conditions on the exponents of L(y) (cf. (Kovacic, 8; Singer and Ulmer, )) and is in general much easier than the computation of the minimal polynomial F C(x)[Y ] of a solution of L(y) = 0. The procedure to compute F C(x)[Y ] given in (Singer and Ulmer, b) is based on the computation of values of semi-invariants of G and involves solving high symmetric powers. In (Fakler, ; Hoeij et al., ; van Hoeij and Weil, 7; Weil, ) the approach was simplified using syzygies, systems and standard constructors (Hessian, Jacobian, bordered Hessian) for the invariants. The computation of F C(x)[Y ] is still difficult and the method using the above constructors only works up to third order equations. In the following we show how strong necessary conditions can be derived for each of the twelve finite groups and how the computation of F C(x)[Y ] can be simplified. We present the results for third order equations, but the generalisation to higher order is straightforward... Ramification data Let L(y) be an n-th order linear differential equation over C(x) with finite differential Galois group G GL(n, C). According to (Singer and Ulmer, ) Lemma. () all singular points c of L(y) must be regular singular points. From (Singer and Ulmer, ) Lemma. we get that the monodromy matrix M c,l corresponding to a small loop γ around c containing no other singularities Note that in (van Hoeij, 00) it is shown how to construct the minimal polynomial of a solution from the minimal polynomial of the Riccati equation.

17 Felix Ulmer: Liouvillian solutions of third order equations 7 of L(y) is conjugate to e π α e π α e π α n () where α,..., α n Q are the exponents of L(y) at c. The eigenvalues of M c,l determine the exponents at c up to integers. In other words, writing α i = a i b i + n i where n i Z and a i b i ]0, ], then a i b i is the part of the exponent that is uniquely determined by M c,l. Definition.: With the above notation we define { a b, a b,..., an b n } as the ramification data of L(y) at the singularity c or, equivalently, of M c,l G. Note that ramification data of conjugate elements of G coincide. Example.: To illustrate the computation of the ramification data we have reproduced a small part of the character table of G in form of an output of the computer algebra system Magma: Class Size Order... p = p =... χ,... χ, J J... χ 0 + J J χ J J J + J... χ where J + J + = 0 and p = j is the p-power map which indicates for σ G in the i-th conjugacy class to which class σ j belongs. Consider a third order differential equation L(y) with G = G whose character is χ. If the monodromy matrix M c,l at a given singularity c belongs to the class number 0, then, according to the character table we get trace(m c,l ) = + J. Using

18 Felix Ulmer: Liouvillian solutions of third order equations 8 the p-power maps we also get trace ( (M c,l ) ) = J and trace ( (M c,l ) ) =. Therefore if M c,l is conjugate to α α 0, 0 0 α then i= α i = + J, i= α i = J and i= α i =. Using the newton formulae we get α i are roots of ( ) ( ) ( ) X (J + )X JX = X e π X e π X e π. The ramification data of the character χ for a monodromy matrix in the conjugacy class number 0 is {,, }. Clearly such a computation is possible for any conjugation class and any character. In the application we don t know the class of M c,l, but only few conjugacy classes correspond to a given ramification data. For example the conjugacy class 0 is the only possible for the ramification data {,, { }, while for,, } there are possible conjugacy classes in G. Suppose L(y) is of order n and denote Sol(L) and Sol(L s m ) the solution space of L(y) = 0 and L s m = 0. From a basis Y,..., Y n of Sol(L) we get a basis Y m, Y my,..., Yn m of Symm (Sol(L)) and the symmetrization of the representation ρ gives a morphism ρ m : GL(Sol(L)) GL(Sym m (Sol(L))) a a, a,n m, a, m a, a m,n ma, m a, ma n,n a m,n. a n, a n,n a m n, an, m a n, a m n,n Let Φ m be the G-morphism obtained by restriction of the evaluation morphism Φ to the vector space of homogeneous forms of degree m of C[Y,, Y n ] ((Singer and Ulmer, a) Lemma.). By composition we get a representation ψ of G on Sol(L s m ). If L i is a right factor of L s m, then Sol(L i ) is a G-submodule of Sol(L s m ).

19 Felix Ulmer: Liouvillian solutions of third order equations ρ ρ G m GL(Sol(L)) GL(Sym m (Sol(L))) ψ Φ m GL(Sol(L s m )) ψ i GL(Sol(L i )) We denote resp. χ, χ s m, χ L s m and χ Li the character of resp. ρ, (ρ m ρ), ψ and (ψ i ψ). Lemma.: Let L(y) be a irreducible linear differential equation whose differential Galois group G is finite. Then, using the above notation, we have. χ L s m is a summand of χ s m.. If c is a non apparent singularity of L s m (y), then c is a singularity of L(y). There exists a σ G such that the monodromy matrices of L and of L s m at c are conjugated respectively to ρ(σ) and ψ(σ).. Suppose that L s m (y) is the least common left multiple of irreducible right factors L (y),..., L t (y). If L i (y) is a right factor of L s m (y) then χ Li is a summand of χ s m. There exists a σ G such that the monodromy matrices of L and of L i at c are conjugated respectively to ρ(σ) and (ψ i ψ)(σ). Proof:. The first assertion follows from the complete reducibility of G.. Since y m, ym y,..., yn m span the solution space of L s m (y), a non apparent singularity of L s m (y) must be a singularity of L(y). We now suppose that the basis y,..., y n chosen so that the monodromy matrix M c,l of L at c is diagonal and given by (). Note that there exists σ G such that ρ(σ) = M c,l. For this basis a continuation along a small loop around c transforms y i into e π α i y i and its affect on the generating set y m, ym y,..., yn m corresponds to the action of ρ m(m c,l ) = ρ m (ρ(σ)) on the basis Y m, Y m Y,..., Yn m of Sym m (V ). Since the monodromy matrix M c,l s m is conjugated to the image of ρ m (ρ(σ)) = ψ(σ) in GL(Sol(L s m )) we get the result.. Since G is completely reducible we get from (Singer, ) that L s m (y) is the least common left multiple of irreducible right factors L (y),..., L t (y) whose solution spaces by the above are G-submodules of Sym m (V ). The result now follows like in the previous cases.

20 Felix Ulmer: Liouvillian solutions of third order equations 0 The fact that the monodromy matrices are images of elements of the same conjugation class of G allows to link the exponents of L(y) and those of a right factor L i of L s m as shown in the following example: Example.: We continue with the previous example. Suppose that L(y) is a third order differential equation with G = G and that χ is the character of the representation of G on the solution space V = C of L(y) = 0. The character χ, is a summand of χ s and thus possibly (the order of L s (y) may be less than 0) the character of a right factor L (y) of order of L s (y). Suppose that at a singularity c of L(y) the ramification data (given by the exponents) is {,, }, then the ramification data of L (y) at c must be { }. If at a singularity c of L(y) the ramification data (given by the exponents) is {,, }, then the ramification data of L (y) at c must be { }, { } or. We will illustrate the use of the ramification data in the case of the imprimitive group G in the following subsection and keep the same notation for the other groups... The group G... Ramification table Suppose the differential Galois group G of L(y) = 0 being a unimodular three dimensional imprimitive representation of G with character χ. Decomposing the character χ s (cf. (Singer and Ulmer, a), Section.) we get χ s = χ, + χ, + χ + χ, where χ, is a linear character of order, χ, is a linear character of order, χ is an irreducible character of degree and χ is an irreducible character of degree. In the table below the top part contains all possible ramification data of G for χ. For each ramification data of χ we included all possible ramification data for some characters in the decomposition of χ s (which are possible characters of factors of L s (y)). Note that a ramification data like {,, } of χ may behave in different ways for possible characters of factors of L s (y), in which case all possible cases have been included. The identity, corresponding to an apparent singularity, has not been included since it gives no information. Ramification data of χ, which are identical for several conjugation classes of G and which give the same ramification data for all characters in the table will be merged into one single column.

21 Felix Ulmer: Liouvillian solutions of third order equations χ χ, χ, χ χ The information given in a ramification table of the above type gives stronger necessary conditions on the exponents of L(y) for G to be the group under consideration, than those of the type given in ((Singer and Ulmer, ), Necessary conditions for case ). Example.: The following differential operator was constructed using the method given in (van der Put and Ulmer, 000): L(y) = d y dx 8x x (x ) d y dx + x x x (x ) dy dx 780 x 800 x + 7 x 0 8x (x ) y The operator is irreducible and L s (y) is of order. From Corollary. we get that G is either G 7, G, G 8, G or F. Since L s (y) = 0 has no rational solution the groups G 7 and G 8 are not possible. At 0, and the exponents are {,, }, {, 7 8, 8 } and {, 0, }, corresponding to the ramification indices {,, }, {, 8, 7 8 } and {,, }. Since neither G, nor F contain elements of order 8 (see ramification table of those groups), the group must be G.... Ramification data and exponential solutions For G we want to find the value (up to multiples) of the two semi-invariants of degree. The fact that one value will be a square root with character χ,

22 Felix Ulmer: Liouvillian solutions of third order equations (the value of a semi-invariant that factors into linear forms) and the other value will be a cube root will allow us to identify them. If L s (y) is of order, then the cube root must vanish, since a product of non zero solutions cannot be zero. To compute an exponential solution g of L s (y) = 0 which is a j-th root of a rational function (i.e. which is the value of a semi-invariant of degree and order j) we write g in the form g = f j c i (x c i ) e i where the product is taken over all finite singular points c i of L s (y) (which are also singular points of L(y)), e i are exponents of L s (y) at c i and f C[x]. Since g is a j-th root, the exponents e i above must be of the form m i with m j i Z. The degree m of f must be of the form ( s ) m = j e i + e N () i= where e is an exponent of L s (y) at infinity (cf. (Singer and Ulmer, ) Theorem.). Example.: In example. the ramification data at 0, and is {,, }, {,, 7} and {,, }. From the ramification table of G 8 8 we get that the only possible ramification data at 0,, of a factor of order of L s (y) = 0 with character χ,, must be { }, { }, {}. This shows that the exponents at 0,, of an exponential solution g of L s (y) = 0 whose square is rational, must be of the form [ + n, + n, n ] with n i Z. Therefore g must be of the form g = f x n (x ) n with f C[x] such that the degree m of the polynomial f is a positive integer of the form m = ( + n + + n ) + e for some integer exponent e at of L s (y) = 0. In the above example the actual exponents of L s (y) at 0, and are {,, 7, 8,,,,, 7 } {8,, 7,,,, 7,, } {, 0,,,,,,, }. Among them we must find exponents at 0,, of the form + n, + n, n with the property ( + n + + n ) + n = m N. It turns out that in this example, the only possible exponents for g at 0, and are [,, ] giving m = 0. Therefore the exponential solution must be g = x / (x ) /.

23 Felix Ulmer: Liouvillian solutions of third order equations In general the exponents will not determine the solution completely, but will narrow down the possible ramifications at the singularities and thus simplify the computation of an exponential solution. If we don t know priorly that G = G, then we could suppose this to be true and do the above in order to find g. If none of the possible exponents produces a solution g of L s (y) = 0, then we can conclude that G = G. This gives further necessary conditions for G (cf. Example.).... Ramification data and factors in symmetric powers We will use the following facts about Fuchsian linear differential equations L(y) of order n. The number of apparent singularities can be bounded by µ (µ )n(n ) j= n e j,i () where n is the order of the equation under consideration, µ the number of non apparent singularities (possibly including ) and the e i,j are the exponents at the non apparent singularities ((Cormier et al., 00), relation ()). Fuchs s relation states that the sum of all exponents at all ν singularities of L(y) (including apparent singularities and possibly including ) must be i= (ν )n(n ) () The integer exponents at the apparent singularities can be bounded using this relation. If L(y) has ν singular points (including apparent singularities and possibly including ), then L(y) is determined by its exponents up to (n )(nν n ) () accessory parameters ((Ince, ) Section.). Suppose L(y) is a third order equation with imprimitive differential Galois group G = G SL(, C). We want to compute a second order right factor L (y) of L s (y) whose character is χ. Again we use the information of the ramification table to get the ramification type of the solutions of L (y), corresponding to the character χ, at those singularities which are also singularities of L(y). New apparent singularities (whose exponents are all positive integers) may also appear in L (y). An example where apparent singularities appear is given in Example..

24 Felix Ulmer: Liouvillian solutions of third order equations Example.7: In example. the ramification data at 0, and is {,, }, {,, 7 } and {,, }. From the ramification table of G 8 8 we get that the only possible ramification data at 0, of a right factor L (y) is {, }, {, } and at it could be {, }, {, }, {, } or {, }. Thus the form of the exponents of L (y) is {n 0,, / + n 0, } at 0, { + n,, + n,} at and {n,, n, }, { + n,, + n,}, { + n,, + n,} or { + n,, + n,} at where n i,j Z. According to the list of exponents of L s (y) given in the previous section, the smallest possible exponents at 0,, which satisfy these conditions are {, /}, {/, 7/}, {, /}. Relation () implies that for these exponents (and thus also for any larger exponents) there are no apparent singularities in L (y). Since the smallest possible exponents satisfy Fuchs s relation (), the exponents cannot be larger. A second order equation with singular points (here 0,, ) is uniquely determined by its exponents and can be obtained using the formula from (Ince, ) Section.: L (y) = d y 7 (8 x ) dx x (x ) dy dx + 0 x 8 x (x ) y x The character table will restrict the possible ramification data at the singularities of L and therefore simplifies the factorisation of L s (y). But in general, several possible sets of exponents might satisfy Fuchs s relation and apparent singularities might appear. The accessory parameters and the apparent singularities must satisfy polynomial relations obtained by setting the remainder of a right division of L s (y) by L (y) to zero. The gain of the approach depends on the order of the factor we are looking for, the number of singularities and on arithmetic conditions of the exponents. But in any case the information should be used in the algorithms.... Precomputation of the invariants We now investigate the computation of the minimal polynomial F C(x)[Y ] of a solution of L(y) for G = G along the lines of (Singer and Ulmer, b) Section.. The new contribution presented here is the use of invariants of factors of some constructions L s m (y) to improve the computation of values of invariants which are needed to express the minimal polynomial of a solution of L(y). The group G has a unique normal subgroup N of order 7 which is the kernel of χ. Thus the group G /N of order 8 is the differential Galois group of the, up to multiples, unique second order right factor L (y) of L s (y). Our strategy will be to compute L (y) and its invariants and to relate them to invariants of G. We denote Y, Y, Y a basis of V = C corresponding to the representation of G given in section.. The subspace of Sym (V ) corresponding to the representation ρ (G ) of degree of G with character χ is spanned by X = Y Y and X = Y Y. In the basis X, X the group ρ (G ) is