﻿ From Braided Infinitesimal Bialgebras to Braided Lie Bialgebras

Vol.07 No.07(2017), Article ID:77682,9 pages
10.4236/apm.2017.77023

From Braided Infinitesimal Bialgebras to Braided Lie Bialgebras

Shengxiang Wang1,2

1Department of Mathematics, Nanjing University, Nanjing, China

2School of Mathematics and Finance, Chuzhou University, Chuzhou, China    Received: May 22, 2017; Accepted: July 14, 2017; Published: July 17, 2017

ABSTRACT

The present paper is a continuation of  , where we considered braided infinitesimal Hopf algebras (i.e., infinitesimal Hopf algebras in the Yetter-Drin- feld category ${}_{H}^{H}\mathcal{Y}\mathcal{D}$ for any Hopf algebra H), and constructed their Drinfeld double as a generalization of Aguiar’s result. In this paper we mainly investigate the necessary and sufficient condition for a braided infinitesimal bialgebra to be a braided Lie bialgebra (i.e., a Lie bialgebra in the category ${}_{H}^{H}\mathcal{Y}\mathcal{D}$ ).

Keywords:

Braided Infinitesimal Bialgebra, Braided Lie Bialgebra, Yetter-Drinfeld Category, Balanceator 1. Introduction

An infinitesimal bialgebra is a triple $\left(A,m,\Delta \right)$ , where $\left(A,m\right)$ is an associative algebra (possibly without unit), $\left(A,\Delta \right)$ is a coassociative coalgebra (possibly without counit) such that

$\Delta \left(xy\right)=x{y}_{1}\otimes {y}_{2}+{x}_{1}\otimes {x}_{2}y,x,y\in A.$

Infinitesimal bialgebras were introduced by Joni and Rota in  , called infini- tesimal coalgebra there, in the context of the calculus of divided differences  . In combinatorics, they were further studied in    . Aguiar established the basic theory of infinitesimal bialgebras in   by investigating several examples and the notions of antipode, Drinfeld double and the associative Yang- Baxter equation keeping close to ordinary Hopf algebras. In  , Yau introduced the notion of Hom-infinitesimal bialgebras and extended Aguiar’s main results in   to Hom-infinitesimal bialgebras.

One of the motivations of studying infinitesimal bialgebras is that they are closely related to Drinfeld’s Lie bialgebras (see  ). The cobracket $\Delta$ in a Lie bialgebra is a 1-cocycle in Chevalley-Eilenberg cohomology, which is a 1-cocycle in Hochschild cohomology (i.e., a derivation) in a infinitesimal bialgebra. So the compatible condition in a infinitesimal bialgebra can be seen as an associative analog of the cocycle condition in a Lie bialgebra.

Motivated by  , in which we considered infinitesimal Hopf algebras in the Yetter-Drinfeld categories, called braided infinitesimal Hopf algebras, the natural idea is whether we can obtain braided Lie bialgebras (called generalized H-Lie bialgebras in   ) from braided infinitesimal Hopf algebras. This becomes our motivation of writing this paper.

To give a positive answer to the question above, we organize this paper as follows.

In Section 1, we recall some basic definitions about Yetter-Drinfeld modules and braided Lie bialgerbas. In Section 2, we introduce the notion of the balanceator of a braided infinitesimal bialgerba and show that a braided infinitesimal bialgerba gives rise to a braided Lie bialgerba if and only if the balanceator is symmetric (see Theorem 2.3).

2. Preliminaries

In this paper, k always denotes a fixed field, often omitted from the notation. We use Sweedler’s (  ) notation for the comultiplication: $\Delta \left(h\right)={h}_{1}\otimes {h}_{2}$ , for all $h\in H$ . Let H be a Hopf algebra. We denote the category of left H-modules by ${}_{H}\mathcal{M}$ . Similarly, we have the category ${}^{H}\mathcal{M}$ of left H-comodules. For a left H- comodules $\left(M,\rho \right)$ , we also use Sweedler’s notation: $\rho \left(m\right)={m}_{\left(-1\right)}\otimes {m}_{0},$ for all $m\in M$ .

A left-left Yetter-Drinfeld module M is both a left H-module and a left H- comodule satisfying the compatibility condition

${h}_{1}{m}_{\left(-1\right)}\otimes {h}_{2}\cdot {m}_{0}={\left({h}_{1}\cdot m\right)}_{\left(-1\right)}{h}_{2}\otimes {\left({h}_{1}\cdot m\right)}_{0}$ (2.1)

for all $h\in H$ and $m\in M$ . The equation (1.1) is equivalent to

$\rho \left(h\cdot m\right)={h}_{1}{m}_{\left(-1\right)}S\left({h}_{3}\right)\otimes \left({h}_{2}\cdot {m}_{0}\right).$ (2.2)

By   , the left-left Yetter-Drinfeld category ${}_{H}^{H}\mathcal{Y}\mathcal{D}$ is a braided monoi- dal category whose objects are Yetter-Drinfeld modules, morphisms are both left H-linear and H-colinear maps, and its braiding ${C}_{-,-}$ is given by

${C}_{M,N}\left(m\otimes n\right)={m}_{\left(-1\right)}\cdot n\otimes {m}_{\left(0\right)},$

for all $m\in M\in {}_{H}{}^{H}\mathcal{Y}\mathcal{D}$ and $n\in N\in {}_{H}{}^{H}\mathcal{Y}\mathcal{D}$ .

Let $A$ be an object in ${}_{H}{}^{H}\mathcal{Y}\mathcal{D}$ , the braiding $\tau$ is called symmetric on $A$ if the following condition holds:

$\left({\left({a}_{\left(-1\right)}\cdot b\right)}_{\left(-1\right)}\cdot {a}_{0}\right)\otimes {\left({a}_{\left(-1\right)}\cdot b\right)}_{0}=a\otimes b,$ (2.3)

which is equivalent to the following condition:

${a}_{\left(-1\right)}\cdot b\otimes {a}_{0}={b}_{0}\otimes {S}^{-1}\left({b}_{\left(-1\right)}\right)\cdot a,$ (2.4)

for any $a,b\in A.$

In the category ${}_{H}{}^{H}\mathcal{Y}\mathcal{D}$ , we call an (co)algebra simply if it is both a left H- module (co)algebra and a left H-comodule (co)algebra. For more details about (co)module-(co)algebras, the reader can refer to   .

A braided Lie algebra (  ) in ${}_{H}{}^{H}\mathcal{Y}\mathcal{D}$ , called generalized H-Lie algebra there, is an object L in ${}_{H}{}^{H}\mathcal{Y}\mathcal{D}$ together with a bracket operation $\left[,\right]:L\otimes L\to L$ , which is a morphism in ${}_{H}{}^{H}\mathcal{Y}\mathcal{D}$ satisfying

(1) H-anti-commutativity: $\left[l,{l}^{\prime }\right]=-\left[{l}_{\left(-1\right)}\cdot {l}^{\prime },{l}_{0}\right],l,{l}^{\prime }\in L,$

(2) H-Jacobi identity:

$\left\{l\otimes {l}^{\prime }\otimes {l}^{″}\right\}+\left\{\left(\tau \otimes 1\right)\left(1\otimes \tau \right)\left(l\otimes {l}^{\prime }\otimes {l}^{″}\right)\right\}+\left\{\left(1\otimes \tau \right)\left(\tau \otimes 1\right)\left(l\otimes {l}^{\prime }\otimes {l}^{″}\right)\right\}=0,$

for all $l,{l}^{\prime },{l}^{″}\in L$ , where $\left\{l\otimes {l}^{\prime }\otimes {l}^{″}\right\}$ denotes $\left[l,\left[{l}^{\prime },{l}^{″}\right]\right]$ and $\tau$ the braiding for L.

Let A be an associative algebra in ${}_{H}{}^{H}\mathcal{Y}\mathcal{D}$ . Assume that the braiding is symmetric on A. Define

$\left[a,b\right]=ab-\left({a}_{\left(-1\right)}\cdot b\right){a}_{0},a,b\in A.$

Then $\left(A,\left[,\right]\right)$ is a braided Lie algebra (see  ).

A braided Lie coalgebra (  ) $\Gamma$ is an object in ${}_{H}{}^{H}\mathcal{Y}\mathcal{D}$ together with a linear map $\delta :\Gamma \to \Gamma \otimes \Gamma$ (called the cobracket), which is also a morphism in ${}_{H}{}^{H}\mathcal{Y}\mathcal{D}$ subject to the following conditions:

(1) H-anti-cocommutativity: $\delta =-\tau \delta ,$

(2) H-coJacobi identity:

$\left(id+\left(id\otimes \tau \right)\left(\tau \otimes id\right)+\left(\tau \otimes id\right)\left(id\otimes \tau \right)\right)\left(id\otimes \delta \right)\delta =0,$

where $\tau$ denotes the braiding for L.

Dually, let $\left(C,\Delta \right)$ be a coassociative coalgebra in ${}_{H}{}^{H}\mathcal{Y}\mathcal{D}$ . Assume that the braiding on C is symmetric. Define $\delta :C\to C\otimes C,$ by

$c↦{c}_{1}\otimes {c}_{2}-{c}_{1\left(-1\right)}\cdot {c}_{2}\otimes {c}_{10},c\in C.$

Then $\left(C,\delta \right)$ is a braided Lie coalgebras in ${}_{H}{}^{H}\mathcal{Y}\mathcal{D}$ (see  ).

A braided Lie bialgebra (  ) is $\left(L,\left[,\right],\delta \right)$ in ${}_{H}{}^{H}\mathcal{Y}\mathcal{D}$ , where $\left(L,\left[,\right]\right)$ is a braided Lie algebra, and $\left(L,\delta \right)$ is a braided Lie coalgebra, such that the compatibility condition holds:

$\delta \left[x,y\right]=\left(\left(\left[,\right]\otimes id\right)\left(id\otimes \delta \right)+\left(id\otimes \left[,\right]\right)\left(\tau \otimes id\right)\left(id\otimes \delta \right)\right)\left(id\otimes id-\tau \right)\left(x\otimes y\right),x,y\in L,$

where $\tau$ denotes the braiding for L.

3. Main Results

In this section, we will study the relation between braided infinitesimal bialge- bras and braided Lie bialgebras as a generalization of Aguiar’s result in  .

Let $\left(A,m,\Delta \right)$ be a braided e-bialgebra in ${}_{H}{}^{H}\mathcal{Y}\mathcal{D}$ . For any $x,y,z\in A$ , define an action of A on $A\otimes A$ by

$x\to \left(y\otimes z\right)=xy\otimes z-{x}_{\left(-1\right)}\cdot y\otimes \left({x}_{0\left(-1\right)}\cdot z\right){x}_{00}.$

Then the action ® is a morphism in ${}_{H}{}^{H}\mathcal{Y}\mathcal{D}$ . In fact, for any $x,y,z\in A$ and $h\in H$ , we have

$\begin{array}{l}{h}_{1}\cdot x\to {h}_{2}\cdot \left(y\otimes z\right)={h}_{1}\cdot x\to \left({h}_{2}\cdot y\otimes {h}_{3}\cdot z\right)\\ =\left({h}_{1}\cdot x\right)\left({h}_{2}\cdot y\right)\otimes \left({h}_{3}\cdot z\right)-{\left({h}_{1}\cdot x\right)}_{\left(-1\right)}\cdot {h}_{2}\cdot y\otimes \left({\left({h}_{1}\cdot x\right)}_{0\left(-1\right)}\cdot {h}_{3}\cdot z\right){\left({h}_{1}\cdot x\right)}_{00}\\ =\left({h}_{1}\cdot x\right)\left({h}_{2}\cdot y\right)\otimes \left({h}_{3}\cdot z\right)-{h}_{11}{x}_{\left(-1\right)}S\left({h}_{13}\right)\cdot {h}_{2}\cdot y\otimes \left({\left({h}_{12}\cdot {x}_{0}\right)}_{\left(-1\right)}\cdot {h}_{3}\cdot z\right){\left({h}_{12}\cdot {x}_{0}\right)}_{0}\\ =\left({h}_{1}\cdot x\right)\left({h}_{2}\cdot y\right)\otimes \left({h}_{3}\cdot z\right)-{h}_{1}{x}_{\left(-1\right)}\cdot y\otimes \left({\left({h}_{2}\cdot {x}_{0}\right)}_{\left(-1\right)}\cdot {h}_{3}\cdot z\right){\left({h}_{2}\cdot {x}_{0}\right)}_{0}\\ =\left({h}_{1}\cdot x\right)\left({h}_{2}\cdot y\right)\otimes \left({h}_{3}\cdot z\right)-{h}_{1}{x}_{\left(-1\right)}\cdot y\otimes \left({h}_{21}{x}_{0\left(-1\right)}S\left({h}_{23}\right)\cdot {h}_{3}\cdot z\right)\left({h}_{22}\cdot {x}_{00}\right)\\ =\left({h}_{1}\cdot x\right)\left({h}_{2}\cdot y\right)\otimes \left({h}_{3}\cdot z\right)-{h}_{1}{x}_{\left(-1\right)}\cdot y\otimes \left({h}_{2}{x}_{0\left(-1\right)}\cdot z\right)\left({h}_{3}\cdot {x}_{00}\right)\\ ={h}_{1}\cdot \left(xy\right)\otimes \left({h}_{2}\cdot z\right)-{h}_{1}{x}_{\left(-1\right)}\cdot y\otimes {h}_{2}\cdot \left(\left({x}_{0\left(-1\right)}\cdot z\right){x}_{00}\right)\\ =h\cdot \left(xy\otimes z-{x}_{\left(-1\right)}\cdot y\otimes \left({x}_{0\left(-1\right)}\cdot z\right){x}_{00}\right).\end{array}$

So ® is left H-linear. To show the left H-colinearity of the action ®, we compute

$\begin{array}{l}\rho \left(x\to \left(y\otimes z\right)\right)=\rho \left(xy\otimes z-{x}_{\left(-1\right)}\cdot y\otimes \left({x}_{0\left(-1\right)}\cdot z\right){x}_{00}\right)\\ ={x}_{\left(-1\right)}{y}_{\left(-1\right)}{z}_{\left(-1\right)}\otimes {x}_{0}{y}_{0}\otimes {z}_{0}-{\left({x}_{\left(-1\right)}\cdot y\right)}_{\left(-1\right)}{\left({x}_{0\left(-1\right)}\cdot z\right)}_{\left(-1\right)}{x}_{00\left(-1\right)}\otimes {\left({x}_{\left(-1\right)}\cdot y\right)}_{0}\otimes {\left({x}_{0\left(-1\right)}\cdot z\right)}_{0}{x}_{000}\\ ={x}_{\left(-1\right)}{y}_{\left(-1\right)}{z}_{\left(-1\right)}\otimes {x}_{0}{y}_{0}\otimes {z}_{0}-{x}_{\left(-11\right)}{y}_{\left(-1\right)}S\left({x}_{\left(-13\right)}\right){x}_{\left(-1\right)4}{z}_{\left(-1\right)}S\left({x}_{\left(-1\right)6}\right){x}_{\left(-1\right)7}\otimes {x}_{\left(-1\right)2}\cdot {y}_{0}\otimes \left({x}_{\left(-1\right)5}\cdot {z}_{0}\right){x}_{0}\\ ={x}_{\left(-1\right)}{y}_{\left(-1\right)}{z}_{\left(-1\right)}\otimes {x}_{0}{y}_{0}\otimes {z}_{0}-{x}_{\left(-11\right)}{y}_{\left(-1\right)}{z}_{\left(-1\right)}\otimes {x}_{\left(-1\right)2}\cdot {y}_{0}\otimes \left({x}_{\left(-1\right)3}\cdot {z}_{0}\right){x}_{0},\end{array}$

and

$\begin{array}{l}\left(id\otimes \to \right)\rho \left(x\otimes y\otimes z\right)=\left(id\otimes \to \right)\left({x}_{\left(-1\right)}{y}_{\left(-1\right)}{z}_{\left(-1\right)}\otimes {x}_{0}\otimes {y}_{0}\otimes {z}_{0}\right)\\ ={x}_{\left(-1\right)}{y}_{\left(-1\right)}{z}_{\left(-1\right)}\otimes {x}_{0}{y}_{0}\otimes {z}_{0}-{x}_{\left(-1\right)}{y}_{\left(-1\right)}{z}_{\left(-1\right)}\otimes {x}_{0\left(-1\right)}\cdot {y}_{0}\otimes \left({x}_{00\left(-1\right)}\cdot {z}_{0}\right){x}_{000}\\ ={x}_{\left(-1\right)}{y}_{\left(-1\right)}{z}_{\left(-1\right)}\otimes {x}_{0}{y}_{0}\otimes {z}_{0}-{x}_{\left(-1\right)1}{y}_{\left(-1\right)}{z}_{\left(-1\right)}\otimes {x}_{\left(-1\right)2}\cdot {y}_{0}\otimes \left({x}_{\left(-1\right)3}\cdot {z}_{0}\right){x}_{0},\end{array}$

as desired.

Definition 2.1. Let $\left(A,m,\Delta \right)$ be a braided infinitesimal bialgebra and $\tau$ the braiding of A. The map $B:A\otimes A\to A\otimes A$ defined by

$B\left(x,y\right)=x\to \tau \Delta \left(y\right)+\tau \left(y\to \tau \Delta \left(x\right)\right),x,y\in A,$ (3.1)

is called the balanceator of A. The balanceator B is called symmetric if $B=B\circ \tau$ . The braided infinitesimal bialgebra A is called balanced if $B\equiv 0$ on A.

Condition (2.1) can be written as follows:

$\begin{array}{l}B\left(x,y\right)=x\left({y}_{1\left(-1\right)}\cdot {y}_{2}\right)\otimes {y}_{10}-{x}_{\left(-1\right)}{y}_{1\left(-1\right)}\cdot {y}_{2}\otimes \left({x}_{0\left(-1\right)}\cdot {y}_{10}\right){x}_{00}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\left({x}_{\left(-1\right)}\cdot y\right)}_{\left(-1\right)}\cdot {x}_{01}\otimes {\left({x}_{\left(-1\right)}\cdot y\right)}_{0}{x}_{02}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\left({\left({x}_{\left(-1\right)}\cdot y\right)}_{\left(-1\right)}\cdot {x}_{01}\right){\left({x}_{\left(-1\right)}\cdot y\right)}_{0}\otimes {x}_{02}\end{array}$

Obviously,

$\begin{array}{c}B\left({x}_{\left(-1\right)}\cdot y,{x}_{0}\right)=\left({x}_{\left(-1\right)}\cdot y\right)\left({x}_{0\left(-1\right)}\cdot {x}_{02}\right)\otimes {x}_{010}-\left({x}_{\left(-1\right)}\cdot {y}_{1}\right){x}_{0}\otimes {y}_{2}+{x}_{\left(-1\right)}\cdot {y}_{1}\otimes {x}_{0}{y}_{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-{\left({x}_{\left(-1\right)}\cdot y\right)}_{\left(-1\right)}{x}_{01\left(-1\right)}\cdot {x}_{02}\otimes \left({\left({x}_{\left(-1\right)}\cdot y\right)}_{0\left(-1\right)}\cdot {x}_{010}\right){\left({x}_{\left(-1\right)}\cdot y\right)}_{00}.\end{array}$

Lemma 2.2. Let $\left(A,m,\Delta \right)$ be a braided infinitesimal bialgebra and $x,y\in A$ . Assume that the braiding $\tau$ on A is symmetric. Then the following equations hold:

(1) $\left({\left({x}_{\left(-1\right)}\cdot y\right)}_{\left(-1\right)}\cdot {x}_{01}\right)\otimes \left({\left({x}_{\left(-1\right)}\cdot y\right)}_{0\left(-1\right)}\cdot {x}_{02}\right){\left({x}_{\left(-1\right)}\cdot y\right)}_{00}={x}_{1}\otimes {x}_{2}y,$

(2) $\left({\left({x}_{\left(-1\right)}\cdot y\right)}_{\left(-1\right)}{x}_{01\left(-1\right)}\cdot {x}_{02}\right){\left({x}_{\left(-1\right)}\cdot y\right)}_{0}\otimes {x}_{010}={x}_{1\left(-1\right)}\cdot \left({x}_{2}y\right)\otimes {x}_{10},$

(3) ${\left({x}_{\left(-1\right)1}\cdot {y}_{1}\right)}_{\left(-1\right)}\cdot \left(\left({x}_{\left(-1\right)2}\cdot {y}_{2}\right){x}_{0}\right)\otimes {\left({x}_{\left(-1\right)1}\cdot {y}_{1}\right)}_{0}=\left({x}_{\left(-1\right)}{y}_{1\left(-1\right)}\cdot {y}_{2}\right){x}_{0}\otimes {y}_{10}.$

Proof. (1) Since the braiding $\tau$ on A is symmetric, for all $x,y\in A$ , we have ${\left({x}_{\left(-1\right)}\cdot y\right)}_{\left(-1\right)}\cdot {x}_{0}\otimes {\left({x}_{\left(-1\right)}\cdot y\right)}_{0}=x\otimes y$ , then

$\left(id\otimes m\right)\left(\Delta \otimes id\right)\left({\left({x}_{\left(-1\right)}\cdot y\right)}_{\left(-1\right)}\cdot {x}_{0}\otimes {\left({x}_{\left(-1\right)}\cdot y\right)}_{0}\right)=\left(id\otimes m\right)\left(\Delta \otimes id\right)\left(x\otimes y\right)$

that is,

${\left({x}_{\left(-1\right)}\cdot y\right)}_{\left(-1\right)}\cdot {x}_{01}\otimes \left({\left({x}_{\left(-1\right)}\cdot y\right)}_{0\left(-1\right)}\cdot {x}_{02}\right){\left({x}_{\left(-1\right)}\cdot y\right)}_{00}={x}_{1}\otimes {x}_{2}y.$

So (1) holds.

(2) To show the Equation (2.2), we need the following computation:

$\begin{array}{l}\left({\left({x}_{\left(-1\right)}\cdot y\right)}_{\left(-1\right)}{x}_{01\left(-1\right)}\cdot {x}_{02}\right){\left({x}_{\left(-1\right)}\cdot y\right)}_{0}\otimes {x}_{010}\\ =\left({\left({x}_{1\left(-1\right)}{x}_{2\left(-1\right)}\cdot y\right)}_{\left(-1\right)}{x}_{10\left(-1\right)}\cdot {x}_{20}\right){\left({x}_{1\left(-1\right)}{x}_{2\left(-1\right)}\cdot y\right)}_{0}\otimes {x}_{100}\\ =\left({x}_{1\left(-1\right)1}{x}_{2\left(-1\right)1}{y}_{\left(-1\right)}S\left({x}_{2\left(-1\right)3}\right)S\left({x}_{12\left(-1\right)3}\right){x}_{10\left(-1\right)}\cdot {x}_{20}\right)\left({x}_{1\left(-1\right)2}{x}_{2\left(-1\right)2}\cdot {y}_{0}\right)\otimes {x}_{100}\\ =\left({x}_{1\left(-1\right)1}{x}_{2\left(-1\right)1}{y}_{\left(-1\right)}S\left({x}_{2\left(-1\right)3}\right)\cdot {x}_{20}\right)\left({x}_{1\left(-1\right)2}{x}_{2\left(-1\right)2}\cdot {y}_{0}\right)\otimes {x}_{10}\\ =\left({x}_{1\left(-1\right)1}{\left({x}_{2\left(-1\right)}\cdot y\right)}_{\left(-1\right)}\cdot {x}_{20}\right)\left({x}_{1\left(-1\right)2}{\left({x}_{2\left(-1\right)}\cdot y\right)}_{0}\right)\otimes {x}_{10}\\ ={x}_{1\left(-1\right)}\cdot \left(\left({\left({x}_{2\left(-1\right)}\cdot y\right)}_{\left(-1\right)}\cdot {x}_{20}\right){\left({x}_{2\left(-1\right)}\cdot y\right)}_{0}\right)\otimes {x}_{10}={x}_{1\left(-1\right)}\cdot \left({x}_{2}y\right)\otimes {x}_{10}.\end{array}$

The last equality holds since $\tau$ is symmetric on A. Hence (2) holds.

(3) Finally, we check the Equation (2.3) as follows:

$\begin{array}{l}{\left({x}_{\left(-1\right)1}\cdot {y}_{1}\right)}_{\left(-1\right)}\left(\left({x}_{\left(-1\right)2}\cdot {y}_{2}\right){x}_{0}\right)\otimes {\left({x}_{\left(-1\right)1}\cdot {y}_{1}\right)}_{0}\\ =\left({x}_{\left(-1\right)11}{y}_{1\left(-1\right)}S\left({x}_{\left(-1\right)13}\right)\right)\cdot \left(\left({x}_{\left(-1\right)2}\cdot {y}_{2}\right){x}_{0}\right)\otimes {x}_{\left(-1\right)12}\cdot {y}_{10}\\ =\left({x}_{\left(-1\right)1}{y}_{1\left(-1\right)}S\left({x}_{\left(-1\right)3}\right)\right)\cdot \left(\left({x}_{\left(-1\right)4}\cdot {y}_{2}\right){x}_{0}\right)\otimes {x}_{\left(-1\right)2}\cdot {y}_{10}\\ =\left({x}_{\left(-1\right)11}{y}_{1\left(-1\right)1}S\left({x}_{\left(-1\right)32}\right){x}_{\left(-1\right)4}\cdot {y}_{2}\right)\left({x}_{\left(-1\right)12}{y}_{1\left(-1\right)2}S\left({x}_{\left(-1\right)31}\right)\cdot {x}_{0}\right)\otimes {x}_{\left(-1\right)2}\cdot {y}_{10}\\ =\left({x}_{\left(-1\right)11}{y}_{1\left(-1\right)1}\cdot {y}_{2}\right)\left({x}_{\left(-1\right)12}{y}_{1\left(-1\right)2}S\left({x}_{\left(-1\right)3}\right)\cdot {x}_{0}\right)\otimes {x}_{\left(-1\right)2}\cdot {y}_{10}\\ =\left({x}_{\left(-1\right)1}{y}_{1\left(-1\right)}\cdot {y}_{2}\right)\left({\left({x}_{\left(-1\right)2}\cdot {y}_{10}\right)}_{\left(-1\right)}\cdot {x}_{0}\right)\otimes {\left({x}_{\left(-1\right)2}\cdot {y}_{10}\right)}_{0}\\ =\left({x}_{\left(-1\right)}{y}_{1\left(-1\right)}\cdot {y}_{2}\right)\left({\left({x}_{0\left(-1\right)}\cdot {y}_{10}\right)}_{\left(-1\right)}\cdot {x}_{00}\right)\otimes {\left({x}_{0\left(-1\right)}\cdot {y}_{10}\right)}_{0}=\left({x}_{\left(-1\right)}{y}_{1\left(-1\right)}\cdot {y}_{2}\right){x}_{0}\otimes {y}_{10}.\end{array}$

The last equality holds since $\tau$ is symmetric on A. Hence (3) holds as required. ,

Therem 2.3. Let $\left(A,m,\Delta \right)$ be a braided infinitesimal bialgebra. Assume that the braiding $\tau$ on A is symmetric. Then $\left(A,\left[,\right]=m-m\tau ,\delta =\Delta -\tau \Delta \right)$ is a braided Lie bialgebra if and only if $B=B\circ \tau$ .

Proof. Since $\left(A,m\right)$ is an associative algebra and $\left(A,\Delta \right)$ is a coassociative coalgebra in ${}_{H}{}^{H}\mathcal{Y}\mathcal{D}$ , $\left(A,\left[,\right]=m-m\tau \right)$ is a braided Lie algebra and $\left(A,\delta =\Delta -\tau \Delta \right)$ is a braided Lie coalgebra. Therefore it remains to check the compatible condition:

$\delta \left[x,y\right]=\left(\left(\left[,\right]\otimes id\right)\left(id\otimes \delta \right)+\left(id\otimes \left[,\right]\right)\left(\tau \otimes id\right)\left(id\otimes \delta \right)\right)\left(id\otimes id-\tau \right)\left(x\otimes y\right),$

for all $x,y\in A$ . In fact, on the one hand, we have

$\begin{array}{l}\delta \left[x,y\right]=\delta \left(xy-\left({x}_{\left(-1\right)}\cdot y\right){x}_{0}\right)\\ =\left(1-\tau \right)\Delta \left(xy\right)-\left(1-\tau \right)\Delta \left(\left({x}_{\left(-1\right)}\cdot y\right){x}_{0}\right)\\ =\left(1-\tau \right)\left({x}_{1}\otimes {x}_{2}y+x{y}_{1}\otimes {y}_{2}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\left(1-\tau \right)\left(\left({x}_{\left(-1\right)1}\cdot {y}_{1}\right)\otimes \left({x}_{\left(-1\right)2}\cdot {y}_{2}\right){x}_{0}+\left({x}_{\left(-1\right)}\cdot y\right){x}_{01}\otimes {x}_{02}\right)\\ ={x}_{1}\otimes {x}_{2}y+x{y}_{1}\otimes {y}_{2}-{x}_{1\left(-1\right)}\cdot \left({x}_{2}y\right)\otimes {x}_{10}-{\left(x{y}_{1}\right)}_{\left(-1\right)}\cdot {y}_{2}\otimes {\left(x{y}_{1}\right)}_{0}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\left({x}_{\left(-1\right)}\cdot y\right){x}_{01}\otimes {x}_{02}-\left({x}_{\left(-1\right)1}\cdot {y}_{1}\right)\otimes \left({x}_{\left(-1\right)2}\cdot {y}_{2}\right){x}_{0}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\left({x}_{\left(-1\right)1}\cdot {y}_{1}\right)}_{\left(-1\right)}\cdot \left(\left({x}_{\left(-1\right)2}\cdot {y}_{2}\right){x}_{0}\right)\otimes {\left({x}_{\left(-1\right)1}\cdot {y}_{1}\right)}_{0}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\left(\left({x}_{\left(-1\right)}\cdot y\right){x}_{01}\right)}_{\left(-1\right)}\cdot {x}_{02}\otimes {\left(\left({x}_{\left(-1\right)}\cdot y\right){x}_{01}\right)}_{0}.\end{array}$

On the other hand, we have

$\begin{array}{l}\left(\left(\left[,\right]\otimes id\right)\left(id\otimes \delta \right)+\left(id\otimes \left[,\right]\right)\left(\tau \otimes id\right)\left(id\otimes \delta \right)\right)\left(id\otimes id-\tau \right)\left(x\otimes y\right)\\ =\left(\left(\left[,\right]\otimes id\right)\left(id\otimes \delta \right)+\left(id\otimes \left[,\right]\right)\left(\tau \otimes id\right)\left(id\otimes \delta \right)\right)\left(xy-\left({x}_{\left(-1\right)}\cdot y\right){x}_{0}\right)\\ =x{y}_{1}\otimes {y}_{2}-\left({x}_{\left(-1\right)}\cdot {y}_{1}\right){x}_{0}\otimes {y}_{2}-x\left({y}_{1\left(-1\right)}\cdot {y}_{2}\right)\otimes {y}_{10}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left({x}_{\left(-1\right)}{y}_{1\left(-1\right)}\cdot {y}_{2}\right){x}_{0}\otimes {y}_{10}-\left({x}_{\left(-1\right)}\cdot y\right){x}_{01}\otimes {x}_{02}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left({\left({x}_{\left(-1\right)}\cdot y\right)}_{\left(-1\right)}\cdot {x}_{01}\right){\left({x}_{\left(-1\right)}\cdot y\right)}_{0}\otimes {x}_{02}+\left({x}_{\left(-1\right)}\cdot y\right)\left({x}_{01\left(-1\right)}\cdot {x}_{02}\right)\otimes {x}_{010}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+{x}_{\left(-1\right)}\cdot {y}_{1}\otimes {x}_{0}{y}_{2}-\left({\left({x}_{\left(-1\right)}\cdot y\right)}_{\left(-1\right)}{x}_{01\left(-1\right)}\cdot {x}_{02}\right){\left({x}_{\left(-1\right)}\cdot y\right)}_{0}\otimes {x}_{010}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-{x}_{\left(-1\right)}\cdot {y}_{1}\otimes \left({x}_{0\left(-1\right)}\cdot {y}_{2}\right){x}_{00}-{x}_{\left(-1\right)}{y}_{1\left(-1\right)}\cdot {y}_{2}\otimes {x}_{0}{y}_{10}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+{x}_{\left(-1\right)}{y}_{1\left(-1\right)}\cdot {y}_{2}\otimes \left({x}_{0\left(-1\right)}\cdot {y}_{10}\right){x}_{00}-{\left({x}_{\left(-1\right)}\cdot y\right)}_{\left(-1\right)}\cdot {x}_{01}\otimes {\left({x}_{\left(-1\right)}\cdot y\right)}_{0}{x}_{02}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\left({x}_{\left(-1\right)}\cdot y\right)}_{\left(-1\right)}\cdot {x}_{01}\otimes \left({\left({x}_{\left(-1\right)}\cdot y\right)}_{0\left(-1\right)}\cdot {x}_{02}\right){\left({x}_{\left(-1\right)}\cdot y\right)}_{00}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\left({x}_{\left(-1\right)}\cdot y\right)}_{\left(-1\right)}{x}_{01\left(-1\right)}\cdot {x}_{02}\otimes {\left({x}_{\left(-1\right)}\cdot y\right)}_{0}{x}_{010}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-{\left({x}_{\left(-1\right)}\cdot y\right)}_{\left(-1\right)}{x}_{01\left(-1\right)}\cdot {x}_{02}\otimes \left({\left({x}_{\left(-1\right)}\cdot y\right)}_{0\left(-1\right)}\cdot {x}_{010}\right){\left({x}_{\left(-1\right)}\cdot y\right)}_{00}.\end{array}$

According to Lemma 2.2, we have

$\begin{array}{l}x{y}_{1}\otimes {y}_{2}+\left({x}_{\left(-1\right)}{y}_{1\left(-1\right)}\cdot {y}_{2}\right){x}_{0}\otimes {y}_{10}-\left({x}_{\left(-1\right)}\cdot y\right){x}_{01}\otimes {x}_{02}\\ -\left({\left({x}_{\left(-1\right)}\cdot y\right)}_{\left(-1\right)}{x}_{01\left(-1\right)}\cdot {x}_{02}\right){\left({x}_{\left(-1\right)}\cdot y\right)}_{0}\otimes {x}_{010}\\ -\text{\hspace{0.17em}}{x}_{\left(-1\right)}\cdot {y}_{1}\otimes \left({x}_{0\left(-1\right)}\cdot {y}_{2}\right){x}_{00}-{x}_{\left(-1\right)}{y}_{1\left(-1\right)}\cdot {y}_{2}\otimes {x}_{0}{y}_{10}\\ +{\left({x}_{\left(-1\right)}\cdot y\right)}_{\left(-1\right)}\cdot {x}_{01}\otimes \left({\left({x}_{\left(-1\right)}\cdot y\right)}_{0\left(-1\right)}\cdot {x}_{02}\right){\left({x}_{\left(-1\right)}\cdot y\right)}_{00}\\ +{\left({x}_{\left(-1\right)}\cdot y\right)}_{\left(-1\right)}{x}_{01\left(-1\right)}\cdot {x}_{02}\otimes {\left({x}_{\left(-1\right)}\cdot y\right)}_{0}{x}_{010}\\ =x{y}_{1}\otimes {y}_{2}+{\left({x}_{\left(-1\right)1}\cdot {y}_{1}\right)}_{\left(-1\right)}\cdot \left(\left({x}_{\left(-1\right)2}\cdot {y}_{2}\right){x}_{0}\right)\otimes {\left({x}_{\left(-1\right)1}\cdot {y}_{1}\right)}_{0}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\left({x}_{\left(-1\right)}\cdot y\right){x}_{01}\otimes {x}_{02}-{x}_{1\left(-1\right)}\cdot \left({x}_{2}y\right)\otimes {x}_{10}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-{x}_{\left(-1\right)1}\cdot {y}_{1}\otimes \left({x}_{\left(-1\right)2}\cdot {y}_{2}\right){x}_{0}-{\left(x{y}_{1}\right)}_{\left(-1\right)}\cdot {y}_{2}\otimes {\left(x{y}_{1}\right)}_{0}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+{x}_{1}\otimes {x}_{2}y+{\left(\left({x}_{\left(-1\right)}\cdot y\right){x}_{01}\right)}_{\left(-1\right)}\cdot {x}_{02}\otimes {\left(\left({x}_{\left(-1\right)}\cdot y\right){x}_{01}\right)}_{0}\\ =\delta \left[x,y\right].\end{array}$

Therefore,

$\begin{array}{l}\left(\left(\left[,\right]\otimes id\right)\left(id\otimes \delta \right)+\left(id\otimes \left[,\right]\right)\left(\tau \otimes id\right)\left(id\otimes \delta \right)\right)\left(id\otimes id-\tau \right)\left(x\otimes y\right)\\ =\delta \left[x,y\right]-x\left({y}_{1\left(-1\right)}\cdot {y}_{2}\right)\otimes {y}_{10}+\left({\left({x}_{\left(-1\right)}\cdot y\right)}_{\left(-1\right)}\cdot {x}_{01}\right){\left({x}_{\left(-1\right)}\cdot y\right)}_{0}\otimes {x}_{02}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+{x}_{\left(-1\right)}{y}_{1\left(-1\right)}\cdot {y}_{2}\otimes \left({x}_{0\left(-1\right)}\cdot {y}_{10}\right){x}_{00}-{\left({x}_{\left(-1\right)}\cdot y\right)}_{\left(-1\right)}\cdot {x}_{01}\otimes {\left({x}_{\left(-1\right)}\cdot y\right)}_{0}{x}_{02}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left({x}_{\left(-1\right)}\cdot y\right)\left({x}_{01\left(-1\right)}\cdot {x}_{02}\right)\otimes {x}_{010}-\left({x}_{\left(-1\right)}\cdot {y}_{1}\right){x}_{0}\otimes {y}_{2}+{x}_{\left(-1\right)}\cdot {y}_{1}\otimes {x}_{0}{y}_{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-{\left({x}_{\left(-1\right)}\cdot y\right)}_{\left(-1\right)}{x}_{01\left(-1\right)}\cdot {x}_{02}\otimes \left({\left({x}_{\left(-1\right)}\cdot y\right)}_{0\left(-1\right)}\cdot {x}_{010}\right){\left({x}_{\left(-1\right)}\cdot y\right)}_{00}\\ =\delta \left[x,y\right]-B\left(x,y\right)+B\left({x}_{\left(-1\right)}\cdot y,{x}_{0}\right)\\ =\delta \left[x,y\right]-B\left(x,y\right)+B\circ \tau \left(x,y\right),\end{array}$

as desired. We complete the proof. ,

Corollary 2.4. Let $\left(A,m,\Delta \right)$ be a braided infinitesimal bialgebra. Assume that the braiding $\tau$ on A is symmetric and the balanceator $B=0$ . Then $\left(A,\left[,\right]=m-m\tau ,\delta =\Delta -\tau \Delta \right)$ is a braided Lie bialgebra.

Proof. Straightforward from Theorem 2.3. ,

Example 2.5. Let q be an 2th root of unit of k and G the cyclic group of order 2 generated by g, $H=kG$ be the group algebra in the usual way. We consider the algebra ${A}_{4}=k\left[x\right]/\left({x}^{4}\right)$ . By  , ${A}_{4}$ is a infinitesimal bialgebra equipped with the comultiplication:

$\Delta \left(1\right)=0,\Delta \left(x\right)=x\otimes {x}^{2}-1\otimes {x}^{3},\Delta \left({x}^{2}\right)={x}^{2}\otimes {x}^{2},\Delta \left({x}^{3}\right)={x}^{3}\otimes {x}^{2}.$

Define the left-H-module action and the left-H-comodule coaction of A by

${g}^{i}\cdot {x}^{j}={q}^{ij}{x}^{j},\rho \left({x}^{j}\right)={g}^{j}\otimes {x}^{j},\text{\hspace{0.17em}}i=0,1,\text{\hspace{0.17em}}j=0,1,2,3.$

It is not hard to check that the multiplication and the comultiplicaition are both H-linear and H-colinear, therefore ${A}_{4}$ is a braided infinitesimal bialgebra. Since $B\left(x,x\right)=2{x}^{2}\otimes {x}^{2}-qx\otimes {x}^{2}-q{x}^{2}\otimes x-q{x}^{3}\otimes x-x\otimes {x}^{3}$ and $\tau \left(x\otimes x\right)=\left({x}_{\left(-1\right)}\cdot x\right){x}_{0}=\left(g\cdot x\right)x=qx\otimes x,$ it is clear that $B\left(x,x\right)=B\tau \left(x,x\right)$ if and only if $q=1$ . If $q=1$ , it is not hard to check that the balanceator is symmetric on ${A}_{4}$ . By Theorem 2.3, $\left({A}_{4},\left[,\right]=m-m\tau ,\delta =\Delta -\tau \Delta \right)$ is a braided Lie bialgebra.

Example 2.6. Let q be a 4th root of unit of k. Consider the Hopf algebra $H=kG$ , where G is a cyclic group of order 4 generated by g. Recall from  that $A={M}_{2}\left(k\right)$ is a braided infinitesimal bialgebra in ${}_{H}{}^{H}\mathcal{Y}\mathcal{D}$ equipped with the comultiplication:

$\Delta \left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=\left(\begin{array}{cc}0& a\\ 0& c\end{array}\right)\otimes \left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)-\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\otimes \left(\begin{array}{cc}c& d\\ 0& 0\end{array}\right)$

and the H-module action, the H-comodule coaction:

${g}^{k}\cdot {E}_{ij}={q}^{2k\left(i+j\right)}{E}_{ij},\rho \left({E}_{ij}\right)={g}^{2\left(i+j\right)}\otimes {E}_{ij},\text{\hspace{0.17em}}k=0,1,2,3,\text{\hspace{0.17em}}i,j=1,2.$

Since

$B\left({E}_{11},{E}_{21}\right)=2\left({E}_{12}\otimes {E}_{22}-{E}_{11}\otimes {E}_{12}\right),$

$B\left({E}_{{11}_{\left(-1\right)}}\cdot {E}_{21},{E}_{{11}_{0}}\right)=B\left({E}_{21},{E}_{11}\right)=2\left({E}_{22}\otimes {E}_{12}-{E}_{11}\otimes {E}_{11}\right),$

we claim that the balanceator is not symmetric. By Theorem 2.3, $\left({M}_{2}\left(k\right),\left[,\right]=m-m\tau ,\delta =\Delta -\tau \Delta \right)$ is not a braided Lie bialgebra, where m is the multiplication of A.

Let ${A}_{1}=\left\{\left(\begin{array}{cc}a& b\\ 0& a\end{array}\right)|a,b\in k\right\}\subset {M}_{2}\left(k\right)$ . It is clear that ${A}_{1}$ is both H-stable and

H-costable, hence ${A}_{1}$ is also a braided infinitesimal bialgebra contained in A. One can check easily that the balanceator $B=0$ on ${A}_{1}$ . By Corollary 2.4, $\left({A}_{1},\left[,\right]=m-m\tau ,\delta =\Delta -\tau \Delta \right)$ is a braided Lie bialgebra.

Acknowledgements

The paper is partially supported by the China Postdoctoral Science Foundation (No. 2015M571725), the Key University Science Research Project of Anhui Province (Nos. KJ2015A294 and KJ2016A545), the outstanding top-notch talent cultivation project of Anhui Province (No. gxfx2017123) and the NSF of Chuzhou University (No. 2015qd01).

Cite this paper

Wang, S.X. (2017) From Braided Infinitesimal Bialgebras to Braided Lie Bialgebras. Advances in Pure Mathematics, 7, 366-374. https://doi.org/10.4236/apm.2017.77023

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