Algebraic Aspects of Linear Differential and Difference Equations [PhD Thesis ed.]
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Algebraic Aspects of Linear Di erential and Di erence Equations Peter A. Hendriks
University of Groningen Department of Mathematics P.O. Box 800 9700 AV Groningen The Netherlands e{mail [email protected]
Rijksuniversiteit Groningen
Algebraic Aspects of Linear Di erential and Di erence Equations Proefschrift ter verkrijging van het doctoraat in de Wiskunde en Natuurwetenschappen aan de Rijksuniversiteit Groningen op gezag van de Rector Magni cus Dr. F. van der Woude, in het openbaar te verdedigen op vrijdag 29 november 1996 des namiddags te 4.15 uur door
Peter Anne Hendriks geboren op 9 september 1968 te Groningen
Promotor: Prof.dr. M. van der Put Referent: Dr. F. Beukers
Aan mijn ouders
Contents 1 Introduction 2 Galois Action on Solutions of a Di erential Equation
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 The universal eld of k((x)) . . . . . . . . . . . . . . . . 2.3 The k-structure on the space of solutions . . . . . . . . . 2.3.1 Remarks . . . . . . . . . . . . . . . . . . . . . . . 2.4 Rational solutions of Riccati . . . . . . . . . . . . . . . . 2.5 Algebraic solutions of the Riccati equation and examples 2.5.1 Equations of order two . . . . . . . . . . . . . . . 2.5.2 Equations of order three . . . . . . . . . . . . . . 2.5.3 Remarks . . . . . . . . . . . . . . . . . . . . . . . 2.6 Symmetries . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.1 Symmetries of L . . . . . . . . . . . . . . . . . . 2.6.2 Transforming algebraic solutions of Riccati . . . 2.6.3 Forms of a di erential operator . . . . . . . . . . 2.6.4 Construction of special di erential equations . . . 2.6.5 Examples . . . . . . . . . . . . . . . . . . . . . .
3 Shidlovskii irreducibility 3.1 3.2 3.3 3.4 3.5
Introduction . . . . . . . Preliminaries . . . . . . Siegel normality . . . . . Shidlovskii irreducibility Examples . . . . . . . .
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3 7
7 8 12 13 13 14 16 17 19 19 20 22 22 24 24
29
29 30 34 37 43
4 An algorithm determining the di erence Galois group of second order linear di erence equations 49 4.1 4.2 4.3 4.4
Introduction . . . . . . . . . . . . . . . . Preliminaries on di erence Galois theory First order di erence equations . . . . . Second order di erence equations . . . . 4.4.1 The Riccati equation . . . . . . . 1
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49 50 52 53 53
4.4.2 is reducible . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.3 is imprimitive . . . . . . . . . . . . . . . . . . . . . . . 4.4.4 contains (2 Q ) . . . . . . . . . . . . . . . . . . . . . . 4.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5.1 Sequences spaces . . . . . . . . . . . . . . . . . . . . . . . 4.5.2 The case where an extension of the constant eld is needed G
G
G
Sl
;
58 60 62 63 63 70
5 An algorithm for computing a standard form for second order linear -di erence equations 71 5.1 5.2 5.3 5.4
q
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A short introduction to algebraic aspects of -di erence equations. First order -di erence equations . . . . . . . . . . . . . . . . . . Second order -di erence equations . . . . . . . . . . . . . . . . . 5.4.1 The Riccati equation . . . . . . . . . . . . . . . . . . . . . 5.4.2 is reducible . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.3 is imprimitive . . . . . . . . . . . . . . . . . . . . . . . 5.4.4 contains (2 C) . . . . . . . . . . . . . . . . . . . . . . 5.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.1 Sequences spaces . . . . . . . . . . . . . . . . . . . . . . . 5.5.2 Hypergeometric -di erence equations . . . . . . . . . . . q
q
q
G
G
G
Sl
;
q
6 On the classi cation of a class of -di erence equations 6.1 6.2 6.3 6.4
q
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . Picard{Vessiot Theory . . . . . . . . . . . . . . . . . . . Classi cation of -Di erence Modules . . . . . . . . . . . Tannakian Categories and the {Di erence Galois Group q
q
2
. . . .
. . . .
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. . . .
71 71 75 76 76 81 83 84 85 85 87
91
91 91 92 97
Chapter 1 Introduction This thesis is about some algebraic and algorithmic aspects of linear di erential, di erence and -di erence equations over the eld of rational functions C( ). The starting-point for a great part of this thesis was Kovacic's algorithm [Kov86]. Kovacic's algorithm computes a Liouvillian solution of the second order linear di erential equation 00 + 0 + = 0, where 2 C( ), provided a Liouvillian solution exists. We will de ne the notion Liouvillian solution. A di erential eld extension C( ) is called a Liouvillian extension if there is a nite sequence of di erential elds q
z
y
L
C( ) = z
such that
Li
=
Li
1
ay
by
a; b
z
z
L0 L1 Ln
1
Ln
=
L
( ), where one of the following three cases holds: ai
1.
ai
is algebraic over
2.
ai
satis es a di erential equation 0 = , where
3.
ai
satis es a di erential equation 0 = , where
Li
1 y
y
by
b
b 2 Li
b 2 Li
1 1
A solution of a linear di erential equation is called Liouvillian if lies in some Liouvillian extension. Kovacic's algorithm is based on the results of di erential Galois theory. For a short introduction of di erential Galois theory we refer to the sections 2.1 and 3.2 in this thesis. Many interesting properties of linear di erential equations can be characterized in terms of the di erential Galois group and its action on the vector space of solutions. For instance a linear di erential equation is irreducible (that is the corresponding di erential operator does not factor over C( ) ) if and only if the di erential Galois group associated to this equation acts irreducibly on the vector space of solutions. All solutions of a linear di erential equation are algebraic if and only if the di erential Galois group is nite. And all the solutions of a linear f
f
z
G
G
3
di erential equation are Liouvillian if and only if the identity component 0 of the di erential Galois group is solvable. Two other interesting properties of linear di erential equations are being Siegel normal and being Shidlovskii irreducible. These properties play a role in some part of transcendental number theory. In chapter 3 these properties will be characterized in terms of the standard representation of the di erential Galois group . Further there will be explained how these characterizations can be used to verify Siegel normality or Shidlovskii irreducibility in some concrete practical examples. Chapter 3 is published before [Hen94]. Back to Kovacic's algorithm. For any solution 6= 0 of the di erential equation + + = 0 the element = yy satis es the associated Riccati equation + 2 + + = 0. The di erential equation + + = 0 has a Liouvillian solution if and only if the associated Riccati equation + 2 + + = 0 has an algebraic solution. The proof of this statement is based on di erential Galois theory and the classi cation of the algebraic subgroups of (2 C). Kovacic's algorithm tries to compute a solution of the Riccati equation that is algebraic and of minimal degree over C( ). It is shown in [Kov86] that this minimal degree can be 1 2 4 6 or 12. For analogous results for third order linear di erential equations we refer to [SU93]. Suppose now that 2 Q( ). In general the algorithm described in [Kov86] generates algebraic numbers in a rather chaotic way. Therefore in some practical cases the algorithm did not work very well. The coecients of the monic irreducible polynomial of a solution of the Riccati equation that is algebraic and of minimal degree over C( ) are in ( ) where is a nite algebraic extension of Q. In chapter 2 bounds are given for the degree of the extension Q. Analogous results are obtained for third order di erential equations. These bounds help to make Kovacic's algorithm more ecient. Chapter 2 is published before [HP95]. Recently Galois theory of di erence equations is developed by M. van der Put and M.F. Singer [PS96]. In chapters 4 and 5 algorithms are presented for determining the di erence Galois group and the {di erence Galois group of second order di erence and -di erence equations respectively. These algorithms can be considered as the analogue for di erence and {di erence equations of Kovacic's algorithm for di erential equations. Further the notion of Liouvillian solutions is introduced for di erence and -di erence equations. If the di erence or -di erence Galois group is not too big (i.e does not contain the group (2 C)) then it is possible to compute two linearly independent solutions in a certain sequences space. In some respects di erence and -di erence equations are more dicult to treat than di erential equations, due to the fact that the Picard{Vessiot rings associated to di erence or {di erence equations are in general not integral domains but only reduced algebras. But on the other hand the nite primitive groups that cause the most troubles in the di erential case do not occur as di erG
G
G
y
y
0
u
u
00
ay
0
0
by
au
u
b
y
00
ay
0
by
0
u
u
au
Sl
n
;
;
b
;
z
;
a; b
z
z
C z
C
C
q
q
q
q
q
Sl
q
q
4
;
ence or q{di erence Galois group. Chapter 4 is accepted for publication [Hen96]. In chapter 6 q-di erence equations over the eld C(z) are classi ed, where q is an m th root of unity. In this case there is not a unique Picard{Vessiot ring for every system of q-di erence equations. Therefore it is not possible to de ne the q{di erence Galois group in the usual way. If q is a root of unity then we will use the theory of Tannakian categories for a suitable de nition of the q{di erence Galois group.
5
6
Chapter 2 Galois Action on Solutions of a Di erential Equation 2.1
Introduction
Consider a second order di erential equation of the form + + = 0 with 2 Q( ). If one works over the algebraic closure Q of Q then di erential Galois theory is available. This theory tells us that there exists a Picard{Vessiot eld Q( ) satisfying y
a; b
ay
0
by
x
F
1.
00
x
F
Q( ) is an extension of di erential elds. x
2. The eld of constants of
F
is Q.
3. The equation has two Q-linear independent solutions in . F
4.
F
is minimal with respect to the conditions (2) and (3).
The theory continues as follows: 1.
F
is unique up to d-isomorphism.
2. Let be the group of d-isomorphism of -invariant elements of is Q( ). G
G
F
F
over Q( ). Then the set of x
x
3. Let be the set of solutions of the di erential equation. Then is a two-dimensional vector space over Q. The group acts faithfully on and the image of in Aut( ) is a linear algebraic subgroup. V
F
V
G
G
V
V
A solution of the Riccati equation is an 2 of the form vv with 2 and 6= 0. Kovacic's algorithm tries to compute the solutions of the Riccati equation. The rst part of this algorithm tries to nd solutions = vv 2 Q( ). This means that Q is a -invariant line in . u
0
F
v
v
u
v
G
V
7
0
x
V
The origin of this paper is the question to nd out for which eld solution u lies in C (x).
C
Q the
Suppose that the group has the following form: has a basis such that consists of all 2 Aut( ) with ( ) = and ( ) = with v 2 Q . Then has only two -invariant lines and i = v = 1 2 are the only solutions of the Riccati which lie in Q( ). G
G
V
g
V
V
c
g v1
cv1
G
g v2 0
i
u
i
;
i
v1 ; v2 1 c v2
;
;
x
The intuitive reasoning is now the following: A Galois automorphism of Q can at most permute f g. Hence lie in ( ) where [ : Q] 2. An explicit p example of this situation is the equation 00 = 2 . The eld is here Q( 2). u1 ; u2
u1 ; u2
C x
y
C
x
4
y
C
In the next two sections we will develop the theory justifying this intuitive idea. We note that there is no suitable di erential Galois theory which works over a constant eld of characteristic 0 which is not algebraically closed. The next sections provides some theory in a special case. In later sections we will summarize and correct the results of [HP93] and we will give some more applications. Further a theory of symmetries and forms of a di erential operator is developed in order to construct di erential operators with special features with respect to the eld of constants which is needed for the Riccati equation. We would like to thank M. F. Singer and F. Ulmer for pointing out a mistake in [HP93]. Unfortunately, it was too late to correct the paper before publication.
2.2 The universal eld of k((x)) Let be a eld of characteristic 0 with algebraic closure . One considers differential equationsPover the di erential eld = (( )) with the di erentiation 0 = d given by ( n n )0 = P n n . We want to make a universal eld extendx sion for which every di erential equation over has a full set of solutions. More precisely one de nes as follows: k
k
K
x
a x
k
K
1.
x
na x
K
is an extension of d- elds. 2. The eld of constants of is . K
k
3. For every (homogeneous linear) di erential equation y
(d)
+
a1 y
(d
1)
+ + :::
ad
with coecients in a nite extension of vector space over of dimension . k
d
8
1y
K
(1)
+
a0 y
=0
the space of solutions in is a
4. is minimal with respect to the conditions (2) and (3). As is well known condition (3) is equivalent to (3) Every matrix di erential equation v0 = Av, where the matrix A has coecients in a nite extension K has a fundamental matrix with coecients in .
Theorem 2.2.1 with the above properties exists and is unique up to isomorphism.
Proof. The eld must certainly contain the following things: 1. k.
2. For every m 1 an element x m1 . 3. A solution of y0 = 1, since has two linearly independent solutions of y00 = 0. 4. A non zero solution of y0 = qy for every q 2 [m1 x 1=m k[x 1=m ]. One would like to see the solutions of the equations above as certain functions. This is however a priori not possible, since we would have to explain then what the product of a formal Laurent series with a solution of say y0 = 1 means. Therefore we will build up in a very algebraic way. Let Q denote the additive group (or k-vector space) [m1 k[x 1=m ]. Further, K denotes the algebraic 1closure of K . We note that K is generated over K by k and the elements x m ; m 1. We remark that, in general, k((x)) is not algebraic1 over K , but every element of K can be represented by a Laurent series in k((x m )) for some m 1. We start with a polynomial ring in many variables R0 := K [L; E (q)]q2Q and we de ne a di erentiation on R0 by: 1. For f = P2Q cx 2 K one has f 0 = P2Q cx 2 K . This is the unique extension of the di erentiation 0 = x dxd to K . 2. L0 = 1.
3. E (q)0 = qE (q). It is easily veri ed that the above de nes a unique di erentiation on R0. Let I R0 denote the ideal generated by E (q1 + q2 ) E (q1 )E (q2) (all q1 ; q2 2 Q) and E () x (all 2 Q). The ideal I is invariant under di erentiation. Indeed, (E (q1 + q2 ) E (q1)E (q2 ))0 = (q1 + q2 )(E (q1 + q2 ) E (q1)E (q2 )), and 9
(E () x)0 = (E () x ) Let R := R0=I and let l and e(q) denote the images of L and E (q) in R. Since the ideal I is invariant under di erentiation one nds a di erentiation on R and one has the following properties: 1. l0 = 1. 2. e(q)0 = qe(q) for all q 2 Q. 3. e(q1 + q2 ) = e(q1 )e(q2) for all q1 ; q2 2 Q. 4. e() = x for all 2 Q.
2
Lemma 2.2.2 The ring R has no zero divisors and there is no prime ideal (6= 0; R) in R which is invariant under di erentiation.
Proof. Choose q1 ; :::; qs 2 Q such that 1; q1 ; :::; qs are linearly independent over Q. Consider the subring A := K [l; e(q1); e( q1 ); :::; e(qs); e( qs )] of R
Every nite subset of R lies in such a subring of A. One can see that there are only the trivial relations e(q1 )e( q1 ) = 1; :::; e(qs)e( qs ) = 1 among the generators of A over K . That means that A is isomorphic to the ring K [X; T1 ; T1 1; :::; Ts; Ts 1]. This ring is a localization of the polynomial ring K [X; T1 ; :::; Ts]. In particular, A and R have no zero divisors. It is easily seen that A can have no non-trivial ideal that is invariant under di erentiation. This is implies the same statement for R. 2
Lemma 2.2.3 Let denote the eld of fractions of R, then is a d- eld con-
taining K and the eld of constants of is k.
Proof. Let u 2 satisfy u0 = 0 and u 6= 0. For suitable q1 ; :::; qs 2 Q such that 1; q1; :::; qs are linearly independent over Q one can write u = fg where f; g 2 A := K [l; e(q1 ); :::; e(qs)]. The generators of A over K have no relations and so A is an ordinary \free" polynomial ring over K in s + 1 generators. One may assume that g.c.d.(f; g) = 1. Now u0 = 0 implies that f 0g = fg0 and in particular f divides f 0 and gPdivides g0. Write f as a nite sum q2Nq1+:::+Nqs aq e(q) with eachPaq 2 K [l] and where N denotes the set of non-negative integers. Then f 0 = (a0q + qaq )e(q) has a total degree which is less or equal to the total degree of f . Hence f 0 = Bf for some B 2 K and also g0 = Bg. For each q one nds the relation a0q + qaq = Baq . We will need the following result.
10
If the equation a0 + ha = 0 with h 2 K has a non-zero solution in K [l], then h 2 Q + [n1 x1=n k[[x1=n ]].
Lemma 2.2.4
Proof. Write a = a0 + a1 l + ::: + ad ld with all ai
l0
2 K and with ad 6= 0. Using
= 1 and looking at the coecient of l , one nds a0d + had = 0. Write ad = P cx (1 + >0 c x ) with c 2 k c 6= 0 and 2 Q. Then it is clear that h has the form speci ed in the lemma. 2 We continue the proof of Lemma 2.2.3. If aq 6= 0 then B must have the form B = q + h where h 2 Q + [n1x1=n k[[x1=n ]]. This can happen for only one q since 1; q1; :::; qs are supposed to be linearly independent over Q. Hence f = aq e(q) and similarly g = br e(r) for some r 2 Nq1 + ::: + Nqs. However q r 2 Q + [n1x1=n k[[x1=n ]] implies that q = r. Since g.c.d.(f; g) = 1 it follows that q = r = 0 and f; g 2 K [l] and B 2 Q + [n1 x1=n k[[x1=n ]]. We may assume that g is monic as polynomial in l. Then g divides g0 implies that g = 1 since the degree of g0 is less then the degree of g. Hence u = f 2 K [l] and f 0 = 0. Write f = f0 + f1 l + ::: + fdld with f0 ; :::; fd 2 K and fd 6= 0. Looking at the coecient of ld in f 0 = 0 one nds that fd = c 2 k and c 6= 0. If d 1 then the coecient of ld 1 gives the equation fd0 1 = dc. It is easily seen that no element in K satis es this relation. Hence d = 0 and u 2 k. This nishes the proof of Lemma 2.2.3. 2 Lemma 2.2.5 Every di erential equation over a nite extension of K has a basis of solutions in . Proof. We will use here the language of di erential modules over K . Such a module (M; ) is a nite dimensional vector space M over K with an operator on it, satisfying: is additive and (km) = k0 m + k(m) for k 2 K and m 2 M . It is well known how to translate a homogeneous linear di erential equation (or a di erential equation in matrix form) into a di erential module. In the following we will replace K by nite extensions of K without changing the notation. We use now the formal classi cation of di erential equations (cf. [Lev75]). After a nite eld extension of K , the di erential module M is a direct sum of submodules N . For each N there is a q 2 Q such that the operator q is nilpotent. Then one easily sees that the eld K (l; e(q)) contains d linearly independent solutions of N , where d denotes the dimension of N . Hence has the desired property. 2 Lemma 2.2.6 The eld is the universal extension of K . Proof. The minimality of has already been discussed. The other properties are contained in Lemmas 2.2.3 and 2.2.5. Thus we have shown the existence of a universal eld. Let 0 be another universal eld. Then there is a di erential homomorphism : R ! 0 . According to lemma 2.2.2, the kernel of is 0 and extends to a di erential homomorphism of to 0 . The minimality of 0 implies that is an isomorphism. 2 d
11
Corollary 2.2.7 Let G+ be the group of all automorphisms of the eld such
that is the identity on K and commutes with the di erentiation on . Let G denote the normal subgroup of G+ consisting of the which are the identity on k. Then the following sequence is split exact. 0 ! G ! G+ ! Gal(k=k) ! 0
Proof . Let 2 GalP(k=k). Extend rst to an automorphism of K by P ( Q ax ) = ( a )x . One de nes an automorphism + 2 G+ extending by +l = l and +e(q) = e( q). It is easily seen that + 2 G+ and that 7! + is a homomorphism of groups. This is the required splitting. 2 0
0
2
0
0
2.3
The
k -structure
on the space of solutions
A k-structure on a vector space V over k is a k-linear subspace W of V such that V = k P k W . Such a k-structure induces a Gal(k=k)-action on V by P ( i fiwi) = i( fi )wi for fi 2 k and wi 2 W . Consider a di erential equation M of order d over k(x). Fix an element a 2 k. This gives an embedding k(x) k((x a)) with the universal eld of k((x a)). The space of solutions V = V (M ) of M is a vector space over k with dimension d. Let G and G+ be as in Corollary 2.2.7. The space V is invariant under G+ and the splitting of Corollary 2.2.7 gives an action of Gal(k=k) on V . Let Va V denote the set of Gal(k=k)-invariant elements. Lemma 2.3.1 V = k k Va and the action of Gal(k=k) on V is induced by this k-structure on V . Proof. We note that any element ! 2 needs only nitely many elements of k for its de nition. Hence the subgroup of Gal(k=k) xing ! is open. Some open normal subgroup N of Gal(k=k) xes a basis fv1; :::; vd g of V . Let W denote the set of N -invariant elements of V and let k be the eld of N -invariant elements of k. On W = k v1 + ::: + k vd acts the Galois group Gal(k =k) = Gal(k=k)=N . For 2 Gal(k =k) one denotes by A( ) 2 Gl(d; k ) the matrix which expresses the vectors (v1 ); : : : ; (vd ) in the basis fv1 ; : : : ; vdg. For ; 2 Gal(k =k) one nds the formula A( ) = A() (A( )). In other words, 7! A( ) is a 1-cocycle. Its image in H 1(Gal(k =k); Gl(d; k )) is trivial since proposition 3 on page 159 of [Ser68] states that the set H 1(Gal(k =k); Gl(d; k )) is trivial. This means that W also has a basis w1; : : : ; wd over k such that (wi ) = wi for every i. One nds that W = k k W with W := kw1 + : : : + kwd. It follows at once that W is equal to Va, the set of Gal(k=k)-invariant elements of V . This proves the lemma. 2 A Picard{Vessiot eld of the equation M over k(x) is the eld F
generated over k(x) by a basis of V . The di erential Galois group G(M ) of 0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
12
0
M over k(x) is the group of the automorphisms of F=k(x) commuting with the di erentiation on F . Fix a basis fv1 ; :::; v g of V . Then G(M ) can be identi ed with an algebraic subgroup of Gl(d; k). This algebraic subgroup is unique up to conjugation with an element in Gl(d; k). The coordinate ring of Gl(d; k) is k[X ; 1 ], where D = det(X ). Let I k[X ; 1 ] denote the (radical) ideal which de nes G(M ). d
i;j
a
i;j
D
Corollary 2.3.2
i;j
D
G(M ) is de ned over the eld k.
For any 2 Gal(k=k) and g 2 G(M ) one has g 1 2 G(M ). This implies that the ideal I is invariant under . A standard argument implies that I is generated by J := I \ k[X ; 1 ]. Let G(M ) denote the algebraic group de ned by J . Then G(M ) = G(M ) k. 2 Proof.
i;j
D
a
k
a
2.3.1 Remarks
1. For two elements a; b 2 k, the k-structures V and V of V seem to be unrelated. It is not clear how the two groups G(M ) and G(M ) are related. a
b
a
b
2. The map ! : M 7! V (M ) from di erential equations over k(x) to vector spaces over k commutes with \all constructions of linear algebra". In more sophisticated terms (see [DM82]) it is a faithful, exact, k-linear, -functor of the rigid abelian category of left k(x)[@ ]-modules of nite dimension over k(x) to the category of nite dimensional vector spaces over k. If we restrict ! to the subcategory generated by a xed M , then it can be seen that the ane group scheme Aut (!) over k coincides with the algebraic group G(M ) over k de ned above. a
a
3. Let Z V = V (M ) be an algebraic subset that is invariant under the di erential Galois group G(M ) of M . For any 2 Gal(k=k) the set Z is again an algebraic subset of V that is invariant under G(M ). A case of special interest is Z is a ( nite union of) G(M )-invariant subspace.
2.4
Rational solutions of Riccati
One considers a linear homogeneous di erential equation of order n over k(x) or k((x)). In the rst case we x an a 2 k and an embedding k(x) k((x a)) . This induces the k-structure on the solution space V of the di erential equation. In the second case the k-structure on V is given in Section 2.3. The di erential 13
Galois group of the equation is denoted by G. A rational solutions u of the Riccati equation determines a line ky V , where y 2 V satis es u = yy , which is invariant under G. Conversely, if ky V is a line, invariant under G, then u = yy is a rational solution of the Riccati equation. In this section we give estimates for the degree of the eld extension of k over which a solution of the Riccati equation is de ned. 0
0
Lemma 2.4.1 Suppose that there is a rational solution of the Riccati equation, i.e. a solution in k(x) or k k k((x)). 1. If there are only nitely many solutions, then each one of them is de ned over a eld k0 k with degree n. 2. If there are in nitely many solutions of the Riccati equation, then there is a solution which is de ned over a eld k0 with degree n2 . Proof. For every algebraic character : G ! k one de nes the vector space VP := fv 2 V j g(v) = (g)v for all g 2 Gg. For the with V 6= 0 one has V V . Any 2 Gal(k=k) acts on the algebraic characters by 7! where (g) = ( g 1). For every non zero V one considers the stabilizer H Gal(k=k) of V . The index of this stabilizer is dimn V . Hence V is de ned over a eld k0 k of degree dimn V . Inside V one can take a line which is also de ned over k0. The solution y corresponding to this line gives a solution u = yy , which lies in k0(x) or k0((x)), of the Riccati equation. If there are only nitely many solutions of the Riccati equation, then all the non-zero V have dimension 1. Hence any choice of of an G-invariant line de nes a eld extension of degree n over k. In the second case, some V has dimension 2. A suitable line in V is therefore de ned over a eld k0 with [k0 : k] n2 . Furthermore, V contains lines which are de ned over arbitrary big extensions of k. 2 0
2.5 Algebraic solutions of the Riccati equation and examples One considers as before a di erential equation over the eld K which is either k(x) or k((x)). The vector space V over k of solutions is given a k-structure as in section 2.3. The di erential Galois group is called G. A solution u = yy of the Riccati equation, algebraic of degree m, corresponds to a line L = ky V such that its stabilizer H G is a subgroup of index m. The element u satis es a monic irreducible equation P of degree m over kK . The collection of all zeroes of P corresponds to the orbit GL of the line L. We want to nd the smallest eld extension k0 of k such that the coecients of P are in 0
14
k0 K . This is equivalent to nding the smallest eld extension k0 k such that the orbit GL is de ned over k. Let us de ne an m-line to be a G-invariant subset of V consisting of m lines through 0, such that G acts transitively on the m lines. The collection of all m-lines is denoted by Lm. This set can be in nite. Let us de ne an m-group to be the conjugacy class [H ] of an algebraic subgroup H of index m such that H is the stabilizer of at least one line in V . The collection of all m-groups is denoted by Hm . This set is nite. The map : Lm ! Hm is de ned as follows. The image of an m-line L1 [ ::: [ Lm is the conjugacy class of the stabilizer of L1 (or any Li ). The map is by de nition surjective. More precisely, let H be an algebraic subgroup of index m such that the collection
X := fL j L is a line with stabilizer gHg 1 for some g 2 Gg is not empty. Then 1([H ]) is the set of G-orbits of X . On Lm the Galois group Gal(k=k) acts in the obvious way. On Hm we let any 2 Gal(k=k) act by ([H ]) = [ 1 H ]. The map is equivariant for the de ned actions.
Proposition 2.5.1 Let u be an algebraic solution of the Riccati equation and let
k0 be the minimal extension of k such that the minimal monic equation of u over kK is de ned over k0K . Let L be the line in V corresponding to u and let H G denote the stabilizer of L. Then [k0 : k] (#Gal(k=k)[H ])(# 1([H ])) Proof. Since k0 is also the minimal extension for which the m-line [L] := GL is de ned over k0, the inequality follows at once from the obvious inequality
(#Gal(k=k)[L]) (#Gal(k=k)[H ])(# 1([H ])):
2
Remark 2.5.2 For m = 1 the last proposition gives the inequality [k0 : k] the number of lines xed by G: This is in accordance with lemma 2.4.1.
15
2.5.1
Equations of order two
One considers an equation of order two over the eld k(x). This equation can be normalized and has then the form y00 = ry, with r 2 k(x). The corresponding Riccati equation is u0 + u2 = r. The di erential Galois group G is an algebraic subgroup of Sl(2; k) and is unique up to conjugation. According to [Kov86] one has the following result:
Theorem 2.5.3 The following statements hold.
1. If there is no algebraic solution over k(x) of the Riccati equation, then G = Sl(2; k). 2. If there is an algebraic solution of the Riccati equation, then the minimal degree m of such an equation can be 1; 2; 4; 6; 12 and ! c d (a) If m = 1 then G f 0 c 1 j c 2 k ; d 2 kg. (b) (c) (d) (e)
If m = 2 then G = D1 or G = Dn with n 3. (The dihedral groups). If m = 4 then G is the tetrahedral group. If m = 6 then G is the octahedral group. If m = 12 then G is the icosahedral group.
Using the inequality in Proposition 2.5.1 and Lemma 2.4.1 one nds Theorem 2.5.4 Assume that the Riccati equation u0 + u2 = r 2 k(x) has a solution which is algebraic over k(x). Let m be the minimal degree, as in Theorem 2.5.3. Then the coecients of the monic minimal polynomial of u over k(x) lie in a eld k0(x) with [k0 : k] 3. More precisely: 1. If m = 1 and G is the multiplicative group Gm or a nite cyclic group of order > 2 then [k0 : k] 2. 2. If m = 2 and G = D4 , then [k0 : k] = 3 or k0 = k. 3. If m = 4 and G is the tetrahedral group, then [k0 : k] 2. 4. In all other cases k0 = k. Proof. We will give the proof case by case. 1. If m = 1 and there are in nitely many solutions or precisely one solution of Riccati in k(x), then by Lemma 2.4.1 one has k0 = k. In the remaining case [k0 : k] 2. 2. If m = 2 and G 6= D4, then G has precisely one algebraic subgroup H of index 2. Hence #Gal(k=k)[H ] = 1. Further # 1 ([H ]) = 1. 16
2
3. If m = 2 and G = D4 , then H2 has three elements. For each such [H ] 2 H2 one has # 1([H ]) = 1. 4. For the tetrahedral group one has #H4 = 1 and # 1 ([H ]) = 2. 5. For the octahedral group one has #H6 = 1 and # 1([H ]) = 1. 6. For the icosahedral group one has #H12 = 1 and # 1([H ]) = 1.
2.5.2
Equations of order three
A third-order equation can be normalized in the form y000 + py0 + qy = 0 with p; q 2 k(x). The di erential Galois group G is an algebraic subgroup of Sl(3; k). The Riccati equation reads u00 + 3uu0 + u3 + pu + q = 0. The analogue Theorem 2.5.5 of Theorem 2.5.3 is proved in [SU93]. We will use their terminology and description of nite primitive groups. In particular, a subgroup H of Sl(3; k) is called 1-reducible if the elements of H have a common eigenvector.
Theorem 2.5.5 The following statements hold.
1. If there is no algebraic solution over k(x) of the Riccati equation then G satis es one of the following properties: (a) G = Sl(3; k); (b) G=Z (G) = PSL(2; k) where Z (G) denotes the center of G; (c) G is reducible but not 1-reducible. 2. If there is an algebraic solution of the Riccati equation then the minimal degree m of such an equation can be 1; 3; 6; 9; 21; 36 and (a) If m = 1 then G is a 1-reducible group. (b) If m = 3 then G is an imprimitive group. (c) If m = 6 then G=Z (G) is isomorphic to F36 or A5 . (d) If m = 9 then G=Z (G) is isomorphic to H72 or H216 . (e) If m = 21 then G=Z (G) = G168 . (f) If m = 36 then G=Z (G) = A6 .
Using Lemma 2.4.1 and Proposition 2.5.1 one nds Theorem 2.5.6 Suppose that the Riccati equation has an algebraic solution and let m be as in Theorem 2.5.5. There exists an algebraic solution of the Riccati equation of degree m such that the coecients of the monic minimal equation over k(x) lie in k0(x) with 17
1. [k0 : k] 3 if the equation is 1-reducible. 2. [k0 : k] 4 if G is imprimitive. 3. [k0 : k] 2 in case G=Z (G) = F36. 4. k0 = k in all other cases. Proof.
1. If G is 1-reducible one applies lemma 2.4.1 and thus part (1) of the theorem is proved. 2. Let G be imprimitive. A system of imprimitivity consists of three lines kei; i = 1; 2; 3 such that fe1; e2; e3 g is a basis of the solution space V and such that G permutes the three lines. In particular, this is a 3-line for G. If G has only one system of imprimitivity then this system is de ned over k and k0 = k holds. If there is more than one system of imprimitivity then one easily sees that Go = 1 and the group G is nite. In particular, G is completely reducible. For any 3-line fL1 ; L2 ; L3g for G the space L1 + L2 + L3 V is G-invariant. If this space has dimension 2 then G has also a one-dimensional invariant subspace. This contradicts our assumption. Hence L3 is the set of all systems of imprimitivity of G. According to [Bli17] there are at most four systems of imprimitivity or in our language #L3 4. This proves part (2) of the theorem. We remark that, according to Blichtfeldt (1917), there are only two imprimitive groups with four systems of imprimitivity. The groups have orders 27 and 54 respectively. The corresponding subgroups of PGl(3) have orders 9 and 18. They are sometimes called H9 and H18. 3. Suppose now that G is a nite primitive group and m the number de ned in theorem 2.5.5. Using the description of such groups given in [SU93], one can verify that any 1-reducible subgroup H of index m in G is non-abelian. Therefore H can only x one line L in the solution space V . It follows that [H ] (the set of conjugates of H in G) consists of m elements. Moreover, for any [H ] 2 Hm the set 1([H ]) consists of one element. Therefore [k0 : k] #Hm = m1 #T , where T denotes the set of 1-reducible subgroups of index m in G. Counting gives #T = m except in one case. This is the case G=Z (G) = F36 . In this case, m = 6 and #T = 12. This proves parts (3) and (4) of the theorem.
2 18
2.5.3 Remarks
1. Consider a di erential equation of order n over the eld K = k((x)). According to [Lev75], there is a eld extension K K of degree n such that the equation over K has a one-dimensional factor. Let e denote the rami cation index of K over K and f the degree of the residue eld exten1 sion. Then ef = [K : K ] n. The extension K K (x e ) is unrami ed and has degree n. This can be reformulated as follows: 0
0
0
0
0
0
The minimal number m such that the Riccati equation has a solution u which is algebraic over kK is n. The minimal monic equation of u over kK has its coecients in a eld k K with [k : k] n. 0
0
2. Let k be a sub eld of the eld of complex numbers. Then we replace the eld of formal Laurent series, used in remark (1) above, by the eld K of convergent Laurent series. Let G be the di erential Galois group of the equation of order n and let Go be the component of 1 in G. One knows that the group G=Go does not change if one makes the step from convergent Laurent series to formal Laurent series. This implies that this group is cyclic. We note that there are in general no algebraic solutions of the Riccati equation! Any algebraic solution u of the Riccati equation corresponds to a line in the solution space with stabilizer H G0 . The group H is uniquely determined by its index in G and in particular invariant under conjugation with any element in Gal(k=k). Using Lemma 2.4.1 one nds the following: If the Riccati equation has an algebraic solution then there is an algebraic solution u such that its monic minimal equation over the eld kK has coecients in a eld k K with [k : k] n. 0
0
2.6 Symmetries The aim of this section is to construct di erential equations over the eld k(x) such that the algebraic solutions of degree m of the corresponding Riccati equation are de ned over a proper extension of the base eld k. There are two ingredients in the construction: 1. The group of symmetries of the equation. 2. Forms of a di erential equation over k(x). 19
Let k be a eld of characteristic 0 with algebraic closure k. The ring of di erential operators k(x)[@x ] is the skew polynomial ring de ned by the relation @xx = x@x + 1. We note that k(x) is the unique maximal sub eld of k(x)[@x ].
Lemma 2.6.1 The k-algebra homomorphisms : k(x)[@x ] ! k(t)[@t ] are given by
(x) = a and (@x) = a10 @t + b with a 2 k(t) n k; a0 := dtd a and b 2 k(t). Proof. is determined by (x) and (@x). These two elements of k(t)[@t ] satisfy the relation (@x) (x) (x) (@x) = 1. A straightforward calculation with this relation yields the lemma. 2
Corollary 2.6.2 Let Aut(k(x)[@x ]) denote the group of the k-algebra automorphisms of k(x)[@x ]. There is a split exact sequence of groups
0 ! k(x) ! Aut(k(x)[@x ]) ! PGl(2; k) ! 1 Proof. The element a = (x) must satisfy k(x) = k(a). Therefore, any automorphism induces an automorphism of k(x). This explains the map of Aut(k(x)[@x ]) ! PGl(2; k). Clearly this map is surjective and the kernel consists of the automorphisms with (x) = x and (@x) = @x + b where b 2 k(x) is arbitrary. The splitting s is given by: if 2 PGl(2; k) then s()(x) = (x) and s()(@x) = (1x) @x. This proves the corollary. 2 0
2.6.1 Symmetries of L
Let L 2 k(x)[@x ] n k(x). The group Symk (L) of the symmetries of L consists of the automorphisms of k(x)[@x ] satisfying (L) = L. The map Symk (L) ! PGl(2; k) is injective as one easily sees. In some interesting cases L itself has very few symmetries but there is a non-trivial nite group H of automorphisms of k(x)[@x ] satisfying (L) = fL for some f 2 k(x) . In this situation one can change L into gL for a certain g 2 k(x) such that H is contained in the group of symmetries of gL. We will call this normalizing L. That normalization is always possible can be seen as follows: Write f ( ) for the element of k(x) satisfying (L) = f ( )L. Then 2 H 7! f ( ) 2 k(x) is a 1-cocycle. The corresponding cohomology group is H (H; k(x) ). We will show that this group is trivial. Indeed, let K k(x) denote the sub eld of the H -invariant elements of k(x). Then K k(x) is a Galois extension with Galois group H . By Hilbert 90 the group H (H; k(x)) = 1. Therefore the 1-cocycle 7! f ( ) is trivial. This implies that we can choose a 1
1
20
g 2 k(x) such that (gL) = gL for every 2 H . In general, the group of symmetries of an operator L is trivial. In many cases a study of the singular points of the operator shows that this group is a nite subgroup of PGl(2; k). We will explain this and introduce some terminology. Let the operator L have the form f (@xn + an 1@xn 1 + : : :+ a0 ). Then p 2 k is called a singularity of L if some ai has a pole at p. The point p = 1 is called a singular point if L expressed in the local parameter t = x 1 is singular for t = 0. We note that an automorphism of k(x)[@x ] not only changes the position of the singular points but singularities can disappear and new ones can appear. We introduce the notion of essential singularity in order to distinguish among the singular points. A singular point p 2 k [ f1g of L is called essential if there is no automorphism , with (x) = x, such that p is a regular point of (L). If p is an essential singular point of L then (p) is an essential singular point of (L). Indeed, it suces to verify this statement for three types of automorphisms: 1. (x) = x. By de nition, this preserves the essential singular points. 2. (x) = x + with 2 k and (@x) = @x. This case is trivial. 3. (x) = x 1 and (@x) =
1
@ . One can verify the statement by hand.
x2 x
One can explain the notion of essential singularity as follows. Let L = f (@xn + an 1 @xn 1 + : : : + a0) be a di erential operator and let p 2 k. We perform on L the following operation: rst, delete the term f ; second, make the unique transformation x 7! x and @x 7! @x ann 1 which kills the coecient of @ n 1 of L. The result is the di erential operator M = @ n + bn 2 @xn 2 + : : : + b0 . It is not dicult to see that L has an essential singularity at p if and only if p is a pole of some of the bi. Any L has only nitely many essential singular points. If this number is 3 then Symk(L) is obviously a nite group. Let H be a nite group of automorphisms of k(x)[@x ]. The set of H -invariant elements of k(x)[@x ] can be identi ed with k(t)[@t ]. Indeed, choose t such that k(t) is the sub eld of k(x) consisting of the H -invariant elements. One can choose b 2 k(x) such that @x := t10 @x + b is H -invariant. This follows from H 1(H; k(x)) = 0. Let : k(t)[@t ] ! k(x)[@x] denote the inclusion. A di erential operator L 2 k(x)[@x ] has H as group of symmetries, if and only if L = (M ) for some M 2 k(t)[@t ]. This shows that any nite subgroup of PGl(2; k) occurs as group of symmetries of some di erential operator. 21
2.6.2 Transforming algebraic solutions of Riccati
Let L 2 k(x)[@x ] n k(x) and a k-algebra homomorphism : k(x)[@x ] ! k(t)[@t ] be given. We extend to a morphism of k(x)[@x] ! k(t)[@t ] by extending the homomorphism of elds k(x) ! k(t) to the algebraic closures k(x) and k(t) of those elds. An algebraic solution u 2 k(x) of the Riccati equation means that L factors as M (@x u) in k(x)[@x]. Then (L) factors as (M )( (@x) (u)). Write (@x) = (1x) @t + b, then (x)0 ( (u) b) is a solution of the Riccati equation of (L). Let P (x; T ) 2 k(x)[T ] be the monic irreducible polynomial of degree m satis ed by u then (x)0( (u) b) satis es the monic polynomial ( (x)0)m P ( (x); (1x) T + b) over k(t). We will denote this polynomial by P . Let G be a group of symmetries of L. Suppose that the collection Rm (L) of the algebraic solutions of degree m over k(x) of the Riccati equation is nite. Then there are distinct monic irreducible polynomials Pi(x; T ); 1 i s of degree m over k(x) such that Rm (L) is equal to the set of solutions of P1:::Ps. For every g 2 G and i the polynomial gPi belongs to fP1; :::; Psg. Hence G acts as a group of permutations on fP1; :::; Psg. 0
0
2.6.3 Forms of a di erential operator
Let L 2 k(x)[@x ] n k(x) be a di erential operator. A di erential operator M 2 k(x)[@x ] of the form M = (L) with 2 Aut(k(x)[@x ]) is called a form of L. Two forms M1 ; M2 are equivalent if there exists a 2 Aut(k(x)[@x ]) with (M1 ) = M2 . We let the group Gal(k=k) act on k(x)[@x ] by its natural action on k and as the identity on k(x)[@x ]. Further Gal(k=k) acts on Symk (L) by (g) = g 1 (with 2 Gal(k=k) and g 2 Symk (L)). In the following we will construct an injective map : Forms(L) ! H 1(Gal(k=k); Symk (L)) , where Forms(L) is the set of equivalence classes of forms of L. For a form M = (L) of L and a 2 Gal(k=k) one has that a( ) := 1 2 Symk (L). Moreover 7! a( ) is a 1-cocycle, i.e. a( 1 2 ) = a( 1 ) 1 (a( 2 )) for all 1; 2 2 Gal(k=k). The class of 7! a( ) in H 1(Gal(k=k); Symk (L)) does not depend on the choice of . Indeed, any other choice for has the form g with g 2 Symk (L). The new 1-cocycle has the form 7! g 1 1 g 1 = g 1( 1 1 ) g 1 . By de nition this is an equivalent 1cocycle. We will verify that the resulting map : Forms(L) ! H 1(Gal(k=k); Symk(L)) is injective. In general will not be surjective. Consider the following two maps: 1
: H 1(Gal(k=k); Symk(L)) ! H 1(Gal(k=k); Aut(k(x)[@x ]))
: H 1(Gal(k=k); Aut(k(x)[@x ]) ! H 1(Gal(k=k); PGl(2; k)) 22
The trivial element of those sets (i.e. the image of the trivial 1-cocycle) is denoted by 1.
Lemma 2.6.3 The following statements hold. 1. 2. 3. 4. 5.
is injective.
The image of is 1 (f1g).
is surjective.
1 (f1g) = f1g. is bijective if either H 1(Gal(k=k); PGl(2; k)) is trivial or if the subgroup Sym (L) of PGl(2; k) lifts to a subgroup of Gl(2; k). Proof. We will prove this theorem case by case. 1. If the M = (L) for i = 1; 2 induce the same element of H 1(Gal(k=k); Sym (L)), then there is a g 2 Sym (L) with 2 1 2 1 = g 1( 1 1 1 1 ) g 1 ; for all 2 Gal(k=k). This implies that 1g 2 1 = 1 g 2 1 for all and so 1g 2 1 2 Aut(k(x)[@ ]). Then 1g 2 1(M2 ) = M1 proves that M1 and M2 are equivalent. 2. Clearly, the image of is f1g. On the other hand, let a be a 1-cocycle for Sym (L) with (a) = 1. Then there exists a 2 Aut(k(x)[@ ]) with a( ) = 1 ( ) = 1 1 for all 2 Gal(k=k). Put M := (L). Then (M ) = 1 (L) = a( )(L) = (L) = M . Hence M 2 k(x)[@ ] is a form of L and M induces the 1-cocycle a. k
i
i
k
k
x
x
k
x
3. This follows from the splitting of the group homomorphism Aut(k(x)[@ ]) ! PGl(2; k). x
4. Let a be a 1-cocycle for Sym (L) which has trivial image in H 1(Gal(k=k); PGl(2; k)). Then a is equivalent to a 1-cocycle for the normal subgroup k(x) of Aut(k(x)[@ ]). Since k(x) = k k(x) one has that H 1(Gal(k=k); k(x)) = 0. Hence a is a trivial 1-cocycle. k
x
k
2
5. This follows from statements (1)-(4) and the well known fact that H 1(Gal(k=k); Gl(2; k)) = 1. 23
2.6.4 Construction of special di erential equations
One xes a di erential equation L 2 k(x)[@x] and a number m 1. One assumes that the set Rm (L), as de ned in Section 2.6.2, is nite and that the monic polynomials P1; :::; Ps actually lie in k(x)[T ]. Let G Symk(L) be some nite group which lifts to a subgroup of Gl(2; k). This group acts as a permutation group on fP1 ; :::; Psg. Let h : Gal(k=k) ! G be a continuous homomorphism. Then one can see h as an element of H 1(Gal(k=k); Symk(L)). There is an automorphism of k(x)[@x ] such that M := (L) 2 k(x)[@x ] such that = h( ) for all 2 Gal(k=k). The set f P1 ; :::; Psg are the monic irreducible polynomials of degree m over k(x) with as set of zeroes all the solutions of the Riccati equation of M which are algebraic over k(x) of degree m. Since = h( ) and Pi = Pi for all , one nds that 2 Gal(k=k) permutes the set f P1; :::; Psg in the same way as h( ) permutes the fP1 ; :::; Psg. We can formulate the results above as follows.
Theorem 2.6.4 Let L 2 k(x)[@x] be a di erential equation with a nite group of symmetries G Symk(L) which lifts to a subgroup of Gl(2; k). Let m 1 and
suppose that the set of algebraic solutions of degree m over k(x) of the Riccati equation of L is nite and given as the set of zeroes of P1 :::Ps where the Pi 2 k(x)[T ] are distinct monic polynomials of degree m and irreducible as elements of k(x)[T ]. Let a continuous homomorphism h : Gal(k=k) ! G be given. Then there is a form M = (L) 2 k(x)[@x ] of L with = h( ) for all 2 Gal(k=k). The polynomials Qi := Pi 2 k(x)[T ]; 1 i s are monic and irreducible. The set of zeroes of Q1 :::Qs coincides with the set of algebraic solutions of degree m over k(x) of the Riccati equation of M . The action of Gal(k=k) on fQ1 ; :::; Qsg coincides with the homomorphism Gal(k=k) !h G ! Perm(fP1; :::; Psg), where Perm(fP1; :::; Psg) denotes the group of permutations of fP1 ; :::; Psg.
2.6.5 Examples
We will give here examples to show that the inequalities in Theorem 2.5.4 and Theorem 2.5.6, part 1 are sharp.
Cyclic groups
Consider the di erential operator L = (x@x )2 a with a 2 k. There are two singular points 0 and 1. They are essential singular points. The symmetry group of L is the group of all the 2 PGl(2; k) for which f0; 1g is invariant. In particular, given by (x) = x 1 and (x@x ) = x@x , is a symmetry. The p d 0 two solutions of the Riccati equation (with respect to = x dx ) are a. The symmetry permutes these two solutions. 24
p
p
1. If a 62 Q then the di erential Galois group is Gm. If, moreover, a 62 k then the Riccati equation de nes a degree two extension of k. 2. If ( nt ) = a with t; n 2 Z; t; n 1 and relative prime, then the di erential Galois group is cyclic of order n. The two solutions of the Riccati equation are in k(x) and since they are permuted by , one can apply Theorem 2.6.4. Hence there is for every quadratic extension k0 k a form M of L such that the solutions of the Riccati equation of M are in k0(x) n k(x). In the next subsection, we will give an explicit example M . 2
Cyclic groups again
Let k0 = k() with 2 k and 62 k. Take u ; u 2 k(x) and u 6= 0 and write u = u + u . The equation r = u0 + u 2 k(x) is equivalent to u = 1=2 uu11 and r = u + u0 + u . Clearly u + u and u u are two solutions of the Riccati equation of y00 = ry. The equation y0 = ( 1=2 uu11 + u )y has a di erential Galois group (over k0(x) or k(x)) which can be nite or the multiplicative group Gm. For a general choice of u 2 k(x) the di erential Galois group will be Gm . If one chooses u = b x2a 2 with a; b 2 Z n 0; b > 1 and g.c.d.(a; b)=1 then 1 a 1 a y = (x ) 2 2b (x + ) 2 2b satis es the equation y 0 = ( 1=2 uu11 + u )y . Therefore the di erential Galois group has order b if a; b are odd and has order 2b if a or b is even. 2
0 2 0
0
1
1
2
1
2
0
2 1
0
1
0
0
0
0
1
1
1
1
( +
(
)
)
(
0
)
1
The group D
4
The standard di erential operator for this group is (see [BD79] with in their notation the case e = e = e = 2) 1
2
L = @x2
3
(
3
16x(x 1)
3 16x
3
2
16(x 1)
2
)
The singular points are f0; 1; 1g. They are essential singular points. The symmetry group of L is therefore contained in the subgroup of P Gl(2; Q) which leaves f0; 1; 1g invariant. This group is isomorphic to S and lifts to a subgroup of Gl(2; Q). 3
The operator L is actually not invariant under S . One has to normalize L into x (1 x )L. The px latter has S as group of symmetries. One can calcux late that u = x x is a solution of the Riccati equation. The corresponding p eld extension of Q(x) is Qp( x). The otherq solutions of the Riccati equation give the eld extensions Q( x 1) and Q( x(x 1)). It follows that the six 2
3
2
3
3 1+ 4 ( 1)
25
quadratic solutions of the Riccati are the roots of three irreducible monic polynomials P ; P ; P 2 Q(x)[T ] and that the symmetry group of L acts as S on fP ; P ; P g. Hence we can apply Theorem 2.6.4. Consider L as given over some eld k Q and let k0 k be a Galois extension with Galois group isomorphic to S . Then there is a form M of L with coecients in k(x) such that the Galois group Gal(k0 =k) acts as S on fQ ; Q ; Q g, where the Qi are the monic irreducible polynomials of degree 2 over k(x) which have as zeroes the six solutions of the Riccati equation of M . The minimal eld ki k such that Qi 2 ki(x)[T ] has degree 3 over k. 1
1
2
2
3
3
3
3
3
1
2
3
One can make this example more explicit as follows. Write k0 = k(a ; a ; a ) such that the Galois group of k0 =k acts as S on fa ; a ; a g. Let the equation of the ai over k be X e X + e X e = 0. De ne : k(x)[@x ] ! k(x)[@x ] by (x) = aa22 aa13 xx aa13 and a suitable (@x ) such that (L) takes the form f (@x r) for some f; r 2 k(x). The term f disappears after a normalization of L. A long calculation yields the expression for r: 1
3
1
3
2
3 (3e e 16
2
1
2
3
3
3
2
e22 ) + (e1 e2 9e3 )x + (3e2 (x3 e1 x2 + e2 x e3 )2
1 3
2
e21 )x2
The special choice: a ; a ; a are the three rootspof X2i 2, gives the nice equation @x + x3 x 2 . The Galois group of F := Q( 3 2; e 3 ) acts as the group S on the three quadratic irreducible minimal polynomials fQ ; Q ; Q g F (x)[T ] corresponding to six quadratic solutions of the Riccati equation. 2
1
27
8(
2
3
3
3
2)
1
2
3
The tetrahedral group
The standard example for the tetrahedral group is the di erential operator (See [BD79] with in their notation e = 2; e = e = 3). 3 2 ) 3 L := @x ( 16x(x 1) 16x 9(x 1) 1
2
3
2
2
2
The three singular points f0; 1; 1g are essential singular points. Any symmetry of L has to permute those three points. Further the singularities at 0 and 1 are isomorphic and the singularity at 1 is not isomorphic to the one at 0. Therefore the only symmetry, 6= 1, is g given by g(x) = x x and g(@x) = (x 1) @x +(x 1). Actually, one has to normalize L into x L and then g(x L) = x L. We note that [BD79] contains a mistake at this point. There are eight algebraic solutions of degree 4 of the Riccati equation and there are two irreducible monic polynomials P ; P 2 Q(x)[T ] of degree 4 such that the set of zeros of P P is the set of those eight solutions. Actually, one of the polynomials, say P , is computed in [Kov86]. This polynomial turns out to lie in Q(x)[T ]. A calculation shows that gP 6= P 2
4
1
1
4
2
4
1
2
1
1
26
1
and so g P1 = P2 . Therefore P1; P2 2 Q(x)[T ] and g permutes them. It seems possible to make a proof of this without the explicit knowledge of P1 . Applying Theorem 2.6.4, one nds a form (M ) of L over k and a quadratic extension k0 k such that the Galois group of k0=k permutes the corresponding polynomials f P1; P2 g of M . Let k0 = k() with 2 2 k and 62 k. Then one can take for the automorphism given by (x) = 1+22x and suitable (@x). A calculation shows that M has the form 32 322 @x2 + 9(1 42x2 )2 4(1 42 x2)
Order three with a 1-reducible Galois group Let T 3 + pT + q 2 k[T ] denote an irreducible polynomial with Galois group S3.
Then the di erential equation y000 + py0 + qy = 0 has as basis for the solutions yi := e x where 1 ; 2 ; 3 are the three roots of the polynomial T 3 + pT + q . The only relation over the eld k(x) satis ed by the yi is y1y2y3 = 1. The di erential Galois group is therefore a maximal torus in Sl(3; k) and there are precisely three solutions of the Riccati equation, namely 1; 2 ; 3. This shows that the inequality in Theorem 2.5.6 part 1 is sharp. i
Remarks In the two remaining cases part 2 and 3 of Theorem 2.5.6 we are unable to give explicit examples to show that the inequalities are sharp.
27
Chapter 3 Shidlovskii irreducibility 3.1
Introduction
In this paper the notions Siegel normality and Shidlovskii irreducibility will be discussed. Being Siegel normal and being Shidlovskii irreducible are interesting properties of systems of ordinary linear di erential equations, which arise in transcendental number theory. The aim of this article is to show how these properties can be characterized in terms of D-modules and the standard representation of the di erential Galois group and how these characterizations can be used to verify Siegel normality or Shidlovskii irreducibility in some concrete practical examples. The notion Siegel normality has been studied by F. Beukers, W.D. Brownawell and G. Heckman. Their important paper [BBH88] was the main source of inspiration for the research concerning the notion Shidlovskii irreducibility which led to this paper. Some interesting remarks on the notion Shidlovskii irreducibility are made in Bertrands paper [Ber90]. The following theorem is a fundamental theorem in the branch of transcendental number theory, which has been developed by C.L. Siegel and A.B. Shidlovskii. The theorem was proved by the latter one in 1959 (see [Shi89], Chapter 3). Theorem 3.1.1 Consider the n n system of linear di erential equations
0 y d BB ..1 dz @ y.
(A) :
n
1 0 CC BB A=@
a11 ...
a1 n
10 a1 C B y1 ... C B ... A@ y a n
nn
1 CC A
n
where a 2 C(z ) for all i; j . Assume that: 1. (f1 (z ); : : : ; f (z )) is a solution of the system (A). 2. The component functions f1 (z ); : : : ; f (z ) are all E -functions. 3. The component functions f1 (z ); : : : ; f (z) are homogeneous algebraic independent over C(z). ij
n
t
n
n
29
4. is a nonzero algebraic number and is not a pole of one of the aij 's. Then the numbers f1 ( ); : : : ; fn( ) are homogeneous algebraic independent over Q.
There exists also a more quantitative version of this theorem, namely if one adds the assumption that system (A) is homogeneous algebraic Siegel normal or that system (A) is homogeneous algebraic Shidlovskii irreducible, then one may conclude that the numbers f1 ( ); : : : ; fn( ) are homogeneous algebraic independent with an e ective measure of homogeneous algebraic independence. So this quantitative version of Shidlovskii's fundamental theorem motivates the study of Siegel's normality criterion and Shidlovskii's weaker irreducibility criterion. The paper is organized in the following manner. In section 3.2 some notation will be xed and some basic facts from D-modules and di erential Galois theory will be recalled. Section 3.3 contains a summary of the results of Beukers, Brownawell and Heckman. In section 3.4 the results concerning the notion Shidlovskii irreducibility will be discussed. Some nice examples will be given in section 3.5. Finally I wish to thank M. van der Put for his advice and interest. And I also would like to express my gratitude to F. Beukers, who called my attention to the book [Shi89] and gave me some ideas.
3.2 Preliminaries De nition 3.2.1 A di erential eld (K; ) is a eld K equipped with a derivation : K ! K . That is (a + b) = a + b and (ab) = (a)b + a (b). The eld of constants of K is C = CK = fc 2 K j c = 0g.
From now on we will assume that char K = char C = 0 and that C is algebraically closed. For the purposes mentioned in the introduction one should take K = C(z ), C = C and = dzd .
K and L K be di erential elds. A eld isomorphism : L ! M is a di erential K -isomorphism, if (a) = a for all a 2 K and (a) = (a) for all a 2 L. If M = L then is a di erential K -automorphism.
De nition 3.2.2 Let M
Consider the system of linear di erential equations (A) : y = Ay, where A is a n n-matrix with coecients in K . De nition 3.2.3 Di erential eld (L; L ) is called a Picard-Vessiot extension of
K associated with (A) if 1. L K and L jK = . 2. CL = CK = C .
30
3. (A) has n linear independent solutions over C in Ln . 4. L is minimal with respect to the conditions 1,2 and 3 or equivalently if U = (uij )i;j=1;:::;n 2 Lnn is a fundamental matrix of the system (A) then L = K (u11; : : : ; unn).
Theorem 3.2.4 For every system of linear di erential equations (A) there exists
a Picard-Vessiot extension L and this extension is unique up to di erential K isomorphism.
De nition 3.2.5 The di erential Galois group DGal(L=K ) is the group consisting of all the di erential K -automorphisms of L.
If U 2 Lnn is a fundamental matrix of system (A) and 2 DGal(L=K ), then it's obvious that also (U ) is a fundamental matrix of system (A). Hence 1 0 1 0 u ( u 11 ) (u1n ) 11 u1n B .. ... C ... C CA T ; CA = BB@ ... (U ) = B @ . (un1) (unn) un1 unn where T 2 Gl(n; C ). So the elements of the di erential Galois group act as C -linear maps on the space of solutions V = fc1u1 + + c1 un j c1; : : : ; cn 2 C g. (ui = (u1i; : : : ; uni)t 2 Ln ). But even a stronger statement holds. Theorem 3.2.6 DGal(L=K ) is a linear algebraic group over the eld of constants C and dimC DGal(L=K ) = deg trK L = deg trK (u11; : : : ; unn): (By deg tr we mean degree of transcendence.) Further the Galois correspondence is of importance. Theorem 3.2.7 Suppose G = DGal(L=K ). Then the following statements holds: 1. (8 2 G : (a) = a) ) a 2 K: 2. If H is an algebraic subgroup of G such that K = fa 2 L j 8 2 H : (a) = ag then H = G. 3. There is a 1-1 correspondence between algebraic subgroups H and di erential sub elds M . H = DGal(L=M ) , M = fa 2 L j 8 2 H : (a) = ag: 4. Under this correspondence normal subgroups H G correspond to PicardVessiot extensions M K and vice versa. And then we have DGal(M=K ) = G=H:
31
The rst proofs of the last two theorems and the existence and uniqueness up to di erential K -isomorphism of the Picard-Vessiot extension were given by E.R. Kolchin. (See [Kol 73], and [Lev90].) For more information about di erential Galois theory we refer to [Kap57] and [Sin89]. Let D = K h@ i be a skew polynomial ring consisting of all expressions P a @ =0 with a 2 K for i = 1; : : : ; n. The multiplication in D is completely xed by the relation @a = a@ + a if a 2 K . To a system of linear di erential equations (A) : y = Ay with A 2 K we associate a left D-module M = K (In the sequel we mean by D-module M a left D-module M with dim M < 1.) in the following manner. If m 2 M then we de ne k
i
i
i
i
n
n
n
K
X
( a @ ) m := k
i
X k
i
i
=0
a ( A) m: i
i
i
=0
Conversely it's possible to associate a system of linear di erential equations to a D-module M with a xed K -base E = fe1 ; : : : ; e g in a natural way. Namely if E = fe1 ; : : : ; e g is a K -base of D-module M then there exist a 2 K such that @e = P a e for j = 1; : : : ; n. Let A = (a ) =1 . If m = P m e 2 M n
n
ij
n
n
i
=1
ij
j
ij
n
i
=1
;:::;n
P P a m e . Hence =1 =1 1 0 0 1 0 m ( m 1 1) BB .. CC BB .. CC BB m.. 1 @ . A 7 ! @ . A A@ . m (m ) m
then @ (m) = P (m )e j
i;j
i
n
=1
i
i
n
i
ij
i
i
i
j
j
n
n
n
1 CC A
describes @ in coordinates. Now the system (A) corresponding to the matrix A is the system of linear di erential equations associated to D-module M with xed K -base E . Let (A~) be a system of linear di erential equations associated to the same D-module with an other xed K -base E~ = fe~1; : : : ; e~ g and let A~ 2 K be the corresponding matrix. Suppose e~ = P t e for i = 1; : : : ; n, where t 2 K =1 for all i; j . Then the matrix T = (t ) =1 2 K is invertible and we have the following relation. A~ = TAT 1 + (T )T 1: The systems (A) and (A~) corresponding to the matrices A and A~ are de ned to be equivalent if the above relation holds for a certain invertible matrix T 2 K . In that case if U 2 L is a fundamental matrix of (A) then TU 2 L is a fundamental matrix of the system (A~). Now it's obvious that the solution spaces V , V~ of the systems (A), (A~) are equivalent as representation spaces of the di erential Galois group DGal(L=K ). And it is also clear that two Picard-Vessiot extensions L, L~ associated with two equivalent systems (A); (A~) are di erential n
n
n
n
ij
i
j
ij
j
ij
i;j
;:::;n
n
n
n
n
n
n
32
n
n
K -isomorphic. Hence it is allowed to write DGal(M ) instead of DGal(L=K ) if L is a Picard-Vessiot extension associated with D-module M . Recapitulating, we can view systems of linear di erential equations as Dmodules with a xed K -base and D-modules as an equivalence class of systems of linear di erential equations. We have introduced D-modules, because we prefer to study some properties of systems of linear di erential equations K -base independently. Let L = LM be a Picard-Vessiot extension associated with the D-module M . Then L K M becomes a Lh@ i-module if we de ne @ (a m) = L(a) m+a @ (m). Then V = VM = ker(@; L K M ) is the vector space of solutions on which the di erential Galois group G = DGal(M ) = DGal(L=K ) faithfully acts. We note that L C V = L K M , because if the elements v ; : : : ; vn 2 V are linear independent over C , then it is not dicult to demonstrate that they are also linear independent over L. The di erential Galois group acts on LM C V by (a v) = (a) (v) if 2 G. By taking the G-invariant elements of LM C V we recover M , i.e. M = (LM C V )G. Moreover there is a 1-1 correspondence between D-submodules M~ M and G-stable subspaces V~ V . M~ 7 ! L K M~ 7 ! ker(@; L K M~ ) = V~M 1
~
V~ 7 ! LM C V~ 7 ! (LM C V~ )G = M~ V Because of this correspondence it's possible to replace M , N , D-(sub)modules and K by V , W , G-stable (sub)spaces in the next de nitions and lemma's of this section. We denote M ' N if M and N are isomorphic as D-modules and V ' W if V and W are equivalent as representation spaces of G. De nition 3.2.8 D-module M is simple if there is no D-submodule M~ such that f0g M~ M . ~
~
~
(In this paper we use the symbol exclusively to denote a strict inclusion.) De nition 3.2.9 D-module M is undecomposable if there exist no D-submodules M , M such that f0g M ; M M and M = M M . 1
2
1
2
1
2
The next two classical lemmas will be used in this paper. Sometimes even tacitly!
Lemma 3.2.10 (Krull-Schmidt). Let M be a D-module, M = iL Mi and M = k
Ll M~j , where the Mi and the M~j are nontrivial undecomposable D-submodules. =1
j =1
Then k = l and there is a permutation 2 Sk such that Mi ' M~ (i) for i = 1; : : : ; k .
33
If N , M are D-modules and N M , then there exist a sequence of D-modules N = N N Nk = M such that Ni=Ni is a simple D-module for i = 1; : : : ; k. Such a sequence is called a Jordan-Holder sequence from N to M . Lemma 3.2.11 (Jordan-Holder). Let M be a D-module and let f0g = M M Mk = M and f0g = M~ M~ M~ l = M be JordanHolder sequences. Then l = k and there is a permutation 2 Sk such that Mi =Mi ' M~ i =M~ i for i = 1; : : : ; k 0
1
1
0
1
0
1
( )
( )
1
1
De nition 3.2.12 Let f0g = M M Mk = M be a JordanHolder sequence from f0g to M and let S be a simple D-module then we de ne multS M := #fi 2 f1; : : : ; kg j Mi =Mi ' S g. 0
1
1
The correctness of this de nition is an immediate consequence of the JordanHolder lemma. Suppose M is a D-module. Let M denote its dual space, that is the the vector space consisting of all the K -linear maps l : M ! K . It is possible to give M a D-module structure. De ne (@ l)(m) = (l(m)) l(@ (m)) for all m 2 M if l 2 M . De nition 3.2.13 D-modules M and M~ are cogredient if there exists a one dimensional D-module N such that M ' N K M~ and they are contragredient if there exists a one dimensional D-module N such that M ' N K M~ .
3.3 Siegel normality This section contains a brief description of the results concerning Siegel normality in [BBH88]. The de nition of the notion of Siegel normality is rather subtle. Consider the square matrix A := (aij )i;j ;:::;n 2 K nn . Sometimes the matrix A = (aij )i;j ;:::;n splits into r submatrices =1
=1
At = (aij )i;j
=1;:::;nt
;
t = 1; : : : ; r;
n + + nr = n: 1
That is the square submatrices At are located along the main diagonal of A and all of the entries outside these submatrices are zero: 0 A 0 0 1 BB 0 A 0 CC A=B B@ ... ... . . . ... CCA : 0 0 Ar 1
2
We denote by (A) the system of di erential equations which corresponds to the matrix A. 34
De nition 3.3.1 System (A) is called linear Siegel normal if for any solution f = (f1 ; : : : ; f ) of (A) , f = (f 1 ; : : : ; f ) and any p 2 K the relation p1 f1 + + p f = 0 implies for each i = 1; : : : ; r that either p = 0 or f = 0. t
r
i
i
ini
t
r r
i
If f = (f1; : : : ; f ) is a solution of system (A) then we let f 1 tuple consisting of all the N 1 monomials r
ni
i
t
l ;:::;lr
i
denote a N 1
l ;:::;lr
-
l ;:::;lr
f1111 j
f1 1 11 f2121 j n
j
n
f2 2 22 j n
n
f 11 jr
r
f
jrnr
rnr
in the component functions of the solution f = (f1 ; : : : ; f ) with P j = l for =1 i = 1; : : : ; r: (The order of components in this N 1 -tuple doesn't matter.) r
ni
t
ik
i
k
l ;:::;lr
De nition 3.3.2 System (A) is called homogeneous algebraic Siegel normal if for any N 1 and any solution f = (f1 ; : : : ; f ) of (A) and any p 1 2 K 1
X
l1
r
p1
l ;:::;lr
+ + =
lr
t
Nl ;:::;lr
l ;:::;lr
f1
l ;:::;lr
=0
N
implies for all r-tuples l1 ; : : : ; l with l1 + : : : + l = N that either p 1 f1 = 0. r
r
= 0 or
l ;:::;lr
l ;:::;lr
These de nitions can be translated easily into terms of D-modules. De nition 3.3.3 Let E = fe1 ; : : : ; e
be a K -base of the D-module M . Say , where the M are D-submodules with M = span (E ), E E . =1 We assume that r is taken as large as possible. Now M is called linear Siegel normal with respect to this K -base E if for all v 2 V = ker(; L M ), say v = P v 2 V with v 2 L M for i = 1; : : : ; r and for all K -linear maps =1 l : M ! K extended as L-linear map (L M ) ! L (i.e. l 2 (L M ) ) we have l(v) = 0 implies for all i: v = 0 or l(E ) = f0g.
M = LM
ng
r
i
i
i
K
i
i
i
K
r
i
i
K
i
i
K
i
K
G
i
De nition 3.3.4 Let E = fe1 ; : : : ; e
be a K -base of the D-module M . Then M is called homogeneous algebraic Siegel normal with respect to the K -base E if S M (t-th symmetric tensor power) is linear Siegel normal with respect to the K -base S E = fe11 e22 e j t 2 N 0; P t = tg for each t 1. ng
t
t
t
t
s
s
s
tn n
n
i
i
=1
i
Unfortunately being Siegel normal is a property, which is not invariant under base transformations, therefore the following base independent de nition is added. De nition 3.3.5 D-module M is called linear (homogeneous algebraic resp.) Siegel normal if there exists a K -base E of M such that M is linear (homogeneous algebraic resp.) Siegel normal with respect to this K -base E .
35
Theorem 3.3.6 Let M be a D-module. Then the following statements are equivalent: 1. M is linear Siegel normal. 2. The D-submodules Mi are simple for i = 1; : : : ; r and Mi 6' Mj if i 6= j . 3. The G-stable subspaces Vi = ker(@; L K Mi ) are simple for i = 1; : : : ; r and Vi 6' Vj if i 6= j .
Proof. The equivalence 2 , 3 is an immediate consequence of the 1-1 correspondence between D-submodules M~ M and G-stable subspaces V~ V . A proof of 1 , 2 will be given. 1 ) 2 . Suppose Mi is not simple, then there exists a D-submodule M~ i with f0g M~ i Mi. Let l be a K -linear map M ! K with M~ i = ker(l). Extend l as L-linear map L K M ! L. Let V~i = ker(@; L K M~ i) and let v~i 2 V~i nf0g. Then one has l(~vi ) = 0, v~i 6= 0 and l(Mi ) 6= 0. Hence M is not linear Siegel normal if Mi is not simple. Suppose Mi ' Mj . Then there exists a K -linear map : Mi ! Mj such that 8q 2 D : (q:mi ) = q: (mi). Extend as L-linear map L K Mi ! L K Mj . Let li be a K -linear map M ! K extended as L-linear map L K M ! L such that li(Mk ) = f0g if k 6= i and li(vi) 6= 0 for a vi 2 Vi. Let vj = (vi). De ne lj = li and l = li lj . Then l(vi + vj ) = 0, vi 6= 0, vj 6= 0, l(Mi ) 6= f0g and l(Mj ) 6= f0g. Hence M is not linear Siegel normal if Mi ' Mj and i 6= j . 2 ) 1 : Let l : M ! K be a K -linear map. Extend l as a L-linear map (L K M ) = (L C V ) ! L and de ne W = fv 2 V j l(v) = 0g. Clearly, W is a G-stable subspaceLof V . Consider also the corresponding D-submodule N = (L C W )G. N = Mi with I f1; : : : ; rg, because the D-modules Mi i2 I are simple for i = 1; : : : ; r and Mi 6' Mj if i 6= j . Of course l(N ) = f0g and also r l(Mi ) = f0g for i 2 I . Suppose v 2 V , v = P vi with vi 2 Vi = ker(; L K Mi) i=1 for i = 1; : : : ; r. If l(v) = 0 then we have vi = 0 for i 2 f1; : : : ; kg n I and l(Mi ) = 0 for i 2 I . Hence M is linear Siegel normal if the Mi are simple and mutual non-equivalent. 2 The main results in [BBH88] are the next two theorems. Theorem 3.3.7 Let M be a D-module. Suppose dimK M = n 2. Further let G = DGal(M ). Then the following statements are equivalent: 1. M is simple and homogeneous algebraic Siegel normal. 2. G contains Sl(n; C ) or Sp(n; C )
Theorem 3.3.8 Let M be a D-module and let G = DGal(M ). Then the following statements are equivalent:
36
1. M is homogeneous algebraic Siegel normal.
2. M = ( L Mi ) L( L Nj ), where the Mi are are non-cogredient and nonj =1 i=1 contragredient simple linear Siegel normal D-modules with dimK Mi 2 and the Nj are one dimensional D-modules satisfying the following condition: r
s
S k1 N1
Ps
S ks Ns Ntriv '
implies either k1 ; : : : ; ks = 0 or ki 6= 0. (Here Ntriv denotes the trivial i=1 D-module, i.e. N = Ke with @e = 0, and if k 2 Z 1 then S k N denotes the D-module S k N .) 3.4
Shidlovskii irreducibility
Consider the n n system of linear di erential equations 0 1 0 10 1 y1 C y a 1 11 a1n B BB .. CC BB .. C d . . .. C (A) : dz @ . A = @ . A B@ .. CA yn yn an1 ann where aij 2 K for all i; j . De nition 3.4.1 System (A) is called linear Shidlovskii irreducible if for any solution f = (f1 ; : : : ; fn)t of (A) and any pi 2 K the relation p1 f1 + + pnfn = 0 implies for each i = 1; : : : ; n that either pi = 0 or fi = 0. In other words system (A) is linear Shidlovskii irreducible if the nonzero components of every solution f = (f1 ; : : : ; fn )t are linear independent over K . De nition 3.4.2 System (A) is called homogeneous algebraic Shidlovskii irreducible if the nonzero components of any solution are homogeneous algebraic independent over K . Compare these de nitions of Shidlovskii irreducibility with the corresponding de nitions of Siegel normality. It is immediately clear that being linear (homogeneous algebraic resp.) Shidlovskii irreducible is a weaker property of systems of linear di erential equations than being linear (homogeneous algebraic resp.) Siegel normal. Translating above de nitions into terms of D-modules we get: De nition 3.4.3 Let E = fe1 ; : : : ; en g be a K -base of the D-module M . M is called linear Shidlovskii irreducible with respect to the K -base E if for all K -linear maps l : M ! K extended as L-linear map L K M ! L (i.e. l 2 (L K M )G ) n and all v = P vi ei 2 V = ker(@; L K M ) we have l(v) = 0 ) 8i : vi = 0 or i=1 l(ei ) = 0. 37
De nition 3.4.4 Let =
be a K -base of the D-module M . Then M is called homogeneous algebraic Shidlovskii irreducible with respect to the K base E if S t M is linear Shidlovskii irreducible with respect to the K -base S t E for each t 1. E
n
fe1 ; : : : ; e g
And now the base independent de nition: De nition 3.4.5 is called linear (homogeneous algebraic resp.) Shidlovskii irreducible if there exists a -base such that is linear (homogeneous algebraic resp.) Shidlovskii irreducible with respect to this -base . M
K
E
M
K
Theorem 3.4.6 Let
E
be a D-module. Equivalent statements are: 1. M is linear Shidlovskii irreducible. 2.
M
has a K -base E = fe1 ; : : : ; eng such that for every D-submodule N M there exists a subset EN E such that N = spanK EN .
M
3. There are only nitely many D-submodules N M . 4.
has a C -base F = ff1 ; : : : ; fng such that for for every G-stable subspace W V there exists a subset FW F such that W = spanC FW . V
5. There are only nitely many G-stable subspaces W
V
.
Proof. The equivalence 3 , 5 can be proved easily by the 1-1 correspondence between D-submodules M~ M and G-stable subspaces V~ V . The proofs of 2 , 3 and 4 , 5 are completely analogous. A proof of 1 , 2 , 3 will be given. 1 ) 2 . Assume that D-module M is linear Shidlovskii irreducible with n respect to the K -base E = fe1 ; : : : ; eng. Suppose further that v = P viei 2 V i=1 and E (v) = fei 2 E j vi 6= 0g. We associate to v a D-submodule M (v) which is the smallest D-submodule of M such that L K M (v) contains v. We note that
( )=(
M v
Consider a -linear map : !
C ! . Clearly, ( ) = 0 ) ( K
L
V
l
L
C
L
M
K
span
extended as -linear map K )) = f0g. So we get: C C( L
l L
l v
G C (G:v ))
span
L
( ( )) = f0g , ( ) = 0 , 8e 2E v ( i) = 0 , (
l M v
l v
( )l e
i
M
=
G:v
l span
K E (v )) = 0:
The second equivalence holds because is linear Shidlovskii irreducible with respect to the -base . Hence ( ) = K ( ) for all 2 . Further if is a -submodule then = ( ) + + ( s ) for certain s2 . From this we get that = f such that for each n g is a -base of -submodule we have = K N for a subset N . M
K
N
M
V
D
E
M v
D
N
E
N
N
span
e1 ; : : : ; e
span
E v
M v1
E
38
v
M v
K
v1 ; : : : ; v
M
E
V
E
) 1 . We assume that E = fe1; : : : ; eng is a K -base of M which satis es the property of statement 2 : For each v 2 V let M (v) = (L C spanC (G:v))G. By assumption there exists a subset E (v) E such that M (v) = spanK E (v). If n v = P viei then 8i : ei 2 E n E (v) ) vi = 0 and if l is a K -linear map l : M ! K i=1 extended as L-linear map L K M ! L then we have l(v) = 0 ) l(M (v)) = 0 ) l(ei ) = 0 if ei 2 E (v): 2
Hence M is linear Shidlovskii irreducible. 2 ) 3 . Trivial. 3 ) 2 . De ne N = fN j N M is a D submodule g. Successively will be proved: ~ Q 2 N with N Q and N~ Q we have: N = N~ if and only A. For any N; N; ~ ) for all simple D-modules S . if multS (N=Q) = multS (N=Q B. (N ; +; \) is a nite distributive lattice. C. There exists a K -base E = fe1; : : : eng of M such that for all N 2 N we have N = spanK EN , where EN E . A. ()) Trivial. (() We will prove this by induction with respect to the length k of the Jordan-Holder sequence from Q to N . If k = 0 there is nothing left to prove. Assume the statement holds for k = l 1. Let Q = N0 N1 Nl = N and Q = N~0 N~1 N~l = N~ be two JordanHolder sequences from Q to N respectively from Q to N~ . Let q be minimal with respect to the condition that N~q =N~q 1 ' N1 =N0. The existence of such a q is ~ ) = multN1 =N0 (N=Q) 1. From the Jordanevident because multN1 =N0 (N=Q Holder lemma we get N~q 1 + N1 N~q 1, because multN1=N0 (N~q 1=Q) = 0 < 1 = multN1 =N0 ((N~q 1 + N1)=Q). So N~q 1 + N1 = N~q because the niteness of the number of submodules of M implies that any quotient of M cannot contain two copies of the same submodule. Hence N N1 ; N~ N1 and for all simple D~ 1). Now we can apply the induction modules we have multS (N=N1) = multS (N=N hypothesis and conclude N = N~ . B. We have to prove the distributivity of the lattice N . First we will prove
multS (N + N~ ) = max(multS N; multS N~ ) if S is a simple D-module and N; N~ 2 N . We will prove multS (N + N~ ) = max(multS N; multS N~ ) by induction with respect to the the length h of the Jordan-Holder sequences from f0g to N . If h = 0 everything is clear. Assume the statement holds for h = k 1. Let f0g = N0 N1 Nk = N be a Jordan-Holder sequence from f0g to N . As a consequence of the proof of 39
statement A) we have N1 N~ if multN1 N~ 1. Hence multS (N + N~ ) = multS (N + (N~ + N1 )) = multS (N=N1 + (N~ + N1)=N1) + SN1 = max(multS N=N1; multS (N~ + N1 )=N1) + SN1 = max(multS N; multS (N~ + N1 )) = max(multN1 N; multN1 N~ ); where SN1 = 0 if S 6' N1 and SN1 = 1 if S ' N1 . The proof of multS (N \ N~ ) = min(multS N; multS N~ ) for each simple Dmodule S and any N; N~ 2 N is dual analogous. (In that case Jordan-Holder sequences from N to M and from N~ to M have to be considered.) ~ N^ 2 N . Then we have for each simple D-module S Let N; N; multS N \ (N~ + N^ ) = min(multS N; multS (N~ + N^ )) ~ ordS N^ )) = min(multS N; max(multS N; = max(min(multS N; multS N~ ); min(multS N; multS N^ )) = max(multS (N \ N~ ); multS (N \ N^ )) = multS ((N \ N~ ) + (N \ N^ )) ~ N^ 2 N : N \ (N~ + N^ ) = Now we get as a consequence from A) that 8N; N; ~ N^ 2 (N \ N~ )+(N \ N^ ). In the same way it is possible to demonstrate that 8N; N; N : N + (N~ \ N^ ) = (N + N~ ) \ (N + N^ ). Hence (N ; +; \) is a nite distributive lattice. C. Suppose N = fM0 ; M1 : : : ; Msg where Mi Mj ) i < j . Thus in particular M0 = f0g and Ms = M . For any i 2 f0; : : : ; sg a set Ei satisfying the next two conditions will be constructed. i.
Ei is a K -base of D-module
Pi Mj .
j =0
ii. 8j 2 f0; : : : ; ig : Mj := spanK (EMj ) with EMj Ei. De ne E0 := ;. Of course E0 satis es the above conditions. Suppose i 1. Now we assume that for all j 2 f0; : : : ; i 1g a set Ej satisfying the conditions i) and ii) has been constructed. If Mi = Mk + Ml with 0 < k; l < i, then de ne Ei := Ei 1. Obviously, Ei satis es the conditions i) and ii) if Ei 1 satis es these conditions. If Mi 6= Ml +Mk for all k; l with 0 < k; l < i, then there exists a unique h 2 f0; : : : ; ig such that Mh Mi and Mi=Mh is a simple D-module. Suppose that EMi = EMh [ E~i (disjunct union) is a K -base of Mi , where EMh Ei 1 is a K -base of Mh. De ne Ei := Ei 1 [ E~i . Of course Ei satis es condition ii). Now 40
we assume that Ei doesn't satisfy condition i) and derive a contradiction. In any i case we have spanK Ei = P Mj . If the elements of Ei satisfy a linear dependence j =0 relation over K then there must be a m 2 (spanK Ei 1 \ spanK E~i) n f0g. In i 1 other words 0 6= m 2 ( P Mj ) \ Mi . But m 62 spanK EMh \ spanK E~i. So j =0 i 1
If P (Mj \ Mi) = Mi then P Mj Mi and so there j =0 j =0 j =0 exists k; l : 0 < k < l < i such that Mi = Mk + Ml (It is possible to choose Mk and Ml in such a way that Mi=Mk and Mi=Ml are simple D-modules.), but this is contradictory to an earlier assumption. Summarizing: if condition i) is not ful lled then we have
Mh ( P Mj ) \ Mi . i
1
i
i X 1
Mh ( Mj ) \ Mi = (distributivity!) j =0
1
i X (Mj \ Mi ) Mi; 1
j =0
contradicting that Mi =Mh is a simple D-module. Now we have obtained the desired contradiction and thus Ei satis es condition i). We conclude that it is possible to construct successively sets Ei satisfying conditions i) and ii). Hence the set E = Es constructed this way is a K -base of M , which has the required property that 8N 2 N : N = spanK EN with EN E . This nishes the proof of the theorem. 2
Theorem 3.4.7 Let M be a D-module. Equivalent statements are: 1. M is homogeneous algebraic Shidlovskii irreducible. 2. M has a K -base E = fe1 ; : : : ; eng such that for all t 1 and for each Dmodule N S t M there exists a subset EN S t E such that N = spanK EN . 3. The vector space of solutions V has a C -base F = ff1 ; : : : ; fng such that for all t 1 and for each G-stable subspace W S t V there exists a subset FW S tF . such that W = spanC FW .
Theorem 3.4.8 Let E = fe1 ; : : : ; eng be a K -base of D-module M . Let l1 ; : : : ; ln be K -linear maps M ! K , extended as L-linear maps L K M ! L, such that li (ej ) = ij for i; j = 1; : : : ; n. Then we have:
1. The following statements are equivalent: (a) M is linear Shidlovskii irreducible with respect to the K -base E . (b) The vector space of solutions V has a C -base F = ff1 ; : : : ; fng such that: i. for every G-stable subspace W V there exists a subset FW F such that W = spanC (FW ).
41
ii. for every G-stable subspace W lj (fi ) = 0.
V
,we have fi
j
2 W; f
62 W )
2. The following statements are equivalent: (a)
is homogeneous algebraic Shidlovskii irreducible with respect to the -base E . (b) The vector space of solutions V has a C -base F = ff1 ; : : : ; fng such that: i. for all t 1 for every G-stable subspace W S t V there exists a subset FW S t F such that W = spanC (FW ). ii. for every G-stable subspace W V ,we have fi 2 W; fj 62 W ) lj (fi ) = 0. M
K
Proof. We will prove only statement 1. (a) ) (b). Because of the 1-1 correspondence between -stable subspaces and -modules we can choose and number 1 2 n in such a way that together with condition i) the following condition is satis ed: 8 f1 g : is a K f i gi2I -module , f g is a -stable subspace. Let be a -stable C i i2 I subspace, then there exists a -module such that = ( K ). Suppose = f g with f 1 g . Then = K i i2I C f i gi2I . If 62 then j ( ) = 0 and thus also j ( K ) = 0 and j ( ) = 0. From this we get j ( i) = 0 if 2 and 62 . (b) ) (a). We assume that has a -base = f 1 n g, which satis es the conditions i) and ii). Let 2 and suppose that )= C( . Let be a -linear map ! , extended as -linear map f g C i i2Iv
K ! such that ( ) = 0. From ii) we get j ( ) = 0 if 62 v . Consider ( ) = ( C ))G. ( ) is # v -dimensional and j ( ( )) = 0 if C( 62 v . So ( ) = K f i gi2Iv and ( ( )) = 0 ) ( i ) = 0 for 2 v . We conclude that is linear Shidlovskii irreducible with respect to the -base =f 1 n g, because j ( ) = 0 if 62 v and ( i ) = 0 if 2 v . 2 It is possible to formulate a theorem analogous to theorem 3.4.8 for Siegel normality. Theorem 3.4.9 Let be a -module and let f0g = 0 1 r = be a Jordan-Holder sequence. Then we have: G
W
V
D
N
M
f ;f ;:::;f
I
D
span
f
j
span
I
l
l
e
N
I
N
i
I
j
L
C
v
l
f
L
M
M v j
K
L
span
M v
F
f ;:::;f
span
K
l
G:v
span
M v
v
j
I
l
l M v
e
l e
l
M
2.
Lr
M =M
L
M =M
i=1 r
i
i
1
I
M v
i
I
K
e ;:::;e
1.
G:v
L
M
E
N
f
W
V
M
l v
L
I
span
I
V
span
l
M
G
ker @; L
W
N
e
V
W
;:::;n
l
f
span W
D
N
;:::;n
G
v
j
I
D
l e
i
M
M
I
M
M
is linear Siegel normal ) M is linear Shidlovskii irreducible.
is homogeneous algebraic Siegel normal ) M is homogeneous algebraic Shidlovskii irreducible.
i=1
i
i
1
42
. Suppose M is not linear Shidlovskii irreducible then it is not dicult to show that there exist a quotient of M whichr contains two copies of the same simple submodule. Hence the graded module L Mi=Mi must contain i r two copies of the same simple submodule. But then L M =M is not linear Proof.
1
1
=1
i
i=1
i
1
Siegel normal because of theorem 3.3.6. r L Mi=Mi is homogeneous algebraic Siegel normal. Let ti 2 N 2. Suppose i for i = 1; : : : ; r. Then according to lemma 2.1 in [BBH88] D-module St1 ;:::;t = r S t1 (M =M ) S t (Mr =Mr ) is simple and further St1 ;:::;t 6' St1 ;:::;t if P ti = i Pr ~ ti and (t ; : : : ; tr ) 6= (t~ ; : : : ; t~r ). Obviously if S is a simple D-module then i r mult S t ( L M =M ) = mult S t M for all t 2 Z . Hence S tM is homogeneous 1
=1
r
1
r
0
1
=1
S
i=1
1
~
r
~r
=1
1
i
i
0
S
1
r algebraic Shidlovskii irreducible because S t ( L Mi=Mi ) = PL St1 ;:::;t . 2 i t t 1
=1
r
i=
We wish to remark that the converse of this theorem does not hold. For instance consider the C(z)h@ i-module M = C(z)e + C(z)e , with @e = 0 and @e = z e and its submodule M = C(z)e . It is not dicult to verify that M is linear and homogeneous algebraic Shidlovskii irreducible, but M M=M is neither linear nor homogeneous algebraic Siegel normal. 2
1
1
1
1
2
1
1
1
1
3.5 Examples In this section the results of the previous sections will be applied to get some concrete examples of systems of linear di erential equations which are homogeneous algebraic Shidlovskii irreducible but which are not homogeneous algebraic Siegel normal. We restrict ourselves to systems of linear di erential equations on which Shidlovskii's fundamental theorem is applicable (See section 1). Let 0 p < q, ; : : : ; p 2 C and ; : : : ; q 2 C n Z . Then we de ne the generalized hypergeometric function 1 X 1 ;:::; 1 ;:::; (z) = (( ))n ((p))n ( q z p ) q p n; n q n n where ( ) = 1 and ( )n = ( + 1) ( + n 1). The function 1 ;:::; 1 ;:::; satis es the q-th order di erential equation ! q p Y Y r ( y) ( + r(i 1)) z ( + ri) y = rq ( 1) (q 1) ; 1
1
p;
0
1
q
(
)
1
=0
0
p;
i=1
q
1
i=1
where r = q p and = z dzd . (See [Shi89], chapter 5, x1.) If ; : : : ; p 2 Q and ; : : : ; q 2 Q n Z then 1;:::; 1;:::; is an E -function 1
1
0
p;
q
43
First we consider the special case p = 0 and q = 1. Suppose 2 C n Z then 1 (z) = P n zn satis es the linear di erential equation 0
n=0
1 ( )
y0 = z + 1z y + z 1
d ): (0= dz
Theorem 3.5.1 Let 2 QnZ . Suppose is a nonzero algebraic number. Then ( ) is a transcendental number with an e ective measure of transcendence. 0
Proof. Consider the system of linear di erential equations ! z+1 0 ! y ! d y 1 1 z (A^) : : y = y 0 0
dz
2
2
! 0 ^ Let (z) = z Then U = 0 1 is a fundamental matrix of the system (A^) . Let L = C(z)( ). Then L is a Picard-Vessiot extension of C(z) associated ! ( ) 0 ^ with the system (A^). If 2 DGal(L=C(z)) then (U^ ) = U: 0 1 where is an character of the di erential Galois group. For all m 1 we have m 6= 1 because m 62 C(z). Hence the system (A^) is homogeneous algebraic Siegel normal as a consequence of theorem 3.3.8 and so the system of linear di erential equations ! z ! y ! d y z z ~ (A) : dz y = y : 0 0 is homogeneous algebraic Shidlovskii irreducible because of theorem 3.4.9 . Further ( ; 1)t is a solution of the system (A~) consisting of nonzero E -functions. Hence the numbers ( ); 1 are homogeneous algebraic independent with an effective measure of homogeneous algebraic independence and so ( ) is transcendental with an e ective measure of transcendence. 2 Theorem 3.5.2 Let ; : : : ; p 2 Q, ; : : : ; p 2 Q n Z, 0 p < q. Suppose that q 2 and that at least one of the following conditions holds. 1. i j 62 Z for 1 i p, 1 j q and the sums i + j (1 i j q) are all distinct modulo Z. 2. p = 0, q = 2 or q is odd and there is not a permutation 2 Sq and a divisor d > 1 of q such that i = i + d (mod Z) for i = 1; : : : ; q. Let be a nonzero algebraic number. Then the numbers 1
ez .
+1
1
1
2
1 2
1
1
( )
1
1 ;:::;p 1 ;:::;q ( ); : : : ; p1;:::;p 1;:::;q ( ) (
;
1)
;
are algebraic independent with an e ective measure of algebraic independence.
44
Proof. Consider the homogeneous linear di erential equation ! p q Y Y r ( + r(i 1)) z ( + ri) y = 0 i=1
i=1
The q q system corresponding to this di erential equation is simple and homogeneous algebraic Siegel normal. For a proof of this statement we refer to proposition 4.4 and the proof of theorem 4.5 in [BBH88]. Hence the (q + 1) (q + 1) system of di erential equations corresponding to (y) is homogeneous algebraic Shidlovskii irreducible. And so the numbers 1 ;:::;p ;1 ;:::;q ( ); : : : ; 1 ;:::;p ;1 ;:::;q ( ) (p
1)
are algebraic independent with an e ective measure of algebraic independence.
2
Let ; 2 C n Z . Then we de ne the function 1 X ( 1)n z n K; (z ) = ( ) ; n ( + 1)n ( + 1)n 2 1
2
=0
which satis es the non-homogeneous second order di erential equation 2 + 2 + 1 y0 + (1 + 4 )y = 4 y 00 + z
z2
z2
It is useful to de ne also the function K; (z) = K;(z), if ; 2 C n Z and 2 C. If ; 2 Q n Z and is an algebraic number then K; is an E -function. (See [Shi89], chapter 5, x1.) The function K; satis es the equation 2 + 2 + 1 y0 + ( + 4 )y = 4 : y 00 + ;
1
1
;
;
2
z
z2
z2
Lemma 3.5.3 Consider the corresponding homogeneous di erential equation (y)
y 00 +
2 + 2 + 1 y0 + ( + 4 )y = 0: 2
z
z2
Let G be the di erential Galois group over C(z ) associated with this di erential equation. If + 21 62 Z and 6= 0 then G contains Sl(2; C). Proof. Let C(z ) be the algebraic closure of C(z ). We will prove that G = y C(z)) ' Sl(2; C). Hence G contains Sl(2; C). After transforming the di erential equation by the substitution x = z+ y we get a new equation 1 ( ) 2 00 0 2 (z) x + x + ( )y = 0: 2
DGal(( );
z
x
45
The di erential equations (y) and (z) are equivalent over C(z). It is known that under the conditions of this lemma DGal((z); C(z)) ' Sl(2; C). (See [Kol68].) Sl(2; C) is connected and [DGal((z); C(z)) : DGal((z); C(z))] < 1. Hence DGal((z); C(z)) ' Sl(2; C) and thus G ' Sl(2; C). 2 ! 0 1 Let A; = . Then the 2 2 system of linear z2 z di erential equations (A; ) : y0 = A; y is the system corresponding to the second order linear di erential equation (y). ;
4
2
2
2
1
;
;
Lemma 3.5.4 Consider the sytems of linear di erential equations (A ; ) and (A ; ). Suppose that i i + 62 Z, i = 1; 2 and ; = 6 0. Suppose further 2
1
1 2
2; 2
1
1; 1
2
that the systems (A1 ;1 ;1 ) and (A2 ;2;2 ) are cogredient or contragredient. Then 12 = 22 and either (1 1) + (2 2) 2 Z or (1 1) (2 2) 2 Z. ! ! f g f g 2 2 1 1 Proof. Let U1 = f 0 g0 and U2 = f 0 g0 be fundamental matrices of 2 2 1 1 the systems (A1 ;1 ;1 ) and (A2 ;2 ;2 ). Then
3 deg trC z (f ; f 0 ; g ; g0 ; f ; f 0 ; g ; g0 ) 4; ( )
1
1
1
1
2
2
2
2
because the systems (A1 ;1 1 ) and (A2 ;2 2 ) are cogredient or contragredient and deg trC z (f ; f 0 ; g ; g0 ) = dimCDGal((A1 ;1 ;1 ); C(z)) = 3: ! ~ f g ~ i i Let f~i = z fi and g~i = z gi, i = 1; 2 and let Vi= f~0 g~0 , i = 1; 2, then i i ! ! 1 0 z 0 Vi = Mi Ui, where Mi = 0 1 if i+i = 0 and else Mi = z z . It is easy to verify that Vi is a fundamental matrix! of the system of di er0 1 2 y: According to [Kol68] ential equations (Bi ) : y0 = i z2 z deg trQ z (f~ ; f~0 ; g~ ; g~0 ; f~ ; f~0 ; g~ ; g~0 ) < 6 implies = and either ( ) + ( ) 2 Z or ( ) ( ) 2 Z. 2 ;
1
( )
;
1
1
i+ i
1
i+ i
i+ i
i+ i
2
( )
2
1
1
2
1
1
1
2
2
2
1
( i
i)
i+ i
1
2 1
2
2
1
2 2
1
1
2
Theorem 3.5.5 Suppose that i; i 2 Q n Z ; i = 1; : : : ; n; n 1; satisfy the conditions i i + 62 Z; and (i i ) + (i i ) 62 Z; (i i ) (i i ) 62 Z; i ; i = 1; : : : ; n ; i = 6 i . Let ; : : : ; m; m 1; be nonzero algebraic numbers such that i = 6 j if i =6 j . Then the 2mn numbers K ; (j ); K0 ; (j ); 1
1 2
2
1
2
2
1
2
1
1
2
2
2
1
1
2
1
i
i
i = 1; : : : ; n j = 1; : : : ; m; are algebraic independent with an e ective measure of i
i
algebraic independence.
Proof. Consider the (2mn + 1) (2mn + 1) system of linear di erential
46
equations
0 0 0 BB A1 ;.. 1;1 . . . ... .. . BB . B 0 Ai ;i;j 0 (A^) : y = B BB .. . ... . .. .. BB . 0 An ;n;m @ 0 0
0
0
0
1
0 ... C CC CC 0C y: ... C CC 0C A 0
System (A^) is homogeneous algebraic Siegel normal ! because of lemma 3.5.3, lemma 3.5.4 and theorem 3.3.8. Let Bi;j = 40 Then the system i i
0 0 0 B1;1 BB A1 ;.. 1 ;1 . . . . ... .. .. . BB . B 0 Ai;i ;j 0 Bi;j (A) : y = B BB .. . . ... . .. .. .. BB . 0 An ;n;m Bn;m @ 0 0
0
0
0
0
1 CC CC CC CC y CC A
is homogeneous algebraic Shidlovskii irreducible because of theorem 3.4.9. The 2mn + 1-tuple (K1;1 ;1 ; : : : ; Ki;i ;j ; : : : ; Kn;n ;m ; 1)t is a solution of the system (A) consisting of nonzero E -functions. Hence the numbers
K1 ;1;1 (1); : : : ; Ki;i ;j (1); : : : ; Kn ;n;m (1); 1 are homogeneous algebraic independent with an e ective measure of homogeneous algebraic independence and so the numbers
K1 ;1 (1); : : : ; Ki;i (j ); : : : ; Kn;n (m) are algebraic independent with an e ective measure of algebraic independence.
2
47
Chapter 4 An algorithm determining the di erence Galois group of second order linear di erence equations 4.1
Introduction
Let K = k(z), where k is a nite algebraic extension of Q and z is a transcendental variable. Furthermore, let be the k-linear automorphism given by (z) = z +1. In this article we will describe an algorithm for determining the di erence Galois group of the second order linear homogeneous di erence equation 2y +a y +by = 0 with a; b 2 K . If the di erence Galois group of an second order di erence equation does not contain the group Sl(2; Q ) then one can compute two linear independent Liouvillian solutions in a sequence space for this equation. This will be explained in the section which is devoted to examples. This algorithm can be considered as an analogue for second order linear di erence equations of Kovacic's algorithm for second order linear di erential equations (See [Kov86]). In [Pet92] an algorithm for nding rational solutions of the Riccati equation of arbitrary order is given. So the algorithm is not completely new. However the rationality result for solutions of the Riccati equation (Theorem 4.4.3) seems to be new. In section 4.2 we summarize the results of [PS96] concerning di erence Galois theory which are needed for our purposes. In section 4.3 we discuss rst order homogeneous linear di erence equations. The algorithm for determining the difference Galois group of second order homogeneous linear di erence equations is described in section 4.4. Section 4.5 is devoted to examples. 49
4.2 Preliminaries on di erence Galois theory
Let Q denote the algebraic closure of Q and let K^ = Q (z). Let be the Q -linear automorphism of K^ given by (z) = z + 1. Consider the system of di erence equations (A) : y = Ay, where A 2 Gl(n; K^ ). (We restrict ourselves to equations with A 2 Gl(n; K^ ) in order to guarantee that we get n independent solutions.) De nition 4.2.1 A ring R together with a xed automorphism R : R ! R is called a Picard-Vessiot extension of K^ associated with (A) if 1. R is a commutative ring, R K^ and R jK = . 2. The only R-invariant ideals are 0 and R. 3. There exists a matrix U 2 Gl(n; R) such that R(U ) = AU . (Such a matrix U is called a fundamental matrix for the system (A).) 4. R is minimal with respect to the conditions 1,2 and 3 or equivalently if U = (uij ) 2 Gl(n; R) is a fundamental matrix for the system (A) then R = K^ [u ; : : : ; unn; det U ]. ^
1 (
11
)
De nition 4.2.2 Consider the di erence rings R and R together with their 1
2
automorphisms R1 and R2 respectively. A ring homomorphism : R1 ! R2 is called a di erence{homomorphism if R1 = R2 .
Theorem 4.2.3 For every system of linear di erence equations (A) there exists
a Picard-Vessiot extension R and this extension is unique up to di erence K^ . isomorphism. Furthermore, we have (r) = r implies r 2 Q
From now on we will denote the extended automorphism R of a Picard{Vessiot extension R also by . De nition 4.2.4 The di erence Galois group DGal(R=K^ ) is the group consisting of all the di erence K^ -automorphisms of R. If U 2 GL(n; R) is a fundamental matrix of system (A) and 2 G = DGal(R=K^ ), then it is obvious that also (U ) is a fundamental matrix of system (A). Hence
0 BB (u.. ) (U ) = @ . (un ) 11
1
1 0 (u n) C B u ... C = B ... A @ un (unn)
11
1
1 un C ... C T ; A unn 1
1
where T 2 Gl(n; Q ). So the elements of the di erential Galois group act as Q -linear maps on the space of solutions V = fc u + + c un j c ; : : : ; cn 2 Q g. (ui = (u i; : : : ; uni)t 2 Rn). But even a stronger statement holds. 1
1
50
1
1
1
Theorem 4.2.5 G = DGal(R=K^ ) is a linear algebraic group over the eld of . constants Q If G is an algebraic subgroup of Gl(n; Q ), then we denote by G(K^ ) the sub-
group of Gl(n; K^ ) which is de ned by the same equations. That is G(K^ ) consists of the K^ {valued points of the algebraic group G. Theorem 4.2.6 Suppose G = DGal(R=K^ ). Then the following statements holds: ^ 1. (8 2 G : (a) = a) ) a 2 K: 2. If H is an algebraic subgroup of G such that K^ = fa 2 R j 8 2 H : (a) = ag then H = G. Two systems (A) and (B ) corresponding to matrices A; B 2 Gl(n; K^ ) are de ned to be equivalent if there exists a T 2 Gl(n; K^ ) such that B = (T )AT 1. In this case if U 2 Gl(n; R) is a fundamental matrix of system (A) then TU is a fundamental matrix of system (B ) and it is obvious that the solution spaces VA; VB of the systems (A) and (B ) are equivalent as representation spaces of the di erence Galois group. Conversely if the Picard-Vessiot extensions associated to the systems (A) and (B ) are di erence K^ -isomorphic and the solution spaces VA and VB are equivalent as representation spaces of the di erence Galois group then (A) and (B ) are equivalent systems of di erence equations. Theorem 4.2.7 Let G Gl(n; Q ) be the di erence Galois group associated to the system of di erence equations (A). Let G0 denote the identity component of G. Then the following statements hold. 1. G=G0 is a nite cyclic group. 2. There exists a B 2 G(K^ ) such that (A) and (B ) are equivalent systems. It is a conjecture that for every linear algebraic group G Gl(n; Q ) with G=G0 nite cyclic there exists a system of di erence equations which has G as di erence Galois group. One can show that this conjecture is true for n = 1; 2 by giving explicit examples for every possible group. Theorem 4.2.8 If A 2 G(K^ ) with G an algebraic subgroup of GL(n; Q ) such that for any proper algebraic subgroup H and for any T 2 G(K^ ) one has that (T )AT 1 62 H (K^ ) then G is the di erence Galois group of the system (A). If A satis es the conditions of theorem above then we say that system (A) is in standard form. This standard form is in general not unique. But one can read o fairly easily the di erence Galois group of a system of di erence equations if one knows already that this system is in standard form. In fact we will present an 51
algorithm which computes for a given equation an equivalent system in standard form. The standard form is also very suitable for nding closed form solutions if the di erence Galois group is not too big, that is does not contain the group Sl(2; Q ). We note that de nitions and theorems 4.2.1 up to 4.2.6 could have been stated for (more) general di erence rings. However theorems 4.2.7 and 4.2.8 are not valid for general di erence rings. For the proofs and much more information about di erence Galois theory we refer to [PS96].
4.3 First order di erence equations Let K = k(z), where k is a nite algebraic extension of Q. We consider the rst order homogeneous linear di erence equation (a) : y = ay, where a 2 K . Any equivalent equation is given by y = ff ay, where f 2 K^ . The di erence Galois group of (a) is a nite cyclic group of order n if there is an f 2 K^ such that ff a is a primitive nth-root of unity, otherwise the di erence Galois group is the multiplicative group Gm = Q . This statement is a consequence of theorems 4.2.7 and 4.2.8. Suppose a 2 K is given. We describe an algorithm which produces an element b 2 K such that the di erence equations (a) and (b) : y = by are equivalent and b = QP with P; Q 2 k[z] satisfying the condition that for all m 2 Z we have gcd(P; mQ) = 1. If b satis es these conditions then we call b reduced and the di erence equation (b) is automatically in standard form. The algorithm avoids factorization in k[z]. We assume that a = QP with P; Q 2 k[z] and gcd(P; Q) = 1. There exists an r > 0 such that the zeros of P and Q are in the disk jzj r. We can express r in terms of the coecients of P and Q. For instance if P = Pczc + + P z + P c d and Q = Qd zd + + Q z + Q then we can take r = [1 + max( P j PPci j; P j QQdj j)]. i j m For m 2 Z with jmj 2r we have gcd(P; Q) = 1. We start with m = 1 and compute g = gcd(P; (Q)). If g 6= 1 then one can write a = 1g g QP and we can go further with a~ = QP instead of a. This gives a reduction of the problem with respect to degrees of P and Q. If g = 1 then we try m = 1 and so forth. The algorithm stops if a is reduced to a constant or if all m 2 Z with jmj 2r are checked o . The above algorithm can easily be modi ed so that we also nd an f 2 K that insures that b = ff a is reduced. For another approach using the resultant of P (z) and Q(z + m) we refer to [Pet92]. This approach works for more general elds. ( )
( )
1
0
1
1 1
=0
=0
~ ( ) ~
~
~
( )
52
0
4.4 Second order di erence equations Consider the second order linear homogeneous di erence equation 2y+a y+by = 0, where a; b 2 K and b 6= 0. As usual we will identify this equation with the 2 2 system of di erence equations (A) : y = ( 0b 1a )y. Throughout this section we will denote the di erence Galois group of above system by G and the vector space of solutions on which G faithfully acts by V . We present an algorithm which produces a matrix B 2 Gl(2; K^ ) so that the systems (A) and (B ) are equivalent and (B ) is in standard form. Further a matrix T is computed such that B = (T )AT 1. Lemma 4.4.1 The algebraic subgroups G of Gl(2; Q ) that can occur as di erence Galois group are ) with G=G0 nite cyclic. 1. Any reducible subgroup of Gl(2; Q 2. Any in nite imprimitive subgroup of Gl(2; Q ) with G=G0 nite cyclic. 3. Any algebraic group containing SL(2; Q ). Proof. This is a consequence of theorem 4.2.7 part 1 and the well known classi cation of the algebraic subgroups of Gl(2; Q ). The algorithm is arranged in the following manner. Given a second order di erence equation we test which of the three cases above holds. After that we compute an appropriate equivalent system in standard form. 4.4.1
The Riccati equation
To a second order linear di erence equation 2y + a y + by = 0, where a; b 2 K and b 6= 0 we can associate a rst order non-linear di erence equation (y) : u (u) + au + b = 0. This equation is called the Riccati equation. If u is a solution of the Riccati equation then the di erence operator 2 + a + b factors as ( ub )( u). A rational solution of the Riccati equation corresponds to a line in the solutionspace that is xed by the di erence Galois group. Therefore we have the following theorem.
Theorem 4.4.2 The following statements hold: 1. If the Riccati equation has no solutions u 2 K^ then G is irreducible. 2. If the Riccati equation has exactly one solution u 2 K^ then G is reducible but not completely reducible.
3. If the Riccati equation has exactly two solutions u1 ; u2 2 K^ then G is completely reducible but G is not an algebraic subgroup of fc:Id j c 2 Q g.
53
4. If the Riccati equation has more than two solutions in K^ then the Riccati equation has in nitely many solutions and G is an algebraic subgroup of fc:Id j c 2 Q g. Proof. We can prove this theorem can very directly using theorems 4.2.7 and 4.2.8.
1. If G is reducible then according to theorem 4.2.7 there exists an upper triangular matrix B 2 Gl(2; K ) and a matrix T = ( tt tt ) 2 Gl(2; K^ ) such that B = (T )( 0 1 )T : One can easily verify that t 6= 0 and
b
11
12
21
22
1
a
22
2 K^ satis es the Riccati equation.
t21 t22
2. If u is a solution of the Riccati equation then let T = ( 1 uu 11 ). We have B = (T )( 0b 1a )T = ( u0 ) is an upper triangular matrix. Hence G is reducible because of theorem 4.2.8. If G is completely reducible then according to theorem 4.2.7 there exists a diagonal matrix D 2 Gl(n; K^ ) and a matrix T = ( tt tt ) 2 Gl(2; K ) such that D = (T )( 0 1 )T : One can easily verify that 1
t ; t ; t ; t 6= 0 and 11
12
21
course
22
6=
t21 t22
t11 t12
b
11
12
21
22
1
a 2 K satisfy the Riccati equation. Of
; tt1211 because det(T ) 6= 0. t 21 t22
u2
1
3. If u ; u are the solutions of the Riccati equation then let T = ( u1u1u2 u1 u2 ). u1 u2 u1 u2 0 1 u 0 We have B = (T )( b a )T = ( 0 u ) is a diagonal matrix. Hence G is completely reducible because of theorem 4.2.8. If G is an algebraic subgroup of fc:Id j c 2 Q g then there exists a matrix D = ( u0 u0 ) with u 2 K and a matrix T = ( tt tt ) 2 Gl(2; K^ ) such that D = (T )( 0 1 )T : One can easily verify that ct21 dt11 2 K^ 1
2
1
1
1
2
b
12
21
22
( + ) (ct22 +dt12 )
1
a
11
is a solution of the Riccati equation for c; d 2 Q with c 6= 0 or d 6= 0. And c1 t21 d1 t11 = cc22 tt2122 dd22 tt1112 if and only if c = c = 0 or dc11 = dc22 . c1 t22 d1 t12 (
+
)
(
+
)
( (
+ +
) )
1
2
4. If u ; u ; u are three distinct solutions of the Riccati equation then the system (A) : y = Ay is equivalent with the systems (B ); (B ) and 1
2
3
1
54
2
(B ) corresponding to the matrices ( u0 u0 ); ( u0 u0 ) and ( u0 u0 ) respectively. From the fact that (B ) and (B ) are equivalent it follows that u = ff u . Hence system (B ) is equivalent to the system (C ) : (y) = ( u0 u0 )y and G is an algebraic subgroup of fc:Id j c 2 Q g according to theorem 4.2.8. From the proof of statement 3 it follows that the Riccati equation has in nitely many solutions. Our task is to make an algorithm which computes all solutions u 2 K^ of the Riccati equation. The algorithm consists of a few steps. 2
1
1
3
( )
2
3
3
3
2
1
2
3
2
2
1. First we compute the rst two terms of all possible local formal solutions at 1. Let t = 1=z. Let be the automorphism of Q ((t)) given by (t) = t t . Then coincides with our former on Q (t) = Q (z). For the moment we consider a and b as Laurent series in t with coecients in k. We de ne the discrete valuation v : Q ((t)) ! Z in the usual way by 1 v( P cntn) = minfn j cn 6= 0g. In particular v(0) = 1. Elementary n 1 properties of v are v(u u ) = v(u )+v(u ) and v(u +u ) min(v(u ); v(u )). Hence if v(u ) 6= v(u ) then v(u + u ) = min(v(u ); v(u )). Another property of v is that v(u) = v( u). We distinguish two cases. +1
=
1
1
2
2
1
2
1
2
1
1
2
1
2
2
(a) 2v(a) < v(b). In this case there are exactly two solutions u; u~ 2 k((t)) of the Riccati equation. The rst solution u satis es v(au) = v(u (u)) < v(b) and v(u) = v(a) and the other solution u~ satis es v(au~) = v(b) < v(~u (~u)) and v(~u) = v(b) v(a). We can compute succesively the coecients of u and u~ by simply solving linear equations with coecients in k. Hence no eld extension of k is needed. (b) v(b) 2v(a). If v(b) is odd then there is clearly no solution of the Riccati equation. If v(b) is even and u 2 k((t)) satis es the Riccati equation then we have v(b) = v(u (u)) v(au) and v(u) = v(b). Let 1 1 l = v(b) We write a = P an tn (al = 0 is allowed), b = P bntn and 1 2
1 2
n l n l 1 P n u = unt and try to solve the equation (y) : u (u) + au + b = 0. n l =
=2
=
The (2l)th-coecient of (y) gives us the equation ul + al ul + b l p= 0. Let D = al 4b l . If D 6= 0 then we nd two solutions ul 2 k( D) of the quadratic equation. If D = 0 then we nd exactly one solution ul 2 k of the quadratic equation. The (2l + 1)th-coecient of (y) gives us the equation (2ul + al )ul lul + al ul + b l = 0. 2
2
2
2
2
+1
+1
2 +1
55
If D 6= 0 then p 2ul + al 6= 0. And we nd for both values of ul one value ul 2 k( D). If D = 0 then 2ul + al = 0. In this case if lul + al ul + b l 6= 0 there is no solution u 2 Q ((t)) of the Riccati equation. If lul + b l + al ul = 0 then we can compute one or two values for ul in at worst a quadratic eld extension of k by solving the quadratic equation ul ((2l +1)ul al )ul +( l(l +1)ul + al )ul + b l = 0 which is given by the (2l + 2)th-coecient of (y). Let Su be the set of all possibilities for the rst two terms of u of its Laurent expansion in t. We discovered that #Su 2, even if the Riccati equation has in nitely many solutions u 2 k(t). If we needed a quadratic extension in our computations then we will denote this quadratic extension for future use by k , otherwise k = k. 2. In this part of the algorithm we will determine the solutions u 2 k (z) of the Riccati equation. We rewrite the Riccati equation as F u (u)+ Gu + H = 0, where F; G; H 2 k[z] and gcd(F; G; H ) = 1. Write u = c QP with c 2 Q , gcd(P; Q) = 1 and P; Q monic. Let R denote the greatest monic divisor of Rp , where p; q are Q such that (R) divides P . Then one can write u = c Rq monic polynomials, gcd(p; (q)) = 1 and gcd( (R)p; Rq) = 1. The Riccati equation reads now: +1
2
2 +1
+1
2 +1
+1
2
+1
2 +1
+1
2
+1
1 2
+2
2 +2
2
2
( )
(z) : c F (R) (p)p + cG (R) (q)p + HR (q)q = 0: 2
2
Hence p is a monic divisor of H and q is a monic divisor of (F ). We de ne the sets Sp = fp 2 k [z] j p monic and pjH g and Sq = fq 2 k [z] j q monic and q j (F )g. Let e be the degree of the polynomial R. Writing RR as a Laurent series in t we get RR = 1 + et + O(t ), where O(t ) stands for the higher order terms. RR must be equal to uq cp . So for every possible combination u 2 Su , p 2 Sp and q 2 Sq with v (u) + v (q ) v (p) = 0 we compute an e 2 k from uq e e + +r z +r cp = 1+ et + O (t ). If e 2 Z then we write R = z + re z with indeterminates r ; r ; : : : re . After substituting this R in (z) we are left with a set of linear equations in the ri. In this way we nd all possible solutions u 2 k (z) of the Riccati equation. This ends this part of the algorithm. 3. In step 2 we have determined all solutions u 2 k (z) of the Riccati equation. Now we have to face the problem whether it is possible that there exist solutions u 2 Q (z) if there are no solutions u 2 k (z). A priori this is possible because F and H need not split in linear factors in k [z]. But it is also clear from the construction in step 2 that if the Riccati equation 1
2
1
2
( )
( )
2
2
( )
2
0
0
1
1
2
1
1
1
2
2
2
1
2
56
0
has a solution in Q (x), there is also a solution in l(x), where l k is the smallest eld which contains the splitting elds of F and H . But we have even a stronger rationality result. (Compare with [HP94], where a similar result is derived for di erential equations.) 2
1
Theorem 4.4.3
If the Riccati equation has a solution in is also a solution in a eld k~(z ) with [k~ : k] 2.
Q (z) then there
Let U be the set of solutions of the Riccati equation in l(x). The group Gal(l=k) acts on the set U . If #U = 1, say U = fug, then u is invariant under Gal(l=k). Hence u 2 k(z). If #U = 2 then the kernel of the map Gal(l=k) ! Aut(U ) is a subgroup of index 2. Hence u 2 k~(z), where k~(z) is a eld extension such that [k~ : k] 2. Suppose that #U = 1. According to Theorems 4.4.2 part 4 and 4.2.8 there exists a T 2 Gl(2; Q (z)) such that (T )( 0b 1a )T = ( u0 u0 ) for some u 2 Q (z). Suppose 2 Gal(Q =k) then ((T ))( 0b 1a )((T )) = ( (0u) (0u) ). There is a g 2 Q (z) such that (u) = gg u: This g is uniquely determined if we require that the numerator and denominator of g are monic. The map 7! g is a 1-cocycle. Hilbert 90 states that this 1-cocycle is trivial. See for instance [Ser68]. So we can replace u by u~ = hh u for a certain h 2 K^ such that u~ is invariant under Gal(Q =k). Hence u~ 2 k(z). Let T~ = hT . Then (T~)( 0b 1a )T~ = ( u0~ u0~ ). Suppose that C T~ in Gl(2; Q (z)) also satis es (C T~)( 0b 1a )(C T~) = ( u0~ u0~ ). Then (C ) = C and so C 2 Gl(2; Q ). For 2 Gal(Q =k) one has therefore (T~) = C T~ for some C 2 Gl(2; Q ). The map 7! C is a 1-cocycle for Gal(Q =k) acting on Gl(2; Q ). According to [Ser68] this 1-cocycle is trivial. This means that S = C T~ is invariant under Gal(Q =k) for a certain C 2 Gl(2; Q ). It follows that S 2 Gl(2; k(z)). In particular the Riccati equation has (in nitely many) solutions in k(z). 2 So if k = k, then we want to determinepall elds k~ l with [k~ : k] 2. It is obvious that k~ must be of the form k( s), where s is an algebraic integer in l. Recall that l is the splitting eld of the polynomial FH . After a suitable linear substitution z 7! nz , where n 2 Z we can assume that FH is monic and that the coecients of FH are algebraic integers. s must be a divisor of the discriminant of this p polynomial. This gives us a nite set of possible quadratic extensions k( s) of k in l. For each possible quadratic extension k~ we determine the sets Sp = fp 2 k~[z] j p monic and pjH g and Sq = fq 2 k~[z] j q monic and qj (F )g. After that we proceed as in 2. Proof.
1
1
(
)
( )
1
1
1
1
1
1
1
2
1
1
57
1
As we will see in Subsection 4.5.2 step 3 of the algorithm is not super uous. That means, if we did not need a quadratic extension of our eld of constants in step 1, it is possible that we really need a quadratic extension of our eld of constants in step 3 for solving the Riccati equation. We will provide an explicit example in Subsection 4.5.2. 4.4.2
G is reducible
If the Riccati equation has a solution in K^ , then we can compute an equivalent system (A) : y = Ay, where A has the form ( a0 db ). Moreover we can assume that b = 0 if the Riccati equation has two or in nitely many solutions. We denote by U the subgroup of Gl(2; Q) consisting of upper triangular matrices and by D the subgroup of diagonal matrices. Lemma 4.4.4 The reducible subgroups G Gl(2; Q ) with G=G nite cyclic are 1. The zero{dimensional groups Dk;l;e = f( 0 0 ) 2 D j k = 1; l = 1; eg g = 1g 0
and the one{dimensional groups
Uk;l;e = f( 0 ) 2 U j k = 1; l = 1; eg g = 1g for l; k; e 2 Z1 with gcd(k; l; e) = 1, where g = gcdk(k;l) and g = gcd(lk;l) . 2. The one{dimensional groups
Dm;n = f( 0 0 ) 2 D j mn = 1g and the two{dimensional groups
Um;n = f( 0 ) 2 U j mn = 1g for n; m 2 Z 3. The two{dimensional group D and the three{dimensional group U . Proof. If G D is a zero-dimensional subgroup with G=G0 nite cyclic then G is a subgroup of D~ k;l = f( 0 0 ) 2 D j k = 1; l = 1g with k; l 2 Z1 . Note that D~ k;l is isomorphic to Z=lZ Z=kZ as an abstract group. Assume that k; l are
58
minimal with respect to the condition that G D~ k;l. Then G = Z=lcm(k; l)Z because G is cyclic and G contains elements of order l and of order k but G does not contain elements of order greater than lcm(l; k). The subgroups of D~ k;l which are isomorphic to Z=lcm(k; l)Z as abstract groups are the groups Dk;l;e = f( 0 0 ) 2 D j k = 1; l = 1; eg g = 1g with l; k; e 2 Z and gcd(k; l; e) = 1, where g = gcdkk;l and g = gcd lk;l . If G a one-dimensional algebraic subgroup of D then G is equal to a group Dm;n = f( 0 0 ) 2 D j mn = 1g with m; n 2 Z. For all m; n 2 Z we have Dm;n=Dm;n is nite cyclic. If G is a two-dimensional subgroup of D then G = D. The classi cation of the reducible but not completely reducible subgroups G U with G=G nite cyclic is analogous to the classi cation of the completely reducible subgroups G D with G=G nite cyclic. 2 1
(
)
(
)
0
0
0
The algorithm continues as follows. Suppose the Riccati equation has exactly one solution in K^ . Then the di erence Galois group G is reducible but not completely reducible. If G is a proper subgroup of U , then according to Theorem 4.2.7 there must be a matrix T = ( f0 hg ) 2 U (K^ ) such that B = (T )AT = f ) 2 G(K^ ) . (a f 0 d hh First we want to determine whether the di erence Galois group is one of the one{dimensional groups described in part 1 of Lemma 4.4.4. We apply the algorithm for rst di erence equations to a and d. This yields a new equivalent system A~ = ( a0~ d~ ), where a~ and d~ are reduced. If a~ and d~ are constant then the system A~ is already in standard form. If both a~ and d~ are roots of unity, then the di erence Galois group is one of the groups Uk;l;e. Otherwise G is one of the groups described in part 2 or part 3 of lemma 4.4.4. Suppose now that that a~ or d~ is not constant. Then the di erence Galois group is not a group described in part 1 of Lemma 4.4.4. If a~ is constant and d~ is not constant then the system (A~) is already in standard form. The di erence Galois group G is the group Un; , if a~ is an n th primitive root of unity. Otherwise the di erence Galois group G is the group U . If d~ is constant and a~ is not constant then the system (A~) is also already in standard form. The di erence Galois group G is the group U ;n, if d~ is an n th primitive root of unity. Otherwise the di erence Galois group G is the group U . Suppose now that both a~ and d~ are reduced but not constant. The di erence Galois group is one of the groups Um;n if and only if there exists r; s 2 Z n f0g 1
( )
( )
0
0
59
with gcd(r; s) = 1 and an f 2 K such that a~r d~s ff is a root of unity. Without loss of generality we can assume that r > 0. Let Na and Nd denote the degree of the numerator of a~ and d~ respectively and let Da and Dd denote the degree of the denominator of a~ and d~ respectively. Suppose rst that there exist r; s 2 Z with gcd(r; s) = 1 and an f 2 K^ such that a~r d~s ff is constant. Then we must have rDa = sNd and rNa = sDd, because a~ and d~ are reduced. If rDa = sNd and rNa = sDd holds then we apply the algorithm for rst order di erence equations to a~r d~s to investigate whether there exist an f 2 K^ such that v a~r d~s ff is constant. If there exist such an f then we let T = ( f0 f0w ), where ~ is in standard form. v; w are chosen such that vr + ws = 1. Then B = (T )AT Suppose now that there exists an r 2 Z and an s 2 Z with gcd(r; s) = 1 and an f 2 K^ such that a~r d~s ff is constant. Then we must have rNa = sNd and rDa = sDd , because a~ and d~ are reduced. If rNa = sNd and rDa = sDd holds then we apply the algorithm for rst order di erence equations to a~r d~s to investigate whether there exist an f 2 K^ v such that a~r d~s ff is constant. If there exist such an f then we let T = ( f0 f0w ), ~ is in standard where v; w are chosen such that vr + ws = 1. Then B = (T )AT form. If none of the two cases above holds then the system (A~) is already in standard form and the di erence Galois group is the upper triangular group U . ( )
1
( )
( )
1
1
( )
1
( )
1
We will not discuss the completely reducible case because this case is analogous to the reducible but not completely reducible case.
4.4.3 G is imprimitive
If we did not nd a solution u 2 K^ for the Riccati equation then the di erence Galois group is irreducible. A group G is called imprimitive if G F = f( 0 0 ) j 6= 0g [ f( 0 0 ) j 6= 0g. In this subsection we determine whether G is imprimitive if we know already that G is not reducible.
Lemma 4.4.5 Suppose that the di erence Galois group G of the di erence equa-
tion (#) : 2 y + a y + by = 0 is irreducible. Then the following statements are equivalent. 1. G is an imprimitive group. 2. There exists an r 2 K^ such that the equations (#) and 2 y + ry = 0 are equivalent.
60
Proof. 1 ) 2: If G is imprimitive and not reducible then according to Theorems 4.2.7 and 4.2.8 there is a system (B ) : y = B y where B is of the form ( g0 f0 ) such that (B ) and (#) are equivalent. Let T = ( 10 f0 ). Then (T )BT 1 = ( (f0 )g 01 ). This matrix corresponds to a second order equation of the right form. 2 ) 1. The di erence equation 2y + ry = 0 is obviously imprimitive because of theorem 4.2.8. 2
Suppose now that the equations (#) : y + a y + by = 0 with a 6= 0 and ^ such that y is a solution y + ry = 0 are equivalent. Then there exist c; d 2 K of (#) if and only if cy + d y is a solution of y + ry = 0. In this case dc y + y satis es the equation y + 2dd ry = 0. Theorem 4.4.6 Suppose that the di erence Galois group G of the equation (#) : y + a y + by = 0, a 6= 0 is irreducible. Then G is imprimitive if and only if there exists a solution E 2 K^ of the Riccati equation (##) : (E )E + ( ( b ) (a) + (b) )E + (b)b = 0: 2
2
2
2
( )
2
2
2
a
a
a2
And if E 2 K^ is a solution of this Riccati equation then y satis es the equation (#) if and only if Dy + y satis es the equation 2y + ry = 0, where D = E + ab and r = a (a) + (b) + a 2 (D). Proof. If G is imprimitive and y satis es the equation (#) then there exists an ^ such that Dy + y satis es an equation of the form 2y + ry = 0, where D2K ^ . We have 2(Dy + (y)) = ( (a)b b 2 (D))y +(a (a) (b) a 2(D)) (y). r2K The group G is irreducible. Hence y and (y) are linearly independent over K^ . Therefore we must have (a)b b 2(D) = rD and a (a) (b) a 2(D) = r. After eliminating r we get the equation 2(D)D ab 2(D) (a)D + (ab) D + (aa)b = 0 for D. Let E = D ab . Then E satis es the Riccati equation 2(E )E +( 2( ab ) (b)b (b) (a) + a )E + a2 = 0. Conversely if E satis es the Riccati equation 2(E )E +( 2( ab ) (a)+ (ab) )E + (b)b 2 a2 = 0 then Dy + y satis es the equation y + ry = 0 if y satis es the equation (#) where D = E + ab and r = a (a) + (b) + a 2 (D). 2 Remark 4.4.7 We can nd the rational solutions of equation (##) as follows. Let : K^ ! K^ be the Q {linear map given by z 7! 2z . Then 2 = . If we apply to the equation (##) then we get the equation b (b) ( (E )) (E ) + ( 2 ( ) (a) + ) (E ) + ( (b2)b ) = 0: a
a
61
a
We can nd all the rational solutions (E ) of this equation by applying the algorithm described in Subsection 4.4.1. Then we apply 1 to these solutions and we nd the actual rational solutions of the equation (##) appearing in Theorem 4.4.6.
Lemma 4.4.8 The imprimitive subgroups G Gl(2; Q ) with G=G nite cyclic 0
are
1. F = f( 0 0 ) j 6= 0g [ f( 0 0 ) j 6= 0g .
2. Hn = f( 0 0 ) j ( )n = 1g [ f( 0 0 ) j ( )n = 1g for n 2 Z1 . 3. Hn+ = f( 0 0 ) j ( )n = 1g [ f( 0 0 ) j ( )n = 1g for n odd. Proof. If G is an imprimitive group with G=G0 nite cyclic then G \ D = D or G \ D = Dm;n for m; n 2 Z n f0g (See Subsection 4.4.2) . If ( 0 0 ) 2 G and ( 0 0 ) 2 G then also ( 0 0 )( 0 0 )( 0 0 ) 1 = ( 0 0 ). So if G \ D = Dm;n for m; n 2 Z n f0g then m = n or m = n. A calculation shows that the second possibility does not yield any imprimitive subgroup with G=G0 nite cyclic and the rst possibility yields the groups Hn for n 2 Z1 and Hn+ for n odd. 2 Consider the system (y) = ( d0 0c )y. We can apply the algorithm for rst order di erence equations to cd. Then we will nd a f 2 K^ such that cd (ff ) is reduced. Let T = ( f0 01 ). Then B = (T )( d0 0c )T 1 is in standard form. 4.4.4
G contains S l
) (2; Q
Consider the system (y) = ( 0b 1a )y. If the di erence Galois group G of this system is neither reducible nor imprimitive , then G = Gl(2; Q ) or G = Sl(2; Q )n = f( ) j ( )n = 1g. We apply the algorithm for rst order di erence equations to b = det( 0b 1a ) and nd an f 2 K^ such that ff b is reduced. Let T = ( f0 01 ). Then B = (T )( 0b 1a )T is in standard form. This nishes the algorithm. ( )
1
62
4.5 Examples In subsection 4.5.1 sequences spaces are used in order to make the Picard{Vessiot rings associated to a system of di erence equations more concrete. The explicit solutions which can be computed are certain sequences which will be called Liouvillian. Some concrete examples will be given. In subsection 4.5.2 we will demonstrate how one can construct examples where a quadratic eld extension of the eld of constants is needed for solving the Riccati equation. 4.5.1
Sequences spaces
Let F be any eld. Consider the set of sequences F N. This set forms a ring by using coordinate-wise addition and multiplication. Let J be the ideal of sequences with at most nitely many non-zero terms. Then we de ne SF = F N=J . The map : SF ! SF given by ((a1 ; a2; a3; : : :)) = (a2; a3 ; : : :) is well de ned on equivalence classes. Moreover, is an automorphism of the ring SF . For simplicity we will identify an element a = (a1 ; a2; a3; : : :) with its equivalence class. Let C be any eld of characteristic zero. (We are interested in the case that C = Q or C = k, where k is an algebraic number eld). The rational function eld C (z) can be embedded in SC by mapping a rational function f to the sequence sf = (s1; s2 ; : : :) where sn = f (n) for all but nitely many n 2 N. We will denote this map f 7! sf by and, following [Pet92], we will denote the image of in SC by RC . Lemma 4.5.1 The map : C (z) ! RC is an isomorphism of di erence elds. Proof. Two rational functions that agree on in nitely many arguments are identical. Further more, it is obvious that = . 2 From now on we will identify rational functions f 2 C (z) and their images sf in RC . Theorem 4.5.2 Consider the system of di erence equations (A) : (y) = Ay, where A 2 Gl(n; Q (z )). There is a di erence subring R SQ such that R is a Picard{Vessiot extension associated to the system (A). Moreover, if A 2 Gl(n; k(z)), where k is an algebraic number eld, then there exists already a basis of the solution space consisting of elements in (Sk )n. Proof. We refer to chapter 4 of [PS96]. 2. Remark 4.5.3 Above theorem shows that there is a natural rational k{structure on the vectorspace of solutions. Hence it is possible to develop Galois theory of di erence equations over a rational function eld with a constant eld which is not 63
algebraically closed. Compare with [HP95], where a similar theory is developed for the di erential case.
From now on we will write for simplicity S instead of SQ and instead of . De nition 4.5.4 We de ne the ring of Liouvillian sequences L recursively. L is the smallest subring of S such that 1. K^ L. 2. a 2 L implies that (a) 2 L 3. a 2 K^ implies that b 2 L if bn+1 = an bn for all but nitely many n 2 N1 . 4. a 2 L implies that b 2 L if bn+1 = an + bn for all but nitely many n 2 N1 . 5. a 2 L implies that b 2 L if there exist a j 2 N1 such that bjn = an and bn = 0 if n 6 0 mod j . In this case we call b an j {interlacing of a with zeroes.
Theorem 4.5.5 Suppose a . Then the following statements are equivalent. 1. a 2. The sequence a satis es a linear di erence equation over K^ so that the 2 S
2 L
di erence Galois group G associated to this equation is solvable.
Proof. 2 ) 1. Consider the system of di erence equations (A) : (y) = Ay, where A 2 Gl(n; K^ ). Let VA = fy 2 S n j (y) = Ayg be the solution space for the system of di erence equations (A). We will prove that if the di erence Galois group associated to this system is solvable then VA is an n{dimensional Q {linear subspace of Ln. This statement is slightly more general then the 2 ) 1 part of the theorem. Suppose now that the di erence Galois group is solvable and G=G0 is nite cyclic of order m. We assume that system (A) is already in standard form, that is A 2 G(K^ ). This is without loss of generality because if (A~) : (y) = A~y is an equivalent system of di erence equations then we have VA Ln if and only if VA~ Ln. Let B = m 1 (A) (A)A. Obviously B 2 G0(K^ ). Because of the LieKolchin theorem we can assume that B is an upper triangular matrix. Namely if B is not yet an upper triangular then according to the Lie-Kolchin theorem there exists a matrix T 2 Gl(n; Q ) so that B~ = T 1BT is an upper triangular matrix and then we continue with B~ instead of B . Consider the system of di erence equations (B ) : m (y) = B y. The solution space VB = fy 2 S n j m (y) = B yg of this system of di erence equations is an nm{dimensional Q - linear space and it is obvious that VA VB . We will prove that VB Ln. Let Si = fa 2 S j aj =
64
0 if j 6 i mod mg for i = 0; : : : ; m 1. Then we have S = S Sm . Let VBi = VB \ Sin for i = 0; : : : ; m 1. Then VBi is an n-dimensional Q -linear space for i = 0; : : : ; m 1 and VB = VB VBm . Let : K^ ! K^ be the Q -linear map given by z 7! mz. Then m = . Let C = B 2 Gl(n; K^ ). Note that C is also an upper triangular matrix. Consider the sytem of linear di erence equations (C ) : (y) = C y. The solution space VC is n-dimensional. We have v 2 VC if and only if w 2 VB , where w is an m-interlacing of v with zeroes. We will show by incomplete induction that there exists an upper triangular fundamental matrix U = (uij ) 2 Gl(n; L) for the system of di erence equations (C ). Let uii 2 S be a nonzero solution of the rst order di erence equation (y) = ciiy for i = 1; : : : ; n. Obviously uii 2 L for i = 1; : : : ; n. Moreover it is obvious that uii is an invertible element of L for i = 1; : : : ; n. If t 2 S satis es (t) t = cc1112 uu1122 then also t 2 L. Let u = tu . Then u satis es (u ) = c u + c u and u 2 L. In more or less the same way one can construct uij 2 L for all i < j so that U = (uij ) is a fundamental matrix for system (C ). We conclude that VC Ln and therefore VB Ln. By similar constructions one can show that VBi Ln for i = 1; : : : ; m 1. Hence VB Ln and VA Ln. 1 ) 2. We will prove this part of the theorem case by case. 0
1
1
0
0
12
12
12
11
12
12
22
11
12
0
1. Suppose a 2 K^ n f0g. Then a satis es the rst order linear di erence equation (y) = cy, where c = aa . The di erence Galois group associated to this di erence equation is the trivial group. ( )
2. Suppose a 2 S n f0g satis es a linear di erence equation then (a) satis es an equivalent equation of the same order. Hence the di erence Galois groups associated to both equations coincide. 3. Suppose b 2 S n f0g satis es the rst order di erence equation (y) = ay, where a 2 K^ n f0g. Then the di erence Galois group G associated to this di erence equation is an algebraic subgroup of the multiplicative group Gm = Q . Hence the di erence Galois group G is solvable. 4. Suppose that L is an n th order linear di erence operator with coecients in K^ so that La = 0. Let G be the di erence Galois group associated to the di erence equation Ly = 0. Let b 2 S satisfy (b) b = a. Then L( 1)b = 0. Hence b satis s an n + 1 th order di erence equation and the di erence Galois group G~ associated to this equation is a subgroup of the semidirect product of G and Q n. (Note that G acts on Q n). Hence the group G~ is solvable if the group G is solvable. 5. Suppose that a 2 S satis es the n th order linear di erence equation n(y) + cn n (y) + + c y = 0, where c ; c ; : : : ; cn 2 K^ and c 6= 0. 1
1
0
0
65
1
1
0
As usual we will identify this di erence equation with the system of di er0 1 0 1 0 0 BB 0 0 1 0 CC B . . CC. . . . ... C BB .. .. ence equations (y) = C y, where C = B C B@ 0 0 0 1 C A c c c cn Let G Gl(n; Q ) be the di erence Galois group associated to this di erence equation, then according to theorem 4.2.7 there exists a matrix T 2 Gl(n; K^ ) so that B = (T )CT 2 G(K^ ). Now let b be an m-interlacing of a with zeroes. Then b satis es the nm th order linear di erence equation nm(y) + ( cn ) n m (y) + + ( c )y = 0, where : K^ ! K^ is the Q -linear map given by z 7! mz. Note that = m . The nm th order linear di erence equation for b can be identi ed0with the nmnm 1sysBB 00 I0 I0 00 CC B . . CC 2 . . . ... C BB .. .. tem of di erence equations (y) = C~ y, where C~ = B C B@ 0 0 0 IC A C 0 0 0 m ^ ~ Gl(nm; K ). Let T = diag 0 ( T; ( T ); : : : ; 1 ( T )) 2 Gl(nm; K^ ). BB 00 I0 I0 00 CC B . . CC. Hence according to the. . . ... C BB .. .. Then B~ = (T~)C~ T~ = B C B@ 0 0 0 IC A B 0 0 0 orem 4.2.8 the di erence Galois group G~ associated to the linear di erence equation for b is a subgroup of a cyclic extension of order m of m copies of G. Hence if G is a solvable group then G~ is a solvable group. We still have to show that if a and b satisfy linear di erence equations over K^ so that the di erence Galois groups associated to these equations are solvable then also a b and ab satisfy linear di erence equations over K^ so that the di erence Galois groups associated to these equations are solvable. Suppose that the sequence a satis es an n th order linear di erence equation over K^ and the sequence b satis es an m th order linear di erence equation. Let Ra and Rb be the Picard{Vessiot extensions associated to the di erence equations for a and b respectively. Let Va and Vb be the solution spaces for these equations and let Ga = DGal(Ra=K^ ) and Gb = DGal(Rb =K^ ). The di erence Galois groups Ga and Gb act faithfully on the solution spaces Va and Vb respectively. Let W = fv + v j v 2 Va; v 2 Vbg. Then W is a Q -linear space of dimension less than or equal than n + m. One can construct a linear di erence equation whose solution space is W . Let R be the Picard Vessiot extension associated to this equation and let G = DGal(R=K^ ) be the di erence Galois group associated 0
1
2
1
1
1
1
(
1)
1
1
1
1
1
1
2
1
2
66
0
1
1
1
1
to this equation. Then G acts on Va and Vb and G acts faithfully on W . The space W can be considered as a quotient space of Va Vb. Hence G acts faithfully on Va Vb. From this it is clear that G is a subgroup of Ga Gb. Hence the group G is solvable if the groups Ga and Gb are solvable. Let W~ = spanQ fv1 v2 j v1 2 Va; v2 2 Vb g. Then W~ is a Q {linear space of dimension less than or equal than nm. One can construct a linear di erence equation whose solution space is W~ . Let R~ be the Picard-Vessiot extension associated to this equation. Obviously R~ R. Hence the di erence Galois group ~ K^ ) is solvable if the group G is solvable. 2 G~ = DGal(R= All the solutions of a second order di erence equation 2y +a y +by = 0 are Liouvillian if the di erence Galois group is reducible or irreducible and imprimitive. The solutions of a second order di erence equation with an imprimitive di erence Galois group are not considered to be \closed form" solutions in [Pet92]. Hence our notion of Liouvillian solutions does really extend the notion of \closed form" solutions given in [Pet92]. If the di erence Galois group G contains Sl(2; Q ), then there are no Liouvillian solutions of the second order di erence equation 2y + a y + by = 0. Now we give some concrete examples.
Example 1 Consider the equation 2y + z y + zy = 0. First we want to determine the set Su. Let t = z1 . The Riccati equation at 1 reads u (u)+ 1t u+ 1t =
0. We have v( 1t ) = 1. So we are in the case 2v(a) < v(b). A simple calculation shows that Su = f 1t + 1 + O(t); 1 t + O(t2)g. Further we have Sp = f1; zg and Sq = f1g. There are two combinations u 2 Su, p 2 Sp and q 2 Sq such that v(u) + v(q) v(p) = 0. Namely u = 1t + 1 + O(t), p = z, q = 1 and u = 1 t+O(t2 ), p = 1, q = 1. The rst combination gives us uqp = 1 t+O(t2 ). So e = 1. Therefore this combination does not yield a solution u 2 Q(z) of the Riccati equation. The second combination gives us uqp = 1+t+O(t2). Hence e = 1. After substituting c = 1, p = 1, q = 1, R = z + r in (z) we get the equation z + r +2 z(z + r +1)+ z(z + r) = r +2 = 0. So r = 2. And u = zz 12 is the only solution of the Riccati equation. Let A = ( 0z 1z ) and T1 = ( 1 uu 11 ) = z 3 6z 2 +9z 3 z 1 2z 3 1 1 z 2 (z 1)(z 2) z 2 ( z 1 1 ). Then B1 = (T1 )AT1 = ( ). The entries z (z 2) 0 z 2 z 1 1 on the diagonal of B1 are not yet reduced. Therefore let T2 = ( z 02 z 0 2 ). z 3 6z 2 +9z 3 1 1 (z 1)2 (z 2)2 ) . Now the system (y) = B y Then B2 = (T2 )B1T2 = ( 2 0 z is in standard form. Hence the di erence Galois group is the reducible group G = f( 0 ) j 2 = 1g: 67
A fundamental matrix for system (B2 ) with entries in SQ is U2 = ( a0 bd ), where an = ( 1)n, dn = ( 1)n(n 1)! and bn = ( 1)n~bn where the sequence b~ = 3 2 n 3 (~b1 ; ~b2 ; : : :) satis es the inhomogeneous equation ~bn+1 ~bn = (n 1)! n(n 61)n2 (+9 n 2)2 n 1 k 3 for all but nitely many n. Hence we can take ~bn = P (k 1)! k(k3 61)k22(+9 k 2)2 for all k=3 n 3. Now T1 1 T2 1 U2 is a fundamental matrix for the equation 2 y + z y + zy = 0. Two linearly independent solutions are u; v 2 SQ , where of this equation nP1 ( n 1)! k3 6k2 +9k 3 n n 2) (k 1)! (k 1)2 (k 2)2 . un = ( 1) (n 2) and vn = ( 1) n 2 + (n
Example 2. Consider the equation 2(y)
k=3
z (y ) + z2+9 y = 0. We want 1 to determine the set Su. Let t = z . The Riccati equation at 1 reads u (u) 3+14t 2 3+14t 2 1+9t u + 1+9t = 0. We have v ( 1+9t ) = v ( 1+9t ) = 0. Hence we are in the case that v(b) 2v(a) and v(b) is even. We nd Su = f1 5t + O(t2); 2 8t + O(t2)g. Further we have Sp = f1; zg and Sq = f1; z + 8g. There are four combinations u 2 Su , p 2 Sp and q 2 Sq such that v (u) + v (q ) v (p) = 0. The rst two combinations are p = 1, q = 1 and u = 1 5t + O(t2 ) or u = 2 8t + O(t2 ). These two combinations does not yield any solution u 2 Q(z) of the Riccati equation. The third combination is p = z,q = z +8 and u = 1 5t + O(t2). This combination gives us uqp = 1 + 3t + O(t2). So e = 3. After substituting c = 1, p = z, q = z + 8 and R = z3 + r2 z2 + r1z + r0 in (z) we are left with four linear equations in the three indeterminates r0 ; r1 and r2. There is a unique solution r0 = 1680, r1 = 392 and r2 = 33. So R = z3 +33z2 +392z +1680 = (z +12)(z2 +21z +140). The fourth combination is p = z,q = z + 8 and u = 2 8t + O(t2). This combination gives us uq 2 4 3 2p = 1+4t + O (t ). After substituting c = 2, p = z , q = z +8 and R = z + r3 z + r2 z 2 +r1 z +r0 in (z) we are left with ve linear equations in the four indeterminates r0 ; r1 ; r2 and r3 . There is a unique solution r0 = 840, r1 = 386, r2 = 83 and r3 = 10. So R = z 4 10z23 + 83z 2 386z + 840. The4 solutions of the Riccati z (z +13)(z +23z +162) z (z 6z 3 +59z 2 246z +528) equation are u1 = (z+8)( and u = 2 2 z +12)(z 2 +21z +140) (z +8)(z 4 10z 3 +83z 2 386z +840) . Let
T1 = (
u2 u1 u2 u1 u1 u2
(
1
u1 u2 1
u1 u2
3z +14 z +9
). Then B1 = (T1)( 02z
z +9
z (z +13)(z 2 +23z +162) (z +8)(z +12)(z 2 +21z +140)
0
2
1 )T 1 = 1
3z +14 z +9
0
z (z 4 6z 3 +59z 2 246z +528) (z +8)(z 4 10z 3 +83z 2 386z +840)
):
The entries on the diagonal of B1 are not yet reduced. Therefore we apply z (z +13)(z 2 +23z +162) the algorithm for rst order di erence equations to (z+8)( z +12)(z 2 +21z +140) and 4 3 2 z (z 6z +59z 246z +528) 2 (z+8)( z 4 10z 3 +83z 2 386z +840) . In this way we nd a matrix T2 = (
(z +12)(z 2 +21z +140) z (z +1)(z +2)(z +3)(z +4)(z +5)(z +6)(z +7)
0
68
0
z 4 10z 3 +83z 2 386z +840 z (z +1)(z +2)(z +3)(z +4)(z +5)(z +6)(z +7)
)
such that B2 = (T2 )B1T2 1 = ( 10 standard form. Hence the di erence group G = f( 10 0 ) j 2 Q g:
0 ). Now the system (y) = B y is in 2 2 Galois group is the completely reducible
A fundamental matrix for system (B2 ) with entries in SQ is U2 = ( a0 d0 ), where an = 1 and dn = 2n. Now T1 1T2 1U2 is a fundamental matrix for the +14 z y = 0. Two linearly independent solutions of equation 2(y) 3zz+9 (y ) + z2+9 this equation are u; v 2 SQ where (n+12)(n2 +21n+140) n4 10n3 +83n2 386n+840 n un = n(n+1)(n+2)( n+3)(n+4)(n+5)(n+6)(n+7) and vn = n(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)(n+7) 2 .
Remark. In example 1 and 2, two special cases of the so-called hypergeometric di erence equations are discussed. For an extensive study of hypergeometric di erence equations we refer to [Bat67]. Example 3. Now consider2 5the 4equation 2 y + a y + by = 0, where a = 3 2 2 (z +1)(5z +8z +4) z (z +7z +19z +25z +16z +3) . We have v(a) = 3 and (z +2)(z 5 +2z 4 +z 3 1) and b = (z +2)(z 5 +2z 4 +z 3 1) v (b) = 1. (See Subsection 4.4.1.) Hence we are in the case that v (b) 2v (a) and v(b) is odd. So there does not exist a rational solution of the Riccati equation. Hence the di erence Galois group G is irreducible. We want to determine whether the di erence Galois group is imprimitive. According to theorem 4.4.6. G is imprimitive if and only if there exists a solution E of the Riccati equation b (b) 2 (E )E + ( 2 ( ) (a) + )E + (b)b = 0: a
a
a2
+88z +48)(z +12z +57z +134z +156z +71) Here we have 2( ab ) (a)+ (ab) = (z+2)(10z +51(zz+3)(5 z 2 +8z +4)(5z 2 +28z +40) (b)b z 2 (z +2)(z 5 +2z 4 +z 3 1)(z 5 +12z 4 +57z 3 +134z 2 +156z +71) and a2 = . Using the algorithm (z +3)(5z 2 +8z +4)2 z 2 (z 5 +2z 4 +z 3 1) for the Riccati equation we nd two solutions E1 = (z+1)(5z2 +8z+4) and E2 = (z +2)2 (z 5 +2z 4 +z 3 1) . Let D = E2 + ab = z(z1+1) and r = a (a) + (b) + z (5z 2 +8z +4) 3 z 5 +12z 4 +57z 3 +134z 2 +156z +71) a 2 (D) = z (z+1)( . Then y is a solution of the equation (z +2)2 (z +3)(z 5 +2z 4 +z 3 1) 2 y + a y + by = 0. if and only if Dy + y satis es 2 y + ry = 0. We have B1 = ( 0r 01 ) (T1)( 0b 1a )T1 1, where T1 = ( Db (D1) a ). 2 Now r is not yet reduced. Therefore let T2 = ( f0 (0f ) ), where f = z5 +2z z(z4++1)z3 1 Then B2 = (T2)B1T2 1 = ( z0 10 ). The system (y) = B2 y is in standard form. Hence the di erence Galois group is the imprimitive group G = f( 0 0 ) j 6= 0g [ f( 0 0 ) j 6= 0g. 3
69
2
5
4
3
2
A fundamental matrix for system (B2 ) with entries in SQ is U2 = ( (cc) (dd) ),
n Q where cn = 2k if n is even and cn = 0 if n is odd and dn = Q (2k 1) if n k=1 k=1 n 1 2
2
1
is odd and dn = 0 if n is even. Now T1 1T2 1U2 is a fundamental matrix for the equation 2y + a y + by = 0. And we nd two linearly independent solutions u and v of this equation, where un = n1 cn + ncn+1 and vn = n1 dn + ndn+1.
4.5.2 The case where an extension of the constant eld is needed In this subsection we will construct second order di erence equations, for which a quadratic eld extension of the constant eld is needed for solving the Riccati equation. In particular we will give an example to show that step 3 in the algorithm for solving the Riccati equation is not super uous. Let k~ = k() with 2 2 k and 62 k. Take u0; u1 2 k(z) with u1 6= 0. And let u = u0 + u1 . Then (u)u + au + b = 0 is equivalent to the equations u0 (u0 ) + 2 (u1 )(u1 ) + au0 + b = 0
and
(u0)u1 + u0 (u1 ) + au1 = 0:
Hence a and b are uniquely determined if we x u0; u1 2 k(x) with u1 6= 0. Obviously u0 + u1 and u0 u1 are two solutions of the Riccati equation associated to the di erence equation 2y + a y + by = 0. If the di erence Galois group g, that is if G of this equation is not an algebraic subgroup of fc:Id j c 2 Q u0 +u1 is not of the form (f ) with f 2 k ~(z), then u0 + u1 and u0 u1 are the u0 u1 f only solutions of the Riccati equation and we really need the quadratic extension k~ k for solving the Riccati equation. If we take u0 = z2 and u1 = 1 then we get the second order equation 2 + ( 2z2 2z 1) y +(z4 2)y = 0. Clearly in this case we do not need a quadratic extension of our constant eld k for computing the rst two terms of the local u1 formal solution at in nity of the associated Riccati equation. Furthermore uu00 +u 1 is not of the form (ff ) with f 2 k(z). Hence we need step 3 of the algorithm described in section 4.4.1 in order to nd the quadratic eld extension k~ needed for solving the Riccati equation. Two linearly independent solutions in SQ of this second order di erence equan 1 n 1 tion are c; d, where cn = Q (j 2 + ) and dn = Q (j 2 ). If one wants to have j =1 j =1 a basis consisting of two k{rational solutions then one can take 21 (c + d) and 1 d). 2 (c 70
Chapter 5 An algorithm for computing a standard form for second order linear q-di erence equations 5.1 Introduction In this article we will present an algorithm for computing a standard form for second order -di erence equations over the rational function eld ( ), where denotes a nite algebraic extension of Q( ) and is a transcendental variable. This standard form will be de ned in section 5.2. One can read o fairly easily the -di erence Galois group of a system of -di erence equations if one knows that the system is in standard form. Moreover, if the -di erence Galois group of a second order -di erence equation is not too big (i.e. does not contain (2 C) ) then one can compute two linearly independent Liouvillian solutions for this equation. For analogous algorithms for second order di erential equations and second order di erence equations we refer to [Kov86] and [Hen96] respectively. In section 5.2 a standard form for -di erence equations will be de ned. Further some notation will be xed. In section 5.3 we will discuss rst order homogeneous linear -di erence equations. The algorithm for second order homogeneous -di erence equations will be explained in section 5.4. Section 5.5 is devoted to examples. q
k z
q
q
k
z
q
q
q
Sl
;
q
q
q
5.2 A short introduction to algebraic aspects of q -di erence equations. We will use the following notation.
q
2
C, but is not a root of unity and =6 0. q
q
71
q ; q ; q ; : : : is a sequence of complex numbers satisfying q = q and qii = qi . 1
2
3
We de ne recursively K = C(z ) and Ki = C(zi ), where zii = zi. Instead of K and z we will write K and z. 1 K1 = S Ki . i 1
1
+1 +1
1
1
+1
+1
+1 +1
1
=1
denotes the C-linear automorphism of K1 given by (zi) = qi zi. Obviously can be restricted to the elds Ki. We will denote these restrictions also by .
We are interested in the Galois theory of q-di erence equations over the eld K1, because if one works over this eld there is a standard form available analogous to the standard form which is de ned for ordinary di erence equations in [Hen96]. This is not the case if one works over the smaller eld K = C(z). Consider the system of di erence equations (A) : y = Ay, where A 2 Gl(n; K1). (We restrict ourselves to equations with A 2 Gl(n; K1) in order to guarantee that we get n independent solutions.)
De nition 5.2.1 A ring R together with an automorphism R : R ! R is called a Picard-Vessiot extension over K1 associated with system (A) if 1. R is a commutative ring, R K1 and RjK1 = . 2. The only R -invariant ideals are 0 and R. 3. There exists a matrix U 2 Gl(n; R) such that R(U ) = AU . (Such a matrix U is called a fundamental matrix for the system (A).) 4. R is minimal with respect to the conditions 1, 2 and 3 or equivalently if U = (uij ) 2 Gl(n; R) is a fundamental matrix for the system (A) then R = K1[u11; : : : ; unn; det1(U ) ].
From now on we will denote R also by .
Theorem 5.2.2 (Existence and unicity) For every system of linear q-di erence equations (A) with A 2 Gl(n; K1) there exist a Picard-Vessiot extension R K1. This extension is unique up to q-di erence K1-isomorphism, that is up to K1{linear isomorphisms commuting with . Further we have that (r) = r implies r 2 C if r 2 R.
De nition 5.2.3 The q-di erence Galois group G = DGal(R=K1) is the group consisting of all the q-di erence K1-automorphisms of R.
72
The vector space of solutions V is the set fy 2 Rn j (y) = Ayg. If U 2 GL(n; R) is a fundamental matrices for the system (A) then the columns of U form a base for V over C Suppose that 2 G = DGal(R=K1), then it is obvious that (U ) is also a fundamental matrix for the sytem (A). Hence (U ) = UT , where T ; T1 2 Gl(n; C). So the elements of the q-di erence Galois groups act as Clinear maps on the C-linear space of solutions V . But even a stronger statement holds. Theorem 5.2.4 (Algebraicity) G = DGal(R=K1) is a linear algebraic groups over the eld of constants C.
Theorem 5.2.5 (Galois correspondence) 1. If a 2 R then (8 2 G : (a) = a) ) a 2 K1 . 2. If H is an algebraic subgroup of G such that K1 = fa 2 R j 8 2 H : (a) = ag then H = G.
Theorems 5.2.2, 5.2.4 and 5.2.5 are special cases of theorems for more general di erence elds which are proved in chapter 1 of [PS96]. If G is an algebraic subgroup of Gl(n; C), then we denote by G(K1) the subgroup of Gl(n; K1) which is de ned by the same equations. Two systems (A) and (B ) corresponding to matrices A; B 2 Gl(n; K1) are de ned to be equivalent over K1 if there exists a T 2 Gl(n; K1) such that B = (T )AT 1. In this case, if U 2 Gl(n; R) is a fundamental matrix of system (A) then TU is a fundamental matrix of system (B ) and it is obvious that the solution spaces VA; VB of the systems (A) and (B ) are equivalent as representation spaces of the q-di erence Galois group G. Conversely if the Picard-Vessiot extensions associated to the systems (A) and (B ) are q-di erence K1-isomorphic and the solution spaces VA and VB are equivalent as representation spaces of the q-di erence Galois group G then (A) and (B ) are equivalent systems of q-di erence equations. The algorithms in section 3 and section 4 are based on theorem 5.2.6 and the converse theorem 5.2.7. These theorems are not valid for general di erence elds.
Theorem 5.2.6 Let G Gl(n; C) be the q-di erence Galois group associated to the system of di erence equations (A). Then the following statements hold. 1. G=G0 is a nite cyclic group. 2. There exists a B 2 G(K1 ) such that (A) and (B ) are equivalent systems.
It is a conjecture that for every linear algebraic group G Gl(n; C) with G=G0 nite cyclic there exists a system of di erence equations which has G as di erence Galois group. One can show that this conjecture is true for n = 1; 2
by giving explicit examples for every possible group. 73
Theorem 5.2.7 If A 2 G(K1) with G an algebraic subgroup of GL(n; C) such that for any proper algebraic subgroup H and for any T 2 G(K1) one has that (T )AT 1 62 H (K1) then G is the di erence Galois group of the system (A). De nition 5.2.8 If A satis es the conditions of theorem 5.2.7 then we say that
system (A) is in standard form.
This standard form is in general not unique. But one can read o fairly easily the q-di erence Galois group of a system of q-di erence equations if one knows already that this system is in standard form. The standard form is also very suitable for nding Liouvillian solutions. This will be explained in section 5.5. In fact we will present an algorithm which computes for a given equation an equivalent system in standard form. The proofs of theorems 5.2.6 and 5.2.7 are completely analogous to the proofs of propositions 1.20 and 1.21 in [PS95] if one applies the next two lemmas.
Lemma 5.2.9 K1 is a quasi-algebraically closed (or C1) eld. Proof. A eld L is called quasi-algebraically closed if it satis es the condition that if P is a homogeneous polynomial of degree d 6= 0 in n variables with coecients in L whose only zero in Ln is (0; : : : ; 0), then d n. Let P be a homogeneous polynomial of degree d 6= 0 in n variables with coecients in K1. Assume that (0; : : : ; 0) is the only zero of P in K1n . There exists an m 2 Z1 such that the coecients of P are in Km. The elds Ki are extensions of transcendence degree 1 of an algebraically closed eld. Such elds are well known to be quasi-algebraically closed. Hence d n. 2
A consequence of lemma 5.2.9 is that if G is a connected group then H 1(Gal(K 1=K1); G(K 1)) = 0 , where K 1 denotes the algebraic closure of K1. See [Ser64].
Lemma 5.2.10 The automorphism does not extend to any proper nite exten-
sion L of K1.
Proof. Suppose L is a nite algebraic extension of K1 . Then L = K1 ( ) for a certain 2 L. The coecients of the monic irreducible polynomial for over K1 are in Km for an m 2 Z. Let denote the morphism of algebraic curves X ! P(C) corresponding to Km ( ) Km . If extends to Km( ) then will permute the nite rami cation points of . The automorphism of P(C) leaves no nite subset of P(C) invariant. This implies that is a nite cyclic covering whose only rami cation points are 0 and 1. Hence Km( ) K1. 2
74
The equations that we look at in the next sections are de ned over a small eld k(z) C(z), where k denotes a nite algebraic extension of Q(q). The algorithm works in the eld K1, but in reality we will use only a small part of this eld. In the main part of the algorithm (section 5.4.1) it suces to work over elds l(z ), where l C is a degree 2 extension of k. 2
5.3 First order q-di erence equations We consider the rst order homogeneous linear di erence equation (a) : y = ay with a 2 k(z) , where k is a nite algebraic extension of Q(q). Any equivalent equation is given by y = ff ay, where f 2 K1 . The q-di erence Galois group G of (a) is a nite cyclic group of order n if there is a f 2 K1 such that f a is a primitive nth -root of unity, otherwise the di erence Galois group is the f multiplicative group Gm = C. This statement is a consequence of Theorems 5.2.6 and 5.2.7. Suppose a 2 k(z) . Then we can write a = czm QP , where c 2 k, m 2 Z and P; Q 2 k[z] are monic polynomials with gcd(P; Q) = gcd(P; z) = gcd(Q; z) = 1. Notice that c; m; P; Q are uniquely determined by a. De nition 5.3.1 Suppose that a = cz m PQ 2 k(z ) . Then a is reduced if ( )
( )
n Z we have gcd(P; nQ) = 1. r 2. If c = q ! where r Q and ! is a root of unity then r = 0. If a is reduced then the equation (a) : (y) = ay is automatically in standard form. Suppose a K is given. We describe an algorithm which produces an element b K such that the di erence equations (a) and (b) : (y) = by are equivalent and b is reduced. We assume that a = czm QP , where c k, m Z and P; Q k[z] are monic polynomials with gcd(P; Q) = gcd(P; z) = gcd(Q; z) = 1. If c = qr ! where r Q and ! is a root of unity then we replace c by ! = zz c. Write P = P P Ps and Q = Q Q Qt where the Pi and Qj are monic and irreducible in k[z]. First we take P and we test wether there exist a Qj and an n Z such that nP = Qj . If nP = Qj then we can go further with a~ = a QP1 . This gives a reduction of the problem with respect to the number of irreducible factors of P and Q. And then we proceed with a~ instead of a and so forth. There is also an algorithm which avoids factorization in k[z] if q = 1. We will present this algorithm here. If q = 1 there exists an R > 0 such that the zeros of P and Q are in the annulus R z R. We can express R in terms of the coecients of P and Q. For instance if P = Pczc + + P z + P and Q = Qd zd + 1. For all
2
2
2
2
2
2
2
(
1
2
1
2
r)
r
2 1
2
1
1
j
j j 6
j j 6 1 j j
75
1
0
+ Q z + Q then we can take R = 1 + max(iP jjPP jj ; jP jjQQ jj ; iP jjPP jj ; jP jjQQ jj ). For m 2 Z with jqmj > R or jqmj < R we have gcd(P; mQ) = 1. We start with m = 1 and compute g = gcd(P; (Q)). If g = 6 1 then one can write a = g g PQ c
1
0
1
=0
i c
d
1
=0
c
j
d
=1
i
0
d
=1
j
0
1
2
2
~
1( ) ~
and we can go further with a~ = QP instead of a. This gives a reduction of the problem with respect to the degrees of P and Q. If g = 1 then we try m = 1 and so forth. The algorithm stops if a is reduced to a constant or if all m 2 Z with R2 jqmj R are checked o . The above algorithms can easily be modi ed so that we alos nd an f 2 K1 that b = ff a is reduced. ~ ~
1
2
( )
5.4 Second order q-di erence equations Consider the second order linear homogeneous q-di erence equation y + a y + by = 0, where a; b 2 k(z) and b 6= 0. As usual we will identify this equation with the 2 2 system of q-di erence equations (A) : y = ( 0b 1a )y. Throughout this section we will denote the q-di erence Galois group over K1 of above system by G. We present an algorithm which produces a matrix B 2 Gl(2; K1) so that the systems (A) and (B ) are equivalent and (B ) is in standard form. Further a matrix T is computed such that B = (T )AT . Lemma 5.4.1 The algebraic subgroups G of Gl(2; C) that can occur as q-di erence Galois group over K1 are 1. Any reducible subgroup of Gl(2; C) with G=G nite cyclic. 2. Any in nite imprimitive subgroup of Gl(2; C) with G=G nite cyclic. 3. Any subgroup containing the group Sl(2; C). Proof. This is a consequence of theorem 5.2.6. and the well known classi cation of the algebraic subgroups of Gl(2; C). 2 The algorithm is arranged in the following manner. Given a second order q-di erence equation we test which of the three cases above holds. After that we compute an appropriate equivalent system in standard form. 2
1
0
0
5.4.1
The Riccati equation
To a second order linear q-di erence equation y + a y + by = 0, where a; b 2 k(z) and b 6= 0 we can associate a rst order non-linear q-di erence equation (y) : u (u) + au + b = 0. This equation is called the Riccati equation. If u is a solution of the Riccati equation then the q-di erence operator + a + b factors 2
2
76
as ( ub )( u). A solution in K1 of the Riccati equation corresponds to a line of the solution space that is xed by the q{di erence Galois group G. Therefore we have the following theorem. See also [Hen96].
Theorem 5.4.2 The following statements hold: 1. If the Riccati equation has no solutions u 2 K1 then G is irreducible. 2. If the Riccati equation has exactly one solution u 2 K1 then G is reducible but not completely reducible.
3. If the Riccati equation has exactly two solutions u1; u2 2 K1 then G is completely reducible but G is not an algebraic subgroup of fc:Id j c 2 C g. 4. If the Riccati equation has more than two solutions in K1 then the Riccati equation has in nitely many solutions and G is an algebraic subgroup of fc:Id j c 2 C g.
In this section we will describe an algorithm which computes all solutions u 2 K1 of the Riccati equation. The algorithm consists of a few steps. In the rst three steps we will compute all solutions u 2 K of the Riccati equation. In the rst step of the algorithm we will compute the rst term of all possible local formal solutions at 0 and 1. In the next two steps we try to glue these local formal solutions to a global solution. In the fourth step we will prove that if there exists a solution u 2 K1 then there is also a solution u 2 K2. After that we will describe how to nd the solutions u 2 K2 if we did not nd any solution u 2 K . 1. Let be the automorphism of C((z)) given by (z) = qz. Then coincides with our former on K = C(z). 1 We de ne the discrete valuation v0 : C((z)) ! Z by v0( P cizi ) = i= 1 minfi j ci 6= 0g. In particular v0(0) = 1. Elementary properties of v0 are v0 (u1u2) = v0 (u1) + v0(u2) and v0(u1 + u2) min(v0 (u1); v0(u2)). Hence if v0 (u1) 6= v0 (u2) then v0 (u1 + u2) = min(v0 (u1); v0(u2)). Another property of v0 is that v0(u) = v0 ( u). Now we consider a and b as Laurent series in z. We distinguish two cases. (a) 2v0(a) < v0 (b). In this case there are two solutions u; u~ 2 k((z)) of the Riccati equation. The rst solution u satis es v0 (au) = v0 (u (u)) < v0(b) and v0(u) = v0 (a) and the other solution u~ satis es v0(au~) = v0(b) < v0 (~u (~u)) and v0 (~u) = v0 (b) v10 (a). Suppose v01(a) = l and v0(b) = k. Then we will write a = P aizi and b = P bi zi . We i=l i=k can compute succesively the coecients of u and u~ by simply solving 77
linear equations with coecients in k. Hence no eld extension of k is needed. We nd u = aqll zl + O(zl ), where O(zl ) stands for the higher order terms and u~ = bakl zk l + O(zk l ). (b) v (b) 2v (a). If u 2 C((z)) satis es the Riccati equation then we have v (b) = v (u (u)) v (au). Hence if v (b) is odd there is obviously no solution of the Riccati equation in the eld C((z)). 1 Suppose now v (b) is even. Let n = v (b). Write b = P bi zi and +1
+1
+1
0
0
0
0
0
0
1 2
0
0
i n 1 1 P i a = ai z . (Note that it is possible that an = 0). Write u = P bi zi . =2
i=n
i=n
Then we try to solve the Riccati equation (y) : u (u) + au + b = 0. The (2n)th coecient of this expression gives us the equation qnun + an un + b n = 0. LetpD = apn 4b nqn. If D 6= 0 then we nd two solutions un = an qn D0 2 k( D ) of the quadratic equation. p So u = an qn D0 zn + O(zn ). If D = 0 then we nd one solution u = qann zn + O(zn ). 2
2
2
0
0
0
2
+1
2
2
0
+1
2
Let S be the set of all possibilities for the rst term of u expressed as pD ) with Laurent series in z . The coecient of this term lies in the eld k ( p [k( D ) : k] 2. We discovered that #S 2, even if the Riccati equation has in nitely many solutions u 2 K1. In the same way we can determine the set S1 of all possibilities for the rst term of u expressed as Laurent pD ). series in z with a coecient in at worst a quadratic eld extension k ( 1 p p ~ We also have #S1 2. For future use we will denote k = k( D ; D1). 2. In this part of the algorithm we will determine the solutions u 2 k~(z) of the Riccati equation. We rewrite the Riccati equation as Fu (u) + Gu + H = 0, where F; G; H 2 k[z] and gcd(F; G; H ) = 1. Write u = czm PT with c 2 k~, m 2 Z, P; T 2 k~[z], where P and T are monic polynomials and and gcd(P; T ) = gcd(P; z) = gcd(T; z) = 1. Let R denote the greatest monic divisor of T such that (R) divides P . Then one can write u = czm RtR p , where p; t are monic polynomials, gcd(p; (t)) = 1 and gcd( (R)p; Rt) = 1. Further gcd(R; z) = gcd(p; z) = gcd(t; z) = 1. The Riccati equation reads now: 0
0
0
0
1
0
( )
(z) : c qmz m (R) (p)pF + czm (R) (t)pG + R (t)tH = 0: 2
2
2
Hence p is a monic divisor of H with v (p) = 0 and t is a monic divisor of (F ) with v (t) = 0. We de ne the sets Sp = fp 2 k~[z] j pjH; v (p) = 0 and p monicg and St = ft 2 k~[z] j tj (F ); v (t) = 0 and t monicg. Writing RR as a Laurent series in z we get RR = 1 + O(z), where O(z) stands for the higher order terms. RR must be equal to czutmp . Hence for 0
1
0
0
1
( )
0
( )
( )
78
every possible combination u 2 S , p 2 Sp and t 2 St we compute a c 2 k~ and a m 2 Z from the equation czum0tp = 1 + O(z). Let e be the degree of R. Writing RR as a Laurent serie in z we get RR = qe + O( z ). So we must have v1(u) degz (t) + m + degz (p) = 0. Hence if there are no terms u1 2 S1 with v1(u1) = degz (t) degz (p) m then there is no solution u 2 K . For all u1 2 S1 satisfying v1(u1) = degz (t) degz (p) m we compute a d 2 k~ from czu1mtp = d + O( z ). If d = qe with e 2 Z then we write R = ze + re + + r z + r with indeterminates r ; r ; : : : ; re . After substituting this R in (z) we are left with a set of equations in the ri. In this way we will nd all possible solutions u 2 k~(z) of the Riccati equation. 3. In step 2 we have determined all solutions u 2 k~(z) of the Riccati equation. Now we have to face the problem whether it is possible that there exist solutions u 2 K if there are no solutions u 2 k~(z). A priori this is possible because F and G need not split in linear factors in k~[z]. But it is also clear from the construction in step 2 that if the Riccati equation has a solution in K , there is also a solution in l(z), where l k~ is the smallest eld which contains the splitting elds of F and G. But we have even a stronger rationality result. Theorem 5.4.3 If the Riccati equation has a solution in K then there is also a solution in a eld k(z ) with [k : k] 2. 0
0
( )
( )
1
1
1
1
1
0
0
0
1
1
1
1
This proof is completely analogous to the proof of theorem 4.3. in [Hen96]. 2 So we want to determine all elds k l with [k : k] 2. If q is algebraic p then k must be of the form k( s), where s is an algebraic integer in l. Recall that l is the splitting eld of the polynomial FG. After a suitable linear substituion z 7! nz , where n 2 Z we can assume that FG is monic and that the coecients of FG are algebraic integers. s must be a divisor of the discriminant of this polynomial. This gives us a nite set of possible p quadratic extensions k( s) of k in l. If q is transcendental then k must p be of the form k( s) where s is a polynomial in qi for a certain i with algebraic integer coecients. After a suitable linear substitution we can assume that FG is monic and the coeents of FG are polynomials in qi (Choose i minimal.) with algebraic integer coecients. Now s must be a divisor of the discriminant of the polynomial FG which is a polynomial in qi with algebraic integer coecients. Again this gives us a nite set of p possible quadratic extensions k( s) of k in l. For each possible quadratic extension k we determine the sets Sp = fp 2 k[z] j pjH; v (p) = 0 and p monicg and St = ft 2 k[z] j tj (F ); v (t) = 0 and t monicg. After that we proceed as in 2. Proof.
1
1
0
79
0
4. In steps 1, 2 and 3 we have determined all solutions u 2 K = C(z) of the Riccati equation. If we did not nd any solution u 2 K then we try to nd solutions u 2 K1 n K . We need the following theorem.
Theorem 5.4.4
If the Riccati equation has a solution in
also a solution in
K .
K1 then there is
2
Let U be the set of solutions of the Riccati equation in K1 . The group Gal(K1 =K ) acts on the set U . If #U = 1, say U = fug, then u is invariant under Gal(K1 =K ). Hence u 2 K . If #U = 2, then the kernel of the map Gal(K1 =K ) ! Aut(U ) is a subgroup of index 2. Hence u is an element of K , because K is the only quadratic eld extension of K in K1. Suppose that #U = 1. According to theorems 5.4.2.4 and 5.2.6 there exists a T 2 Gl(2; K1) such that (T )( 0b 1a )T = ( u0 u0 ) for some u 2 K1. There is a l 2 Z such that T 2 Gl(2; Kl ) and u 2 Kl . Suppose 2 Gal(Kl =K ) then ((T ))( 0b 1a )((T )) = ( (0u) (0u) ). Hence there is a g 2 Kl such that (u) = gg u: This g is uniquely determined if we require that the numerator and denominator of g are monic. The map 7! g is a 1-cocycle. Hilbert 90 states that this 1-cocycle is trivial. See [Ser68]. We can replace u by u~ = hh u for a certain h 2 Kl such that u~ is invariant under Gal(Kl =K ). Hence u~ 2 K . Let T = hT . Then (T )( 0b 1a )T = ( u0~ u0~ ). We have C T 2 Gl(2; K1) satis es (C T )( 0b 1a )(C T ) = ( u0~ u0~ ) if and only if (C ) = C , that is C 2 Gl(2; C). For 2 Gal(Kl =K ) one has therefore (T ) = C T for some C 2 Gl(2; C). The map 7! C is a homomorphism Gal(Kl =K ) ! Gl(2; C). We have Gal(Kl =K ) is a nite cyclic group. So there exists an M in Gl(2; C) such that M C M is a diagonal matrix for all 2 Gal(Kl =K ). Let T = M T = ( tt tt ). Then tt2221 and tt1112 satisfy the Riccati equation. Moreover tt2122 and tt1112 are xed by Gal(Kl =K ) and therefore tt2221 , tt1211 2 K . 2
Proof.
2
2
1
1
(
)
( )
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
2
1
1
11
12
21
22
Now we consider a; b as elements of k(z ). We apply steps 1, 2 and 3 of the algorithm with z and q in the role of z and q respectively. In this way we will nd all solutions u 2 K of the Riccati equation. This nishes the algorithm. 2
2
2
2
80
5.4.2
G is reducible
If the Riccati equation has a solution in K1, then the q-di erence Galois group G is reducible. According to Theorem 5.2.6 we can compute an equivalent system (A) : y = Ay, where A has the form ( a0 db ). Moreover we can assume that b = 0 if the Riccati equation has two or in nitely many solutions. We denote by U the subgroup of Gl(2; C) consisting of upper triangular matrices and by D the subgroup of diagonal matrices. Lemma 5.4.5 The reducible subgroups G Gl(2; C) with G=G nite cyclic are 1. The zero{dimensional groups 0
Dk;l;e = f( 0 0 ) 2 D j k = 1; l = 1; eg g = 1g and the one{dimensional groups
Uk;l;e = f( 0 ) 2 U j k = 1; l = 1; eg g = 1g for l; k; e 2 Z1 with gcd(k; l; e) = 1, where g = gcdk(k;l) and g = gcd(lk;l) . 2. The one{dimensional groups
Dm;n = f( 0 0 ) 2 D j nm = 1g and the two{dimensional groups
Um;n = f( 0 ) 2 U j nm = 1g for m; n 2 Z 3. The two{dimensional group D and the three{dimensional group U . Proof. See [Hen96]. 2
The algorithm continues as follows. Suppose the Riccati equation has exactly one solution in K1. Then the di erence Galois group G is reducible but not completely reducible. If G is a proper subgroup of U , then according to Theorem 5.2.7 there must be a matrix T = ( f0 hg ) 2 U (K1) such that B = (T )AT = f (a f ) 2 G1(K1) . 0 d hh 1
( )
( )
81
First we want to determine whether the q-di erence Galois group G is one of the one{dimensional groups described in part 1 of Lemma 5.4.4. We apply the algorithm for rst order q{di erence equations to a and d. This yields a new equivalent system A~ = ( a0~ d~ ), where a~ and d~ are reduced.
Suppose rst that a~ and d~ are constant. Then the system A~ is already in standard form if a~ or d~ is a root of unity. If both a~ and d~ are roots of unity, then the di erence Galois group is one of the groups Uk;l;e. The q-di erence Galois group G is equal to Um; , if a~ is a primitive m-th root of unity and d~ is not a root a unity. And the q-di erence Galois group G is equal to U ;n, if d~ is a primitive n-th root of unity and a~ is not a root a unity. If a~ and d~ are constant but not roots of unity then if there exist n; m 2 Z such that a~m d~n = qr with r 2 Q the q-di erence Galois group G is one of the groups described in part 2 of lemma 5.4.4. Moreover, suppose that m is minimal and positive. In that case G is equal to the group Um;n. If r = 0 then the system is already in standard form. Suppose now that r 6= 0. It is obvious that there exists an f 2 K1 such that ( ff )n = q r . ~ is in standard form. Let T = ( f0 01 ). Then B = (T )AT Suppose now that that a~ and d~ are not constant. Then the q-di erence Galois group is not a group described in part 1 of Lemma 5.4.4. If a~ is constant and reduced and d~ is not constant and reduced then the system (A~) is already in standard form. The q-di erence Galois group G is the group Um; , if a~ is an m{th primitive root of unity. Otherwise the q-di erence Galois group G is the group U . If d~ is constant and reduced and a~ is not constant and reduced then the system ~ (A) is also already in standard form. The q-di erence Galois group G is the group U ;n, if d~ is an n{th primitive root of unity. Otherwise the q-di erence Galois group G is the group U . Suppose now that both a~ and d~ are reduced but not constant. The q-di erence Galois group is one of the groups Um;n if and only if there exists r; s 2 Z n f0g with gcd(r; s) = 1 and an f 2 K1 such that a~r d~s ff is a root of unity. Without loss of generality we can assume that r > 0. Let Na and Nd denote the degree of the numerator of a~ and d~ respectively and let Da and Dd denote the degree of the denominator of a~ and d~ respectively. Suppose rst that there exist r; s 2 Z with gcd(r; s) = 1 and an f 2 K1 such that a~r d~s ff is constant. Then we must have rDa = sNd and rNa = sDd, because a~ and d~ are reduced. If rDa = sNd and rNa = sDd holds then we apply the algorithm for rst order q{di erence equations to a~r d~s to investigate whether there exist an f 2 K1 such that a~r d~s ff is constant and reduced. If there exist such an f then we let T = 0
0
( )
1
0
0
( )
1
( )
( )
82
v ~ ( f0 f0w ), where v; w are chosen such that vr + ws = 1. Then B = (T )AT is in standard form. Suppose now that there exists an r 2 Z and an s 2 Z with gcd(r; s) = 1 and an f 2 K1 such that a~r d~s ff is constant and reduced. Then we must have rNa = sNd and rDa = sDd , because a~ and d~ are reduced. If rNa = sNd and rDa = sDd holds then we apply the algorithm for rst order q-di erence equations to a~r d~s to investigate whether there exist an f 2 K1 such that a~r d~s ff is constant and reduced. If there exist such an f v then we let T = ( f0 f0w ), where v; w are chosen such that vr + ws = 1. Then ~ is in standard form. B = (T )AT If none of the two cases above hold then the system (A~) is already in standard form and the q-di erence Galois group is the upper triangular group U . 1
1
( )
1
( )
1
We will not discuss the completely reducible case because this case is analogous to the reducible but not completely reducible case.
5.4.3 G is imprimitive
If we did not nd a solution u 2 K1 for the Riccati equation then the q-di erence Galois group G is irreducible. A group G is called imprimitive if G F = f( 0 0 ) j 6= 0g [ f( 0 0 ) j 6= 0g. In this subsection we determine whether G is imprimitive if we know already that G is irreducible.
Lemma 5.4.6 Suppose that the q-di erence Galois group G of the di erence equation (#) : 2 y + a y + by = 0 is irreducible. Then the following statements are equivalent. 1. G is an imprimitive group. 2. There exists an r 2 K1 such that the equations (#) and 2 y + ry = 0 are equivalent. Proof. See [Hen96]. 2 Suppose now that the equations (#) : 2y + a y + by = 0 with a 6= 0 and 2y + ry = 0 are equivalent. Then there exist c; d 2 K1 such that y is a solution of (#) if and only if cy + d y is a solution of 2y + ry = 0. In this case dc y + y satis es the equation 2y + 2d(d) ry = 0.
Theorem 5.4.7 Suppose that the q-di erence Galois group G of the equation (#) : y + a y + by = 0, a = 6 0 is irreducible. Then G is imprimitive if and only 2
83
if there exists a solution E 2 K1 of the Riccati equation b (b) (b)b 2 (E )E + ( 2 ( ) (a) + ) E + 2 = 0: a a a And if E 2 K1 is a solution of this Riccati equation then y satis es the equation (#) if and only if Dy + y satis es the equation 2y + ry = 0, where D = E + ab and r = a (a) + (b) + a 2 (D).
(##) :
Proof. See [Hen96]. 2
Remark 5.4.8 We can nd the rational solutions of equation (##) by applying the algorithm in subsection 5.4.1 with q replaced by q2 .
Lemma 5.4.9 The imprimitive subgroups G Gl(2; C) with G=G nite cyclic 0
are
1. F = f( 0 0 ) j 6= 0g [ f( 0 0 ) j 6= 0g . 2. Hn = f( 0 0 ) j ( )n = 1g [ f( 0 0 ) j ( )n = 1g for n 2 Z1 . 3. Hn+ = f( 0 0 ) j ( )n = 1g [ f( 0 0 ) j ( )n = 1g for n odd. Proof. See [Hen96]. 2 Consider the system (y) = ( d0 0c )y. We can apply the algorithm for rst order q{di erence equations to cd. Then we will nd an f 2 K1 such that cd (ff ) is reduced. Let T = ( f0 01 ). Then B = (T )( d0 0c )T 1 is in standard form. 5.4.4
G contains S l (2; C)
Consider the system (y) = ( 0b 1a )y. If the q-di erence Galois group G of this system is neither reducible nor imprimitive, then G = Gl(2; C) or G = Sl(2; C)n = f( ) j ( )n = 1g. We apply the algorithm for rst order q {di erence equations to b = det( 0 1 ) and nd an f 2 K such that f b b
a
is reduced. Let T = ( f0 01 ). Then B = (T )( 0b 1a )T form. This nishes the algorithm. 84
1
1
( )
f
is in standard
5.5 Examples In this section sequences spaces are used in order to make the Picard{Vessiot rings associated to a system of q{di erence equations more concrete. The explicit solutions which can be computed are certain sequences which will be called Liouvillian. The second order q-di erence equations which will be treated in subsection 5.5.2 are the so-called hypergeometric q-di erence equations. 5.5.1
Sequences spaces
Let F be any eld. Consider the set of sequences F N. This set forms a ring by using coordinate-wise addition and multiplication. Let J be the ideal of sequences with at most nitely many non-zero terms. Then we de ne SF = F N=J . The map : SF ! SF given by ((a1 ; a2; a3; : : :)) = (a2; a3 ; : : :) is well de ned on equivalence classes. Moreover, is an automorphism of the ring SF . For simplicity we will identify an element a = (a1 ; a2; a3; : : :) with its equivalence class. Consider the elds Ki = C(zi ). We can embed the elds Ki in SC by mapping a rational funtion f (zi ) to the sequence sf = (s1; s2; : : :), where sn = f (qin) for all but nitely many n 2 N. We will denote the homomorphisms f 7! sf by . We have (zi) = (zi+1 )i+1. Hence the embeddings of the elds Ki induce an embedding of the eld K1. It is obvious that = . The map : K1 ! SC is an injective homomorphism because K1 is a eld and (1) = 1. Therefore from now on we will identify an element f 2 K1 with its image sf in SC.
Theorem 5.5.1 Consider the system of q-di erence equations (A) : (y) = Ay, where A 2 Gl(n; K1). There exist a q-di erence subring R SC such that (R; ) is a Picard{Vessiot extension associated to the system (A) over the eld K1 . Proof. For the proof of this theorem we refer to chapter 4 of [PS96]. 2
De nition 5.5.2 We de ne the ring of Liouvillian sequences L recursively. L is the smallest subring of SC such that 1. K1 L. 2. a 2 L implies that (a) 2 L 3. a 2 K1 implies that b 2 L if bn+1 = an bn for all but nitely many n 2 N1 . 4. a 2 L implies that b 2 L if bn+1 = an + bn for all but nitely many n 2 N1. 5. a 2 L implies that b 2 L if there exist a j 2 N1 such that bjn = an and bn = 0 if n 6 0 mod j . In this case we call b an interlacing of a with zeroes.
85
Theorem 5.5.3 Suppose a 2 SC . Then the following statements are equivalent. 1. a 2 L 2. The sequence a satis es a linear q-di erence equation over K1 so that the q{di erence Galois group G associated to this equation is solvable.
Proof. For the proof we refer to [Hen96]. 2
A consequence of the proof of the previous theorem is that all the solutions of a second order q{di erence equation 2y + a y + by = 0 are Liouvillian if and only if the di erence Galois group G is reducible or irreducible and imprimitive. If the q{di erence Galois group G contains Sl(2; C), then there are no Liouvillian solutions of the second order q-di erence equation 2y + a y + by = 0.
Example. Let q = 4 and q2 = 2. Let a = 6z and b = 2z2 2z. Consider
the q-di erence equation 2y + a (y) + by = 0 and the corresponding Riccati equation (u)u + au + b = 0. We are in the case that v0 (b) 2v0(a) and v0 (b) is odd. Therefore the Riccati equation has not a solution in K1 . Now we have to apply step 4. of the algorithm to nd out whether there are solutions in K2 . From now on we will consider a and b as elements in Q(z2 ). The Riccati equation reads (u)u 6z22 u + 2z24 2z22 = 0. A calculation shows that S0 = fz2 + O(z22); z2 + O(z22)g and S1 = fz22 + O(z2); 21 z22 + O(z2)g. Further we have Sp = f1; z2; z2 1; z2 +1; z22; z22 z2 ; z22 +z2; z22 1; z23 z22 ; z23 +z22 ; z23 z2 ; z24 z22 g and St = f1g. We must have m = v0 (u)+ v0(p) v0 (t) and m = v1(t) v1(p) v1 (u). Hence v0 (p) + v1(p) = 1. Therefore p = z2 1 and m = 1. Suppose now u0 = z2 + O(z22), u1 = z22 + O(z2) and p = z2 + 1 then we nd u0 t = 1c + O(z2). Hence c = 1. And we nd czu12mtp = 1 + O( z12 ). Now qe must cz2m p be equal to 1. Hence e = 0. R = 1 satis es the equation (z) and one nds that u1 = z2 p = z22 + z2 is a solution of the Riccati equation in K2 . Suppose now u0 = z2 + O(z22), u1 = z22 + O(z2) and p = z2 1 then we nd u0 t = 1c + O(z2). Hence c = 1. And we nd czu12mtp = 1 + O( z12 ). Now qe must cz2m p be equal to 1. Hence e = 0. R = 1 satis es the equation (z) and one nds that u2 = z2 p = z22 z2 is a solution of the Riccati equation in K2. A calculation shows that the other combinations of u0, u1 and p do not yield any solution of the Riccati equation in K2. The di erence equation 2y + a y + by = 0 is equivalent with the system y = ( u01 u0 )y. This system is already 2 a 0 in standard form. Hence G = f( 0 d ) j a 2 C; d 2 Cg. Two linearly independent elementary solutions of the q-di erence equation in n 1 n 1 the ring SC are c and d, where cn = Q (22n + 2n) and dn = Q (22n 2n). The k =1
86
k =1
sequence c 2 SC satis es the rst order q-di erence equation (y) = u y and the sequence d 2 SC satis es the rst order q-di erence equation (y) = u y. 1
2
5.5.2 Hypergeometric q-di erence equations
In this section we apply the algorithm to some cases of the hypergeometric q{ di erence equation which is given by
(y) + ( 4 + q +zq )z1 + 3 q (y) + (2 q )(2 z q )1z 2 + q y = 0: As usual we will identify this equation with the system (A) : y = ( 0 1 )y, 1
1
2
where a =
(
4+
q +q )z +3 q z 1
1
and b =
(2
q )(2 q )z z 1
2+
q
1
.
b
a
Case 1. We will analyse the case where ; and are rational parameters
and q is transcendental. Note that the hypergeometric equation is symmetric in the parameters and . This restricts the number of possibilities we have to consider. Theorem 5.5.4 Let G denote the q{di erence Galois group of the hypergeometric q -di erence equation. 1. If = 0, = 0 and = 1 then G = f( 10 1b ) j b 2 Cg: 2. If ( = 0 and = 1) or ( = 1 and = 0) and 6= 1 then G = f( 10 d0 ) j d 2 Cg: 3. If ( = 0 and 6= 1) or ( 6= 1 and = 0) then G = f( 10 db ) j
b 2 C; d 2 C g:
4. If ( 6= 0 and = 1) or ( = 1 and 6= 0) then G = f( a0 db ) j
b 2 C; a; d 2 C g: 5. If 6= 0, 6= 1, 6= 0 and 6= 1 then G = Gl(2; C). Proof. Consider the rather special cases that ( = 0 and = 1) or ( = 1 and = 0). In these cases the hypergeometric q-di erence equation simpli es to the equation (y) (3 q ) (y) + (2 q )y = 0. If moreover = 1, then we get the equation (y) 2 (y) + y = 0. In this case u = 1 is the only solution of the corresponding Riccati equation. Hence the q-di erence Galois group is reducible but not completely reducible. Let T = 2
1
1
2
87
( 01 11 ) and B = (T )AT = ( 10 11 ). Obviously the equivalent system (B ) : (y) = B y is in standard form. Hence the q{di erence Galois group G is equal to the group f( 10 1b ) j b 2 Cg: Suppose now that 6= 0. Clearly u = 1 and u = 2 q are the only solutions of the corresponding Riccati equation. Hence the q-di erence Galois group G is completely irreducible if 6= 0. In that case the q{di erence equation is equivalent to the system of di erence equations y = ( 10 2 0q )y. This system is in standard form. The q{di erence Galois group G is equal to the group f( 10 d0 ) j d 2 Cg, because q is transcendental and is rational. From now on we will exclude the cases that ( = 0 and = 1) or ( = 1 and = 0). Consider the Riccati equation (y) : (u)u + au + b = 0. A simple calculation shows that S = f1+O(z); 2 q +O(z)g and S1 = f2 q +O( z ); 2 q +O( z )g. q Further we have Sp = f1; z q q g and St = f1; z q g. For all choices u 2 S , p 2 Sp and t 2 St we nd m = 0. For all u1 2 S1 we have v1(u1) = 0. Hence we must have degz (p) = degz (t). Now we have to consider four di erent possibilities. The rst combination is u = 1 + O(z), p = 1, t = 1 and u1 = 2 q + O( z ). We must have ucpt = 1+ O(z). Hence c = 1. And we compute ucp1t = 2 q + O( z ). So d = 2 q . We have d is of the form qe with e 2 Z if and only if = 0. In that case e = 0 and R = 1 turns out to be a solution of the equation (z). Therefore u = c RtR p = 1 is a solution of the Riccati equation if = 0. The second combination is u = 2 q + O(z), p = 1, t = 1 and u1 = 2 qa + O( z ). We must have ucpt = 1 + O(z). Hence c = 2 q . And we compute ucp1t = q q + O( z ). So d = q q . We have d is of the form qe with e 2 Z if and only if = 1. In that case e = 0 and R = 1 turns out to be a solution of the equation (z). Therefore u = c RtR p = 2 q is a solution of the Riccati equation if = 1. q The third combination is u = 1 + O(z), p = z q q , t = z q and u1 = 2 q + O( z ). We must have ucpt = 1 + O(z). Hence c = q q q q . And we compute ucp1t = q q q q + O( z ). So d = q q q q . Now d is not of the form qe with e 2 Z , unless we have = 0 and = 1. But we have excluded this situation. Hence this combination does not yield a rational solution of the Riccati equation. q The fourth combination is u = 2 q + O(z), p = z q q , t = z q u t and u1 = 2 q + O( z ). We must have cp = 1 + O(z). Hence c = q q q . And we compute ucp1t = q(2 q ) (2 q ) + O( z ). So d = q(2 q ) (2 q ). 1
1
1
1
1
0
1
2
(2
0
1
)
)(2
0
1
0
1
0
0
( )
1
0
1
1
0
2
2
1
1
2
1
2
0
( )
1
1
2
0
(2
1
)(2
)
)2 (2
(2
1
2
1
2
0
)
)2 (2
(2
1
(2
)
1 )(2
)
)
1
2
)(2
0
1
(2
0
1
2
88
1
2
1
0
)(2
)
(2
2
Now d is of the form qe with e 2 Z if and only if = = 0. Then e = 1. We substitute R = z + r in (z). (See subsection 5.4.1). We get the equation
)+(z + r ) (t)t(z (2 q ) = q2 (q z + r ) (p)p(z 1)+ q (qz + r ) (t)p( 2z +3 q 0. After simplifying this equation we get two linear equations for r that contradict each other. Hence this combination does not yield any solution of the Riccati equation. We found that there is exactly one solution of the Riccati equation in K if = 0 and 6= 1 or = 1 and 6= 0 or = 0 and 6= 1 or = 1 and 6= 0. Applying step 4 of the algorithm for solving the Riccati equation we nd that all the solutions of the Riccati equation in K1 are already in K . This computation is similar to the computations above. Suppose that = 0 and 6= 1. Let T = ( 01 11 ).Then we get B = 0
1
1
2
(2
)
(2
)
1
1
1)
q z q (T )AT = ( 1 q zz q 1 ). System (B ) is equivalent to system (A). 0 z Moreover system (B ) is in standard form because q is transcendental and 6=
1. The di erential Galois group G = f( 10 db ) j b 2 C; d 2 Cg. 2 1 ).Then we get Suppose that = 1 and 6= 0. Let T = ( 12 + + 2 1
1
1 q q z q
z ). System (B ) is equivalent to B = (T )AT = ( 2 q q z 0 z system (A). Moreover system (B ) is in standard form because q is transcendental and 6= 0. The q-di erence Galois group G = f( a0 db ) j b 2 C; a; d 2 Cg. If 6= 0, 6= 1, 6= 0 and 6= 1 then the q-di erence Galois group is irreducible. By using the algorithms of section 5.4.3 and 5.4.4 one can show that in this case the q-di erence Galois group G is the group Gl(2; C). 2 1
(2 1 (2 1
)
1
1
(1+
1
1
)
1 )
(2
1
1
It is possible to give two linearly independent Liouvillian solution for the hypergeometric q{di erence equations in all those cases where the q{di erence Galois group G is not the group Gl(2; C). If q is not transcendental then the situation can be di erent. For instance let = = = 1. Then the equation simpli es to (y)+ z q z (y)+ q 2z y = 0. If q is transcendentalpthen the q{di erence Galois group G is the z group Gl(2; C). But if q = 7, that is q is a root of the polynomial x + x + 2, then the di erence Galois group p G turns out to be reducible. From now on we will assume that q = 7. Consider the corresponding Riccati equation (y) : (u)u + au + b = 0, where q z q 2z a= z and b = z . We nd S = f1 + O(z)g and S1 = f2 q + Case 2.
(2
)
2
1
1 2
2
1 2
1 2
4+2 ) +2 1
4+2 ) +2 1
1
(
(
(2
)
1 2
1
0
1
89
O( z1 )g. Further we have Sp = f1; z (2 1q)2 g and St = f1; z q g. We must have degz (p) = degz (t). Hence we have to consider two combinations. The rst combination is u0 = 1 + O(z), p = 1, t = 1, and u1 = 2 q + O( 1z ). We must have ucp0t = 1 + O(z). Hence c = 1. We compute ucp1t = 2 q + O( z1 ). So d = 2 q . We have 2 q = q 3 . Hence e = 3. We substitute R = z 3 + r2 z 2 + r1 z + r0 in the equation (z). And we nd the solution R = z3 3q16+1 z2 + 5q64+7 z + 5q64+7 = (z 1 4 q )(z2 + 3 167q z 1 163q ). Now u = (RR) is a solution of the Riccati equation. The second combination is u0 = 1 + O(z), p = z (2 1q)2 , t = z q, and u1 = 2 q + O( 1z ). We must have ucp0 t = 1 + O(z ). Hence c = q(2 1 q)2 . We compute ucp1t = q(2 q)3 + O( 1z ). So d = q(2 q)3 = q10. Hence e = 10. We substitute R = z10 + r9 z9 + + r1z + r0 in (z). After simplifying we get eleven linear equations for the ten variables ri. There are no solutions. We conclude
that the second combination does not yield any solution of the Riccati equation. q q We have u = RR with R = (z )(z + q z ) is the only solution of the Riccati equation in K1. Let T = ( 1 uu 11 ). Then we get 0 1 )T = ( u 1 u + b=u ). System (B ) is not yet in stanB = (T )( b a 0 b=u R dard form. Note that u = R and b = ss , where s = (q z 1)(q z 1) (qz 1)(z 1). Let S = ( R0 R0 ) Then we get C = (S )BS = ( 10 f1 ), where ( )
1
4
2
3
7 16
1
3 16
1
( )
( )
1
f=
5
4
1
s
. System (C ) is in standard form. Therefore the q-di erence Galois group G is equal to the group f( 10 1b ) j b 2 Cg: (u
u2 +b)s (R)2
90
Chapter 6 On the classi cation of a class of q -di erence equations 6.1
Introduction
Let q be an m th root of unity. The aim of this article is to classify the qdi erence equations over the eld K = C(z). The classi cation is rather similar to the classi cation of di erential and di erence equations in characteristic p. See [Put95] and [PS96]. Further some Picard{Vessiot theory will be developed for these equations. 6.2
Picard{Vessiot Theory
Let : C(z) ! C(z) be an C{linear automorphism given by (z) = qz, where q is an m th root of unity. Consider the system of di erence equation (A) : y = Ay, where A 2 Gl(n; C(z)). (We restrict ourselves to equations with A 2 Gl(n; C(z)) in order to guarantee that we get n independent solutions. De nition 6.2.1 A ring R together with an automorphism R : R ! R is called a Picard{Vessiot extension over C(z) associated with system (A) if
1. R is a commutative ring, R C(z ) and RjC(z) = . 2. The only R {invariant ideals of R are 0 and R 3. There exists a matrix U 2 Gl(n; R) such that R (U ) = AU . (Such a matrix U is called a fundamental matrix for the system (A).) 4. R is minimal with respect to the conditions 1, 2 and 3 or equivalently if U = (uij ) 2 Gl(n; R) is a fundamental matrix for the matrix for the system (A) then R = C(z )[u11 ; : : : ; unn; det1(U ) ].
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From now on we will denote R also by . For a given system of q-di erence equations (A) : y = Ay one can construct a Picard{Vessiot extension in the following way. Let (xij ) denote a matrix of indeterminates and let det denote the determinant of this matrix. On the C(z){algebra C(z)[xij ; det1 ] one extends the automorphism by setting (xij ) = A(xij ). Let I be an ideal of C(z)[xij ; det1 ], which is maximal among the proper {invariant ideals. Then C(z)[xij ; det1 ]=I is a Picard{Vessiot ring for the system of equations (A). Any Picard{Vessiot ring is of this form. Unlike the case, where q is not a root of unity, there is no uniqueness of Picard{ Vessiot rings, due to the fact that in our case the eld of constants C(zm ) C(z) is not algebraically closed. For instance assume that q = 1 and consider the rst order equation (y) = y. Then Rc := C(z)[u; u 1]=(u2 c) is a Picard{Vessiot ring for all c 2 C(z) , because if c 2 C(zp) is a square then the only nontrivial p ideals are generated by the cosets of u c and u + c and these ideals are not {invariant. We have Rc1 = Rc2 if and only if cc21 2 C(z) is a square and this is not always the case. If q is not a root of unity, then there is a unique (unique up to C{linear isomorphisms commuting with ) Picard{Vessiot extension associated to every system of q{di erence equations. See [PS96]. Then the q-di erence Galois group of an equation is the group of C{linear automorphisms of the Picard{Vessiot ring commuting with . Unfortunately in our case there is not a unique Picard{ Vessiot ring for every system of q-di erence equations. Hence it is not possible to de ne a q{di erence Galois group of an equation in this way. However in section 4 we will use the theory of Tannakian categories for a suitable de nition of the q{di erence Galois group. The idea is to compare q{di erence modules over C(z) with modules over the ring Z = L[t; t 1 ], where L is the eld of constants C(zm ), because modules over Z = L[t; t 1 ] are easier to understand. To a system of q-di erence equations one can associate a q-di erence module. If a di erence equation y = Ay with A 2 Gl(n; C(z)) is given then one de nes a q{di erence module structure on C(z)n by setting y = A 1 y. We have ay = (a)y if a 2 C(z). In the next section we will classify the q{di erence modules over the eld C(z) if q is a root of unity.
6.3 Classi cation of q-Di erence Modules Let q be an m th root of unity and let D be the skew Laurent polynomial rinq C(z)[; 1 ], where the multiplication is xed by the rule z = qz. In this section we will classify the left D{modules which have nite dimension over C(z). Lemma 6.3.1 Let Z be the center of D. Then 1. Z = C(z m )[m ; m ] is a commutative Laurent polynomial ring. 2. D is a free Z {module of rank m2 . 92
3. Let Quot(Z ) denote the eld of quotients of Z . Then Quot(Z ) Z D is a skew eld with center Quot(Z ) and dimension m2 over its center. Proof.
1. For all integers j 2 Z one has j z = qj zj . This implies that j 2 Z if and only if j is a multiple of m. Further one has zj = qj zj . Hence zj 2 Z if and only if j is a multiple of m. So we have Z C(zm )[m ; m]. P fij zi j with all fij 2 Any f 2 D can be written uniquely as f = 0i;j