Transcendental Numbers and Hopf Algebras

Transcription

1 Transcendental Numbers and Hopf Algebras Michel Waldschmidt Deutsch-Französischer Diskurs, Saarland University, July 4,

2 Algebraic groups (commutative, linear, over Q) Exponential polynomials Transcendence of values of exponential polynomials Hopf algebras (commutative, cocommutative, of finite type) Algebra of multizeta values Deutsch-Französischer Diskurs, Saarland University, July 4,

3 Commutative linear algebraic groups over Q G = G d 0 a G d 1 m d = d 0 + d 1 G(Q) = Q d0 (Q ) d 1 (β 1,..., β d0, α 1,..., α d1 ) Deutsch-Französischer Diskurs, Saarland University, July 4,

4 Commutative linear algebraic groups over Q G = G d 0 a G d 1 m d = d 0 + d 1 G(Q) = Q d0 (Q ) d 1 exp G : T e (G) = C d G(C) = C d 0 (C ) d 1 (z 1,..., z d ) (z 1,..., z d0, e z d 0 +1,..., e z d) Deutsch-Französischer Diskurs, Saarland University, July 4,

5 Commutative linear algebraic groups over Q G = G d 0 a G d 1 m d = d 0 + d 1 G(Q) = Q d0 (Q ) d 1 exp G : T e (G) = C d G(C) = C d 0 (C ) d 1 (z 1,..., z d ) (z 1,..., z d0, e z d 0 +1,..., e z d) For α j and β i in Q, exp G (β 1,..., β d0, log α 1,..., log α d1 ) G(Q) Deutsch-Französischer Diskurs, Saarland University, July 4,

6 Baker s Theorem. If β 0 + β 1 log α β n log α n = 0 with algebraic β i and α j, then 1. β 0 = 0 2. If (β 1,..., β n ) (0,..., 0), then log α 1,..., log α n are Q- linearly dependent. 3.If (log α 1,..., log α n ) (0,..., 0), then β 1,..., β n are Q- linearly dependent. Deutsch-Französischer Diskurs, Saarland University, July 4,

7 Example: (3 2 5) log log 9 log 27 = 0. Deutsch-Französischer Diskurs, Saarland University, July 4,

8 Example: (3 2 5) log log 9 log 27 = 0. Corollaries. 1. Hermite-Lindemann (n = 1): transcendence of e, π, log 2, e 2. Deutsch-Französischer Diskurs, Saarland University, July 4,

9 Example: (3 2 5) log log 9 log 27 = 0. Corollaries. 1. Hermite-Lindemann (n = 1): transcendence of e, π, log 2, e Gel fond-schneider (n = 2, β 0 = 0): transcendence of 2 2, log 2/ log 3, e π. Deutsch-Französischer Diskurs, Saarland University, July 4,

10 Example: (3 2 5) log log 9 log 27 = 0. Corollaries. 1. Hermite-Lindemann (n = 1): transcendence of e, π, log 2, e Gel fond-schneider (n = 2, β 0 = 0): transcendence of 2 2, log 2/ log 3, e π. 3. Example with n = 2, β 0 0: transcendence of 1 0 dx 1 + x 3 = 1 3 log 2 + π 3 3 Deutsch-Französischer Diskurs, Saarland University, July 4,

11 Strong Six Exponentials Theorem. If x 1, x 2 are two complex numbers which are Q-linearly independent, if y 1, y 2, y 3 are three complex numbers which are Q-linearly independent and if β ij are six algebraic numbers such that e x iy j β ij Q for i = 1, 2, j = 1, 2, 3, then x i y j = β ij for i = 1, 2, j = 1, 2, 3. Deutsch-Französischer Diskurs, Saarland University, July 4,

12 Strong Six Exponentials Theorem. If x 1, x 2 are two complex numbers which are Q-linearly independent, if y 1, y 2, y 3 are three complex numbers which are Q-linearly independent and if β ij are six algebraic numbers such that e x iy j β ij Q for i = 1, 2, j = 1, 2, 3, then x i y j = β ij for i = 1, 2, j = 1, 2, 3. Corollary 1 (Six Exponentials Theorem). the six numbers One at least of is transcendental. e x iy j (i = 1, 2, j = 1, 2, 3) Deutsch-Französischer Diskurs, Saarland University, July 4,

13 Strong Four Exponentials Conjecture. If x 1, x 2 are two complex numbers which are Q-linearly independent, if y 1, y 2, are two complex numbers which are Q-linearly independent and if β 11, β 12, β 21, β 22, are four algebraic numbers such that the four numbers e x 1y 1 β 11, e x 1y 2 β 12, e x 2y 1 β 21, e x 2y 2 β 22 are algebraic, then x i y j = β ij for i = 1, 2, j = 1, 2. Deutsch-Französischer Diskurs, Saarland University, July 4,

14 Strong Four Exponentials Conjecture. If x 1, x 2 are Q- linearly independent, if y 1, y 2, are Q-linearly independent and if β 11, β 12, β 21, β 22, are four algebraic numbers such that e x 1y 1 β 11, e x 1y 2 β 12, e x 2y 1 β 21, e x 2y 2 β 22 are algebraic, then x i y j = β ij for i = 1, 2, j = 1, 2. Special case (Four Exponentials Conjecture). of the four numbers One at least e x 1y 1, e x 1y 2, e x 2y 1, e x 2y 2 is transcendental. Deutsch-Französischer Diskurs, Saarland University, July 4,

15 Corollary 2 (Strong Five Exponentials Theorem). If x 1, x 2 are Q-linearly independent, if y 1, y 2 are Q-linearly independent and if α, β 11, β 12, β 21, β 22, γ are six algebraic numbers such that e x 1y 1 β 11, e x 1y 2 β 12, e x 2y 1 β 21, e x 2y 2 β 22, e (γx 2/x 1 ) α are algebraic, then x i y j = β ij for i = 1, 2, j = 1, 2 and also γx 2 = αx 1. Deutsch-Französischer Diskurs, Saarland University, July 4,

16 Corollary 2 (Strong Five Exponentials Theorem). If x 1, x 2 are Q-linearly independent, if y 1, y 2 are Q-linearly independent and if α, β 11, β 12, β 21, β 22, γ are six algebraic numbers such that e x 1y 1 β 11, e x 1y 2 β 12, e x 2y 1 β 21, e x 2y 2 β 22, e (γx 2/x 1 ) α are algebraic, then x i y j = β ij for i = 1, 2, j = 1, 2 and also γx 2 = αx 1. Proof. Set y 3 = γ/x 1, β 13 = γ, β 23 = α, so that x 1 y 3 β 13 = 0 and x 2 y 3 β 23 = (γx 2 /x 1 ) α. Deutsch-Französischer Diskurs, Saarland University, July 4,

17 Corollary 2 (Strong Five Exponentials Theorem). If x 1, x 2 are Q-linearly independent, y 1, y 2 are Q-linearly independent and α, β 11, β 12, β 21, β 22, γ are algebraic numbers such that e x 1y 1 β 11, e x 1y 2 β 12, e x 2y 1 β 21, e x 2y 2 β 22, e (γx 2/x 1 ) α are algebraic, then x i y j = β ij for i = 1, 2, j = 1, 2. Five Exponentials Theorem. If x 1, x 2 are Q-linearly independent, y 1, y 2 are Q-linearly independent and γ is a non zero algebraic number, then one at least of the five numbers is transcendental. e x 1y 1, e x 1y 2, e x 2y 1, e x 2y 2, e γx 2/x 1 Deutsch-Französischer Diskurs, Saarland University, July 4,

18 More general result (D. Roy). If x 1, x 2 are Q-linearly independent and if y 1, y 2, y 3 are Q-linearly independent, then one at least of the six numbers is not of the form x i y j (i = 1, 2, j = 1, 2, 3) β ij + l h=1 β ijh log α h. Deutsch-Französischer Diskurs, Saarland University, July 4,

19 Very Strong Four Exponentials Conjecture. If x 1, x 2 are Q- linearly independent and if y 1, y 2, are Q-linearly independent, then one at least of the four numbers x 1 y 1, x 1 y 2, x 2 y 1, x 2 y 2 does not belong to the Q-vector space { } l β 0 + β h log α h ; α i and β j algebraic h=1 spanned by 1 and exp 1 (Q ). Deutsch-Französischer Diskurs, Saarland University, July 4,

20 Values of exponential polynomials Proof of Baker s Theorem. Assume β 0 + β 1 log α β n 1 log α n 1 = log α n (B 1 ) (Gel fond Baker s Method) Functions: z 0, e z 1,..., e z n 1, e β 0z 0 +β 1 z 1 + +β n 1 z n 1 Points: Z(1, log α 1,..., log α n 1 ) C n Derivatives: / z i, (0 i n 1). Deutsch-Französischer Diskurs, Saarland University, July 4,

21 Values of exponential polynomials Proof of Baker s Theorem. Assume β 0 + β 1 log α β n 1 log α n 1 = log α n (B 1 ) (Gel fond Baker s Method) Functions: z 0, e z 1,..., e z n 1, e β 0z 0 +β 1 z 1 + +β n 1 z n 1 Points: Z(1, log α 1,..., log α n 1 ) C n Derivatives: / z i, (0 i n 1). n + 1 functions, n variables, 1 point, n derivatives Deutsch-Französischer Diskurs, Saarland University, July 4,

22 Another proof of Baker s Theorem. Assume again β 0 + β 1 log α β n 1 log α n 1 = log α n (B 2 ) (Generalization of Schneider s method) Functions: z 0, z 1,..., z n 1, e z 0 α z 1 1 αz n 1 n 1 = exp{z 0 + z 1 log α z n 1 log α n 1} Points: {0} Z n 1 + Z(β 0,,..., β n 1 ) C n Derivative: / z 0. Deutsch-Französischer Diskurs, Saarland University, July 4,

23 Another proof of Baker s Theorem. Assume again β 0 + β 1 log α β n 1 log α n 1 = log α n (B 2 ) (Generalization of Schneider s method) Functions: z 0, z 1,..., z n 1, e z 0 α z 1 1 αz n 1 n 1 = exp{z 0 + z 1 log α z n 1 log α n 1} Points: {0} Z n 1 + Z(β 0,,..., β n 1 ) C n Derivative: / z 0. n + 1 functions, n variables, n points, 1 derivative Deutsch-Französischer Diskurs, Saarland University, July 4,

24 Proof of the strong six exponentials Theorem Assume x 1,..., x a are Q-linearly independent, y 1,..., y b are Q-linearly independent and β ij are algebraic numbers such that with ab > a + b. e x iy j β ij Q for i = 1,..., a,, j = 1,..., b Functions: z i, e x i(z a+1 +z 1 ) z i (1 i a) Points: (β 1j,..., β aj, y j β 1j ) C a+1 (1 j b) Derivatives: / z i (2 i a) et / z a+1 / z 1. Deutsch-Französischer Diskurs, Saarland University, July 4,

25 Proof of the strong six exponentials Theorem Assume x 1,..., x a are Q-linearly independent, y 1,..., y b are Q-linearly independent and β ij are algebraic numbers such that with ab > a + b. e x iy j β ij Q for i = 1,..., a,, j = 1,..., b Functions: z i, e x i(z a+1 +z 1 ) z i (1 i a) Points: (β 1j,..., β aj, y j β 1j ) C a+1 (1 j b) Derivatives: / z i (2 i a) and / z a+1 / z 1. 2a functions, a + 1 variables, b points, a derivatives Deutsch-Französischer Diskurs, Saarland University, July 4,

26 Linear Subgroup Theorem G = G d 0 a G d 1 m, d = d 0 + d 1. W T e (G) a C-subspace which is rational over Q. Let l 0 be its dimension. Y T e (G) a finitely generated subgroup with Γ = exp(y ) contained in G(Q) = Q d0 (Q ) d 1. Let l 1 be the Z-rank of Γ. V T e (G) a C-subspace containing both W and Y. Let n be the dimension of V. Hypothesis: n(l 1 + d 1 ) < l 1 d 1 + l 0 d 1 + l 1 d 0 Deutsch-Französischer Diskurs, Saarland University, July 4,

27 n(l 1 + d 1 ) < l 1 d 1 + l 0 d 1 + l 1 d 0 d 0 + d 1 is the number of functions d 0 are linear d 1 are exponential n is the number of variables l 0 is the number of derivatives l 1 is the number of points Deutsch-Französischer Diskurs, Saarland University, July 4,

28 d 0 d 1 l 0 l 1 n Baker B 1 1 n n 1 n Baker B 2 n 1 1 n n Six exponentials a a a b a + 1 Deutsch-Französischer Diskurs, Saarland University, July 4,

29 Baker: d 0 d 1 l 0 l 1 n Baker B 1 1 n n 1 n Baker B 2 n 1 1 n n Six exponentials a a a b a + 1 n(l 1 + d 1 ) = n 2 + n l 1 d 1 + l 0 d 1 + l 1 d 0 = n 2 + n + 1 Deutsch-Französischer Diskurs, Saarland University, July 4,

30 d 0 d 1 l 0 l 1 n Baker B 1 1 n n 1 n Baker B 2 n 1 1 n n Six exponentials a a a b a + 1 Baker: n(l 1 + d 1 ) = n 2 + n l 1 d 1 + l 0 d 1 + l 1 d 0 = n 2 + n + 1 Six exponentials: a + b < ab n(l 1 + d 1 ) = a 2 + ab + a + b l 1 d 1 + l 0 d 1 + l 1 d 0 = a 2 + 2ab Deutsch-Französischer Diskurs, Saarland University, July 4,

31 duality: (d 0, d 1, l 0, l 1 ) (l 0, l 1, d 0, d 1 ) ( ) s d ( z t e xz) ( ) t d dz = ( z=y z s e yz) dz. z=x Deutsch-Französischer Diskurs, Saarland University, July 4,

32 Fourier-Borel duality: (d 0, d 1, l 0, l 1 ) (l 0, l 1, d 0, d 1 ) ( ) s d ( z t e xz) ( ) t d dz = ( z=y z s e yz) dz. z=x ( ) s d L sy : f f(y). dz f ζ (z) = e zζ, L sy (f ζ ) = ζ s e yζ. ( ) t d L sy (z t f ζ ) = L sy(f ζ ). dζ Deutsch-Französischer Diskurs, Saarland University, July 4,

33 For v = (v 1,..., v n ) C n, set D v = v v n z 1 z n Let w 1,..., w l0, u 1,..., u d0, x and y in C n, t N d 0 s N l 0. For z C n, write and Then (uz) t = (u 1 z) t1 (u d0 z) t d 0 and D s w = D s 1 w 1 D s l 0 w l0. D s w( (uz) t e xz) z=y = D t u( (wz) s e yz) z=x Deutsch-Französischer Diskurs, Saarland University, July 4,

34 Hopf Algebras (over k = C or k = Q) Algebras A k-algebra (A, m, η) is a k-vector space A with a product m : A A A and a unit η : k A which are k-linear maps such that the following diagrams commute: (Associativity) A A A Id m m Id A A A A m A m (Unit) k A η Id A A Id η A k m A = A = A Deutsch-Französischer Diskurs, Saarland University, July 4,

35 Coalgebras A k-coalgebra (A,, ɛ) is a k-vector space A with a coproduct : A A A and a counit ɛ : A k which are k-linear maps such that the following diagrams commute: (Coassociativity) A A A Id A A Id A A A (Counit) A = A = A k A A A A k ɛ Id Id ɛ Deutsch-Französischer Diskurs, Saarland University, July 4,

36 Bialgebras A bialgebra (A, m, η,, ɛ) is a k-algebra (A, m, η) together with a coalgebra structure (A,, ɛ) which is compatible: and ɛ are algebra morphisms (xy) = (x) (y), ɛ(xy) = ɛ(x)ɛ(y). Deutsch-Französischer Diskurs, Saarland University, July 4,

37 Hopf Algebras A Hopf algebra (H, m, η,, ɛ, S) is a bialgebra (H, m, η,, ɛ) with an antipode S : H H which is a k-linear map such that the following diagram commutes: H H H H H Id S η ɛ S Id H H m H m H H Deutsch-Französischer Diskurs, Saarland University, July 4,

38 In a Hopf Algebra the primitive elements (x) = x x satisfy ɛ(x) = 0 and S(x) = x; they form a Lie algebra for the bracket [x, y] = xy yx. Deutsch-Französischer Diskurs, Saarland University, July 4,

39 In a Hopf Algebra the primitive elements (x) = x x satisfy ɛ(x) = 0 and S(x) = x; they form a Lie algebra for the bracket [x, y] = xy yx. The group-like elements (x) = x x, x 0 are invertible, they satisfy ɛ(x) = 1, S(x) = x 1 and form a multiplicative group. Deutsch-Französischer Diskurs, Saarland University, July 4,

40 Example 1. Let G be a finite multiplicative group, kg the algebra of G over k which is a k vector-space with basis G. The mapping extends the product of G by linearity. The unit maps 1 to 1 G. m : kg kg kg (x, y) xy η : k kg Deutsch-Französischer Diskurs, Saarland University, July 4,

41 Define a coproduct and a counit : kg kg kg and ɛ : kg k by extending (x) = x x and ɛ(x) = 1 for x G by linearity. The antipode S : kg kg is defined by S(x) = x 1 for x G. Deutsch-Französischer Diskurs, Saarland University, July 4,

42 Since (x) = x x for x G this Hopf algebra kg is cocommutative. It is a commutative algebra if and only if G is commutative. The set of group like elements is G: one recovers G from kg. Deutsch-Französischer Diskurs, Saarland University, July 4,

43 Example 2. Again let G be a finite multiplicative group. Consider the k- algebra k G of mappings G k, with basis δ g (g G), where δ g (g ) = { 1 for g = g, 0 for g g. Define m by m(δ g δ g ) = δ g δ g. Hence m is commutative and m(δ g δ g ) = δ g for g G. The unit η : k k G maps 1 to g G δ g. Deutsch-Französischer Diskurs, Saarland University, July 4,

44 Define a coproduct : k G k G k G and a counit ɛ : k G k by (δ g ) = δ g δ g and ɛ(δ g ) = δ g (1 G ). g g =g The coproduct is cocommutative if and only if the group G is commutative. Define an antipode S by S(δ g ) = δ g 1. Remark. One may identify k G k G and k G G with δ g δ g = δ g,g. Deutsch-Französischer Diskurs, Saarland University, July 4,

45 Duality of Hopf Algebras The Hopf algebras kg from example 1 and k G from example 2 are dual from each other: kg k G k (g 1, δ g2 ) δ g2 (g 1 ) The basis G of kg is dual to the basis (δ g ) g G of k G. Deutsch-Französischer Diskurs, Saarland University, July 4,

46 Example 3. Let G be a topological compact group over C. Denote by R(G) the set of continuous functions f : G C such that the translates f t : x f(tx), for t G, span a finite dimensional vector space. Define a coproduct, a counit ɛ and an antipode S on R(G) by f(x, y) = f(xy), ɛ(f) = f(1), Sf(x) = f(x 1 ) for x, y G. Hence R(G) is a commutative Hopf algebra. Deutsch-Französischer Diskurs, Saarland University, July 4,

47 Example 4. Let g be a Lie algebra, U(g) its universal envelopping algebra, namely T(g)/I where T(g) is the tensor algebra of g and I the two sided ideal generated by XY Y X [X, Y ]. Define a coproduct, a counit ɛ and an antipode S on U(g) by for x g. (x) = x x, ɛ(x) = 0, S(x) = x Hence U(g) is a cocommutative Hopf algebra. The set of primitive elements is g: one recovers g from U(g). Deutsch-Französischer Diskurs, Saarland University, July 4,

48 Duality of Hopf Algebras (again) Let G be a compact connected Lie group with Lie algebra g. Then the two Hopf algebras R(G) and U(g) are dual from each other. Deutsch-Französischer Diskurs, Saarland University, July 4,

49 Abelian Hopf algebras of finite type 1. H = k[x], (X) = X X, ɛ(x) = 0, S(X) = X. Deutsch-Französischer Diskurs, Saarland University, July 4,

50 Abelian Hopf algebras of finite type 1. H = k[x], (X) = X X, ɛ(x) = 0, S(X) = X. k[x] k[x] k[t 1, T 2 ], X 1 T 1, 1 X T 2 P (X) = P (T 1 + T 2 ), ɛp (X) = P (0), SP (X) = P ( X). Deutsch-Französischer Diskurs, Saarland University, July 4,

51 Abelian Hopf algebras of finite type 1. H = k[x], (X) = X X, ɛ(x) = 0, S(X) = X. k[x] k[x] k[t 1, T 2 ], X 1 T 1, 1 X T 2 P (X) = P (T 1 + T 2 ), ɛp (X) = P (0), SP (X) = P ( X). G a (K) = Hom k (k[x], K), k[g a ] = k[x] k[g a ] is an abelian Hopf algebra of finite type. Deutsch-Französischer Diskurs, Saarland University, July 4,

52 Abelian Hopf algebras of finite type 2. H = k[y, Y 1 ], (Y ) = Y Y, ɛ(y ) = 1, S(Y ) = Y 1. Deutsch-Französischer Diskurs, Saarland University, July 4,

53 Abelian Hopf algebras of finite type 2. H = k[y, Y 1 ], (Y ) = Y Y, ɛ(y ) = 1, S(Y ) = Y 1. H H k[t 1, T 1 1, T 2, T 1 2 ], Y 1 T 1, 1 Y T 2 P (Y ) = P (T 1 T 2 ), ɛp (Y ) = P (1), SP (Y ) = P (Y 1 ). Deutsch-Französischer Diskurs, Saarland University, July 4,

54 Abelian Hopf algebras of finite type 2. H = k[y, Y 1 ], (Y ) = Y Y, ɛ(y ) = 1, S(Y ) = Y 1. H H k[t 1, T 1 1, T 2, T 1 2 ], Y 1 T 1, 1 Y T 2 P (Y ) = P (T 1 T 2 ), ɛp (Y ) = P (1), SP (Y ) = P (Y 1 ). G m (K) = Hom k (k[y, Y 1 ], K), k[g m ] = k[y, Y 1 ], k[g m ] is an abelian Hopf algebra of finite type. Deutsch-Französischer Diskurs, Saarland University, July 4,

55 Abelian Hopf algebras of finite type 3. H = k[x 1,..., X d0, Y 1, Y 1 1,..., Y d1, Y 1 d 1 ] k[x] d 0 k[y, Y 1 ] d 1 Primitive elements: k-space kx kx d0, dimension d 0. Group-like elements: multiplicative group Y 1,..., Y d1, rank d 1. G = G d 0 a G d 1 a k[g] = H, G(K) = Hom k (H, K). Deutsch-Französischer Diskurs, Saarland University, July 4,

56 Abelian Hopf algebras of finite type 3. H = k[x 1,..., X d0, Y 1, Y 1 1,..., Y d1, Y 1 d 1 ] k[x] d 0 k[y, Y 1 ] d 1 The category of commutative linear algebraic groups over k G = G d 0 a G d 1 m is anti-equivalent to the category of Hopf algebras of finite type which are abelian (commutative and cocomutative) H = k[g]. The vector space of primitive elements has dimension d 0 while the rank of the group-like elements is d 1. Deutsch-Französischer Diskurs, Saarland University, July 4,

57 Other examples If W is a k-vector space of dimension l 0, Sym(W ) is an abelian Hopf algebra of finite type, anti-isomorphic to k[g l 0 a ]: For a basis 1,..., l0 of W, Sym(W ) k[ 1,..., l0 ]. Deutsch-Französischer Diskurs, Saarland University, July 4,

58 Other examples If W is a k-vector space of dimension l 0, Sym(W ) is an abelian Hopf algebra of finite type, anti-isomorphic to k[g l 0 a ]. If Γ is a torsion free finitely generated Z-module of rank l 1, then the group algebra kγ is again an abelian Hopf algebra of finite type, anti-isomorphic to k[g l 1 m]: For a basis γ 1,..., γ l1 of Γ, kγ k[γ 1, γ 1 1,..., γ l 1, γ 1 l 1 ]. Deutsch-Französischer Diskurs, Saarland University, July 4,

59 Other examples If W is a k-vector space of dimension l 0, Sym(W ) is an abelian Hopf algebra of finite type, anti-isomorphic to k[g l 0 a ]. If Γ is a torsion free finitely generated Z-module of rank l 1, then the group algebra kγ is again an abelian Hopf algebra of finite type, anti-isomorphic to k[g l 1 m]. The category of abelian Hopf algebras of finite type is equivalent to the category of pairs (W, Γ) where W is a k-vector space and Γ is a finitely generated Z-module: H = Sym(W ) kγ. Deutsch-Französischer Diskurs, Saarland University, July 4,

60 Interpretation of the duality in terms of Hopf algebras following Stéphane Fischler Let C 1 be the category with objects: (G, W, Γ) where G = G d 0 a G d 1 m, W T e (G) is rational over Q and Γ G(Q) is finitely generated morphisms: f : (G 1, W 1, Γ 1 ) (G 2, W 2, Γ 2 ) where f : G 1 G 2 is a morphism of algebraic groups such that f(γ 1 ) Γ 2 and f induces a morphism such that df(w 1 ) W 2. df : T e (G 1 ) T e (G 2 ) Deutsch-Französischer Diskurs, Saarland University, July 4,

61 Let H be an abelian Hopf algebra over Q of finite type. Denote by d 0 the dimension of the Q-vector space of primitive elements and by d 1 the rank of the group of group-like elements. Let H be another abelian Hopf algebra over Q of finite type, l 0 the dimension of the space of primitive elements and l 1 the rank of the group-like elements. Let : H H Q be a bilinear product such that x, yy = x, y y and xx, y = x x, y. Deutsch-Französischer Diskurs, Saarland University, July 4,

62 Let C 2 be the category with objects: (H, H, ) pair of Hopf algebras with a bilinear product as above. morphisms: (f, g) : (H 1, H 1, 1 ) (H 2, H 2, 2 ) where f : H 1 H 2 and g : H 2 H 1 are Hopf algebras morphisms such that x 1, g(x 2) 1 = f(x 1 ), x 2 2. Deutsch-Französischer Diskurs, Saarland University, July 4,

63 Stéphane Fischler: The categories C 1 and C 2 are equivalent. Further, Fourier-Borel duality amounts to permute H and H. Consequence: estimates. interpolation lemmas are equivalent to zero Deutsch-Französischer Diskurs, Saarland University, July 4,

64 Stéphane Fischler: The categories C 1 and C 2 are equivalent. For R C[G], 1,..., k W and γ Γ, set R, γ 1... k = 1... k R(γ). Conversely, for H 1 = C[G] and H 2 = Sym(W ) kγ, consider Γ G(C) γ ( R R, γ ) and W T e (G) ( R R, ) Deutsch-Französischer Diskurs, Saarland University, July 4,

65 Stéphane Fischler: The categories C 1 and C 2 are equivalent. Further, Fourier-Borel duality amounts to permute H and H. Open Problems: Define n associated with (G, Γ, W ) in terms of (H, H, ) Deutsch-Französischer Diskurs, Saarland University, July 4,

66 Stéphane Fischler: The categories C 1 and C 2 are equivalent. Further, Fourier-Borel duality amounts to permute H and H. Open Problems: Define n associated with (G, Γ, W ) in terms of (H, H, ) Extend to non linear commutative algebraic groups (elliptic curves, abelian varieties, and generally semi-abelian varieties) Deutsch-Französischer Diskurs, Saarland University, July 4,

67 Stéphane Fischler: The categories C 1 and C 2 are equivalent. Further, Fourier-Borel duality amounts to permute H and H. Open Problems: Define n associated with (G, Γ, W ) in terms of (H, H, ) Extend to non linear commutative algebraic groups (elliptic curves, abelian varieties, and generally semi-abelian varieties) Extend to non abelian Hopf algebras (of finite type to start with) Deutsch-Französischer Diskurs, Saarland University, July 4,

68 Stéphane Fischler: The categories C 1 and C 2 are equivalent. Further, Fourier-Borel duality amounts to permute H and H. Open Problems: Define n associated with (G, Γ, W ) in terms of (H, H, ) Extend to non linear commutative algebraic groups (elliptic curves, abelian varieties, and generally semi-abelian varieties) Extend to non abelian Hopf algebras (of finite type to start with) (?) Transcendence results on non commutative algebraic groups Deutsch-Französischer Diskurs, Saarland University, July 4,

69 Algebra of multizeta values Denote by S the set of sequences s = (s 1,..., s k ) N k with k 1, s 1 2, s i 1 (2 i k). The weight s of s is s s k, while k is the depth. For s S set ζ(s) = n 1 > >n k 1 n s 1 1 n s k k. Depth 1: values of Riemann zeta function at positive integers (Euler). Deutsch-Französischer Diskurs, Saarland University, July 4,

70 Algebra of multizeta values Denote by S the set of sequences s = (s 1,..., s k ) N k with k 1, s 1 2, s i 1 (2 i k). The weight s of s is s s k, while k is the depth. For s S set ζ(s) = n 1 > >n k 1 n s 1 1 n s k k. Let Z denote the Q-vector space spanned in C by the numbers (2iπ) s ζ(s) (s S). Deutsch-Französischer Diskurs, Saarland University, July 4,

71 For s and s in S, the product ζ(s)ζ(s ) is in two ways a linear combination of numbers ζ(s ). The product of series is one way (quadratic relations arising from the series expansions) - for instance: ζ(s)ζ(s ) = ζ(s, s ) + ζ(s, s) + ζ(s + s ). n m = n>m + m>n + n=m Hence Z is a Q-subalgebra of C bifiltered by the weight and by the depth. Deutsch-Französischer Diskurs, Saarland University, July 4,

72 For a graded Lie algebra C denote by UC its universal envelopping algebra and by UC = n 0(UC) n its graded dual, which is a commutative Hopf algebra. Conjecture (Goncharov). There exists a free graded Lie algebra C and an isomorphism of algebras Z UC filtered by the weight on the left and by the degree on the right. Deutsch-Französischer Diskurs, Saarland University, July 4,

73 Reference: Goncharov A.B. Multiple polylogarithms, cyclotomy and modular complexes. Math. Research Letter 5 (1998), Deutsch-Französischer Diskurs, Saarland University, July 4,

74 Let X = {x 0, x 1 } be an alphabet with two letters. Consider the free monoid (concatenation product) X = {x ɛ1 x ɛk ; ɛ i {0, 1}, (1 i k), k 0} on X (non commutative monomials = words on the alphabet X) including the empty word e. Deutsch-Französischer Diskurs, Saarland University, July 4,

75 Let X = {x 0, x 1 } be an alphabet with two letters. Consider the free monoid (concatenation product) X = {x ɛ1 x ɛk ; ɛ i {0, 1}, (1 i k), k 0} on X (non commutative monomials = words on the alphabet X) including the empty word e. For s S set Hence y s = x s x 1 x s k 1 0 x 1. y s = x s 1 0 x 1, y s = y s1 y sk. Deutsch-Französischer Diskurs, Saarland University, July 4,

76 Let X = {x 0, x 1 } be an alphabet with two letters. Consider the free monoid (concatenation product) X = {x ɛ1 x ɛk ; ɛ i {0, 1}, (1 i k), k 0} on X (non commutative monomials = words on the alphabet X) including the empty word e. For s S set y s = x s x 1 x s k 1 0 x 1. This is a one-to-one correspondance between S and the set x 0 X x 1 of words which start with x 0 and end with x 1. The depth k is the number of x 1, the weight s is the number of letters. Deutsch-Französischer Diskurs, Saarland University, July 4,

77 For s = (s 1,..., s k ) S set p = s and define (ɛ 1,..., ɛ p ) in {0, 1} p by y s = x ɛ1 x ɛp. Hence ɛ 1 = 0 and ɛ p = 1. Deutsch-Französischer Diskurs, Saarland University, July 4,

78 For s = (s 1,..., s k ) S set p = s and define (ɛ 1,..., ɛ p ) in {0, 1} p by y s = x ɛ1 x ɛp. Let ω 0 (t) = dt t and ω 1 (t) = dt 1 t Integral representation of multizeta values: ζ(s) = ω ɛ1 (t 1 ) ω ɛp (t p ). 1>t 1 > >t p >0 Deutsch-Französischer Diskurs, Saarland University, July 4,

79 Examples: y 3 = x 2 0x 1 : ζ(3) = 1 0 dt 1 t 1 t1 0 dt 2 t 2 t2 0 dt 3 1 t 3 Deutsch-Französischer Diskurs, Saarland University, July 4,

80 Examples: y 3 = x 2 0x 1 : ζ(3) = 1 0 dt 1 t 1 t1 0 dt 2 t 2 t2 0 dt 3 1 t 3 y 21 = y 2 y 1 = x 0 x 2 1: ζ(2, 1) = 1 0 dt 1 t 1 t1 0 dt 2 1 t 2 t2 0 dt 3 1 t 3 Deutsch-Französischer Diskurs, Saarland University, July 4,

81 Examples: y 3 = x 2 0x 1 : ζ(3) = 1 0 dt 1 t 1 t1 0 dt 2 t 2 t2 0 dt 3 1 t 3 y 21 = y 2 y 1 = x 0 x 2 1: ζ(2, 1) = 1 0 dt 1 t 1 t1 0 dt 2 1 t 2 t2 0 dt 3 1 t 3 Remark. From (t 1, t 2, t 3 ) (1 t 3, 1 t 2, 1 t 1 ) one deduces ζ(2, 1) = ζ(3). Deutsch-Französischer Diskurs, Saarland University, July 4,

82 Quadratic relations arising from the integral representation The product of two such integrals is a linear combination of similar integrals. Indeed the integral ω ɛ1 (t 1>t 1 > >tp>0 1 ) ω ɛp (t p )ω ɛ 1 (t 1) ω ɛp (t p ) 1>t 1 > >t p >0 is the integral over 1 > u 1 > > u p+p > 0 of the shuffle of ω ɛ1 ω ɛp with ω ɛ 1 ω ɛ p. Deutsch-Französischer Diskurs, Saarland University, July 4,

83 Example: From (x 0 x 1 )x(x 0 x 1 ) = 2x 0 x 1 x 0 x 1 + 4x 2 0x 2 1 one deduces ζ(2) 2 = 2ζ(2, 2) + 4ζ(3, 1). Deutsch-Französischer Diskurs, Saarland University, July 4,

84 Example: From (x 0 x 1 )x(x 0 x 1 ) = 2x 0 x 1 x 0 x 1 + 4x 2 0x 2 1 one deduces ζ(2) 2 = 2ζ(2, 2) + 4ζ(3, 1). For y s x 0 X x 1, set ˆζ(y s ) = ζ(s). This defines ˆζ : x 0 X x 1 R. Then for w and w in x 0 X x 1 we have ˆζ(w)ˆζ(w ) = ˆζ(wxw ) Deutsch-Französischer Diskurs, Saarland University, July 4,

85 Let H be the free algebra Q X over X (non commutative polynomials in x 0 and x 1, Q-vector space with basis X, concatenation product). The subalgebra H 0 = Qe + x 0 Hx 1 of H is also the free algebra Q Y over Y = {y 2, y 3,...}. The shuffle x endows both H and H 0 with commutative algebra structures H x and H 0 x: for x and y in X, u and v in X, xuxyv = x(uxyv) + v(xuxv). Extend ˆζ as a Q-linear map H 0 R. Then ˆζ : H 0 x R is a morphism of commutative algebras. Deutsch-Französischer Diskurs, Saarland University, July 4,

86 A non commutative but cocommutative Hopf algebra structure on H is given by the coproduct P = P (x x 0, x x 1 ) the counit ɛ(p ) = P e and the antipode S(x 1 x n ) = ( 1) n x n x 1 for n 1 and x 1,..., x n in X. Concatenation (or Decomposition) Hopf algebra: (H,, e,, ɛ, S) Deutsch-Französischer Diskurs, Saarland University, July 4,

87 Writing we have P = (P u)u u X P = (P uxv)u v. u,v X Friedrichs Criterion. The set of primitive elements in H is the free Lie algebra Lie(X) on X. Hence P Lie(X) (P uxv) = 0 for all u, v in X \ {e}. Deutsch-Französischer Diskurs, Saarland University, July 4,

88 Dual of the concaténation product: Φ : H H H defined by Φ(w) u v = uv w. Hence Φ(w) = u v. u,v X uv=w Shuffle (or factorization) Hopf algebra: (H, x, e, Φ, ɛ, S). Commutative, not cocommutative. Deutsch-Französischer Diskurs, Saarland University, July 4,

89 Harmonic Algebra M. Hoffmann The quasi-shuffle (or stuffle) product: with : H 0 H 0 H 0 y s u y t v = y s (u y t v) + y t (y s u y t v) + y s+t (u v). endows H 0 with a commutative algebra structure H 0 and the quadratic relations arising from series expansions show that ˆζ : H 0 R is a morphism of commutative algebras. Deutsch-Französischer Diskurs, Saarland University, July 4,

90 Cocommutative quasi-shuffle Hopf algebra Q Y : (y i ) = y i e + e y i, ɛ(p ) = P e, S(y s1 y sk ) = ( 1) k y sk y s1. Deutsch-Französischer Diskurs, Saarland University, July 4,

91 Remark. Combining both quadratic relations yields ˆζ(wxw w w ) = 0 for w and w in x 0 X x 1. From and y 1 xy 2 = x 1 xx 0 x 1 = x 1 x 0 x 1 + 2x 0 x 2 1 = y 1 y 2 + 2y 2 y 1 one deduces y 1 y 2 = y 1 y 2 + y 2 y 1 + y 3 y 1 xy 2 y 1 y 2 = y 2 y 1 y 3. As we know ζ(2, 1) = ζ(3), hence y 1 xy 2 y 1 y 2 lies in the kernel of ˆζ. But y 1 x 0 X x 1. Deutsch-Französischer Diskurs, Saarland University, July 4,

92 Conjecture (Ihara-Kaneko). The kernel of ˆζ as a Q-linear map H 0 R is spanned by the regularized double shuffle relations. Results on the formal algebra by J. Écalle. Deutsch-Französischer Diskurs, Saarland University, July 4,