Differential Groups and Differential Relations. Michael F. Singer Department of Mathematics North Carolina State University Raleigh, NC USA

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1 Differential Groups and Differential Relations Michael F. Singer Department of Mathematics North Carolina State University Raleigh, NC USA singer

2 Theorem: (Hölder, 1887) The Gamma function Γ(x + 1) = xγ(x) satisfies no polynomial differential equation. Goal: Prove this using differential algebraic groups and generalize Ex. If y 1 (x) and y 2 (x) are lin. indep. solutions of y(x + 2) xy(x + 1) + y(x) = 0 then y 1 (x), y 1 (x + 1) and y 2 (x) satisfy no polynomial differential equation. 1

3 To study an object X, study its group of symmetries G The size of G, measures the size of X The relations defining G give us the relations on X. 2

4 Galois Theory of Polynomial Equations Galois Theory of Difference Equations Linear Differential Algebraic Groups Differential Galois Theory of Difference Equations Differential Relations Among Solutions of Linear Difference Equations Final Comments 3

5 Galois Theory of Polynomial Equations f(y) = 0, f k[y] of degree n and irreducible Galois group = the group of transformations of the roots of f that preserve all algebraic relations among them. More formally: Splitting Ring: K = k[y 1,..., y n, ( i<j (y i y j )) 1 ]/M = k[α 1,..., α n ], M a max ideal containing (f(y 1 ),..., f(y n )) Note: K is a field and all such are isomorphic. Galois group = Gal(K/k) = {σ : K K σ is a k-autom. } 4

6 K = k[α 1,..., α n ] α = (α 1,..., α n ), V = {σ(α) σ Gal(K/k)} K n V is a variety, inv. under Gal(k/k) V defined over k Gal(K/k) acts trans. and freely on V V is a Gal(K/k)-torsor K = k[α 1,..., α n ] = coordinate ring of V K Gal(K/k) = k Gal(K/k) = V = dim k K The size of Gal(K/k) measures relations among the roots. Ex. Gal(K/k) = deg(f) all roots are expressed in terms of one. 5

7 Galois Theory of Difference Equations k - field, σ - an automorphism Ex. C(x), σ(x) = x + 1, σ(x) = qx Difference Equation: σ(y ) = AY A GL n (k) 1 Splitting Ring: k[y, det(y ) ], Y = (y i,j) indeterminates, σ(y ) = AY, M = max σ-ideal 1 R = k[y, det(y ) ]/M = k[z, 1 ] = σ-picard-vessiot Ring det(z) M is radical R is reduced If C = k σ = {c k σc = c} is alg closed R is unique and R σ = C Ex. k = C σ(y) = y R = C[y, 1 y ]/(y2 1) 6

8 σ-galois Group: Gal σ (R/k) = {φ : R R φ is a σ k-automorphism} Ex. k = C σ(y) = y R = C[y, 1 y ]/(y2 1) Gal σ (R/k) = Z/2Z Ex. k = C(x), σ(x) = x + 1 ( ) 0 1 σ 2 y xσy + y = 0 σy = Y 1 x R = k[y, 1 det(y ) ]/(det(y ) 1), Gal σ = SL 2 (C) Ex. σ(y) y = f, f k σ ( 1 y 0 1 ) = ( 1 f 0 1 φ Gal σ φ(y) = y + c φ, c φ C Gal σ = (C, +) or {0} ) ( 1 y 0 1 ) 7

9 φ Gal σ, σ(z) = AZ φ(z) = Z[φ], [φ] GL n (C) Gal σ GL n (C) and the image is Zariski closed Gal σ = G(C), G a lin. alg. gp. /C. R = coord. ring of a G-torsor R Gal σ = k dim(g) = Krull dim. k R ( trans. deg. of quotient field) 8

10 The size of Gal(K/k) measures algebraic relations among the solutions. Ex. σ 2 y x σy + y = 0 σy = Gal σ = SL 2 (C) ( x ) Y 3 = dim SL 2 (C) = tr. deg. k k(y 1, y 2, σ(y 1 ), σ(y 2 )) y 1, y 2, σ(y 1 ) alg. indep. over k 9

11 Ex. f 1,..., f n k, k a difference field w. alg. closed const. Picard-Vessiot ring = k[y 1,..., y n ] Prop. σ(y 1 ) y 1 = f 1. σ(y n ) y n = f n y 1,..., y n alg. dep. over k if and only if g k and a const coeff. linear form L s.t. L(y 1,..., y n ) = g (equiv., c 1 f c n f n = σ(g) g) Proof. Gal σ (C, +) n. ) Alg. dependent Gal σ (C, +) n L s.t. Gal σ {(c 1,..., c n ) L(c 1,..., c n ) = 0} φ Gal σ, φ(l(y 1,..., y n )) = L(y 1 + c 1,..., y n + c n ) = L(y 1,..., y n ) + L(c 1,..., c n ) = L(y 1,..., y n ) So, L(y 1,..., y n ) = g k. Ex. y(x + 1) y(x) = 1 y(x) is not alg. over C(x). x 10

12 Linear Differential Algebraic Groups P. Cassidy- Differential Algebraic Groups Am. J. Math. 94(1972), more papers, book by Kolchin, papers by Buium, Pillay et al., Ovchinnikov (k, δ) = a differentially closed differential field Definition: A subgroup G GL n (k) k n2 is a linear differential algebraic group if it is Kolchin-closed in GL n (k), that is, G is the set of zeros in GL n (k) of a collection of differential polynomials in n 2 variables. Ex. Any linear algebraic group defined over k, that is, a subgroup of GL n (k) defined by (algebraic) polynomials, e.g., GL n (k), SL n (k) Ex. Let C = ker δ and let G(k) be a linear algebraic group defined over k. Then G(C) is a linear differential algebraic group (just add {δy i,j = 0} n i,j=1 to the defining equations!) 11

13 ( 1 z Ex. Differential subgroups of G a (k) = (k, +) = { 0 1 ) z k} The linear differential subgroups are all of the form G L a = {z k L(z) = 0} where L is a linear homogeneous differential polynomial. For example, if m = 1, G δ a = {z k δ(z) = 0} = G a(c)} Ex. Differential subgroups of G n a (k) = (kn, +) The linear differential subgroups are all of the form G L a = {(z 1,..., z n ) k n L(z 1,..., z n ) = 0} where L is a linear homogeneous differential polynomial. 12

14 Ex. H a Zariski-dense proper differential subgroup of SL n (k) g SL n (k) such that ghg 1 = SL n (C), C = ker(δ). In general if H a Zariski-dense proper differential subgroup of G GL n (k), a simple algebraic group defined over C g GL n (k) such that ghg 1 = G(C), C = ker(δ). 13

15 Differential Galois Theory of Difference Equations k - field, σ - an automorphism δ - a derivation s.t. σδ = δσ Ex. C(x) : σ(x) = x + 1, δ = d dx σ(x) = qx, δ = x d dx C(x, t) : σ(x) = x + 1, δ = t Difference Equation: σ(y ) = AY A GL n (k) 1 Splitting Ring: k{y, det(y ) } = k[y, δy, δ2 Y,..., Y = (y i,j ) differential indeterminates 1 det(y ) ] σ(y ) = AY, σ(δy ) = A(δY ) + (δa)y,... M = max σδ-ideal 1 R = k{y, det(y ) }/M = k{z, 1 } = σδ-picard-vessiot Ring det(z) 14

16 R = k{z, k - σδ field σ(y ) = AY, A GL n (k) 1 } - σδ-picard-vessiot ring det(z) R is reduced If C = k σ = {c k σc = c} is differentially closed R is unique and R σ = C 15

17 σδ-galois Group: Gal σδ (R/k) = {φ : R R φ is a σδ k-automorphism} φ Gal σδ σ(z) = AZ φ(z) = Z[φ], [φ] GL n (C) Gal σδ GL n (C) and the image is Kolchin closed Gal σδ = G(C), G a lin. differential alg. gp. /C. Gal σδ is Zariski dense in Gal σ R = coord. ring of a G-torsor - R Gal σδ = k - Assume G connected. Then diff. dim. C (G) = diff. tr. deg k F where F is the quotient field of R. 16

18 Ex. k = C σ(y) = y R = k[y, 1 y ]/(y2 1) Gal σδ (R/k) = Z/2Z Ex. σ(y) y = f, f k, Gal σδ G a Gal σδ = {c R σ L(c) = 0} for some L R σ [δ]. Ex. k = C(x), σ(x) = x + 1, δ(x) = 1 ( ) 0 1 σ 2 y xy + y = 0 σy = Y 1 x Will show: R = k{y, 1 }/{det(y ) 1} det(y ) Gal δσ = SL 2 ( C) 17

19 Differential Relations Among Solutions of Linear Difference Equations Groups Measure Relations k - σδ - field, C = k σ differentially closed. Differential subgroups of G n a (k) = (kn, +) are all of the form G L a = {(z 1,..., z n ) k n L(z 1,..., z n ) = 0} where L is a linear homogeneous differential polynomial. Proposition. Let R be a σδ-picard-vessiot extension of k containing z 1,..., z n such that σ(z i ) z i = f i, i = 1,..., n. with f i k. Then z 1,..., z n are differentially dependent over k if and only if there is a homogeneous linear differential polynomial L over C such that L(z 1,..., z n ) = g g k Equivalently, L(f 1,..., f n ) = σ(g) g. 18

20 Corollary. Let f 1,..., f n C(x), σ(x) = x + 1, δ = d dx and let z 1,..., z n satisfy σ(z i ) z i = f i, i = 1,..., n. Then z 1,..., z n are differentially dependent over F(x) (F is the field of 1- periodic functions) if and only if there is a homogeneous linear differential polynomial L over C such that L(z 1,..., z n ) = g g C(x) Equivalently, L(f 1,..., f n ) = σ(g) g. - Similar result for q-difference equations. Also for σy i = f i y i - C. Hardouin proved (using difference Galois theory) similar result for σy i = f i y i and gave criterion in terms of divisors of the f i. Simplified by M. van der Put 19

21 The Gamma function is hypertranscendental. z(x) = Γ (x)/γ(x) satisfies σ(z) z = 1 x. If z(x) satisfies a polynomial differential equation, then L C[ d dx ], g(x) C(x) s.t. L(1 ) = g(x + 1) g(x) x L( 1 ) has a pole g(x) has a pole. x If g(x) has a pole then g(x + 1) g(x) has at least two poles but L( 1 x ) has exactly one pole. 20

22 If H a Zariski-dense proper differential subgroup of G GL n (k), a simple algebraic group defined over C g GL n (k) such that ghg 1 = G(C), C = ker(δ). Proposition. Let A GL n (k) and assume that the σ-galois group of σ(y ) = AY is a simple (noncommutative) linear algebraic group G of dimension t. Let R = k{z, } be the σδ-pv ring. 1 det Z The differential trans. deg. of R over k is less than t B gl n (k) s.t. σ(b) = ABA 1 + δ(a)a 1 (in which case, (δz BZ)Z 1 gl n (k σ )) 21

23 Ex. k = C(x), σ(x) = x + 1 ( ) 0 1 σ 2 y x σy + y = 0 σy = Y 1 x R = k[y, 1 det(y ) ]/(det(y ) 1), Gal σ = SL 2 (C) σ ( a b c d y 1 (x), y 2 (x) linearly independent solutions. y 1 (x), y 2 (x), y 1 (x + 1) are differentialy dependent over C(x) ) = ( x ( ) a b gl c d 2 (C(x)) s.t. ) ( ) 1 ( x 1 x ) ( a b c d ) ( x This 4 th order inhomogeneous equation has no such solutions y 1 (x), y 2 (x), y 1 (x + 1) are differentialy independent over C(x) ) 1 22

24 Final Comments q-hypergeometric functions 2 φ 1 (a, b; c; x) satisfy φ(q 2 (a b)x (1 + c/q) x) φ(qx)+ x 1 abx c/q abx c/q φ(x) = 0 Classify differential dependence among these. (J. Roques has calculated the (tannakian) Galois groups.) Nonlinear equations Isomonodromic deformations of difference equations Inverse problem Relation to model theory of σδ-fields (R. Bustamante) 23