A note about rational representations of differential Galois groups

0.17 MB·English

by  Marc Reversat

A NOTE ABOUT RATIONAL REPRESENTATIONS OF
DIFFERENTIAL GALOIS GROUPS.
MARC REVERSAT
7
0 Abstra
t. We give a des
ription of the rational representations
0 of the di(cid:27)erential Galois group of a Pi
ard-Vessiot extension.
2

n
a
J

9
Contents
2

] 1. On representations of Galois groups of Pi
ard-Vessiot
S
extensions. 1
D
2. A
orrespondan
e. 6
.
h Referen
es 8
t
a
m
[

1 We give a des
ription of the rational representations of the di(cid:27)eren-
v
tial Galois group of a Pi
ard-Vessiot extension (theorems 1.1 and 2.1).
4
4 This gives a new des
ription of the di(cid:27)erential Galois
orresponden
e.
8 More results will be given for abelian di(cid:27)erential extensions in a forth-
1

oming paper [1℄, espe
ially the analog of the Artin
orresponden
e.
0
7
0
/ 1. On representations of Galois groups of Pi
ard-Vessiot
h
extensions.
t
a
m Inallthisnote we
onsiderdi(cid:27)erential(cid:28)eldswithalgebrai
ally
losed
C
: (cid:28)elds of
onstants, denoted by . The derivative will be denoted by a
v
i dash.
X K n ≥ 1 Mn(K)
Let be a di(cid:27)erential (cid:28)eld. Let be an integer, and
r GLn(K) n×n
a are the usual notationsfor algebraand group of matri
es
K GL (K) M (K)
n n
with entries in . The group a
ts on by the following
rule:
GL (K)×M (K) −→ M (K)
n n n
(U,A) 7−→ U′U−1 +UAU−1
U = (u ) U′ = (u′ )
i,j 1≤i,j≤n i,j 1≤i,j≤n
where if , then . This a
tion
an be
de(cid:28)ned in an other way, maybe more
omprehensible. Consider the
group
H (K) := M (K)×GL (K),
n n n
Date: February 2, 2008.
1
2 MARC REVERSAT
A,B M (K)
n
the law being de(cid:28)ned by the following formula: for all in
F,G ∈ GL (K)
n
and all
(A,F)(B,G) = (A+FBF−1,FG).
It admits the subgroups
∆ (K) := (U′U−1,U) / U ∈ GL (K) ,
n n
(cid:8) (cid:9)
{0}×GL (K) M (K)×{1}
n n
and , this last one being normal. We set
Z (K) := ∆ (K)\H (K)/({0}×GL (K)).
n n n n
GL (K) M (K)×{1}
n n
The group a
ts on by the following rule:
GL (K)×(M (K)×{1}) −→ (M (K)×{1})
n n n
(U,(A,1)) 7−→ (U′U−1,U)(A,1)(0,U−1)
= (U′U−1 +UAU−1,1)
M (K) = (M (K)×{1})
n n
With the identi(cid:28)
ation and in
lusion
(M (K)×{1}) ⊂ H (K)
n n
, it indu
es a
anoni
al bije
tion
GL (K)\M (K) ≃ ∆ (K)\H (K)/({0}×GL (K)) = Z (K).
n n n n n n
(1)
Z (K)
n
We will use below these two de(cid:28)nitions of .
A M (K) F GL (K) [(A,F)]
n n
Let be in and be in , we denote by ,
[A] Z (K) (A,F) ∈ H (K) A ∈ M (K)
n n n
resp. , the
lass in of , resp. of .
L/K H (K) ⊆
n
Let be a Pi
ard-Vessiot extension. The in
lusion
H (L) α(L/K) : Z (K) → Z (L)
n n n
gives rise to a map . We set
Z (L/K) := {a ∈ Z (K) / α(L/K)(a) = [0]}.
n n
G Rep (G)
n
For any group we denote by the set of equivalent
lasses
G GL (C) G = dGal(L/K)
n
of representations of in , if isthe di(cid:27)erential
L K Rep (L/K) := Rep (G)
n n
group of over , we set .
L/K
Theorem 1.1. Let be a Pi
ard-Vessiot extension, then there ex-
Z (L/K) Rep (L/K)
n n
ists a natural bije
tion between and .
Proof. First of all we re
all some fa
ts that are of main importan
e in
our proof.
c Y′ =
A
1.0.1. The representation . Consider the di(cid:27)erential equation
AY A ∈ M (K) E/K
m
, with , and let be a
orresponding Pi
ard-Vessiot
E K
extension, i.e. is generated over by the
oe(cid:30)
ients of a funda-
F c
A A
mental matrix of the equation. The rational representation is
dGal(E/K) −→ GL (C)
m
σ 7−→ c (σ)
A
c (σ) σ(F ) = F c (σ) c
A A A A A
where [Ais]su
(hAt,h1a)t Z (K) . NBot=e tUh′aUt−1+dUepAeUn−d1s only
m
Uon∈thGeL
las(sK) of in ,be
auseif Y′ = BY UwFith
m A
, a fundamental matrix of the equation is
c = c
B A
and we see that . Note also that an other fundamemtal matrix
F γ γ ∈ GL (C)
A n
is of the form , with , then it gives the representation
γ−1c γ c
A A
equivalent to .
DIFFERENTIAL GALOIS GROUP. 3
c c c
A [A] E
We will write equivalently , or for this
lass of representa-
tions.
dGal(E/K) A ∈ M (K)
m
1.0.2. The Galois group . Let and
R = K[(X ) ,(det)−1]/q = K[(x ) ]
i,j 1≤i,j≤m i,j 1≤i,j≤m
K Y′ = AY
be aPi
ard-Vessiotringover fortheequation . In these for-
X K[(X ) ]
i,j i,j 1≤i,j≤m
mulasthe areindetermin(Xate′s),thering= A(X ) isdeetquipped
i,j 1≤i,j≤m i,j 1≤i,j≤m
by the derivation satisfying , (cid:16) (cid:17) is the
(X ) q
i,j 1≤i,j≤m
deterxminantofthematriXx E =,Quisota(Rm)aximaUld=i(cid:27)edrGenatli(aEl/idKea)l
i,j i,j
and is the image of . Let and ,

onsider
K (X ) ,(det)−1 ⊆ E (X ) ,(det)−1
i,j 1≤i,j≤m i,j 1≤i,j≤m
(cid:2) (cid:3) (cid:2) (cid:3)
= E (Y ) ,(det)−1 ⊇ C (Y ) ,(det)−1 ,
i,j 1≤i,j≤m i,j 1≤i,j≤m
(cid:2) (cid:3) (cid:2) (cid:3)
(Y ) (X ) = (x ) (Y )
i,j 1≤i,j≤m i,j 1≤i,j≤m i,j 1≤i,j≤m i,j 1≤i,j≤m
where Y′ = 0 isde(cid:28)nedby .
i,j
Note that . We know that
U = dGal(E/K) = SpecC (Y ) ,(det)−1 /J
i,j 1≤i,j≤m
(cid:2) (cid:3)
J = qE[(Y ) ,(det)−1]∩C[(Y ) ,(det)−1]
i,j 1≤i,j≤m i,j 1≤i,j≤m
where ([2℄proof
y
i,j
of prop. 1.24 or the beginning of Ÿ1.5). We denote by the image of
Y
i,j
, then we have
U = dGal(E/K) = SpecC[(y ) ]
i,j 1≤i,j≤m
U = dGal(E/K)
1.0.3. T = SpecRas a torsor. We T
ontinueUwith the prKevious
notations. Set , we know that is an -torsor over ([2℄
theorem 1.30), moreover, we know that there exists a (cid:28)nite extension
K K T× K = Spec R⊗ K U
K (cid:16) K (cid:17)
of su
h that is a trivial -torsor over
Ke e e b ∈ T(K)
([2℄
or. 1.31), this means that there exists su
h that the
e K e
following map is an isomorphism of -s
hemes
ψ : U× K −→e T × K
C K
(c ) e 7−→ b(c ) e
i,j 1≤i,j≤m i,j 1≤i,j≤m
b
(
an be seen as a matrix, on the right this is a produ
t of matri
es;
R
see the de(cid:28)nition of above).
σ U = dGal(E/K)
1.0.4. Galois a
tions. Let be an element of , the
σ R x
i,j
a
tion of on is given by the images of the , whi
h are de(cid:28)ned
(σ(x )) = (x )c (σ) σ♭
i,j i,j E
by the matrix formula . We denote by the
σ T T× K U
K
morphismindu
es by on or on , this isthe a
tionof whi
h
e T(K)
de(cid:28)nes the torsor stru
ture. An element of
an be represented by
a = (a ) a K e σ♭(a) = ac (σ)
i,j 1≤i,j≤m i,j E
a matrixσ U λwith in e, its image isU σ .
σ
For any in denote by the right translation on by , i.e.
λ : U → U
σ
τ 7→ τσ
4 MARC REVERSAT
ψWrite again λσ for λσ ×IdKe : U×C Ke → U×C Ke, σth∈enUthe morphism
of (1.0.3) is equivariant, this means that for any the following
diagram is
ommutative
U× K →ψ T × K
C K
λ ↓e ↓ σ♭e
σ
U× K →ψ T × K
C K
e e
Z (L/K) → R (L/K)
n n
Proof of the theorem, the map .
A ∈ M (K) [A] ∈ Z (L/K) U ∈
n n
Let su
h that , then there exists
GL (L)
n
su
h that
A = U′U−1,
U Y′ = AY
this means that is a fundamental matrix of the equation ,
L E L
as it is with entries in , it exists a di(cid:27)erential subYex′t=enAsiYon of
whi
h is a Pi
ard-Vessiot extension for the equation . Denote
ρ
A
by the representation
ρ : dGal(L/K) res−tr→iction dGal(E/K) −c→A GL (C).
A n
(2)
ρ
A
Now we prove that this representation does not depend on the
A Z (K) U ∈ GL (L) A =
n n

Ula′Uss−1of in and of the
hoi
e of su
h that
.
B ∈ M (K) [B] = [A] Z (K)
n n
Let su
h that in , then there exists
W,T ∈ GL (K)
n
su
h that
(B,1) = (W′W−1,W)(A,1)(0,T),
B = W′W−1 + WAW−1 WU
it follows that , this means that is a
Y′ = BY ρ = ρ
A B
fundamental matrix of the equation . We see that ,
ρ
[A]
and we denote this representation by .
V ∈ GL (L) A = U′U−1 = V′V−1
n
(VL−e1tU)′ = 0 su
h that γ ∈ GL (), then we Usee=thVaγt
n
, this meansthat there exists su
h that
and the two representations de(cid:28)ne as before in (2) are
onjugate.
[A] Z (L/K)
n
Thentoea
h element of we have asso
iatedtheelement
ρ Rep (L/K)
[A] n
of .
R (L/K) → Z (L/K)
n n
Proof of the theorem, the map .
ρ : dGal(L/K) → GL (C)
n
E Let kerρ beUa=radtGioanla(El r/eKpr)esentation. Let
be theρ(cid:28)xed (cid:28)eld of ,Uw֒e→seGtL (C) and we denote
n
again by the representation
oming from the given one.
E
TYh′ e=(cid:28)AelYd is aAPi∈
aMrd-V(Kes)siot extension
orresponding to an equatioAn
m
, with . Our aim is to prove that one
an
hose
M (K) m = n [A] 7→ ρ
n [A]
in , i.e. , andthatthisgivestheinversemaEpoRf U T .
We uGseLth(eC)pr=evSiopuescCn[o(tTati)ons and,(ddeest
)r−i1p]tions of , , , et
.
n r,s 1≤r,s≤n
We set , let
ρ♯ : C (T ) ,(det)−1 −→ C (Y ) ,(det)−1 /J
r,s 1≤r,s≤n i,j 1≤i,j≤m
(cid:2) (cid:3) (cid:2) (cid:3)
DIFFERENTIAL GALOIS GROUP. 5
ρ : U ֒→ GL (C) ρ♯ I = ker(ρ♯)
n
be the
omorphism of ; is onto. Set ,
ρ♯
then we have an isomorphism indu
es by
ρ¯: C (T ) ,(det)−1 /I ≃ C (Y ) ,(det)−1 /J.
r,s 1≤r,s≤n i,j 1≤i,j≤m
(cid:2) (cid:3) (cid:2) (cid:3)
t T
r,s r,s
Let be the image of in the quotient on the left, and re
all that
y Y
i,j i,j
are that of in the quotient on the right, then the pre
eding
formula
an be written
ρ¯: C[(t ) ] ≃ C[(y ) ].
r,s 1≤r,s≤n i,j 1≤i,j≤m
V = Spec(C[(t ) ]) GL (C)
r,s 1≤r,s≤n n
Set U ,thisisanalgebrai
sρ¯ubgroupof ,
it is isomorpρhi:
Uto≃ Vvia the morphism indu
es by , denoted by abuse
of language .
The
omposed morphism (see (1.0.3))
ϕ : T ⊗ K −ψ→−1 U× K ρ−×→IdKe V× K
K C C
(3)
K e e e U
iVs an isomorphism of e-s
heσmes,Uequivariantϕf◦orσt♭h=e aλ
tion◦sϕof and
ρ(σ)
, this means that for any in we have , where,
λ V× K
ρ(σ) C
as before, ρi(sσt)he enVdomorphism of e
oming from the right
translation by on (1.0.4).
ϕ♯ ϕ
rL,esm=m1a,··1·.2,n. Let z =beϕth♯(et
o)morphism ofV =(sSepeec(C2)[()tand) for an]y
r,s r,s r,s 1≤r,s≤n
σs∈etU (re
all thaat(σ) ∈ GL (C) ).
n
Then, for all , there exists a matrix su
h that we
(σ(z )) = (z ) a(σ)
r,s 1≤r,s≤n r,s 1≤r,s≤n
have the equality of matri
es: .
♯
λ ρ(σ)
Proof. Denote by ρ(σ) the
omorphismof the right translation by
V× K
C
on , we have the equalities of matri
es
e (σ(z )) = σ ϕ♯(t )
r,s 1≤r,s≤n (cid:0) (cid:0) r,s (cid:1)(cid:1)1≤r,s≤n
= ϕ♯ λ♯ (t ) ,
(cid:16) (cid:16) ρ(σ) r,s (cid:17)(cid:17)
1≤r,s≤n
ϕ
be
ause is equivariant, and
λ♯ (t ) = (t ) a(ρ(σ))
(cid:16) ρ(σ) r,s (cid:17) r,s 1≤r,s≤n
1≤r,s≤n
τ ∈ V a(τ) GL (C)
n
where for any the matrix is in and is su
h that
(τ(t )) = (t ) a(τ)
the formula r,s 1≤r,s≤n r,s 1≤r,s≤n de(cid:28)nes the images of the
t λ♯ V τ
r,s τ
by the
omorphism of the right translation on by . We have
(cid:28)nd
(σ(z )) = (z ) a(ρ(σ))
r,s 1≤r,s≤n r,s 1≤r,s≤n
a(ρ(σ)) GL (C) (cid:3)
n
with in .
ϕ R⊗ K
K
The fa
t that is an isomorphism implies that is generated
K z 1 ≤ r,s ≤ n R⊗ K e K
r,s K
over by the , , indeed is generated over
e C V := Cz e e
by the -spa
e P1≤r,s≤n r,s and the lemma shows that this
V
spa
e is (globally) invariant under the a
tion of the Galois group
U Gal(K/K)
. The (ordinary) Galois group a
ts as usual on the right
e
6 MARC REVERSAT
R⊗ K
K
hand fa
tor of and trivially on the left one, then we see that
R Ke z 1 ≤ r,s ≤ n
r,s
is generated over by the , .
Another
onsequen
e of the previous lemma is that the matrix
D d=ef z′ (z )−1 ,
(cid:0) r,s(cid:1)1≤r,s≤n r,s 1≤r,s≤n
M (K) ϕ♯ R
n
is in , then, be
ause is an isomorphism, the ring Yi′s=geDneYr-
ated by the entries of a fundamental matrix of the equation ,
R R
we know also that is a simple di(cid:27)erential ring. It follows that ,
E K
resp. , is the Pi
ard-Vessiot ring, resp. (cid:28)eld, over of this equation.
ρ : dGAl(L/K) → GL (C)
n
To a rational representation we have
[D] Z (L/K)
n
ass[oA
]ia7→tedρan element of , this is
learly the inverse ma(cid:3)p
[A]
of .
2. A
orrespondan
e.
Kdiff K Gdiff =
Let be auniversalPi
ard-Vessiotextension of andset
dGal(Kdiff/K) GL (C) = GL(Cn)
n
.We
hooseon
eofallanidenti(cid:28)
ation .
Rep (Gdiff) Gdiff GL (C)
n
Let n be the
ategoryofrepresentations of in :
ρ : Gdiff → GL (C) f : ρ → ρ
n 1 2
the obje
ts are morphisms , an arrow
C Cn g ∈ Gdiff
is a -linear map from into itself su
h that, for any , the
following diagram is
ommutative
Cn −ρ1→(g) Cn
f ↓ ↓ f
Cn −ρ2→(g) Cn
Z (K)
n
To de(cid:28)ne the
ategory we need the following remarks. Let
M N M (K)
n
and be two elements of , we say that they are equivalent
U V GL (K) N = VMU
n
if there exists and in su
h that . We denote
M M A ∈ M (K) i = 1,2
i n
by the equivalent
lass of . Let , and let
M ∈ M (K)
n
su
h that
M′ = A M −MA .
2 1
(4)
B ∈ [A ] U ∈ GL (K)
i i i n
Let , let su
h that
A = U′U−1 +U B U−1,
i i i i i i
then an easy
al
ulation shows that
(U−1MU )′ = B (U−1MU )−(U−1MU )B .
2 1 2 2 1 2 1 2
M ∈ GL (K)
n
Suppose that and satis(cid:28)es (4), then
(M−1)′ = A M−1 −M−1A .
1 2
Z (K)
n
Now we
an de(cid:28)ne the
ategory . Its obje
ts are elements of
Z (K) [A ] → [A ] A A
n 1 2 1 2
(see (1)), an arrow , where and are elements
M (K) M M (K)
n n
of , is an equivalen
e
lass in su
h that there exists
M ∈ M
satisfying (4). The two pre
eding formulas show that this
A [A ] i = 1,2
i i
de(cid:28)nition does not depend on the
hoi
e of in , , and
DIFFERENTIAL GALOIS GROUP. 7
Z (K)
n
that invertible arrows in
orrespond to equivalen
e
lasses of
M :
invertible matri
es. We explain the
omposition of arrows. Let
[A ] → [A ] N : [A ] → [A ] Z (K) M ∈ M
1 2 2 3 n
and two arrows of ,
hoose ,
N ∈ N
su
h that
M′ = A M −MA N′ = A N −NA ,
2 1 3 2
and
then we see that
(NM)′ = A NM −NMA .
3 1
N◦M = NM
The
omposed arrow is , fora good
hoi
e of representing
elements of the di(cid:27)erent
lasses of matri
es.
Z (K)
n
Then is a
ategory, indeed it is easily to see that it is an
additive
ategory.
Z (K) Rep (Gdiff)
Theorem 2.1. The two
ategories n and n are equiva-
[A] 7→ c
[A]
lent. On obje
ts, this equivalen
e is (see (1.0.1)).
c L/K
[A]
Proof. Note that here towrite isan abuse of notation,if isthe
Kdiff
PYi′
=ardA-YVessiotextension (
onctainedin ) asso
iatedtotheequation
[A]
, we denote always the representation
Gdiff res−tr→iction dGal(L/K) −c[→A] GL (C).
n
[A] 7→ c
[A]
The map on obje
ts of the
ategories has been
onstru
ted
[A ] [A ]
1 2
in the previous theorem, it is one to one. Let and be two
Z (K) M : [A ] → [A ] M ∈ M
n 1 1
obje
ts ofM′ = A aMnd−MA F ,F ∈beGaLn(aKrrdoiffw), sele
t
2 1 1 2 n
su
h that . Let Y′ = A Y Yb′e=fuAndYamental
1 2
mF′a=triA
esFforir=esp1,e2
tivelyfth=e eFq−u1aMtioFns f and GL (K) . Then
i i i, . Let 2 1, a priori is in n , but
f′ = (F−1)′MF +F−1M′F +F−1MF′
2 1 2 1 2 1
= (−F−1A )MF +F−1(A M −MA )F
2 2 1 2 2 1 1
+F−1MA F = 0.
2 1 1
f = F−1MF GL (C) f
2 1 n
Then is in . Now we prove that is a morphism
c c g Gdiff g
from [A1] to [A2]. Let be an element of . Applying to the
f = F−1MF
relation 2 1 we (cid:28)nd
f = g(F−1)Mg(F ) = g(F−1)F fF−1g(F ) = c (g)−1fc (g),
2 1 2 2 1 1 [A2] [A1]
g f : c → c
(see (1.0.1)) for all . This means that [A1] [A2] is a map in
Rep (Gdiff)
n .
f : ρ → ρ Rep (Gdiff)
1 2
Conversely let be an arrow of n , then we
an
f C A
i
see as a matrixwith
oe(cid:30)
ient in . We know that there exists in
M (K) ρ = c i = 1,2 F
n su
h that i [Ai], . Let as before i be a fundamental
Y′ = A Y M = F fF−1
matrix for the equation i . Set 2 1 .
M M (K) f
n
- We prove that is in . The fa
t that is a morphism of
g Gdiff
representations means that for all in we have
fc (g) = c (g)f,
[A1] [A1]
8 MARC REVERSAT
whi
h is equivalent to
fF−1g(F ) = F−1g(F )f,
1 1 2 2
then F fF−1 = g(F )fg(F−1) = g(F fF−1).
2 1 2 1 2 1
M K
This prove that the entrMies′ =ofA Mare−inMA.
2 1
- We prove the formula . We have
M′ = F′fF−1 +F f(F−1)′ = A F fF−1 +F f(−F−1A )
2 1 2 1 2 2 1 2 1 1
(cid:3)
whi
h is the expe
ted formula.
Referen
es
[1℄ vanderPUTMarius-REVERSATMar
.Arithmeti
problemsfordi(cid:27)ren-
tial equations (forth
oming).
[2℄ van der PUT Marius - SINGER Mi
hael F. Galois theory of linear dif-
ferential equations. Grundlehren der Mathematis
he Wissens
haften 328,
Springer Verlag 2003.
Intitut de Mathématiques de Toulouse, Université Paul Sabatier,
118 route de Narbonne, 31062 Toulouse
edex 9, Fran
e.
E-mail address: mar
.reversatmath.ups-tlse.fr