On the Galois Group of Linear Difference-Differential ... · Galois group of R over k: Gal(R=k) ,f˙ -k-automorphism of Rg Gal(R=k) is a linear algebraic group over the constant ﬁeld

On the Galois Group ofLinear Difference-Differential Equations

Ruyong Feng

KLMM, Chinese Academy of Sciences, China

Ruyong Feng (KLMM, CAS) Galois Group 1 / 19

Contents

1 Basic Notations and Concepts

2 Problem Statement

3 Ring of Sequences

4 Main Results

5 Future Work


1. Basic Notations and Concepts

In this talk, all fields are of characteristic zero.

σ: shift operator; δ: differential operator (σδ = δσ)

k : σδ-field with alg. closed constant field.

Example: C(x , t) with σ(x) = x + 1 and δ = ddt

Difference-differential equations:σ(Y ) = AY ,δ(Y ) = BY

where Y = (y1, y2, · · · , yn)T , A ∈ GLn(k),B ∈ gln(k).

Integrable condition:σ(B)A = δ(A) + AB.



Example: Tchebychev polynomial

Tn(t) =n2

[ n2 ]∑

m=0

(−1)m(n −m − 1)!

m!(n − 2m)!(2t)n−2m.

Let Y = (Tn(t),Tn+1(t))T , thenY (n + 1, t) =

(0 1−1 2t

)Y (n, t),

dY (n,t)dt =

((n−1)t1−t2 − n−1

1−t2

n1−t2 − nt

1−t2

)Y (n, t).

(Hermite polynomial, Legendre polynomial, Bessel polynomial, · · · )



R: σδ-Picard Vessiot extension of k w.r.t. σ(Y ) = AY , δ(Y ) = BY if

R is a simple σδ-ring (no nontrivial σδ-ideals);

∃ Z ∈ GLn(R) s.t. σ(Z ) = AZ and δ(Z ) = BZ ;

R = k [Zi,j ,1

det(Z ) ] where Z = (Zi,j ).

Galois group of R over k :

Gal(R/k) , σδ-k -automorphism of R

Gal(R/k) is a linear algebraic group over the constant field of k .

Reference: Hardouin, C. and Singer, M. F., Differential Galois Theory ofLinear Difference Equations, Math. Ann., 342(2), 333-377, 2008.


2. Problem Statement

Let k = C(x , t).σ(Y ) = A(x , t)Y

δ(Y ) = B(x , t)Y Gσδ : Galois group over C(x , t)

c ∈ C, s.t. A(x , c),B(x , c) are well-defined and det(A(x , c)) 6= 0

σ(Y ) = A(x , c)Y Gσc : Galois group over C(x)

` ∈ Z, s.t. A(`, t),B(`, t) are well-defined and det(A(`, t)) 6= 0

δ(Y ) = B(`, t)Y Gδ` : Galois group over C(t)

Question: What are the relations among Gσδ,Gδ` and Gσ

c ?

Note:σ(B)A = δ(A) + AB ⇒ δ(Y ) = B(`, t)Y ∼ δ(Y ) = B(m, t)Y ⇒ Gδ

` = Gδm.



Let k = C(x , t).σ(Y ) = A(x , t)Y







c ?


` = Gδm.



Let k = C(x , t).σ(Y ) = A(x , t)Y







c ?


` = Gδm.



Let k = C(x , t).σ(Y ) = A(x , t)Y







c ?


` = Gδm.



Let k = C(x , t).σ(Y ) = A(x , t)Y







c ?


` = Gδm.



Example: Consider σ(y) = ty

δ(y) = xt y

Gσδ = C∗

σ(y) = cy , c ∈ C∗ Gσc =

ξ ∈ C|ξm = 1 cm = 1

C∗ otherwise

δ(y) =`

ty , ` ∈ Z Gδ

` = 1

Gσδ = Gσc Gδ

` , if c is not a root of unity.




δ(y) = xt y

Gσδ = C∗

σ(y) = cy , c ∈ C∗ Gσc =

ξ ∈ C|ξm = 1 cm = 1

C∗ otherwise

δ(y) =`

ty , ` ∈ Z Gδ

` = 1

Gσδ = Gσc Gδ





δ(y) = xt y

Gσδ = C∗

σ(y) = cy , c ∈ C∗

Gσc =

ξ ∈ C|ξm = 1 cm = 1

C∗ otherwise

δ(y) =`

ty , ` ∈ Z Gδ

` = 1

Gσδ = Gσc Gδ





δ(y) = xt y

Gσδ = C∗

σ(y) = cy , c ∈ C∗ Gσc =

ξ ∈ C|ξm = 1 cm = 1

C∗ otherwise

δ(y) =`

ty , ` ∈ Z Gδ

` = 1

Gσδ = Gσc Gδ





δ(y) = xt y

Gσδ = C∗

σ(y) = cy , c ∈ C∗ Gσc =

ξ ∈ C|ξm = 1 cm = 1

C∗ otherwise

δ(y) =`

ty , ` ∈ Z

Gδ` = 1

Gσδ = Gσc Gδ





δ(y) = xt y

Gσδ = C∗

σ(y) = cy , c ∈ C∗ Gσc =

ξ ∈ C|ξm = 1 cm = 1

C∗ otherwise

δ(y) =`

ty , ` ∈ Z Gδ

` = 1

Gσδ = Gσc Gδ





δ(y) = xt y

Gσδ = C∗

σ(y) = cy , c ∈ C∗ Gσc =

ξ ∈ C|ξm = 1 cm = 1

C∗ otherwise

δ(y) =`

ty , ` ∈ Z Gδ

` = 1

Gσδ = Gσc Gδ



.

I will present partial results on the relations among Gσδ,Gδ` and Gσ

c .

To describe the relations among these groups, we would need to embedPicard Vessiot extensions of the above systems into the ring of sequences.


3. Ring of Sequences

F : differential field with derivation δ.

The ring of sequences over F :

SF := a = (a0,a1, · · · )|ai ∈ F/ ∼

where a ∼ b⇔ ∃ d ∈ Z≥0, s.t . ai = bi for all i ≥ d .

Define

a + b = (a0 + b0,a1 + b1, · · · ),ab = (a0b0,a1b1, · · · ),

σ((a0,a1, · · · , )) = (a1,a2, · · · , ),δ((a0,a1, · · · , )) = (δ(a0), δ(a1), · · · , ).

SF is a σδ-ring.



Define σ|F = 1F and σ(x) = x + 1. Then F (x) becomes a σδ-field.

F (x) can be σδ-embedded into SF :

F (x) −→ SF

f (x) −→ (0, · · · ,0, f (N), f (N + 1), · · · , )

where f (i) is well-defined for all i ≥ N.

In particular,

F −→ SF

b −→ (b,b,b, · · · , ).

SF is a σδ-extension ring of F (x).



F (x): σδ-field with alg. closed constant field.

A(x , t) ∈ GLn(F (x)),B(x , t) ∈ gln(F (x)).

Let ` ∈ Z>0 satisfy for all i ≥ `,

A(i , t),B(i , t) are well-defined;

det(A(i , t)) 6= 0.

K : quotient field of δ-PV extension of δ(Y ) = B(`, t)Y over F .

U: fundamental matrix of δ(Y ) = B(`, t)Y in GLn(K ).



Define V = (V0,V1, · · · , ) ∈ GLn(SK ) as

V0 = · · · = V`−1 = 0,V` = U,V`+1 = A(`+ 1, t)V`,V`+2 = A(`+ 2, t)V`+1, · · · .

Theorem: F (x)[V ,1/det(V )] is a σδ-Picard Vessiot extension ofσ(Y ) = A(x , t)Y ,δ(Y ) = B(x , t)Y

over F (x).

Note: F (x)[V ,1/det(V )] is a σδ-subring of SF .



Define V = (V0,V1, · · · , ) ∈ GLn(SK ) as

V0 = · · · = V`−1 = 0,V` = U,V`+1 = A(`+ 1, t)V`,V`+2 = A(`+ 2, t)V`+1, · · · .

Theorem: F (x)[V ,1/det(V )] is a σδ-Picard Vessiot extension ofσ(Y ) = A(x , t)Y ,δ(Y ) = B(x , t)Y

over F (x).

Note: F (x)[V ,1/det(V )] is a σδ-subring of SF .


4. Main Results

Let F = C(t).

Lemma: Gδ` is an algebraic subgroup of Gσδ (under isomorphism).

Proof:ψ : Gδ

` = Gal(K/C(t)) −→ σδ-Aut(SK/C(x , t))

ρ −→ ψ(ρ)

ψ(ρ)(a) = (ρ(a))

ψ(Gδ` ) −→ Gσδ

ψ(ρ) −→ ψ(ρ)|C(x,t)[V ,1/ det(V )]


4. Main Results

Let F = C(t).


Proof:ψ : Gδ


ρ −→ ψ(ρ)

ψ(ρ)(a) = (ρ(a))


ψ(ρ) −→ ψ(ρ)|C(x,t)[V ,1/ det(V )]


4. Main Results

Let F = C(t).


Proof:ψ : Gδ


ρ −→ ψ(ρ)

ψ(ρ)(a) = (ρ(a))


ψ(ρ) −→ ψ(ρ)|C(x,t)[V ,1/ det(V )]


4. Main Results

Ω: differentially closed field containing C(t).

GσδΩ : Galois group of

σ(Y ) = A(x , t)Yδ(Y ) = B(x , t)Y

over Ω(x).

Lemma: GσδΩ is a normal algebraic subgroup of Gσδ (under isomorphism).

Theorem: Gσδ = GσδΩ Gδ

` .

Gσt : Galois group of σ(Y ) = A(x , t)Y over C(t)(x).

Theorem: Gσt (Ω) is conjugate over Ω to Gσδ

Ω (Ω).

Remark: Under the conjugation, Gσt is an algebraic group defined over C and

Gσt (C) = Gσδ

Ω . In this sense, Gσδ = Gσt (C)Gδ

` .


4. Main Results



σ(Y ) = A(x , t)Yδ(Y ) = B(x , t)Y

over Ω(x).



` .



Ω (Ω).


Gσt (C) = Gσδ


` .


4. Main Results



σ(Y ) = A(x , t)Yδ(Y ) = B(x , t)Y

over Ω(x).



` .



Ω (Ω).


Gσt (C) = Gσδ


` .


4. Main Results



σ(Y ) = A(x , t)Yδ(Y ) = B(x , t)Y

over Ω(x).



` .



Ω (Ω).


Gσt (C) = Gσδ


` .


4. Main Results



σ(Y ) = A(x , t)Yδ(Y ) = B(x , t)Y

over Ω(x).



` .



Ω (Ω).


Gσt (C) = Gσδ


` .


4. Main Results

Example: Y (n + 1, t) =

(0 1−1 2t

)Y (n, t),

dY (n,t)dt =

(n−1)t1−t2 − n−1

1−t2

n1−t2 − nt

1−t2

Y (n, t).

Gσδ =(

ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C

⋃(0 ξη 0

)∣∣∣ ξη = 1ξ, η ∈ C

Gσδ

Ω =(

ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C


4. Main Results


(0 1−1 2t

)Y (n, t),

dY (n,t)dt =

(n−1)t1−t2 − n−1

1−t2

n1−t2 − nt

1−t2

Y (n, t).

Gσδ =(

ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C

⋃(0 ξη 0

)∣∣∣ ξη = 1ξ, η ∈ C

GσδΩ =

(ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C


4. Main Results


(0 1−1 2t

)Y (n, t),

dY (n,t)dt =

(n−1)t1−t2 − n−1

1−t2

n1−t2 − nt

1−t2

Y (n, t).

Gσδ =(

ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C

⋃(0 ξη 0

)∣∣∣ ξη = 1ξ, η ∈ C

Gσδ

Ω =(

ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C


4. Main Results


(0 1−1 2t

)Y (n, t),

dY (n,t)dt =

(n−1)t1−t2 − n−1

1−t2

n1−t2 − nt

1−t2

Y (n, t).

Gσδ =(

ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C

⋃(0 ξη 0

)∣∣∣ ξη = 1ξ, η ∈ C

Gσδ

Ω =(

ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C

Gσt =

(ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C(t)

Gσ

t (C) =(

ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C


4. Main Results


(0 1−1 2t

)Y (n, t),

dY (n,t)dt =

(n−1)t1−t2 − n−1

1−t2

n1−t2 − nt

1−t2

Y (n, t).

Gσδ =(

ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C

⋃(0 ξη 0

)∣∣∣ ξη = 1ξ, η ∈ C

Gσδ

Ω =(

ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C

Gσ

t =(

ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C(t)

Gσ

t (C) =(

ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C


4. Main Results


(0 1−1 2t

)Y (n, t),

dY (1,t)dt =

(1−1)t1−t2 − 1−1

1−t2

11−t2 − 1t

1−t2

Y (1, t).

Gσδ =(

ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C

⋃(0 ξη 0

)∣∣∣ ξη = 1ξ, η ∈ C

Gσδ

Ω =(

ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C

Gσ

t =(

ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C(t)

Gσ

t (C) =(

ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C

Gδ1 =

(1 00 1

),(

0 11 0

)


4. Main Results


(0 1−1 2t

)Y (n, t),

dY (1,t)dt =

(1−1)t1−t2 − 1−1

1−t2

11−t2 − 1t

1−t2

Y (1, t).

Gσδ =(

ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C

⋃(0 ξη 0

)∣∣∣ ξη = 1ξ, η ∈ C

Gσδ

Ω =(

ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C

Gσ

t =(

ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C(t)

Gσ

t (C) =(

ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C

Gδ

1 =(

1 00 1

),(

0 11 0

)


4. Main Results


(0 1−1 2t

)Y (n, t),

dY (n,t)dt =

(n−1)t1−t2 − n−1

1−t2

n1−t2 − nt

1−t2

Y (n, t).

Gσδ =(

ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C

⋃(0 ξη 0

)∣∣∣ ξη = 1ξ, η ∈ C

Gσδ

Ω =(

ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C

Gσ

t =(

ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C(t)

Gσ

t (C) =(

ξ 00 η

)∣∣∣ ξη = 1ξ, η ∈ C

Gδ

1 =(

1 00 1

),(

0 11 0

)Gσδ = Gσδ

Ω Gδ1 = Gσ

t (C)Gδ1


5. Future Work

Gσt : Galois group of σ(Y ) = A(x , t)Y over C(t)(x)

Gσc : Galois group of σ(Y ) = A(x , c)Y over C(x)

To give the complete results, one need to solve

Problem: What are the relations between Gσt and Gσ

c ?


Documents

On the Galois Group of Linear Difference-Differential ... · Galois group of R over k: Gal(R=k) ,f˙ -k-automorphism of Rg Gal(R=k) is a linear algebraic group over the constant ﬁeld