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:

32

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

35

’ 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