Proof of Theorem 4.1.5. In what follows, G ⊆ SL3 is an admissible subgroup; G = Ru P is

a Levi decomposition with P (resp., Ru ) a maximal reductive subgroup (resp., the unipotent

radical) of G; H = (P, P ) is semisimple and T = Z(P )—¦ is a torus. Let G = [G]E0 , where

E0 = {e1 , e2 , e3 } is a ¬xed basis of the vector space V C 3 and G ⊆ SL(V ). Also let

Ru = [Ru ]E0 , P = [P]E0 , H = [H]E0 , and T = [T]E0 .

By Lemma 4.2.3, there exist a basis F = {f1 , f2 , f3 } and a unique subgroup T ⊆ SL3

such that T = [T]F is one of the subgroups listed in Lemma 4.2.3. By Lemma 4.3.4, there

exist a basis B = {b1 , b2 , b3 } and a unique subgroup Ru ⊆ SL3 such that Ru = [Ru ]B is one

of the subgroups listed in Lemma 4.3.4.

We study the admissible subgroups by cases according to decomposition, as follows:

1. H 1.

(a) T 1. In this case, the zero-defect property of admissible groups implies that

86

G 1.

C—.

(b) T

C — is as described in Lemma 4.2.3.

i. Ru 0. Here, G

C — . See Lemma 4.5.3.

C. Here, G C

ii. Ru

C — . See Lemma 4.5.3.

C 2 . Here, G C2

iii. Ru

C — . See Lemma 4.5.7.

U3 . Here G U3

iv. Ru

C— — C—.

(c) T

C — — C — is conjugate to D3 © SL3 , cf. Lemma 4.2.3.

i. Ru 0. Here, G

(C — — C — ). See Lemma 4.5.5.

C. Here, G C

ii. Ru

(C — — C — ). See Lemma 4.5.5.

C 2 . Then G C2

iii. Ru

U3 (C — —C — ). In this case, it is clear that G is conjugate

U3 . Then G

iv. Ru

to [G]A = T3 © SL3 for some suitable basis A. The G-invariant subspaces are

a1 and a1 , a2 .

SL2 .

2. H

(a) T 1.

SL2 , and it follows from Lemma 4.5.9 that there exists a

i. Ru 0. Here, G

basis A of V such that

±

0

t13 = t23 =

(tij ) ∈ SL3 : t31 = t32

[G]A = .

= 0

1

t33 =

By Lemma 4.5.9, the G-invariant subspaces are a1 , a2 and a3 .

C. Here, by Lemma 4.5.9, the only possible conjugation action of SL2

ii. Ru

on C is the trivial action. Therefore, by hypothesis on the defect of G, we

conclude that this case does not arise.

C 2 . See Lemma 4.5.8.

iii. Ru

U3 . In this case, we have G U3 SL2 . Then, it is not hard to show

iv. Ru

U3 (T2 © SL2 ). It is also straightforward

that G has Borel subgroup B

to show that the unipotent radical R of B is isomorphic to U3 U2 . If G

can be embedded in SL3 , then (by Kolchin™s theorem) R can be embedded

in U3 . But R has dimension 4; we obtain a contradiction and conclude that

this case does not arise.

87

C — or T C — — C — . By Item 3 of Lemma 4.5.9, V has exactly two nontrivial

(b) T

proper H-invariant subspaces V1 and V2 , where V = V1 + V2 (direct sum) and Vr

has dimension r for r = 1, 2. Since T commutes with H, we have that the image

of V1 (resp., of V2 ) under T is an H-invariant subspace of V, which must be either

0 or V1 or V2 . Moreover, T is a group of automorphisms of V ; thus, for r = 1, 2,

we have that T maps Vr to a subspace of equal dimension; it follows that T maps

Vr onto Vr . Item 2 of Lemma 4.5.9 then implies that T acts as scalar multiples

of the identity on V1 (resp., on V2 ). Since T is connected, it then follows that

T|V2 ⊆ GL(V2 ) is either 1 or the full group of scalar-multiple operators on V2 .

Note that T must act nontrivially on one of V1 , V2 .

C — . If T|V2 is trivial, then we see that T|V1 C — ; this implies

Suppose T

SL2 — C — , an impossibility by Item 4 of Lemma 4.5.9. Assume instead that

P

T acts as the full group of scalar-multiple operators on V2 . In this case, we see

that P |V2 = GL(V2 ) and therefore P is conjugate to the subgroup

{(tij ) ∈ SL3 : t13 = t23 = t31 = t32 = 0} GL2 . (4.7)

C — — C — . Since H © T is ¬nite, we see that the Levi subgroup

Now suppose T

P = HT contains a 3-dimensional torus. This contradicts the fact that any torus

embedded in SL3 has dimension at most 2. We conclude that this case does not

arise.

GL2 :

We consider the following cases for Ru , assuming P

GL2 conjugate to the group described in

i. Ru 0. Here, we have G = P

(4.7). If A is a suitable basis, then one checks that the G-invariant subspaces

are a1 , a2 and a3 .

C. Here, we have G C GL2 . Then G includes a copy of C SL2 .

ii. Ru

By Item 1 of Lemma 4.5.9, SL2 can only act trivially on C; it follows that G

includes a copy of C — SL2 . But (by Item 4 of Lemma 4.5.9) C — SL2 cannot

be embedded in SL3 ; we obtain a contradiction and conclude that this case

does not arise.

C 2 . See Lemma 4.5.8.

iii. Ru

U3 . Here, it follows from our result in the case (H SL2 , T = 1, Ru

iv. Ru

U3 ) that this case does not arise.

PSL2 . In this case, it follows from Lemma 4.5.9 and its proof that T = 1.

3. H

88

PSL2 . Lemma 4.5.9 implies that, for some suitable basis A,

(a) Ru = 0. Here, G

ˆ

the subgroup G = [G]A has a faithful parametrization

®

® ®

2 2

ab b

a

a b a b

° » ’ 2ac 2bd , ° » ∈ PSL2 .

ad + bc

° »

c d c d

c2 2

cd d

A calculation shows that this subgroup is cut out of SL3 by the equations t2 =

12

t11 t13 , t2 = 4t11 t31 , t2 = 4t13 t33 , t2 = t31 t33 , (t22 + 1)2 = 4t11 t33 , and

21 23 32

(t22 ’ 1)2 = 4t13 t31 . It follows from Lemma 4.5.9 that there are no nontrivial

proper G-invariant subspaces in this case.

(b) Ru = C or Ru = C 2 or Ru = U3 . In this case, we have that G PSL2 .

Ru

Let V be the vector group Ru /(Ru , Ru ), and note that V has dimension at

most 2. Now consider the representation Ψ : PSL2 ’ GL(V ) induced by the

conjugation action of PSL2 on Ru . Recall that Item 1 of Lemma 4.5.9 implies

that all nontrivial irreducible representations of PSL2 have dimension at least

3. Since PSL2 is reductive, it follows that V is built from copies of the trivial

representation of PSL2 , i.e., that Ψ is the trivial map. This contradicts the

hypothesis that G has defect zero, and we conclude that this case does not arise.

SL3 . In this case it is clear that G = SL3 and that there are no nontrivial proper

4. H

G-invariant subspaces.

This exhausts the possibilities for H, T and Ru , and we conclude that our list of admissible

subgroups of SL3 is complete.

Computing the group of D3 + aD + b, a, b ∈ C[x]

4.6

In this section, unless and until speci¬ed otherwise, C is a computable, algebraically closed

constant ¬eld of characteristic zero with factorization algorithm; k = C(x) unless otherwise

speci¬ed; and D = k[D].

fi Di ∈ D, de¬ne the adjoint of L (see [Sin96]) to be

Given an operator L = i

We have that adj : D ’ D is an anti-automorphism of D and

ii

adj L = i (’1) D fi .

(D/DL)— as D-modules.

that D/D adj L

89

Given an equation L(y) = 0, we may de¬ne the associated Riccati equation Ricc L (see

[Sin96]) as follows: Formally substitute the formula y = exp( u) into L(y) = 0, then divide

by exp( u). The resulting nonlinear equation in u is Ricc L. We see that the following are

equivalent for a given f ∈ k :

1. f is a solution of Ricc L.

2. F is a solution of L(y) = 0, where F /F = f.

3. D ’ f is a right factor of L.

Given L ∈ C[x] of order 3. By computing rational solutions of Ricc L (resp., Ricc(adj L)),

we can compute n1 , the number of ¬rst-order right factors (resp., n2 , the number of ¬rst-

order left factors ” which is also the number of second-order right factors) of L. Corol-

lary 4.1.3 implies that the group of L is one of the groups listed in the appropriate category

in Theorem 4.1.5 or, equivalently, the appropriate cell in Table 4.1. The following results

enable us to compute the group of a given operator L once n1 and n2 are known, in those

cases in which there is more than one entry in Table 4.1. Below, Lemma 4.6.1 addresses the

case where n1 = n2 = 0. The remaining lemmas in this section address the case in which

GL is solvable.

Lemma 4.6.1 Let L ∈ D be a third-order di¬erential operator and let G be the group of

the equation L(y) = 0. Suppose G is known to be isomorphic to either SL3 or PSL2 . If L 2

PSL2 ; otherwise G SL3 .

has order 5 or factors, then G

Proof. This is a special case of Theorem 4.7 in [SU93].

Before proceeding, remark that if

D3 + a2 D2 + a1 D + a0

L=

(D ’ r3 ) —¦ (D ’ r2 ) —¦ (D ’ r1 ),

=

where a2 , a1 , a0 , r1 , r2 , r3 ∈ C(x), then a straightforward calculation yields r3 = ’a2 ’r1 ’r2 .

In particular, suppose GL ⊆ SL3 , so that a2 = h /h for some h ∈ C(x) by Lemma 4.1.2;

then r3 = ’r1 ’ r2 + h /h.

Lemma 4.6.2 Given L = (D + r1 + r2 ’ h /h) —¦ (D ’ r2 ) —¦ (D ’ r1 ) ∈ D, r1 , r2 , h ∈ k.

90

1. The Picard-Vessiot extension KL /k is generated by elements R1 , R2 , y2 , ξ, y3 , where:

(a) Ri = ri Ri for i = 1, 2

(b) y2 = r1 y2 + R2

’1 ’1

(c) ξ = r2 ξ + R1 R2 h

(d) y3 = r1 y3 + ξ.

The full solution space VL has ordered basis F(r1 ,r2 ) = {y1 , y2 , y3 } , where y1 = R1 .

2. Let K1 = k(R1 , R2 ) ⊆ KL . Then K1 /k is a Picard-Vessiot extension for the operator

Lred = LCLM(D ’ r1 , D ’ r2 , D + r1 + r2 ’ h /h), so that we may write KLred = K1 .

3. De¬ne T = {σ ∈ GL : σ(yi )/yi ∈ C — for i = 1, 2, 3} ⊆ GL . Then T is a maximal torus

of GL . We have