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..

 . . .
.
. . . »
°
···
0 0 B

One checks that a linear transformation φ ∈ Homk (M1 , M2 ) is a morphism of connec-
tions if and only if Hom (φ) = 0.
Suppose φ : M1 ’ M2 is a vector space isomorphism. Let E1 (resp., E2 ) be a basis of
M1 (resp., M2 ). Then one checks that φ is an isomorphism of connections if and only if

A2 = P P ’1 + P A1 P ’1 , (3.6)

where P = [φ]E1 ,E2 and Ai = [ i ]Ei for i = 1, 2.
Given an equation L(y) = 0, where

L = Dn + an’1 Dn’1 + · · · + a1 D + a0 ∈ D, (3.7)

we associate the the D-module M = (D/DL)— and de¬ne to be the induced connection
L

operator. Let E = 1, D, D2 , . . . , Dn’1 be a basis of D/DL, where n = ord(L). Then one
checks that [ L ]E — = AL , where
® 
···
0 1 0 0
 
 
···
 
0 0 1 0
AL =  . (3.8)
 
. . . .
..
 
. . . .
.
. . . .
° »
’a1 ’a2 ··· ’an’1
’a0

We see that ((D/DL)— , (k n , AL ). So, L corresponds to a ¬rst-order system Y = AL Y.
L)

Note that this is the system obtained from L(y) = 0 by Y = (y1 , y2 , . . . , yn )T , where y1 = y
and yi = yi’1 for 2 ¤ i ¤ n. We call AL the companion matrix of L.
Given a connection (M, ), it is a fact ([Kat87]) that if Ck k, then M contains a
cyclic vector, i.e., an element u ∈ M such that the set

E = u, (u), 2 n’1
(u), . . . , (u)

is a basis of M, where n = dim(M ). Applying this fact to (M— , —
), let u— be an element
and
F = u— , (u— ), (u— ), . . . , (u— )
2 n’1



14
a basis of M— . Then, one checks that
® 
···
0 0 0 a0
 
 
’1 ···
 
0 0 a1
 
 

’1 0 ···
[ ]F =  
0 a2
 
 
. . . .. .
 
. . . .
.
. . . .
° »
···
0 0 0 an’1


]F . It follows that (M—— , ——
for some a0 , . . . , an’1 ∈ k. Let B = [ ) (M, ) has matrix
[ ]F — = ’B T = AL , where L is given by (3.7); that is, [ ]F — is the companion matrix of
L. We note that there are algorithms to ¬nd a cyclic vector for a given system; one of these
is given in [Kat87].
Given a homogeneous equation L(y) = 0, L ∈ D, and a system Y = AY, A ∈ k n—n ,
we say that L(y) = 0 and Y = AY are equivalent to each other over k if (D/DL)— and
(k n , A) are isomorphic connections. Evidently L(y) = 0 and Y = AL Y are equivalent to
each other. If L(y) = 0 and Y = AY are equivalent to each other, then the extension K/k
is a Picard-Vessiot extension for L(y) = 0 if and only if it is a Picard-Vessiot extension for
Y = AY.
We say two operators L1 , L2 ∈ D are equivalent over k if the D-modules D/DL1 and
D/DL2 are isomorphic.

Proposition 3.1.1 Given L ∈ D with ord(L) > 0. Then, there exist r ∈ k, L1 , L2 , . . . , Lm ∈
˜˜ ˜ ˜˜
D, Li monic and irreducible for 1 ¤ i ¤ m, such that L = rL1 L2 · · · Lm . If L = rL1 L2 · · · Lm
is another such factorization, then r = r, m = m, and there exists a permutation σ ∈ Sm
˜ ˜
˜
such that Li is equivalent over k to Lσ(i) for all i, 1 ¤ i ¤ m.

Proof. This is Proposition 2.11 of [Sin96].




We say that the ¬rst-order systems Y = A1 Y and Y = A2 Y are equivalent over k if
they have the same order n and the connections (k n , and (k n ,
A1 ) A2 ) are isomorphic.

Proposition 3.1.2 Given L1 , L2 ∈ D, A1 , A2 ∈ k n—n . Suppose the equation Li (y) = 0 is
equivalent to the system Y = Ai Y for i = 1, 2. Let Ki /k (resp., Vi ) be the Picard-Vessiot
extension (resp., the full solution space) of Li (y) = 0 for i = 1, 2. Then, the following are
equivalent:

15
1. L1 and L2 are equivalent operators.

2. There exist operators R, S ∈ D of orders less than n such that

GCRD(R, L1 ) = 1, L2 R = SL1 . (3.9)


3. If K/k is a Picard-Vessiot extension containing KL1 and KL2 , then V1 V2 as G-
modules, where G = Gal(K/k).

4. Y = A1 Y and Y = A2 Y are equivalent systems.

5. There exists a matrix P ∈ GLn (k) such that (3.6) holds.

Proof. Equivalence of the ¬rst three statements is proved in Corollary 2.6 of [Sin96]. Observe
that if (3.9) holds, then the map 1 ’ R yields an isomorphism from D/DL2 to D/DL1 .
Equivalence of the ¬rst and fourth statements follows from de¬nitions. Equivalence of the
fourth and ¬fth statements follows from the discussion given immediately before and after
(3.6).




We de¬ne reducibility over k in various settings as follows: An operator L ∈ D is
reducible over k if there exist operators L1 , L2 of lower order such that L = L1 L2 . A system
Y = AY is reducible over k if it is equivalent over k to a system of the form
® 
B1 0
Y =° » Y.
B2 B3

A module is reducible if it has a nontrivial proper submodule. A connection (M, ) is
reducible if it includes a proper nontrivial subconnection, i.e., a vector subspace N ⊆ M
that is closed under . In each setting, we suppress the phrase “over k” when k is clear
from context.

Proposition 3.1.3 Let Y = AY be a ¬rst-order system over k. Let K/k be a Picard-
Vessiot extension for this system and G = Gal(K/k). Let L ∈ D be an operator that is
equivalent to this system. Then, the following are equivalent:

1. The connection (k n , A) contains a proper nonzero subconnection.

2. The D-module k n , where DY = A (Y ), is reducible.

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3. The D-module D/DL is reducible.

4. L is reducible over k.

5. Y = AY is reducible over k.

6. The solution space of Y = AY in K is a reducible G-module.

Proof. This is Proposition 2.1 of [CS99].




We de¬ne complete reducibility over k in various settings as follows: An operator L ∈ D
is completely reducible over k if is the least common left multiple of irreducible operators.
A system Y = AY is completely reducible over k if it is equivalent over k to a system of
the form
Y = diag(B1 , B2 , . . . , Bt )Y,

where the system Z = Bi Z is irreducible for 1 ¤ i ¤ t. A module is completely reducible if
it is the direct sum of irreducible submodules. In each setting, we suppress the phrase “over
k” when k is clear from context.

Proposition 3.1.4 Let Y = AY be a ¬rst-order system over k. Let K/k be a Picard-
Vessiot extension for this system and G = Gal(K/k). Let L ∈ D be an operator that is
equivalent to this system. Then, the following are equivalent:

1. The connection (k n , A) is completely reducible.

2. The D-module k n , where DY = A (Y ), is completely reducible.

3. The D-module D/DL is completely reducible.

4. L is completely reducible over k.

5. Y = AY is completely reducible over k.

6. The solution space of Y = AY in K is a completely reducible G-module.

7. G is a reductive group.

Proof. This is Proposition 2.2 of [CS99].



17
Computing the group of L(y) = b, L completely re-
3.2
ducible
ˆ
Consider the inhomogeneous equation L(y) = b, b ∈ k. Let L = (D ’ b /b) —¦ L. De¬ne the
Picard-Vessiot extension KI (resp., the Galois group GI ) of L(y) = b to be KL (resp., GL ).
ˆ ˆ

ˆ
Note that L is a right factor of L, so that VL ⊆ KI includes a full solution set of L(y) = 0.
ˆ

Thus, we may write KH ⊆ KI and VL ⊆ VL . Moreover, if f ∈ VL \ VL , then there exists a
ˆ ˆ

nonzero constant c such that f0 = cf and L(f0 ) = b. Since any two solutions of L(y) = b
di¬er by an element of VL , we see that the full solution set of L(y) = b is f0 + VL . Moreover,
if E is a basis of VL , then E ∪ {f } is a basis of VL . It follows that KI /k is the minimal
ˆ

di¬erential ¬eld extension containing the full solution set of L(y) = b.

Proposition 3.2.1 Given:

1. k a di¬erential ¬eld, Ck algebraically closed of characteristic zero

2. L ∈ D = k[D] a completely reducible operator

3. b ∈ k

4. KI /k (resp., VI , GI ) is the Picard-Vessiot extension (resp., full solution space, group)
of L(y) = b

5. KH /k (resp., VH , GH ) is the Picard-Vessiot extension (resp., full solution space, group)
of L(y) = 0

6. L1 , L0 ∈ D are monic operators satisfying the following conditions:

(a) L1 (y) = b has a k-rational solution.

(b) L = L1 L0 for some L0 ∈ D.

(c) L1 is of maximal order.

Then GI has Levi decomposition GI W GH , where W VL0 as vector groups. In
addition, the pair (L1 , L0 ) given in Item 6 above is unique.

Proof. This is Proposition 2.1 of [BS99]; it is an adaptation of Th´or`me 1 of [Ber92]. We
ee

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