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We are now ready to prove Theorem 4.1.5.
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

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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.

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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

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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

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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.

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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

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