Irreducible Linear Di erential Equations of Prime ... - Semantic Scholar

J. Symbolic Computation (1995) 11, 1{000

Irreducible Linear Dierential Equations of Prime Order FELIX ULMERy IRMAR Universite de Rennes I Campus de Beaulieu F-35042 Rennes Cedex Email: [email protected] (Received 11 February 1994)

With the exception of a nite set of nite dierential Galois groups, if an irreducible linear dierential equation L(y) = 0 of prime order with unimodular dierential Galois group has a Liouvillian solution, then all algebraic solutions of smallest degree of the associated Riccati equation are solutions of a unique minimal polynomial. If the coecients of L(y) = 0 are in Q()(x)  Q(x) this unique minimal polynomial is also de ned over Q()(x). In the nite number of exceptions all solutions of L(y) = 0 are algebraic and in each case one can apriori give an extension Q( )(x) over which the minimal polynomial of an algebraic solution of L(y) = 0 can be computed.

1. Introduction In this paper we consider linear dierential equations of the form dp y + a dp?1 y + : : : + a y; a 2 k (1.1) L(y) = dx p?1 dxp?1 0 i p where p is a prime and k is a dierential eld whose eld of constants is algebraically closed of characteristic 0. We also assume that there exists b 2 k such that b0 =b = ap?1 (i.e. the dierential Galois group is unimodular), which can always be achieved by a suitable variable transformation without altering the Liouvillian character of the solutions (cf. Kaplansky (1957), p. 41). We will also only consider irreducible equations L(y) (i.e. whose dierential Galois group are irreducible linear groups), leaving the case of a reducible equation as an induction case (cf. Singer and Ulmer (1993b), Section 2). We note that if L(y) = 0 y This paper was prepared during the author's visit at the Department of Mathematics at Cornell University during the Fall semester of 1993. This work was supported in part by the David and Lucile Packard Foundation 0747{7171/90/000000 + 00 $03.00/0

c 1995 Academic Press Limited

2

Felix Ulmer

is reducible of order > 3, the bounds used in the algorithms (cf. Singer (1981), Singer and Ulmer (1993b), Ulmer (1992)) are not as good as in the irreducible case and that a factorisation should be computed rst. The current algorithms to compute Liouvillian solutions (see e.g. Kovacic (1986), Singer and Ulmer (1993b) or Ulmer (1992) for de nitions) of L(y) = 0 require to consider the coecients ai of L(y) in a dierential eld whose eld of constants is algebraically closed. Thus, even if the coecients are in Q()(x)  Q(x), one has to consider them as being in Q(x). This means that an algebraic extension Q(; ) of Q() might be needed in the computations and in the result. It is likely that an appriori knowledge of this extension would simplify the computation, in particular in the cases where one knows beforehand that no such extension is needed. If L(y) = 0 has a Liouvillian solution, then L(y) = 0 has a solution z whose logarithmic derivative w = z 0 =z is algebraic over k (cf. Singer (1981)). The computation of the Liouvillian solutions can thus be reduced to the computation of the minimal polynomial of some algebraic solution of the Riccati equation R(u) = 0 associated with L(y) = 0. For imprimitive linear groups, this is the only known method. Recently some new results Hendriks and Van der Put (1993) and Zharkov (1993) have been obtained concerning the problem of rationality for this computation. We denote by PR = fP1; P2; : : :g the set of all minimal polynomials of an algebraic solution of minimal degree of R(u) = 0. We have PR 6= fg if and only if L(y) = 0 has a Liouvillian solution (cf. Singer (1981)). If the coecients of L(y) = 0 belong to Q()(x)  Q(x), then the degree of the algebraic extension Q(; )=Q() can be bounded by the number of elements of PR (cf. Hendriks and Van der Put (1993)). In Hendriks and Van der Put (1993) and Zharkov (1993) the number of elements of PR has been bounded for the primes p = 2 and p = 3. However, the results do not produce the extension Q( )=Q or characterize the linear dierential equations, via its coecients, for which such an extension might be needed. We show that for an irreducible equation of prime order and unimodular dierential Galois group one can always distinguish the two following cases: 1 PR contains exactly one element. If the coecients of L(y) = 0 belong to Q()(x)  Q(x), then the coecients of the unique element of PR are in Q()(x). 2 The dierential Galois group of L(y) = 0 belongs to a nite set of nite groups. In this case all solutions of L(y) = 0 are algebraic and the minimal polynomial of a solution of L(y) = 0 can be constructed according to Singer and Ulmer (1993b). If the coecients belong to Q()(x)  Q(x) one can, for each group, appriori give an algebraic extension Q(; ) of Q() so that all computations can be done in Q(; )y. In particular the algebraic extension needed during the computation is not only of bounded degree but known in advance. Considering rst the nite set of possible nite groups, one can then assume that PR contains exactly one element with coecients in Q()(x). We note that in the existing algorithms (Kovacic (1986), Singer and Ulmer (1993b)) one already has to consider separately a nite set of nite groups (the primitive linear groups). For second order equations only

y The result, i.e. the coecients, may not be expressible over Q( ), but the nal Grobner basis computation will yield polynomials whose roots generate an extension F of Q such that the coecients belong to F (x) (cf. Section 5.2)

Linear Dierential Equations of Prime Order

3

the group of quaternions and for third order equations only 2 imprimitive groups of order resp. 27 and 54 must be added to the list of nite primitive groups. For each prime p the approach also produces examples of an imprimitive subgroup of SL(p; C ) for which the number of elements of PR is exactly p + 1. The paper is organised as follows: In the rst section we derive properties of monomial groups. In the second we connect maximal normal subgroups and elements of PR . We then show that the above case distinction is always possible. In the fourth section we show how a minimal polynomial of a solution can be computed directly for a known nite dierential Galois group and apply this method to an example. We also derive necessary conditions in terms of exponents which must be satis ed, if a given nite group is the dierential Galois group of a given equation.

2. Abelian normal subgroups of monomial groups The aim of this section is to determine the structure of a monomial group containing two distinct maximal abelian normal subgroups. We will show in the next section that those subgroups correspond to distinct elements Pi 2 PR , i.e. to cases where PR has more than one element. De nition & Theorem: (see e.g. Dixon (1971), Theorem 4.2B) Let G be a subgroup of GL(n; C ) acting irreducibly, i.e. G is a linear group acting irreducibly on the vector space V of dimension n over C . Then G is called imprimitive if, for k > 1, there exist subspaces V1;


; Vk such that V = V1 


 Vk and S = fV1; : : :; Vk g is a transitive G-set. This gives a homomorphism  of G onto a transitive subgroup TS of k elements. All the Vi have the same dimension n=k and the set S is called a system of imprimitivity of G. The stabilizer of Vi is denoted Gi and \i Gi is a normal subgroup of G. An irreducible group G  GL(n; C ) which is not imprimitive is called primitive. If all the subspaces Vi are one-dimensional, then G is called monomial. In this case \iGi is a maximal abelian normal subgroup of G. If \iGi  Z(G) we say that G is central-monomial of degree n. There are only nitely many central-monomial subgroups of SL(n; C ), since the order of such a group divides n
(n!). A central-monomial group of degree n is a central extension of a transitive permutation group of degree n. If G is monomial but not centralmonomial, then \iGi is a non-central maximal abelian normal subgroup of G. We note that an irreducible representation of degree n of a group G is imprimitive if and only if the representation is induced by a representation of degree m of a subgroup H of index k with n = k
m and 1  k < n (cf. Issacs (1976), pp. 65-66). This gives a constructive method to test if a given representation is imprimitive. If  is a character of H, we denote the induced character of G by  G . Let V be a nite dimensional C -vectorspace, G  GL(V ) and H a subgroup of G. Let W be a minimal H-invariant subspace of V . Then the homogeneous component VW of H associated with W is the H-invariant subspace of V formed as the sum of all H-invariant subspaces W 0 which are isomorphic with W as H-modules (cf. Dixon (1971), x4.2). Imprimitivity is closely related to the existence of certain normal subgroups (cf. Dixon (1971), x4.2). In particular, if an irreducible group G  GL(n; C ) has a non-central normal abelian subgroup, then G is imprimitive (see e.g. Dixon (1971), Corollary 4.2A). However, the converse is false, i.e. not any monomialgroup has a non-central normal abelian subgroup:

4

Felix Ulmer

Example. - Let G be the alternating group on 5 letters. Then G has a unique irreducible faithful character  of degree 5. The subgroup H generated by the permutations (1; 3; 4) and (1; 5; 3) of index 5 is isomorphic to A4 and thus has exactly two non-trivial linear characters, say i ; i 2 f1; 2g. Clearly, by Frobenius' reciprocity, the trivial character of G is not a constituent of iG . Since G has ve irreducible characters the degrees of which are respectively 1, 3, 3, 4 and 5, we get that iG = . In particular  is the character of an imprimitive and central-monomial representation of G. 

We note that there are no central-monomial subgroups of SL(2; C ). The order of such a group must divide 4 and the group would be abelian. Also, there are no central-monomial subgroups of SL(3; C ). Such a group is a central extension of A3 or S3 which both have a trivial Schur-Multiplicator (see e.g. Issacs (1976) or Ulmer (1992)). Since neither A3 nor S3 have an irreducible representation of degree 3 we get that such a group does not exist. The above example shows that SL(5; C ) has a central-monomial subgroup. Example. - For p = 2 there exist two non-abelian groups of order p3 = 8, the dihedral group D4 and the quaternion group Q8 . Since D4 contains non-central elements of order 2, the group has no faithful irreducible representation in SL(2; C ). An irreducible representation of Q8 in SL(2; C ) is given in section 5.1.1. For a prime p > 2 there exist two non-abelian groups of order p3: The extra-special group Ep;1 of order p3 and exponent p and the extra-special group Ep;2 of order p3 of exponent p2 (cf. Huppert (1983), Ch. 1, Theorem 14.10). Both groups have p ? 1 irreducible representations of degree p ( Huppert (1983), Ch. V, Theorem 17.13) which must be monomial since the groups are nilpotent ( Huppert (1983), Ch. V, Theorem 18.5). Since we will restrict our attention to unimodular groups we note that:

1 The representations of degree p of Ep;1 must be unimodular. Consider the subgroup generated by an arbitrary non-central element g and the center. This abelian group is of index p and normal. Since the group is not central all homogeneous components are one-dimensional. In particular all eigenvalues of g are distinct p-th roots of unity and their product, since p > 2, is 1. 2 Any representation of degree p of Ep;2 has a non-central element g of order p2 with det(g) 6= 1. To see this note that gp 6= 1 is a central element of order p and thus gp is a scalar multiplication by a primitive p-th root of unity ". Using the same argument as above we get that all eigenvalues of g are distinct p-th roots of " 6= 1 and that their product is ". Thus none of the monomial representations of degree p of Ep;2 is unimodular. This allows the following Definition 2.1. We denote Np the (up to isomorphism) unique non-abelian subgroup of

order p3 of SL(p; C ). For p = 2 this is the group of quaternions and for p  3 it is the extra-special group of exponent p. Theorem 2.1. Let p be prime and G  SL(p; C ) be an imprimitive linear group. If G has

two distinct maximal normal abelian subgroups, then:

Linear Dierential Equations of Prime Order

5

1 G is isomorphic to a split extension C o Np of Np and a cyclic group C of order dividing p ? 1. In particular the order of G divides p3(p ? 1). 2 There are at most p + 1, for G  = Np exactly p + 1, distinct maximal normal abelian subgroups of G.

1 We denote by A1 and A2 the two distinct maximal normal abelian subgroups of G. The groups Ai must both properly contain the center Z(G) of G which, since G is unimodular, is of order 1 or p. Put N = A1 A2 and note that N is a non-abelian normal subgroup of G. In particular, N is not contained in Z(G). If N was reducible, then the dimension of a homogeneous component of N would be 1 or p; in any case N would be abelian, a contradiction. Hence N is irreducible. For the commutator subgroup N 0 of N we have 1 6= N 0  A1 \ A2  Z(N). By Dixon (1971), Theorem 4.3 the group N=Z(N) must be abelian of order p2 . Since N is non-abelian and acts irreducibly we have jZ(N)j = p which implies jN j = p3 and jN : Ai j = p. In particular N  = Np . The p distinct homogeneous components of Ai form a system of imprimitivity for G. This gives a homomorphism of G=Ai onto a transitive subgroup TA of Sp which shows that the order of G=Ai must divide (p!). Since G=Ai divide (p!) and jAi j = p2, the largest power of p dividing the order of G is at most p3. Thus N is a p-Sylow subgroup of G. Since N is a normal subgroup of G, it is the unique subgroup of order p3 of G. From the above we have that the order of G divides p2
(p!) and thus for p = 2 the order must be 23 and we must have G = N  = N2 , the group of quaternions. We now consider the case p  3. Since both Ai and N are normal subgroups of G, the group N=Ai is a normal subgroup of G=Ai = TA  Sp , which shows that the transitive permutation group TA also has a normal p-Sylow subgroup. Thus TA is generated by an element P of order p and an element Q whose order divides p ? 1 (cf. Huppert (1983), Ch. 5, Theorem 21.1). From Huppert (1983), Ch. 1, Theorem 18.1 we get that N  = Np has a complement in G which must be isomporphic to the cyclic group generated by Q. This gives the semi-direct product representation stated in the theorem. 2 Assume that G has a third maximal normal abelian subgroup A3 . Using the above reasoning for, say, A2 and A3 we get that A2A3 is a non-abelian normal subgroup of G of order p3. Since N is the only subgroup of G of order p3, A2 A3 and thus A3 must be contained in N. In particular any maximal normal abelian subgroup of G is a subgroup of N of order p2 . The number  of such subgroups is congruent to 1 mod p ( Huppert (1983), Ch. 1, Theorem 7.2). Since there are at least two such groups and jN j = p3, we get that there are exactly p + 1 maximal abelian normal subgroups of N and thus at most p + 1 abelian normal subgroups of G. If G is of order p3, then G  = Np has exactly p + 1 abelian normal subgroups ( Huppert (1983), Ch. 3, Theorem 13.7.(f)).

Proof.

i

i

i

i

The Theorem shows that there is only a nite number of unimodular monomial groups of prime degree containing more than one maximal abelian normal subgroup and that those groups are all nite. It also shows how those groups can be constructed. One way of constructing the groups is to note that G = C o Np is a subgroup of the wreath product Np o C (cf. Huppert (1983), Ch. 1, Theorem 15.12). For p = 3 we get that N3 o (Z=2Z) has up to congugation only one subgroup of order 33
2 which we denote by Z=2Zo N3 (cf. Blichfeld (1917) p. 105). An irreducible representation in SL(3; C ) of Z=2Zo N3 is given in Section 5.1.2.

6

Felix Ulmer

For prime degree p a worst case, i.e. an unimodular group with a maximal number of non-central maximal abelian normal subgroups, is always given by Np .

3. Normal abelian subgroups and the Riccati In this section we connect non-central normal abelian subgroups and algebraic solutions of the Riccati. We consider L(y) = 0 given in (1.1) and associate to L(y) = 0 a Picard-Vessiot extension K and a dierential Galois group G (K=k) consisting of all dierential eld automorphims of K=k (see e.g. Kaplansky (1957), Singer and Ulmer (1993b) or Ulmer (1992) for de nitions). We also denote G (K=k) by G (L). The action of G (L) on the solution space of L(y) = 0 gives a faithful representation of degree p of G (L) over the eld of constants of k. Unless otherwise mentioned, this will be the representation of G (L) in what follows. Theorem 3.1. Consider an equation (1.1) of prime degree p whose dierential Galois group

G (L) is an imprimitive subgroup of SL(p; C ). Then

1 PR is non-empty and consists of polynomials of degree p. 2 If G (L) has no faithful central-monomial representation of degree p, then PR contains at most p+1 elements and there is a bijection between maximal normal abelian subgroups of G (L) and elements of PR . 3 If G (L) has no faithful central-monomial representation of degree p and is not isomorphic to a split extension C o Np of Np and a cyclic group C of order dividing p ? 1 (in particular if the order of G (L) does not divide p
(p!) or p3 (p ? 1)), then PR contains exactly one element. Proof. Since G (L) is a monomial group, G (L) has a subgroup H of index p which has a commoneigenvector z, i.e. which is 1-reducible ( Ulmer (1992), Lemma 4.2). The logarithmic derivative of z is left xed by H and thus is algebraic of degree p. The group G (L)  SL(p; C ) cannot have a 1-reducible subgroup H 0 of index < p whithout being reducible, since the orbit of the eigenvector of H 0 would generate a non-trivial G (L)-invariant subspace. Thus p is the minimal index of a 1-reducible subgroup of G (L). This shows that PR is non-empty and that all polynomials in PR are of degree p (see Singer and Ulmer (1993b), Lemma 3.1). We now assume that G (L) has no central-monomial representation of degree p. Let Pi be an element of PR of degree p and w a root of Pi which is the logarithmic derivative of a solution z. The one dimensional subspaces Vi generated by the p vectors f(z)j 2 G (L)g (i.e. the conjugates of z under G (L)) form a system of imprimitivity for G (L). We set Gi = f 2 G (L)j(Vi ) = Vi g. Then Ai = \iGi is a maximal normal abelian subgroup of G (L). Since G (L) has no central-monomial representation, Ai must be a maximal noncentral normal abelian subgroup of G (L). Since Ai is a non-central abelian normal subgroup of G (L), there are p homogeneous components of Ai and thus up to multiples a unique basis of eigenvectors of Ai whose logarithmic derivatives are the solutions of Pi . Since Ai is a maximal abelian subgroup, we get that, to each Pi corresponds a unique maximal normal abelian subgroup Ai . Conversely, the logarithmic derivatives of a basis of eigenvectors of a maximal (and thus non-central) normal abelian subgroup Ai corresponds to the p solutions of an element Pi of PR , which gives a bijection between elements of PR and maximal normal abelian subgroups of G (L). From Theorem 2.1 we get that an imprimitive group has at most

Linear Dierential Equations of Prime Order

7

p + 1 maximal normal abelian subgroups. From Theorem 2.1 we get that if G (L) is not isomorphic to a split extension C o Np of Np and a cyclic group C of order dividing p ? 1, then the imprimitive group G (L) has exactly one maximal normal abelian subgroup. Since G (L) has no faithful central-monomial representation then there is a bijection between maximal normal abelian subgroups of G (L) and elements of PR , which shows that PR contains exactly one element. 2 Let L(y) = 0 with coecients in Q()(x)  Q(x). Considering K as a dierential eld extension of Q()(x) (with new constants) we get the group G (K=Q()(x)) of dierential eld automorphisms of K=Q()(x). We denote G(Q=Q()) the classical Galois group of Q=Q(). In Hendriks and Van der Put (1993) it is shown that the following sequence 1 ?! G (K=Q(x)) ,! G (K=Q()(x)) ?! G(Q=Q()) ?! 1 is split exact. Choose a point a 2 Q which is a regular point of L(y) = 0 and consider K as a sub eld of Q((x ? a)). We denote by s P the splitting homomorphism and P1 given in Hendriks i i Van der Put (1993) and de ned by s( 1 i=k ai (x ? a) ) = ? i=k (ai )(x
? a) . It has the property that the elements of Q(x) which are left xed by s G(Q=Q()) are in Q()(x). From ?the splitting
one gets, that G (K=Q()(x)) is the semi-direct product of G (K=Q(x)) and s G(Q=Q()) . Corollary 3.2. Consider an equation (1.1) of prime degree p with coecients in Q()(x) 

k = Q(x) whose dierential Galois group G (L) is an imprimitive subgroup of SL(p; C ). If G (K=k) has no faithful central-monomial representation of degree p and is not isomorphic to a split extension C o Np of Np and a cyclic group C of order dividing p ? 1, then the unique element of PR has coecients in Q()(x). Proof. Theorem 3.1 shows that PR contains exactly one element P and that G (K=k) has

exactly one maximal normal abelian subgroup A. The solutions ui of P are the logarithmic derivatives of a basis of eigenvectors yi for A. The group G (K=Q(x)) permutes the ui's and thus ? leaves the
coecients of P, which are in Q(x), invariant. We denote  an element of s G(Q=Q()) . Since yi is an eigenvector of A, we get that (yi ) is an eigenvector of A = A?1 which is thus also a non-central abelian normal subgroup of G (K=Q(x)) isomorphic to A. Since A is the unique maximal non-central normal abelian subgroup ? of G(K=Q
(x)) we have A = A. In particular (yi ) is a multiple of some yj and thus s G(Q=Q()) also permutes the ui's and leaves the coecients of P xed. The coecients must belong to Q()(x). 2 The number of elements in PR is only an upper bound for degree of the algebraic extension needed to represent the elements of PR . Even if PR contains more than one polynomial, all its elements are in some cases de ned over the coecient eld of L(y) (cf. Ulmer and Weil (1994), pp. 15-16). Corollary 3.3. (cf. Hendriks and Van der Put (1993) and Zharkov (1993)y ) Let L(y) =

y The corresponding results in Hendriks and Van der Put (1993) and Zharkov (1993) contain a mistake which has been corrected by the authors in later preprints (cf. Hendriks and Van der Put (1993b)). The

8

Felix Ulmer

0 be a linear dierential equation of 2 (resp. 3) with coecients in Q()(x) and whose dierential Galois group G (K=Q(x)) is an imprimitive subgroup of SL(p; C ). If G (K=Q(x)) is not isomorphic to N2 = Q8 (resp. to N3 or Z=2Zo N3), then PR contains exactly one element with coecients in Q()(x). If G (K=Q(x)) is isomorphic to Q8 (resp. to N3 or Z=2Zo N3 ), then PR contains exactly 3 (resp. 4) elements. Proof. It only remains to show that Z=2Zo N3 has 4 normal abelian subgroups of order 9. This can be computed directly (e.g. using Cayley) or by noting that elements of order

2 of N act invertingly on the quotient N3 =Z(N3 ). 2

4. The two possible cases Theorem 4.1. (Jordan's Theorem. See e.g. Ulmer (1992), Section 3) There exists a func-

tion f : N 7! N depending only on n, such that any nite subgroup of GL(n; C ) has a normal abelian subgroup of index  f(n).

Several bounds for f(n) are known. One has f(2) = 60, f(3) = 360, f(4) = 25920,... (see e.g. Ulmer (1992) Section 3 for further references). From Jordan's Theorem we get that any nite primitive subgroup of SL(n; C ) is of order at most n
f(n) and thus that there are at most nitely many such groups. Theorem 4.2. Let L(y) = 0 be an irreducible linear dierential equation over k of prime

order p with dierential Galois group G (L)  SL(p; C ). If L(y) = 0 has a Liouvillian solution over k, then one (or bothy ) of the following holds:

1 PR contains exactly one element. 2 G (L) belongs to a nite set of nite groups. The group G (L) is a nite primitive group or a central-monomial group or isomorphic to a split extension C o Np of Np and a cyclic group C of order dividing p ? 1 (in particular the order of G (L) is either less than p
f(p) or divides p3(p ? 1) or divides p
(p!)). Proof. The group G (L) is either imprimitive or primitive. If G (L) is imprimitive, then the result follows from Theorem 3.1. If G (L) is primitive, then L(y) = 0 has a Liouvillian solution if and only G (L)  SL(p; C ) is a nite group (cf. Ulmer (1992), Corollary 3.7). The result follows from Jordan's Theorem. 2

The fact that there are only a nite set of nite groups where PR may contain more than one element, allows to consider those cases separately. If G (L) is one of those nite groups, then all solutions of L(y) = 0 must be algebraic and one can use the method presented in Singer and Ulmer (1993b) to compute the minimal polynomial of a solution of L(y) = 0 (instead of the minimal polynomial of an algebraic solution of the Riccati). This will be presented in the next section. If, after having tried the nitely many nite groups approach via systems of imprimitivity used in this paper allowed M.F. Singer and the author to nd the error in Hendriks and Van der Put (1993) for the third order case. y For some of the nite groups considered, PR has only one element

Linear Dierential Equations of Prime Order

9

no solution is found, then we can assume that PR contains at most one element and, if k = Q(x), that no algebraic extension will be needed to represent the polynomial in PR .

5. Case of a known nite dierential Galois group In this section we review the method presented in Singer and Ulmer (1993b) to nd the minimalpolynomial of an algebraic solution of L(y) = 0 (instead of the minimal polynopmial of a logarithmic derivative) if G (L) is a given nite subgroup of SL(p; C ). In the following we assume that, for linear dierential equations over the eld k, algorithms for computing solutions that are in k exists (cf. Bronstein (1992) and the references given in Singer and Ulmer (1993b), Section 1). In order to apply the method presented in Section 4 of Singer and Ulmer (1993b) we start with a group G (L)  SL(p; C ) given in terms of matrices over a basis corresponding to unknown solutions which we denote symbolically fy1 ; y2;


; yp g and proceed as follows:

P

1 Compute a maximal subgroup Hz having the common eigenvector z = i iyi . In practice one can choose z so that Hz is of maximal order. 2 Compute a maximal subgroup StabG (L) (z) leaving z invariant (the stabiliser of z in G (L)) and denote by i the index of StabG (L) (z) in Hz . 3 Compute a set of left coset representatives T of Hz in G (L). The minimal polynomial P(Y ) of z is given by: Y? i
P(Y ) = Y ? (z)i 2Hz

whose coecients will be in k and thus invariant under G (L). 4 Compute a basis I1 ;


II of the ring of invariants (cf. Sturmfels ( 1993)) of the nite group G (L)  SL(p; C ) and decompose the coecients of P(Y ) in this basis. The above computations do not depend on the particular equation and needs to be done only once for each nite group in the list. The above decomposition does not depend on the coecient eld of L(y), but only on the choice of the matrix representation of G (L) and on the choice of the common eigenvector z. In order to nd the minimal polynomial of a solution of a given particular equation L(y) = 0 via the above decomposition we use symmetric powers: Definition 5.1. Let fy1 ; : : :; yng be a fundamental set of solutions of the linear dierential equation L(y) = 0. The mth symmetric power L s m (y) of L(y) is the unique dierential equation of smallest order whose the solution space is spanned by fy1m ; y1m?1 y2 ; : : :; ynm g.

An algorithm to construct the equation L s m (y) is given in Singer (1980) and Singer and Ulmer (1993), Section 3.2.2. Since L s m (y) is obtained by solving a linear sytem over the eld of coecients of L(y) = 0, the coecients of L s m (y) belong also to this eld. In the following we use the fact that rational solutions of L s m (y) = 0 are homomorphic images of invariants of degree m of G (L)  SL(p; C ) (cf. Singer and Ulmer (1993b), Lemma 1.6). This allows to connect the above decomposition to a particular given linear dierential equation L(y):

10

Felix Ulmer

1 Collect the invariants I1 ;


II into set Sd = fId1 ;


; Id g of equal total degree d. Compute (cf. Bronstein (1992)) a basis Qd;1 ;


Qd;r of the space of rational P solutions (i.e. solutions in k) of d-th symmetric power L s d (y) = 0 and set Id = rt=1 cd;j;t Qd;t, where cd;j;t are unknown P constants. 2 Substitute Id by rt=1 cd;j;tQd;t in the decomposed expression of P(Y ) and compute for s 2 f2;


; pg the corresponding expressions for the derivatives Y (s) of Y in terms of the unknown constants cd;j;t . 3 Substitute the values Y (s) depending on the cd;j;t in L(y) = 0, which gives a system of polynomial equations for the variables cd;j;t. 4 Compute a lexicographical Groebner basis (cf. Sturmfels ( 1993)) for the above system of polynomial equations. 5 If a value for the cd;j;t can be found for which the polynomial P(Y ) is square free, then P(Y ) is the minimal polynomial of a solution of L(y) = 0. i

j

j

If the above methode does not produce a square free polynomial P(Y ), then the chosen nite subgroup of SL(p; C ) is not the dierential Galois group of L(y) = 0 and another group must eventually be considered. The connection with the rationallity problem is that, for k = Q(x) all computations can be done in an extension of the coecient eld where the above decomposition of the coecients of P(Y ) in terms of invariants is possible. The correctness of the above method is proven in Singer and Ulmer (1993b) where it is applied to the nite primitive subgroups of SL(2; C ) and SL(3; C ). In the following we applied the method to the imprimitive subgroups N2 = Q8 , N3 and Z=2Zo N3 of SL(2; C ) and SL(3; C ) where PR has more than one element. We rst compute the above decomposition for those groups and then apply the method to an example. In a nal subsection we derive necessary condition which allows to test if a given nite group is the dierential Galois group of L(y) = 0. 5.1. Precomputations Theorem 5.1. Let L(y) be a linear dierential equation of prime degree p over k with imprimitive dierential Galois group G (L)  SL(p; C ). If PR has more than one element, then:

1 If p = 2 then G (L)  = N2 and L(y) = 0 has a solution which is algebraic over k and whose minimal polynomial is of the form:

P(Y) = Y8 ? I1 Y4 + (I2 )2 where I1 and I2 are solutions in k of L s 4 (y) = 0. 2 If p = 3 then G (L)  = N3 or G (K=k)  = (Z=2Zo N3 ). (a) If G (K=k)  = N3 , then L(y) = 0 has an algebraic solution whose minimal polynomial is of the form:

?
P(Y) = Y9 ? I2 Y6 + 21 (I2 )2 ? I3 Y3 ? (I1 )3

where I1 and I2 are solutions in k of L s 3 (y) = 0 and I3 is a solution in k of L s 6(y) = 0.

Linear Dierential Equations of Prime Order

11

(b) If G (K=k)  = (Z=2Zo N3), then L(y) = 0 has an algebraic solution whose minimal polynomial is of the form:

1



P(Y) = Y ? I4 Y + 4 (I4 ) ? 21 I2 I4 ? 2I1I3 + 41 (I2 )2 Y6 ? (I1 )3 where I1 , I2 , I3 and I4 are solutions in k of L s 6(y) = 0. 18

12

2

The Theorem is proven in the following subsections. We note that alltrough all decompositions given in the Theorem are over Q, an extension of Q is sometimes necessary, due to the fact that the invariants of G (L) are not de ned over Q (the representation of G (L) is usually not de ned over Q) or that the choosen eigenvector does not have coordinates in Q. This occurs for some primitive nite subgroups of SL(2; C ) and SL(3; C ) (cf. Singer and Ulmer (1993b), Section 4). 5.1.1. Second order equations

From Corollary 3.3 we get that for second order equations the quaternion group N2 = Q8 is the only imprimitive group where PR has more than one element. We now determine the algebraic degree of a solution of L(y) = 0 and decompose the coecients of its minimal polynomial in terms of invariants of G (L). An irreducible representation of degree 2 of Q8 (unique up to equivalence) is generated by: i 0   0 ?1  r = 0 ?i t= 1 0 ; We denote by fy1 ; y2g the basis of the solution space corresponding to the above representation. The group Hy1 generated by r is of order 4 and is a maximal subgroup of Q8 having a common eigenvector which is y1 . The degree of the extension [Q(x)(y1 ) : Q(x)] must divide the order 8 = jQ8j. Since Q8 does not have a faithful transitive representation of degree 4 or 2, we get that the degree of the minimal polynomial P(Y ) of y1 must be 8 and that the stabiliser StabG (L) (y1 ) of y1 is trivial. Thus i = [Hy1 : StabG (L) (y1 )] = 4. A transversal of Hy1 in Q8 is given by T = fid; tg which gives the following formula for P(Y ): Y? 4

?
? P(Y) = Y ? ((y))4 = Y4 ? y14 Y4 ? y24 = Y8 ? (y14 + y24 )Y4 + (y1 y2)4 2T

In order to decompose the above coecients in terms of the invariants of Q8 we need to compute a basisy of the ring of invariants C [y1 ; y2]Q8 of Q8 . This can be done using the algorithms described in Sturmfels ( 1993). The expansion of the Hilbert series G (z) of the ring C [y1 ; y2]Q8 is X z 4 ? z 2 + 1 = 1 + 2z 4 + z 6 + 3z 8 + O(z 9 ) 1 G (z) = jQ1 j = 6 4 2 8 g2Q8 det(id ? zg) z ? z ? z + 1 This shows that there are two fundamental invariants I1 and I2 of degree 4 and one fundamental invariant of degree 6. In order to decompose P(Y ) we will only need the two

y Not all the invariants of Q8 are needed, only those whose degree is less than the maximal degree of the homogenous forms appearing in the coecients

12

Felix Ulmer

fundamental invariants of degree 4 which are I1 = (y14 + y24 ) and I2 = (y1 y2 )2 . Using a Grobner basis (cf. Sturmfels ( 1993)) or an Ansatz one can decompose the coecients of P(Y ) in terms of those invariants: P(Y) = Y8 ? I1 Y4 + (I2 )2 5.1.2. Third order equations

According to Blichfeld (1917), pp. 105-106 we de ne 01 0 0 1 00 0 11 0 ?1 0 0 1 S1 = @ 0 ! 0 A T = @ 1 0 0 A R = @ 0 0 ?1 A 0 0 !2 0 1 0 0 ?1 0 where !3 = 1 and ! 6= 1. The commutator S2 = [S1 ; T] corresponds to the scalar multiplication by !. The group N3 is generated by S1 and T and the group Z=2Zo N3 is generated by S1 , T and R (cf. Blichfeld (1917), pp. 105-106). According to Corollary 3.3 those are, up to isomorphism, the only two imprimitive subgroups of SL(3; C ) for which PR has more than one element. To decompose the minimal polynomial of a solution of L(y) = 0 we proceed as in the previous section using that: 1 For G (L) =< S1 ; T > = N3 the group Hy1 =< S1 ; S2 > of order 9 has y1 as a common eigenvector. The stabiliser of y1 is < S1 > and i = [Hy1 : StabG (L) (y1 )] = 3. The set T = fid; T; T 2g is a set of left coset representative of Hy1 in G (L). One can decompose the minimal polynomial of y1 using the invariants I1 = y1 y2 y3 ; I2 = y13 + y23 + y33 ; I3 = y16 + y26 + y36 2 For G (L) =< S1 ; T; R > = (Z=2Zo N3 ) the group Hy1 =< S1 ; S2; R > of order 18 has y1 as a common eigenvector. The stabiliser of y1 is < S1 > and i = [Hy1 : StabG (L) (y1 )] = 6. The set T = fid; T; T 2g is a set of left coset representative of Hy1 in G (L). One can decompose the minimal polynomial of y1 using the invariants: I1 = (y1 y2 y3 )2 ; I2 = (y13 + y23 + y33 )2 ; I3 = y1 y2 y3 (y13 + y23 + y33 ); I4 = y16 + y26 + y36 The fact that the faithful representations of N3 and of Z=2Zo N3 in SL(3; C ) are conjugated under the Galois automorphism ! 7! !2 shows that the given decompositions, which are invariant under this automorphism, hold for any such representation. 5.2. Application to an example

To illustrate the method we apply it to the following irreducible linear dierential equation given in Hendriks and Van der Put (1993b): d2y + 27x y = 0 L(y) = dx 2 8(x3 ? 2)2 The group G (L)  SL(2; C ) is the quaternion group N2 = Q8 , since the rational solution space (computed using the algorithm described in Bronstein (1992)) of d5y + 135x d3y + 405(5x3 + 2) d2y L s 4 (y) = dx 5 2(x3 ? 2)2 dx3 4(x3 ? 2)3 dx2

Linear Dierential Equations of Prime Order 2 (x3 + 2) dy ? 405x(7x6 + 35x3 + 10) y + 3645x 2(x3 ? 2)4 dx (x3 ? 2)5

13

is generated by x3 ? 2 and x(x3 ? 2) and thus is of dimension 2 (cf. Ulmer and Weil (1994), Lemma 6). This example was constructed in Hendriks and Van der Put (1993b) in order to prove that an algebraic extension of degree 3 of the coecient eld is sometime necessary to construct an element of PR if G (L) = Q8. Using the method presented in Ulmer and Weil (1994) we get the 3 elements P (U) of PR : 4 2 3 2 2 x ? 6x2 ? x ? 42 ; P (U) = U 2 ? 2x x+3 ? x2 ?  U + 8x + 8 6 8x ? 32x3 + 32 where 23 + 1 = 0. Which shows that a cubic extension of the coecient eld Q(x) is necessary to represent the elements of PR in this case. We now use the method presented in Singer and Ulmer (1993b) to compute the minimal polynomial of a solution of L(y) = 0. From Theorem 5.1 we get that if G (L) = Q8 , then there is a solution of L(y), whose minimal polynomial is of the form P(Y) = Y8 ? I1 Y4 + (I2 )2 where I1 and I2 are solutions in k of the 4-th symmetric power L s 4 (y) = 0. We set
?
?

? ? I2 = c3 x3 ? 2 + c4 x4 ? 2x ?I1 = c1 x3 ? 2 + c2 x4 ? 2x and get ? ?
?

? ?
?

P(Y) = Y8 + c1 x3 ? 2 + c2 x4 ? 2x Y4 + c3 x3 ? 2 + c4 x4 ? 2x 2 This equation is square free if and only if its discriminant Disc(P(Y)) = 65536(x3 ? 2)14(c4x + c3 )6 ((c2 + 2c4)x + (c1 + 2c3 ))4 ((c2 ? 2c4 )x + (c1 ? 2c3))4 is not zero. Following Singer and Ulmer (1993b) (Section 5) we compute the expression for Y 00 which we substitute into L(y) = 0. This gives a polynomial in Y and x whose coecients, which are polynomials in c1; : : :; c4, must be zero. We get a set of polynomial equations for the constants c1; : : :; c4. Computing a lexicographical Grobner basis (cf. Sturmfels ( 1993)) for c1 > c2 > c3 > c4 we get that c1; : : :; c4 must, among others, satisfy the following polynomials: 1 c2 c4) (c64 ? 41 c22c44 ? 41 c63 + 16 1 3 c3(c1 c34 ? 12 c2 c3c24 ? 18 c32 c3) c34(c1 c34 ? 12 c2 c3c24 ? 18 c32 c3) 1 c2) c3(c4 ? 21 c2)c4 (c4 + 12 c2 )(c24 + 12 2 (c4 ? 12 c2 )c24 (c4 + 12 c2)(c34 + 21 c33) Form the last equation we get that c4 equals 0,  21 c2 or ? p312 c3.

14

Felix Ulmer

1 Let c4 = 0. In this case c3 must be non zero for Disc(P(Y)) to be non-zero. From the second polynomial we get that c2 is zero and from the rst that c3 =  21 c1. But then Disc(P(Y)) = 0. 2 Let c4 = 21 c2 6= 0 (resp. c4 = ? 21 c2 6= 0). The rst polynomial implies that c3 = 0 or c3 = 12 c1 (resp. c3 = ? 21 c1 ). If c3 = 0, then the third polynomial implies c1 = 0. In any case Disc(P( p Y)) = 0. p 3 Let c3 = ? 3 2c4 6= 0.p From p3 the fourth equation we get that c2 = 2 ?3c4 and from the this always gives second that c1 = 2 ?3 2c4 . Since Disc(P(Y)) 6= 0 p p p free p a square polynomial. One veri es that for c4 6= 0 the point (2 ?3 3 2c4 ; 2 ?3c4; ? 3 2c4; c4) belongs to the variety de ned by the above Grobner Basis, and thus that p p ?
?
 P(Y) = Y8 + 2c4 ?3 3 2 x3 ? 2 + x4 ? 2x Y4

 p?


?

+ (c4 )2 ? 3 2 x3 ? 2 + x4 ? 2x is the minimal polynomial of a solution of L(y) = 0.




2

If the coecient eld of L(y) is a sub eld of Q(x), then all computations including the nal Grobner basis in c1; : : :; c4 can be done over the eld containing the coecient eld and over which the decomposition of P(Y ) in terms of invariants can be done. The nal Grobner basis gives equations for the algebraic extension needed to represent P(Y ). The minimal polynomials found above have roots whose logarithmic derivatives, by the choice of y1 as eigenvector, are roots of an element of PR . The example shows that a p eld extension (here ?3) is sometime needed to get p3 from the minimal polynomial of a logarithmic derivative u, de ned in this case over Q ( 2)(x),pto pthe minimal polynomial of R the algebraic solution exp( u), de ned in this case over Q( 3 2; ?3)(x). 5.3. Necessary conditions for finite groups

If the coecients of (1.1) are in C (x), we can use Theorem 3.5 of Singer and Ulmer (1994) to derive necessary conditions for G (L) to be a given nite subgroup of SL(p; C ). The goal of those necessary conditions is to exclude, with little computation, a group from the list of possible dierential Galois groups of a given equation L(y). Such conditions are given in Singer and Ulmer (1994) for the nite primitive groups and will be extended now to the imprimitive nite groups N2 = Q8, N3 and Z=2Zo N3 : 1 Let L(y) be an irreducible second order linear dierential equation with coecients in C (x) whose dierential Galois group is the quaternion group Q8. Then L(y) = 0 must be a dierential equation of fuchsian type having at any singularity 2 distinct rational exponents e1 ; e2 such that: (a) each ei is of the form ai =mi , with (ai ; mi ) = 1, ai ; mi 2 Zand lcm(m1 ; m2) 2 f1; 2; 4g, and (b) there exist non-negative integers n1; n2, such that n1 +n2 = 2 and 2(n1e1 +n2 e2 ) 2 Z. We also have the condition that L s 4(y) = 0 has a two dimensional solution space of rational solutions (i.e. solutions in k).

Lemma 5.2.

Linear Dierential Equations of Prime Order

15

2 Let L(y) be an irreducible third order linear dierential equation with coecients in C (x) whose dierential Galois group is isomorphic to N3 or Z=2Zo N3. Then L(y) = 0 must be a dierential equation of fuchsian type having at any singularity 3 distinct rational exponents e1 , e2 , e3 such that: (a) If G (L)  = N3 , then i each ei is of the form ai=mi , with (ai ; mi ) = 1, ai; mi 2 Zand lcm(m1 ; m2; m3 ) 2 f1; 3g and P ii therePexist non-negative integers n1 ; n2; n3, such that 3i=1 ni = 3 and such that 3i=1 ni ei 2 Z. (b) If G (L)  = (Z=2Zo N3 ), then i each ei is of the form ai=mi , with (ai ; mi ) = 1, ai; mi 2 Zand lcm(m1 ; m2; m3 ) 2 f1; 2; 3; 6g, and P3 n = 2 and such ii there exist non-negative integers n 1 ; n2 ; n3 , such that i=1 i P that 2( 3i=1 ni ei ) 2 Z. We also have the condition that L s 3 (y) = 0 has a non-trivial solution in k for G (L)  = N3 and no non-trivial solution in k for G (L)  = (Z=2Zo N3 ). Proof. The proof follows from Theorem 3.5 of Singer and Ulmer (1994) and is similar to

the proof of the Necessary Conditions 3 in Singer and Ulmer (1994).

1 The representation of Q8 is given in Section 5.1.1. To conclude as in Singer and Ulmer (1994) we need that the elements of Q8 are of order 1, 2 or 4 and that y1 y2 is a semiinvariant of order 2 ( i.e. (y1 y2 )2 is an invariant). Also the two invariants (y1 y2 )2 and y14 + y24 generate a two dimensional space of solutions in k of L s 4(y) = 0 (according to Lemma 3.5 of Singer and Ulmer (1993) the homomorphic image of those invariants is never be 0). 2 We consider the representation of N3 and Z=2Zo N3 given in Section 5.1.2 (to which the other representations are conjugate under ! 7! !2 ). The rst assertion follows from the fact that the order of the elements of N3 is in f1; 3g and that those of Z=2Zo N3 is in f1; 2; 3; 6g. The second assertion follows from the fact that y1 y2 y3 is an invariant of N3 and a semi-invariant of order 2 of Z=2Zo N3 . If G (L)  = N3 , then y1 y2 y3 is a non-zero solution in k of L s 3(y) = 0. Since Z=2Zo N3 has no invariant of degree 3, L s 3 (y) = 0 cannot have a solution in k if G (L)  = Z=2Zo N3 .

2 Example. - We consider the following dierential equationy :

2 3 ? 27x + 27 d y ? 64x3 ? 81x2 + 135x ? 54 y = 0 L(y) = ddxy3 + 32x 36x2(x ? 1)2 dx 72x3(x ? 1)3

y This equation is the second symmetric power of a second order linear dierential equation whose dierential Galois group is the tetrahedral group (cf. Singer and Ulmer (1993), Section 5, p. 31). The dierential Galois group G (K=k) of L(y) = 0 is thus isomorphic to A4 , the alternating group of 4 elements

16

Felix Ulmer

and want to use necessary conditions on the exponents to show, with little computation, that the unimodular group G (L) is an imprimitive group which is not isomorphic to N3 or Z=2Zo N3 and thus that PR has exactly one element whose coecients belong to Q(x) (Corollary 3.3) The equation is of fuchsian type (cf. Singer and Ulmer (1994)) and has 3 regular singular points at x = 0, x = 1 and x = 1. The exponents at those singularities are: f1; 21 ; 32 g, f1; 23 ; 43 g and f?1; ? 32 ; ? 34 g. 1 The equation is irreducible, since it is not possible to nd an exponent P ai at each nite singular point such that for some exponent e1 at 1, the sum ( i ai ) + e1 is a non-positive integer (cf. Corollary 3.3 of Singer and Ulmer (1994)). 2 The group G (K=k) is an imprimitive subgroup of SL(3; C ), since x2(x ? 1)2 is a solution of L s 3 (y) = 0 (cf. Singer and Ulmer (1993)), Theorem 4.6). 3 We now use Lemma 5.2: G (K=k) cannot be N3 , since we have exponents, e.g. 21 , whose denominators do not divide any element of f1; 3g. The group G (K=k) cannot be Z=2Zo N3 , since L s 3 (y) = 0 has a solution in k = Q(x), e.g. x2(x ? 1)2.

6. Conclusion For an irreducible linear dierential equation L(y) = 0 of prime degree, one can always reduce the computation of a Liouvillian solution to one of the following: 1 The computation of a solution of L(y) = 0 where G (K=k) belongs to a nite set of nite groups. 2 The computation of the unique minimal polynomial P(Y ) in PR whose coecients, if the coecients of L(y) = 0 are in Q()(x)  Q(x), are also in Q()(x). It is likely that this result simpli es the computation of P(Y ) in the second case, but no result in this direction is curently known to the author.

Acknowledgement:I would like to thank M.F. Singer for many useful comments, B. Sturm-

fels for his invitation to Cornell and his support during my visit, W. Lempken for pointing out that the complement of Np in Theorem 2.1 must be cyclic, J.A. Weil for computing the 3 elements of PR given in Section 5.2 and the referees for helpful suggestions.

References Blichfeld, H.F. (1917). Finite Collineation Groups . Chicago: The University of Chicago Press. Bronstein, M. (1992). On solutions of linear dierential equation in their coecient eld. J. Symb. Comp. 13. Dixon, J. D. (1971). The Structure of Linear Groups . London: Van Nostrand Reinhold. Hendriks, P.A., Van der Put, M. (1993). A rationality result for Kovacic's algorithm. Proceedings of the 1993 International Symposium on Symbolic and Algebraic Computation, ACM Press . Hendriks, P.A., Van der Put, M. (1993b). Galois action on solutions of a dierential equation. Preprint University Groningen . Huppert, B. (1983). Endliche Gruppen I . Grundlehren der mathematischen Wissenschaften Band 134, Berlin: Springer-Verlag. Issacs, M. (1976). Character Theory of Finite Groups . New York: Academic Press. Kaplansky, I. (1957). Introduction to dierential algebra . Paris: Hermann.

Linear Dierential Equations of Prime Order

17

Kovacic, J. (1986). An algorithmfor solving second order linear homogeneous dierentialequations. J. Symb. Comp. 2. Singer, M.F. (1980). Algebraic solutions of n-th order linear dierential equations. Proceedings of the 1979 Queens Conference on Number Theory, Queens Papers in Pure and Applied Mathematics 54. Singer, M.F. (1981). Liouvillian Solutions of nth Order Linear Dierential Equations. Am. J. Math. 103. Singer, M.F., Ulmer, F. (1993). Galois groups of second and third order linear dierential equations. J. Symb. Comp. 16. Singer, M.F., Ulmer, F. (1993b). Liouvillian and algebraic solutions of second and third order linear dierential equations. J. Symb. Comp. 16. Singer, M.F., Ulmer, F. (1994). Necessary Conditions for Liouvillian Solutions of (Third Order) Linear Dierential Equations. J. Appl. Alg. in Eng., Comm. and Comp. 6. Sturmfels, B. ( 1993). Algorithms in Invariant Theory . Texts and Monographs in Symbolic Computation, Wien: Springer-Verlag. Ulmer, F. (1992). On Liouvillian solutions of dierentialequations. J. Appl. Alg. in Eng., Comm. and Comp. 2. Ulmer, F., Weil, J.A. (1994). Note on Kovacic's algorithm. Prepublication 94-13, IRMAR, Universite de Rennes 1 . Zharkov, A. (1993). On algebraic solutions of rst order Riccati equation. Proceedings of the 1993 International Symposium on Symbolic and Algebraic Computation, ACM Press.