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vector group. Moreover, we have Gal(K/k) W Gal(F/k). The action of Gal(F/k)
on W is induced from the action of Gal(E/k) on W by the equality σ.w = (σ|E ) .w
for σ ∈ Gal(F/k).

Proof. The ¬rst two statements follow immediately from the discussion preceding the
Lemma.
We prove the third statement as follows: Complete reducibility of the systems Y = A1 Y
and Y = A2 Y implies that Gal(F/k) is reductive; it follows that Gal(E/k) is reductive as
well. The ¬rst part of the third statement now follows from Proposition 3.3.1.
To prove the second part, we consider the following diagram:

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F(˜) =K
·
 
 
F E(˜)
·
rr 

E

k

˜ ˜
De¬ne W = Gal(K/F ). De¬ne a map ψ : W ’ Gal(E(˜)/E) by composing the inclusion
·
˜
map W ’ Gal(K/E) with the restriction map Gal(K/E) Gal(E(˜)/E). We claim that
·
˜
ψ maps W isomorphically onto W, so that

˜
W = Gal(F (˜)/F )
· Gal(E(˜)/E)
· W.

To prove this claim, it su¬ces to show that F © E(˜) = E (see Lemma 5.10 of [Kap76] and
·
W is abelian. It follows that Gal(E(˜)/(F © E(˜)) is
its proof). Note that Gal(E(˜)/E)
· · ·
a normal subgroup of W and therefore that (F © E(˜))/E is a Picard-Vessiot extension. We
·
have that Gal((F © E(˜))/E) is a quotient of W and so is unipotent. Since Gal(F © E(˜)/E)
· ·
is also a quotient of the reductive group Gal(F/k) it is also reductive and therefore must be
trivial. Therefore F © E(˜) = E.
·
˜ ˜
Since Gal(K/F )/W G(F/k) is reductive, we have that W is the unipotent radical of
˜
Gal(K/k). It follows that G = Gal(K/k) has Levi decomposition G = W P (semidirect prod-
Gal(F/k) maps P ⊆ G onto Gal(F/k).
uct of subgroups), where the restriction map G
Note also that Gal(E(˜)/k) has Levi decomposition Gal(E(˜)/k) = W Q (semidirect prod-
· ·
uct of subgroups), where Q ⊆ Gal(E(˜)/k) is the image of P under the restriction map
·
P/PE given
G Gal(E(˜)/k). Moreover, we have natural isomorphisms Q
· Gal(E/k)
by restriction maps.
Let PE = P © Gal(K/E), so that the image of PE under the restriction map G
Gal(F/k) is the subgroup Gal(F/E) ⊆ Gal(F/k). We see that PE is normal in P and
so is reductive. The image of PE under the restriction homomorphism Gal(K/E)
Gal(E(˜)/E)
· W is a unipotent group and is therefore trivial. This means that PE
˜
¬xes · . Since an automorphism in W = Gal(F (˜)/F ) is determined by its action on · ,
˜ · ˜
˜
we see that PE commutes elementwise with W and therefore that the conjugation action
˜ ˜
of PE on W is trivial. It follows that the map ψ : W ’ W is an isomorphism not only
of P -modules but of P/PE -modules. The desired result now follows after considering the
natural isomorphism P/PE Gal(E/k).



33
This last result and its proof tell us how to compute the Galois group of equation (3.22)
when k = C(x). Here is the algorithm, followed by examples, as stated in [BS99]:
Algorithm II
Input: A system of linear di¬erential equations (3.22) where Y = A1 Y and Y = A2 Y are
completely reducible with A1 ∈ C(x)n—n , A2 ∈ C(x)m—m .
Output: A system of equations in m + n variables de¬ning the Galois group G(F/k) ‚
GLn+m (C) of the Picard-Vessiot extension corresponding to the system (3.26), an integer t
and a rational homomorphism ¦ : G(F/k) ’ GLt (C) such that the Galois group of (3.22) is
Ct G(F/k) where the action of G(F/k) on C t by conjugation is given by ¦.

˜
1. One ¬rst calculates the Galois group G of equation (3.26) using the results of [CS99].
This Galois group will be represented as matrices acting on diag(Y1 , Y2 ) where Y1
is a fundamental solution matrix of Y = A1 Y and Y2 is a fundamental solution
matrix of Y = A2 Y . One can easily calculate the action of G(F/k) on (Y1’1 )T — Y2
and so calculate the Galois group G(E/k) of equation (3.27) as well as the map
G(F/k) ’ G(E/k).

ˆ
2. Find a scalar equation L(y) = 0 equivalent to the equation (3.27) as well as an element
ˆ ∈ k so that equation (3.25) is equivalent to L(y) = ˆ (an algorithm to do this is
ˆ
b b
presented in [Kat87]; in the examples below ad hoc methods are used). Using the
˜ ˜
transformation of Y = AL Y to V = (’AT — In + Im — A2 )V allows us to calculate
ˆ 1
ˆ
the action of G(E/k) on the solution space of L(y) = 0.

3. Proposition 3.2.1 allows us to calculate a vector group W so that the Galois group of
L(y) = ˆ (and so of equation (3.25)) is W
ˆ b G(E/k).

4. Lemma 3.3.3 now tells us that the Galois group of equation (3.22) is the group
W G(F/k) where the action of G(F/k) on W (i.e., the homomorphism ¦) can
be calculated from the information we have.

Remark: As in the case of the equation L(y) = b, the algorithms of [CS99] can be
combined with the above to give a presentation of the Picard-Vessiot extension corresponding
to L1 (L2 (y)) = 0.
We will now give three examples of this method. In these examples we will start with an
equation of the form L1 (L2 (y)) = 0 with coe¬cients in k = C(x). The Galois group G(F/k)
” that is the Galois group of equation (3.26) in Lemma 3.3.3 ” is the same as the Galois

34
group of LCLM(L1 , L2 ). In the examples we shall apply ad hoc methods to calculate this
Galois group. We will then calculate a scalar equation equivalent to the system (3.27) as
well as the matrix B de¬ning this equivalence. This will allow us to ¬nd a scalar equation
L(y) = ˆ equivalent to the system (3.25). We then apply the methods of Section 3.2 to
ˆ b
calculate the vector space W.

Example 3.3.4 Consider the equation L(y) = 0, where L = L1 —¦ L2 , L1 = D2 ’ x, L2 =
1
D2 + x D + 1.
˜
The Galois group of this equation is an extension of the Galois group GL of L =
˜

LCLM(L1 , L2 ). Since L1 and L2 are both known to have Galois group isomorphic to
SL2 (C) (L1 is a form of Airy™s equation and L2 is a Bessel equation), GL is a subgroup
˜

of SL2 (C) — SL2 (C).
We claim that if GL is a proper subgroup of SL2 (C)—SL2 (C), then the operators L1 2 and
˜

L2 2 are equivalent over C(x). We prove this claim as follows: Let {y1 , y2 } (resp., {z1 , z2 }) be
a basis for VL1 (resp., VL2 ). Before proceeding, we make the following auxiliary calculations
2 2
related to the basis z1 , z1 z2 , z2 of VL 2 . For i = 1, 2, we have
2


2
(zi ) = 2zi zi

’ (zi )
2
2(zi zi + (zi )2 )
=
1
2(zi (’ zi ’ zi ) + (zi )2 ),
=
x

from which a straightforward simpli¬cation yields

12 12
(zi )2 = zi +
2
(zi ) + (zi ) . (3.28)
2x 2

A similar set of computations involving the ¬rst and second derivatives of z1 z2 yields

1 1
z 1 z 2 = z1 z2 + (z1 z2 ) + (z1 z2 ) . (3.29)
2x 2

Now suppose GL is a proper subgroup of SL2 (C) — SL2 (C). Then, according to ([Kol68], p.
˜

˜ ˜
1158), there exist a quadratic extension k/C(x), an element ± ∈ k, and a matrix S = (sij ) ∈
GL2 (C(x)) such that ±2 ∈ C(x) and

Wr(y1 , y2 ) = diag(±, ±) · S · Wr(z1 , z2 ).

For i = 1, 2, we have

= ±(s11 zi + s12 zi )
yi

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’ yi
2
= ±2 (s2 zi + 2s11 s12 zi zi + s2 (zi )2 )
2
11 12
12 12
= ±2 (s2 zi + s11 s12 (zi ) + s2 (zi +
2 2 2
(zi ) + (zi ) )) by (3.28)
11 12
2x 2
12 1
±2 (s2 + s2 )zi + ±2 (s11 s12 +
2
s12 )(zi ) + ±2 s2 (zi ) ,
2 2
= (3.30)
11 12 12
2x 2
±2 (s2 z1 z2 + s11 s12 (z1 z2 + z1 z2 ) + s2 z1 z2 )
y 1 y2 = 11 12
1 1
±2 (s2 z1 z2 + s11 s12 (z1 z2 ) + s2 (z1 z2 +
= (z1 z2 ) + (z1 z2 ) )) by (3.29)
11 12
2x 2
12 1
±2 (s2 + s2 )z1 z2 + ±2 (s11 s12 + s12 )(z1 z2 ) + ±2 s2 (z1 z2 ) .
= (3.31)
11 12 12
2x 2
From (3.30) and (3.31), it follows that the linear operator
1 22 2 12
s12 )D + ±2 (s2 + s2 ) ∈ C(x)[D]
± s12 D + ±2 (s11 s12 + 11 12
2 2x
2 2 2 2
maps the ordered basis z1 , z1 z2 , z2 of VL onto the ordered basis y1 , y1 y2 , y2 of VL 2 .
2
2 1

The claim now follows immediately.
In our case, we claim that L1 2 and L2 2 are inequivalent (and therefore that GL =
˜

SL2 (C) — SL2 (C)). The expanded version of DEtools developed by Mark van Hoeij for
MapleV.5 allows one to calculate symmetric powers, LCLM™s and a basis of the ring of D-
module endomorphisms of D/DL for an operator L ∈ D = C(x)[D]. Using this we proceed
as follows. A calculation shows that

M = LCLM(L1 2 , L2 2 )

is of order 4. If L1 2 and L2 2 were equivalent then D/DM would be the direct sum of two
isomorphic D-modules. The endomorphism ring of D/DM would therefore have dimension
4. Using the eigenring command in DEtools one sees that this ring has dimension 2 and
the desired result follows.
We now consider the equation

˜ ˜ ˜
V = H V + C,

where

’AT — I2 + I2 — A2 ,
H =
®1 
01
=° »,
A1
x0
® 
0 1

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