Connected Linear Groups as Differential Galois Groups

JOURNAL OF ALGEBRA ARTICLE NO.

184, 333]361 Ž1996.

0263

Connected Linear Groups as Differential Galois Groups C. Mitschi* Institut de Recherche Mathematique A¨ ancee, ´ ´ Uni¨ ersite´ Louis Pasteur et C.N.R.S., 7 rue Rene´ Descartes, 67084 Strasbourg Cedex, France

and M. F. Singer † Department of Mathematics, North Carolina State Uni¨ ersity, Box 8205, Raleigh, North Carolina 27695-8205 Communicated by Wilberd ¨ an der Kallen Received October 16, 1995

1. INTRODUCTION In this paper we give a proof of the following THEOREM 1.1. Let C be an algebraically closed field of characteristic zero, G a connected linear algebraic group defined o¨ er C, and k a differential field containing C as its field of constants and of finite, nonzero transcendence degree o¨ er C; then G can be realized as the Galois group of a Picard]Vessiot extension of k. Previous work by Kovacic w17, 18x reduced this problem to the case of powers of a simple connected linear algebraic group. Our contribution is to show that one is able to realize any connected semisimple group as a Galois group and, when k s C Ž x ., x9 s 1, to control the number and types of singularities when one constructs a system Y 9 s AY, A g MnŽ C Ž x .., realizing an arbitrary connected linear algebraic group as its *The first author thanks the Institute for Advanced Study, Princeton, NJ, where this paper was completed. † The second author thanks the Institut de Recherche Mathematique Avancee, ´ ´ Universite´ Louis Pasteur et C.N.R.S., for its hospitality and support during the preparation of this paper. The preparation was also partially supported by NSF Grant CCR-93222422. 333 0021-8693r96 $18.00 Copyright Q 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.

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Galois group. More precisely, we give a constructive, purely algebraic proof of the following. Let G be a connected linear algebraic group and let R u be its unipotent radical and P a Levi factor w27x. The group R urŽ R u , R u ., where Ž R u , R u . is the Žclosed. commutator subgroup, is a commutative unipotent group and so is isomorphic to a vector group C n. The group P acts on R u by conjugation and this action factors to an action on R urŽ R u , R u .. Therefore we may write R urŽ R u , R u . s U1n1 [ ??? [ Usn s , where each Ui is an irreducible P-module. We shall assume that U1 is the trivial one-dimensional P-module Žand so allow the possibility that n1 s 0.. We may write P s T ? H, where T is a torus and H is a semisimple group. Let m i s n i if the action of H on Ui is trivial and let m i s n i q 1 if the action of H on Ui is nontrivial. Let N s 0 if H is trivial and N s 1 if H is nontrivial. We define the defect dŽ G . of G to be the number n1 and the excess eŽ G . of G to be max N, m 2 , . . . , m s 4 . We note that any two Levi factors are conjugate so that these numbers are independent of the choice of P. Furthermore, one can show that dŽ G . is the dimension of R urŽ G, R u . Žsee the Appendix.. THEOREM 1.2. Let G be a connected linear algebraic group defined o¨ er an algebraically closed field C of characteristic zero. Then G is the Galois group of a Picard]Vessiot extension of C Ž x . corresponding to a system of the form Y9 s

ž

A1 x y a1

q...q

A dŽG. x y a dŽG.

/

q A` Y

where A i , i s 1, . . . , dŽ G ., are constant matrices and A` is a matrix with polynomial entries of degree at most eŽ G .. In particular, the only possible singularities of this system are dŽ G. regular singular points in the finite plane and a Ž possibly irregular . singular point at infinity. For example, this result implies that for any connected reductive group G, there exists constant matrices A and B such that G is the Galois group of an equation of the form Y 9 s Ž A q xB .Y. We now give a brief history of work on the inverse problem in differential Galois theory. An early contribution to this problem is due to Bialynicki-Birula w3x who showed that, for any differential field k of characteristic zero with algebraically closed field of constants C, if the transcendence degree of k over C is finite and nonzero then any connected nilpotent group is a Galois group over k. This result was generalized by Kovacic, who showed that the same is true for any connected solvable group. In w17, 18x Kovacic introduced powerful machinery to solve the inverse problem. In particular, he made extensive use of the logarithmic derivative Žsee below. and he developed an inductive technique that gave criteria to lift a

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335

solution of the inverse problem for GrR u to a solution for the full group G. Using this, Kovacic showed that to give a complete solution of the inverse problem, one needed only solve the problem for reductive groups Žnote that GrR u is reductive.. He was able to solve the problem for tori and so could give a solution when GrR u is such a group Ži.e., when G is solvable.. He also reduced the problem for reductive groups to the problem for powers of simple groups. We will use logarithmic derivatives and the inductive technique Žin a very simple explicit form. below. When one considers specific fields, more is known. If K s C x 4w xy1 x, the quotient field of convergent series, Kovacic w17x showed that a necessary and sufficient condition for a connected solvable group G to be a Galois group over K is that the unipotent radical of the center of GrŽ R u , R u . have dimension at most 1, where R u is the unipotent radical of G. Using analytic techniques, Ramis showed that any connected semisimple group is a Galois group over K Žcf. w28x.. Recently, Ramis extended this result to show that a necessary and sufficient condition for a linear algebraic group to be a Galois group over K is that it have a local Galois structure Žcf. w29x., a condition expressed in terms of the Lie algebra of the group. This condition was restated in w26x in more group theoretic terms: a necessary and sufficient condition for a linear algebraic group G to be a Galois group over K is that Ži. GrG 0 is cyclic, Žii. dŽ G 0 . F 1, and Žiii. GrG 0 acts trivally on R urŽ R u , G 0 .. For connected groups this reduces to the condition that dŽ G . F 1. Ramis also shows how solving the inverse problem over K is equivalent to solving the inverse problem on the sphere where the differential equation has a regular singularity at 0 and an arbitrary singularity at `. Tretkoff and Tretkoff w37x have shown that any linear algebraic group is a Galois group over C Ž x . when C s C, the field of complex numbers. Their result depends on the solution of a weak version of the 21st Hilbert Problem.1 For arbitrary C, Singer w34x showed that a class of linear algebraic groups Žincluding all connected groups and large classes of nonconnected linear algebraic groups. are Galois groups over C Ž x .. Singer’s proof used the result of Tretkoff and Tretkoff and a transfer 1

Tretkoff and Tretkoff needed the fact that given a finitely generated group H of matrices, there exists a homomorphism of the fundamental group of the sphere minus a finite set of points onto this group and that this representation is the monodromy representation of a Fuchsian equation. Letting one of the generators be the identity matrix, and mapping the homotopy classes of loops around distinct points to distinct generators, one can use the classical solutions of Plemelj or Birkhoff to produce a system with simple poles Žthey need one of the matrices to be diagonalizable to insure that all poles are simple yet a careful examination of their techniques shows that their results still yield a Fuchsian system, without any hypothesis on the matrices Žcf. w1x.. This is not in conflict with the recent work of Bolibruch w1x.

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principle to go from C to any algebraically closed field of characteristic zero. Recently, Magid w22x claimed to have proved that all connected groups having no subgroups of codimension one must be Galois groups over C Ž x .. In particular, this would have implied that any connected semisimple group not having PSLŽ2, C . as a quotient could be realized as a Galois group over C Ž x .. Regretably, Magid’s proof is flawed.2 Finally, Ramis w29x has shown that any linear algebraic group defined over C is the Galois group of a Picard]Vessiot extension of CŽ x .. Ramis’s proof relies on his solution of the inverse problem for the local field C x 4w xy1 x and a technique for gluing solutions of the local problem to form a solution of the global problem. Ramis is also able to bound the number and type of singularities. Another approach to the inverse problem was given by Goldman and Miller. In w10x, Goldman developed the notion of a generic differential equation with group G analogous to what E. Noether did for algebraic equations. He showed that many groups have such an equation. In his thesis w23x, Miller developed the notion of a differentially Hilbertian differential field and gave a sufficient condition for the generic equation of a group to specialize over such a field to an equation having this group as Galois group. Regrettably, this condition gave a stronger hypothesis than in the analogous theory of algebraic equations. This condition made it difficult to apply the theory and Miller was unable to apply this to any groups that were not already known to be Galois groups. Finally, many groups have been shown to appear as Galois groups for classical families of linear differential equations. The family of generalized hypergeometric equations has been particularly accessible to computation, either by algebraic methods as in Beukers and Heckmann w2x, Katz w14x, and Boussel w6x, or by mixed analytic and algebraic methods as in Duval and Mitschi w8x or Mitschi w24, 25x. These equations in particular provide classical groups and the exceptional group G 2 . Other examples were treated algorithmically, as in Duval and Loday-Richaud w7x or Ulmer and Weil w38x using the Kovacic algorithm for second order equations, or in Singer and Ulmer w35, 36x using a new algorithm for third order equations. The rest of this paper is organized as follows. In Section 2 we recall some of the results of differential Galois theory and Kovacic’s program for solving the inverse problem. In Section 3 we present proofs of the Theorems 1.1 and 1.2. We thank J. Kovacic for pointing out mistakes in a previous attempt at proving Theorem 1.2 and for showing us that Theorem 1.2 implies Theo2 We had previously given a ‘‘proof’’ of theorem 1.2 based on Magid’s ideas and, in particular, on Proposition 7.13 of w22x. Kovacic w20x showed us a counterexample to this proposition invalidating our previous argument.

DIFFERENTIAL GALOIS GROUPS

337

rem 1.1. We also thank Jean-Pierre Ramis for stimulating conversations concerning the inverse problem and for sharing his ideas with us.

2. DIFFERENTIAL GALOIS THEORY AND KOVACIC’S PROGRAM The material of this section comes from w17x and w18x. Let k be a differential field of characteristic zero with algebraically closed field of constants C. Given a connected C-group G we wish to construct Picard]Vessiot extensions of k of the form k Ž g . where g is a point of G. Such a field is called a G-primiti¨ e extension of k. We say that a G-primitive extension is full if g is a k-generic point of G. Let V be a finite dimensional faithful G-module. We denote by G ; glŽ V . the Lie algebra of G ; GLŽ V .. For any field extension K of C we denote by GŽ K . the group of K-points of G and by GK the Lie algebra Lie K Ž G . s Lie C Ž G . mC K of G over K. The following summarizes some of the material found in Chapters I and II of w17x. PROPOSITION 2.1.

Let G, V, and G be as abo¨ e.

1. If k Ž g . is a G-primiti¨ e extension of k, then g 9gy1 g Gk and the Galois group H of k Ž g . o¨ er k is an algebraic subgroup of G. The action of H on k Ž g . is gi¨ en by g ¬ gh for any C-rational point h of H. In particular, k Ž g . is a full G-primiti¨ e extension if and only if its Galois group o¨ er k is G. 2. If A g Gk ; glŽ V . mC k, then there exists a G-primiti¨ e extension k Ž g . of k such that g is a solution of Y 9 s AY. Kovacic refers to the element g 9gy1 as the logarithmic deri¨ ati¨ e of g and denotes it as l d Ž g .. He, in fact, shows that if k is a universal field over Ck , the field of constants of k, and g g GŽ k ., then the map g ¬ l d Ž g . maps GŽ k . onto Gk . We shall only need the fact as stated in Proposition 2.1. Note that l d Ž gh. s l d Ž g . q gl d Ž h. gy1 . We shall need the following corollary of this result in the next section. COROLLARY 2.2. Let k s C Ž x . and G, V, and G be as abo¨ e. Let A g Gk . If k Ž g . is a G-primiti¨ e extension of k ha¨ ing connected Galois group H ; G and g 9 s Ag, then there exists an element ˜ g g GŽ k . such that

˜g w A x s ˜g 9g˜y1 q ˜gAg˜y1 g Hk where H ; G is the Lie algebra of H.

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Proof. Since k is a cohomologically trivial field, one knows w15, Corollary 1, p. 426x that any Picard]Vessiot extension of k with connected Galois group H is an H-primitive extension of k. Therefore, k Ž g . s k Ž h. for some point h of H. The Galois group leaves ˜ g s hgy1 fixed so ˜ g is a y1 C Ž x .-rational point of G. Calculating Ž ˜ gg .9Ž ˜ gg . we have y1 ˜g w A x s ˜g 9g˜y1 q ˜gAg˜y1 s Ž ˜gg . 9 Ž ˜gg . s h9hy1 .

Proposition 2.1.1 implies that h9hy1 g Hk . We now recall Kovacic’s program. Any connected linear algebraic group can be written as a semidirect product R u i P where R u denotes the unipotent radical of G and P is a maximal reductive subgroup of G. This is called a Le¨ i decomposition of G and P is called a Le¨ i factor. It is known that any connected reductive subgroup is contained in a Levi factor and that any two Levi factors are conjugate Žcf. w27x.. Kovacic showed that one can make several reductions of the inverse problem. First, he showed that one can assume that R u is commutative and so must be of the form C m. Since P is reductive and acts on C m , one can further write C m as a sum of irreducible P-modules. Kovacic further reduced the inverse problem to the case where R u is the direct sum of copies of a unique irreducible P-module. Second, Kovacic gave a method for selecting a solution of the inverse problem for P that will lift to a solution of the inverse problem for G s R u i P. In general terms, he showed that a solution of the inverse problem for P will lift provided that an appropriately chosen ‘‘inhomogeneous inverse problem’’ for R u can be solved. Finally, Kovacic described the obstructions to solving this inhomogeneous inverse problem for R u . Since we will need to keep a careful accounting of the singularities we introduce in each of these reduction steps in order to prove Theorem 1.2, we shall now describe these reductions in detail. It is convenient to have the following: DEFINITION 2.3. A connected linear algebraic group G defined over an algebraically closed field C of characteristic zero is realizable over a differential field k whose field of constants is C if there exists a full G-primitive extension k Ž g . of k. Let A s l d Ž g . g Gk . We say that A realizes G. Note that Proposition 2.1 implies that given A g Gk there is always a G-primitive extension k Ž g . such that l d Ž g . s A. The above definition of realizability requires that the Galois group be the full group G. 2.1. Reduction to Commutati¨ e R u Let G be a connected linear algebraic group defined over an algebraically closed field C and let G s GrŽ R u , R u ., where Ž R u , R u . denotes

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339

the Žclosed. subgroup of commutators of R u . Let G be the Lie algebra of G, G be the Lie algebra of G, and dp : G ª G be the canonical map. Using the fact that the only algebraic subgroup of G that maps surjectively onto G is G, Kovacic shows Žcf. w17, Lemma 2; 18, Lemma 7, Proposition 18x: PROPOSITION 2.4. Let A1 , . . . , A t g G ; f 1 , . . . , f t g k; and A1 , . . . , A t g G satisfy dp Ž A i . s A i . If Ý A i m f i realizes G o¨ er k, then Ý A i m f i realizes G o¨ er k. 2.2. Lifting Solutions of the In¨ erse Problem We shall assume that G is a connected linear algebraic group defined over C and that the unipotent radical of G, which we denote here by U, is abelian. For any Levi decompositions G s U i P we will describe Kovacic’s method for lifting a solution of the inverse problem for the reductive group P to a solution of the inverse problem for G. Let us first consider what happens when we have a full G-primitive extension K s k Ž g . of a differential field k Žwith algebraically closed field of constants .. We may write g s up with u g U, p g P. Since U is normal in G and k Ž GrU . s k Ž P . is the fixed field of U, we can use Galois correspondence and Proposition 2.1 to show that k Ž p . is a full P-primitive extension of k. Let G , U , P be the Lie algebras of G, U, P respectively. Note that l d Ž g . s A G g Gk and that l d Ž p . s A P g Pk . Calculating we find l d Ž g . s Ž up . 9 Ž up .

y1

s Ž p ? py1 up . 9 Ž p ? py1 up .

y1

s p9 py1 q p Ž l d Ž py1 up . . py1 . Kovacic shows that if ˜ p is an element of P such that k Ž ˜ p . is a P-primitive extension of k with l d Ž ˜ p . s l d Ž p ., then ˜ pŽ l d Ž ˜ py1 up ˜.. ˜py1 s pŽ l d Ž py1 up.. py1 . We therefore define l A pd Ž u . s p Ž l d Ž py1 up . . py1 where p is any element of P such that l d Ž p . s A P and the constants of k Ž p . are the same as the constants of k. Note that U is left fixed by any automorphism of G so l A P d Ž u. g U . The following result of Kovacic shows that one can reverse this process Žcf. w17, Proposition 13; 18, Proposition 19x.. PROPOSITION 2.5. Let G be a connected linear algebraic group defined o¨ er an algebraically closed field C and let k be a differential field with field of

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constants C. Assume that the unipotent radical U of G is abelian and let G s U i P be a Le¨ i factor decomposition. Let A P g Pk realize P and AU g Uk be such that 1. there exists an element p g P with l d Ž p . s A P such that k Ž p . is a full P-primiti¨ e extension of k, 2. there exists an element u g U with l A pd Ž u. s AU , such that the field of constants of k Ž u, p . is C, 3. the map s ¬ py1 uy1s Ž u. p is a C-isomorphism from the differential Galois group GalŽ k Ž up.rk Ž p .. onto UŽ C .. Then l d Ž up. s AU q A P and k Ž up. is a full G-primiti¨ e extension of k. The condition l A P d Ž u. s AU can be described in a simple way. To do this note that U being a commutative unipotent group is the isomorphic image, via the exponential map, of the vector group of its Lie algebra U Žcf. w4, 7.3x. and it is easy to show that Žexp g .9 s g 9 exp g for any g g Uk . This implies that l d Žexp g . s g 9 and if we identify U with the vector group of U via the exponential map, we have that l d Ž g . s g 9 for all g g UŽ k .. Via the identification of U with U the action of P on U by conjugation induces the Žadjoint. representation r : P ª GLŽU . and the corresponding representation d r : P ª glŽU . on the Lie algebras. For AU , A P , u, p as in Proposition 2.5, we have l A pd Ž u . s p Ž l d Ž py1 up . . py1 s r Ž p . ? Ž r Ž py1 . ? u . 9 s u9 y l d Ž r Ž p . . ? u s u9 y d r Ž A P . ? u. EXAMPLE 1. Let k s C Ž x . where x9 s 1 and C is algebraically closed. Let G s C i C* s

½ž

b

a

0

by1

5

/

b / 0, a g C .

We identify the Lie algebra U of C with

½ž

0 0

a 0

/

agC

5

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and the Lie algebra P of C* with b 0

½ž

0 yb

/

5

bgC .

We therefore seek elements AU s Ž 00 0a . and A P s Ž 0b such that:

0 yb

., a, b g C Ž x .

0 . 1. There is a full C*-primitive extension k Ž p ., p s Ž b0 by1 where b 0 l d Ž p . s Ž 0 yb .. This is equivalent to demanding that the Galois group of k Ž b . over k is C* where b 9 s bb .

2. There exists an element u s Ž 10 a1 . with IA P d Ž u. s AU , such that the field of constants of k Ž u, p . is C. Note that l A pd Ž u . s

s

s

ž ž ž

b

0

0

by1

b

0

0

by1

b

0

0

by1

b 2 Ž by2a . 9 0

a 9 y 2 ba . 0

s

ž

0 0

s

ž

0 0

/ žž / žž /ž ld

ld

y1

b

0

0

by1

1 0

by2a 1

0 0

Ž by2a 0

/

ž

//ž . /ž 9

1 0

a 1

/ž 0

0

by1 0

0

by1

0

0

by1

// ž

b

0

0

by1

y1

/

y1

b

b

b

/ y1

/

/

/

Therefore the condition l A P d Ž u. s AU is equivalent to a 9 y 2 ba s a. 3. The map s ª Ž b0

0 y1 1 a y1 b y1 0 1

.

Ž

. s ŽŽ 10

a 1

..Ž b0

0 by 1

. is a C-isomorphism

0 .. from GalŽ k ŽŽ 10 a1 .Ž b0 by1 rk Ž b .. onto UŽ C .. This latter condition is equivalent to the condition that the map s ¬ s Ž a . y a is a C-isomorphism of GalŽ k Ž ab .rk Ž b .. onto C.

Therefore to realize G as a Galois group over C Ž x ., we must find a, b g C Ž x . and Ž a , b . such that

b 9 y bb s 0 a 9 y 2 ba s a

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and such that k Ž b . is a Picard]Vessiot extension of k with Galois group C* and k Ž ab . is a Picard]Vessiot extension of k Ž b . with Galois group C. 2 2 2 If we let b s x and a s 1 then b s e x r2 and a s e x H eyx . We will show in the next section that this selection meets our needs and discuss how this choice was made. Note that in this example the representation r is given by r Ž b . s b 2 . Kovacic is able to refine Proposition 2.5 in the following way. Since P is reductive we may write U as a sum of irreducible P-modules. Grouping isomorphic copies, we write U s U1r 1 [ ??? [ Usr s , where the Ui are nonisomorphic P-modules. Let r i : P ª GLŽUi . be the representation of P on each simple module and d r i : P ª glŽUi . be the representations on their Lie algebras. We denote by r ir i and d r ir i the representations on their powers. As before, we shall identify each Ui and its Lie algebra Ui with some C n i . Kovacic shows w18, Proposition 19x: PROPOSITION 2.6. Let G be a connected linear algebraic group defined o¨ er an algebraically closed field C and let k be a differential field with constants C. Assume that G s ŽU1r 1 [ ??? [ Usr s . i P as abo¨ e. Let A P g Pk and A i g Ž Ui r i . k be such that 1. there exists an element p g P with l d Ž p . s A P such that k Ž p . is a full P-primiti¨ e extension of k, 2. there exists an element u i g Ui r i with uXi y d r ir i Ž A P . u i s A i , such that the field of constants of k Ž u i , p . is C, 3. the map s ¬ r ir i Ž py1 . ? Ž s Ž u i . y u i . is a C-isomorphism from GalŽ k Ž u i p .rk Ž p .. onto Ui r i Ž C ... Then for u s u1 q ??? qu s , l d Ž up. s A1 q ??? qA s q A P and k Ž up. is a full G-primiti¨ e extension. 2.3. Selecting AU and A P Let G s U i P as above. We shall first show that the inverse problem can be reduced to the case where P is a direct product T = H where T is a torus and H is a semisimple group. We note that any connected reductive linear algebraic group P is of the form T ? H where T is a torus and coincides with the center of P, H is semisimple, and T l H is finite. We therefore have a homomorphism p : T = H ª P with finite kernel. Let T , H , P be the Lie algebras of T, H, P respectively and let dp : T = H ª P be the Lie algebra homomorphism associated to p . PROPOSITION 2.7. If AU q AT , H realizes a semidirect product U i ŽT = T . where AU g Uk , AT , H g Ž T [ H . k , then AU q dp Ž AT , H . realizes U i P o¨ er k.

DIFFERENTIAL GALOIS GROUPS

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The proof of this proposition is contained in the proof of Proposition 9 in w17x Žnote that the condition concerning the kernel of p in this latter proposition is not needed in the part of the proof that verifies the above proposition.. This proposition allows us to assume that our group is of the form U i ŽT = H .. The next proposition shows that to realize T = H, we need only realize T and H separately Žcf. w17, Proposition 12x.. PROPOSITION 2.8. Let G1 and G 2 be connected groups such that the only common homomorphic image of both groups is the tri¨ ial group. If A1 and A 2 realize G1 and G 2 respecti¨ ely, then A1 q A 2 realizes G1 = G 2 Ž here we identify the Lie algebra of G1 = G 2 with the direct sum of the Lie algebras of G1 and G 2 .. COROLLARY 2.9. With notation as abo¨ e, if A1 realizes T and A 2 realizes H, then A1 q A 2 realizes T = H. Proof. T and H have no common homomorphic image other than the trivial group. We do not know a general criterion for realizing a semisimple group over an arbitrary differential field, but in the next section we will show how to realize such a group over C Ž x .. The following is a criterion for realizing tori over any differential field Žcf. w17, Proposition 15x.. It is a consequence of the Kolchin]Ostrowski theorem Žcf. w16x.. If T is a torus defined over C, we identify T Ž C . with the m-fold product C* = ??? = C* and the Lie algebra TC of T Ž C . with the m-fold sum C [ ??? [ C. With this identification, the logarithmic derivative of an element in T becomes X l d Ž a 1 , . . . , a m . s Ž a 1X ra 1 , . . . , a m ra m .. PROPOSITION 2.10. Let T be as abo¨ e and F a differential field containing C. Let Ž a1 , . . . , a m . g TF s F m . A necessary and sufficient condition that Ž a1 , . . . , a m . realize T o¨ er F is that there exists no relation of the form n1 a1 q ??? qn m a m s f 9rf with n i g Z not all zero and f g F. Finally, Kovacic gives a criterion for finding elements satisfying Proposition 2.6. Let L A P , r : k m ª k m be the map defined by L A p , r Ž ¨ . s ¨ 9 y d r Ž A P . ? ¨ and let p : k m ª k m rL A p , r Ž k m . be the quotient homomorphism of C-vector spaces. Kovacic shows Žcf. w18, Proposition 20x.: PROPOSITION 2.11. With notations as in Proposition 2.6, assume that s s 1. If A p g Pk satisfies condition 1 abo¨ e, then A1 s Ž a1 , . . . , a r . g U1r satisfies conditions 2 and 3 if and only if p a1 , . . . , p a r are linearly independent o¨ er C. EXAMPLE 1 Žbis.. Propositions 2.10 and 2.11 allow us to verify the claims at the end of the exposition of Example 1. Using the notation of that example, we apply Proposition 2.10 to b s x. Let nx s f 9rf for some

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f g C Ž x .. For any finite zero or pole g of f, f 9rf will have a simple pole at g . Since nx has no poles, we must have that f is a constant and so n s 0. Therefore x realizes the torus C*. To use Proposition 2.11 we must show that a 9 y 2 x a s 1 has no rational solutions a g C Ž x .. Let a g C Ž x . be a solution. At any finite pole of a , the order of a 9 is larger than the order of 2 x a , so there can be no cancellation. Therefore a must be a polynomial. Let n be the degree of this polynomial. The degree of 2 x a is larger than the degree of a so the degree of the right-hand side of this equation is larger than the degree of the left hand side, a contradiction.

3. PROOFS OF THE THEOREMS 3.1. Reducti¨ e Groups We start by showing that any connected semisimple group can be realized over C Ž x . by a system Y9 s AY where the entries of A are polynomials of degree at most one. LEMMA 3.1. Let G be a connected semisimple linear group defined o¨ er C. There exists a faithful finite-dimensional G-module V such that 1. V contains no one-dimensional G-modules. 2. Any proper connected closed subgroup H ; G lea¨ es a one-dimensional subspace WH ; V in¨ ariant. Proof. Recall that Chevalley’s theorem Žcf. w12, p. 80x. states that for any proper algebraic subgroup H ; G there is a G-module V such that H is the stabilizer of a line L H in V. Since G is semisimple, we can write V as the sum of irreducible G-modules. The projection of L H into one of these irreducible components Žof dimension necessarily greater than 1. must be non-trivial, so we can assume that H stabilizes a line in some irreducible G-module V. Note that any subgroup conjugate to H will also stabilize some line in V. Dynkin’s theorem w9x implies that there are only a finite number of conjugacy classes of maximal proper connected closed subgroups in G. Selecting a Vi as above for each of these we can now let V s V1 [ ??? [ Vm . If need be, we can take the direct sum of this with a faithful irreducible G-module of dimension greater than 1 to assume that V is faithful. We will refer to a faithful G-module satisfying the conclusions of the above lemma as a Che¨ alley module for G. For example, if G s SLŽ2., then any irreducible G-module is a Chevalley module for G since any proper connected subgroup is solvable and will leave some line invariant.

DIFFERENTIAL GALOIS GROUPS

345

LEMMA 3.2. Let G be a connected semisimple linear group o¨ er C with Lie algebra G and let V be a Che¨ alley module for G. Let A g G mC C w x x. If C Ž x .Ž g . is a G-primiti¨ e extension of C Ž x . with g 9 s Ag whose Galois group is a proper subgroup of G then there exist ¨ g V, ¨ / 0, c g C Ž x ., and ˜g g GŽ C Ž x .. such that

˜g w A x ? ¨ s c ? ¨ . Proof. Note that since A has polynomial entries the monodromy group is trivial. Gabber’s Lemma Žcf. w14, Proposition 1.2.5x. implies that the Galois group is connected.3 Corollary 2.2 implies that there exists a ˜ ggG such that ˜ g w A x g Hk , where H is the Lie algebra of H. Since V is a Chevalley module for G, there exists a line left invariant by H and therefore by H . The conclusion now follows. LEMMA 3.3. Let G be a connected semisimple linear group o¨ er C with Lie algebra G and let V be a Che¨ alley module for G. Let A g G mC C w x x and let t be the maximum degree of the polynomials appearing in A. If C Ž x .Ž g . is a G-primiti¨ e extension of C Ž x . with g 9 s Ag whose Galois group is a proper subgroup of G then there exists w g V mC C w x x, w / 0, and c g C w x x with deg x c F t such that w9 y Ž A y cI . w s 0. Proof. By Lemma 3.2 we know that there exists ¨ g V, ¨ / 0, c g C Ž x ., and ˜ g g GŽ C Ž x .. such that Ž ˜ g 9g˜y1 q ˜ gAg˜y1 . ¨ s c¨ or, equivalently,

Ž ˜g 9g˜y1 q Ž ˜gAg˜y1 y cI . . ¨ s 0. If we differentiate ˜˜ ggy1 s I we get that ˜ g 9g˜y1 s yg˜Ž ˜ gy1 .9. Therefore we have

Ž yg˜Ž ˜gy1 . 9 q Ž ˜gAg˜y1 y cI . . ¨ s 0. If we let w s yg˜y1 ¨ we get

Ž ˜gw9 y Ž ˜gA y cIg˜. . w s 0. Multiplying through by ˜ gy1 gives us an equation of the form w9 y Ž A y cI . w s 0. 3

Strictly speaking this is not an algebraic argument but it can be replaced by a tedious algebraic proof.

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We now must show that there exists an equation of this form with w g V mC C w x x, w / 0, and c g C w x x with deg x c F t. We first will show that we can assume that c is a polynomial. Let a be a finite pole of c and write n

c s c n Ž x y a . q c nq1 Ž x y a . m

nq1

w s wm Ž x y a . q wmq1 Ž x y a .

q...,

mq1

q...,

where n F y1, wi g V, and c i g C. Substituting these expressions into w9 y Ž A y cI . w s 0 and comparing leading terms, we see that n s y1 and c n s ym. Therefore we have that yma cs Ý qp xya where p is a polynomial. Letting w1 s w ?

ŁŽ xya.

ym a

we have wX1 y Ž A ypI . w 1 s wX ? Ł Ž x ya .

ym a

q w ? Ł Ž z ya .

žÝ

yma xya

/

ym a

ŁŽ xya. ym s Ž w9 y Ž A y cI . w . Ł Ž x y a . y Ž A y pI . w ?

ym a

a

s 0. Therefore we may replace w by w 1 and assume that c is a polynomial. Again by comparing leading terms, we see that the entries of w have no finite poles. Let c s c 0 q c1 x q ??? qc m x m and w s w 0 q w 1 x q ??? qwn x n. If m ) t then the term of highest power in x appearing in w9 y Ž A y cI . w is c m wn x mq n, which is not zero. Therefore, m F t. When t F 1, we can state the conclusion of this result even more concretely. Let A s A 0 q A1 x and c s c 0 q c1 x be as above. Any prospective solution w of w9 y Ž A y cI . w s 0 is of the form w s wm x m q ??? qw 0 where the wi are in V. Substituting into w9 y Ž A y cI . w s 0 we have 0 s w9 y Ž A y cI . w s mwm x my 1 q Ž m y 1 . wmy1 x my 2 q . . . qw1 y Ž A 0 q xA1 y Ž c 0 q c1 x . I . Ž wm x m q . . . qw 0 . s Ž c1 I y A1 . wm x mq 1 q Ž c 0 I y A 0 . wm q Ž c1 I y A1 . wmy1 x m q Ž c 0 I y A 0 . wmy1 q Ž c1 I y A1 . wmy2 q mwm x my 1

DIFFERENTIAL GALOIS GROUPS

347

.. . q Ž c 0 I y A 0 . w 1 q Ž c 1 I y A1 . w 0 q 2 w 2 x q Ž c0 I y A 0 . w 0 q w1 . Therefore Lemma 3.3 states that if the Galois group is a proper connected subgroup of G then there exists a nonnegative integer m and constants c0 , c1 such that the above system has a solution Ž wm , . . . , w 0 . in V mq 1 with wm / 0. EXAMPLE 2. Consider the usual representation of SLŽ2. in GLŽ2.. We have already noted that this makes C 2 into a Chevalley module for SLŽ2.. Let A1 s

ž

A0 s

1 0

ž

0 y1

0 1

/

1 . 0

/

For a given m let us consider the coefficients of x mq 1, x m , and x my 1 in the above system. The coefficient of x mq 1 must vanish so c1 is an eigenvalue and wm is an eigenvector of A1. Let us assume that c1 s 1 Žthe proof when c1 s y1 is similar.. Since we can multiply w by any nonzero element of C, we can furthermore assume that wm s Ž1, 0.T . Let wmy1 s Ž u, ¨ .T . Setting the coefficient of x m equal to zero, we have

ž

c0 y1

y1 c0

/ž

1 0 q 0 0

/ ž

0 2

u q 0 . 0

/ž / ž / ¨

Equating entries, we get c 0 s 0 and ¨ s 12 . Let wmy2 s Ž y, z .T and, setting the coefficient of x my 1 equal to zero, we have

ž

0 y1

y1 0

/ž

u 1 2

q

0 0

0 2

1 0 y qm s . z 0 0

/ ž /ž / ž / ž /

This implies that m s 21 , contradicting the fact that m is an integer. Therefore we can conclude that the Galois group of the differential system Y9 s is SLŽ2..

ž

x 1

1 Y yx

/

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MITSCHI AND SINGER

Note that in the above example A1 is a regular element of a Cartan subalgebra of SLŽ2. and A 0 is the sum of generators of the root subspaces of SLŽ2. with respect to this Cartan algebra. We shall now show how one can generalize this example to deal with any connected semisimple linear group. Let G be a semisimple Lie algebra and let GsH[

@ Ga

where H is a Cartan subalgebra and Ga are the root spaces. Let V be a G-module and let s

Vs

[V is1

bi

where the bi are distinct weights of H on V and the Vb i are the corresponding weight spaces. Let  ¨ 1 , . . . , ¨ r 14 be a basis of Vb 1, and w ¨ r q1 , . . . , ¨ r 4 be a basis of Vb for i s 2, . . . , s Žso dimŽ Vb . s ri y riy1 iy 1 i i i where r 0 s 0.. We then have that  ¨ i 41 F i F dimŽV . is a basis of V. LEMMA 3.4.

Let a be a root of G and Xa g Ga , Xa / 0.

1. If b is a weight of H on V and ¨b is an element of Vb then Xa Ž ¨b . g Vaq b . 2. The matrix Ž x i j . of Xa with respect to  ¨ i 4 has the property that x i, j s 0 for each l, 0 F l F s, r l F i F r lq1 , and r l F j F r lq1 , that is, Ž x i j . has s blocks of zeroes along the diagonal. Proof. Ž1. We note that for any A g H we have a Ž A. Xa s w A, Xa x. Therefore a Ž A . Xa Ž ¨b . s AXa Ž ¨b . y Xa AŽ ¨b . s AXa Ž ¨b . y b Ž A . Xa Ž ¨b .. This implies that AXa Ž ¨b . s Ž a Ž A. q b Ž A.. Xa Ž ¨b .. Ž2. We know by Ž1. that Xa Ž ¨ i, j . lies in Vaq b for every 1 F l F s, l r l F i, j F r lq1. Since a is a root, Vaq b l / Vb l . Therefore, the Vb l-component of Xa Ž ¨ i, j . must be 0. We now select two special generators of G . Let A 0 s Ý Xa where we sum over all roots of G and each Xa is a nonzero element of G0 . Let V be a G-module. Lemma 3.4 implies that the matrix of A 0 with respect to the basis  ¨ i 4 of V, as before, has s blocks of zeroes along the diagonal.

349

DIFFERENTIAL GALOIS GROUPS

In glŽ V .

¡ A0 s

0 .. . 0

... .. . ...

0 .. . 0

a1 , r 1q1 .. . a r 1 , r 1q1

... .. . ...

... .. . ...

... .. . ...

... .. . ...

a r 1q1 , 1 .. . ar 2 , 1 .. . an , 1

??? .. . ... .. . ...

a r 1q1 , r 1 .. . ar 2 , r1 .. . ...

0 .. . 0 .. . ...

??? .. . ... .. . ...

0 .. . 0 .. . ...

a r 1q1 , r 2q1 .. . a r 2 , r 2q1 .. . ...

??? .. . ... .. . ...

¢

a1 , n .. . ar1 , n

¦

a r 1q1 , n . .. . ar 2 , n .. . 0

§

Let A1 be a regular element of H such that distinct roots take distinct, non-zero values on A1 and furthermore distinct weights Žof H in V . take distinct values on A1. The set of such elements is Zariski dense in H . Note that there are d i g C such that A1 s diagŽ d1 , . . . , d n . where d1 s ??? s d r 1, d r 1q1 s ??? s d r 1qr 2 , etc. We shall refer to such a pair Ž A 0 , A1 . as a regular pair of generators of G . The fact that A 0 and A1 generate G is shown in w5, Chap. 8, Sect. 2, Ex. 8, p. 221x. PROPOSITION 3.5. Let G be a connected semisimple linear algebraic group and G its Lie algebra. Then there exists a faithful G-module V and a regular pair of generators Ž A 0 , A1 . of the Lie algebra G ; glŽ V . such that Y9 s Ž A 0 q A1 x . Y has Galois group G. Proof. Let V be a Chevalley module for G and let Ž A 0 , A1 . be a regular pair of generators of G . We will show that there exist non-zero t g C such that the Galois group of Y9 s Ž tA 0 q A1 x .Y is G. Let A s tA 0 q A1 x, with t / 0. Lemma 3.3 implies that if the Galois group of Y 9 s AY is not G, then there exists x s c 0 q c1 x, c i g C, and w s wm x m q ??? qw 0 where the wi are in V and wm / 0, such that w9 y Ž A y cI . w s 0. Substituting into this equation we have 0 s w9 y Ž A y cI . w s mwm x my 1 q Ž m y 1 . wmy1 x my 2 q ??? qw1 y Ž tA 0 q xA1 y Ž c 0 q c1 x . I . Ž wm x m q ??? qw 0 . s Ž c1 I y A1 . wm x mq 1 q Ž c 0 I y tA 0 . wm q Ž c1 I y A1 . wmy1 x m q Ž c0 I y tA 0 . wmy1 q Ž c1 I y A1 . wmy2 q mwm x my 1

350

MITSCHI AND SINGER

.. . q Ž c 0 I y tA 0 . w 1 q Ž c1 I y A1 . w 0 q 2 w 2 x q Ž c 0 I y tA 0 . w 0 q w 1 . Setting the coefficients of x mq 1 , x m , and x my 1 equal to zero, we have

Ž c1 I y A1 . wm s 0

Ž 1.

Ž c0 I y tA 0 . wm q Ž c1 I y A1 . wmy1 s 0

Ž 2.

Ž c0 I y tA 0 . wmy1 q Ž c1 I y A1 . wmy2 q mwm s 0.

Ž 3.

Equation Ž1. implies that wm is an eigenvector of A1. We will assume that c1 s d1. Since any constant multiple of w is again a solution of w9 y Ž A y cI . w s 0, we can assume that wm s Ž1, 0, . . . , 0.T . Substituting this into Eq. Ž2. and letting A 0 be as above, wmy1 s Ž u1 , . . . , u n .T , and r s r 1 , we have

¡

¦¡¦ 1

c0 0 .. .

0 c0 .. .

... ... .. .

0 0 .. .

yta1, rq1 yta2, rq1 .. .

... ... .. .

yta1 n yta2, n .. .

0 ytarq1, 1 .. .

0 ... .. .

... ... .. .

c0 ... .. .

ytar , rq1 ... .. .

... ... .. .

yta1 n ta rq1, n .. .

...

...

...

...

...

ta n , n

... .. .

0 .. .

0 .. .

... .. .

0 .. .

... ... .. .

0 0 .. .

0 d1 y d rq1 .. .

... ... .. .

0 0 .. .

...

...

...

0

¢ yta ¡0. n1

.. 0 q 0 .. .

¢0

§¢§ ¦¡ u. ¦ 1

s s .. . s s .. . s

.. ur s 0. u rq1 .. . un

§¢ §

d1 y d n

This yields the system of equations: c 0 q 0 u1 0 q 0 u2 .. . 0 q 0 ur ytarq1, 1 q Ž d1 y d rq1 . u rq1 .. . ytan , 1 q Ž d1 y d n . u n

0 .. . .. . .. . .. . 0

0 0 .. . 0 0 .. . 0.

351

DIFFERENTIAL GALOIS GROUPS

This implies that c0 s 0. Since each d1 y d i / 0 for i ) r, we also have that ta i1 ui s d1 y d i for i s r q 1, . . . , n. We now consider Eq. Ž3.

¡

¦¡u ¦

0 0 .. .

0 0 .. .

... 0 .. .

0 ... .. .

yta1, rq1 yta2, rq1 .. .

... ... .. .

yta1 n yta2, n .. .

0 ytarq1, 1 .. .

0 ... .. .

... ... .. .

0 ... .. .

ytar , rq1 ... .. .

... ... .. .

yta1 n ta rq1, n .. .

...

...

...

...

...

ta n , n

... .. .

0 .. .

0 .. .

... .. .

0 .. .

... ... .. .

0 0 .. .

0 d1 y d rq1 .. .

... ... .. .

0 0 .. .

...

...

...

0

¢ yta ¡0. n1

.. 0 q 0 .. .

¢0

§¢ § ¦ ) ¡m¦

d yd§ n

1

1

u2 .. . .. . .. . .. . un

.. . ) q ) .. . )

0

0 .. . .. . .. .

s 0.

¢0§

The first row of this matrix equation is yta1, rq1 u rq1 y ??? yta1 n u n q m s 0. Substituting the values for u i we have yt 2 a1, rq1

ž

a rq1, 1 d1 y d rq1

q ??? qa1 n

a n1 d1 y d n

/

q m s 0.

If a1, rq1

a rq1, 1 d1 y d rq1

q ??? qa1 n

a n1 d1 y d n

s 0,

then we must have that m s 0. This means that w g V. Since 0 s w9 y Ž A y cI . w s Ž cI y tA 0 y A1 x . w, this implies that A 0 and A1 have a common eigenvector and since A 0 and A1 generate G , this contradicts the fact that V is a Chevalley module. Therefore, we have that a1 , rq1

a rq1, 1 d1 y d rq1

q ??? qa1, n

a n1 d1 y d n

/0

352

MITSCHI AND SINGER

and again we get a contradiction if we select t g C such that yt 2 a1, rq1

ž

a rq1, 1 d1 y d rq1

q ??? qa1, n

an , 1 d1 y d n

/

is not an integer. The set of such t in Q, the algebraic closure of the rationals, is a dense open set. For these t the Galois group of Y 9 s Ž tA 0 q A1 x . must be G. The above result gives a very simple system that realizes a connected semisimple group. In order to realize a Levi factor of a connected linear algebraic group in such a way that we may utilize Propositions 2.5 and 2.11, we may need a system with polynomial entries of higher degree. PROPOSITION 3.5. Let G be a connected semisimple linear algebraic group and G its Lie algebra. Let n ) 1 be an integer. Then there exists a faithful G-module V and a regular pair of generators Ž A 0 , A1 . of the Lie algebra G ; glŽ V . such that Y9 s Ž A 0 q A1 x n . Y has Galois group G. Proof. We again let V be a Chevalley module for G and Ž A 0 , A1 . be a regular pair of generators of G . If the Galois group of Y 9 s Ž A 0 q A1 x n .Y is not G, then Lemma 3.3 implies that there exists cŽ x . s c n x n q ??? q c 0 g C w x x and W Ž x . s wm x m q ??? qw 0 g V m C w x 4 such that w9 y Ž A 0 q A1 x n y cI . w s 0. We shall deal with three cases: n ) m: Setting the coefficients of x mq n, . . . , x n equal to zero we have:

Ž c n I y A1 . wm s 0 Ž c n I y A1 . wmy1 q c ny1 Iwm s 0 Ž c n I y A1 . wmy2 q c ny2 Iwmy1 q c ny1 Iwm s 0 .. .

Ž c n I y A . w1 q c ny1 Iw 2 q . . . qc nymq1 Iwm Ž c n I y A1 . w 0 q c ny1 Iw 1 q . . . qc nym Iwm

.. . s s

.. . 0 0.

The first equation implies that c n is an eigenvalue of A1 and wn is an eigenvector of A1. Let W denote the c n-eigenspace of A1. We have that the image of Ž c n I y A1 . intersects W trivially. Therefore the second equation implies that c ny 1 s 0 and that wmy1 g W. Considering each equation in turn, we have that c ny 1 s ??? s c nym s 0 and wm , . . . , w 0 g

353

DIFFERENTIAL GALOIS GROUPS

W. We now set the coefficients of x ny 1 , . . . , x m equal to zero: c ny 1 Iw 0 q . . . qc nymy1 Iwm s 0 .. .. .. . . . c mq 1 Iw 0 q . . . qc1 Iwm s 0 c m Iw 0 q . . . q Ž c 0 I y A 0 . wm s 0. Considering all but the last of these equations in order, we see that c ny my1 s ??? s c1 s 0. The final equation then implies that wm is an eigenvector of A 0 . This implies that A1 and A 0 have a common eigenvector, a contradiction. n - m: Setting the coefficients of x mq n, . . . , x m equal to zero, we have:

Ž c n I y A1 . wm s 0 Ž c n I y A1 . wmy1 q c ny1 Iwm s 0 Ž c n I y A1 . wmy2 q c ny2 Iwmy1 q c ny1 Iwm s 0 .. .

Ž c n I y A1 . wmyn q c ny1 Iwmynq1 q . . . q Ž c0 I y A 0 . wm

.. . s

.. . 0.

The first equation implies that c n is an eigenvalue of A1 and wn is an eigenvector of A1. Let W denote the c n-eigenspace of A1. Arguing as above we see that c ny 1 s ??? s c1 s 0 and wm , . . . , wnym g W. Using the form of A 0 established in Lemma 3.4, the last equation implies that c 0 s 0 as well. Note that unlike the situation in Proposition 3.5, we can conclude at this point that wmy 1 g W. We use here the fact that n ) 1. Setting the coefficient of x my 1 equal to zero we have that mwm q Ž c n I y A1 . wmyny1 q Ž c 0 I y A 0 . wmy1 s 0. Since wmy 1 g W and c 0 s 0, we have that the W-component of Ž c 0 I y A 0 . wmy1 s A 0 wmy1 is 0. Furthermore, the W-component of Ž c n I y A1 . wmyny1 must also be 0. Therefore this latter equation implies that m s 0. This contradicts the fact that m ) n ) 1. m s n: Setting the coefficients of x mq n, . . . , x m equal to zero we have:

Ž c n I y A1 . wm s 0 Ž c n I y A1 . wmy1 q c ny1 Iwm s 0 Ž c n I y A1 . wmy2 q c ny2 Iwmy1 q c ny1 Iwm s 0 .. .

Ž c n I y A1 . w 0 q c ny1 Iw 1 q . . . q Ž c0 I y A 0 . wm

.. . s

.. . 0.

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MITSCHI AND SINGER

As before we can conclude that c ny 1 s ??? s c 0 s 0 and wm , . . . , w 0 g W. Setting the coefficient of x my 1 equal to zero we have that mwm q Ž c 0 I y A 0 . wmy1 s 0. Since c 0 s 0 and the W-component of A 0 wmy1 is 0, we must have that m s 0. This contradicts the fact that m s n ) 1. We now turn to the case of a torus. Let T be a torus of dimension r and T its Lie algebra. We identify T Ž C . with Ž C*. r and TC with C r. PROPOSITION 3.7. Let  c1 , . . . , c r 4 ; C be linearly independent o¨ er Q. For any n G 0, the element Ž c1 x n, . . . , c r x n . g TCŽ x . realizes T o¨ er C Ž x .. Proof. By Proposition 2.10, it is enough to show that if n1 c1 x n q . . . qn r c r x n s f 9rf with n i g Z and f g C Ž x ., then all the n i s 0. If such a relation exists, let f s ŁŽ x y d j . m j , d j g C, m j g Z. We then have that f 9rf s Ý Ž m jrŽ x y d j ... Therefore, each m j s 0. Since the c i are Q-linearly independent, we have that each n i s 0. We now combine Propositions 3.6 and 3.7 with Proposition 2.4, Proposition 2.7, and Corollary 2.9 to yield: PROPOSITION 3.8. Let G be a connected reducti¨ e linear algebraic group. Let m ) 1 be an integer. Then there exists a differential equation of the form Y 9 s Ž A q Bx m . Y , with A, B constant matrices, with Galois group G. For later use, we recall that if G s T = H, where T is a torus and H is a semisimple group, then we can select B s D q E where Ž A, D ., for some A, is a regular pair of generators of the Lie algebra of H and E is a diagonal matrix whose non-zero entries are linearly independent over Q. Note that we can select a basis so that both D and E Žand therefore B . are diagonal. 3.2. Proof of Theorem 1.2 In order to use the lifting results of Kovacic we will need the following technical lemma. LEMMA 3.9. Let A, B be s = s square matrices with entries in an algebraically closed field C. Assume that 1. 2. for any 3.

B is diagonal. If Ž C s . 0 is the eigenspace of B corresponding to eigen¨ alue 0, then w g Ž C s . 0 the Ž C s . 0-component of Aw is zero. A and B share no nonzero eigen¨ ectors.

DIFFERENTIAL GALOIS GROUPS

355

Then for any integer m ) 0 there exists a nonzero subspace V of C s such that if c my 2 , . . . , c 0 g V and Y 9 y Ž A q Bx m . Y s c my 2 x my 2 q ??? qc0

Ž 4.

has a solution T g Ž C Ž x .. s, then c my 2 s ??? s c 0 s 0. Proof. Note that by comparing the orders of poles, one can see that any solution Y of Ž4. must lie in Ž C w x x. s. Now assume that Ž C s . 0 s  04 and let V s C s. Let Y s wn x n q ??? qw 0 , wi g C s, be a solution of Ž4.. Comparing coefficients of x mq n we see that Bwn s 0 so wn s 0. Therefore Y s 0 and so c my 2 s ??? s c 0 s 0. Now assume that Ž C s . 0 /  04 and let V s Ž C s . 0 . Let Y s wn x n q ??? qw 0 , wi g C s, wn / 0 be a solution of Ž4.. We will deal with three cases: m ) n: Comparing coefficients of x nq m in Ž4. we see that Bwn s 0 so wn g Ž C s . 0 . Comparing coefficients of x n in Ž4., we see that Awn s yc n . Since c n g Ž C s . 0 and the Ž C s . 0-component of Awn is 0, we have that c n s 0 so Awn s 0. Therefore wn would be a common eigenvector of A and B, a contradiction. m s n: Comparing coefficients of x mq n s x 2 m in Ž4. we see that Bwm s 0. Similarly, comparing coefficients of x mq ny1 s x 2 my1 in Ž4. we see that Bwmy 1 s 0. Comparing coefficients of x my 1 in Ž4. we see that Awmy 1 q mwm s 0. Since wm g Ž C s . 0 and the Ž C s . 0-component of Awmy1 is 0, we must have m s 0, a contradiction since m ) 0. m - n: Considering the coefficients of x mq n and x mq ny1 as above, we see that wn , wny1 g Ž C s . 0 . Comparing coefficients of x ny 1 , we see that nwn q Bwnymy1 q Awny1 s 0. Since the Ž C s . 0-component of Bwnymy1 and Awny 1 is 0, we have that n s 0, contradicting that n ) m ) 0. We now give the proof of Theorem 1.2. Let G be a connected linear algebraic group defined over C with defect d and excess e. Proposition 2.4 implies that we may assume that R u is commutative Žnote that the defect and excess of the groups G and GrŽ R u , R u . are the same.. Furthermore, Proposition 2.7 implies that we may assume that G s U i ŽT = H . where U is a commutative unipotent group, T is a torus, and H is a semisimple group Žnote that e and d are unchanged under id = p .. We identify the Lie algebra TC of T Ž C . with C l and let Ž c1 , . . . , c l . g TC , where the c i are Q-linearly independent. Proposition 3.7 implies that Ž c1 x e, . . . , c l x e . realizes T over C Ž x .. If Ž A 0 , A1 . is a regular pair of generators of the Lie algebra HC of H Ž C ., then Propositions 3.5 and 3.6 imply that A 0 q x eA1 realize H. Therefore Corollary 2.9 implies that A P s Ž c1 x e, . . . , c l x e . q A 0 q x eA1 realizes T = H. We shall now use Propositions 2.6 and 2.11 to realize G. Let U s U1r 1 [ ??? [ Usr s be the unipotent radical where each Ui

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is an irreducible P s T = H-module and assume that U1 is the one-dimensional trivial module. Fix some i and let U s Ui , r s ri , and r s r i . Proposition 2.11 implies that it is sufficient to show that for the above r Ž . choice of A P , there exists Ž a1 , . . . , a r . g UCŽ x . such that if ¨ 9 y d r A P ? ¨ s a 1 a1 q ??? qa r a r has a solution ¨ g C Ž x ., for some a j g C, then all a j must be zero. Let us first assume that i s 1, i.e., that U is the trivial one-dimensional module and r s d. In this case d r ' 0. Let g 1 , . . . , gd be distinct elements of C. If any ¨ g C Ž x . satisfies ¨9 s

a1 x y g1

q ??? q

ad x y gd

,

then clearly each a j s 0. Therefore A1 s Ž1rŽ x y g 1 ., . . . , 1rŽ x y gd .. satisfies Proposition 2.11. We now consider the case where i ) 1. In this case U is a nontrivial irreducible T = H-module. Since T is diagonalizable and H commutes with T, we have that H preserves each eigenspace of T. Therefore, each element of T acts as a constant matrix on U. This furthermore implies that U is an irreducible H-module. We now consider two subcases. The first is the case when U is the one-dimensional trivial H-module. In this case, d r Ž A 0 . s d r Ž A1 . s 0 and d r Ž c1 x e, . . . , c l x e . s Ž q1 c1 q ??? qql c l . x e for some integers qi , not all zero. Note that our assumptions on the c i imply that this latter expression is nonzero. Let a1 s 1, . . . , a r s x ry1 and note that r y 1 - e. We must show that if ¨ 9 y Ž q1 c1 q ??? qql c l . x e ? ¨ s a 1 q ??? qa r x ry1

has a solution ¨ g C Ž x ., then all the a j s 0. To see this note that a solution ¨ cannot have finite poles and so must be a polynomial. Comparing degrees we get r s 0 so A i s Ž1, . . . , x ry1 . satisfies Proposition 2.11. The second case is the case when the action of H on U is non-trivial. In this case, both d r Ž A 0 . and d r Ž A1 . are non-zero and the matrices A s A 0 and B s d r ŽŽ c1 , . . . , c l .. q d r Ž A. satisfy the hypotheses of Lemma 3.9. Let a1 s 1, . . . , a r s x ry1. Since r y 1 F e y 2, Lemma 3.9 implies that AUi s Ž1, . . . , x ry1 . satisfies Proposition 2.11. In all cases we have introduced no more than d finite simple poles and polynomials of degree at most e. EXAMPLE 3. Consider the group G s Ž C = C . i C*, where the action of G on the first factor of C = C is trivial and the action of c g C* on the

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second factor is given by multiplication by c 2 . In concrete terms this is the group:

¡1

~



0 0

¢0

b1 1 0

0 0 a

0 0 b2

0

0

ay1

0

¦

¥

a, b1 , b 2 g C, a / 0 .

§

The unipotent radical is C = C and d s 1 and e s 1. We can let A1 s 1, A 2 s 1, and A P s x. In terms of matrices, we see that 0 0 Y9 s 0 0



1 0 0 0

0 0 x 0

0 0 Y 1 yx

0

realizes this group. The solution of this equation is log x 1 2 Ys . x 2 r2 e H eyx

 0 eyx

2

r2

We end this section by showing that the bound of dŽ G . q 1 singular points is sharp. We do not know if the bound on the degrees of the entries of A` is sharp. PROPOSITION 3.10. If G s C r , then dŽ G . s r and G cannot be realized by a differential equation o¨ er C Ž x . with fewer than r q 1 singular points. Proof. Assume not and let k s C Ž x . and K be a Picard]Vessiot extension corresponding to the system. We know that K s k Ž G . so we can write k Ž G . s k Ž t 1 , . . . , t r . where the action of the Galois group is given by s Ž t i . s t i q c i g for s g G and c i g g C. This implies that each tXi is left invariant by the Galois group G and so must be in k. Therefore each t i is of the form t i s Ý j c i j logŽ f j . q g i , where f j , g i g k. The assumption on the singularities implies that any element in k Ž G . has singularities Žexcept for poles. in some fixed set  p1 , . . . , pry1 , pr 4 . Making a change of variables we Ž can assume that pr s `. Therefore we may write logŽ f j . s Ý ry1 is1 n i j log x y pi .. This gives us r linear equations in the r y 1 logarithms and so the t i will be algebraically dependent over C Ž x ., a contradiction. 3.3. Proof of Theorem 1.1 Theorem 1.1 follows from the following proposition, whose statement and proof are due to Kovacic w19x. In w18x he attacked the inverse problem

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for arbitrary connected Žnot necessarily linear. algebraic groups. He showed that one could reduce this problem to the inverse problem for connected linear groups and for abelian varieties. Kovacic also showed that if G is an abelian variety and F is a differential field of finite but nonzero transcendence degree over its algebraically closed field of constants, then there exists a strongly normal extension of F whose differential Galois group is G Ž strongly normal . generalizes the notion of Picard]Vessiot, cf. w15, 17, 18x; a strongly normal extension whose Galois group is linear is a Picard]Vessiot extension.. The following proposition is stated and proved in such a way that it holds in this generality but so that readers unfamiliar with strongly normal extensions can restrict themselves to Picard]Vessiot extensions and follow the proof as well. In general, a connected algebraic group is realizable over F if there is a strongly normal extension K of F whose Galois group is isomorphic to the given group. PROPOSITION 3.11. Let H be a connected C-group such that for any natural number n, H n is realizable o¨ er C Ž x .. Then for any differential field F of finite but non-zero transcendence degree o¨ er C, the group H is realizable o¨ er F. Proof. Choose any x in F not in C and note that x is transcendental over C. Let d be the derivation d s Ž dx .y1 d, where d is the derivation on F. Then any d-extension K Ža differential extension of F, with derivation d . is also a d-extension K. We also have that K is strongly normal if and only if K is, and GalŽ KrF . s GalŽ KrF .. Thus we may replace d by d and thereby assume that C Ž x . ; F. Let n be the transcendence degree of F over C. Choose an extension K of C Ž x . that realizes H n Ž K need not contain F .. Since C Ž x . is cohomologically trivial, there exists a s Ž a1 , . . . , a n . g H n such that K s C Ž x .Ž a. and l d a g Lie CŽ x .Ž H n .. Note that trdegŽ C Ž x .Ž a.rC Ž x .. s n dim H. Choose b s Ž b1 , . . . , bn . g H n such that l d b s l d a and such that the field of constants of F Ž b . is C. Then there exists c in H Ž L. n Žwhere L is the field of constants of some universal differential field. with b s ac. Since GalŽ C Ž x .Ž a.rC Ž x .. s H n, there is a differential isomorphism, s , of C Ž x .Ž a. over C Ž x . such that c s ay1s Ž a.. So b s s Ž a.. Thus tr degŽ C Ž x .Ž b .rC Ž x .. s n dim H. Evidently, for each i, F Ž bi . is a strongly normal extension of F. We claim that GalŽ F Ž bi .rF . s H for at least one i. Suppose not. Then tr deg F Ž bi .rF F dim H y 1, for all i s 1, . . . , n. So tr deg F Ž b .rF F n dim H y n and tr deg F Ž b .rC Ž x . F n dim H y 1 Žsince tr deg FrC Ž x . s n y 1.. But C Ž x .Ž b . ; F Ž b . and tr deg C Ž x .Ž b .rC Ž x . s n dim H. This contradiction proves the theorem. COROLLARY 3.12. Let F be of finite, but non-zero, transcendence degree o¨ er C. Let G be any connected C-group. Then G is realizable o¨ er F.

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359

APPENDIX Let G be a connected linear algebraic group and let R u be its unipotent radical and P a Levi factor. We shall verify that dŽ G . is the dimension of R urŽ G, R u .. We shall use the notation of the Introduction and write R urŽ R u , R u . s U1n1 [ ??? [ Usn s where each Ui is an irreducible P-module and U1 is the trivial one-dimensional P-module. Since Ž R u , R u . ; Ž G, R u . we have a canonical surjective homomorphism p : R urŽ R u , R u . ª R urŽ G, R u .. We shall show that the kernel of this homomorphism is U2n 2 [ ??? [ Usn s and so dŽ G . s n1 s dim R urŽ G, R u .. First note that the kernel of p is Ž G, R u .rŽ R u , R u . and that this latter group is Ž P, R u .rŽ R u , R u .. To see this second statement note that for g g G we may write g s pu, p g P, u g R u . For any w g R u , gwgy 1 w y 1 s puwuy 1 py 1 w y 1 s p Ž uwuy 1 . py 1 Ž uw y 1 uy 1 .Ž uwuy 1 w y 1 .. Therefore Ž G, R u .rŽ R u , R u . ; Ž P, R u .rŽ R u , R u .. The reverse inclusion is clear. Next we will show that Ž P, R u .rŽ R u , R u . s U2n 2 [ ??? [ Usn s . The group Ž P, R u .rŽ R u , R u . is the subgroup of U1n1 [ . . . [ Usn s generated by the elements p¨ py1 y ¨ where p g P, ¨ g U1n1 [ ??? [ Usn s . Since the action of P on U1 via conjugation is trivial, we see that any element p¨ py1 y ¨ as above must lie in U2n 2 [ ??? [ Usn s . Furthermore, note that for each i, the image of the map P = Ui ª Ui given by Ž p, u. ¬ pupy1 y u generates a P-invariant subspace of Ui . For i G 2, this image is nontrivial so, since Ui is an irreducible P-module, we have that the image generates all of Ui . This implies that the elements p¨ py1 y ¨ , where p g P, ¨ g U1n1 [ ??? [ Usn s , generate all of U2n 2 [ ??? [ Usn s , and completes the proof.

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8. A. Duval and C. Mitschi, Matrices de Stokes et groupe de Galois des ´ equations hypergeometriques confluentes geneneralisees, ´ ´ ´ ´ ´ Pacific J. Math. 138 Ž1989., 25]56. 9. E. Dynkin, Maximal subgroups of the classical groups. Trudy Mosko¨ . Mat. Obshch. 1, 39]160; Amer. Math. Soc. Transl. Ser. 2 6 Ž1957., 245]378. 10. L. Goldman, Specializations and the Picard]Vessiot theory, Trans. Amer. Math. Soc. 85 Ž1957., 327]356. 11. M. Hall, Jr., ‘‘The Theory of Groups,’’ Macmillan, New York, 1959. 12. J. E. Humphreys, ‘‘Linear Algebraic Groups,’’ 2nd ed., Springer-Verlag, New York, 1981. 13. I. Kaplansky, ‘‘An Introduction to Differential Algebra,’’ 2nd ed., Hermann, Paris, 1976. 14. N. Katz, On the calculation of some differential Galois groups, In¨ ent. Math. 87 Ž1987., 13]61. 15. E. R. Kolchin, ‘‘Differential Algebra and Algebraic Groups,’’ Academic Press, New York, 1973. 16. E. R. Kolchin, Algebraic groups and algebraic dependence, Amer. J. Math. 90 Ž1968., 1151]1164. 17. J. Kovacic, The inverse problem in the Galois theory of differential fields, Ann. of Math. 89 Ž1969., 583]608. 18. J. Kovacic, On the inverse problem in the Galois theory of differential fields, Ann. of Math. 93 Ž1971., 269]284. 19. J. Kovacic, e-mail communication, May 2, 1995. 20. J. Kovacic, e-mail to Magid, Mitschi, Singer, et al., March 2, 1995. 21. A. H. M. Levelt, Differential Galois theory and tensor products, Indag. Math. Ž N.S.. 1 Ž1990., 439]450. 22. A. Magid, ‘‘Lectures on Differential Galois Theory,’’ University Lecture Series, American Mathematical Society, Volume 7, Providence, RI, 1994. 23. J. Miller, ‘‘On Differentially Hilbertian Differential Fields,’’ Ph.D. thesis, Columbia University, 1970. 24. C. Mitschi, Groupe de Galois des ´ equations hypergeometriques confluentes generalisees, ´ ´ ´ ´ ´ C. R. Acad. Sci. Paris 309, No. I Ž1989., 217]220. 25. C. Mitschi, Differential Galois groups of generalized hypergeometric equations: an approach using Stokes multipliers, Pacific J. Math., to appear. 26. C. Mitschi and M. F. Singer, On Ramis’s solution of the local inverse problem of differential Galois theory, J. Pure Appl. Alg. 110 Ž1996., 185]194. 27. G. D. Mostow, Fully reducible subgroups of algebraic groups, Amer. J. Math. 78 Ž1956., 200]221. 28. J. P. Ramis, About the solution of some inverse problems in differential Galois theory by Hamburger equations, in ‘‘Differential Equations, Dynamical Systems, and Control Science,’’ ŽElworthy, Everett, and Lee, Eds.. Lecture Notes in Pure and Applied Mathematics, Vol. 157, pp. 277]300, Springer-Verlag, New YorkrBerlin, 1994. 29. J. P. Ramis, About the inverse problem in differential Galois theory, preprint, 1995. 30. M. Rosenlicht, Integration in finite terms, Am. Math. Monthly ŽNov. 1977., 963]972. 31. J. P. Serre, ‘‘Cohomologie Galoisienne,’’ Springer-Verlag, New York, 1964. 32. M. F. Singer, Algebraic relations among solutions of linear differential equations, Trans. Am. Math. Soc. 295, No. 2 Ž1986., 753]763. 33. M. F. Singer, Algebraic relations among solutions of linear differential equations: Fano’s theorem, Am. J. Math. 110 Ž1988., 115]144. 34. M. F. Singer, Moduli of linear differential equations on the Riemann sphere with fixed Galois groups, Pacific J. Math. 106, No. 2 Ž1993., 343]395. 35. M. F. Singer and F. Ulmer, Liouvillian and algebraic solutions of second and third order linear differential equations, J. Symbolic Comput. 16 Ž1993., 37]74.

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