The structures of Hopf $\ast$-algebra on Radford algebras

TL;DR

This paper classifies all possible Hopf $\ast$-algebra structures on Radford algebras over the complex numbers, providing explicit descriptions and an equivalence classification.

Contribution

It explicitly determines and classifies all Hopf $\ast$-structures on Radford algebras over $\mathbb{C}$, a novel comprehensive analysis.

Findings

All $*$-structures on Radford algebras are explicitly characterized.

Hopf $\ast$-algebra structures are classified up to equivalence.

Provides a complete description of these structures over $\mathbb{C}$.

Abstract

We investigate the structures of Hopf $\ast$-algebra on the Radford algebras over $\mathbb {C}$. All the $*$-structures on $H$ are explicitly given. Moreover, these Hopf $*$-algebra structures are classified up to equivalence.

Equations

$$\psi ({\lambda }_1{v}_1+{\lambda }_2{v}_2)=\overline{{\lambda }_1}\psi (v_1)+\overline{{\lambda }_2}\psi (v_2),\forall v_1,v_2\in V,\forall {\lambda }_1,{\lambda }_2\in \mathbb{C}.$$

$$\psi (a{a}^{\prime })=\psi (a)\psi (a^{\prime }),\psi (1)=1,\forall a,a^{\prime }\in A,$$

$$\psi (a{a}^{\prime })=\psi (a^{\prime })\psi (a),\psi (1)=1,\forall a,a^{\prime }\in A.$$

$$\sum \psi (c{)}_1\otimes \psi (c{)}_2=\sum \psi (c_1)\otimes \psi (c_2),\epsilon (\psi (c))=\overline{\epsilon (c)},\forall c\in C,$$

$$\sum \psi (c{)}_1\otimes \psi (c{)}_2=\sum \psi (c_2)\otimes \psi (c_1),\epsilon (\psi (c))=\overline{\epsilon (c)},\forall c\in C.$$

$$\begin{array}[]{ll}(h^{\ast})^{\ast}=h,&(hl)^{\ast}={l}^{\ast}{h}^{\ast},\\ \sum{(h^{\ast})_{1}\otimes(h^{\ast})_{2}}=\sum{(h_{1})^{\ast}\otimes(h_{2})^{\ast}},&S(S(h^{*})^{*})=h,\\ \end{array}$$

$$\epsilon (h^{\ast })=\overline{\epsilon (h)},\forall h\in H.$$

$$\begin{array}[]{lllll}g^{n}=1,&x^{n}=y^{n}=0,&xg=\omega gx,&gy=\omega yg,&xy=\omega yx.\\ \end{array}$$

$$\begin{array}[]{lll}\triangle(g)=g\otimes g,&\varepsilon(g)=1,&S(g)=g^{n-1},\\ \triangle(x)=x\otimes g+1\otimes x,&\varepsilon(x)=0,&S(x)=-xg^{n-1},\\ \triangle(y)=y\otimes g+1\otimes y,&\varepsilon(y)=0,&S(y)=-yg^{n-1}.\end{array}$$

$$g^{\ast }=g,x^{\ast }=\alpha x,y^{\ast }=\beta y,$$

$$\begin{array}[]{rl}S(S(g^{\ast})^{*})&=S(S(g)^{\ast}))=S((g^{-1})^{\ast})=S((g^{\ast})^{-1})=S(g^{-1})=g,\\ S(S(x^{\ast})^{\ast})&=S(S(\alpha x)^{*})=S((-\alpha xg^{n-1})^{\ast})=S(-\overline{\alpha}(g^{n-1})^{\ast}x^{\ast})\\ &=S(-\overline{\alpha}\alpha g^{n-1}x)=-S(x)S(g^{n-1})=xg^{n-1}g=x,\\ \end{array}$$

$$A=\left(\begin{array}[]{cc}\alpha_{11}&\alpha_{12}\\ \alpha_{21}&\alpha_{22}\\ \end{array}\right)\in M_{2}(\mathbb{C}),$$

$$\overline{A}=\left(\begin{array}[]{cc}\overline{\alpha_{11}}&\overline{\alpha_{12}}\\ \overline{\alpha_{21}}&\overline{\alpha_{22}}\\ \end{array}\right)\in M_{2}(\mathbb{C}).$$

$$g^{\ast }=g,x^{\ast }={\alpha }_{11}x+{\alpha }_{12}y,y^{\ast }={\alpha }_{21}x+{\alpha }_{22}y.$$

$$\begin{array}[]{rl}(x^{*})^{*}&=(\alpha_{11}x+\alpha_{12}y)^{*}=\overline{\alpha_{11}}x^{*}+\overline{\alpha_{12}}y^{*}\\ &=\overline{\alpha_{11}}(\alpha_{11}x+\alpha_{12}y)+\overline{\alpha_{12}}(\alpha_{21}x+\alpha_{22}y)\\ &=(\overline{\alpha_{11}}\alpha_{11}+\overline{\alpha_{12}}\alpha_{21})x+(\overline{\alpha_{11}}\alpha_{12}+\overline{\alpha_{12}}\alpha_{22})y=x.\\ \end{array}$$

$$\begin{array}[]{rl}S(S(g^{*})^{*})&=S(S(g)^{*}))=S(g^{*})=S(g)=g,\\ S(S(x^{*})^{*})&=S(S(\alpha_{11}x+\alpha_{12}y)^{*})=S((-\alpha_{11}xg-\alpha_{12}yg)^{*})\\ &=S((\alpha_{11}gx+\alpha_{12}gy)^{*})=S(\overline{\alpha_{11}}x^{*}g^{*}+\overline{\alpha_{12}}y^{*}g^{*})\\ &=S(\overline{\alpha_{11}}(\alpha_{11}x+\alpha_{12}y)g+\overline{\alpha_{12}}(\alpha_{21}x+\alpha_{22}y)g)\\ &=S(g)S((\overline{\alpha_{11}}\alpha_{11}+\overline{\alpha_{12}}\alpha_{21})x+(\overline{\alpha_{11}}\alpha_{12}+\overline{\alpha_{12}}\alpha_{22})y)\\ &=S(g)S(x)=g(-xg)=x,\\ \end{array}$$

$$△(y^r{x}^s{g}^l)=\sum _{i=0}^r\sum _{j=0}^s{\omega }^{-(r-i)j}{\left(\genfrac{}{}{0ex}{}{r}{i}\right)}_{\omega }{\left(\genfrac{}{}{0ex}{}{s}{j}\right)}_{{\omega }^{-1}}{y}^{r-i}{x}^{s-j}{g}^l\otimes y^i{x}^j{g}^{l+s-j+r-i}.$$

$$(x\otimes g)(1\otimes x)={\omega }^{-1}(1\otimes x)(x\otimes g),(y\otimes g)(1\otimes y)=\omega (1\otimes y)(y\otimes g),$$

$$\begin{array}[]{ll}\triangle(x)^{s}=(1\otimes x+x\otimes g)^{s}=\sum_{j=0}^{s}{\binom{s}{j}}_{\omega^{-1}}{x^{s-j}}\otimes{x^{j}g^{s-j}},\\ \triangle(y)^{r}=(1\otimes y+y\otimes g)^{r}=\sum_{i=0}^{r}{\binom{r}{i}}_{\omega}{y^{r-i}}\otimes{y^{i}}g^{r-i}.\\ \end{array}$$

$$\begin{array}[]{rl}\triangle({y^{r}x^{s}g^{l}})=&\triangle(y)^{r}\triangle(x)^{s}\triangle(g)^{l}\\ =&(1\otimes y+y\otimes g)^{r}(1\otimes x+x\otimes g)^{s}(g\otimes g)^{l}\\ =&\sum_{i=0}^{r}\sum_{j=0}^{s}\binom{r}{i}_{\omega}\binom{s}{j}_{\omega^{-1}}y^{r-i}x^{s-j}g^{l}\otimes y^{i}g^{r-i}x^{j}g^{l+s-j}\\ =&\sum_{i=0}^{r}\sum_{j=0}^{s}\omega^{-(r-i)j}\binom{r}{i}_{\omega}\binom{s}{j}_{\omega^{-1}}y^{r-i}x^{s-j}g^{l}\otimes y^{i}x^{j}g^{l+s-j+r-i}.\\ \end{array}$$

$$\{y^r{x}^s{g}^l\otimes y^{r_1}{x}^{s_1}{g}^{l_1}|0⩽r,r_1,s,s_1,l,l_1<n\}.$$

$$h=\sum _{0⩽r,s,l<n}{\lambda }_{r,s,l}{y}^r{x}^s{g}^l$$

$$h=\sum _{0⩽r,s,l,r_1,s_1,l_1<n}{\lambda }_{r,s,l,r_1,s_1,l_1}{y}^r{x}^s{g}^l\otimes y^{r_1}{x}^{s_1}{g}^{l_1}$$

$$y^r{x}^s{g}^l\otimes y^{r_1}{x}^{s_1}{g}^{l_1}.$$

$$h=\sum _{0⩽r,s,l<n}{\lambda }_{r,s,l}{y}^r{x}^s{g}^l\in G(H),$$

$$△(h)=\sum _{r,s,l}{\lambda }_{r,s,l}△(y^r{x}^s{g}^l),$$

$$y^{r_1}{x}^{s_1}{g}^{l_1}\otimes y^{r_1}{x}^{s_1}{g}^{l_1}$$

$$△(h)=h\otimes g+1\otimes h.$$

$$h_{r,s}=\sum _{l=0}^{n-1}{\lambda }_{r,s,l}{y}^r{x}^s{g}^l.$$

$$△(h_{r,s})=h_{r,s}\otimes g+1\otimes h_{r,s},\forall r,s.$$

Full text

2010 Mathematics Subject Classification. 16G99, 16T05

1. Introduction

Woronowicz studied compact matrix pseudogroup in [14], which is a generalization of compact matrix group. Using the language of $C^{*}$-algebra, Woronowicz described compact matrix pseudogroups as $C^{*}$-algebras endowed with some comultiplications. This induces the concept of Hopf $*$-algebras. In [14, 15, 16], Woronowicz exhibited Hopf $*$-algebra structures on quantum groups in the framework of $C^{*}$-algebras. It was shown that $GL_{q}(2)$, $SL_{q}(2)$ and $U_{q}(sl(2))$ are Hopf $*$-algebras, see [2, 5]. Van Deale [13] studied the Harr measure on a compact quantum group. Podleś [8] studied coquasitriangular Hopf $\ast$-algebras. Tucker-Simmons [12] studied the $*$-structure of module algebras over a Hopf $\ast$-algebra. Recently, we investigated the Hopf $\ast$-algebra structures on $H(1,q)$ over ${\mathbb{C}}$ and classified these $*$-structures up to the equivalence [6].

Radford [9] constructed for every integer $n>1$ a finite dimensional unimodular Hopf algebra with antipode of order 2n and proved that for every even integer there is a finite dimensional Hopf algebra $H$. For more details, the reader is directed to [3, 9, 10].

In this paper, we study the structures of Hopf $\ast$-algebra on the Radford algebra $H$ over the complex number field ${\mathbb{C}}$. This paper is organized as follow. In Section 2, we recall some basic notions about the Hopf $\ast$-algebra, and the Radford algebra $H$. In Section 3, we first describe all structures of Hopf $\ast$-algebra on Radford algebra. It is shown that when $n>2$, a Hopf $*$-algebra structure on $H$ is uniquely determined by a pair $(\alpha,\beta)$ of elements in $\mathbb{C}$ with $|\alpha|=|\beta|=1$, and that when $n=2$, a Hopf $*$-algebra structure on $H$ is uniquely determined by a $2\times 2$-matrix $A$ over $\mathbb{C}$ with $\overline{A}A=I_{2}$. Then we classify the Hopf $*$-algebra structures up to equivalence. It is shown that any two $*$-structures on $H$ are equivalent when $n>2$. When $n=2$, the two $*$-structures determined by two matrices $A$ and $B$, respectively, are equivalent if and only if there exists an invertible $2\times 2$-matrix $\Lambda$ over $\mathbb{C}$ such that $A\overline{\Lambda}=\Lambda B$.

2. Preliminaries

Throughout, let $\mathbb{Z}$, $\mathbb{N}$, $\mathbb{R}$ and $\mathbb{C}$ denotes the all integers, all nonnegative integers, the field of real numbers, and the field of complex numbers, respectively. Let $i\in\mathbb{C}$ be the imaginary unit. For any $\lambda\in\mathbb{C}$, let $\overline{\lambda}$ denote the conjugate complex number of $\lambda$, and let $|\lambda|$ denote the norm of $\lambda$. For a Hopf algebra $H$, we use $\triangle$, $\varepsilon$ and $S$ denote the comultiplication, the counit, and the antipode of $H$ as usual. For the theory of quantum groups and Hopf algebras, we refer to [2, 4, 7, 10, 11]. Let $G(H)$ denote the set of group-like elements in a Hopf algebra $H$, which is a group.

Let $V$ and $W$ be vector spaces over $\mathbb{C}$. A mapping $\psi:V\rightarrow W$ is said to be conjugate-linear (or antilinear) if

[TABLE]

Let $A$ and $B$ be $\mathbb{C}$-algebras. A conjugate-linear map $\psi:{A}\rightarrow{B}$ (resp., $A$) is said to be a conjugate-linear algebra map (resp., a conjugate-linear algebra endomorphism) if

[TABLE]

and $\psi$ is said to be a conjugate-linear antialgebra map (resp., a conjugate-linear antialgebra endomorphism) if

[TABLE]

Let $C$ and $D$ be two coalgebras over $\mathbb{C}$. A conjugate-linear map $\psi:C\rightarrow D$ (resp., $C$) is said to be a conjugate-linear coalgebra map (resp., a conjugate-linear coalgebra endomorphism) if

[TABLE]

and $\psi$ is said to be a conjugate-linear anticoalgebra map (resp., a conjugate-linear anticoalgebra endomorphism) if

[TABLE]

Definition 2.1**.**

Let $H$ be a Hopf algebra over $\mathbb{C}$. A $\ast$-structure on $H$ is a conjugate-linear map $\ast:H\rightarrow H$ such that the following conditions are satisfied:

[TABLE]

where $h,l\in H$. If $H$ is equipped with a $\ast$-structure, then we call $H$ a Hopf $\ast$-algebra. Two $\ast$-structures ${\ast^{\prime}}$ and ${\ast^{\prime\prime}}$ on $H$ are said to be equivalent if there exists a Hopf algebra automorphism $\psi$ of $H$ such that $\psi(h^{*^{\prime}})=\psi(h)^{*^{\prime\prime}}$ for all $h\in H$.

Let $H$ be a Hopf $\ast$-algebra. Then it is not difficult to check that

[TABLE]

Hence, the map $\ast$ is an antilinear coalgebra endomorphism of $H$. $\mathbb{C}=\mathbb{C}1_{H}$ is a subalgebra of $H$. In this case, $\lambda^{*}=\overline{\lambda}$ for any $\lambda\in\mathbb{C}\subseteq H$.

Fix a positive integer $n>1$ and let $\omega\in\mathbb{C}$ be a root of unity of order $n$. The Radford algebra $H$ over $\mathbb{C}$ is generated, as a $\mathbb{C}$-algebra, by $g$, $x$ and $y$ subject to the relations:

[TABLE]

Then $H$ is a Hopf algebra with the coalgebra structure and the antipode given by

[TABLE]

Note that $H$ has a canonical basis $\{g^{l}x^{r}y^{s}|0\leqslant l,r,s<n\}$ over $\mathbb{C}$. For the details, the reader is directed to [3, 9, 10].

3. The structres of Hopf $\ast$-algebras on $H$

Throughout this section, let $H$ be the Radford algebra over $\mathbb{C}$ described in the last section. In this section, we study the $*$-structures on the Hopf algebra $H$. Let $Z(H)$ denote the center of $H$. Note that $H$ is generated, as an algebra over $\mathbb{R}$, by ${g,x,y}$, and $i$ subject to the relations given in the last section together with ${i^{2}}=-1$ and $i\in Z(H)$. In the following, let ${H^{op}}$ denote the opposite algebra of $H$. For any $h,l\in H^{op}$, let $h\cdot l$ denote the product of $h$ and $l$ in $H^{op}$, i.e., $h\cdot l=lh$.

Lemma 3.1**.**

Let $\alpha,\beta\in\mathbb{C}$ with $|\alpha|=|\beta|=1$. Then $H$ is a Hopf $\ast$-algebra with the $\ast$-structure given by

[TABLE]

Proof.

We first prove that the relations given in the lemma together with ${i^{\ast}}={-i}$ give rise to a real antialgebra endomorphism of $H$, i.e., a real algebra map from $H$ to $H^{op}$. Since $|\omega|$=$1$, we have $\omega^{\ast}=\overline{\omega}=\omega^{-1}$. Hence in $H^{op}$, we have $(g^{*})^{n}=g^{n}=1$, $(x^{\ast})^{n}=(\alpha x)^{n}=0$, $x^{\ast}\cdot g^{\ast}=\alpha x\cdot g=\alpha gx=\alpha\omega^{-1}xg=\alpha\omega^{-1}g\cdot x=\omega^{\ast}g^{\ast}\cdot x^{\ast}$ and $x^{\ast}\cdot y^{\ast}=\alpha\beta x\cdot y=\alpha\beta yx=\alpha\beta\omega^{-1}xy=\alpha\beta\omega^{-1}y\cdot x=\omega^{\ast}y^{\ast}\cdot x^{\ast}$. Similarly, one can check that $(y^{*})^{n}=0$ and $g^{\ast}\cdot y^{\ast}=\omega^{\ast}y^{\ast}\cdot g^{\ast}$. We also have $i^{\ast}=-i\in Z(H^{op})$ and $(i^{\ast})^{2}=(-i)^{2}=-1$. This shows that the relations given in the lemma together with ${i^{\ast}=-i}$ determine a real algebra map $\ast:H\rightarrow{H^{op}}$. Then it follows that $\ast$ is a conjugate-linear antialgebra endomorphism of $H$. Hence the composition $\ast\circ\ast$ is a complex algebra endomorphism of $H$. It is not difficult to check that $(h^{\ast})^{\ast}=h$, $\forall h\in\{g,x,y\}$, and so $(h^{\ast})^{\ast}=h$, $\forall h\in{H}$. Thus, $\ast$ is an involution of $H$. Note that both $\triangle\circ\ast$ and $(\ast\otimes\ast)\circ\triangle$ are conjugate-linear antialgebra maps from $H$ to $H\otimes H$. It is easy check that $\triangle(h^{\ast})=\sum(h_{1})^{\ast}\otimes(h_{2})^{\ast}$ for any $h\in\{g,x,y\}$. It follows that $\triangle(h^{\ast})=\sum(h_{1})^{\ast}\otimes(h_{2})^{\ast}$ for all $h\in H$. Similarly, we have $\varepsilon(h^{*})=\overline{\varepsilon(h)}$ for all $h\in H$. Finally, since $S$ is a complex antialgebra endomorphism of $H$ and $\ast$ is a conjugate-linear antialgebra endomorphism of $H$, the map $H\rightarrow H$, $h\mapsto{S(S(h^{\ast})^{\ast})}$ is a complex algebra endomorphism of $H$. Now we have

[TABLE]

and similarly $S(S(y^{\ast})^{\ast})=y$. It follows that $S(S(h^{\ast})^{\ast})=h$ for all $h\in{H}$. ∎

Let $M_{2}(\mathbb{C})$ be the matrix algebra of all $2\times 2$-matrices over $\mathbb{C}$. For a matrix

[TABLE]

let

[TABLE]

Lemma 3.2**.**

Assume that $n=2$ and let A=\left(\begin{array}[]{cc}\alpha_{11}&\alpha_{12}\\ \alpha_{21}&\alpha_{22}\\ \end{array}\right)\in M_{2}(\mathbb{C}) with $\overline{A}A=I_{2}$, the $2\times 2$ identity matrix. Then $H$ is a Hopf $\ast$-algebra with the $\ast$-structure given by

[TABLE]

Proof.

Assume that $n=2$. Then $\omega=-1$. We first prove that the relations given in the lemma together with ${i^{\ast}}={-i}$ give rise to a real antialgebra endomorphism of $H$, i.e., a real algebra map from $H$ to $H^{op}$. In $H^{op}$, we have $(g^{\ast})^{2}=g^{2}=1$, $(x^{*})^{2}=(\alpha_{11}x+\alpha_{12}y)^{2}=\alpha_{11}^{2}x^{2}+\alpha_{11}\alpha_{12}xy+\alpha_{12}\alpha_{11}yx+\alpha_{12}^{2}y^{2}=0$ and $x^{*}\cdot g^{*}=(\alpha_{11}x+\alpha_{12}y)\cdot g=\alpha_{11}gx+\alpha_{12}gy=-\alpha_{11}xg-\alpha_{12}yg=-g\cdot(\alpha_{11}x+\alpha_{12}y)=-g^{*}\cdot x^{*}$. We also have $x^{*}\cdot y^{*}=(\alpha_{21}x+\alpha_{22}y)(\alpha_{11}x+\alpha_{12}y)=\alpha_{21}\alpha_{11}x^{2}+\alpha_{21}\alpha_{12}xy+\alpha_{22}\alpha_{11}yx+\alpha_{22}\alpha_{12}y^{2}=(\alpha_{21}\alpha_{12}-\alpha_{22}\alpha_{11})xy$ and $y^{*}\cdot x^{*}=(\alpha_{11}x+\alpha_{12}y)(\alpha_{21}x+\alpha_{22}y)=\alpha_{11}\alpha_{21}x^{2}+\alpha_{11}\alpha_{22}xy+\alpha_{12}\alpha_{21}yx+\alpha_{12}\alpha_{22}y^{2}=(\alpha_{11}\alpha_{22}-\alpha_{12}\alpha_{21})xy$, which implies that $x^{*}\cdot y^{*}=-y^{*}\cdot x^{*}$. Similarly, one can check that $(y^{*})^{2}=0$ and $g^{\ast}\cdot y^{\ast}=-y^{\ast}\cdot g^{\ast}$. We also have $i^{\ast}=-i\in Z(H^{op})$ and $(i^{\ast})^{2}=(-i)^{2}=-1$. This shows that the relations given in the lemma together with ${i^{\ast}=-i}$ determine a real algebra map $\ast:H\rightarrow{H^{op}}$. Then it follows that $\ast$ is a conjugate-linear antialgebra endomorphism of $H$. Hence the composition $\ast\circ\ast$ is a complex algebra endomorphism of $H$. Clearly, $(g^{*})^{*}=g$. Since $\overline{A}A=I_{2}$, $\overline{\alpha_{i1}}\alpha_{1j}+\overline{\alpha_{i2}}\alpha_{2j}=\delta_{ij}$ for $1\leqslant i,j\leqslant 2$. Hence we have

[TABLE]

Similarly, we also have $(y^{*})^{*}=y$. It follows that $(h^{*})^{*}=h$ for all $h\in{H}$. Thus, $\ast$ is an involution of $H$. Note that both $\triangle\circ\ast$ and $(\ast\otimes\ast)\circ\triangle$ are conjugate-linear antialgebra maps from $H$ to $H\otimes H$. It is easy check that $\triangle(h^{\ast})=\sum(h_{1})^{\ast}\otimes(h_{2})^{\ast}$ for any $h\in\{g,x,y\}$. It follows that $\triangle(h^{\ast})=\sum(h_{1})^{\ast}\otimes(h_{2})^{\ast}$ for all $h\in H$. Similarly, we have $\varepsilon(h^{*})=\overline{\varepsilon(h)}$ for all $h\in H$. Finally, since $S$ is a complex antialgebra endomorphism of $H$ and $\ast$ is a conjugate-linear antialgebra endomorphism of $H$, the map $H\rightarrow H$, $h\mapsto{S(S(h^{\ast})^{\ast})}$ is a complex algebra endomorphism of $H$. Now, we have

[TABLE]

and similarly $S(S(y^{\ast})^{\ast})=y$. It follows that $S(S(h^{\ast})^{\ast})=h$ for all $h\in{H}$. ∎

The following proposition follows similarly to [1, Lemma 2.7].

Proposition 3.3**.**

*For any $r,s\in{\mathbb{N}}$ and $l\in{\mathbb{Z}}$,

[TABLE]

Proof.

Since

[TABLE]

it follows from [2, Proposition IV.2.2] that

[TABLE]

Now, since $\triangle$ is an algebra map, we have

[TABLE]

∎

Note that $\{y^{r}x^{s}g^{l}|0\leqslant r,s,l<n\}$ is a canonical basis of $H$ over $\mathbb{C}$. Hence,

[TABLE]

is a basis of $H\otimes H$ over $\mathbb{C}$. For an element

[TABLE]

in $H$, if $\lambda_{r,s,l}\neq 0$, then we say that ${y^{r}x^{s}g^{l}}$ is a term of $h$. Moreover, $r$ (resp., $s$) is called the degree of $y$ (resp., $x$) in the term ${y^{r}x^{s}g^{l}}$. Similarly, for an element

[TABLE]

in $H\otimes H$, if $\lambda_{r,s,l,{r_{1}},{s_{1}},{l_{1}}}\neq 0$, then we say that ${y^{r}x^{s}g^{l}}\otimes y^{r_{1}}x^{s_{1}}g^{l_{1}}$ is a term of $h$. Moreover, ${r+{r_{1}}}$ (resp., ${s+{s_{1}}}$) is called the total degree of $y$ (resp., $x$) in the term

[TABLE]

Lemma 3.4**.**

$G(H)=\{g^{l}|0\leqslant l<n\}$.

Proof.

Obviously, $g^{l}\in G(H)$ for all $0\leqslant l<n$. Conversely, let

[TABLE]

where ${\lambda_{{r},{s},{l}}\in\mathbb{C}}$. Assume that ${r_{1}}$ is the highest degree of $y$ in the terms of $h$, that is, there is a nonzero coefficient ${\lambda_{{r_{1}},{s_{1}},{l_{1}}}\neq 0}$ in the above expression of ${h}$ such that $\lambda_{{r},{s},{l}}\neq 0$ implies $r\leqslant r_{1}$. From Proposition 3.3, one knows that the total degree of $y$ in each term of the expression of $\triangle(y^{r}x^{s}g^{l})$ is $r$. Then from

[TABLE]

one gets that the highest total degree of $y$ in the terms of $\triangle(h)$ is ${r_{1}}$. However,

[TABLE]

is a term of $h\otimes h$ with the nonzero coefficient ${{\lambda^{2}}_{{r_{1}},{s_{1}},{l_{1}}}\neq 0}$. It follows from $\triangle(h)=h\otimes h$ that $0\leqslant{2r_{1}}\leqslant{r_{1}}$, which implies ${r_{1}}=0$. Thus, if $r>0$ then ${\lambda_{r,s,l}}=0$. Similarly, one can show that ${\lambda_{r,s,l}}=0$ for any $s>0$. Therefore, $h\in{\rm span}\{g^{l}|0\leqslant l<n\}$. Since $G(H)$ is linearly independent over ${\mathbb{C}}$ and $\{g^{l}|0\leqslant l<n\}\subseteq G(H)$, we have $h=g^{l}$ for some $0\leqslant l<n$. Hence $G(H)\subseteq\{g^{l}|0\leqslant l<n\}$, and so $G(H)=\{g^{l}|0\leqslant l<n\}$. ∎

Lemma 3.5**.**

Let $h\in H$. If $\triangle(h)={h\otimes g+1\otimes h}$, then $h=\lambda_{1}x+\lambda_{2}y+\lambda_{3}(1-g)$ for some $\lambda_{1},\lambda_{2},\lambda_{3}\in{\mathbb{C}}$.

Proof.

Let $h=\sum_{0\leqslant r,s,l<n}\lambda_{r,s,l}{y^{r}x^{s}g^{l}}$ with $\lambda_{r,s,l}\in{\mathbb{C}}$ such that

[TABLE]

Then $\varepsilon(h)=0$. For any $0\leqslant r,s<n$, let

[TABLE]

Then by Proposition 3.3 and the proof of Lemma 3.4, one knows that

[TABLE]

Hence, one may assume that

[TABLE]

for some $\lambda_{l}\in{\mathbb{C}}$, where $r$ and $s$ are fixed integers with $0\leqslant r,s<n$. Now, by Proposition 3.3, we have

[TABLE]

and

[TABLE]

By the paragraph before Lemma 3.4, $H\otimes H$ has a canonical basis over ${\mathbb{C}}$

[TABLE]

Now by comparing the coefficients of the basis element $g^{l}\otimes y^{r}x^{s}g^{l}$ in the two expressions of $\triangle(h)$ given above, one gets that ${\lambda_{l}=0}$ if $l>1$, and that $\lambda_{1}=0$ if $(r,s)\neq(0,0)$. Hence, $h=\lambda_{0}y^{r}x^{s}$ when $r+s\neq 0$, and $h=\lambda_{0}+\lambda_{1}g$ when $r=s=0$.

If $h=\lambda_{0}+\lambda_{1}g$, then $\lambda_{0}+\lambda_{1}=0$ by $\varepsilon(h)=0$, and so $h=\lambda_{0}(1-g)$. Now assume $h=\lambda_{0}y^{r}x^{s}$ with $r+s\neq 0$. Then (3.1) and (3.2) becomes

[TABLE]

and

[TABLE]

respectively. If both $r>0$ and $s>0$, then by comparing the coefficients of the basis element $y^{r}\otimes x^{s}g^{r}$ in the two expressions of $\triangle(h)$ given above, one gets that $\lambda_{0}=0$, and hence $h=0$, a contradiction. Hence either $r>0$ and $s=0$, or $r=0$ and $s>0$. If $r>0$ and $s=0$, then $h=\lambda_{0}y^{r}$, and (3.3) and (3.4) becomes

[TABLE]

and

[TABLE]

respectively. If $r>1$, then by comparing the coefficients of the basis element $y^{r-1}\otimes yg^{r-1}$ in the two expressions of $\triangle(h)$ given above, one gets that $\lambda_{0}=0$, and hence $h=0$, a contradiction. Hence $r=1$ and so $h=\lambda_{0}y$. Similarly, one can check that if $r=0$ and $s>0$ then $h=\lambda_{0}x$. This completes the proof. ∎

Lemma 3.6**.**

Let $h\in H$ with $\triangle(h)=h\otimes g^{w}+1\otimes h$ for some $1<w<n$. Then $h=\lambda(1-g^{w})$ for some $\lambda\in\mathbb{C}$.

Proof.

It is similar to the proof of Lemma 3.5. We only need to consider the case of

[TABLE]

for some $\lambda_{l}\in\mathbb{C}$, where $r$ and $s$ are fixed integers with $0\leqslant r,s<n$. Then we have

[TABLE]

and

[TABLE]

If $r\neq 0$ and $s\neq 0$, then by comparing the coefficients of the basis element $y^{r}g^{l}\otimes x^{s}g^{l+r}$ in the two expressions of $\triangle(h)$ given above, one gets that ${\lambda_{l}=0}$ for all $0\leqslant l<n$, and hence $h=0$, a contradiction. So $r=0$ or $s=0$. Assume that $r\neq 0$. Then $s=0$. In this case, by comparing the coefficients of the basis element $g^{l}\otimes y^{r}g^{l}$ in the two expressions of $\triangle(h)$ given above, one gets that ${\lambda_{l}=0}$ if $l>0$. Hence $h=\lambda_{0}y^{r}$, and (3.7) and (3.8) become

[TABLE]

and

[TABLE]

respectively. Then by comparing the coefficients of the basis element $y^{r}\otimes g^{r}$ in the both expressions of $\triangle(h)$ given in (3.9) and (3.10), one gets that $r=w>1$ since $h=\lambda_{0}y^{r}\neq 0$. Now by comparing the coefficients of the basis element $y^{r-1}\otimes yg^{r-1}$ in the both expressions of $\triangle(h)$ given in (3.9) and (3.10), one finds that $\lambda_{0}=0$, a contradiction. This shows that $r=0$. Similarly, one can show that $s=0$. Hence $h=\sum_{l}\lambda_{l}g^{l}\neq 0$. Then it is easy to see that $h=\lambda(1-g^{w})$ for some $\lambda\in{\mathbb{C}}$. ∎

Theorem 3.7**.**

$(1)$* If $n>2$, then Lemma 3.1 gives all Hopf $\ast$-algebra structures on $H$.

$(2)$ If $n=2$, then Lemma 3.2 gives all Hopf $\ast$-algebra structures on $H$.*

Proof.

Assume that $H$ has a Hopf $\ast$-algebra structure $\ast$. Then

[TABLE]

Hence $g^{*}\in G(H)$. By Lemma 3.4, $g^{*}=g^{w}$ for some $0\leqslant w<n$. Since $*$ is an involution and $1^{*}=1$, $g^{*}\neq 1$. Hence $w\neq 0$, and so $0<w<n$. We also have

[TABLE]

If $w\neq 1$, then it follows from Lemma 3.6 that $x^{*}=\lambda(1-g^{w})$ for some $\lambda\in\mathbb{C}$. Since $*$ is an involution and a conjugate-linear antialgebra endomorphism of $H$, we have $x=(x^{*})^{*}=(\lambda(1-g^{w}))^{*}=\overline{\lambda}(1-g^{w^{2}})$. This is impossible. Hence $w=1$, and so $g^{*}=g$ and $\triangle(x^{\ast})=x^{\ast}\otimes g+1\otimes x^{\ast}$. Then by Lemma 3.5, $x^{\ast}=\alpha_{11}x+\alpha_{12}y+\alpha_{13}(1-g)$ for some $\alpha_{11},\alpha_{12},\alpha_{13}\in\mathbb{C}$. Similarly, one can show that $y^{\ast}=\alpha_{21}x+\alpha_{22}y+\alpha_{23}(1-g)$ for some $\alpha_{21},\alpha_{22},\alpha_{23}\in\mathbb{C}$. Then by $xg=\omega gx$, one gets that $(xg)^{*}=(\omega gx)^{*}$. However, $(xg)^{*}=g^{*}x^{*}=g(\alpha_{11}x+\alpha_{12}y+\alpha_{13}(1-g))=\alpha_{11}gx+\alpha_{12}gy+\alpha_{13}(g-g^{2})$ and $(\omega gx)^{*}=\overline{\omega}x^{*}g^{*}=\omega^{-1}(\alpha_{11}x+\alpha_{12}y+\alpha_{13}(1-g))g=\omega^{-1}\alpha_{11}xg+\omega^{-1}\alpha_{12}yg+\omega^{-1}\alpha_{13}(g-g^{2})=\alpha_{11}gx+\omega^{-2}\alpha_{12}gy+\omega^{-1}\alpha_{13}(g-g^{2})$. It follows that $\alpha_{12}=\omega^{-2}\alpha_{12}$ and $\alpha_{13}=\omega^{-1}\alpha_{13}$. Hence $\alpha_{12}(1-\omega^{2})=0$ and $\alpha_{13}=0$ by $\omega\neq 1$. Similarly, from $(gy)^{*}=(\omega yg)^{*}$, one gets that $\alpha_{21}(1-\omega^{2})=0$ and $\alpha_{23}=0$.

(1) Assume that $n>2$. Then $\omega^{2}\neq 1$, and hence $\alpha_{12}=\alpha_{21}=0$ by $\alpha_{12}(1-\omega^{2})=0$ and $\alpha_{21}(1-\omega^{2})=0$. Thus, $x^{*}=\alpha_{11}x$ and $y^{*}=\alpha_{22}y$. Then we have $x=(x^{*})^{*}=(\alpha_{11}x)^{*}=\overline{\alpha_{11}}x^{*}=\overline{\alpha_{11}}\alpha_{11}x$, which implies that $|\alpha_{11}|=1$. Similarly, one can show that $|\alpha_{22}|=1$. This shows Part (1).

(2) Assume that $n=2$. Then $x^{*}=\alpha_{11}x+\alpha_{12}y$ and $y^{*}=\alpha_{21}x+\alpha_{22}y$. Hence we have $x=(x^{*})^{*}=(\alpha_{11}x+\alpha_{12}y)^{*}=\overline{\alpha_{11}}x^{*}+\overline{\alpha_{12}}y^{*}=\overline{\alpha_{11}}(\alpha_{11}x+\alpha_{12}y)+\overline{\alpha_{12}}(\alpha_{21}x+\alpha_{22}y)=(\overline{\alpha_{11}}\alpha_{11}+\overline{\alpha_{12}}\alpha_{21})x+(\overline{\alpha_{11}}\alpha_{12}+\overline{\alpha_{12}}\alpha_{22})y$ and $y=(y^{*})^{*}=(\alpha_{21}x+\alpha_{22}y)^{*}=\overline{\alpha_{21}}x^{*}+\overline{\alpha_{22}}y^{*}=\overline{\alpha_{21}}(\alpha_{11}x+\alpha_{12}y)+\overline{\alpha_{22}}(\alpha_{21}x+\alpha_{22}y)=(\overline{\alpha_{21}}\alpha_{11}+\overline{\alpha_{22}}\alpha_{21})x+(\overline{\alpha_{21}}\alpha_{12}+\overline{\alpha_{22}}\alpha_{22})y$. It follows that

[TABLE]

This shows Part (2). ∎

Theorem 3.8**.**

If $n\geqslant 3$, then up to equivalence, there is a unique Hopf $\ast$-algebra structure on $H$ given by

[TABLE]

Proof.

Assume that $n\geqslant 3$. Then by Lemma 3.1, the relations given in the theorem determine a Hopf $*$-algbera structure on $H$, denoted by $*^{\prime}$. Now let $*$ be any Hopf $*$-algebra structure on $H$. Then by Lemma 3.1 and Theorem 3.7(1), there exist elements $\alpha,\beta\in\mathbb{C}$ with $|\alpha|=|\beta|=1$ such that

[TABLE]

Pick up two elements $\lambda_{1},\lambda_{2}\in\mathbb{C}$ with $\lambda_{1}^{2}=\alpha$ and $\lambda_{2}^{2}=\beta$. Then $|\lambda_{1}|=|\lambda_{2}|=1$ by $|\alpha|=|\beta|=1$, and hence $\lambda_{1}^{-1}=\overline{\lambda_{1}}$ and $\lambda_{2}^{-1}=\overline{\lambda_{2}}$. It is easy to see that there is a Hopf algebra automorphism $\phi$ of $H$ such that $\phi(g)=g$, $\phi(x)=\lambda_{1}x$ and $\phi(y)=\lambda_{2}y$. Then $\phi(g^{*^{\prime}})=\phi(g)=g=g^{*}=\phi(g)^{*}$, $\phi(x^{*^{\prime}})=\phi(x)=\lambda_{1}x=\lambda_{1}^{-1}\alpha x=\overline{\lambda_{1}}x^{*}=(\lambda_{1}x)^{*}=\phi(x)^{*}$ and $\phi(y^{*^{\prime}})=\phi(y)=\lambda_{2}y=\lambda_{2}^{-1}\beta y=\overline{\lambda_{2}}y^{*}=(\lambda_{2}y)^{*}=\phi(y)^{*}$. Hence $\phi(h^{*^{\prime}})=\phi(h)^{*}$ for all $h\in H$, and so $*$ is equivalent to $*^{\prime}$. ∎

Throughout the following, assume that $n=2$. In this case, $\omega=-1$.

Let A=\left(\begin{array}[]{cc}\alpha_{11}&\alpha_{12}\\ \alpha_{21}&\alpha_{22}\\ \end{array}\right) and B=\left(\begin{array}[]{cc}\beta_{11}&\beta_{12}\\ \beta_{21}&\beta_{22}\\ \end{array}\right) be two matrices in $M_{2}(\mathbb{C})$ with $\overline{A}A=\overline{B}B=I_{2}$, and let $*_{A}$ and $*_{B}$ be the corresponding Hopf $*$-algebra structures on $H$ determined by $A$ and $B$ as in Lemma 3.2, respectively. Then we have the following proposition.

Proposition 3.9**.**

$*_{A}$* and $*_{B}$ are equivalent $*$-structures on $H$ if and only if there exists an invertible matrix $\Lambda$ in $M_{2}(\mathbb{C})$ such that $A\Lambda=\overline{\Lambda}B$, i.e., $\overline{\Lambda}^{-1}A\Lambda=B$.*

Proof.

Suppose that $*_{A}$ and $*_{B}$ are equivalent. Then there exists a Hopf algebra automorphism $\phi$ of $H$ such that $\phi(h^{*_{A}})=\phi(h)^{*_{B}}$ for all $h\in H$. By Lemma 3.4 and $n=2$, one can see that $\phi(g)=g$. Then by Lemma 3.5, a straightforward computation shows that there exists a matrix \Lambda=\left(\begin{array}[]{cc}\lambda_{11}&\lambda_{12}\\ \lambda_{21}&\lambda_{22}\\ \end{array}\right) in $M_{2}(\mathbb{C})$ such that $\phi(x)=\lambda_{11}x+\lambda_{12}y$ and $\phi(y)=\lambda_{21}x+\lambda_{22}y$. Since $\phi$ is an isomorphism, one can check that $\Lambda$ is an invertible matrix in $M_{2}(\mathbb{C})$. Now we have

[TABLE]

and

[TABLE]

Hence it follows from $\phi(x^{*_{A}})=\phi(x)^{*_{B}}$ that $\alpha_{11}\lambda_{11}+\alpha_{12}\lambda_{21}=\overline{\lambda_{11}}\beta_{11}+\overline{\lambda_{12}}\beta_{21}$ and $\alpha_{11}\lambda_{12}+\alpha_{12}\lambda_{22}=\overline{\lambda_{11}}\beta_{12}+\overline{\lambda_{12}}\beta_{22}$. Similarly, from $\phi(y^{*_{A}})=\phi(y)^{*_{B}}$, one gets that $\alpha_{21}\lambda_{11}+\alpha_{22}\lambda_{21}=\overline{\lambda_{21}}\beta_{11}+\overline{\lambda_{22}}\beta_{21}$ and $\alpha_{21}\lambda_{12}+\alpha_{22}\lambda_{22}=\overline{\lambda_{21}}\beta_{12}+\overline{\lambda_{22}}\beta_{22}$. Thus, we have $A\Lambda=\overline{\Lambda}B$.

Conversely, suppose that there exists an invertible matrix \Lambda=\left(\begin{array}[]{cc}\lambda_{11}&\lambda_{12}\\ \lambda_{21}&\lambda_{22}\\ \end{array}\right) in $M_{2}(\mathbb{C})$ such that $A\Lambda=\overline{\Lambda}B$. Then it is straightforward to check that there is a Hopf algebra automorphism $\phi$ of $H$ uniquely determined by $\phi(g)=g$, $\phi(x)=\lambda_{11}x+\lambda_{12}y$ and $\phi(y)=\lambda_{21}x+\lambda_{22}y$. Obviously, $\phi(g^{*_{A}})=\phi(g)^{*_{B}}=g$. From the computation above, one gets that $\phi(x^{*_{A}})=\phi(x)^{*_{B}}$ and $\phi(y^{*_{A}})=\phi(y)^{*_{B}}$. It follows that $\phi(h^{*_{A}})=\phi(h)^{*_{B}}$ for any $h\in H$. This shows that $*_{A}$ and $*_{B}$ are equivalent. ∎

Acknowledgements This work supported by the National Natural Science Foundation of China (Grant No.11571298, 11711530703).