0 0 1

We claim that a = 0 and c = 0 : Indeed, this claim follows easily from Lemma 4.3.4

1

and the fact that clos(M0 ) = U(1,1) .

Suppose [T ]F(r1 ,r2 ) = T(m1 ,m2 ) is such that T(m1 ,m2 ) = T(1,0) . Then we claim that

[Ru ]F(r1 ,r2 ) = U3 , so that the algorithm is correct in this case. Indeed, if [Ru ]F(r1 ,r2 )

¯

U3 , then Ru Qd for d = 1 or 2, and we see by Item 5 of Lemma 4.3.3 that

[Ru ]F(r1 ,r2 ) ⊆ clos(M ) · U(0,0) . Applying Item 2 of Lemma 4.5.1 with

1

Q = diag(tm1 , tm2 , t’m1 ’m2 ),

t some nonroot of unity, we obtain T(m1 ,m2 ) = T(1,0) (note that T(’1,0) = T(1,0) ),

contradicting hypothesis. This proves the claim.

¯

Finally, suppose T(m1 ,m2 ) = T(1,0) . Then one checks that r2 = g /g for some g ∈ Q(x).

Correctness now follows from Lemma 4.6.5.

This exhausts the list of subgroups listed in Theorem 4.1.5 satisfying n1 = n2 = 1.

• (1, ∞) (resp., (∞, 1); (2, ∞); (∞, 2)): Correctness follows from the fact that there is

exactly one subgroup listed in Theorem 4.1.5 having these values for n1 and n2 .

111

4.7 Examples

Before presenting examples related to this algorithm, we state the following result:

Theorem 4.7.1 Let G be a connected linear algebraic group de¬ned over an algebraically

closed ¬eld C of characteristic zero. Let d(G) (resp., e(G)) be the defect (resp., the excess

” cf. [MS96]) of G. Then G is the Galois group of a Picard-Vessiot extension of C(x)

corresponding to a system of the form

Ad(G)

A1

+ ··· +

Y= + A∞ Y,

x ’ ±1 x ’ ±d(G)

where Ai is a constant matrix for i = 1, . . . , d(G), and A∞ is a matrix with polynomial

entries of degree at most e(G). In particular, the only possible singularities of this system

are d(G) regular singular points in the ¬nite plane and a (possibly irregular) singular point

at in¬nity.

Proof. This is Theorem 1.2 in [MS96].

This theorem, suitably rewritten in terms of operators, implies that each algebraic group

named in Theorem 4.1.5 arises as the Galois group of a third order operator whose only

singularities in the ¬nite plane are apparent singularities.

For each group G ⊆ SL3 named in Theorem 4.1.5, we would like to ¬nd an operator of

¯

the form L = D3 + aD + b, a, b ∈ Q[x], such that GL G. Unfortunately, it is not known

whether such an operator exists in each case. Below, for each such subgroup G ⊆ SL3 , we

G; L has nonzero D2 term and one or more

name a third-order operator L such that GL

apparent singularities in the ¬nite plane in general. In each case, we apply either Algorithm

IV or a modi¬ed version of Algorithm IV. For each example, ni denotes the number of

ith-order right factors of L for i = 1, 2.

1. Examples satisfying H 1.

1. Let L = D3 . Applying Algorithm IV, we compute

(a) Examples satisfying T

¯

n1 = n2 = ∞, and 1, x, x2 is a basis of Q(x)-rational solutions. Algorithm IV

then returns the trivial group.

Q— .

¯

(b) Examples satisfying T

112

i. Examples satisfying Ru 0. Let

= D3 + (’a2 ’ a1 a2 ’ a2 )D + (a2 a2 + a1 a2 )

L 1 2 1 2

= (D + a1 + a2 ) —¦ (D ’ a2 ) —¦ (D ’ a1 ),

a1 , a2 ∈ Z \ {0} , GCD(a1 , a2 ) = 1, a1 + a2 = 0,

a1 = ’a1 ’ a2 , a2 = ’a1 ’ a2 .

Q— . Varying

¯

One checks that n1 = n2 = 3. Moreover, we easily see that GL

a1 and a2 yields the di¬erent representations of Q— described in Lemma 4.2.3.

¯

Q— ,

¯ ¯ ¯

Q. Here, we seek examples in which G Q

ii. Examples satisfying Ru

as in Items 1-4 of Lemma 4.5.2 and Item 1 of Lemma 4.5.3.

A. Let

LCLM (D ’ a2 x) —¦ (D ’ a1 x), D + (a1 + a2 )x ,

L=

a1 , a2 ∈ Z \ {0} , GCD(a1 , a2 ) = 1, a1 + a2 = 0,

a1 = ’a1 ’ a2 , a2 = ’a1 ’ a2 .

It can be shown that n1 = n2 = 2, that applying Lemma 4.6.9 yields

[T ]F(r1 ,r2 ) = T(a1 ,a2 ) , and that n1 = 1, n2 = 2; we conclude that Return

Q— with (Int t)(u) = ta1 ’a2 u.

¯ ¯

Q

GL

Alternatively, let Ri ∈ KL be such that Ri /Ri = ai x for i = 1, 2.

Let y2 ∈ KL be such that y2 ’ a1 xy2 = R2 . Then one checks that

’1 ’1

F= R1 , y2 , R1 R2 is an ordered basis of VL . One further checks

1

that [GL ]F has unipotent radical U(1,0) and maximal torus T(a1 ,a2 ) . A

Q— , with (Int t)(u) = ta1 ’a2 u.

¯ ¯

Q

computation then shows that GL

Note that by varying a1 and a2 , one can obtain any of the semidirect

product structures described in Item 1 of Lemma 4.5.2.

B. Let L = D3 +(’x2 ’2)D ’x = (D +x)—¦D —¦(D ’x). Maple computations

show that n1 = n2 = 1, that r2 = g /g for g = 1 and that the equations

¯

given in Item 3(d) and 4(d) of Lemma 4.6.5 admit Q(x)-rational solu-

Q— with (Int t)(u) = tu as in Item 2

¯ ¯

Q

tions. We conclude that GL

of Lemma 4.5.2.

C. Let L = D3 +(’3x2 +3)D+(2x3 ’6x) = (D’x)—¦(D’x)—¦(D+2x). Maple

Q— ,

¯ ¯

computations yield n1 = 2, n2 = ∞, and we conclude that GL Q

113

with (Int t)(u) = t’3 u as in Item 4 of Lemma 4.5.2. Taking adj L yields

the subgroup given in Item 3 of Lemma 4.5.2.

¯

Q2 .

iii. Examples satisfying Ru

A. Let

LCLM(D ’ a1 x, D ’ a2 x) —¦ (D + (a1 + a2 )x),

L= (4.25)

a1 , a2 ∈ Z \ {0} , GCD(a1 , a2 ) = 1, a1 + a2 = 0,

a1 = ’a1 ’ a2 , a2 = ’a1 ’ a2 .

Here, it can be shown that n1 = 1 and n2 = 2 and that applying

Lemma 4.6.9 yields [T ]F(r1 ,r2 ) = T(’a1 ’a2 ,a1 ) . We conclude that GL

Q— with (Int t)(u, v) = (ta1 ’a2 u, t2a1 +a2 v).

¯ ¯

Q2

Alternatively, let Ri ∈ KL be such that Ri /Ri = ai x for i = 1, 2. Let

yj ∈ KL be such that yj + (a1 + a2 )xyj = Rj’1 for j = 2, 3. Then

’1 ’1

one checks that F = R1 R 2 , y 2 , y 3 is an ordered basis of VL . One

2

further checks that [GL ]F has unipotent radical U(1,0) and maximal torus

Q— , with

¯ ¯

Q2

T(’a1 ’a2 ,a1 ) . A computation then shows that GL

(Int t)(u, v) = (t’2a1 ’a2 u, t’a1 ’2a2 v)

for t ∈ Q— , u, v ∈ Q.

¯ ¯

By varying a1 and a2 , one can obtain any of the semidirect product

structures described in Item 5 of Lemma 4.5.2.

¯

B. Let L = adj L, where L is as de¬ned in (4.25). Arguments similar to

Q— is as described in

¯ ¯

Q2

those used in computing GL show that GL

¯

Item 6 of Lemma 4.5.2.

C. Let L = D3 + (’x4 ’ 5x)D + (’3x3 ’ 3) = (D + x2 ) —¦ (D + 1/x) —¦ (D ’

x2 ’1/x). Maple computations show that n1 = n2 = 1, that r2 = g /g for

g = 1/x, that the equation given in Item 3(d) of Lemma 4.6.5 admits a

¯

Q(x)-rational solution but the equation given in Item 4(d) of that lemma

Q— , with (Int t)(u, v) = (tu, t2 v)

¯ ¯

Q2

does not. We conclude that GL

as in Item 7 of Lemma 4.5.2.

D. Let

L = D3 + (’3x4 + 6x)D + (2x6 ’ 12x3 + 4)

(D ’ x2 ) —¦ (D ’ x2 ) —¦ (D + 2x2 ).

= (4.26)

114

Maple computations show that n1 = 1, n2 = ∞, and we conclude that

Q— , with (Int t)(u, v) = (t’3 u, t’3 v) as described in Item 8 of

¯ ¯

Q2

GL

Lemma 4.5.2.

¯

E. Let L = adj L, where L is as de¬ned in (4.26). Arguments similar to

Q— is as described in

¯ ¯

Q2

those used in computing GL show that GL

¯

Item 9 of Lemma 4.5.2.

U3 .

iv. Examples satisfying Ru

A. Let

(D + (a1 + a2 )x) —¦ (D ’ a2 x) —¦ (D ’ a1 x),

L=

a1 , a2 ∈ Z \ {0} , GCD(a1 , a2 ) = 1, a2 = 0.

It can be shown that n1 = n2 = 1 and that applying Lemma 4.6.9 yields

U3 · T(a1 ,a2 ) .

[T ]F(r1 ,r2 ) = T(a1 ,a2 ) . We conclude that GL

Alternatively, it can be shown that KL /k is generated by elements R1 ,

R2 , y2 , ξ, y3 ∈ KL satisfying Ri /Ri = ai x for i = 1, 2, y2 ’ a1 xy2 =

’1 ’1

R2 , ξ ’ a2 xξ = R1 R2 and y3 ’ a1 xy3 = ξ. It can moreover be shown

that VL has ordered basis B = {R1 , y2 , y3 } ; that [GL ]B has maximal torus

T(a1 ,a2 ) and unipotent radical U3 . These facts yield GL U3 · T(a1 ,a2 ) .

Varying a1 and a2 yields the di¬erent conjugation actions described in

Item 1 of Lemma 4.5.6.

B. Let

D3 + (’x4 ’ 2x3 ’ x2 ’ 5x ’ 3)D + (’3x3 ’ 5x2 ’ 2x ’ 3)

L =

(D + x2 + x) —¦ (D + 1/x) —¦ (D ’ x2 ’ x ’ 1/x).

=

Maple computations show that n1 = n2 = 1, that r2 = g /g for g = 1/x,

and that the equation given in Item 3(d) of Lemma 4.6.5 fails to admit

Q— , with

¯ ¯

a Q(x)-rational solution. We conclude that U3

GL

«®

®

tb t2 c

1bc 1

¬

·