48

ˆ ˜

• The action of Q ∈ GH on v ∈ C r i mi

˜

is given by Q.v = Qv (matrix-by-vector

i

multiplication), where

˜ ˜ ˜

Q= diag(Q1 , . . . , Qs ),

˜ ¯ ¯

Qi = diag(Q1 , . . . , Q1 ) (˜i copies),

r

¯

where Qi is as described above.

We prove the correctness of this algorithm as follows:

Proof. First, we recall that equivalent systems have identical Picard-Vessiot extensions

(resp., Galois groups). This algorithm computes the group of (3.33) and thus group of the

original system.

Computing GH is accomplished by [CS99], so by Proposition 3.3.1 we need to compute

the unipotent radical U and the action of GH on U. Write U = Ui as in Lemma 3.4.7.

i

˜ ˜

r

˜

VMii as modules over the group of Y = AY and thus

Then Lemma 3.4.8 implies that Ui

over GH . Correctness of Algorithm III is now immediate.

We now present examples of this algorithm.

Example 3.4.10 Consider the equation L(y) = b, where

L = D2 ’ 4xD + (4x2 ’ 2) = (D ’ 2x) —¦ (D ’ 2x)

and b ∈ C(x) as® Example 3.2.2. This equation is equivalent to the system Y = AY +

in

0 1

B, where A = ° » and B = (0, b)T . A computation shows that another

’4x2 + 2 4x

˜ ˜˜ ˜ ˜ ˜

equivalent system is Y = AY + B, where A = diag(2x, 2x) and B = (’xb, b)T . The

˜

transformation from one system to the other is obtained by writing Y = P Y, where P =

®

1 + 2x2 ’x

° » . Applying Lemma 3.4.8, we see that U C 2’r , where

’2x 1

dimC (c1 , c2 ) : y = 2xy ’ c1 xb + c2 b

r =

admits a C(x)-rational solution .

This computation is essentially identical to the one obtained in the ¬rst remark at the end

of Section 3.2. Thus, for this particular example, Algorithm III essentially coincides with

Algorithm I.

49

Example 3.4.11 Consider the ¬rst-order system

Y = diag(M, M, M, M, M )Y + (0, x2 , 0, x, 0, 1, 0, 1/x, 0, 1/x2 )T ,

®

01

where M = ° » . In this case, we see that GH is the group of the equation y ’xy = 0.

x0

SL2 . To compute U, we consider the

From Example 3.3.4 above, we conclude that GH

equation

y ’ xy = c1 x2 + c2 x + c3 + c4 /x + c5 /x2 .

ˆ

Applying the algorithm given in the proof of Lemma 3.4.9, we consider the equation L(y) =

0, where

ˆ

L = LCLM(D ’ 2/x, D ’ 1/x, D, D + 1/x, D + 2/x) —¦ (D2 ’ x).

ˆ

A ratsols computation in Maple shows that the space of rational solutions of L(y) = 0 is

spanned by the elements 1 and x, which furthermore fail to satisfy y ’ xy = 0. Applying

C6 SL2 , where

Lemma 3.4.8, we now see that GI

T

Q.(v1 , v2 , v3 )T = (Qv1 )T , (Qv2 )T , (Qv3 )T

T T T

for Q ∈ SL2 , v1 , v2 , v3 ∈ C 2 , and Qvi is the standard matrix-by-vector product.

Example 3.4.12 Consider the matrix equation

diag(A1 , A2 )Y + (B1 , B2 )T ,

T T

Y =

®

0 1

»,

diag(M, M, M, M, M ), M = °

A1 =

x0

A2 = diag(2x, 2x),

(0, x2 , 0, x, 0, 1, 0, 1/x, 0, 1/x2 )T ,

B1 =

(’x, 1)T .

B2 =

Evidently GH is the group of the equation L(y) = 0, where L = LCLM(D2 ’ x, D ’ 2x).

We see that GH is a subgroup of SL2 — C — that projects surjectively onto each factor. As in

Example 3.3.5, the Theorem of [Kol68] yields GH = SL2 — C — . Next, Lemma 3.4.7 implies

U1 • U2 (direct sum of GH -modules), where U1 (resp., U2 ) is the unipotent radical

that U

of the system given in Example 3.4.11 (resp., Example 3.4.10 with b = 1). After applying

(SL2 — C — ), where

C7

the results of Example 3.4.11 and 3.2.2, we conclude that GI

T

(Q, t).(v1 , v2 , v3 , w)T = (Qv1 )T , (Qv2 )T , (Qv3 )T , tw

50

for (Q, t) ∈ SL2 — C — , v1 , v2 , v3 ∈ C 2 , w ∈ C.

Note that applying the method of Algorithm I to this system (or rather, more precisely,

to an equivalent inhomogeneous scalar equation) would involve computing all factorizations

of a twelfth-order completely reducible operator; this step alone would require solving a very

large system of equations, in contrast with the simple steps performed above.

51

52

Chapter 4

Computing the group of

D3 + aD + b, a, b ∈ C[x]

4.1 De¬nitions and main results

In this chapter, except where otherwise speci¬ed, C is an algebraically closed constant ¬eld

of characteristic zero.

We de¬ne the defect of a connected group, using the de¬nition given in [Sin99], as

follows: Assume G is a connected linear algebraic group with Levi decomposition G = Ru P

(semidirect product of subgroups), where Ru is the unipotent radical and P a Levi subgroup

of G. Note that the group Ru /(Ru , Ru ) is commutative and unipotent, hence isomorphic to a

vector group C n for some n. We see that the conjugation action of P on Ru leaves (Ru , Ru )

invariant and therefore induces a representation of P on the vector group Ru /(Ru , Ru ).

U1 1 • · · · • Us s , where each Ui is an irreducible P -module. Suppose

n n

Write Ru /(Ru , Ru )

U1 is the trivial one-dimensional P -module. Then the defect of G is the number n1 .

Proposition 4.1.1 Given L ∈ C(x)[D] of order three such that all singularities of L in the

¬nite plane are apparent singularities and the group GL of L(y) = 0 over C(x) is included

in SL3 . Then GL is connected and has defect zero.

Proof. This result follows from Proposition 11.20 and Theorem 11.21 of [dPS].

53

The following well-known result allows us to apply Proposition 4.1.1 to operators of the

form L = D3 + aD + b, a, b ∈ C[x].

Lemma 4.1.2 Given L = Dn + an’1 Dn’1 + · · · + a1 D + a0 ∈ D. Then the group GL is

isomorphic to a subgroup of SLn if and only if an’1 = f /f for some f ∈ C(x).

Proof. Let B = {y1 , . . . , yn } be a basis of VL . Then the fundamental solution matrix

(i’1)

associated to B is ZB = (yj ) ∈ GLn . It is well-known (see, e.g., [Mag94]) that if Z = ZB

is a fundamental solution matrix of L, then an’1 is the logarithmic derivative of det Z. Also,

given σ ∈ GL , one checks that σ(det Z) = det([σ]B ) det Z. It then follows from basic Galois

theory that det Z is contained in C(x) if and only if det([σ]B ) = 1 for all σ ∈ GL , i.e.,

[GL ]B ⊆ SLn . The desired result then follows easily.

Corollary 4.1.3 Given L = D3 + aD + b ∈ C(x)[D], a, b ∈ C[x]. Then GL is included in

SL3 , is connected, and has defect zero.

Proof. This result follows easily from Proposition 4.1.1 and Lemma 4.1.2.

In light of the above results, we de¬ne a subgroup G ⊆ SL3 to be admissible if it is

connected and has defect zero. Theorem 4.1.5 below enumerates the admissible subgroups

of SL3 up to conjugation.

Before proceeding, we de¬ne some speci¬c algebraic subgroups of SL3 using certain

parameterizations. Remark that these parameterizations are noncanonical; we use them

in this section for clarity of description only. See Lemmas 4.2.2 and 4.3.1 for equivalent

de¬nitions of these subgroups that do not use parameterizations.

diag(y d1 , y d2 , y ’d1 ’d2 ) : y ∈ C — ,

T(d1 ,d2 ) =

d1 , d2 ∈ Z, d1 ≥ d2 ≥ 0, d1 > 0, GCD(d1 , d2 ) = 1

±®

10x

:x∈C

1

0 1 0

U(0,0) =

°

»

001

54

±®

12

2x

1 x

x :x∈C

1

U(1,1) = 0 1

°

»

0 0 1

±®

1 x y

0 : x, y ∈ C

2

=