On the Galois Group of Linear 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 Ruyong Feng (KLMM, CAS) Galois Group 2 / 19
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 = d dt Difference-differential equations: { σ(y ) = AY, δ(y ) = BY where Y = (y 1, y 2,, y n ) T, A GL n (k), B gl n (k). Integrable condition: σ(b)a = δ(a) + AB. Ruyong Feng (KLMM, CAS) Galois Group 3 / 19
1. Basic Notations and Concepts Example: Tchebychev polynomial T n (t) = n 2 [ n 2 ] m=0 Let Y = (T n (t), T n+1 (t)) T, then Y (n + 1, t) = ( 1) m (n m 1)! (2t) n 2m. m!(n 2m)! ( ( (n 1)t dy (n,t) dt = n 0 1 1 2t ) n 1 nt Y (n, t), ) Y (n, t). (Hermite polynomial, Legendre polynomial, Bessel polynomial, ) Ruyong Feng (KLMM, CAS) Galois Group 4 / 19
1. Basic Notations and Concepts 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 GL n (R) s.t. σ(z ) = AZ and δ(z ) = BZ ; R = k[z i,j, 1 det(z ) ] where Z = (Z i,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 of Linear Difference Equations, Math. Ann., 342(2), 333-377, 2008. Ruyong Feng (KLMM, CAS) Galois Group 5 / 19
2. Problem Statement Let k = C(x, t). σ(y ) = A(x, t)y G σδ : Galois group over C(x, t) δ(y ) = B(x, t)y Ruyong Feng (KLMM, CAS) Galois Group 6 / 19
2. Problem Statement Let k = C(x, t). σ(y ) = A(x, t)y G σδ : Galois group over C(x, t) δ(y ) = B(x, t)y c C, s.t. A(x, c), B(x, c) are well-defined and det(a(x, c)) 0 σ(y ) = A(x, c)y G σ c : Galois group over C(x) Ruyong Feng (KLMM, CAS) Galois Group 6 / 19
2. Problem Statement Let k = C(x, t). σ(y ) = A(x, t)y G σδ : Galois group over C(x, t) δ(y ) = B(x, t)y c C, s.t. A(x, c), B(x, c) are well-defined and det(a(x, c)) 0 σ(y ) = A(x, c)y G σ c : Galois group over C(x) l Z, s.t. A(l, t), B(l, t) are well-defined and det(a(l, t)) 0 δ(y ) = B(l, t)y G δ l : Galois group over C(t) Ruyong Feng (KLMM, CAS) Galois Group 6 / 19
2. Problem Statement Let k = C(x, t). σ(y ) = A(x, t)y G σδ : Galois group over C(x, t) δ(y ) = B(x, t)y c C, s.t. A(x, c), B(x, c) are well-defined and det(a(x, c)) 0 σ(y ) = A(x, c)y G σ c : Galois group over C(x) l Z, s.t. A(l, t), B(l, t) are well-defined and det(a(l, t)) 0 δ(y ) = B(l, t)y G δ l : Galois group over C(t) Question: What are the relations among G σδ, G δ l and Gσ c? Ruyong Feng (KLMM, CAS) Galois Group 6 / 19
2. Problem Statement Let k = C(x, t). σ(y ) = A(x, t)y G σδ : Galois group over C(x, t) δ(y ) = B(x, t)y c C, s.t. A(x, c), B(x, c) are well-defined and det(a(x, c)) 0 σ(y ) = A(x, c)y G σ c : Galois group over C(x) l Z, s.t. A(l, t), B(l, t) are well-defined and det(a(l, t)) 0 δ(y ) = B(l, t)y G δ l : Galois group over C(t) Question: What are the relations among G σδ, G δ l and Gσ c? Note: σ(b)a = δ(a) + AB δ(y ) = B(l, t)y δ(y ) = B(m, t)y G δ l = Gδ m. Ruyong Feng (KLMM, CAS) Galois Group 6 / 19
2. Problem Statement Example: Consider σ(y) = ty δ(y) = x t y Ruyong Feng (KLMM, CAS) Galois Group 7 / 19
2. Problem Statement Example: Consider σ(y) = ty δ(y) = x t y G σδ = C Ruyong Feng (KLMM, CAS) Galois Group 7 / 19
2. Problem Statement Example: Consider σ(y) = ty δ(y) = x t y G σδ = C σ(y) = cy, c C Ruyong Feng (KLMM, CAS) Galois Group 7 / 19
2. Problem Statement Example: Consider σ(y) = ty G σδ = C δ(y) = x t y {ξ C ξ m = 1} c m = 1 σ(y) = cy, c C Gc σ = C otherwise Ruyong Feng (KLMM, CAS) Galois Group 7 / 19
2. Problem Statement Example: Consider σ(y) = ty G σδ = C δ(y) = x t y {ξ C ξ m = 1} c m = 1 σ(y) = cy, c C Gc σ = C otherwise δ(y) = l t y, l Z Ruyong Feng (KLMM, CAS) Galois Group 7 / 19
2. Problem Statement Example: Consider σ(y) = ty G σδ = C δ(y) = x t y {ξ C ξ m = 1} c m = 1 σ(y) = cy, c C Gc σ = C otherwise δ(y) = l t y, l Z Gδ l = 1 Ruyong Feng (KLMM, CAS) Galois Group 7 / 19
2. Problem Statement Example: Consider σ(y) = ty G σδ = C δ(y) = x t y {ξ C ξ m = 1} c m = 1 σ(y) = cy, c C Gc σ = C otherwise δ(y) = l t y, l Z Gδ l = 1 G σδ = G σ c G δ l, if c is not a root of unity. Ruyong Feng (KLMM, CAS) Galois Group 7 / 19
. I will present partial results on the relations among G σδ, G δ l and Gσ c. To describe the relations among these groups, we would need to embed Picard Vessiot extensions of the above systems into the ring of sequences. Ruyong Feng (KLMM, CAS) Galois Group 8 / 19
3. Ring of Sequences F : differential field with derivation δ. The ring of sequences over F : S F := {a = (a 0, a 1, ) a i F }/ where a b d Z 0, s.t. a i = b i for all i d. Define a + b = (a 0 + b 0, a 1 + b 1, ), ab = (a 0 b 0, a 1 b 1, ), σ((a 0, a 1,, )) = (a 1, a 2,, ), δ((a 0, a 1,, )) = (δ(a 0 ), δ(a 1 ),, ). S F is a σδ-ring. Ruyong Feng (KLMM, CAS) Galois Group 9 / 19
3. Ring of Sequences Define σ F = 1 F and σ(x) = x + 1. Then F(x) becomes a σδ-field. F (x) can be σδ-embedded into S F : F(x) S F f (x) (0,, 0, f (N), f (N + 1),, ) where f (i) is well-defined for all i N. In particular, F S F b (b, b, b,, ). S F is a σδ-extension ring of F (x). Ruyong Feng (KLMM, CAS) Galois Group 10 / 19
3. Ring of Sequences F (x): σδ-field with alg. closed constant field. A(x, t) GL n (F(x)), B(x, t) gl n (F(x)). Let l Z >0 satisfy for all i l, A(i, t), B(i, t) are well-defined; det(a(i, t)) 0. K : quotient field of δ-pv extension of δ(y ) = B(l, t)y over F. U: fundamental matrix of δ(y ) = B(l, t)y in GL n (K ). Ruyong Feng (KLMM, CAS) Galois Group 11 / 19
3. Ring of Sequences Define V = (V 0, V 1,, ) GL n (S K ) as V 0 = = V l 1 = 0, V l = U, V l+1 = A(l + 1, t)v l, V l+2 = A(l + 2, t)v l+1,. Ruyong Feng (KLMM, CAS) Galois Group 12 / 19
3. Ring of Sequences Define V = (V 0, V 1,, ) GL n (S K ) as V 0 = = V l 1 = 0, V l = U, V l+1 = A(l + 1, t)v l, V l+2 = A(l + 2, t)v l+1,. Theorem: F(x)[V, 1/ det(v )] is a σδ-picard Vessiot extension of { σ(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 S F. Ruyong Feng (KLMM, CAS) Galois Group 12 / 19
4. Main Results Let F = C(t). Lemma: G δ l is an algebraic subgroup of Gσδ (under isomorphism). Ruyong Feng (KLMM, CAS) Galois Group 13 / 19
4. Main Results Let F = C(t). Lemma: Gl δ is an algebraic subgroup of Gσδ (under isomorphism). Proof: ψ : Gl δ = Gal(K /C(t)) σδ-aut(s K /C(x, t)) ρ ψ(ρ) ψ(ρ)(a) = (ρ(a)) Ruyong Feng (KLMM, CAS) Galois Group 13 / 19
4. Main Results Let F = C(t). Lemma: Gl δ is an algebraic subgroup of Gσδ (under isomorphism). Proof: ψ : Gl δ = Gal(K /C(t)) σδ-aut(s K /C(x, t)) ρ ψ(ρ) ψ(ρ)(a) = (ρ(a)) ψ(gl δ ) Gσδ ψ(ρ) ψ(ρ) C(x,t)[V,1/ det(v )] Ruyong Feng (KLMM, CAS) Galois Group 13 / 19
4. Main Results Ω: differentially closed field containing C(t). { GΩ σδ : Galois group of σ(y ) = A(x, t)y over Ω(x). δ(y ) = B(x, t)y Ruyong Feng (KLMM, CAS) Galois Group 14 / 19
4. Main Results Ω: differentially closed field containing C(t). { GΩ σδ : Galois group of σ(y ) = A(x, t)y over Ω(x). δ(y ) = B(x, t)y Lemma: G σδ Ω Theorem: G σδ = G σδ Ω Gδ l. is a normal algebraic subgroup of Gσδ (under isomorphism). Ruyong Feng (KLMM, CAS) Galois Group 14 / 19
4. Main Results Ω: differentially closed field containing C(t). { GΩ σδ : Galois group of σ(y ) = A(x, t)y over Ω(x). δ(y ) = B(x, t)y Lemma: G σδ Ω Theorem: G σδ = G σδ Ω Gδ l. G σ t is a normal algebraic subgroup of Gσδ (under isomorphism). : Galois group of σ(y ) = A(x, t)y over C(t)(x). Ruyong Feng (KLMM, CAS) Galois Group 14 / 19
4. Main Results Ω: differentially closed field containing C(t). { GΩ σδ : Galois group of σ(y ) = A(x, t)y over Ω(x). δ(y ) = B(x, t)y Lemma: G σδ Ω Theorem: G σδ = G σδ Ω Gδ l. G σ t is a normal algebraic subgroup of Gσδ (under isomorphism). : Galois group of σ(y ) = A(x, t)y over C(t)(x). Theorem: G σ t (Ω) is conjugate over Ω to Gσδ Ω (Ω). Ruyong Feng (KLMM, CAS) Galois Group 14 / 19
4. Main Results Ω: differentially closed field containing C(t). { GΩ σδ : Galois group of σ(y ) = A(x, t)y over Ω(x). δ(y ) = B(x, t)y Lemma: G σδ Ω Theorem: G σδ = G σδ Ω Gδ l. G σ t is a normal algebraic subgroup of Gσδ (under isomorphism). : Galois group of σ(y ) = A(x, t)y over C(t)(x). Theorem: G σ t (Ω) is conjugate over Ω to Gσδ Ω (Ω). Remark: Under the conjugation, Gt σ is an algebraic group defined over C and Gt σ (C) = Gσδ Ω. In this sense, Gσδ = Gt σ(c)gδ l. Ruyong Feng (KLMM, CAS) Galois Group 14 / 19
4. Main Results Example: ( 0 1 Y (n + 1, t) = 1 2t (n 1)t dy (n,t) = dt n n 1 nt ) Y (n, t), Y (n, t). Ruyong Feng (KLMM, CAS) Galois Group 15 / 19
4. Main Results Example: G σδ = {( ) ξ 0 ξη = 1 0 η ξ, η C ( 0 1 Y (n + 1, t) = 1 2t (n 1)t dy (n,t) = dt n n 1 nt } {( ) 0 ξ ξη = 1 η 0 ξ, η C ) Y (n, t), Y (n, t). } Ruyong Feng (KLMM, CAS) Galois Group 15 / 19
4. Main Results Example: ( 0 1 Y (n + 1, t) = 1 2t (n 1)t dy (n,t) = dt n n 1 nt ) Y (n, t), {( ) } G σδ = ξ 0 ξη = 1 {( ) } 0 ξ ξη = 1 0 η ξ, η C η 0 ξ, η C {( ) } GΩ σδ = ξ 0 ξη = 1 0 η ξ, η C Y (n, t). Ruyong Feng (KLMM, CAS) Galois Group 15 / 19
4. Main Results Example: ( 0 1 Y (n + 1, t) = 1 2t (n 1)t dy (n,t) = dt n n 1 nt ) Y (n, t), {( ) } G σδ = ξ 0 ξη = 1 {( ) } 0 ξ ξη = 1 0 η ξ, η C η 0 ξ, η C {( ) } GΩ σδ = ξ 0 ξη = 1 0 η ξ, η C Y (n, t). Ruyong Feng (KLMM, CAS) Galois Group 16 / 19
4. Main Results Example: ( 0 1 Y (n + 1, t) = 1 2t (n 1)t dy (n,t) = dt n n 1 nt ) Y (n, t), Y (n, t). {( ) } G σδ = ξ 0 ξη = 1 {( ) } 0 ξ ξη = 1 0 η ξ, η C η 0 ξ, η C {( ) } GΩ σδ = ξ 0 ξη = 1 0 η ξ, η C {( ) } {( ) } Gt σ = ξ 0 ξη = 1 0 η G ξ, η C(t) t σ(c) = ξ 0 ξη = 1 0 η ξ, η C Ruyong Feng (KLMM, CAS) Galois Group 16 / 19
4. Main Results Example: ( 0 1 Y (n + 1, t) = 1 2t (1 1)t dy (1,t) = dt 1 1 1 1t ) Y (n, t), Y (1, t). {( ) } G σδ = ξ 0 ξη = 1 {( ) } 0 ξ ξη = 1 0 η ξ, η C η 0 ξ, η C {( ) } GΩ σδ = ξ 0 ξη = 1 0 η ξ, η C {( ) } {( ) } Gt σ = ξ 0 ξη = 1 0 η G ξ, η C(t) t σ(c) = ξ 0 ξη = 1 0 η ξ, η C Ruyong Feng (KLMM, CAS) Galois Group 17 / 19
4. Main Results Example: ( 0 1 Y (n + 1, t) = 1 2t (1 1)t dy (1,t) = dt 1 1 1 1t ) Y (n, t), Y (1, t). {( ) } G σδ = ξ 0 ξη = 1 {( ) } 0 ξ ξη = 1 0 η ξ, η C η 0 ξ, η C {( ) } GΩ σδ = ξ 0 ξη = 1 0 η ξ, η C {( ) } {( ) } Gt σ = ξ 0 ξη = 1 0 η G ξ, η C(t) t σ(c) = ξ 0 ξη = 1 0 η ξ, η C {( ) ( )} G1 δ = 1 0 0 1, 0 1 1 0 Ruyong Feng (KLMM, CAS) Galois Group 17 / 19
4. Main Results Example: ( 0 1 Y (n + 1, t) = 1 2t (n 1)t dy (n,t) = dt n n 1 nt ) Y (n, t), Y (n, t). {( ) } G σδ = ξ 0 ξη = 1 {( ) } 0 ξ ξη = 1 0 η ξ, η C η 0 ξ, η C {( ) } GΩ σδ = ξ 0 ξη = 1 0 η ξ, η C {( ) } {( ) } Gt σ = ξ 0 ξη = 1 0 η G ξ, η C(t) t σ(c) = ξ 0 ξη = 1 0 η ξ, η C {( ) ( )} G1 δ = 1 0 0 1, 0 1 1 0 G σδ = G σδ Ω Gδ 1 = Gσ t (C)Gδ 1 Ruyong Feng (KLMM, CAS) Galois Group 18 / 19
5. Future Work G σ t : Galois group of σ(y ) = A(x, t)y over C(t)(x) Gc σ : 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 Gt σ and Gc σ? Ruyong Feng (KLMM, CAS) Galois Group 19 / 19