Hopf subalgebras of the Hopf algebra of rooted trees coming from Dyson-Schwinger equations and Lie algebras of Fa di Bruno type

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1 Hopf subalgebras of the Hopf algebra of rooted trees coming from Dyson-Schwinger equations and Lie algebras of Fa di Bruno type Loïc Foissy Contents 1 The Hopf algebra of rooted trees and Dyson-Schwinger equations The Connes-Kreimer Hopf algebra Gradation of H and completion cocycle of H and Dyson-Schwinger equations Formal series giving a Hopf subalgebras Statement of the main theorem Proof of 1 = Proof of 2 = What is H α,β? FdB Lie algebras Definitions and first properties Case where µ 3 = Case where µ 3 = Dual of enveloping algebras of FdB Lie algebras The corolla Lie algebra The third FdB Lie algebra Introduction The Connes-Kreimer algebra H of rooted trees is introduced in [7]. This graded Hopf algebra is commutative, non cocommutative, and is given a linear basis by the set of rooted forests. A particularly important operator of H is the grafting on a root B +, which satisfies the following equation: B + (x = B + (x 1 + (Id B + (x. In other terms, B + is a 1-cocycle for the Cartier-Quillen cohomology of coalgebras. Moreover, the couple (H, B + satisfies a universal property, see theorem 3 of the present text. We consider here a family of subalgebras of H, associated to the combinatorial Dyson- Schwinger equation [1, 8, 9]: X = B + (f(x, where f = p n h n is a formal series such that p 0 = 1, and X is an element of the completion of H for the topology given by the gradation of H. This equation admits a unique solution X = x n, where x n is, for all n 1, a linear span of rooted trees of weight n, inductively given by: x 1 = p 0, n x n+1 = p B + (x a1... x a. =1 a a =n 1

2 We denote by H f the subalgebra of H generated by the x n s. For the usual Dyson-Schwinger equation, f = (1 h 1. It turns out that, in this case, H f is a Hopf subalgebra. This is not the case in general; we characterise here the formal series f such that H f is Hopf. Namely, H f is a Hopf subalgebra of H if, and only if, there exists (α, β K 2, such that f(h = 1 if α = 0, or f(h = e αh if β = 0, or f(h = (1 αβh 1 β if αβ 0. We obtain in this way a two-parameters family H α,β of Hopf subalgebras of H and we explicitely describe a system of generator of these algebras. In particular, if α = 0, then H α,β = K[ ]; if α 0, then H α,β = H 1,β. The Hopf algebra H α,β is commutative, graded and connected. By the Milnor-Moore theorem [10], its dual is the enveloping algebra of a Lie algebra g α,β. Computing this Lie algebra, we find three isomorphism classes of H α,β s: 1. H 0,1, equal to K[ ]. 2. H 1, 1, the subalgebra of ladders, isomorphic to the Hopf algebra of symmetric functions. 3. The H 1,β s, with β 1, isomorphic to the Faà di Bruno Hopf algebra. Note that non commutative versions of these results are exposed in [5]. In particular, if H α,β is non cocommutative, it is isomorphic to the Faà di Bruno Hopf algebra. We try to explain this fact in the third section of this text. The dual Lie algebra g α,β satisfies the following properties: 1. g α,β is graded and connected. 2. The homogeneous component g(n of degree n of g is 1-dimensional for all n 1. Moreover, if H α,β is not cocommutative, then [g(1, g(n] (0 if n 2. Such a Lie algebra will be called a FdB Lie algebra. We prove here that there exists, up to an isomorphism, only three FdB Lie algebras: 1. The Faà di Bruno Lie algebra, which is the Lie algebra of the group of formal diffeomorphisms tangent to the identity at The Lie algebra of corollas. 3. A third Lie algebra. In particular, with a stronger condition of non commutativity, a FdB Lie algebra is isomorphic to the Faà di Bruno Lie algebra, and this result can be applied to all H 1,β s when β 1. The dual of the enveloping algebras of the two others FdB Lie algebras can also be embedded in H, using corollas for the second, giving in a certain way a limit of H 1,β when β goes to, and the third one with a different construction. Notation. We denote by K a commutative field of characteristic zero. 1 The Hopf algebra of rooted trees and Dyson-Schwinger equations 1.1 The Connes-Kreimer Hopf algebra Definition 1 [12, 13] 1. A rooted tree is a finite graph, connected and without loops, with a special vertex called the root. 2

3 2. The weight of a rooted tree is the number of its vertices. 3. The set of rooted trees will be denoted by T. Examples. Rooted trees of weight 5:,,,,,,,,,,,,,,,,. The Connes-Kreimer Hopf algebra of rooted trees H is introduced in [2]. As an algebra, H is the free associative, commutative, unitary algebra generated by the elements of T. In other terms, a K-basis of H is given by rooted forests, that is to say non necessarily connected graphs F such that each connected component of F is a rooted tree. The set of rooted forests will be denoted by F. The product of H is given by the concatenation of rooted forests, and the unit is the empty forest, denoted by 1. Examples. Rooted forests of weight 4: 1,,,,,,,,,,,,,,,,. In order to mae H a bialgebra, we now introduce the notion of cut of a tree t. A non total cut c of a tree t is a choice of edges of t. Deleting the chosen edges, the cut maes t into a forest denoted by W c (t. The cut c is admissible if any oriented path 1 in the tree meets at most one cut edge. For such a cut, the tree of W c (t which contains the root of t is denoted by R c (t and the product of the other trees of W c (t is denoted by P c (t. We also add the total cut, which is by convention an admissible cut such that R c (t = 1 and P c (t = W c (t = t. The set of admissible cuts of t is denoted by Adm (t. Note that the empty cut of t is admissible; we denote Adm(t = Adm (t {empty cut, total cut}. Example. Let us consider the rooted tree t =. As it as 3 edges, it has 2 3 non total cuts. cut c total Admissible? yes yes yes yes no yes yes no yes W c (t R c (t 1 P c (t 1 The coproduct of H is defined as the unique algebra morphism from H to H H such that, for all rooted tree t T : (t = P c (t R c (t = t t + P c (t R c (t. c Adm (t c Adm(t As H is the polynomial algebra generated by T, this maes sense. Example. ( = The edges of the tree are oriented from the root to the leaves. 3

4 Theorem 2 [2] With this coproduct, H is a bialgebra. The counit of H is given by: { ε : H K F F δ 1,F. The antipode is the algebra endomorphism defined for all t T by: S(t = where n c is the number of cut edges in c. c non total cut of t ( 1 nc W c (t, 1.2 Gradation of H and completion We graduate H by putting the forests of weight n homogeneous of degree n. We denote by H(n the homogeneous component of H of degree n. Then H is a graded bialgebra, that is to say: 1. For all i, j N, H(iH(j H(i + j. 2. For all N, (H( We define, for all x, y H: i+j= H(i H(j. val(x = max n N / x H(, n d(x, y = 2 val(x y, with the convention 2 = 0. Then d is a distance on H. The metric space (H, d is not complete: its completion will be denoted by Ĥ. As a vector space: Ĥ = n N H(n. The elements of Ĥ will be denoted x n, where x n H(n for all n N. The product m : H H H is homogeneous of degree 0, so is continuous. So it can be extended from Ĥ Ĥ to Ĥ, which is then an associative, commutative algebra. Similarly, the coproduct of H can be extended in an application: : Ĥ H H = i,j N H(i H(j. Let f(h = p n h n K[[h]] be any formal series, and let X = x n Ĥ, such that x 0 = 0. The series of Ĥ of term p n X n is Cauchy, so converges. Its limit will be denoted by f(x. In other terms, f(x = y n, with: y n = n =1 a a =n p x a1... x a. Remar. If f(h K[[h]], g(h K[[h]], without constant terms, and X constant terms, it is easy to show that (f g(x = f(g(x. Ĥ, without 4

5 1.3 1-cocycle of H and Dyson-Schwinger equations We define the operator B + : H H, sending a forest t 1... t n to the tree obtained by grafting t 1,..., t n to a common root. For example, B + ( =. This operator satisfies the following relation: for all x H, B + (x = B + (x 1 + (Id B + (x. (1 This means that B + is a 1-cocycle for a certain cohomology, namely the Cartier-Quillen cohomology for coalgebras, dual notion of the Hochschild cohomology [2]. Moreover, (H, B + satisfies the following universal property: Theorem 3 (Universal property Let A be a commutative algebra and let L : A A be a linear application. 1. There exists a unique algebra morphism φ : H A, such that φ B + = L φ. 2. If moreover A is a Hopf algebra and L satisfies (1, then φ is a Hopf algebra morphism. The operator B + is homogeneous of degree 1, so is continuous. As a consequence, it can be extended as an operator B + : Ĥ Ĥ. This operator still satisfies (1. Definition 4 [1, 8, 9] Let f K[[h]]. The Dyson-Schwinger equation associated to f is: where X is an element of Ĥ, without constant term. X = B + (f(x, (2 Proposition 5 The Dyson-Schwinger equation associated to f(h = p n h n admits a unique solution X = x n, inductively defined by: x 0 = 0, x 1 = p 0, n x n+1 = p B + (x a1... x a. =1 a a =n Proof. It is enough to identify the homogeneous components of the two members of (2. Definition 6 The subalgebra of H generated by the homogeneous components x n s of the unique solution X of the Dyson-Schwinger equation (2 associated to f will be denoted by H f. The aim of this text is to give a necessary and suffisant condition on f for H f to be a Hopf subalgebra of H. Remars. 1. If f(0 = 0, the unique solution of (2 is 0. As a consequence, H f = K is a Hopf subalgebra. 2. For all α K, if X = x n is the solution of the Dyson-Schwinger equation associated to f, the unique solution of the Dyson-Schwinger equation associated to αf is α n x n. As a consequence, if α 0, H f = H αf. We shall then suppose in the sequel that p 0 = 1. In this case, x 1 =. Examples. 5

6 1. We tae f(h = 1 + h. Then x 1 =, x 2 =, x 3 =, x 4 =. More generally, x n is the ladder with n vertices, that is to say (B + n (1 (definition 7. As a consequence, for all n 1: (x n = x i x j. i+j=n So H 1+h is Hopf. Moreover, it is cocommutative. 2. We tae f(h = 1 + h + h 2 + 2h 3. Then: x 1 =, x 2 =, x 3 = +, x 4 = Hence: so H f is not Hopf. (x 1 = x x 1, (x 2 = x x 2 + x 1 x 1, (x 3 = x x 3 + x 2 1 x 1 + 3x 1 x 2 + x 2 x 1, (x 4 = x x x 2 1 x 2 + x 3 1 x 1 + 3x 2 x 2 +2x 1 x 2 x 1 + x 3 x 1 + x 1 (8 + 5, We shall need later these two families of rooted trees: Definition 7 Let n The ladder l n of weight n is the rooted tree (B + n (1. For example: l 1 =, l 2 =, l 3 =, l 4 =. 2. The corolla c n of weight n is the rooted tree B + ( n 1. For example: c 1 =, c 2 =, c 3 =, c 4 =. The following lemma is an immediate corollary of proposition 5: Lemma 8 The coefficient of the ladder of weight n in x n is p n 1 1. The coefficient of the corolla of weight n in x n is p n 1. Using (1: Lemma 9 For all n 1: 1. (l n = n l i l n i, with the convention l 0 = 1. i=0 n 1 ( n 1 2. (c n = c n 1 + i c n i. i i=0 6

7 2 Formal series giving a Hopf subalgebras 2.1 Statement of the main theorem Theorem 10 Let f(h K[[h]], such that f(0 = 1. The following assertions are equivalent: 1. H f is a Hopf subalgebra of H. 2. There exists (α, β K 2, such that (1 αβhf (h = αf(h. 3. There exists (α, β K 2, such that f(h = 1 if α = 0, or f(h = e αh if β = 0, or f(h = (1 αβh 1 β if αβ 0. Clearly, the second and the third points are equivalent. 2.2 Proof of 1 = 2 We suppose that H f is Hopf. Lemma 11 Let us suppose that p 1 = 0. Then f(h = 1, so 2 holds with α = 0. Proof. Let us suppose that p n 0 for a certain n 2. Let us choose a minimal n. Then x 1 =, x 2 =... = x n = 0, and x n+1 = p n c n+1. So: (x n+1 = x n x n+1 + n i=1 ( n p n i c n+1 i H f H f. i In particular, for i = n 1, c 2 = H f, so x 2 0: contradiction. We now assume that p 1 0. Let Z : H K, defined by Z (F = δ,f for all F F. This application Z is homogeneous of degree 1, so is continuous and can be extended in an application Z : Ĥ K. We put (Z Id (X = y n, where X is the unique solution of (2. A direct computation shows that y n can be computed by induction with: y 0 = 1, n y n+1 = =1 a a =n + n =1 a a =n ( + 1p +1 B + (x a1... x a p B + (y a1 x a2... x a. As H f is Hopf, y n H f for all n N. Moreover, y n is a linear span of rooted trees of weight n, so is a multiple of x n : we put y n = α n x n. Let us consider the coefficient of the ladder of weight n in y n. By lemma 8, this is α n p n 1 1. So, for all n 1: p n 1 α n+1 = 2p1 n 1 p 2 + p n 1 α n. As α 1 = p 1, for all n 1, α n = p p 2 p 1 (n 1. Let us consider the coefficient of the corolla of weight n in y n. By lemma 8, this is α n p n. So, for all n 1: α n p n = (n + 1p n+1 + np n p 1. Summing all these relations, putting α = p 1 and β = 2 p 2 p 1 1, we obtain (1 αβhf (h = f(h, so (2 holds. 7

8 2.3 Proof of 2 = 1 Let us suppose 2 or, equivalently, 3. We now write H α,β instead of H f. We first give a description of the x n s. Definition Let F F. The coefficient s F is inductively computed by: s t a ta s B+ (t a ta s = 1, where t 1,..., t are distinct elements of T. = a 1!... a!s a 1 t 1... s a = a 1!... a!s a 1 t 1... s a t, t, 2. Let F F. The coefficient e F is inductively computed by: e t a ta e B+ (t a ta e = 1, = (a a! e a 1 t a 1!... a! 1... e a t, = (a a! e a 1 t a 1!... a! 1... e a t, where t 1,..., t are distinct elements of T. Remars. 1. The coefficient s F is the number of symmetries of F, that is to say the number of graph automorphisms of F respecting the roots. 2. The coefficient e F is the number of embeddings of F in the plane, that is to say the number of planar forests which underlying rooted forest is F. We now give β-equivalents of these coefficients. For all N, we put [] β = 1 + β( 1 and [] β! = [1] β... [] β. We then inductively define [s F ] β and [e F ] β for all F F by: [s t a ta [s ] β = 1, ] β = [a 1 ] β!... [a ] β![s t1 ] a 1 β [s B+ (t a ta ] β = [a 1 ] β!... [a ] β![s t1 ] a 1 β [e t a ta [e ] = 1,... [s t ] a β,... [s t ] a β, ] β = [a a ] β! [a 1 ] β!... [a ] β! [e t 1 ] a 1 β... [e t ] a β, [e B+ (t a ta ] β = [a a ] β! [a 1 ] β!... [a ] β! [e t 1 ] a 1 β... [e t ] a β, where t 1,..., t are distinct elements of T. In particular, [s t ] 1 = s t and [e t ] 1 = e t, whereras [s t ] 0 = 1 and [e t ] 0 = 1 all t T. 8

9 Examples. t s t [s t ] β e t [e t ] β (1 + β (1 + β(1 + 2β (1 + β 2 (1 + β Proposition 13 For all n N, in H α,β, x n = α n 1 Examples. x 1 =, t T, weight(t=n [s t ] β [e t ] β s t t. x 2 = α, x 3 = α 2 ( (1 + β 2 +, x 4 = α 3 ( (1 + 2β(1 + β 6 x 5 = α 4 +(1 + β + (1 + β (1 + β + 2 (1+3β(1+2β(1+β 24 + (1+2β(1+β 6 + (1+2β(1+β + (1+β 2 2 +, + (1 + β 2 + (1 + β + (1+β 2 +. Proof. For any t T, we denote by b t the coefficient of t in x weight(t. Then b = 1. The formal series f(h is given by: f(h = α n [n] β! h n. n! If t = B + (t a ta, where t 1,..., t are distinct elements of T, then: n=0 b t = α a a [a a ] β! (a a! The result comes from an easy induction. (a a! b a 1 t a 1!... a! 1... b a t. As a consequence, H 0,β = K[ ], so H 0,β is a Hopf subalgebra. Moreover, H α,β = H 1,β if α 0: we can restrict ourselves to the case α = 1. In order to lighten the notations, we put n t = s t e t and [n t ] β = [s t ] β [e t ] β for all t T. Then: { n = 1, n B + (t 1...t =!n t1... n t, { [n ] β = 1, [n B + (t 1...t ] β = [] β![n t1 ] β... [n t ] β. 9

10 As a consequence: n t = s vertex of t (fertility of s!, [n t ] β = We shall use the following result, proved in [4, 6]: s vertex of t [fertility of s] β!. Lemma 14 For all forests F F, G, H T, we denote by n(f, G; H the coefficient of F G in (H, and n (F, G; H the number of graftings of the trees of F over G giving the tree H. Then n (F, G; Hs H = n(f, G; Hs F s G. Lemma 15 Let, n N. We put, in K[X 1,..., X n ], S = X X n. Then: α α n= n i=1 X i (X i (X i + α i 1 α i = S(S (S + 1.! Proof. By induction on, see [5]. Proposition 16 If α = 1: (X = X 1 + (1 βx n(1/β+1+1 x n. So H 1,β is a Hopf subalgebra. Proof. As for all n 1, x n is a linear span of trees, we can write: (X = X 1 + a F,t F t. Then, if F F, G T : n=1 F F, t T a F,G = H T [n H ] β n(f, G; H = s H H T [n H ] β s F s G n (F, G; H. We put F = t 1... t, and we denote by s 1,..., s n the vertices of the tree G, of respective fertility f 1,..., f n. Let us consider a grafting of F over G, such that α i trees of F are grafted on the vertex s i. Then α α n =. Denoting by H the result of this grafting: [n H ] β = [n G ] β [n t1 ] β... [n t ] β [f 1 + α 1 ] β! [f 1 ] β! Moreover, the number of such graftings is and s = x x n : a F,G = α α n= = [n G] β!![n t1 ] β... [n t ] β s G s F = [n G] β!![n t1 ] β... [n t ] β s G s F... [f n + α n ] β!. [f n ] β!! α 1!... α n!. So, with lemma 15, putting x i = f i + 1/β! 1 [f 1 + α 1 ] β! [n G ] β [n t1 ] β... [n t ] β α 1!... α! s F s G [f 1 ] β! α α n= α α n= = [n G] β!![n t1 ] β... [n t ] β β s G s F α α n= n i=1 n i=1... [f n + α n ] β! [f n ] β! (1 + f i β... (1 + (f i + α i 1β α i! β α i x i(x i (x i + α i + 1 α i! n i=1 = [n G] β!![n t1 ] β... [n t ] β s(s (s + 1 β. s G s F! 10 x i (x i (x i + α i + 1 α i!

11 Moreover, as G is a tree, s = f f n + n/β = n 1 + n/β = n(1 + 1/β 1. so: We now write F = t 1... t = u a ua l l, where u 1,..., u l are distinct elements of T. Then: s F = s a 1 u 1... s a l u l a 1!... a l!,![n t1 ] β... [n t ] β = (a a l! s F a 1!... a l! ( [nt1 ] a1 ( β [ntl ] al β.... S(S (S + 1 As a conclusion, putting Q (S = :! (X = X 1 + (a a l! β a a l Q a a a n 1 t a ta l 1!... a l! l (n(1 + 1/β 1 l F ( [nt1 ] a1 ( β [ntl ] al β t 1... t l [n G ] β! s t1 s tl G s G = X 1 + s t1 G T weight(g=n (1 βx n(1/β+1+1 x n. n=1 So (X H H. Projecting on the homogeneous component of degree n, we obtain (x H H, so H 1,β is a Hopf subalgebra. Remars. 1. For (α, β = (1, 0, f(h = e h and, for all n N, x n = s tl t T weight(t=n 2. For (α, β = (1, 1, f(h = (1 h 1 and, for all n N, x n = 1 s t t. t T weight(t=n 3. For (α, β = (1, 1, f(h = 1 + h and, as [i] 1 = 0 if i 2, for all n N, x n is the ladder of weight n. 2.4 What is H α,β? If α = 0, then H 0,β = K[ ]. If α 0, then obviously H α,β = H 1,β : let us suppose that α = 1. The Hopf algebra H 1,β is graded, connected and commutative. Dually, its graded dual H 1,β is a graded, connected, cocommutative Hopf algebra. By the Milnor-Moore theorem [10], it is isomorphic to the enveloping algebra of the Lie algebra of its primitive elements. We now denote this Lie algebra by g 1,β. The dual of g 1,β is identified with the quotient space: cop rim(h 1,β = H 1,β (1 Ker(ε 2, and the transposition of the Lie bracet is the Lie cobracet δ induced by: (ϖ ϖ ( op, where ϖ is the canonical projection on cop rim(h 1,β. As H 1,β is the polynomial algebra generated by the x n s, a basis of cop rim(h 1,β is (ϖ(x n n N. By proposition 16: ( (ϖ ϖ (X = (ϖ ϖ (1 βx n(1/β+1+1 x n n=1 = n 1 (n(1 + β β ϖ(x ϖ(x n. e t t. 11

12 Projecting on the homogeneous component of degree : As a consequence: (ϖ ϖ (x = i+j= (j(1 + β β ϖ(x i ϖ(x j. δ(ϖ(x = (1 + β(j iϖ(x i ϖ(x j. i+j= Dually, the Lie algebra g 1,β has the dual basis (Z n n 1, with bracet given by: [Z i, Z j ] = (1 + β(j iz i+j. So, if β 1, this Lie algebra is isomorphic to the Faà di Bruno Lie algebra g F db, which has a basis (f n n 1, and its bracet defined by [f i, f j ] = (j if i+j. So H 1,β is isomorphic to the Hopf algebra U(g F db, namely the Faà di Bruno Hopf algebra [3], coordinates ring of the group of formal diffeomorphisms of the line tangent to Id, that is to say: ({ } G F db = an h n K[[h]] / a 0 = 0, a 1 = 1,. Theorem 17 algebra. 1. If α 0 and β 1, H α,β is isomorphic to the Faà di Bruno Hopf 2. If α 0 and β = 1, H α,β is isomorphic to the Hopf algebra of symmetric functions. 3. If α = 0, H α,β = K[ ]. Remar. If β and β 1, then H 1,β and H 1,β shown by considering x 3. are isomorphic but are not equal, as it is 3 FdB Lie algebras In the preceding section, we considered Hopf subalgebra of H, generated in each degree by a linear span of trees. Their graded dual is then the enveloping algebra of a Lie algebra g, graded, with Poincaré-Hilbert formal series: h 1 h = h n. Under an hypothesis of non-commutativity, we show that such a g is isomorphic to the Faà di Bruno Lie algebra, so the considered Hopf subalgebra is isomorphic to the Faà di Bruno Hopf algebra. Remar. The proves of these section were completed using MuPAD pro 4. The noteboo of the computations can be found at Definitions and first properties Definition 18 Let g be a N-graded Lie algebra. For all n N, we denote by g(n the homogeneous component of degree n of g. We shall say that g is FdB if: 1. g is connected, that is to say g(0 = (0. 2. For all i N, g is one-dimensional. 3. For all n 2, [g(1, g(n] (0. n=1 12

13 Let g be a FdB Lie algebra. For all i N, we fix a non-zero element Z i of g(i. By conditions 1 and 2, (Z i i 1 is a basis of g. By homogeneity of the bracet of g, for all i, j 1, there exists an element λ i,j K, such that: The Jacobi relation gives, for all i, j, 1: [Z i, Z j ] = λ i,j Z i+j. λ i,j λ i+j, + λ j, λ j+,i + λ,i λ +i,j = 0. (3 Moreover, by antisymmetry, λ j,i = λ i,j for all i, j 1. Condition 2 is expressed as λ 1,j 0 for all j 2. Lemma 19 Up to a change of basis, we can suppose that λ 1,j = 1 for all j 2 and that λ 2,3 {0, 1}. Proof. We define a family of scalars by: α 1 = 1, α 2 0, α n = λ 1,2... λ 1,n 1 α 2 if n 3. By condition 3, all these scalars are non-zero. We put Z i = α iz i. Then, for all j 2: [Z 1, Z j] = α j λ 1,j Z 1+j = α jλ 1,j α j+1 Z 1+j = Z 1+j. So, replacing the Z i s by the Z i s, we can suppose that λ 1,j = 1 if j 2. Let us suppose now that λ 2,3 0. We then choose: α 2 = λ 1,3λ 1,4 λ 2,3. Then: [Z 2, Z 3] = λ 2,3α 2 α 3 α 5 Z 5 = λ 2,3α 2 λ 1,2 α 2 λ 1,2 λ 1,3 λ 1,4 α 2 Z 5 = Z 5. So, replacing the Z i s by the Z i s, we can suppose that λ 2,3 = 1. i 2 ( i 2 Lemma 20 If i, j 2, λ i,j = ( 1 λ 2,j+. =0 Proof. Let us write (3 with i = 1: λ 1,j λ j+1, + λ j, λ j+,1 + λ,1 λ +1,j = 0. If j, 2, then λ 1,j = λ,1 = λ j+,1 = 1, so: If = 2, this gives the announced formula for i = 3. λ +1,j = λ,j λ,j+1. (4 13

14 Let us prove the result by induction on i. This is obvious for i = 2 and done for i = 3. Let us assume the result at ran i 1. Then, by (4: λ i,j = λ i 1,j λ i 1,j+1 i 3 ( i 3 i 3 ( i 3 = ( 1 λ 2,j+ =0 =0 i 3 ( i 3 i 2 = ( 1 λ 2,j+ + =0 =1 i 3 ( i 2 = λ 2,j + =1 i 2 ( i 2 = ( 1 λ 2,j+. =0 ( 1 λ 2,j+1+ ( i 3 ( 1 λ 2,j+ 1 ( 1 λ 2,j+ + ( 1 i 2 λ 2,j+i 2 So the result is true for all i 2. As a consequence, the λ i,j s are entirely determined by the λ 2,j s. We can ameliorate this result, using the following lemma: Lemma 21 For all 2, λ 2,2 = l=0 ( 2 2 l ( 1 l λ 2,l+3. Proof. Let us write the relation of lemma 20 for (i, j = (3, 2 and (i, j = (2, 3: λ 3,2 = λ 2,2 λ 2,2+1, λ 2,3 = 2 2 ( 2 2 ( 1 l λ 2,3+l l l=0 = λ 2,2+1 (2 2λ 2, l=0 ( 2 2 l ( 1 l λ 2,3+l. Summing these two relations: (2 3λ 2, l=0 ( 2 2 l ( 1 l λ 2,3+l = 0. This gives the announced result. As a consequence, the λ i,j s are entirely determined by the λ 2,j s, with j odd. In order to lighten the notations, we put µ j = λ 2,j for all j odd. Then, for example: λ 2,4 = µ 3, λ 2,6 = 2µ 5 µ 3, λ 2,8 = 3µ 7 5µ 5 + 3µ 3, λ 2,10 = 4µ 9 14µ µ 5 17µ 3, λ 2,12 = 5µ 11 30µ µ 7 255µ µ 3. Moreover, we showed that we can suppose that µ 3 = 0 or 1. 14

15 Remar. The coefficient λ 2,2+4 is then a linear span of coefficients µ 2i+3, 0 i. We put, for all N: λ 2,2+4 = a,i µ 2i+3. We can prove inductively the following results: 1. For all N, a, = + 1. i=0 2. For all 1, a, 1 = 1 ( : up to the sign, this is the sequence A of [11] (pyramidal numbers. 3. For all 2, a, 2 = 1 ( : this is the sequence A of [11]. 4. The sequence ( a,0 is the sequence of signed Genocchi number, A in [11]. It seems that for all i : ( a, i = 22i B 2i+2, i + 1 2i + 1 where the B 2n s are the Bernoulli number (see sequence A of [11]. 3.2 Case where µ 3 = 1 Lemma 22 Suppose that µ 3 = 1. Then µ 5 = 1 or Proof. By relation (3 for (i, j, = (2, 3, 4: 5µ 5 3µ 7 + µ 5 µ 7 3 = 0. If µ 5 = 3, we obtain 12 = 0, absurd. So µ 7 = 5µ 5 3. By relation (3 for (i, j, = (2, 3, 6: µ ( (2µ5 4µ 5 µ µ 5 + µ 2 5 5µ 3 5 = 0. µ 5 3 If µ 5 = 0, we obtain 2 = 0, absurd. If µ 5 = 2, we obtain 66 = 0, absurd. So: Writing relation (3 for (i, j, = (3, 4, 5: µ 9 = 3 7µ 5 + µ 2 5 5µ3 5 (2µ 5 4µ 5. 9(µ (10µ 5 9 µ 5 (µ 5 2(µ = 0. So µ 5 = 1 or µ 5 = Proposition 23 Let us suppose that µ 3 = µ 5 = 1. Then: λ 1,j = 1 if j 2, λ 2,j = 1 if j 3, λ i,j = 0 if i, j 3. 15

16 Proof. Let us first prove inductively on j that λ 2,j = 1 if j 3. This is immediate if j = 3 or 5 and comes from λ 2,4 = µ 3 for j = 4. Let us suppose the λ 2,j = 1 for 3 j < n. If n = 2 is even, then: λ 2,2 = l=0 ( 2 2 l ( 1 l = If n = is odd, write relation (3 for (i, j, = (2, 3, 2 2: l=0 ( 2 2 l ( 1 l = 1. λ 2,3 λ 5,2 2 + λ 3,2 2 λ 2+1,2 + λ 2 2,2 λ 2,3 = 0, 3 ( 3 ( 1 l λ 2,2 2+l λ 2,2 2 + λ 2,2 1 + λ 2,2 2 (λ 2,2 λ 2,2+1 l = 0, l=0 λ 2,2 2 3λ 2, λ 2,2 λ 2, λ 2,2+1 = 0, i 2 ( i 2 so λ 2,2+1 = 1. Finally, if i, j 3, λ i,j = ( 1 = 0. Lemma 24 For all N 2, S N = Proof. Indeed: S N = = = N l=0 N l=0 = λ 2, = 0, ( N ( 1 l (l + 1 l (l + 2(l + 3 = N 1 (N + 3(N + 2(N + 1. N! (l + 3!(N l! ( 1l (l (N + 3(N + 2(N (N + 3(N + 2(N + 1 N+3 j=3 N+3 j=0 ( N + 3 j ( N + 3 j ( 1 j (j 2 2 ( 1 j (j (4 (N + 3 (N + 3(N + 2(N + 1 N 1 = 0 + (N + 3(N + 2(N + 1. Proposition 25 Let us suppose that µ 3 = 1 and µ 5 = 9. Then, for all i, j 1: 10 λ i,j = 6(i j(i 2!(j 2!. (i + j 2! 6(n 2 Proof. We first prove that λ 2,n =. This is immediate for n = 1, 2, 3, 4, 5. Let us (n 1n suppose the result for all j < n, with n 6. If n = 2 is even, using lemma 20: λ 2,2 = l=0 ( 2 2 l ( 1 l l + 1 (l + 2(l + 3. Then lemma 24 gives the result. If n = is odd, let us write the relation (3 with (i, j, = (2, 3, 2: λ 2,3 λ 5,2 + λ 3,2 λ 2+3,2 + λ 2,2 λ 2+2,3 = 0. 16

17 So, with relation (4: λ 2,2 3λ 2, λ 2,2+2 λ 2,2+3 λ 2,2+3 (λ 2,2 λ 2,2+1 + λ 2,2 (λ 2,2+2 λ 2,2+3 = 0, λ 2,2+3 ( 1 2λ 2,2 + λ 2,2+1 + λ 2,2 3λ 2, λ 2,2+2 + λ 2,2 λ 2,2+2 = 0, λ 2,2+3 (2 + 3( 1(2 + 5 (2 + 1( (2 + 5( 1 ( + 1(2 1 = 0, which implies the result. Let us now prove the result by induction on i. This is immediate if i = 1, and the first part of this proof for i = 2. Let us suppose the result at ran i. Then, by relation (4: λ i+1,j = λ i,j λ i,j+1 (j i(i 2!(j 2! (j + 1 i(i 2!(j 1! = 6 6 (i + j 2! (i + j 1! (i 1!(j 2!(j + 1 i = 6. (i + j 1! So the result is true for all i, j. 3.3 Case where µ 3 = 0 Proposition 26 If µ 3 = 0, then λ i,j = 0 for all i, j 2. Proof. We first prove that µ 5 = 0. If not, by (3 for (i, j, = (2, 3, 4, µ 5 µ 7 = 0, so µ 7 = 0. By (3 with (i, j, = (2, 3, 7, 5µ 5 (28µ 5 + 4µ 9 = 0, so µ 9 = 7µ 5. By (3 with (i, j, = (3, 4, 5, 36µ 2 5 = 0: contradiction. So µ 5 = 0. Let us then prove that all the µ 2+1 s, 1, are zero. We assume that µ 3 = µ 5 =... = µ 2 1 = 0, and µ 2+1 0, with l 3. By lemma 21, λ 2,2 =... = λ 2,2 1 = λ 2,2 = 0 and λ 2, By relation (3 for (i, j, = (2, 3, n, combined with (4: λ 2,3 λ 5,n + λ 3,n λ n+3,2 + λ n,2 λ n+2,3 = 0, (λ 2,n λ 2,n+1 λ 2,n+3 + λ 2,n (λ 2,n+2 λ 2,n+3 = 0. For n = 2, this gives λ 2,2+1 λ 2,2+3 = 0, so λ 2,2+3 = 0. For n = 2 + 2: By lemma 21: λ 2,2+2 = ( = λ 2,2+1, (( λ 2,2+4 = With (5: = λ 2,2+2 (λ 2,2+4 2λ 2,2+5 = 0. (5 λ 2,2+1 λ 2,2+3 ( + 1(2 + 1 λ 2, λ 2,2+5 = By relation (3 for (i, j, = (3, 4, 2: ( λ 2, ( + 1(2 + 1 λ 2, λ 3,4 λ 7,2 + λ 4,2 λ 4+2,3 + λ 2,3 λ 2+3,4 = 0. ( λ 2,

18 Moreover, using lemma 20: This gives: λ 3,4 = λ 2,4 λ 2,5 = 0, λ 4,2 = λ 2,2 2λ 2,2+1 + λ 2,2+2, λ 3,4+2 = λ 2,4+2 λ 2,5+2, λ 2,3 = λ 2,2 + λ 2,2+1, λ 3+2,4 = λ 2, λ 2,4+2 λ 2,5+2. λ 2 2,2+1(3 11(2 + 1( + 1 = 0, 12 so λ 2,2+1 = µ 2+1 = 0: contradiction. So all the µ 2+1, 1, are zero. By lemma 21, the λ 2,i s, i 2, are zero. By lemma 20, the λ i,j s, i, j 2, are zero. Theorem 27 Up to an isomorphism, there are three FdB Lie algebras: 1. The Faà di Bruno Lie algebra g F db, with basis (e i i 1, and bracet given by [e i, e j ] = (j ie i+j for all i, j The corolla Lie algebra g c, with basis (e i i 1, and bracet given by [e 1, e j ] = e j+1 and [e i, e j ] = 0 for all i, j Another Lie algebra g 3, with basis (e i i 1, and bracet given by [e 1, e i ] = e i+1, [e 2, e j ] = e j+2, and [e i, e j ] = 0 for all i 2, j 3. Proof. We have first to prove that these are indeed Lie algebras: this is done by direct computations. Let g be a FdB Lie algebra. We showed that three cases are possible: 1. µ 3 = 1 and µ 5 = By proposition 25, putting e i = obtain the Faà di Bruno Lie algebra. 2. µ 3 = µ 5 = 1. By proposition 23, we obtain the third Lie algebra. 3. µ 3 = 0. By proposition 26, we obtain the corolla Lie algebra. Z i 6(i 2! if i 2 and e 1 = Z 1, we Corollary 28 Let g be a FdB Lie algebra, such that if i and j are two distinct elements of N, then [g(i, g(j] (0. Then g is isomorphic to the Faà di Bruno Lie algebra. 4 Dual of enveloping algebras of FdB Lie algebras We realized in the first section the Faà di Bruno Hopf algebra, dual of the enveloping algebra of the Faà di Bruno Lie algebra, as a Hopf subalgebra of H. We now give a similar result for the two other FdB Lie algebra. 4.1 The corolla Lie algebra Definition 29 We denote by H c the subalgebra of H generated by the corollas. Proposition 30 H c is a graded Hopf subalgebra of H. Its dual is isomorphic to the enveloping algebra of the corolla Lie algebra. 18

19 Proof. The subalgebra H c, being generated by homogeneous elements, is graded. By lemma 9, H c is a Hopf subalgebra of H. As it is commutative, its dual is the enveloping algebra of the Lie algebra P rim(h c. The dual of this Lie algebra is the Lie coalgebra cop rim(h c = H c (1 Ker(ε 2, with cobracet δ induced by (ϖ ϖ ( op. As H c is generated by the corollas, a basis of cop rim(h c is (ϖ(c n n 1. Moreover, if n 1: (ϖ ϖ (c n = ϖ(c 1 ϖ(c n 1, δ(c n = ϖ(c 1 ϖ(c n 1 ϖ(c n 1 ϖ(c 1. Let (Z n n 1 be the basis of P rim(hc, dual of the basis (ϖ(c n n 1. By duality, for all i, j N, such that i j: Z 1+j if i = 1, [Z i, Z j ] = Z i+1 if j = 1, 0 otherwise. So P rim(h c is isomorphic to the corolla Lie algebra, via the morphism: { gc P rim(h c e i Z i. Dually, H c is isomorphic to U(g c. Remar. We wor in K[T ][β]. The generators of H 1,β then satisfy: Note that the degree of [n] β! in β is n 1. So: x n+1 = [n] β! c n+1 + O(β n 2. n! lim β n! [n] β! x n+1 = c n+1. In this sense, the Hopf algebra H c is the limit of H 1,β when β goes to infinity. 4.2 The third FdB Lie algebra We consider the following element of Ĥ: ( Y = B (exp = y n. n 1 For example: y 1 = y 2 =, y 3 =, y 4 = 1 3, y 5 = Definition 31 We denote by H 3 the subalgebra of H generated by the y n s. Proposition 32 H 3 is a graded Hopf subalgebra of H. Its dual is isomorphic to the enveloping algebra of the third FdB Lie algebra. 19

20 Proof. The subalgebra H 3, being generated by homogeneous elements, is graded. An easy computation proves that X = is a primitive element of H. As a consequence, in Ĥ, by (1: (X = X X, (exp(x = exp(x X = exp(x 1 exp(1 X = (exp(x 1(1 exp(x = exp(x exp(x, (Y = B + (exp(x = Y 1 + exp(x Y. Moreover, X = y y2 1 + y 1 H 3, so taing the homogeneous component of degree n of (Y, we obtain: (y n = y n 1 + n =1 n l=1 a a l =n 1 l! x a 1... x al y, where x 1 = = y 1, x 2 = 1 2 = y y2 1 and x i = 0 if i 3, so (y n H 3 H 3 and H 3 is a Hopf subalgebra of H. As it is commutative, its dual is the enveloping algebra of the Lie algebra P rim(h3. The dual of this Lie algebra is the Lie coalgebra cop rim(h H 3 3 = (1 Ker(ε 2, with cobracet δ induced by (ϖ ϖ ( op. As H 3 is generated by the y n s, a basis of cop rim(h 3 is (ϖ(y n n 1. Moreover: (ϖ ϖ (Y = ϖ(exp(x ϖ(y = ϖ(x ϖ(y = (ϖ(y 2 + ϖ(y 1 ϖ(y. Taing the homogeneous component of degree n, with the convention y 1 = y 0 = 0: (ϖ ϖ (y n = ϖ(y 2 ϖ(y n 2 + ϖ(y 1 ϖ(y n 1, δ(ϖ(y n = ϖ(y 2 ϖ(y n 2 + ϖ(y 1 ϖ(y n 1 ϖ(y n 2 ϖ(y 2 ϖ(y n 1 ϖ(y 1. Let (Z n n 1 be the basis of P rim(h3, dual of the basis (ϖ(c n n 1. By duality, for all i, j N, such that i 2 and j 3: [Z 1, Z i ] = Z 1+j, [Z 2, Z j ] = Z 2+j, [Z i, Z j ] = 0. So P rim(h3 is isomorphic to third FdB Lie algebra, via the morphism: { g3 P rim(hc e i Z i. Dually, H 3 is isomorphic to U(g 3. References [1] Christoph Bergbauer and Dir Kreimer, Hopf algebras in renormalization theory: locality and Dyson-Schwinger equations from Hochschild cohomology, IRMA Lect. Math. Theor. Phys., vol. 10, Eur. Math. Soc., Zrich, 2006, arxiv:hep-th/

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