Combinatorial Hopf algebras from renormalization

TL;DR

This paper explores the structure of three non-commutative Hopf algebras relevant to quantum field theory renormalization, providing new methods to define their associative products.

Contribution

It introduces the right-sided combinatorial Hopf structures of three specific non-commutative Hopf algebras in renormalization, with two novel approaches to defining their associative products.

Findings

Describes the non-commutative Faà di Bruno Hopf algebra

Details the non-commutative charge renormalization Hopf algebra on planar binary trees

Introduces recursive and grafting/shuffling methods for associative product definition

Abstract

In this paper we describe the right-sided combinatorial Hopf structure of three Hopf algebras appearing in the context of renormalization in quantum field theory: the non-commutative version of the Fa\`a di Bruno Hopf algebra, the non-commutative version of the charge renormalization Hopf algebra on planar binary trees for quantum electrodynamics, and the non-commutative version of the Pinter renormalization Hopf algebra on any bosonic field. We also describe two general ways to define the associative product in such Hopf algebras, the first one by recursion, and the second one by grafting and shuffling some decorated rooted trees.

Equations

$$T^{\ge p}(R)\star T(R)\subset T^{\ge p}(R).$$

$$T_{<p}(R)\star T(R)\subset T_{<p}(R).$$

$$\displaystyle\big{\{}\{x;y_{1},\ldots,y_{n}\};z_{1},\ldots,z_{m}\big{\}}$$

$$M_{1q}(x;y_1\cdots y_q)$$

$$M_{10}(x;1)\equiv M_{10}(x)$$

$$(x_1\cdots x_n)\star (y_1\cdots y_m)$$

$$M_{00}=0\text{mbox}and{M}_{01}=M_{10}=\mathrm{Id},$$

$$\displaystyle M_{0p}=M_{p0}=0\qquad\mbox{for $p>1$},$$

$$\displaystyle M_{pq}=0\qquad\mbox{for $p>1$ and $q\neq 0$},$$

$$x\star y$$

$$=M_{11}(x;y)+M_{10}(x)M_{01}(y)+M_{01}(y)M_{10}(x)=M_{11}(x;y)+xy+yx.$$

$$(xy)\star z$$

$$x\star (yz)$$

$$x\star (y_1\cdots y_m)$$

$$X\star Y$$

$$=\sum _{\genfrac{}{}{0ex}{}{X=X^1{X}^2}{Y=Y^1{Y}^2}}M(X^1;Y^1)(X^2\star Y^2)$$

$$=\sum _{\genfrac{}{}{0ex}{}{X=X^1{X}^2{X}^3}{Y=Y^1{Y}^2{Y}^3}}(X^1\star Y^1)M(X^2,Y^2)(X^3\star Y^3),$$

$$X\star Y$$

$$(x_1\cdots x_n)\star (y_1\cdots y_m)$$

$$\displaystyle\qquad+\sum_{i=1}^{m-1}\big{[}(x_{1}\cdots x_{n-1})\star(y_{1}\cdots y_{i})\big{]}M_{1m-i}(x_{n};y_{i+1}\cdots y_{m}).$$

$$M(M(X;Y);Z)$$

$$\sum _{k\ge 1}\sum _{\genfrac{}{}{0ex}{}{Y=Y^1\cdots Y^k}{Z=Z^1\cdots Z^k}}M(X;M(Y^1;Z^1)\cdots M(Y^k;Z^k)),$$

$$\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt]\text{TC}\ast [linecolor=black,radius=2pt]\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt]\text{TC}\ast [linecolor=black,radius=2pt]\stackrel{d}{⟶}\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=b]x\text{TC}\ast [linecolor=black,radius=2pt][tnpos=l]y\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=r]u\text{TC}\ast [linecolor=black,radius=2pt][tnpos=a]v$$

$$\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=b]x\text{TC}\ast [linecolor=black,radius=2pt][tnpos=l]y\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=r]{\lambda }_1{u}_1+{\lambda }_2{u}_2\text{TC}\ast [linecolor=black,radius=2pt][tnpos=a]v={\lambda }_1\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=b]x\text{TC}\ast [linecolor=black,radius=2pt][tnpos=l]y\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=r]u_1\text{TC}\ast [linecolor=black,radius=2pt][tnpos=a]v+{\lambda }_2\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=b]x\text{TC}\ast [linecolor=black,radius=2pt][tnpos=l]y\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=r]u_2\text{TC}\ast [linecolor=black,radius=2pt][tnpos=a]v.$$

$$R^{\mathcal{T}}=\underset{t\in \mathcal{T}}{⨁}{R}^t$$

$$B_{+}^x:T(R^{\mathcal{T}})⟶R^{\mathcal{T}}$$

$$B_{+}^x\left(\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=b]y\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=b]u\text{TC}\ast [linecolor=black,radius=2pt][tnpos=a]v\right)=\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=b]x\text{TC}\ast [linecolor=black,radius=2pt][tnpos=l]y\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=r]u\text{TC}\ast [linecolor=black,radius=2pt][tnpos=a]v.$$

$$M(B_{+}^x(f_1);f_2)$$

$$M(M(t_1;f_2);f_3)=M(t_1;f_2\star f_3)$$

$$M(M(B_{+}^x(f_1);f_2);f_3)$$

$$=B_{+}^x((f_1\star f_2)\star f_3)\text{(Recursion)}$$

$$=B_{+}^x(f_1\star (f_2\star f_3))$$

$$=M(B_{+}^x(f_1);f_2\star f_3).$$

$$M\left(\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=b]x;\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=b]y\right)=B_{+}^x\left(\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=b]y\right)=\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=b]x\text{TC}\ast [linecolor=black,radius=2pt][tnpos=a]y$$

$$\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=b]x\star \text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=b]y=M\left(\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=b]x;\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=b]y\right)+\left(\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=b]x\star 1\right)M\left(1;\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=b]y\right)+\left(1\star \text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=b]y\right)M\left(\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=b]x;1\right)=\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=b]x\text{TC}\ast [linecolor=black,radius=2pt][tnpos=a]y+\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=b]x\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=b]y+\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=b]y\text{pstree}\text{TC}\ast [linecolor=black,radius=2pt][tnpos=b]x$$

Full text

We also describe two general ways to define the associative product in such Hopf algebras, the first one by recursion, and the second one by grafting and shuffling some decorated rooted trees.

1 Introduction

In the paper [13], J.-L. Loday and M. Ronco gave the definition of a right-sided combinatorial Hopf algebra (CHA) which includes the examples of Hopf algebras describing the renormalization in quantum field theory. The toy example, which inspired the work in [13], is Kreimer’s Hopf algebra on rooted trees introduced in [9]. In this paper, we describe the CHA structure of three other such examples: the non-commutative version of the Faà di Bruno Hopf algebra given in [3], the non-commutative version of the charge renormalization Hopf algebra on planar binary trees given in [1] for quantum electrodynamics, and the non-commutative version of the Pinter renormalization Hopf algebra on any quantum fields, as given in [4].

Since Loday and Ronco work in the context of cofree-coassociative CHAs, we describe our examples, which are all free-associative CHAs, by considering their linear duals. A right-sided cofree-coassociative CHA is completely determined by the brace structure on the set its primitive elements. Therefore, to describe our examples, it will be sufficient to give their brace structure.

In the first section we fix the notation on right-sided cofree-coassociative CHAs and brace algebras from [13]. We also give a recursive definition of the product and show how to construct it starting from the one defined on decorated rooted trees, that is, the product defined on the right-sided CHA induced by the free brace algebra over the space of decorations. Then we fix the notation for the dual Hopf algebras, which are used in the sequel. Each example is then treated in a separate section.

2 Combinatorial Hopf algebras from

dual Hopf algebras

2.1 Right-sided combinatorial Hopf algebras

In this section we recall the definition and main properties of a right-sided combinatorial Hopf algebra (r-s CHA), as given by J.-L. Loday and M. Ronco in [13], in the cofree-coassociative case.

A right-sided cofree-coassociative CHA is, up to an isomorphism, a tensor coalgebra $T(R)=\oplus_{n\geq 0}R^{\otimes n}$ with the deconcatenation coproduct, endowed with an associative product $\star$ which makes it into a bialgebra and which satisfies the right-sided condition: for any $p\geq 0$, the subspace $T^{\geq p}(R)=\oplus_{n\geq p}R^{\otimes n}$ is a right ideal of $T(R)$, that is

[TABLE]

This condition is equivalent to demanding that the product $\star$ induces a right action of $T(R)$ on the quotient space $T_{<p}(R)=T(R)/T^{\geq p}(R)$, that is

[TABLE]

Loday and Ronco proved that a cofree-coassociative CHA is right-sided if and only if the set of its primitive elements $R$ is a brace algebra, that is, it is endowed with a brace product $\{\ ;\,...,\ \}$ acting on $R\otimes T(R)$ with value in $R$, satisfying the brace relation

[TABLE]

for any $x,y_{i},z_{j}\in R$. The brace product restricted to $R\otimes R^{\otimes q}$ is also denoted by $M_{1q}$, if one wishes to specify the number of variables on the right, or to underline that it is a special case of a multibrace product $M_{pq}$ defined on any powers $R^{\otimes p}\otimes R^{\otimes q}$.

The brace product can be found from the associative product $\star$ by projecting the result onto the space $R$ of co-generators. If we denote by $\pi:T(R)\longrightarrow R$ the canonical projection, this means that

[TABLE]

for any $x\in R$ and $y_{1}\cdots y_{q}\in R^{\otimes q}$, where we denote the tensor product in $T(R)$ by the concatenation. In particular, for any $x\in R$ we have

[TABLE]

therefore $M_{10}=\mathrm{Id}$ on $R$.

Conversely, the product $\star$ in $T(R)$ can be reconstructed from the brace product on $R$, using the deconcatenation coproduct to recover $T(R)$ from $R$. The resulting formula of Loday and Ronco for the product is (cf. [13])

[TABLE]

where each map $M_{pq}$ is applied to a block of $p$ generators $x$ and $q$ generators $y$, with the following assumptions:

[TABLE]

and where the operators $M_{1q}$ for $q\neq 0$ of course satisfy the brace relation (2.2).

Applying formula (2.3) requires some computations. Let us take the example of $x\star y$. For $k=1$ we have $p_{1}=q_{1}=1$ and the only term is $M_{11}$. For $k=2$ we have $(p_{1},p_{2})=(1,0)$ or $(p_{1},p_{2})=(0,1)$ and $(q_{1},q_{2})=(1,0)$ or $(q_{1},q_{2})=(0,1)$. This gives us the four terms $M_{11}M_{00}+M_{10}M_{01}+M_{01}M_{10}+M_{00}M_{11}$. Since $M_{00}=0$, we have only two terms left $M_{10}M_{01}+M_{01}M_{10}$. For $k>2$, all the terms are zero because each term contains a factor $M_{00}$. In general, the terms with $k>n+m$ are zero in the product $(x_{1}\cdots x_{n})\star(y_{1}\cdots y_{m})$. Therefore, using (2.1), we get

[TABLE]

Similarly, for $(xy)\star z$, we have two terms for $k=2$: $M_{11}M_{10}+M_{10}M_{11}$ and three terms for $k=3$: $M_{01}M_{10}M_{10}+M_{10}M_{01}M_{10}+M_{10}M_{10}M_{01}$, so that

[TABLE]

For $x\star(yz)$, we have one term for $k=1$: $M_{12}$, two terms for $k=2$: $M_{11}M_{01}+M_{01}M_{11}$ and three terms for $k=3$: $M_{01}M_{01}M_{10}+M_{01}M_{10}M_{01}+M_{10}M_{01}M_{01}$, so that

[TABLE]

In particular then, formula (2.3) for $n=1$ gives

[TABLE]

2.2 Recursive definition of the product.

Formula (2.3) can be given in a recursive way. To do it, let us denote by $M:R\otimes T(R)\longrightarrow R$ the brace product $\{\ ;\ ,\ldots,\ \}$ induced on $R$ by the associative product $\star$ on $T(R)$, and let us denote by $X,Y,Z...$ the generic words in $T(R)$.

Proposition 2.1

Given the brace product $M$ on $R$, the $\star$ product on $T(R)$ can be reconstructed in the following recursive way. For any $X,Y\in T(R)$, with $X,Y\notin T^{0}(R)$:

[TABLE]

where the sums run over all possible factorisations of $X$ and $Y$ with respect to the concatenation.

Proof. With the assumptions (2.1) on $M$, formula (2.3) can be expressed as

[TABLE]

where the sum runs other all the factorisations (for the concatenation) of $X=X^{1}\cdots X^{k}$ and $Y=Y^{1}\cdots Y^{k}$. Note that the factors can be in $T^{0}(R)$ but the condition $M(1;1)=0$ ensures that the sum is finite. It also ensures that formula (2.6) is recursive.

If we then use this formula to express the factor $X^{1}\star Y^{1}$ in line (2.6), we obtain the desired expression for $X\star Y$. $\Box$

In particular, if we expand the words $X$ and $Y$ of $T(R)$ in terms of elements $x_{i}$ and $y_{j}$ of $R$, with the notation of Section 2.1, formula (2.6) becomes

[TABLE]

Corollary 2.2

The brace relation (2.2) is equivalent to the identity

[TABLE]

for any $X\in R$ and $Y,Z\in T(R)$.

Proof. If we call $X=x$, $Y=y_{1}\cdots y_{n}$ and $Z=z_{1}\cdots z_{m}$, the left hand-side of the brace relation (2.2) is simply $M(M(X;Y);Z)$. The right-hand side, instead, becomes

[TABLE]

where we use the assumptions (2.1) on $M$. From (2.7) it then follows that the right-hand side is equal to $M(X;Y\star Z)$. $\Box$

Remark 2.3

The right-sided combinatorial Hopf algebras, as well as the shuffle and the quasishuffle Hopf algebras, are examples of more general cofree-coassociative combinatorial Hopf algebras $T(R)$ introduced by Loday and Ronco, where $R$ is a so-called multibrace algebra. See [13] and the references therein for details. In all these examples, as well as in the general case, the product $\star$ on $T(R)$ can be defined recursively from the multibrace product on $R$. The expressions (2.7), (2.6) and (2.9), with $X\in T(R)$, are valid in the most general case as well as in all the examples.

2.3 Right-sided Hopf algebras and trees

Let $T(R)$ be a right-sided CHA such that the brace algebra $R$ is a finite dimensional or a graded vector space. In the first part of this section we forget the brace structure on $R$ and consider the free brace algebra on $R$ modeled by planar rooted trees. The free brace algebra has already been described in terms of trees by F. Chapoton in [5], as a consequence of his construction of the Brace operad. Here we give a direct description of the free brace algebra within the notations we established in section 2.2.

In the second part of this section we relate the free brace structure on $R$ with the original one, and deduce a way to describe the associative product $\star$ on $T(R)$ starting from the one defined on trees. Again, this is a particular case of the equivalence between the categories of brace and dendriform algebras given by Chapoton in [5]. However it seems to us useful for computations to write an explicit statement.

Let ${\cal{T}}$ denote the set of plane rooted trees and for any non-empty tree $t$ denote by $V(t)$ the set of its vertices and by $|t|$ the cardinality of $V(t)$. If $d$ is a map from $V(t)$ to $R^{\otimes|t|}$, then the pair $(t,d)$ is a plane tree with decorations in $R$. For example, the following tree with 4 vertices is decorated by the given map $d$:

[TABLE]

For any $t\in{\cal{T}}$, we denote by $R^{t}$ the linear span of all the decorated trees of shape $t$, modulo the obvious linear relations coming from $R$, such as:

[TABLE]

Then the vector space

[TABLE]

contains the decorated trees of any shape, and the elements of the tensor space $T(R^{{\cal{T}}})$ are called decorated forests. From now on, we denote by the concatenation the tensor product between trees, and we simply denote by $t$ a decorated tree, omitting the decoration.

We now define the structure of a brace algebra on $R^{{\cal{T}}}$, and consequently that of a right-sided CHA on $T(R^{{\cal{T}}})$, using the results of section 2.2. Therefore, we endow $T(R^{{\cal{T}}})$ with the free (deconcatenation) coproduct and we suppose that the product $\star$ is defined through the brace product $M$ on $R^{{\cal{T}}}$.

To define the brace product, consider, for any $x\in R$, the linear map

[TABLE]

which maps a forest $f=t_{1}\cdots t_{n}$ into the tree $t=B_{+}^{x}(f)$ obtained by joining the ordered trees $t_{1},...,t_{n}$ to a new root decorated by $x$. For instance

[TABLE]

Any tree $t$ can be written as $B_{+}^{x}(f)$. If $t$ is made only of the root, decorated by $x$, the forest $f$ is the formal unit of $T(R^{{\cal{T}}})$, that is $1\in(R^{{\cal{T}}})^{0}$.

Theorem 2.4

Let $M:R^{{\cal{T}}}\otimes T(R^{{\cal{T}}})\longrightarrow R^{{\cal{T}}}$ be the linear map defined on any decorated tree $t=B_{+}^{x}(f_{1})$ and any decorated forest $f_{2}$ by

[TABLE]

where $\star$ is the product induced by $M$ on $T(R^{{\cal{T}}})$ according to Eq. (2.7) and with the assumptions (2.1).

Then $M$ is a brace product on $R^{{\cal{T}}}$, and consequently $\star$ is an associative product on $T(R^{{\cal{T}}})$ which makes it into a right-sided CHA.

Proof. Assuming conditions (2.1) on the maps $M_{pq}$, it suffices to show that $M$ satisfies the brace identity (2.9), that is

[TABLE]

for any tree $t_{1}$ and any forests $f_{2},f_{3}$. If $t_{1}$, $f_{2}$, $f_{3}$ have respectively $n_{1}$, $n_{2}$ and $n_{3}$ vertices, we prove the brace identity by induction on the total number of vertices $n_{1}+n_{2}+n_{3}$.

Since $t_{1}$ is a tree, it has at least one vertex (the root) and therefore $n_{1}+n_{2}+n_{3}\geq 1$. If $n_{1}+n_{2}+n_{3}=1$, only $t_{1}$ is a tree, namely the single root, and $f_{2},f_{3}$ are scalars. Because $1\star 1=1$ and $M_{10}=M_{01}=\mathrm{Id}$, both terms in the equality produce $t_{1}$ and therefore coincide.

Now suppose that the brace identity holds for $n_{1}+n_{2}+n_{3}\leq n$, for a given $n\geq 1$. According to the previous sections, this implies that the product $\star$ is associative up to the same number $n$ of vertices. (In fact, the associativity of $\star$ does not depend on the definition (2.10) of $M$ and was proved in general by Loday and Ronco.) Then consider a tree and two forests with a total number of vertices $n_{1}+n_{2}+n_{3}=n+1$. Using the definition of $M$ and the associativity of $\star$ up to $n$, for $t_{1}=B_{+}^{x}(f_{1})$ we obtain

[TABLE]

$\Box$

Lemma 2.5

The brace product $M$ on $R^{{\cal{T}}}$ can be expressed in terms of graftings of trees: if $t\in R^{{\cal{T}}}$ and $s_{1}\cdots s_{n}\in T(R^{{\cal{T}}})$, then $M(t;s_{1}\cdots s_{n})$ is the sum of the terms obtained by grafting the trees $s_{1},...s_{n}$ onto the vertices of $t$, in all the possible ways which preserve the order of the $s_{i}$’s from left to right.

Similarly, the product $\star$ on $T(R^{{\cal{T}}})$ is the sum of terms mixing all possible shuffles and all possible graftings of the trees on the right-hand side onto the vertices of the trees on the left hand-side.

This result shows that the recursive definition of the brace product (2.10) is equivalent to the definition given in [5] in terms of grafting of trees.

Proof. We prove it by induction on the size of the forests, given by the total number of vertices. For this, we need to treat at the same time $M$ and $\star$, which is given by (2.6).

The induction starts with the first non-trivial operations:

[TABLE]

is the grafting of the second $\bullet$ on the first one. The claim is obtained because this is the only possible grafting. Similarly, according to (2.6), the product

[TABLE]

is exactly made of the only possible grafting and the two allowed shuffles.

Now suppose that $M(t_{1};f_{2})$ is made of all possible ordered graftings of the trees composing the forest $f_{2}$ on the vertices of the tree $t_{1}$, for $t_{1}$ and $f_{2}$ having a total number of vertices $n_{1}+n_{2}\leq n$, for some $n\geq 1$. Similarly, suppose that $f_{1}\star f_{2}$ is made of the terms mixing all possible shuffles of the trees of $f_{1}$ with those of $f_{2}$, and of graftings of the trees of $f_{2}$ over the vertices of those contained in $f_{1}$, for $f_{1}$ and $f_{2}$ having a total number of vertices $n_{1}+n_{2}\leq n$.

Then, consider $t_{1}$ and $f_{2}$ with $n_{1}+n_{2}=n+1$. Applying the definition (2.10) for $t_{1}=B_{+}^{x}(f_{1})$, we see that

[TABLE]

is the grafting of $f_{1}\star f_{2}$ onto the original root. The claim is then guaranteed by the inductive hypothesis on $f_{1}\star f_{2}$.

Similarly, consider $t_{1}$ and $f_{2}$ with $n_{1}+n_{2}=n+1$. Applying formula (2.6) we see that

[TABLE]

contains exactly all possible shuffles mixed with all possible graftings. This is guaranteed by the inductive hypothesis on the factors $f_{1}^{\prime}\star f_{2}^{\prime}$ and $M(f_{1}^{\prime\prime};f_{2}^{\prime\prime})$, and the fact that the sum runs over all decompositions of $f_{1}$ and $f_{2}$ into two factors, including the trivial ones. $\Box$

Here are some examples of graftings and shuffles:

[TABLE]

According to the results of Loday and Ronco, in particular paragraph 3.14 in [13], we can conclude that

Corollary 2.6

The brace algebra $R^{{\cal{T}}}$, endowed with $M$, is the free brace algebra over $R$.

Remark 2.7

The algebras $R^{{\cal{T}}}$ and $T(R^{{\cal{T}}})$, or their duals, or their non-planar and commutative versions, appeared in really many recent works. We quote, but the list is surely not exhaustive, the works by R. Grossman and R. G. Larson [7], F. Chapoton [5], L. Foissy [6], D. Guin et J.-M. Oudom [8], M. Ronco [15] and J.-L. Loday and M. Ronco [13].

Now suppose that $R$ itself is a brace algebra, and therefore $T(R)$ is a right-sided CHA. We are going to show that the $\star$ product defined in $T(R)$ can be found by computing its lifting to $T(R^{{\cal{T}}})$.

Let $\iota$ be the linear morphism from $T(R)$ to $T(R^{{\cal{T}}})$ defined by $\iota(1)=1$ and

[TABLE]

It is clear that $\iota$ preserves the concatenation in $T(R)$ and $T(R^{{\cal{T}}})$, and that it identifies $R$ with $R^{\bullet}\subset R^{{\cal{T}}}$.

Note that the map $\iota$ does not preserve the brace product, in fact

[TABLE]

and therefore it does not preserve the product $\star$ in $T(R)$ and $T(R^{{\cal{T}}})$. We show that $\iota$ has an inverse map which allows to compute the product $\star$ in $T(R)$ starting from that in $T(R^{{\cal{T}}})$.

Theorem 2.8

Let $\mu$ the linear morphism from $T(R^{{\cal{T}}})$ to $T(R)$ defined recursively by

$\mu(1)=1$, 2. 2.

$\mu(t_{1}\ldots t_{n})=\mu(t_{1})\ldots\mu(t_{n})$, 3. 3.

\mu\big{(}B_{+}^{x}(s_{1}\ldots s_{q})\big{)}=M_{1q}\big{(}x;\mu(s_{1}\ldots s_{q})\big{)}.

Then, for any $X,Y\in T(R)$, we have

[TABLE]

Proof. The proof is straightforward and relies on the recursive formulas in brace algebras and on the fact that $\mu\circ\iota=Id_{T(R)}$. If $X=x_{1}\cdots x_{p}$ and $Y=y_{1}\cdots y_{q}$ then

[TABLE]

Since $R$ is a brace algebra, either $X^{2}=1$ or $X^{2}=x_{p}$.

• •

If $X^{2}=1$, then

[TABLE]

• •

If $X^{2}=x_{p}$, then

[TABLE]

Recursively, we get

[TABLE]

$\Box$

This result means that one can compute the $\star$ product in $T(R)$ with the help of the $\star$ product in $T(R^{{\cal{T}}})$. This last product, given by shuffle and grafting of trees, is quite easy to handle. Let us illustrate the application with an example:

[TABLE]

This computation can be compared with that made in section 2.1.

Remark 2.9

The result of theorem (2.8) can be obtained also as a consequence of the equivalence between the categories of brace and dendriform algebras proved by Chapoton in [5]. Since $\mu$ is by assumption a brace homomorphism between $R^{{\cal{T}}}$ and $R$, this equivalence allows to state that for any $f_{1}$ and $f_{2}$ in $T(R^{{\cal{T}}})$ we have $\mu(f_{1})\star\mu(f_{2})=\mu(f_{1}\star f_{2})$. If we then choose $f_{1}=\iota(X)$ and $f_{2}=\iota(Y)$, we obtail formula (2.11).

Remark 2.10

The algebra $T(R^{{\cal{T}}})$ is the dual of Foissy’s Hopf algebra of decorated plane trees, cf. [6]. Foissy’s Hopf algebra on rooted trees has a “universal property”, in the sense that it is endowed with the Hochschild cocycle operator $B_{+}$. Note that the property that we show for $T(R^{{\cal{T}}})$ is not the dual property of that one.

2.4 Right-sided CHAs from dual Hopf algebras

In the next sections we determine some right-sided CHAs $T(R)$ from the dualization of given Hopf algebras $T(V)$ where the vector space $V$ is itself graded. In this case, the vector space $R$ coincides with the dual space $V^{*}$, and we give a general tool to compute the brace structure on $R$ starting from the coproduct on $T(V)$. This tool is then used in the examples presented in the rest of the paper.

Let ${\cal{H}}$ be a graded Hopf algebra that is free as an algebra, finite dimensional in all degrees, and with generators which are themselves graded. If we denote by $V$ be the vector space spanned by all the generators of ${\cal{H}}$ (infinite but countable many), the Hopf algebra can be given as ${\cal{H}}=T(V)=\bigoplus_{n=0}^{\infty}V^{\otimes n}$.

The dual Hopf algebra ${\cal{H}}^{*}$ is defined as follows. As a vector space, it is the graded linear dual vector space of ${\cal{H}}$, that is ${\cal{H}}^{*}=T(V^{*})$, where $V^{*}$ is the graded linear dual vector space of $V$. If we denote by $\{v_{n},n\geq 1\}$ a generic basis of $V$, that is a set of generators of ${\cal{H}}$, the element $v_{n}^{*}$ dual to each $v_{n}$ is defined by the pairing $\langle v_{n}^{*}|v_{m}\rangle=\delta_{n,m}$. Therefore, the dual of a generic basis element $v_{n_{1}}\cdots v_{n_{i}}$ in ${\cal{H}}$ is the element $v^{*}_{n_{1}}\cdots v^{*}_{n_{i}}$ in ${\cal{H}}^{*}$ defined by the pairing

[TABLE]

As a coalgebra, ${\cal{H}}^{*}$ inherits the coproduct $\Delta^{*}$ dual to the product of ${\cal{H}}$, that is, such that

[TABLE]

Since ${\cal{H}}$ is free and unital as an algebra, ${\cal{H}}^{*}$ is cofree and counital as a coalgebra, that is, $\Delta^{*}$ is the deconcatenation coproduct

[TABLE]

and with the counit $\varepsilon=1^{*}$. In this expression, $1$ is not (yet) a unit, it denotes the generator of $(V^{*})^{\otimes 0}$. The set of primitive elements of ${\cal{H}}^{*}$ is then obviously the dual $V^{*}$ of the set of generators of ${\cal{H}}$.

Finally, as an algebra, ${\cal{H}}^{*}$ inherits the product $\star$ dual to the coproduct $\Delta$ of ${\cal{H}}$, that is, the product such that

[TABLE]

The right-sided condition (2.1) for the Hopf algebra $T(V^{*})$ is equivalent to require that the coproduct $\Delta$ on $T(V)$ induces a right coaction of the Hopf algebra $T(V)$ on the subspace $T^{\leq p}(V)=\oplus_{n\leq p}V^{\otimes n}$, that is

[TABLE]

Explicitly, this means that for any element $a=v_{n_{1}}\cdots v_{n_{p}}\in V^{\otimes p}$, the coproduct $\Delta(a)=\sum a_{(1)}\otimes a_{(2)}$ (in Sweedler’s notation) produces terms with $a_{(1)}\in T^{\leq p}(V)$. Since the coproduct is an algebra morphism, and the product is free, it is sufficient that this property holds for $p=1$: for any generator $v\in V$, the coproduct $\Delta(v)=\sum v_{(1)}\otimes v_{(2)}$ produces terms with $v_{(1)}\in T^{\leq 1}(V)=K\oplus V$ (cf. [13], Section 3.6).

Finally, the brace product on $V^{*}$ is the projection of $\star$ onto $V^{*}$. The projection $\pi:T(V^{*})\longrightarrow V^{*}$, in this case, is the map

[TABLE]

for any $a^{*}\in T(V^{*})$. Therefore, the maps $M_{1q}:V^{*}\otimes(V^{*})^{\otimes q}\longrightarrow V^{*}$ are simply given by

[TABLE]

This sum if finite. In fact, let us denote by $|a|$ [resp. $|a^{*}|$] the degree of the elements in the graded Hopf algebra $T(V)$ [resp. $T(V^{*})$]. We recall that if $|v_{n}|$ denotes the degree of a generic basis element of $V$, then the degree of an element $v_{n_{1}}\cdots v_{n_{p}}\in T(V)$ is given by $|v_{n_{1}}\cdots v_{n_{p}}|=|v_{n_{1}}|+\cdots+|v_{n_{p}}|$. Then the sum over $m$ in Formula (2.13) is limited by the fact that $|v^{*}_{m}|=|v_{m}|=|v_{n}^{*}|+|a^{*}|$.

3 The right-sided combinatorial structure

of the Hopf algebra $({\cal{H}}^{\mathrm{dif,nc}})^{*}$

Let ${\cal{H}}^{\mathrm{dif,nc}}$ be the non-commutative Hopf algebra of formal diffeomorphisms, as defined in [16] or [3] (where it was denoted by ${\cal{H}}^{\mathrm{dif}}$). It is the graded and connected Hopf algebra $\mathbb{Q}\langle v_{1},v_{2},...\rangle=T(V)$ on the generators $V=\mathrm{Span}\{v_{1},v_{2},...\}$ graded by $|v_{n}|=n$, considered with the free (tensor) product, and endowed with the coproduct

[TABLE]

where we set $v_{0}=1$, and with the counit $\varepsilon(v_{n})=\delta_{n,0}$. This Hopf algebra is a free-associative CHA, and the formula (3.1) clearly says that the coproduct $\Delta^{\mathrm{dif}}$ satisfies the right-sided condition (2.12).

Let $({\cal{H}}^{\mathrm{dif,nc}})^{*}=T(V^{*})$ be its dual Hopf algebra, as described in section 2, with primitive elements in $V^{*}=\mathrm{Span}\{v^{*}_{1},v^{*}_{2},...\}$. Then $({\cal{H}}^{\mathrm{dif,nc}})^{*}=T(V^{*})$ is a right-sided cofree-coassociative CHA, and moreover a dendriform algebra, and $V^{*}$ is a brace algebra.

Proposition 3.1

The brace product on $V^{*}$ which induces the CHA structure on $({\cal{H}}^{\mathrm{dif,nc}})^{*}$ is given by

[TABLE]

Proof. According to eq. (2.13), for any $v_{n}\in V$ and any $a\in{\cal{H}}^{\mathrm{dif,nc}}$, we have

[TABLE]

If we put $v_{0}=1$, the coproduct of $v_{m}$ becomes [3]

[TABLE]

with

[TABLE]

Therefore,

[TABLE]

We take $a=v_{m_{1}}\dots v_{m_{q}}$ and we evaluate $\langle v_{m_{1}}^{*}\dots v_{m_{q}}^{*}\ |\ Q^{(n)}_{m-n}(v)\rangle$. From the definition of $Q^{(n)}_{m-n}(v)$, we must have $l=q$, $j_{1}=m_{1},\dots,j_{q}=m_{q}$. Therefore, $m-n=j_{1}+\dots+j_{q}=m_{1}+\dots+m_{q}$ and

[TABLE]

$\Box$

4 The right-sided combinatorial structure

of the Hopf algebra $({\cal{H}}^{\alpha,nc})^{*}$