The map σ ’ σ|KLred gives an isomorphism of T onto GLred .

Proof. Items 1 and 2 are proved by straightforward calculations. We prove Item 3 as

follows. First note that VD’r1 = spanC {y1 } and V(D’r2 )—¦(D’r1 ) = spanC {y1 , y2 } are GL -

invariant subspaces of VL , so that [GL ]F(r1 ,r2 ) ⊆ T3 . It is now clear that [GL ]F(r1 ,r2 ) has

Levi decomposition

[GL ]F(r1 ,r2 ) = ([GL ]F(r1 ,r2 ) © U3 )([GL ]F(r1 ,r2 ) © D3 ) semidirect product of subgroups.

We now see that T = [GL ]F(r1 ,r2 ) © D3 is a Levi subgroup of GL . Next, notice that

’1 ’1

y1 = R1 , (D ’ r1 )(y2 ) = R2 , ((D ’ r2 ) —¦ (D ’ r1 ))(y3 ) = R1 R2 h.

Given σ ∈ T, suppose σ(yi ) = ti yi for some ti ∈ C — for i = 1, 2, 3. We compute

= σ((D ’ r1 )(y2 ))

σ(R2 )

(D ’ r1 )(σ(y2 )) = (D ’ r1 )(t2 y2 )

=

= t2 (D ’ r1 )(y2 ) = t2 R2 ,

’1 ’1 ’1 ’1

and a similar computation shows that σ(R1 R2 h) = t3 R1 R2 h; (4.8) follows easily. The

last statement follows from the fact that a member of T (resp., of GLred ) is determined by

’1 ’1

its action on F(r1 ,r2 ) (resp., on R1 , R2 , R1 R2 h ).

91

The following three lemmas will be used to decide the Galois group of an operator

L = D3 + aD + b, a, b ∈ C[x], in case L has a unique monic factor Li of order i for i = 1, 2

such that GL is solvable and acts trivially on VL2 /VL1 . Remark that these results can be

easily generalized to the case in which L is a third-order operator whose only singularities

in the ¬nite plane are apparent singularities and whose Galois group is unimodular.

Lemma 4.6.3 Let k be a di¬erential ¬eld whose constant ¬eld is algebraically closed and

has characteristic zero. Let r ∈ k. Then there is a unique ring automorphism shiftr : k[D] ’

k[D] such that shiftr (D) = D ’ r and shiftr (h) = h for all h ∈ k. Moreover, if k(·)/k is a

Picard-Vessiot extension such that · /· = r, then

shiftr (L) = · —¦ L· ’1 ∈ k(·)[D] for all L ∈ k[D].

Proof. This lemma is an easy exercise using the relevant de¬nitions.

Lemma 4.6.4 Given: L = D3 + aD + b = (D + r1 + r2 ) —¦ (D ’ r2 ) —¦ (D ’ r1 ), a, b ∈

C[x], r1 , r2 ∈ C(x), and g ∈ C(x) \ {0} such that r2 = g /g and the only monic right factors

of L are

1, L1 = D ’ r1 , L2 = (D ’ r2 ) —¦ (D ’ r1 ), and L.

Let y1 , y2 , ξ, y3 be as de¬ned in Lemma 4.6.2.

˜

1. De¬ne L2 = (D + r1 + r2 ) —¦ (D ’ r2 ), the unique monic left factor of L of order 2.

˜

Then g, ξ span a full set of solutions of L2 in KL , so we may write k ⊆ KL2 ⊆ KL .

˜

˜

D ’ r2 is the only monic right factor of L2 of order 1.

2. De¬ne L2 = (D ’ 3r2 ) —¦ (D ’ r1 ’ 2r2 ) = shift2r2 (L2 ). Then g 2 y1 , g 2 y2 span a full set

of solutions of L2 in KL , so we may write k ⊆ KL ⊆ KL . D ’ r1 ’ 2r2 is the only

2

monic right factor of L2 of order 1.

˜ ˜

3. De¬ne L2 = D —¦ (D ’ r1 ’ 2r2 ) = shiftr1 +r2 (L2 ). Then g 2 y1 , gy1 ξ span a full set of

˜

solutions of L2 in KL , so we may write k ⊆ KL ⊆ KL . D ’ r1 ’ 2r2 is the only monic

˜

2

˜

right factor of L2 of order 1.

92

1 ’1 2 ˆ ˆ

and let L = (D ’ 2r2 + r1 ) —¦ L2 . Then B = {y1 , y2 , y3 } is an

4. De¬ne y3 =

ˆ 2 y1 y2 ˆ

ˆ

(ordered) basis of solutions of L in KL , so we may write k ⊆ KL ⊆ KL .

ˆ

Proof. These statements are all veri¬ed by straightforward computations using known facts.

Lemma 4.6.5 Given: L = D3 + aD + b = (D + r1 + r2 ) —¦ (D ’ r2 ) —¦ (D ’ r1 ), a, b ∈

C[x], r1 , r2 ∈ C(x), and g ∈ C(x) \ {0} such that r2 = g /g and the only monic right factors

of L are

1, L1 = D ’ r1 , L2 = (D ’ r2 ) —¦ (D ’ r1 ), and L.

˜ ˜

Let the elements y1 , y2 , ξ, y3 be as de¬ned in Lemma 4.6.2. Let the operators L2 , L2 , L2 and

ˆ

L be as de¬ned in Lemma 4.6.4.

1. There exists a basis A = {·1 , ·2 , ·3 } of VL such that [GL ]A is one of

U(1,1) T(1,0) , U(1,1) T(1,0) , U3 T(1,0) .

1 2

2. We have y1 , y2 , ξ, y3 ∈ C(x).

/

3. The following are equivalent:

d

(a) [GL ]A = U(1,1) T(1,0) for d = 1 or 2.

˜

(b) L2 is equivalent to L2 over C(x).

(c) The equation (D ’ r1 ’ 2r2 )(y) = g 3 + c admits a C(x)-rational solution for some

c ∈ C \ {0} .

(d) The equation D —¦ (D ’ r1 ’ 2r2 ) (y) = r2 g 3 admits a C(x)-rational solution.

4. The following are equivalent:

1

(a) [GL ]A = U(1,1) T(1,0) .

ˆ

(b) L and L are equivalent over C(x).

(c) The equation (D ’ r1 ’ 2r2 ) —¦ (D ’ 2r1 ’ r2 ) (y) = g 3 + c admits a C(x)-rational

solution for some c ∈ C \ {0} .

(d) The equation D —¦(D ’r1 ’2r2 )—¦(D ’2r1 ’r2 ) (y) = r2 g 3 admits a C(x)-rational

solution.

Proof.

93

1. By Corollary 4.1.3 and Theorem 4.1.5, there exists a basis A = {·1 , ·2 , ·3 } of VL such

that [GL ]A is one of the subgroups listed in Theorem 4.1.5 having exactly one invariant

subspace of dimension 1 (resp., 2). Moreover, since L1 maps VL2 onto VD’r2 =

spanC {g} ⊆ C(x), we see that GL acts trivially on VL2 /VL1 . The only subgroups listed

in Theorem 4.1.5 having these properties are U(1,1) T(1,0) , U(1,1) T(1,0) and U3 T(1,0) .

1 2

2. Let A = {·1 , ·2 , ·3 } be as described in Item 1 of the conclusion of the lemma. From

the possibilities for [GL ]A , we see that there exist σ0 , „0 ∈ GL such that

[σ0 ]A = diag(t0 , 1, t’1 ), t0 ∈ C — ,

0

®

and

b0

1 a0

c0 , a0 , b0 , c0 ∈ C, a0 = 0, c0 = 0.

[„0 ]A = 0 1

° »

00 1

One checks that, for all v ∈ VL \{0} , either σ0 (v) = v or „0 (v) = v, so that v lies outside

the ¬xed ¬eld C(x) of GL . This yields y1 , y2 , y3 ∈ C(x). Finally, notice that L1 maps A

/

˜

onto a basis of VL2 and that L1 (·1 ) = 0; it follows that the set A2 = {L1 (·2 ), L1 (·3 )}

®

˜

1 c0

is an (ordered) basis of VL2 . Moreover, we have that [„ ]A2 = ° » with c0 = 0;

˜ ˜

01

it follows that VL2 ⊆ C(x). By Item 2 of Lemma 4.6.4, another basis of VL2 is {g, ξ} .

˜ ˜

Since g ∈ C(x) and VL2 ⊆ C(x), we have ξ ∈ C(x).

/

˜

3. First of all, we show that the ¬rst and second conditions are equivalent, as follows.

Let A2 = {·1 , ·2 } ; it is clear that A2 is an (ordered) basis of VL2 . We also see that

˜

A2 = g 2 ·1 , g 2 ·2 (resp., A2 = {g·1 L1 (·2 ), g·1 L1 (·3 )}) is an (ordered) basis of VL

2

(resp., VL ). Let GL have Levi decomposition GL = Ru T (semidirect product of

˜

2

subgroups). From Item 1 of the conclusion of the lemma, we see that [Ru ]A ⊆ U3 and

[T ]A = T(1,0) . Let σ be a typical element of T, and write

[σ]A = diag(tσ , 1, t’1 ), tσ ∈ C — . (4.9)

σ

One checks that

® ®

1 0 tσ 0

[σ]A2 = ° » , [σ]A2 = [σ] = [σ] ˜ = ° ».

˜ A A

t’1 2 2

0 0 1

σ

It follows that VL and VL are isomorphic as T -modules. Thus, they are isomorphic

˜

2 2

as GL -modules if and only if they are isomorphic as Ru -modules. Let „ be a typical

94

element of Ru , and write

®

1 a„ b„

c„ , a„ , b„ , c„ ∈ C.

[„ ]A = 0 (4.10)

1

° »

0 0 1

A calculation shows that

® ®

1 a„ 1 c„

[„ ]A2 = [„ ]A = ° » , [„ ] ˜ = [„ ] ˜ = ° ».

A2 A

2 2

0 1 0 1

From here, one checks that VL and VL are isomorphic as Ru -modules if and only

˜

2 2

if there exists a nonzero constant ± such that a„ = ±c„ for all „ ∈ Ru . The latter

condition holds (with ± = 1) if Ru ⊆ U(1,1) and fails if Ru = U3 . Equivalence of the

2

¬rst two statements is now clear.

Before proceeding, we make some auxiliary computations:

(D ’ r1 ’ 2r2 )(g 2 y1 ) (g 2 ) y1 + g 2 y1 ’ (r1 + 2r2 )g 2 y1

=

2r2 g 2 y1 + r1 g 2 y1 ’ (r1 + 2r2 )g 2 y1

=

= 0; (4.11)

(D ’ r1 ’ 2r2 )(g 2 y2 ) (g 2 ) y2 + g 2 y2 ’ (r1 + 2r2 )g 2 y2

=