Braided Algebraic Quantum Groups

Abstract

In this paper, we mainly introduce the notion of a braided algebraic group, which unifies the notions of a braided Hopf algebra with an integral, a Hopf group-coalgebra with a group-integral and an algebraic quantum group. For this, we introduce the notion of braided multiplier rings and study their properties. Then we study some structural properties of a braided multiplier algebra with a nontrivial example.

1. Introduction

The representation theory of quantum groups (see [1]) and abstract algebra occupy a central position in science and engineering, with their theories and methods permeating mathematical physics, engineering, computer science, and other disciplines, serving as a bridge between abstract mathematical theory and practical application. In particular, the concept of groups in linear algebra and abstract algebra (see [2]) has very important applications in almost all disciplines, and it has also been developed, for example, through braid groups (see [1]), algebraic quantum groups (see [3]), etc.

In order to obtain high-dimensional topological invariants, V. Turaev in [4] (or [5]) introduced the notion of Hopf group-coalgebras (or Turaev’s group-coalgebras) which were studied by S. Caenepeel and M. De Lombaerde in [6] as Hopf algebras in a symmetric monoidal category, which we call the Turaev category. This means that Turaev’s group-coalgebras are some braided Hopf algebras in a certain braided monoidal category. A similar result holds for Hopf group-algebras. In 2007, A. T. Abd El-hafez et al. in [7] proved that Turaev’s group-coalgebras are also can be regarded as group-cograded multiplier Hopf algebras which are generalized in the setting of weak multiplier Hopf algebras (see [8,9,10]).

Based on the above materials, how do we naturally define a general notion of multiplier Hopf algebras in a braided monoidal category? This becomes the motivation of writing this paper.

The main aim of this article is to introduce and study the concept of a braided quantum group with a nontrivial example based on any braided group in a symmetric category. This generalizes the notions of a braided Hopf algebra with an integral (see [11]), a Hopf group-coalgebra with a group-integral (see [4]) and an algebraic quantum group (see [3]). In a separate article, we specifically study the Pontryagin duality problem of braided quantum groups.

This paper is organized as follows: In Section 1, we recall some basic notions of braided monoidal categories and symmetric categories. In Section 2, we introduce the notion of braided multiplier rings and study their properties (see Propositions 1 and 4).

In Section 3, we mainly introduce the concept of a braided multiplier Hopf algebra in a symmetric category $C$ (see Theorem 2). In the final section, we introduce the concept of a braided quantum group and develop the main result established as in the last section (see Theorem 3).

2. Preliminaries

Let us recall some basic definitions. A monoidal category $C =(C ,$ $\otimes , a, I, l, r)$ is a category $C$ endowed with a tensor product functor $\otimes :C \otimes C ⟶C$, a tensor unit object $I\in C$, natural associativity constraint isomorphisms

$$a=a_{U,V,W}:(U\otimes V)\otimes W\cong U\otimes (V\otimes W)$$

for all $U,V,W\in C$, a left unit constraint $l=l_U:I\otimes U\cong U$ and a right unit constraint $r=r_U:U\otimes I\cong U$ for any $U\in C$ such that the associativity pentagon

$$a_{U,V,W\otimes X}\circ a_{U\otimes V,W,X}=(i{d}_U\otimes a_{V,W,X})\circ a_{U,V\otimes W,X}\circ (a_{U,V,W}\otimes i{d}_X)$$

and

$$(i{d}_U\otimes l_V)\circ (r_U\otimes i{d}_V)=a_{U,I,V}$$

are satisfied. A tensor category $C$ is strict if all the constraints are identities. We say that $C$ is left (resp. right) autonomous when any object has a left (resp. right) dual and that $C$ is autonomous if it is both left and right autonomous. For the basic category theory, we refer to [12].

A monoidal category $C$ is called a braided monoidal category if for every object M in $C$ we have natural isomorphisms ${\tau }_{M,-}:M\otimes -\cong -\otimes M$ and ${\tau }_{-,M}:-\otimes M\cong M\otimes -$ which verify some appropriate coherence conditions, and furthermore, $C$ is symmetric if ${\tau }_{M,-}\circ {\tau }_{-,M}=id$; we refer to [13].

An object A in a monoidal category $C$ is called an algebra if there are two morphisms, ${\mu }_A:A\otimes A⟶A$ and ${\eta }_A:I⟶A$, satisfying the associativity and unitary conditions of usual algebras but expressed in diagrams. We call ${\mu }_A$ the product in A. If $C$ is braided, then the opposite algebra $A^{op}$ of A in $C$ is the object A with the product ${\mu }_A^{op}$ in $A^{op}$ given by ${\mu }_A^{op}={\mu }_A\circ \tau$.

We recall from [1] some notions concerning modules in a monoidal category. Let $C$ be a monoidal category, and let A and B be algebras in $C$. An object M in $C$ is called an A-B-bimodule if there are two morphisms in $C$: $m_A:A\otimes M⟶M$ and $m_B:M\otimes B⟶M$ satisfying the coherence conditions: $m_A\circ (id\otimes m_A)=m_A\circ ({\mu }_A\otimes id)$, $m_B\circ (m_B\otimes id)=m_B\circ (id\otimes {\mu }_B)$ and $m_A\circ (id\otimes m_B)=m_B\circ (m_A\otimes id)$ on $A\otimes M\otimes B$. The bimodule category _A$C$_B consists of objects in $C$ which have an A-B-bimodule structure and morphisms in $C$ which are A-B-bilinear. We write _A$C$ and $C_B$ for the categories _A$C_I$ and _I$C_B$, respectively.

Throughout, $\mathbb{k}$ will denote a fixed field. We denote ${\mathbf{LS}}_{\mathbb{k}}$ by the category of linear spaces and linear maps. Then $({\mathbf{LS}}_{\mathbb{k}},{\otimes }_{\mathbb{k}},\mathbb{k})$ is a symmetric category. Let H be an ordinary Hopf algebra which can be regarded as a braided Hopf algebra in ${\mathbf{LS}}_{\mathbb{k}}$. We will use the simplest Sweedler’s notation (cf. [14]) $\Delta \left(h\right)=\sum h_1\otimes h_2$ for any $h\in H$.

Let A be a bialgebra. A has an associative multiplication $m:A\otimes A⟶A$, a unit element $\mu :\mathbb{k}⟶A$, a coassociative comultiplication $\Delta :A⟶A\otimes A$, and a counit $\epsilon :A⟶\mathbb{k}$ such that $\Delta$ and $\epsilon$ are algebra maps. The set $End\left(A\right)$ of all linear maps of A forms a monoid with respect to the convolution product $f\ast g=m(f\otimes g)\Delta$ and the unit element $\mu \epsilon$.

A bialgebra H is a Hopf algebra if the identity map $id\in End\left(H\right)$ is invertible, that is, if there is an antipode $S\in End\left(H\right)$ satisfying $S\ast id=id\ast S=\mu \epsilon$ as functions on H. Explicitly, if $\Delta \left(h\right)=\sum h_1\otimes h_2$, the S must satisfy

$$(\forall h\in H):\sum S\left(h_1\right)h_2=\epsilon \left(h\right)1=\sum h_{1}S\left(h_2\right).$$

Such an S is unique, and is an algebra and coalgebra antimorphism of H (cf, [14]).

Let H be a Hopf algebra with a bijective antipode S. We will use $\mathcal{D}={}_H\mathcal{M}$ to denote the monoidal category of all left H-modules. An algebra A in $\mathcal{D}={}_H\mathcal{M}$ is equivalent to A being a left H-module algebra. We say that A is nondegenerate if $ab=0$ for all $b\in A$ implies $a=0$ and $ab=0$ for all $a\in A$ implies $b=0$. A linear functional $\omega :A⟶\mathbb{k}$ is said to be faithful if $\omega \left(ab\right)=0$ for all b implies $a=0$ and $\omega \left(ab\right)=0$ for all a implies $b=0$. A left A-module M is called nondegenerate if for all $m\in M$, the equalities $a·m=0$ for all $a\in A$ imply that $m=0$. A left A-module M is called idempotent (or unital) if $M=AM$.

3. Braided Multiplier Rings

In this section we will work on the monoidal category $\mathcal{D}={}_H\mathcal{M}$ of all left H-modules over a Hopf algebra H with a bijective antipode S. We will introduce the notion of braided multiplier rings and study their properties, generalizing the corresponding notions and results in [15].

Let A be an algebra in $\mathcal{D}$. We have the following definitions and Propositions:

Definition 1.

We say that a linear map $\lambda :A⟶A$ is a left multiplier of A if $\lambda \mu =\mu (\lambda \otimes id)$. Similarly, we call a linear $\rho :A⟶A$ a right multiplier of A if $\rho \mu =\mu (id\otimes \rho )$. We denote the space of left and right multipliers by $\mathbb{L}\left(A\right)$ and $\mathbb{R}\left(A\right)$, respectively.

Denote the set of all pairs $(\lambda ,\rho )$ by $\mathbb{M}\left(\mathbb{A}\right)$ where $\lambda \in \mathbb{L}\left(A\right)$ and $\rho \in \mathbb{R}\left(A\right)$ such that $\mu (id\otimes \lambda )=\mu (\rho \otimes id)$ on $A\otimes A$. We call such a pair $(\lambda ,\rho )$ a multiplier.

Proposition 1.

The above $\mathbb{M}\left(\mathbb{A}\right)$ can be viewed as an algebra in $\mathcal{D}$ with multiplication

$$(\lambda ,\rho )({\lambda }^{\prime },{\rho }^{\prime })=(\lambda \circ {\lambda }^{\prime },{\rho }^{\prime }\circ \rho )$$

for all $(\lambda ,\rho ),({\lambda }^{\prime },{\rho }^{\prime })\in \mathbb{M}\left(\mathbb{A}\right)$, and with unit $1_{\mathbb{M}\left(\mathbb{A}\right)}=(id,id)$.

Proof. 

Obviously, the associativity of the multiplication and the unitary are the same as the ones in the usual multiplier algebra. Now, we have to prove that $\mathbb{M}\left(\mathbb{A}\right)$ is an object in $C$. This consists of the following three steps:

Step 1. Define the following linear map ▷:

$$▷:H\otimes \mathbb{L}\left(A\right)⟶\mathbb{L}\left(A\right),(h▷\lambda )\left(a\right)=\sum h_1·\left(\lambda (S\left(h_2\right)·a)\right)$$

for all $a\in A,\lambda \in \mathbb{L}\left(A\right)$ and $h\in H$. In fact, we compute, for all $a,b\in A$,

$$\begin{array}{ccc}\hfill (h▷\lambda )\left(ab\right)& =& \sum h_1·\lambda (S\left(h_2\right)·\left(ab\right))\hfill \\ & =& \sum h_1·\left(\lambda (S\left(h_3\right)·a)(S\left(h_2\right)·b)\right)\hfill \\ & =& \sum (h_1·\lambda (S\left(h_4\right)·a))(h_{2}S\left(h_3\right)·b))=(h▷\lambda )\left(a\right)b\hfill \end{array}$$

and so $h▷\lambda \in \mathbb{L}\left(A\right)$. Moreover, $\mathbb{L}\left(A\right)\in \mathcal{D}$. First, it is easy to see that $\left(\mathbb{L}\right(A),▷)$ is a left H-module and $h▷id=\epsilon \left(h\right)id$ for all $h\in H$. Then we also find that

$$\begin{array}{ccc}\hfill \sum ((h_1▷\lambda )\circ (h_2▷{\lambda }^{\prime }))\left(a\right)& =& \sum (h_1▷\lambda )(h_2·\left({\lambda }^{\prime }(S\left(h_3\right)·a)\right)\hfill \\ & =& \sum h_1·\left(\lambda (S\left(h_2\right)h_3·\left({\lambda }^{\prime }(S\left(h_4\right)·a)\right))\right)\hfill \\ & =& (h▷(\lambda \circ {\lambda }^{\prime }))\left(a\right)\hfill \end{array}$$

for all $\lambda ,{\lambda }^{\prime }\in \mathbb{L}\left(A\right),a\in A$ and $h\in H$, and so

$$h▷(\lambda \circ {\lambda }^{\prime })=\sum (h_1▷\lambda )\circ (h_2▷{\lambda }^{\prime }).$$

Step 2. We define the following linear map ⊢:

$$⊢:H\otimes \mathbb{R}\left(A\right)⟶\mathbb{R}\left(A\right),(h⊢\rho )\left(a\right)=\sum h_2·\left(\rho (S^{-1}\left(h_1\right)·a)\right)$$

for all $a\in A,\rho \in \mathbb{R}\left(A\right)$ and $h\in H$. We check that $h⊢\rho \in \mathbb{R}\left(A\right)$ as follows for all $a,b\in A$

$$\begin{array}{ccc}\hfill (h⊢\rho )\left(ab\right)& =& \sum h_2·\rho (S^{-1}\left(h_1\right)·\left(ab\right))\hfill \\ & =& \sum h_3·\left((S^{-1}\left(h_2\right)·a)\rho (S^{-1}\left(h_1\right)·b)\right)\hfill \\ & =& \sum (h_3{S}^{-1}\left(h_2\right)·a)(h_4·\rho (S^{-1}\left(h_1\right)·b))=a(h⊢\rho )\left(b\right).\hfill \end{array}$$

It is straightforward to check that $\left(\mathbb{R}\right(A),⊢)$ is a left H-module and $h⊢id=\epsilon \left(h\right)id$ for all $h\in H$. However, $\mathbb{R}\left(A\right)$ is not an object in $C$. In fact, we compute

$$\begin{array}{ccc}\hfill \sum ((h_2⊢\rho )\circ (h_1⊢{\rho }^{\prime }))\left(a\right)& =& \sum (h_3⊢\rho )(h_2·{\rho }^{\prime }(S^{-1}\left(h_1\right)·a))\hfill \\ & =& \sum h_4\rho (S^{-1}\left(h_3\right)h_2·{\rho }^{\prime }(S^{-1}\left(h_1\right)·a))\hfill \\ & =& (h⊢(\rho \circ {\rho }^{\prime }))\left(a\right)\hfill \end{array}$$

for all $\rho ,{\rho }^{\prime }\in \mathbb{R}\left(A\right),a\in A$ and $h\in H$, and so

$$h⊢(\rho \circ {\rho }^{\prime })=\sum (h_2⊢\rho )\circ (h_1⊢{\rho }^{\prime }).$$

Step 3. We first define a left H-module action on $\mathbb{M}\left(A\right)$ by the following formula:

$$h·(\lambda ,\rho )=(h▷\lambda ,h⊢\rho )$$

for all $h\in H$ and $(\lambda ,\rho )\in \mathbb{M}\left(A\right)$. First, to check $h·(\lambda ,\rho )\in \mathbb{M}\left(A\right)$, we compute

$$\begin{array}{ccc}\hfill a\left(\right(h▷\lambda \left)\right(b\left)\right)& =& \sum a(h_1·\left(\lambda (S\left(h_2\right)·b)\right))\hfill \\ & =& \sum h_2·\left((S^{-1}\left(h_1\right)·a)\lambda (S\left(h_3\right)·b)\right)\hfill \\ & =& \sum h_2·\left(\rho (S^{-1}\left(h_1\right)·a)(S\left(h_3\right)·b)\right)\hfill \\ & =& \sum (h_2·\rho (S^{-1}\left(h_1\right)·a))(h_{3}S\left(h_4\right)·b)=(h⊢\rho )\left(a\right)b.\hfill \end{array}$$

Finally, we find that for all $(\lambda ,\rho ),({\lambda }^{\prime },{\rho }^{\prime })\in \mathbb{M}\left(A\right)$ and $h\in H$,

$$\begin{array}{ccc}\hfill h·\left((\lambda ,r)({\lambda }^{\prime },{\rho }^{\prime })\right)& =& (h▷\left(\lambda {\lambda }^{\prime }\right),h⊢\left({\rho }^{\prime }\rho \right))\hfill \\ & =& \sum ((h_1▷\lambda )\circ (h_2▷{\lambda }^{\prime }),(h_2⊢{\rho }^{\prime })\circ (h_1⊢\rho ))\hfill \\ & =& \sum ((h_1▷\lambda ),(h_1⊢\rho ))((h_2▷{\lambda }^{\prime }),(h_2⊢{\rho }^{\prime }))\hfill \\ & =& \sum (h_1·(\lambda ,\rho ))(h_2·({\lambda }^{\prime },{\rho }^{\prime }))\hfill \end{array}$$

and $h\in (id,id)=\epsilon \left(h\right)(id,id)$.

This finishes the proof of the Proposition.   □

Proposition 2.

There is natural algebra homomorphism in $\mathcal{D}$

$$\begin{array}{c}{i}_A:A\hookrightarrow \mathbb{M}\left(A\right),i_A\left(a\right)=({\lambda }_a,{\rho }_a),\hfill \\ {{\rm Y}}_A^l:\mathbb{M}\left(A\right)⟶\mathbb{L}\left(A\right),{{\rm Y}}_A^l\left((\lambda ,\rho )\right)=\lambda ,and\hfill \\ {{\rm Y}}_A^r:\mathbb{M}\left(A\right)⟶\mathbb{R}{\left(A\right)}^{op},{{\rm Y}}_A^r\left((\lambda ,\rho )\right)=\rho \hfill \end{array}$$

where ${\lambda }_a\left(b\right)=ab$ and ${\rho }_a\left(b\right)=ba$ for all $a,b\in A$.

Proof. 

Obviously, these maps are algebra homomorphisms, and ${{\rm Y}}_A^l$ and ${{\rm Y}}_A^r$ are morphisms in $\mathcal{D}$.

Then, by the proof of the Proposition, we know that $\mathbb{L}\left(A\right),\mathbb{R}{\left(A\right)}^{op}$ and $\mathbb{M}\left(A\right)$ are algebras in $\mathcal{D}$. For all $a,b\in A$ and $h\in H$, we find that

$$(h▷{\lambda }_a)\left(b\right)=\sum h_1·\left({\lambda }_a(S\left(h_2\right)·b)\right)=\sum (h_1·a)(h_2·(S\left(h_3\right)·b))=(h·a)b={\lambda }_{h·a}\left(b\right).$$

Similarly, we get $(h⊢{\rho }_a)={\rho }_{h⊢a}$.

From these equations we can imply that $i_A$ is a morphism in $\mathcal{D}$.   □

Corollary 1.

A is nondegenerate if and only if $L={{\rm Y}}_A^l\circ i_A$ and $R={{\rm Y}}_A^r\circ i_A$ are injective; in particular, $i_A$ is injective as well.

Definition 2.

Let H be a Hopf algebra. We say that A is a left H-module algebra extension of B if the following conditions hold:

(i) A and B are left H-module algebras;

(ii) $(A,\rightharpoonup ,\leftharpoonup )$ is a B-bimodule in $\mathcal{D}$;

(iii) $(b\rightharpoonup a)a^{\prime }=b\rightharpoonup \left(a{a}^{\prime }\right)$ and $a(a^{\prime }\leftharpoonup b)=\left(a{a}^{\prime }\right)\leftharpoonup b$ for all $a,a^{\prime }\in A$ and $b\in B$;

(iv) $(a\leftharpoonup b)a^{\prime }=a(b\rightharpoonup a^{\prime })$ for all $a,a^{\prime }\in A$ and $b\in B$.

Furthermore, A is said to be a nondegenerate (resp. idempotent) left H-module algebra extension of B if A is nondegenerate (resp. idempotent) as a B-bimodule as well.

Proposition 3.

Let A be a nondegenerate algebra $\mathcal{D}$. Then A is a nondegenerate idempotent left H-module algebra extension of $\mathbb{M}\left(A\right)$ with actions given by

$$z\rightharpoonup a=\lambda \left(a\right)anda\leftharpoonup z=\rho \left(a\right)$$

for any $z=(\lambda ,\rho )\in \mathbb{M}\left(A\right)$ and $a\in A$.

Proof. 

By Proposition 2, we know that A is a left H-module algebra. We prove that ⇀ is a morphism in $\mathcal{D}$. Similarly, we get ↼ as a morphism in $\mathcal{D}$ as well. In fact, we compute

$$\begin{array}{ccc}\hfill \rightharpoonup (h·(z\otimes a\left)\right)& =& \sum (h_1·z)\rightharpoonup (h_2·a)=(h_1▷\lambda )(h_2·a)\hfill \\ & =& \sum h_1·\left(\lambda (S\left(h_2\right)h_3·a)\right)=h·\lambda \left(a\right)=h·(z\rightharpoonup a)\hfill \end{array}$$

for all $h\in H,z\in \mathbb{M}\left(A\right)$ and $a\in A$. The rest are obvious. Also, since for all $(\lambda ,\rho ),({\lambda }^{\prime },{\rho }^{\prime })\in \mathbb{M}\left(A\right)$ and $a,b\in A$, we find that

$$\lambda \left({\rho }^{\prime }\left(a\right)\right)b=\lambda \left({\rho }^{\prime }\left(a\right)b\right)=\lambda \left(a{\lambda }^{\prime }\left(b\right)\right)=\lambda \left(a\right){\lambda }^{\prime }\left(b\right)={\rho }^{\prime }\left(\lambda \left(a\right)\right)b.$$

By the nondegenerateness of A, the above equation implies that A is an $\mathbb{M}\left(A\right)$-bimodule.

Conditions (iii) and (iv) in Definition 2 are obvious.

Finally, we have to prove that A is nondegenerate as an $\mathbb{M}\left(A\right)$-bimodule; e.g., for any $a\in A$ such that $z\rightharpoonup a=0$ for all $z\in \mathbb{M}\left(A\right)$ we want to show that $a=0$. In fact, since $A=A\leftharpoonup \mathbb{M}\left(A\right)$, we can write $b={\sum }_i{b}_i\leftharpoonup z_i$ for $b_i\in A$ and $z_i\in \mathbb{M}\left(A\right)$. Hence we find that $ba=({\sum }_i{b}_i\leftharpoonup z_i)a={\sum }_i{b}_i(z_i\rightharpoonup a)=0$. By the nondegenerateness of A, we conclude that $a=0$.   □

Proposition 4.

Let A and B be algebras in $\mathcal{D}$. Then:

(i) There exists a bijective correspondence between algebra homomorphisms $L:B⟶\mathbb{L}\left(A\right)$ in $\mathcal{D}$ and left B-module structures on A such that ${\mu }_A(ba\otimes a^{\prime })=b{\mu }_A(a\otimes a^{\prime })$ for all $a,a^{\prime }\in A$ and $b\in B$.

(ii) There exists a bijective correspondence between algebra homomorphisms $R:B⟶\mathbb{R}{\left(A\right)}^{op}$ in $\mathcal{D}$ and right B-module structures on A such that ${\mu }_A(a\otimes a^{\prime }b)={\mu }_A(a\otimes a^{\prime })b$ for all $a,a^{\prime }\in A$ and $b\in B$.

(iii) There exists a bijective correspondence between algebra homomorphisms $M:B⟶\mathbb{M}\left(A\right)$ in $\mathcal{D}$ and left H-module algebra extension structures of B on A.

Proof. 

(i) If there is an algebra homomorphism $L:B⟶\mathbb{L}\left(A\right)$ in $\mathcal{D}$, then we define $•:B\otimes A⟶A$ by $b•a=L\left(b\right)\left(a\right)$ for all $b\in B$ and $a\in A$. It is straightforward to check that the action is associative and that ${\mu }_A(b•a\otimes a^{\prime })=b•{\mu }_A(a\otimes a^{\prime })$ for all $a,a^{\prime }\in A$ and $b\in B$. We also have to prove that • is a left H-linear; in fact, we have

$$\begin{array}{ccc}\hfill •(h·(b\otimes a\left)\right)& =& \sum •(h_1·b\otimes h_2·a)=L(h_1·b)(h_2·a)\hfill \\ & =& \sum (h_1▷L\left(b\right))(h_2·a)=h_1·(L\left(b\right)(S\left(h_2\right)h_3·a)\hfill \\ & =& h·\left(L\right(b\left)\right(a\left)\right)=h·(b•a)\hfill \end{array}$$

for all $a\in A,b\in B$ and $h\in H$. Conversely, if A is a left B-module with action ≻ in $C$, then one defines $L:B⟶\mathbb{L}\left(A\right)$ by $L\left(b\right)\left(a\right)=b\succ a$ for all $a\in A$ and $b\in B$. Since ${\mu }_A$ is a left B-linear this map is well-defined. It is not hard to show that L is an algebra map. Finally, it remains to show that L is a morphism in $\mathcal{D}$. Indeed,

$$\begin{array}{ccc}\hfill (h▷L(b\left)\right)\left(a\right)& =& \sum h_1·(L\left(b\right)(S\left(h_2\right)·a)=\sum h_1·(b\succ (S\left(h_2\right)·a))\hfill \\ & =& \sum (h_1·b)\succ (h_{2}S\left(h_3\right)·a)=(h·b)\succ a=L(h·b)\left(a\right)\hfill \end{array}$$

for all $a\in A,b\in B$ and $h\in H$.

(ii) Similar to part (i).

(iii) Suppose that the map $M:B⟶\mathbb{M}\left(A\right)$ in $\mathcal{D}$ exists. Then we get two algebras maps ${{\rm Y}}_A^l\circ M:B⟶\mathbb{L}\left(A\right)$ and ${{\rm Y}}_A^r\circ M:B⟶\mathbb{R}{\left(A\right)}^{op}$ in $\mathcal{D}$. Thus by part (i) and part (ii), A will be a left and right B-module with B-actions given by $b•a={{\rm Y}}_A^l\left(M\left(b\right)\right)\left(a\right)=M\left(b\right)\rightharpoonup a$ and $a·b={{\rm Y}}_A^r\left(M\left(b\right)\right)\left(a\right)=a\leftharpoonup M\left(b\right)$ for all $a\in A$ and $b\in B$, where ⇀ and ↼ are the left and right $\mathbb{M}\left(A\right)$-actions on A (see Proposition 3). Since A is a left H-module algebra extension of $\mathbb{M}\left(A\right)$, it follows that A is a nondegenerate idempotent left H-module algebra extension of B. Conversely, by the first two parts, we have two algebra morphisms $L:B⟶\mathbb{L}\left(A\right)$ and $R:B⟶\mathbb{R}{\left(A\right)}^{op}$ in $\mathcal{D}$. We define $M:B⟶\mathbb{M}\left(A\right)$ by $M\left(b\right)=\left(L\right(b),R(b\left)\right)$ for all $b\in B$. The verifications of the rest such that M is an algebra morphism in $\mathcal{D}$ are straightforward.

This completes the proof of the Proposition.   □

4. Braided Multiplier Hopf Algebras

A Hopf algebra H is said to be quasitriangular provided there exits an invertible element $R=\sum R^{\left(1\right)}\otimes R^{\left(2\right)}\in H\otimes H$ satisfying $(r:=R)$:

(QT1) $\sum {R^{\left(1\right)}}_1\otimes {R^{\left(1\right)}}_2\otimes R^{\left(2\right)}=\sum (R^{\left(1\right)}\otimes 1\otimes R^{\left(2\right)})(1\otimes r^{\left(1\right)}\otimes r^{\left(2\right)})$;

(QT2) $\sum R^{\left(1\right)}\otimes {R^{\left(2\right)}}_1\otimes {R^{\left(2\right)}}_2=\sum (R^{\left(1\right)}\otimes 1\otimes R^{\left(2\right)})(r^{\left(1\right)}\otimes r^{\left(2\right)}\otimes 1),$

(QT3) $\sum {\Delta }^{cop}\left(h\right)=R\Delta \left(h\right)R^{-1},$ with $h\in H$.

Furthermore, when $R^{-1}=\sum R^{\left(2\right)}\otimes R^{\left(1\right)}$, $(H,R)$ is said to be triangular. In this case, the category ${}_H\mathbb{M}^R$ of all left H-modules is symmetric.

In this section we always let $C ={}_H\mathbb{M}^R$ be a symmetric braided monoidal category. We shall introduce the concept of a braided multiplier Hopf algebra in $C$ generalizing the main result in [15].

Let A be an algebra in $C$. Let $\Delta :A⟶M(A\otimes A)$ be a morphism in $C$. Then the left–right, right–left, left–left and right–right entwining maps $E_{lr},E_{rl},E_{ll},E_{rr}:A\otimes A⟶M(A\otimes A)$ for the pair $(\mu ,\Delta )$ be defined by

$$\begin{array}{ccc}& & E_{lr}=(id\otimes {\mu }_A)\circ ({\Delta }_A\otimes id),E_{rl}=({\mu }_A\otimes id)\circ (id\otimes {\Delta }_A),\hfill \\ & & E_{ll}=({\mu }_A^{op}\otimes id)\circ (id\otimes {\Delta }_A)\circ \tau ,E_{rr}=(id\otimes {\mu }_A^{op})\circ ({\Delta }_A\otimes id)\circ \tau .\hfill \end{array}$$

Similarly, we write $E_{lr}^{op},E_{rl}^{op},E_{ll}^{op},E_{rr}^{op}$ for the corresponding entwining maps of the pair $({\mu }^{op},\Delta )$.

Lemma 1.

If Δ is a morphism in $C$, then $E_{lr},E_{rl},E_{ll}$ and $E_{rr}$ are all morphisms in $C$.

Proof. 

The proof of the lemma is straightforward. In fact, for example, for $E_{lr}$, since A is an algebra in $C$, by definition, ${\mu }_A$ is a morphism in $C$. Also, since $\Delta :A⟶M(A\otimes A)$ is a morphism in $C$, therefore, $E_{lr}=(id\otimes {\mu }_A)\circ ({\Delta }_A\otimes id)$ is a morphism in $C$.   □

Definition 3.

Let A be an algebra in $C$. A morphism $\Delta :A⟶M(A\otimes A)$ in $C$ is called a coproduct if it is coassociative in the sense of the following

$$(E_{rl}\otimes id)\circ (id\otimes E_{lr})=(id\otimes E_{lr})\circ (E_{rl}\otimes id).$$

(1)

Since $C$ is braided, the opposite coproduct ${\Delta }^{cop}:A⟶M(A\otimes A)$ in $C$ can be defined by ${\Delta }^{cop}=\tau \otimes \Delta$. We will write $A^{cop}$ for A equipped with the opposite coproduct. Then we write $E_{lr}^{cop},E_{rl}^{cop},E_{ll}^{cop},E_{rr}^{cop}$ for the corresponding entwining maps of the pair $(\mu ,{\Delta }^{cop})$.

We may also reverse both product and coproduct and then get $A^{opcop}=:{\left(A^{op}\right)}^{cop}$. Similarly, we write $E_{lr}^{opcop},E_{rl}^{opcop},E_{ll}^{opcop},E_{rr}^{opcop}$ for the corresponding entwining maps of the pair $({\mu }^{op},{\Delta }^{cop})$.

It is not hard to check the lemma below.

Lemma 2.

Let $\Delta :A⟶M(A\otimes A)$ be a coproduct in the braided monoidal category $C$. Then

$$\begin{array}{ccc}& & E_{lr}^{op}=E_{rr}\circ \tau ,E_{rl}^{op}=E_{ll}\circ \tau ,E_{ll}^{op}=E_{rl}\circ \tau ,E_{rr}^{op}=E_{lr}\circ \tau ,\hfill \\ & & E_{lr}^{cop}=\tau \circ E_{ll},E_{rl}^{cop}=\tau \circ E_{rr},E_{ll}^{cop}=\tau \circ E_{lr},E_{rr}^{cop}=\tau \circ E_{rl},\hfill \\ & & E_{lr}^{opcop}=\tau \circ E_{rl}\circ \tau ,E_{rl}^{opcop}=\tau \circ E_{lr}\circ \tau ,\hfill \\ & & E_{ll}^{opcop}=\tau \circ E_{rr}\circ \tau ,E_{rr}^{opcop}=\tau \circ E_{ll}\circ \tau .\hfill \end{array}$$

Definition 4.

Let $\Delta :A⟶M(A\otimes A)$ be a coproduct in $C$. An algebra homomorphism $\epsilon :A⟶\mathbb{k}$ in $C$ is called a counit if

$$(\epsilon \otimes id)\circ E_{lr}=\mu =(id\otimes \epsilon )\circ E_{rl}.$$

(2)

Definition 5.

Let $\Delta :A⟶M(A\otimes A)$ be a coproduct in $C$ and ε a counit on A. We say that A is a braided multiplier Hopf algebra if $E_{lr}$ and $E_{rl}$ are bijective so that $E_{lr}(A\otimes A),E_{rl}(A\otimes A)\subseteq A\otimes A$. Furthermore, A is called regular if $E_{ll}$ and $E_{rr}$ are bijective such that $E_{ll}(A\otimes A),E_{rr}(A\otimes A)\subseteq A\otimes A$.

Example 1.

Any braided Hopf algebra in $C$ is a braided multiplier Hopf algebra in $C$.

Example 2.

A  ${\mathbb{Z}}_2$-graded group is a group G with a decomposition $G=G_0\bigsqcup G_1$ such that

$$G_i{G}_j\subseteq G_{i+j(mod2)}\mathit{for}i,j\in \{0,1\}.$$

Equivalently, it is a braided group G in the symmetric monoidal category $C ={}_{{\mathbb{Z}}_2}Mod$ of all left ${\mathbb{Z}}_2$-modules, where ${\mathbb{Z}}_2:=\mathbb{Z}/2\mathbb{Z}$ is a triangular Hopf algebra (see [1]).

Example 3.

Recall from [1] that the Majid braid group, introduced by S Majid in the framework of braided group theory and quantum groups, generalizes the classical Artin braid group to the setting of braided monoidal categories, serving as a fundamental combinatorial object for describing braidings in non-commutative and braided algebraic structures.

Let $\mathcal{C}$ be a braided monoidal category with braiding isomorphism

$${\tau }_{X,Y}:X\otimes Y\to Y\otimes X,\forall X,Y\in \mathrm{Ob}\left(\mathcal{C}\right),$$

satisfying the hexagon axioms for braided monoidal categories. For a fixed object $V\in \mathrm{Ob}\left(\mathcal{C}\right)$, the n-strand Majid braid group ${\mathbf{B}}_n^{\mathrm{Maj}}\left(V\right)$ is defined as the group generated by braid generators ${\sigma }_1,{\sigma }_2,\dots ,{\sigma }_{n-1}$ subject to the Majid braid relations, which refine the classical Artin braid relations by incorporating the categorical braiding τ:

Adjacent commutation relation: ${\sigma }_i{\sigma }_j={\sigma }_j{\sigma }_i,|i-j|\ge 2;$

Braided Yang-Baxter relation (Majid’s braid relation):

$${\sigma }_i{\sigma }_{i+1}{\sigma }_i=\tau \left({\sigma }_{i+1}{\sigma }_i{\sigma }_{i+1}\right),1\le i\le n-2,$$

where τ denotes the categorical braiding action permuting the tensor factors corresponding to braid strands.

In the special case where the category $\mathcal{C}$ is the Cartesian monoidal category of sets (with trivial braiding $\tau =\mathrm{id}$), the Majid braid group ${\mathbf{B}}_n^{\mathrm{Maj}}\left(V\right)$ recovers the classical Artin braid group ${\mathbf{B}}_n$. For nontrivial braided categories (e.g., the category of modules over a Hopf algebra with a quasitriangular structure), the Majid braid group encodes nontrivial braid statistics, forming a bridge between combinatorial braid theory, quantum groups, and non-commutative geometry.

Example (1): Trivial Braiding (Classical Artin Braid Group):  Let $\mathcal{C}=\mathbf{Set}$ be the category of sets with the Cartesian product as the monoidal product and trivial braiding ${\tau }_{X,Y}={\mathrm{id}}_{X\times Y}$. The Majid braid relation reduces to the classical Artin braid relation

$${\sigma }_i{\sigma }_{i+1}{\sigma }_i={\sigma }_{i+1}{\sigma }_i{\sigma }_{i+1}.$$

Thus ${\mathbf{B}}_n^{\mathrm{Maj}}\left(V\right)\cong {\mathbf{B}}_n$ is the classical n-strand braid group, which is the free group generated by ${\sigma }_1,\dots ,{\sigma }_{n-1}$ modulo the above relations.

Example (2): Majid Braid Group for Quantum ${\mathfrak{sl}}_2$ (Quasitriangular Hopf Algebra Setting): Let $U_q\left({\mathfrak{sl}}_2\right)$ be the quantum universal enveloping algebra of ${\mathfrak{sl}}_2$, a quasitriangular Hopf algebra with universal R-matrix $\mathcal{R}\in U_q\left({\mathfrak{sl}}_2\right)\otimes U_q\left({\mathfrak{sl}}_2\right)$. The category ${\mathcal{C}}_{U_q\left({\mathfrak{sl}}_2\right)}$ of finite-dimensional $U_q\left({\mathfrak{sl}}_2\right)$-modules is a braided monoidal category with braiding

$${\tau }_{M,N}(m\otimes n)=\mathcal{R}·(m\otimes n),m\in M,n\in N.$$

For the standard two-dimensional fundamental module $V={\mathbb{C}}_q^2$, the Majid braid group ${\mathbf{B}}_n^{\mathrm{Maj}}\left(V\right)$ is generated by ${\sigma }_1,\dots ,{\sigma }_{n-1}$ with relations:

1. 

${\sigma }_i{\sigma }_j={\sigma }_j{\sigma }_i$ for $|i-j|\ge 2$;

2. 

${\sigma }_i{\sigma }_{i+1}{\sigma }_i=\mathcal{R}\left({\sigma }_{i+1}{\sigma }_i{\sigma }_{i+1}\right)$,

where $\mathcal{R}$ acts on the tensor product of module endomorphisms associated with braid crossings. This braid group governs the braid statistics of quantum ${\mathfrak{sl}}_2$ modules and is central to constructing quantum link invariants.

Example (3): Majid Braid Group on a Groupoid (Groupoid-Based Braiding): Let $\mathcal{G}$ be a discrete groupoid with a monoidal product given by composition of morphisms, equipped with a braiding τ induced by a two-cocycle on $\mathcal{G}$. The Majid braid group ${\mathbf{B}}_n^{\mathrm{Maj}}\left(\mathrm{Ob}\left(\mathcal{G}\right)\right)$ describes braidings of morphisms in $\mathcal{G}$, where each generator ${\sigma }_i$ represents a crossing of two parallel morphism strands, and the braid relations are twisted by the groupoid two-cocycle. This example is widely used in Majid’s theory of braided groups and Hopf groupoids.

For any a Majid braided group, $Q={\mathbf{B}}_n^{\mathrm{Maj}}\left(V\right)(V={\mathbb{C}}_q^2,orV=\mathrm{Ob}\left(\mathcal{G}\right))$ as above. Let $A=\mathbb{C}\left(Q\right)$ be the algebra of complex, finitely supported functions on Q. In this case $M\left(\mathbb{C}\right(Q\left)\right)$ consists of all complex functions on Q. Moreover $\mathbb{C}\left(Q\right)\otimes \mathbb{C}\left(Q\right)$ can be identified with finitely supported complex functions on $Q\times Q$ so that $M\left(\mathbb{C}\right(Q)\otimes \mathbb{C}(Q\left)\right)$ is the space of all complex functions on $Q\times Q$. We now define $\Delta :\mathbb{C}\left(Q\right)⟶M\left(\mathbb{C}\right(Q)\otimes \mathbb{C}(Q\left)\right)$ by $\Delta \left(f\right)(p,q)=f\left(pq\right)$ for all $p,q\in Q$, and $\epsilon :\mathbb{C}\left(Q\right)⟶\mathbb{C}$ by ${\epsilon }_A\left(f\right)=f\left(e\right)for all f\in \mathbb{C}\left(Q\right).$ It is straightforward to check that Δ and ε are algebra homomorphisms.

We define:

$$E_{lr}(f\otimes g)(p,q)=\Delta \left(f\right)(1\otimes g)(p,q)=\Delta \left(f\right)(p,q)(1\otimes g)(p,q)=f\left(pq\right)g\left(q\right)$$

and

$$E_{rl}(f\otimes g)(p,q)=(f\otimes 1)\Delta \left(g\right)(p,q)=(f\otimes 1)(p,q){\Delta }^A\left(g\right)\left(pq\right)=f\left(p\right)g\left(pq\right);$$

$E_{lr}(f\otimes g)$ and $E_{rl}(f\otimes g)$ have finite support and belong to $\mathbb{C}\left(Q\right)\otimes \mathbb{C}\left(Q\right)$.

Similarly, we define $E_{lr}$ and $E_{rl}$ by

$$R_{lr}\left(h\right)(p,q)=h(p{q}^{-1},q)and{R}_{rl}\left(h\right)(p,q)=h(p,p^{-1}q),$$

for all $h\in \mathbb{C}\left(Q\right))\otimes \mathbb{C}(Q)$ and $p,q\in Q$. It is easy to check that $E_{lr}$ is bijective with the inverse $R_{lr}$, and similarly for $E_{rl}$ with the inverse $R_{rl}$.

Furthermore, we have $\left[(\epsilon \otimes id)E_{lr}(f\otimes g)\right]\left(p\right)=\left(fg\right)\left(p\right)$ and $\left[(id\otimes \epsilon )E^{rl}(f\otimes g)\right]=fg$ for all $f,g\in \mathbb{C}\left(Q\right)$ and $p\in Q$.

Therefore, $A=\mathbb{C}\left(Q\right)$ is a braided multiplier Hopf algebra in $C$. In particular, when $\mathcal{C}=\mathbf{Set}$ is the category of sets with Cartesian product as the monoidal product and trivial braiding ${\tau }_{X,Y}={\mathrm{id}}_{X\times Y}$, we get a classical multiplier Hopf algebra in [15].

Corollary 2.

Let $(A,\Delta )$ be as before with counit ε. Then

$$\begin{array}{c}\hfill (\epsilon \otimes id)\circ E_{rr}=\mu =(id\otimes \epsilon )E_{ll}.\end{array}$$

Proposition 5.

Let $(A,\mu ,\Delta )$ be as before and assume that ε is a counit. Then:

(1) if ${\epsilon }^{\prime }$ is any linear map from A to $\mathbb{k}$ satisfying $(id\otimes {\epsilon }^{\prime })\circ E_{rl}=\mu$, then $\epsilon ={\epsilon }^{\prime }$.

(2) if ${\epsilon }^{\prime }$ is any linear map from A to $\mathbb{k}$ satisfying $(id\otimes {\epsilon }^{\prime })\circ E_{ll}=\mu$, then $\epsilon ={\epsilon }^{\prime }$.

(3) if ${\epsilon }^{\prime }$ is any linear map from A to $\mathbb{k}$ satisfying $({\epsilon }^{\prime }\otimes id)\circ E_{lr}=\mu$, then $\epsilon ={\epsilon }^{\prime }$.

(4) if ${\epsilon }^{\prime }$ is any linear map from A to $\mathbb{k}$ satisfying $({\epsilon }^{\prime }\otimes id)\circ E_{rr}=\mu$, then $\epsilon ={\epsilon }^{\prime }$.

Proof. 

Note that $\epsilon$ is an algebra map. Assume that ${\epsilon }^{\prime }$ is any linear map from A to $\mathbb{k}$ satisfying $(id\otimes {\epsilon }^{\prime })E_{rl}=m$. We know that $(\epsilon \otimes id)E_{rl}=(\epsilon \otimes id)$ since $\epsilon$ is a counit. Applying ${\epsilon }^{\prime }$ to this equation, we obtain that $\epsilon m=(\epsilon \otimes {\epsilon }^{\prime })$. Because $\epsilon m=(\epsilon \otimes \epsilon )$ and $\epsilon$ can not be trivially zero, we get $\epsilon ={\epsilon }^{\prime }$. This proves part (1). The other parts are proven in a similar way.   □

Lemma 3.

Let $\Delta :A⟶M(A\otimes A)$ be a coproduct in the braided monoidal category $C$. Then

$$\begin{array}{ccc}\hfill \left(1\right)& & E_{lr}\circ (\mu \otimes id)=(id\otimes \mu )\circ (E_{ll}\otimes id)\circ (id\otimes E_{lr}),\hfill \end{array}$$

(3)

$$\begin{array}{ccc}\hfill \left(2\right)& & (\mu \otimes id)\circ (id\otimes E_{lr}^{-1})=(id\otimes \mu )\circ (E_{lr}^{-1}\otimes id)\circ (E_{ll}\otimes id).\hfill \end{array}$$

(4)

Proof. 

(1) We compute as follows

$$\begin{array}{ccc}& & E_{lr}\circ (\mu \otimes id)=(id\otimes \mu )\circ \underline{(\Delta \otimes id)\circ (\mu \otimes id)}\hfill \\ & =& (id\otimes \mu )\circ (\mu \otimes \mu \otimes id)\circ (id\otimes \tau \otimes i{d}^2)\circ (\Delta \otimes \Delta \otimes id)\hfill \\ & =& (id\otimes \mu )\circ (\mu \otimes id\otimes \mu )\circ (id\otimes \tau \otimes i{d}^2)\circ (\Delta \otimes i{d}^3)\circ (id\otimes \Delta \otimes id)\hfill \\ & =& (id\otimes \mu )\circ (\mu \otimes i{d}^2)\circ (id\otimes \tau \otimes id)\circ (\Delta \otimes i{d}^2)\circ (i{d}^2\otimes \mu )\circ (id\otimes \Delta \otimes id)\hfill \\ & =& (id\otimes \mu )\circ \left(\right(\mu \otimes id)\circ (id\otimes \tau )\circ (\Delta \otimes id)\otimes id)\circ (id\otimes (id\otimes \mu )\circ (id\otimes \Delta \left)\right)\hfill \\ & =& (id\otimes \mu )\circ (E_{ll}\otimes id)\circ (id\otimes E_{lr}),\hfill \end{array}$$

as desired.

(2) By the first part (1), we have

$$\begin{array}{ccc}& & (\mu \otimes id)\circ (id\otimes E_{lr}^{-1})=E_{lr}^{-1}\circ \underline{E_{lr}\circ (\mu \otimes id)}\circ (id\otimes E_{lr}^{-1})\hfill \\ & =& E_{lr}^{-1}\circ (id\otimes \mu )\circ (E_{ll}\otimes id)\hfill \\ & =& (id\otimes \mu )\circ (E_{lr}^{-1}\otimes id)\circ (E_{ll}\otimes id).\hfill \end{array}$$

  □

Lemma 4.

Let $\Delta :A⟶M(A\otimes A)$ be a coproduct in the braided monoidal category $C$. Then

$$\begin{array}{ccc}\hfill \left(1\right)& & E_{lr}\circ (id\otimes \mu )=(id\otimes \mu )\circ (E_{lr}\otimes id),\hfill \\ \hfill \left(2\right)& & (id\otimes \mu )\circ (E_{rl}\otimes id)=(\mu \otimes id)\circ (id\otimes E_{lr}),\hfill \\ \hfill \left(3\right)& & E_{rl}\circ (id\otimes \mu )=(\mu \otimes \mu )\circ (id\otimes \tau \otimes id)\circ (E_{rl}\otimes \Delta ).\hfill \end{array}$$

(5)

Proof. 

This is straightforward.   □

Now we prove the main result of this section.

Theorem 1.

Let $\Delta :A⟶M(A\otimes A)$ be a coproduct and ε a counit in the braided monoidal category $C$. Then A is a regular braided multiplier Hopf algebra if and only if there exists a unique isomorphism $S:A⟶A$ which is both an algebra antihomomorphism and a coalgebra antihomomorphism such that

$${\mu }_A\circ (S\otimes id)\circ E_{lr}=\epsilon \otimes id,{\mu }_A\circ (id\otimes S)\circ E_{rl}=id\otimes \epsilon and\epsilon \circ S=\epsilon .$$

Proof. 

The proof is similar to that in [15] or [3]. If A is a braided multiplier Hopf algebra in the braided monoidal category $C$, then we have to construct the morphisms S.

We first define a map $S_l:A⟶End\left(A\right)$ by

$$m(S_l\otimes id)=((\epsilon \otimes id)\circ E_{lr}^{-1}).$$

It is obvious that $S_l$ is a morphism in $C$ and furthermore, it is a left multiplier. We claim that $S_l:A⟶\mathbb{L}\left(A\right)$ is an algebra antihomomorphism. We compute

$$\begin{array}{ccc}& & \mu \circ (id\otimes \mu )\circ (id\otimes S_l\otimes id)\circ (E_{rl}\otimes id)\hfill \\ & =& \mu \circ (id\otimes \epsilon \otimes id)\circ \underline{(id\otimes E_{lr}^{-1})\circ (E_{rl}\otimes id)}\hfill \\ & \stackrel{\left(1\right)}{=}& \mu \circ (\underline{(id\otimes \epsilon )\circ E_{rl}}\otimes id)\circ (id\otimes E_{lr}^{-1})\hfill \\ & \stackrel{\left(2\right)}{=}& \mu \circ (\mu \otimes id)\circ (id\otimes E_{lr}^{-1})=\mu \circ (id\otimes \mu \circ E_{lr}^{-1})\hfill \\ & \stackrel{\left(2\right)}{=}& \mu \circ (id\otimes \epsilon \otimes id)\hfill \end{array}$$

and so we have

$$\mu \circ (id\otimes \mu )\circ (id\otimes S_l\otimes id)\circ (E_{rl}\otimes id)=\mu \circ (id\epsilon \otimes id).$$

(6)

Then,

$$\begin{array}{ccc}& & \mu \circ (id\otimes \mu )\circ (id\otimes S_l\otimes id)\circ (id\otimes \mu \otimes id)\hfill \\ & & \circ (id\otimes \tau \otimes id)\circ (E_{rl}\otimes id)\circ (id\otimes \tau \otimes id)\circ (E_{rl}\otimes i{d}^2)\hfill \\ & =& \mu \circ (id\otimes \mu )\circ (id\otimes S_l\otimes id)\hfill \\ & & \circ \underline{(\mu \otimes \mu \otimes id)\circ (id\otimes \tau \otimes i{d}^2)\circ (i{d}^2\otimes \Delta \otimes id)\circ (E_{rl}\otimes i{d}^2)}\hfill \\ & =& \mu \circ (id\otimes \mu )\circ (id\otimes S_l\otimes id)\circ \underline{(\mu \otimes \mu \otimes id)\circ (id\otimes \tau \otimes i{d}^2)\circ (E_{rl}\otimes \Delta \otimes id)}\hfill \\ & \stackrel{\left(5\right)}{=}& \underline{\mu \circ (id\otimes \mu )\circ (id\otimes S_{l}id)\circ (E_{rl}\otimes id)}\circ (id\otimes \mu \otimes id)\hfill \\ & \stackrel{\left(6\right)}{=}& \mu \circ (id\otimes \epsilon \otimes id)\circ (id\otimes \mu \otimes id)\hfill \\ & =& \underline{\mu \circ (id\otimes \epsilon \otimes id)}\circ (i{d}^2\otimes \epsilon \otimes id)\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{e} \epsilon \mathrm{i}\mathrm{s} \mathrm{a}\mathrm{n} \mathrm{a}\mathrm{l}\mathrm{g}\mathrm{e}\mathrm{b}\mathrm{r}\mathrm{a} \mathrm{m}\mathrm{a}\mathrm{p}\hfill \\ & \stackrel{\left(6\right)}{=}& \underline{\mu \circ (id\otimes \mu )}\circ (id\otimes S_{l}id)\circ (E_{rl}\otimes id)\circ (i{d}^2\otimes \epsilon \otimes id)\hfill \end{array}$$

$$\begin{array}{ccc}& =& \mu \circ (\mu \otimes id)\circ (id\otimes S_l\otimes id)\circ (i{d}^2\otimes \epsilon \otimes id)\circ (E_{rl}\otimes i{d}^2)\hfill \\ & =& \mu \circ (\underline{\mu \circ (id\otimes \epsilon \otimes id)}\otimes id)\circ (i{d}^2{S}_l\otimes id)\circ (id\otimes \tau \otimes id)\circ (E_{rl}\otimes i{d}^2)\hfill \\ & \stackrel{\left(6\right)}{=}& \underbrace{\mu \circ (\mu \otimes id)\circ (id\otimes \mu \otimes id)\circ (id\otimes S_l\otimes i{d}^2)}\circ \underline{(E_{rl}\otimes i{d}^2)}\hfill \\ & & \underline{\circ (i{d}^2\otimes S_l\otimes id)}\circ (id\otimes \tau \otimes id)\circ (E_{rl}\otimes i{d}^2)\hfill \end{array}$$

$$\begin{array}{ccc}& =& \mu \circ (id\otimes \mu )\circ (id\otimes S_l\otimes id)\circ (i{d}^2\otimes \mu )\circ (i{d}^2\otimes S_l\otimes id)\hfill \\ & & \circ \underline{(E_{rl}\otimes i{d}^2)\circ (id\otimes \tau \otimes id)}\circ (E_{rl}\otimes i{d}^2)\hfill \\ & =& \mu \circ (id\otimes \mu )\circ \underline{(id\otimes S_l\otimes id)\circ (i{d}^2\otimes \mu )\circ (i{d}^2\otimes S_l\otimes id)}\hfill \\ & & \circ (id\otimes \tau \otimes id)\circ (id\otimes \tau \otimes id)\circ \left(E_{rl}id\right)\circ (id\otimes \tau \otimes id)\circ (E_{rl}\otimes i{d}^2)\hfill \\ & =& \mu \circ (id\otimes \mu )\circ (id\otimes \mu \otimes id)\circ (id\otimes \tau \otimes id)\circ (id\otimes S_l\otimes S_l\otimes id)\hfill \\ & & \circ (id\otimes \tau \otimes id)\circ (id\otimes \tau \otimes id)\circ \left(E_{rl}id\right)\circ (id\otimes \tau \otimes id)\circ (E_{rl}\otimes i{d}^2),\hfill \end{array}$$

and hence

$$\begin{array}{ccc}& & \mu \circ (id\otimes \mu )\circ (id\otimes S_l\otimes id)\circ (id\otimes \mu \otimes id)\hfill \\ & & =\mu \circ (id\otimes \mu )\circ (id\otimes \mu \otimes id)\circ (id\otimes \tau \otimes id)\circ (id\otimes S_l\otimes S_l\otimes id).\hfill \end{array}$$

Since the product in A is nondegenerate, we get $S_l\left(ab\right)=S_l\left(b\right)S_l\left(a\right)$ for all $a,b\in A$, as required.

Similarly, we define a map $S_r:A⟶End\left(A\right)$ by

$$m(id\otimes S_r)=((id\otimes \epsilon )\circ E_{rl}^{-1}).$$

For $S_r$, by an analogous calculation, it is not hard to see that $S_r$ is an algebra antihomomorphism from A to $\mathbb{R}\left(A\right)$, and we have

$$\mu \circ (\mu \otimes id)\circ (id\otimes S_r\otimes id)\circ (E_{lr}\otimes id)=\mu \circ (id\epsilon \otimes id).$$

(7)

By Equations (6) and (7), we obtain that

$$\mu \circ (id\otimes \mu )\circ (id\otimes S_l\otimes id)=\mu \circ (\mu \otimes id)\circ (id{S}_r\otimes id),$$

which means that $(S_l,S_r)$ is a multiplier of A. We denote $S:A⟶\mathbb{M}\left(A\right)$ by the morphism in $C$ given by $S_l$ and $S_r$. Then one has thus proved so far that $S:A⟶\mathbb{M}\left(A\right)$ is an algebra antihomomorphism.

In a similar way, one defines ${\tilde{S}}_l\in \mathbb{L}\left(A\right)$ and ${\tilde{S}}_r\in \mathbb{R}\left(A\right)$ by

$$m({\tilde{S}}_l\otimes id)=(\epsilon \otimes id)\circ E_{ll}^{-1}\circ \tau \mathrm{a}\mathrm{n}\mathrm{d}m(id\otimes {\tilde{S}}_l)=(id\otimes \epsilon )\circ E_{rr}^{-1}\circ \tau .$$

Based on Lemma 4, we have ${\left(E_{lr}^{cop}\right)}^{-1}=E_{ll}^{-1}\circ \tau$ and ${\left(E_{rl}^{cop}\right)}^{-1}=E_{rr}^{-1}\circ \tau$. The above discussion applied to $A^{cop}$ verifies that ${\tilde{S}}_l$ and ${\tilde{S}}_r$ determine an algebra antihomomorphism $\tilde{S}:A⟶\mathbb{M}\left(A\right)$.

Now we want to show that S and $\tilde{S}$ define morphisms in $C$ from A into A which are inverse to each other. To do this, we compute

$$\begin{array}{ccc}& & \mu \circ (id\otimes \mu )\circ (id\otimes \mu \otimes id)\circ (id\otimes \tilde{S}i{d}^2)\circ (id\otimes \tau \otimes id)\hfill \\ & =& \mu \circ (\mu \otimes id)\circ (id\otimes \tilde{S}id)\circ (i{d}^2\otimes \mu )\circ (id\otimes \tau \otimes id)\hfill \\ & =& \mu \circ (\mu \otimes id)\circ (id\otimes \tilde{S}id)\hfill \\ & & \circ \underline{(i{d}^2\otimes \mu )\circ (id\otimes \tau \otimes id)\circ (id\otimes E_{lr}\otimes id)}\circ (id\otimes E_{lr}^{-1}\otimes id)\hfill \\ & =& \mu \circ (\mu \otimes id)\circ \underline{(id\otimes \tilde{S}id)\circ (id\otimes \mu \otimes id)}\hfill \\ & & \circ (i{d}^3\otimes \mu )\circ (i{d}^2\otimes \tau \otimes id)\circ (id\otimes {\Delta }^{cop}\otimes i{d}^2)\circ (id\otimes E_{lr}^{-1}\otimes id)\hfill \end{array}$$

$$\begin{array}{ccc}& =& \mu \circ (\mu \otimes id)\circ (id\otimes \mu \otimes id)\circ (id\otimes \tilde{S}\otimes \tilde{S}\otimes id)\circ \underline{(id\otimes \tau \otimes id)\circ (i{d}^3\otimes \mu )}\hfill \\ & & \underline{\circ (i{d}^2\otimes \tau \otimes id)\circ (id\otimes {\Delta }^{cop}\otimes i{d}^2)\circ (id\otimes \tau \otimes id)}\circ (id\otimes \tau \circ E_{lr}^{-1}\otimes id)\hfill \\ & & \mathrm{since} \tilde{S} \mathrm{is} \mathrm{an} \mathrm{algebra} \mathrm{antihomomorphism}\hfill \\ & =& \underbrace{\mu \circ (\mu \otimes id)\circ (id\otimes \mu \otimes id)}\circ \underline{(id\otimes \tilde{S}\otimes \tilde{S}\otimes id)}\hfill \\ & & \circ (i{d}^2\otimes E_{lr}^{cop})\circ (id\otimes \tau \circ E_{lr}^{-1}\otimes id)\hfill \\ & =& \underline{\mu \circ (id\otimes \mu )\circ (id\otimes \mu \otimes id)\circ (id\otimes \tilde{S}i{d}^2)}\circ (i{d}^2\otimes \tilde{S}\otimes id)\hfill \\ & & \circ (i{d}^2\otimes E_{lr}^{cop})\circ (id\otimes \tau \circ E_{lr}^{-1}\otimes id)\hfill \\ & =& \mu \circ (id\otimes \mu )\circ (id\otimes \tilde{S}id)\circ \underline{(i{d}^2\otimes \mu )\circ (i{d}^2\otimes \tilde{S}\otimes id)}\hfill \\ & & \underline{\circ (i{d}^2\otimes E_{lr}^{cop})}\circ (id\otimes \tau \circ E_{lr}^{-1}\otimes id)\hfill \end{array}$$

$$\begin{array}{ccc}& =& \mu \circ (id\otimes \mu )\circ (id\otimes \tilde{S}id)\circ \underline{(i{d}^2\otimes \epsilon \otimes id)\circ (id\otimes \tau \circ E_{lr}^{-1}\otimes id)}\hfill \\ & =& \mu \circ (id\otimes \mu )\circ (id\otimes \tilde{S}id)\circ (id\otimes \underline{(\epsilon \otimes id)\circ E_{lr}^{-1}}\otimes id)\hfill \\ & =& \mu \circ (id\otimes \mu )\circ (id\otimes \tilde{S}id)\circ (id\otimes \mu \otimes id)\circ (id\otimes S\otimes i{d}^2)\hfill \end{array}$$

where in the last step of the computations above we use the definitions of $\tilde{S}$ and S.

As a consequence, we get the following equation $\mu \circ (\tilde{S}\otimes id)\circ \tau =\tilde{S}\circ \mu (S\otimes id)$. Here both sides are regarded as morphisms from $A\otimes A$ into $M\left(A\right)$ in $C$. By this equation, we find that $\tilde{S}=(\tilde{S}\otimes \epsilon )=\tilde{S}\circ \mu \circ (S\otimes id)\circ E_{lr}=\mu \circ (\tilde{S}\otimes id)\circ \tau \circ E_{lr}$. This proves that $\tilde{S}$ defines a morphism from A to itself. Similarly, we find that S is also a morphism from A to A. Again, we find that $\mu \circ (\mu \otimes id)\circ (id\otimes \tilde{S}\otimes id)=\mu \circ (\mu \otimes id)\circ (id\otimes \tilde{S}\otimes \tilde{S}\circ S)$. This implies that $\tilde{S}\circ S=id$. Similarly, we get $S\circ \tilde{S}=id$.

By the definition of S, we see that elements of the form $S\left(a\right)b$ span A for any $a,b\in A$. Similarly, $a\tilde{S}\left(b\right)$ span A. Similar to the proof of Proposition 5.2 in [15], we have $S\left(A\right)\subseteq A$ and $\tilde{S}\left(A\right)\subseteq A$.

By Equations (6) and (7), we find that

$${\mu }_A\circ (S\otimes id)\circ E_{lr}=\epsilon \otimes id,{\mu }_A\circ (id\otimes S)\circ E_{rl}=id\otimes \epsilon$$

as desired. Furthermore, these two equations imply the uniqueness of S.

We have to show that S is an antihomomorphism of coalgebras. By Equations (3) and (4), we find

$$(id\otimes \mu )\circ ((id\otimes S)\circ E_{rl}\otimes id)=(id\otimes \mu )\circ (E_{lr}^{-1}\circ E_{ll}\otimes id).$$

This implies that

$$E_{lr}^{-1}\circ E_{ll}=(id\otimes S)\circ E_{rl}.$$

(8)

Similarly, we obtain that

$$E_{ll}^{-1}\circ E_{lr}={\left(E_{lr}^{-1}\right)}^{cop}\circ E_{ll}^{cop}=\left(id{S}^{-1}\right)\circ E_{rl}^{cop}=(id\otimes S^{-1})\circ \tau \circ E_{rr}.$$

(9)

By Equations (8) and (9), we find

$$E_{rr}\circ (S\otimes S)=(S\otimes id)\circ \tau \circ E_{rl}^{-1}\circ (S\otimes id).$$

(10)

From the definition of S and Equation (2), we compute

$$\begin{array}{ccc}& & \underline{(id\otimes \mu )}\circ (E_{rl}^{-1}\otimes id)\circ (\tau \otimes id)\circ (id\otimes S\otimes id)\hfill \\ & =& \underline{(id\otimes \epsilon \otimes id)}\circ (id\otimes E_{lr})\circ \left(E_{rl}^{-1}id\right)\circ (\tau \otimes id)\circ (id\otimes S\otimes id)\hfill \\ & =& (\mu \otimes id)\circ (id\otimes S\otimes id)\circ \underline{(E_{rl}\otimes id)\circ (id\otimes E_{lr})}\hfill \\ & & \circ (E_{rl}^{-1}\otimes id)\circ \underbrace{(\tau \otimes id)\circ (id\otimes S\otimes id)}\hfill \\ & \stackrel{\left(1\right)}{=}& (\mu \otimes id)\circ (id\otimes S\otimes id)\circ \underline{(id\otimes E_{lr})\circ (S\otimes i{d}^2)}\circ (\tau \otimes id)\hfill \end{array}$$

$$\begin{array}{ccc}& =& \underline{(\mu \otimes id)\circ (S\otimes S\otimes id)}\circ (id\otimes E_{lr})\circ (\tau \otimes id)\hfill \\ & =& (S\otimes id)\circ ({\mu }^{op}\otimes id)\circ \underline{(id\otimes E_{lr})}\circ (\tau \otimes id)\hfill \\ & =& (S\otimes id)\circ \underline{({\mu }^{op}\otimes id)\circ (i{d}^2\otimes \mu )}\circ (id\otimes \Delta \otimes id)\circ (\tau \otimes id)\hfill \\ & =& \underline{(S\otimes id)\circ (id\otimes \mu )}\circ (\underbrace{({\mu }^{op}\otimes id)\circ (id\otimes \Delta )\circ \tau }\otimes id)\hfill \\ & =& (id\otimes \mu )\circ (S\otimes i{d}^2)\circ (E_{ll}\otimes id).\hfill \end{array}$$

This implies

$$(S\otimes id)\circ E_{ll}=E_{rl}^{-1}\circ \tau \circ (id\otimes S).$$

(11)

By Equations (10) and (11) we find

$$\begin{array}{ccc}& & \underline{(id\otimes {\mu }^{op})\circ (\Delta \otimes id)}\circ (S\otimes id)\circ (id\otimes S)\hfill \\ & =& E_{rr}\circ (S\otimes S)\circ \tau \hfill \\ & =& (S\otimes id)\circ \tau \circ E_{rl}^{-1}\circ (S\otimes id)\circ \tau \hfill \\ & =& (S\otimes S)\circ \tau \circ E_{ll}\hfill \\ & =& (id\otimes {\mu }^{op})\circ (S\otimes S\otimes id)\circ (\tau \otimes id)\circ (\Delta \otimes id)\circ (id\otimes S).\hfill \end{array}$$

This implies $\Delta \circ S=(S\otimes S)\circ \tau \circ \Delta$ and proves that S is an antihomomorphism of coalgebras in $C$. Finally, since $S:A⟶A^{op}$ is an algebra and coalgebra homomorphism we find that $E_{lr}\circ (S\otimes S)=(S\otimes S)\circ E_{lr}$. Hence one gets

$$(\epsilon \otimes id)\circ (S\otimes S)\circ E_{lr}=(\epsilon \otimes id)\circ E_{lr}\circ (S\otimes S)=\mu \circ \circ (S\otimes S)=S\circ \mu =(\epsilon \otimes S)\circ E_{lr}.$$

From this equation we deduce that $(\epsilon \circ S)\otimes S=\epsilon \otimes S$, because $E_{lr}$ is bijective. Since the product in A is nondegenerate and S is bijective, this shows $\epsilon \circ S=\epsilon$. Therefore, we have shown that there is a unique map S satisfying the desired properties.

Conversely, we define the inverse $E_{lr}^{-1}$ of $E_{lr}$ by

$$E_{lr}^{-1}=(id\otimes \mu )\circ (id\otimes S\otimes id)\circ (\Delta \otimes id)$$

where both sides are viewed as morphisms from $A\otimes A$ to $M(A\otimes A)$ in $C$. In particular, the image of the last map is contained in $A\otimes A$. We calculate

$$\begin{array}{ccc}& & (\mu \otimes \mu )\circ (id\otimes \tau \otimes id)\circ (E_{lr}^{-1}\circ E_{lr}\otimes i{d}^2)\hfill \\ & =& (\mu \otimes \mu )\circ (id\otimes \tau \otimes id)\hfill \\ & & \circ \underline{(id\otimes \mu \otimes i{d}^2)\circ (id\otimes Si{d}^3)}\circ (\Delta \otimes i{d}^3)\circ \underline{(E_{lr}\otimes i{d}^2)}\hfill \\ & =& (\mu \otimes \mu )\circ (id\otimes \tau \otimes id)\circ (id\otimes \epsilon \otimes i{d}^2)\circ (\Delta \otimes i{d}^3)\hfill \\ & =& (\mu \otimes \mu )\circ (id\otimes \tau \otimes id).\hfill \end{array}$$

Similarly, we find that $(\mu \otimes \mu )\circ (id\otimes \tau \otimes id)\circ (E_{lr}^{-1}\circ E_{lr}\otimes i{d}^2)=(\mu \otimes \mu )\circ (id\otimes \tau \otimes id)$, which proves that $E_{lr}$ is an isomorphism in $C$.

Analogously, the other entwining maps can be shown to be isomorphisms in $C$ as well.

Finally, we compute

$$\begin{array}{ccc}\hfill (id\otimes S)E_{rr}& =& (id\otimes S)\left(E_{rl}\right)(id\otimes S)=E_{lr}^{-1}(id\otimes S).\hfill \end{array}$$

Hence

$$E_{rr}=(id\otimes S^{-1})\circ E_{lr}^{-1}\circ (id\otimes S).$$

It follows that $E_{rr}(A\otimes A)\subseteq A\otimes A$. By a similar argument, we find $E_{ll}(A\otimes A)\subseteq A\otimes A$. This proves that A is regular.

This completes the proof of the theorem.   □

As a corollary of Theorem 1, we find the following characterization of a braided Hopf algebra in [11]:

Corollary 3.

H is a braided Hopf algebra in $C$ if and only if the associated $E_{lr}$ and $E_{rl}$ are bijective in $C$.

We also find the following important result:

Theorem 2.

If A is an algebra in $C$ with an identity and Δ a coproduct on A such that $(A,\Delta )$ is a braided Hopf algebra, then $(A,\Delta )$ is a braided multiplier Hopf algebra. Conversely, if $(A,\Delta )$ is a braided multiplier Hopf algebra and if A has an identity, then $(A,\Delta )$ is a braided Hopf algebra.

Proof. 

If $(A,\Delta )$ is a braided Hopf algebra in $C$, then we have $\Delta :A⟶A\otimes A=M(A\otimes A)$. In this case, the structures on a braided Hopf algebra $(A,\Delta )$ make $(A,\Delta )$ into a braided multiplier Hopf algebra. In fact, if S is the antipode, the maps $R_{lr}$, $R_{rl}$ defined on $A\otimes A$ by

$$\begin{array}{ccc}& & R_{lr}=(id\otimes {\mu }_A)(id\otimes S\otimes id)({\Delta }_A\otimes id),\hfill \\ & & R_{rl}=(m_A\otimes id)(id\otimes S\otimes id)(id\otimes {\Delta }_A).\hfill \end{array}$$

are determined to be the inverses of $E_{lr}$ and $E_{rl}$. For example, we can compute

$$\begin{array}{ccc}\hfill R_{lr}\circ E_{lr}& =& (id\otimes {\mu }_A)(id\otimes S\otimes id)({\Delta }_A\otimes id)(id\otimes {\mu }_A)({\Delta }_A\otimes id)\hfill \\ & =& (id\otimes {\mu }_A)(id\otimes {\mu }_A\otimes id)(id\otimes S\otimes id\otimes id)({\Delta }_A\otimes id\otimes id)({\Delta }_A\otimes id)\hfill \\ & =& (id\otimes {\mu }_A)(id\otimes {\mu }_A\otimes id)(id\otimes S\otimes id\otimes id)(id\otimes {\Delta }_A\otimes id)({\Delta }_A\otimes id)\hfill \\ & =& (id\otimes {\mu }_A)[id\otimes \left({\mu }_A(S\otimes id){\Delta }_A\right)\otimes id]({\Delta }_A\otimes id)\hfill \\ & =& (id\otimes {\mu }_A)[id\otimes {\epsilon }_A\otimes id]({\Delta }_A\otimes id)\hfill \\ & =& ({\mu }_A(id\otimes {\epsilon }_A){\Delta }_A\otimes id)=id\otimes id,\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill E_{lr}\circ R_{lr}& =& (id\otimes {\mu }_A)({\Delta }_A\otimes id)(id\otimes {\mu }_A)(id\otimes S\otimes id)({\Delta }_A\otimes id)\hfill \\ & =& (id\otimes {\mu }_A)(id\otimes id\otimes {\mu }_A)(id\otimes id\otimes S\otimes id)(id\otimes {\Delta }_A\otimes id)({\Delta }_A\otimes id)\hfill \\ & =& (id\otimes {\mu }_A)(id\otimes {\mu }_A\otimes id)(id\otimes id\otimes S\otimes id)(id\otimes {\Delta }_A\otimes id)({\Delta }_A\otimes id)\hfill \\ & =& (id\otimes {\mu }_A)[id\otimes \left({\mu }_A(id\otimes S){\Delta }_A\right)\otimes id]({\Delta }_A\otimes id)\hfill \\ & =& (id\otimes {\mu }_A)[id\otimes {\epsilon }_A\otimes id]({\Delta }_A\otimes id)\hfill \\ & =& ({\mu }_A(id\otimes {\epsilon }_A){\Delta }_A\otimes id)=id\otimes id.\hfill \end{array}$$

Therefore, we have $R_{lr}\circ E_{lr}=id\otimes id$ and $E_{lr}\circ R_{lr}=id\otimes id$.

Conversely, if $(A,\Delta )$ is a braided multiplier Hopf algebra and if A has an identity, then we find that $M(A\otimes A)=A\otimes A$ and so $(A,\Delta )$ is a braided Hopf algebra. □

5. Braided Algebraic Quantum Groups

In this section we introduce the concept of a braided algebraic quantum group in a symmetric category $C ={}_H\mathbb{M}^R$ as in Section 4, which is similar to the one in [16], and develop the main result established as in the last section.

Definition 6.

Let Δ be a coproduct on A. Then a non-zero morphism $\phi :A⟶\mathbb{k}$ in $C$ is called left integral if

$$\left(\mathcal{B}\widehat{\otimes }\phi \right)\circ E_{rl}=(id\otimes \phi ).$$

Similarly, a non-zero morphism $\psi :A⟶\mathbb{k}$ in $C$ is called right integral if

$$\left(\psi \widehat{\otimes }\mathcal{B}\right)\circ E_{lr}=(\psi \otimes id).$$

Definition 7.

Let $\Delta :A⟶M(A\otimes A)$ be a coproduct in $C$. We say that A is a braided quantum group if the entwining maps $E_{lr},E_{ll},E_{rl}$ and $E_{rr}$ are bijective so that $E_{lr}(A\otimes A),E_{rl}(A\otimes A),E_{ll}(A\otimes A),E_{rr}(A\otimes A)\subseteq A\otimes A$ and there is a faithful left integral φ on A.

Lemma 5.

Let A be a regular braided multiplier Hopf algebra in $C$. If φ is a left integral on A, then φ is faithful.

Proof. 

The proof of the following lemma is parallel to the one for the regular multiplier Hopf algebra [3] (also see [3]), so we omit it.   □

The following is the main result of this section:

Theorem 3.

Let $\Delta :A⟶M(A\otimes A)$ be a coproduct in the braided monoidal category $C$ such that the images of all associated entwining maps belong to $A\otimes A$. Assume that $\phi :A⟶\mathbb{k}$ is a left integral. Then A is a braided quantum group if and only if there exists a unique counit ε on A and an isomorphism $S:A⟶A$ in $C$ which is both an algebra antihomomorphism and a coalgebra antihomomorphism such that

$${\mu }_A\circ (S\otimes id)\circ E_{lr}=\epsilon \otimes id,{\mu }_A\circ (id\otimes S)\circ E_{rl}=id\otimes \epsilon and\epsilon \circ S=\epsilon .$$

Proof. 

By Theorem 1, Lemma 5 and Proposition 5, we can just construct the map $\epsilon$. Now we define a linear map $\epsilon :A⟶\mathbb{k}$ by $\epsilon =(\phi \circ \mu \circ E_{rl}^{-1})$ for all $a\in A$. Obviously, $\epsilon$ is a morphism in $C$. Now, we define a morphism $E:A⟶\mathbb{L}\left(A\right)$ by $m(E\otimes id)=(\mu \circ E_{lr}^{-1})$. By Lemma 4(1), one can easily see that the morphism E is well-defined. By Equation (1), we compute

$$\begin{array}{ccc}& & (id\otimes \mu )\circ (id\otimes E\otimes id)\hfill \\ & =& \underbrace{(id\otimes \mu )\circ (id\otimes E\otimes id)}\circ (E_{rl}\otimes id)\circ \underline{(E_{rl}^{-1}\otimes id)\circ (id\otimes E_{lr})}\circ (id\otimes E_{lr}^{-1})\hfill \\ & =& (id\otimes \mu )\circ (id\otimes E_{lr}^{-1})\circ (\underline{E_{rl}\otimes id)\circ \left(id{E}_{lr}\right)}\circ (E_{rl}^{-1}\otimes id)\circ (id\otimes E_{lr}^{-1})\hfill \\ & =& (id\otimes \mu )\circ \underbrace{(id\otimes E_{lr}^{-1})\circ (id\otimes E_{lr})}\circ \underline{(E_{rl}\otimes id)\circ (E_{rl}^{-1}\otimes id)}\circ (id\otimes E_{lr}^{-1})\hfill \\ & =& (id\otimes \mu )\circ (id\otimes E_{lr}^{-1})\hfill \\ & =& (id\otimes \mu )\circ \underline{(id\otimes E_{lr}^{-1})\circ (E_{rl}\otimes id)}\circ (E_{rl}^{-1}\otimes id)\hfill \\ & =& \underline{(id\otimes \mu )\circ (E_{rl}\otimes id)\circ (id\otimes E_{lr}^{-1})}\circ (E_{rl}^{-1}\otimes id)\hfill \\ & =& (\mu \otimes id)\circ (E_{rl}^{-1}\otimes id)\mathrm{by} \mathrm{Lemma} 4\left(2\right).\hfill \end{array}$$

This implies

$$(id\otimes \mu )\circ (id\otimes E\otimes id)=\mu \circ E_{rl}^{-1}\otimes id.$$

(12)

On one hand, applying $\phi \otimes id$ to Equation (12), we obtain

$$m(E\otimes id)=m[\phi (\mu \circ E_{rl}^{-1})\otimes ]b=(\epsilon \otimes id).$$

Because the product on A is nondegenerate, we hence have $(\epsilon \otimes id)\circ E_{lr}=\mu \circ (E\otimes id)\circ E_{lr}=\mu$.

On the other hand, by Equation (12), we get $\mu \circ E_{rl}^{-1}=(id\otimes \mu )\circ (id\otimes E\otimes id)=(id\otimes \epsilon \otimes id)$, which implies $(id\otimes \epsilon )\circ E_{rl}=\mu$. This proves that Equation (2) holds.

To conclude that $\epsilon$ is a counit, we have to show finally that $\epsilon$ is an algebra homomorphism. We compute

$$\begin{array}{ccc}& & (id\otimes \epsilon )\circ (id\otimes \mu )\hfill \\ & =& (id\otimes \epsilon )\circ \underline{(id\otimes \mu )\circ (id\otimes \tau )\circ (E_{rl}\otimes id)\circ \left(id\tau \right)}\hfill \\ & & \circ (id\otimes \tau )\circ (E_{rl}^{-1}\otimes id)\circ (id\otimes \tau )\hfill \\ & =& (id\otimes \epsilon )\circ (\mu \otimes \mu )\circ (id\otimes \tau \otimes id)\circ (i{d}^2\otimes \Delta )\hfill \\ & & \circ (id\otimes \tau )\circ (E_{rl}^{-1}\otimes id)\circ (id\otimes \tau )\hfill \\ & =& (id\otimes \epsilon )\circ \underline{(\mu \otimes \mu )\circ (id\otimes \tau \otimes id)\circ (i{d}^2\otimes \Delta )\circ (E_{rl}\otimes id)}\circ (E_{rl}^{-1}\otimes id)\hfill \\ & & \circ (id\otimes \tau )\circ (E_{rl}^{-1}\otimes id)\circ (id\otimes \tau )\hfill \\ & \stackrel{\left(5\right)}{=}& \underline{(id\otimes \epsilon )\circ E_{rl}}\circ \left(id\mu \right)\circ (E_{rl}^{-1}\otimes id)\circ (id\otimes \tau )\circ (E_{rl}^{-1}\otimes id)\circ (id\otimes \tau )\hfill \end{array}$$

$$\begin{array}{ccc}& =& \mu \circ \underline{(\mu \otimes id)}\circ (E_{rl}^{-1}\otimes id)\circ (id\otimes \tau )\circ (E_{rl}^{-1}\otimes id)\circ (id\otimes \tau )\hfill \\ & =& \mu \circ (id\otimes \epsilon \otimes id)\circ \underline{(E_{rl}\otimes id)\circ (E_{rl}^{-1}\otimes id)}\circ (id\otimes \tau )\circ (E_{rl}^{-1}\otimes id)\circ (id\otimes \tau )\hfill \\ & =& \underline{\mu }\circ (id\otimes \epsilon \otimes id)\circ (id\otimes \tau )\circ (E_{rl}^{-1}\otimes id)\circ (id\otimes \tau )\hfill \\ & =& (id\otimes \epsilon )\circ (id\otimes \epsilon \otimes id)\circ (id\otimes \tau )\circ (E_{rl}\otimes id)\circ (id\otimes \tau )\hfill \\ & & \circ (id\otimes \tau )\circ (E_{rl}^{-1}\otimes id)\circ (id\otimes \tau )\hfill \\ & =& (id\otimes \epsilon )\circ (id\otimes \epsilon \otimes id)\hfill \end{array}$$

which implies that $\epsilon \circ m=\epsilon \otimes \epsilon$. Therefore $\epsilon$ is an algebra homomorphism.

This completes the proof of the theorem.   □

6. Conclusions and Further Research

In this paper we study the notion of braided multiplier rings and their properties (see Propositions 1 and 4) and establish a braided multiplier Hopf algebra in a symmetric category $C$ (see Theorem 2). Then we develop the concept of a braided quantum group (see Theorem 3). These results generalize the existing work in the references.

During our research process, we found that certain questions were worth studying, for example, the question of how to construct the Drinfeld double for braided algebraic quantums. In fact, the Drinfeld double plays a very important role in the classical Hopf algebra theory, with a very rich structure, and is also an important source of solutions to the quantum Yang–Baxter equation. As everyone knows, the quasitriangular structure can give rise to the solution of the quantum Yang–Baxter equation. Then, another question was that of how to define a quasitriangular braided algebraic quantum, and so on.

Finally, we believe that the idea of this paper can significantly enhance student s’ competencies in scientific programming, abstract problem-solving, and research-oriented thinking, providing an effective paradigm for synergizing advanced abstract thinking training with talent cultivation in STEM education. We also hope that the braid algebraic quantum groups we study not only have good application value in scientific research, but also help train undergraduate students’ innovative thinking abilities. Next, we will use the main ideas of this article in students’ mathematical modeling and graduation projects. Moreover, this makes our article equivalent to the literature [17] in terms of cultivating students’ creativity.