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

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

Lo¨ıc Foissy

Contents

1 The Hopf algebra of rooted trees and Dyson-Schwinger equations 2

1.1 The Connes-Kreimer Hopf algebra . . . 2

1.2 Gradation ofHand completion . . . 4

1.3 1-cocycle of Hand Dyson-Schwinger equations . . . 5

2 Formal series giving a Hopf subalgebras 7 2.1 Statement of the main theorem . . . 7

2.2 Proof of 1 =⇒2 . . . 7

2.3 Proof of 2 =⇒1 . . . 8

2.4 What isH_α,β ? . . . 11

3 FdB Lie algebras 12 3.1 Definitions and first properties . . . 12

3.2 Case whereµ₃ = 1 . . . 15

3.3 Case whereµ3 = 0 . . . 17

4 Dual of enveloping algebras of FdB Lie algebras 18 4.1 The corolla Lie algebra . . . 18

4.2 The third FdB Lie algebra . . . 19

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 ofH is the grafting on a rootB⁺, 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

pnhⁿ is a formal series such that p0 = 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=P

x_n, wherex_nis, for alln≥1, a linear span of rooted trees of weightn, inductively given by:







x1 = p0q, x_n+1 =

n

X

k=1

X

a1+...+ak=n

p_kB⁺(x_a1. . . x_ak).

We denote by H_f the subalgebra of Hgenerated 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 seriesf such thatH_f is Hopf. Namely,H_f is a Hopf subalgebra of Hif, and only if, there exists (α, β)∈K², such that f(h) = 1 if α = 0, or f(h) = e^αh if β = 0, or f(h) = (1−αβh)^−β1 if αβ 6= 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[q]; if α6= 0, then H_α,β=H_1,β.

The Hopf algebraH_α,βis commutative, graded and connected. By the Milnor-Moore theorem [10], its dual is the enveloping algebra of a Lie algebrag_α,β. Computing this Lie algebra, we find three isomorphism classes ofH_α,β’s:

1. H_0,1, equal toK[q].

2. H_1,−1, the subalgebra of ladders, isomorphic to the Hopf algebra of symmetric functions.

3. The H_1,β’s, withβ 6=−1, isomorphic to the Fa`a 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 Fa`a 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 componentg(n) of degreenof g is 1-dimensional for alln≥1.

Moreover, ifH_α,β is not cocommutative, then [g(1),g(n)]6= (0) ifn≥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 Fa`a di Bruno Lie algebra, which is the Lie algebra of the group of formal diffeomor- phisms tangent to the identity at 0.

2. 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 Fa`a di Bruno Lie algebra, and this result can be applied to all H_1,β’s whenβ6=−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 ofH_1,β whenβ 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 equa- tions

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 theroot.

2. Theweightof a rooted tree is the number of its vertices.

3. The set of rooted trees will be denoted byT. Examples. Rooted trees of weight≤5:

q, qq, ∨^qq^q, qqq

, ∨^qqq^q, ∨^qq^q q

, ∨^qq_q^q , qqqq

,_H^q ∨^qq^{q q}, ∨^qqq^q q

, ∨^qq^q q q

, ∨^q∨^qq^qq , ∨^qq^q

qq , ∨^qq_q^qq

, ∨^qq_q^q q

,∨^{q q}qqq , qqqqq

.

The Connes-Kreimer Hopf algebra of rooted treesH is introduced in [2]. As an algebra, H is the free associative, commutative, unitary algebra generated by the elements of T. In other terms, aK-basis ofHis 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,q,q q, qq,q q q, qq q, ∨^qq^q,qqq

,q q q, qq q q, qq qq, ∨^qq^q _q,qqq

q, ∨^qqq^q, ∨^qq^q q

, ∨^qq_q^q ,qqqq

.

In order to make Ha bialgebra, we now introduce the notion of cut of a treet. A non total cutc of a treetis a choice of edges oft. Deleting the chosen edges, the cut makestinto a forest denoted by W^c(t). The cut c is admissibleif any oriented path¹ 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 andP^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 denoteAdm(t) =Adm∗(t)− {empty cut, total cut}.

Example. Let us consider the rooted treet= ∨^qq^q q

. As it as 3 edges, it has 2³ non total cuts.

cutc ∨^qq^q q

q

∨^{q q} q

q

∨^{q q} q

q

∨^{q q} q

q

∨^{q q} q

q

∨^{q q} q

q

∨^{q q} q

q

∨^{q q} q

total Admissible? yes yes yes yes no yes yes no yes

W^c(t) ∨^qq^q q

qq qq q q∨^{q q} _qq^q_{q q q qq qq q q} qq q q q q q q ∨^qq^q

q R^c(t) ∨^qq^q

q

qq ∨^qq^q _qq^q × q qq × 1

P^c(t) 1 qq q q × qq q q q × ∨^qq^q q

The coproduct ofHis defined as the unique algebra morphism from HtoH ⊗ Hsuch that, for all rooted tree t∈ T:

∆(t) = X

c∈Adm∗(t)

P^c(t)⊗R^c(t) =t⊗1 + 1⊗t+ X

c∈Adm(t)

P^c(t)⊗R^c(t).

AsHis the polynomial algebra generated by T, this makes sense.

Example.

∆( ∨^qq^q q

) = ∨^qq^q q

⊗1 + 1⊗ ∨^qq^q q

+ qq ⊗ qq + q⊗ ∨^qq^q + q⊗ qqq

+ qq q ⊗ q+ q q ⊗ qq.

1The edges of the tree are oriented from the root to the leaves.

Theorem 2 [2] With this coproduct, His a bialgebra. The counit of H is given by:

ε:

H −→ K F ∈ F −→ δ1,F.

The antipode is the algebra endomorphism defined for all t∈ T by:

S(t) =− X

cnon total cut oft

(−1)^ncW^c(t),

where nc is the number of cut edges in c.

1.2 Gradation of H and completion

We graduateHby putting the forests of weightnhomogeneous of degreen. We denote byH(n) the homogeneous component of Hof degree n. Then His a graded bialgebra, that is to say:

1. For alli, j∈N,H(i)H(j)⊆ H(i+j).

2. For allk∈N, ∆(H(k))⊆ X

i+j=k

H(i)⊗ H(j).

We define, for all x, y∈ H:











val(x) = max







n∈N/ x∈M

k≥n

H(k)





 , 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 H. As a vector space:b

Hb= Y

n∈N

H(n).

The elements of Hb will be denoted P

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 H ⊗b Hb toH, which is then an associative, commutative algebra. Similarly, the coproduct ofb H can be extended in an application:

∆ :H −→ Hb ⊗Hb = Y

i,j∈N

H(i)⊗ H(j).

Let f(h) =P

p_nhⁿ∈K[[h]] be any formal series, and letX=P

x_n∈H, such thatb x₀= 0.

The series of Hb of term pnXⁿ is Cauchy, so converges. Its limit will be denoted by f(X). In other terms, f(X) =P

y_n, with:

y_n=

n

X

k=1

X

a1+...+a_k=n

p_kx_a1. . . x_ak.

Remark. If f(h) ∈ K[[h]], g(h) ∈ K[[h]], without constant terms, and X ∈ H, withoutb constant terms, it is easy to show that (f ◦g)(X) =f(g(X)).

1.3 1-cocycle of H and Dyson-Schwinger equations

We define the operatorB⁺:H −→ H, sending a forestt₁. . . t_n to the tree obtained by grafting t1, . . . , tn to a common root. For example, B⁺(qq q) = ∨^qq^q

q

. This operator satisfies the following relation: for allx∈ 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 coho- mology 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 andL satisfies (1), then φis a Hopf algebra morphism.

The operator B⁺ is homogeneous of degree 1, so is continuous. As a consequence, it can be extended as an operatorB⁺ :H −→b H. This operator still satisfies (1).b

Definition 4 [1, 8, 9] Letf ∈K[[h]]. The Dyson-Schwinger equationassociated to f is:

X =B⁺(f(X)), (2)

whereX is an element of H, without constant term.b

Proposition 5 The Dyson-Schwinger equation associated tof(h) =P

p_nhⁿadmits a unique solution X=P

xn, inductively defined by:











x₀ = 0, x1 = p0q, x_n+1 =

n

X

k=1

X

a1+...+ak=n

p_kB⁺(x_a1. . . x_ak).

Proof. It is enough to identify the homogeneous components of the two members of (2).

Definition 6 The subalgebra ofH generated by the homogeneous components x_n’s of the unique solutionX 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 forH_f to be a Hopf subalgebra ofH.

Remarks.

1. Iff(0) = 0, the unique solution of (2) is 0. As a consequence,H_f =Kis a Hopf subalgebra.

2. For allα∈K, ifX =P

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 P

αⁿx_n. As a consequence, ifα 6= 0,H_f =H_αf. We shall then suppose in the sequel that p0 = 1. In this case,x₁ = q.

Examples.

1. We take f(h) = 1 +h. Then x₁ = q,x₂ = qq,x₃ = qqq

, x₄ = qqqq

. 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+j=n

x_i⊗x_j. So H_1+h is Hopf. Moreover, it is cocommutative.

2. We take f(h) = 1 +h+h²+ 2h³. Then:















x1 = q, x₂ = qq, x₃ = ∨^qq^q + qqq

, x4 = 2 ∨^qqq^q + 2 ∨^qq^q

q

+ ∨^qq_q^q + qqqq

. Hence:

∆(x₁) = x₁⊗1 + 1⊗x₁,

∆(x₂) = x₂⊗1 + 1⊗x₂+x₁⊗x₁,

∆(x3) = x3⊗1 + 1⊗x3+x²₁⊗x1+ 3x1⊗x2+x2⊗x1,

∆(x4) = x4⊗1 + 1⊗x4+ 10x²₁⊗x2+x³₁⊗x1+ 3x2⊗x2

+2x1x2⊗x1+x3⊗x1+x1⊗(8 ∨^qq^q + 5qqq ), soH_f is not Hopf.

We shall need later these two families of rooted trees:

Definition 7 Letn≥1.

1. The ladderl_nof weight nis the rooted tree (B⁺)ⁿ(1). For example:

l₁= q, l₂ = qq, l₃ = qqq

, l₄ = qqqq .

2. The corollacn of weightn is the rooted tree B⁺(qⁿ⁻¹). For example:

c₁= q, c₂ = qq, c₃ = ∨^qq^q, c₄= ∨^qqq^q. The following lemma is an immediate corollary of proposition 5:

Lemma 8 The coefficient of the ladder of weight n in xn is pⁿ⁻¹₁ . The coefficient of the corolla of weight nin x_n is pn−1.

Using (1):

Lemma 9 For all n≥1:

1. ∆(l_n) =

n

X

i=0

l_i⊗ln−i, with the convention l₀ = 1.

2. ∆(c_n) =c_n⊗1 +

n−1

X

i=0

n−1 i

qⁱ⊗cn−i.

2 Formal series giving a Hopf subalgebras

2.1 Statement of the main theorem

Theorem 10 Letf(h)∈K[[h]], such thatf(0) = 1. The following assertions are equivalent:

1. H_f is a Hopf subalgebra of H.

2. There exists (α, β)∈K², such that (1−αβh)f⁰(h) =αf(h).

3. There exists (α, β) ∈ K², such that f(h) = 1 if α = 0, or f(h) = e^αh if β = 0, or f(h) = (1−αβh)^−β1 if αβ6= 0.

Clearly, the second and the third points are equivalent.

2.2 Proof of 1 =⇒2 We suppose thatH_f is Hopf.

Lemma 11 Let us suppose that p1 = 0. Then f(h) = 1, so 2 holds withα= 0.

Proof. Let us suppose that pn 6= 0 for a certain n≥2. Let us choose a minimal n. Then x₁ = q,x₂=. . .=x_n= 0, andx_n+1 =p_nc_n+1. So:

∆(x_n+1) =x_n+1⊗1 + 1⊗x_n+1+

n

X

i=1

n i

p_nqⁱ⊗cn+1−i∈ H_f⊗ H_f.

In particular, fori=n−1,c2 = qq ∈ H_f, sox26= 0: contradiction.

We now assume that p1 6= 0. Let Z_q :H −→ K, defined by Z_q(F) = δ_q,F for all F ∈ F. This application Z_q is homogeneous of degree −1, so is continuous and can be extended in an applicationZ_q :H −→b K. We put (Z_q⊗Id)◦∆(X) =P

y_n, whereX is the unique solution of (2). A direct computation shows that yncan be computed by induction with:



















y₀ = 1, y_n+1 =

n

X

k=1

X

a1+...+a_k=n

(k+ 1)p_k+1B⁺(x_a1. . . x_ak) +

n

X

k=1

X

a1+...+a_k=n

kp_kB⁺(y_a1x_a2. . . x_ak).

AsH_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 xn: we put yn=αnxn.

Let us consider the coefficient of the ladder of weight niny_n. By lemma 8, this is α_npⁿ⁻¹₁ . So, for alln≥1:

pⁿ₁α_n+1= 2pⁿ⁻¹₁ p₂+pⁿ₁α_n. Asα₁ =p₁, for all n≥1, α_n=p₁+ 2p₂

p₁(n−1). Let us consider the coefficient of the corolla of weight niny_n. By lemma 8, this is α_np_n. So, for all n≥1:

α_np_n= (n+ 1)p_n+1+np_np₁. Summing all these relations, puttingα=p₁ andβ = 2p₂

p1

−1, we obtain (1−αβh)f⁰(h) =f(h), so (2) holds.

2.3 Proof of 2 =⇒1

Let us suppose 2 or, equivalently, 3. We now writeH_α,βinstead ofH_f. We first give a description of the xn’s.

Definition 12

1. LetF ∈ F. The coefficient s_F is inductively computed by:







s_q = 1, s_t^a1

1 ...t^ak_k = a1!. . . ak!s^a_t1¹. . . s^a_tk^k, s_B+(^ta1¹...t^ak_k ) = a1!. . . ak!s^a_t1¹. . . s^a_tk^k, wheret₁, . . . , t_k are distinct elements ofT.

2. LetF ∈ F. The coefficient e_F is inductively computed by:











e_q = 1, e_t^a1

1 ...t^ak_k = (a₁+. . .+a_k)!

a1!. . . ak! e^a_t1¹. . . e^a_tk^k, e_B+(^ta1¹...t^ak_k ) = (a1+. . .+ak)!

a₁!. . . a_k! e^a_t1¹. . . e^a_tk^k, wheret1, . . . , tk are distinct elements ofT.

Remarks.

1. The coefficient s_F is the number of symmetries of F, that is to say the number of graph automorphisms ofF respecting the roots.

2. The coefficiente_F is the number of embeddings ofF in the plane, that is to say the number of planar forests which underlying rooted forest isF.

We now give β-equivalents of these coefficients. For allk∈N^∗, we put [k]β = 1 +β(k−1) and [k]_β! = [1]_β. . .[k]_β. We then inductively define [s_F]_β and [e_F]_β for allF ∈ F by:







[s_q]_β = 1, [s_t^a1

1 ...t^ak_k ]_β = [a₁]_β!. . .[a_k]_β![s_t1]^a_β¹. . .[s_tk]^a_β^k, [s_B+(^ta1¹...t^ak_k )]^β = [a₁]_β!. . .[a_k]_β![s_t1]^a_β¹. . .[s_tk]^a_β^k,















[e_q] = 1, [e_t^a1

1 ...t^ak_k ]_β = [a1+. . .+ak]β!

[a₁]_β!. . .[a_k]_β! [et1]^a_β¹. . .[et_k]^a_β^k, [e_B+(^ta1¹...t^ak_k )]^β = [a₁+. . .+a_k]_β!

[a₁]_β!. . .[a_k]_β! [et1]^a_β¹. . .[etk]^a_β^k,

where t1, . . . , t_k are distinct elements of T. In particular, [st]1 = st and [et]1 = et, whereras [s_t]₀ = 1 and [e_t]₀= 1 all t∈ T.

Examples.

t st [st]β et [et]β

q 1 1 1 1

qq 1 1 1 1

q

∨^{q q} 2 (1 +β) 1 1 qqq

1 1 1 1

q

∨^qqq 6 (1 +β)(1 + 2β) 1 1 q

∨^{q q} q

1 1 2 (1 +β)

q

∨^q_q^q

2 (1 +β) 1 1

qqqq

1 1 1 1

Proposition 13 For all n∈N^∗, in H_α,β,xn=αⁿ⁻¹ X

t∈T, weight(t)=n

[s_t]_β[e_t]_β st

t.

Examples.

x1 = q, x₂ = αqq, x3 = α²

(1 +β)

2 ∨^qqq + qqq ,

x₄ = α³ (1 + 2β)(1 +β)

6 ∨^qqqq + (1 +β) ∨^qq^q q

+(1 +β) 2

q

∨^q_q^q + qqqq !

,

x₅ = α⁴













(1+3β)(1+2β)(1+β)

24 _H^q ∨^qq^{q q}+(1+2β)(1+β) 2 ∨^qqq^q

q

+ (1 +β)² ∨^q∨^qq^qq

+(1 +β) ∨^qq^q qq

+(1+2β)(1+β) 6

q

∨^q qqq

+^(1+β)₂ ∨^qq^q q q

+ (1 +β) ∨^qq_q^q q

+^(1+β)₂ ∨^{q q}qqq + qqqqq











 .

Proof. For any t∈ T, we denote bybt the coefficient oft in x_weight(t). Then b_q = 1. The formal seriesf(h) is given by:

f(h) =

∞

X

n=0

αⁿ[n]_β! n! hⁿ.

Ift=B⁺(t^a₁¹. . . t^a_k^k), wheret1, . . . , tk are distinct elements ofT, then:

bt=α^a1+...+ak[a₁+. . .+a_k]_β! (a1+. . .+ak)!

(a₁+. . .+a_k)!

a1!. . . ak! b^a_t1¹. . . b^a_tk^k.

The result comes from an easy induction.

As a consequence, H_0,β = K[q], so H_0,β is a Hopf subalgebra. Moreover, H_α,β = H_1,β if α 6= 0: we can restrict ourselves to the case α = 1. In order to lighten the notations, we put n_t=s_te_t and [n_t]_β = [s_t]_β[e_t]_β for all t∈ T. Then:

n_q = 1,

n_B⁺_(t1...tk) = k!nt1. . . nt_k, [n_q]_β = 1,

[n_B+(t1...tk)]_β = [k]_β![n_t1]_β. . .[n_tk]_β.

As a consequence:

n_t= Y

svertex oft

(fertility ofs)!, [n_t]_β = Y

svertex oft

[fertility of s]_β!.

We shall use the following result, proved in [4, 6]:

Lemma 14 For all forests F ∈ F, G, H ∈ T, we denote by n(F, G;H) the coefficient of F⊗Gin ∆(H), and n⁰(F, G;H) the number of graftings of the trees of F overG giving the tree H. Then n⁰(F, G;H)s_H =n(F, G;H)s_Fs_G.

Lemma 15 Let k, n∈N^∗. We put, in K[X₁, . . . , X_n], S =X₁+. . .+X_n. Then:

X

α1+...+αn=k n

Y

i=1

X_i(X_i+ 1). . .(X_i+α_i−1) αi

= S(S+ 1). . .(S+k−1)

k! .

Proof. By induction onk, see [5].

Proposition 16 If α= 1:

∆(X) =X⊗1 +

∞

X

n=1

(1−βX)−n(1/β+1)+1⊗xn. SoH_1,β is a Hopf subalgebra.

Proof. As for alln≥1,x_n is a linear span of trees, we can write:

∆(X) =X⊗1 + X

F∈F, t∈T

aF,tF⊗t.

Then, ifF ∈ F,G∈ T:

aF,G = X

H∈T

[n_H]_β sH

n(F, G;H) = X

H∈T

[n_H]_β sFsG

n⁰(F, G;H).

We putF =t₁. . . t_k, and we denote bys₁, . . . , s_nthe vertices of the treeG, of respective fertility f₁, . . . , f_n. Let us consider a grafting of F over G, such that α_i trees of F are grafted on the vertex si. Then α1+. . .+αn=k. Denoting by H the result of this grafting:

[n_H]_β = [n_G]_β[n_t1]_β. . .[n_tk]_β[f₁+α₁]_β!

[f1]β! . . .[f_n+α_n]_β! [fn]β! . Moreover, the number of such graftings is k!

α₁!. . . α_n!. So, with lemma 15, puttingx_i=f_i+ 1/β and s=x1+. . .+xn:

aF,G = X

α1+...+αn=k

k!

α1!. . . αk! 1 sFsG

[nG]β[nt1]β. . .[ntk]β

[f₁+α₁]_β!

[f1]β! . . .[f_n+α_n]_β! [fn]β!

= [nG]β! s_G

k![nt1]β. . .[ntk]β

s_F

X

α1+...+αn=k n

Y

i=1

(1 +f_iβ). . .(1 + (f_i+α_i−1)β) α_i!

= [n_G]_β! sG

k![n_t1]_β. . .[n_tk]_β sF

X

α1+...+αn=k n

Y

i=1

β^αixi(xi+ 1). . .(xi+αi+ 1) αi!

= [nG]β! s_G

k![nt1]β. . .[ntk]β

s_F β^k X

α1+...+αn=k n

Y

i=1

x_i(x_i+ 1). . .(x_i+α_i+ 1) α_i!

= [nG]β! s_G

k![nt1]β. . .[ntk]β

s_F β^ks(s+ 1). . .(s+k−1)

k! .

Moreover, asG is a tree,s=f₁+. . .+f_n+n/β =n−1 +n/β =n(1 + 1/β)−1.

We now writeF =t1. . . tk =u^a₁¹. . . u^a_l^l, whereu1, . . . , ul are distinct elements ofT. Then:

sF =s^a_u¹₁. . . s^a_u^l_la1!. . . a_l!, so: k![n_t1]_β. . .[n_tk]_β

sF

= (a1+. . .+al)!

a1!. . . a_l!

[n_t1]_β st1

a1

. . .

[n_tl]_β st_l

al

. As a conclusion, puttingQk(S) = S(S+ 1). . .(S+k−1)

k! :

∆(X) = X⊗1 +X

n≥1

X

t^a₁¹...t^al_l ∈F

(a₁+. . .+a_l)!

a1!. . . al! β^a1+...+alQ_a1+...+al(n(1 + 1/β)−1) [n_t1]_β

st1

t1

a1

. . .

[n_tl]_β stl

tl

al

⊗







 X

G∈T weight(G)=n

[n_G]_β! sG

G









= X⊗1 +

∞

X

n=1

(1−βX)−n(1/β+1)+1⊗xn.

So ∆(X) ∈ H⊗H. Projecting on the homogeneous component of degreeb n, we obtain ∆(x) ∈ H ⊗ H, soH_1,β is a Hopf subalgebra.

Remarks.

1. For (α, β) = (1,0), f(h) =e^h and, for alln∈N^∗,xn= X

t∈T weight(t)=n

1 st

t.

2. For (α, β) = (1,1), f(h) = (1−h)⁻¹ and, for all n∈N^∗,xn= X

t∈T weight(t)=n

ett.

3. For (α, β) = (1,−1),f(h) = 1 +hand, as [i]−1= 0 ifi≥2, for alln∈N^∗,xnis the ladder of weightn.

2.4 What is Hα,β ?

If α= 0, then H_0,β =K[q]. Ifα 6= 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 byg_1,β. The dual ofg_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 bracket is the Lie cobracketδ induced by:

($⊗$)◦(∆−∆^op),

where$ is the canonical projection on coP rim(H_1,β). As H_1,β is the polynomial algebra gen- erated by the x_n’s, a basis ofcoP rim(H_1,β) is ($(x_n))n∈N^∗. By proposition 16:

($⊗$)◦∆(X) = ($⊗$)

∞

X

n=1

(1−βX)−n(1/β+1)+1⊗x_n

!

= X

n≥1

(n(1 +β)−β)$(X)⊗$(x_n).

Projecting on the homogeneous component of degreek:

($⊗$)◦∆(x_k) =

k

X

i+j=k

(j(1 +β)−β)$(x_i)⊗$(x_j).

As a consequence:

δ($(x_k)) = X

i+j=k

(1 +β)(j−i)$(xi)⊗$(xj).

Dually, the Lie algebrag_1,β has the dual basis (Zn)n≥1, with bracket given by:

[Z_i, Z_j] = (1 +β)(j−i)Z_i+j.

So, ifβ 6=−1, this Lie algebra is isomorphic to the Fa`a di Bruno Lie algebrag<sub>F dB</sub>, which has a basis (fn)n≥1, and its bracket defined by [fi, fj] = (j−i)fi+j. SoH<sub>1,β</sub> is isomorphic to the Hopf algebra U(g<sub>F dB</sub>)<sup>∗</sup>, namely the Fa`a di Bruno Hopf algebra [3], coordinates ring of the group of formal diffeomorphisms of the line tangent toId, that is to say:

G_{F dB} =nX

a_nhⁿ∈K[[h]]/ a₀= 0, a₁ = 1o ,◦

.

Theorem 17 1. If α 6= 0 and β 6= −1, H_α,β is isomorphic to the Fa`a di Bruno Hopf algebra.

2. If α6= 0 and β =−1, H_α,β is isomorphic to the Hopf algebra of symmetric functions.

3. If α= 0,H_α,β =K[q].

Remark. If β and β⁰ 6=−1, then H_1,β and H_1,β⁰ are isomorphic but are not equal, as it is shown by consideringx₃.

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 algebrag, graded, with Poincar´e-Hilbert formal series:

h 1−h =

∞

X

n=1

hⁿ.

Under an hypothesis of non-commutativity, we show that such a g is isomorphic to the Fa`a di Bruno Lie algebra, so the considered Hopf subalgebra is isomorphic to the Fa`a di Bruno Hopf algebra.

Remark. The proves of these section were completed using MuPAD pro 4. The notebook of the computations can be found at http://loic.foissy.free.fr/pageperso/publications.html.

3.1 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 degreen ofg. We shall say thatg is FdBif:

1. g is connected, that is to sayg(0) = (0).

2. For alli∈N^∗,gis one-dimensional.

3. For alln≥2, [g(1),g(n)]6= (0).

Letgbe a FdB Lie algebra. For alli∈N^∗, we fix a non-zero elementZ_i ofg(i). By conditions 1 and 2, (Z_i)i≥1 is a basis of g. By homogeneity of the bracket ofg, for all i, j≥1, there exists an elementλi,j ∈K, such that:

[Zi, Zj] =λi,jZi+j. The Jacobi relation gives, for alli, j, k ≥1:

λi,jλ_i+j,k+λ_j,kλ_j+k,i+λ_k,iλ_k+i,j = 0. (3) Moreover, by antisymmetry,λ_j,i=−λ_i,j for alli, j≥1. Condition 2 is expressed asλ_1,j 6= 0 for allj 6= 2.

Lemma 19 Up to a change of basis, we can suppose that λ1,j = 1 for all j ≥ 2 and that λ2,3 ∈ {0,1}.

Proof. We define a family of scalars by:







α1 = 1, α2 6= 0,

α_n = λ_1,2. . . λ1,n−1α₂ ifn≥3.

By condition 3, all these scalars are non-zero. We putZ_i⁰=αiZi. Then, for all j ≥2:

[Z₁⁰, Z_j⁰] =α_jλ_1,jZ_1+j = α_jλ_1,j αj+1

Z_1+j⁰ =Z_1+j⁰ . So, replacing theZi’s by the Z_i⁰’s, we can suppose thatλ1,j = 1 if j≥2.

Let us suppose now that λ2,3 6= 0. We then choose:

α2 = λ_1,3λ_1,4 λ2,3

. Then:

[Z₂⁰, Z₃⁰] = λ_2,3α₂α₃ α5

Z₅⁰ = λ_2,3α₂λ_1,2α₂ λ1,2λ1,3λ1,4α2

Z₅⁰ =Z₅⁰.

So, replacing theZi’s by the Z_i⁰’s, we can suppose thatλ2,3 = 1.

Lemma 20 If i, j≥2, λ_i,j =

i−2

X

k=0

i−2 k

(−1)^kλ_2,j+k.

Proof. Let us write (3) with i= 1:

λ1,jλj+1,k+λj,kλj+k,1+λk,1λk+1,j = 0.

Ifj, k≥2, thenλ_1,j =−λ_k,1 =−λ_j+k,1 = 1, so:

λk+1,j =λk,j−λk,j+1. (4)

Ifk= 2, this gives the announced formula fori= 3.

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 ranki−1. Then, by (4):

λi,j = λi−1,j−λi−1,j+1

=

i−3

X

k=0

i−3 k

(−1)^kλ_2,j+k−

i−3

X

k=0

i−3 k

(−1)^kλ_2,j+1+k

=

i−3

X

k=0

i−3 k

(−1)^kλ_2,j+k+

i−2

X

k=1

i−3 k−1

(−1)^kλ_2,j+k

= λ2,j+

i−3

X

k=1

i−2 k

(−1)^kλ2,j+k+ (−1)ⁱ⁻²λ2,j+i−2

=

i−2

X

k=0

i−2 k

(−1)^kλ_2,j+k.

So the result is true for alli≥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 k≥2, λ_2,2k= 1 2k−3

2k−4

X

l=0

2k−2 l

(−1)^lλ_2,l+3.

Proof. Let us write the relation of lemma 20 for (i, j) = (3,2k) and (i, j) = (2k,3):

λ3,2k = λ2,2k−λ2,2k+1, λ_2k,3 =

2k−2

X

l=0

2k−2 l

(−1)^lλ_2,3+l

= λ_2,2k+1−(2k−2)λ_2,2k+

2k−4

X

l=0

2k−2 l

(−1)^lλ_2,3+l.

Summing these two relations:

−(2k−3)λ_2,2k+

2k−4

X

l=0

2k−2 l

(−1)^lλ_2,3+l= 0.

This gives the announced result.

As a consequence, the λi,j’s are entirely determined by the λ2,j’s, with j odd. In order to lighten the notations, we put µ_j =λ_2,j for allj odd. Then, for example:















λ2,4 = µ3, λ2,6 = 2µ5−µ3,

λ_2,8 = 3µ₇−5µ₅+ 3µ₃,

λ2,10 = 4µ9−14µ7+ 28µ5−17µ3,

λ2,12 = 5µ11−30µ9+ 126µ7−255µ5+ 155µ3. Moreover, we showed that we can suppose that µ₃= 0 or 1.

Remark. The coefficient λ_2,2k+4 is then a linear span of coefficients µ_2i+3, 0≤i≤k. We put, for allk∈N:

λ_2,2k+4=

k

X

i=0

a_k,iµ_2i+3. We can prove inductively the following results:

1. For allk∈N,a_k,k =k+ 1.

2. For all k ≥ 1, ak,k−1 = −¹₄ ^2k+2₃

: up to the sign, this is the sequence A000330 of [11]

(pyramidal numbers).

3. For allk≥2,ak,k−2= ¹₂ ^2k+2₅

: this is the sequenceA053132 of [11].

4. The sequence (−a_k,0) is the sequence of signed Genocchi number, A001469 in [11].

It seems that for alli≤k:

ak,k−i= 2²ⁱ⁺²−1 i+ 1 B_2i+2

2k+ 2 2i+ 1

,

where theB2n’s are the Bernoulli number (see sequence A002105 of [11]).

3.2 Case where µ₃ = 1

Lemma 22 Suppose that µ₃= 1. Then µ₅= 1 or 9 10. Proof. By relation (3) for (i, j, k) = (2,3,4):

5µ5−3µ7+µ5µ7−3 = 0.

Ifµ5 = 3, we obtain 12 = 0, absurd. Soµ7 =−5µ₅−3

µ5−3 . By relation (3) for (i, j, k) = (2,3,6):

−2

µ5−3 (2µ₅−4)µ₅µ₉+ 3−7µ₅+µ²₅−5µ³₅

= 0.

Ifµ5 = 0, we obtain 2 = 0, absurd. Ifµ5= 2, we obtain 66 = 0, absurd. So:

µ₉ =−3−7µ₅+µ²₅−5µ³₅ (2µ5−4)µ5

. Writing relation (3) for (i, j, k) = (3,4,5):

−9(µ₅−1)⁵(10µ₅−9) µ5(µ5−2)(µ5−3)² = 0.

Soµ₅ = 1 orµ₅= 9

10.

Proposition 23 Let us suppose that µ3=µ5 = 1. Then:







λ1,j = 1 if j≥2, λ_2,j = 1 if j≥3,

λi,j = 0 if i, j≥3.

Proof. Let us first prove inductively onj thatλ_2,j = 1 ifj≥3. This is immediate ifj= 3 or 5 and comes from λ_2,4 =µ₃ for j= 4. Let us suppose the λ_2,j = 1 for 3≤j < n. If n= 2k is even, then:

λ_2,2k= 1 2k−3

2k−4

X

l=0

2k−2 l

(−1)^l= 1 + 1 2k−3

2k−2

X

l=0

2k−2 l

(−1)^l= 1.

Ifn= 2k+ 1 is odd, write relation (3) for (i, j, k) = (2,3,2k−2):

λ2,3λ5,2k−2+λ3,2k−2λ2k+1,2+λ2k−2,2λ2k,3 = 0,

3

X

l=0

3 l

(−1)^lλ2,2k−2+l−λ2,2k−2+λ2,2k−1+λ2,2k−2(λ_2,2k−λ_2,2k+1) = 0, λ2,2k−2−3λ2,2k−1+ 3λ_2,2k−λ_2,2k+1+ 1−λ_2,2k+1 = 0, 1−3 + 3−2λ_2,2k+1+ 1 = 0, soλ2,2k+1= 1. Finally, ifi, j ≥3,λi,j =

i−2

X

k=0

i−2 k

(−1)^k= 0.

Lemma 24 For all N ≥2, SN =

N

X

l=0

N l

(−1)^l (l+ 1)

(l+ 2)(l+ 3) = N −1

(N + 3)(N + 2)(N + 1). Proof. Indeed:

S_N =

N

X

l=0

N!

(l+ 3)!(N −l)!(−1)^l(l+ 1)²

= 1

(N + 3)(N+ 2)(N+ 1)

N+3

X

j=3

N + 3 j

(−1)^j(j−2)²

= 1

(N + 3)(N+ 2)(N+ 1)

N+3

X

j=0

N + 3 j

(−1)^j(j−2)²

− 1

(N+ 3)(N+ 2)(N + 1)(4−(N + 3))

= 0 + N −1

(N + 3)(N + 2)(N+ 1).

Proposition 25 Let us suppose that µ₃= 1 and µ₅ = 9

10. Then, for all i, j≥1:

λi,j= 6(i−j)(i−2)!(j−2)!

(i+j−2)! . Proof. We first prove thatλ_2,n = 6(n−2)

(n−1)n. This is immediate forn= 1, 2, 3, 4, 5. Let us suppose the result for allj < n, with n≥6. If n= 2kis even, using lemma 20:

λ2,2k= 6 2k−3

2k−4

X

l=0

2k−2 l

(−1)^l l+ 1 (l+ 2)(l+ 3).

Then lemma 24 gives the result. If n = 2k+ 3 is odd, let us write the relation (3) with (i, j, k) = (2,3,2k):

λ2,3λ5,2k+λ3,2kλ2k+3,2+λ2k,2λ2k+2,3 = 0.

So, with relation (4):

λ_2,2k−3λ_2,2k+1+ 3λ_2,2k+2−λ_2,2k+3−λ_2,2k+3(λ_2,2k−λ_2,2k+1) +λ_2,2k(λ_2,2k+2−λ_2,2k+3) = 0, λ2,2k+3(−1−2λ2,2k+λ2,2k+1) +λ2,2k−3λ2,2k+1+ 3λ2,2k+2+λ2,2kλ2,2k+2 = 0,

−λ_2,2k+3(2k+ 3)(k−1)(2k+ 5)

k(2k+ 1)(2k−1) +3(2k+ 5)(k−1)

k(k+ 1)(2k−1) = 0, which implies the result.

Let us now prove the result by induction on i. This is immediate ifi= 1, and the first part of this proof fori= 2. Let us suppose the result at rank i. Then, by relation (4):

λi+1,j = λi,j−λi,j+1

= 6(j−i)(i−2)!(j−2)!

(i+j−2)! −6(j+ 1−i)(i−2)!(j−1)!

(i+j−1)!

= 6(i−1)!(j−2)!(j+ 1−i) (i+j−1)! .

So the result is true for alli, j.

3.3 Case where µ₃ = 0

Proposition 26 If µ3 = 0, then λi,j = 0 for all i, j≥2.

Proof. We first prove that µ5 = 0. If not, by (3) for (i, j, k) = (2,3,4), µ5µ7 = 0, so µ7 = 0. By (3) with (i, j, k) = (2,3,7), −5µ₅(28µ5 + 4µ9) = 0, so µ9 = −7µ₅. By (3) with (i, j, k) = (3,4,5),−36µ²₅ = 0: contradiction. So µ₅= 0.

Let us then prove that all the µ_2k+1’s, k ≥ 1, are zero. We assume that µ3 = µ5 =. . . = µ2k−1 = 0, and µ_2k+1 6= 0, with l ≥ 3. By lemma 21, λ_2,2 = . . . = λ2,2k−1 = λ_2,2k = 0 and λ2,2k+1 6= 0. By relation (3) for (i, j, k) = (2,3, n), combined with (4):

λ_2,3λ_5,n+λ_3,nλ_n+3,2+λ_n,2λ_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.

Forn= 2k, this givesλ_2,2k+1λ_2,2k+3 = 0, soλ_2,2k+3 = 0. Forn= 2k+ 2:

λ_2,2k+2(λ_2,2k+4−2λ_2,2k+5) = 0. (5)

By lemma 21:

λ_2,2k+2 = 1 2k−1

2k 2k−2

λ_2,2k+1

= kλ2,2k+1, λ_2,2k+4 = 1

2k+ 1

2k+ 2 2k

λ_2,2k+3−

2k+ 2 2k−1

λ_2,2k+2+

2k+ 2 2k−2

λ_2,2k+1

= −k(k+ 1)(2k+ 1)

6 λ_2,2k+1. With (5):

λ_2,2k+5 =−k(k+ 1)(2k+ 1)

12 λ_2,2k+1. By relation (3) for (i, j, k) = (3,4,2k):

λ3,4λ7,2k+λ4,2kλ4+2k,3+λ2k,3λ2k+3,4= 0.