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σ(yi )/yi = σ(Ri )/Ri , i = 1, 2, 3, for all σ ∈ T. (4.8)

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 ).

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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.

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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

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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
=

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