Some Hopf Algebras related to $\mathfrak{sl}_2$

TL;DR

This paper introduces a new family of Artin-Schelter Gorenstein Hopf algebras, explores their relationships with known Hopf algebras, and computes their Grothendieck rings, revealing non-isomorphic algebras with identical Grothendieck rings.

Contribution

It defines a new series of Hopf algebras $H_eta$, analyzes their properties, and determines their Grothendieck rings, connecting them to existing Hopf algebras.

Findings

Radford's and Gelaki's Hopf algebras are homomorphic images of $H_eta$.

The Grothendieck ring $G_0(H_eta)$ is explicitly determined.

Non-isomorphic Hopf algebras with isomorphic Grothendieck rings are identified.

Abstract

We define a series of Artin-Schelter Gorenstein Hopf algebras $H_\beta$ with injective dimensions 3. Radford's Hopf algebra and Gelaki's Hopf algebra are homomorphic images of $H_\beta$. We determine its Grothendieck ring $G_0(H_\beta)$. Meanwhile we can obtain Grothendieck rings of Gelaki's Hopf algebras and Radford's Hopf algebras $U_{(N,\nu,\omega)}$ in \cite{R}, and non-isomorphic Hopf algebras with isomorphic Grothendieck rings.

Equations

$$ab=ba,ac=ca,bc=cb,xa=qax,ya=q^{-1}ay,bx=xb,cx=xc,by=yb,cy=yc,$$

$$yx-q^{-n_1}xy={\beta }_3(a^{2n_1}-bc),x^n={\beta }_1(a^{n{n}_1}-b^n),y^n={\beta }_2(a^{n{n}_1}-c^n),$$

$$\Delta (a)=a\otimes a,\Delta (b)=b\otimes b,\Delta (c)=c\otimes c,\Delta (x)=x\otimes a^{n_1}+b\otimes x,\Delta (y)=y\otimes a^{n_1}+c\otimes y,$$

$$\epsilon (a)=\epsilon (b)=\epsilon (c)=1,\epsilon (x)=\epsilon (y)=0$$

$$s(a)=a^{-1},s(b)=b^{-1},s(c)=c^{-1},s(x)=-q^{-n_1}{a}^{-n_1}{b}^{-1}x,s(y)=-q^{n_1}{a}^{-n_1}{c}^{-1}y.$$

$$\{a^i{b}^j{c}^k{x}^u{y}^v|i,j,k\in \mathbb{Z},0\le u,v\le n-1\}.$$

$$\begin{array}[]{lll}\Delta(x)^{n}&=&(x\otimes a^{n_{1}}+b\otimes x)^{n}=x^{n}\otimes a^{n_{1}n}+b^{n}\otimes x^{n}\\ &=&\beta_{1}((a^{n_{1}n}-b^{n})\otimes a^{n_{1}n}+b^{n}\otimes(a^{n_{1}n}-b^{n}))\\ &=&\beta_{1}(a^{n_{1}n}\otimes a^{n_{1}n}-b^{n}\otimes b^{n})=\Delta(x^{n}).\end{array}$$

$$\begin{array}[]{lll}&&\Delta(y)\Delta(x)-q^{-n_{1}}\Delta(x)\Delta(y)\\ &=&(y\otimes a^{n_{1}}+c\otimes y)(x\otimes a^{n_{1}}+b\otimes x)-q^{-n_{1}}(x\otimes a^{n_{1}}+b\otimes x)(y\otimes a^{n_{1}}+c\otimes y)\\ &=&yx\otimes a^{2n_{1}}-q^{-n_{1}}xy\otimes a^{2n_{1}}+bc\otimes yx-q^{-n_{1}}bc\otimes xy\\ &=&\beta_{3}((a^{2n_{1}}-bc)\otimes a^{2n_{1}}+bc\otimes(a^{2n_{1}}-bc))\\ &=&\beta_{3}(a^{2n_{1}}\otimes a^{2n_{1}}-bc\otimes bc)\\ &=&\Delta(yx-q^{-n_{1}}xy).\end{array}$$

$$\Delta (x)\Delta (a)=q\Delta (a)\Delta (x),\Delta (y)\Delta (a)=q^{-1}\Delta (a)\Delta (y),\Delta (x)\Delta (b)=\Delta (b)\Delta (x),$$

$$\Delta (x)\Delta (c)=\Delta (c)\Delta (x),\Delta (y)\Delta (b)=\Delta (b)\Delta (y),\Delta (y)\Delta (c)=\Delta (c)\Delta (y),$$

$$\Delta (a)\Delta (b)=\Delta (b)\Delta (a),\Delta (a)\Delta (c)=\Delta (c)\Delta (a),\Delta (b)\Delta (c)=\Delta (c)\Delta (b).$$

$$(\Delta \otimes 1)\Delta (K)=(1\otimes \Delta )\Delta (K),(1\otimes \epsilon )\Delta (K)=(\epsilon \otimes 1)\Delta (K)=K$$

$$\begin{array}[]{lll}(s(x))^{n}&=&(-q^{-n_{1}})^{n}b^{-n}(a^{-n_{1}}x)^{n}\\ &=&(-1)^{n}b^{-n}q^{\frac{1}{2}n(n-1)n_{1}}a^{-n_{1}n}x^{n}\\ &=&(-1)^{n}q^{\frac{1}{2}n(n-1)n_{1}}a^{-nn_{1}}b^{-n}\beta_{1}(a^{nn_{1}}-b^{n})\\ &=&(-1)^{n+1}q^{\frac{1}{2}n(n-1)n_{1}}\beta_{1}(a^{-nn_{1}}-b^{-n})\\ &=&(-1)^{n+1}q^{\frac{1}{2}n(n-1)n_{1}}s(x^{n}).\end{array}$$

$$(-1{)}^{n+1}{q}^{\frac{1}{2}n(n-1)n_1}=(-1{)}^{n+1}{q}^{\frac{1}{2}n(n-2)n_1+\frac{1}{2}n{n}_1}=(-1{)}^{n+n_1+1}=1.$$

$$\begin{array}[]{lll}&&s(x)s(y)-q^{-n_{1}}s(y)s(x)\\ &=&a^{-n_{1}}b^{-1}xa^{-n_{1}}c^{-1}y-q^{-n_{1}}a^{-n_{1}}c^{-1}ya^{-n_{1}}b^{-1}x\\ &=&a^{-2n_{1}}(bc)^{-1}(q^{-n_{1}}xy-yx)=-\beta_{3}a^{-2n_{1}}(bc)^{-1}(a^{2n_{1}}-bc)\\ &=&\beta_{3}(a^{-2n_{1}}-(bc)^{-1})=s(yx-q^{-n_{1}}xy).\end{array}$$

$$s(a)s(x)=qs(x)s(a),s(a)s(y)=q^{-1}s(y)s(a),s(a)s(b)=s(b)s(a),$$

$$s(c)s(b)=s(b)s(c),s(c)s(a)=s(a)s(c),s(b)s(x)=s(x)s(b),$$

$$s(b)s(y)=s(y)s(b),s(c)s(x)=s(x)s(c),s(c)s(y)=s(y)s(c).$$

$$a^N=1,b=c=1,N\in \mathbb{Z}\text{and}n\mid N,$$

$${\mathcal{U}}_{(N/(N,n_1),N,n_1,{\omega }^{n_1},0,0,\gamma )}\simeq U_{(N,n_1,\omega )}$$

$$U_{(N,n_1,\omega )}={\mathcal{U}}_{(N/(N,n_1),N,n_1,{\omega }^{n_1},0,0,1)}.$$

$$\begin{array}[]{llll}GKdim(H_{\beta})&=&GKdim({\bf K}[a^{\pm n},b^{\pm 1},c^{\pm 1}]),\\ &=&GKdim({\bf K}[a^{n},b,c]),\\ &=&3.\end{array}$$

$$H_{\alpha ,\beta }=H_{\beta }/I.$$

$$a^{mn(n-1)}=1,x^n={\beta }_1(a^{n_{1}n}-a^{mn{n}_2}),y^n={\beta }_2(a^{n_{1}n}-a^{mn{n}_3}),$$

$$xa=qax,ya=q^{-1}ay,yx-q^{-n_1}xy={\beta }_3(a^{2n_1}-a^{mn(n_2+n_3)}).$$

$$\Delta (a)=a\otimes a,\Delta (x)=x\otimes a^{n_1}+a^{mn{n}_2}\otimes x,$$

$$\Delta (y)=y\otimes a^{n_1}+a^{mn{n}_3}\otimes y,\epsilon (a)=1,\epsilon (x)=\epsilon (y)=0.$$

$$s(a)=a^{-1}=a^{N-1},s(x)=-q^{-n_1}{a}^{-mn{n}_2-n_1}x,s(y)=-q^{n_1}{a}^{-mn{n}_3-n_1}y.$$

$$y^{k}x=q^{-k{n}_1}x{y}^k+{\beta }_3{u}_k(q^{-(k-1)n_1}{a}^{2n_1}-a^{mn(n_2+n_3)})y^{k-1},$$

$$x_{01}^n(g^{n^{\prime }{n}_1^{\prime }}-g^{m^{\prime }{n}^{\prime }{n}_2^{\prime }})=g^{n{n}_1}-g^{mn{n}_2}\text{i.e.}(x_{01}^n{g}^{n^{\prime }-n}-1)g^{n{n}_1-mn{n}_2}=x_{01}^n-1.$$

Full text

Key Words: Hopf algebra, irreducible module, Grothendieck ring, generators, relations

The author Jing Wang is supported by NNSFC (No.11901034) and the Fundamental Research Funds for the Central Universities (No.BLX201721). The work of Zhixiang Wu and Yan Tan is sponsored by ZJNSF (No. LY17A010015) and NNSFC (No.11871421).

1. Introduction

The tensor product of representations of a Hopf algebra is important in the representation theory of Hopf algebras and quantum groups. In particular, the decomposition of the tensor product of indecomposable modules into a direct sum of indecomposable modules has received enormous attention. However, little is known about how a tensor product of two indecomposable modules decomposes into a direct sum of indecomposable modules over a Hopf algebra or a quantum group. There are some results for the decompositions of tensor products of modules over a Hopf algebra or a quantum group [3, 5, 8, 9, 16]. Recently, Green rings and their homomorphic images, which are called Grothendieck rings, of various finite-dimensional Hopf algebras have attracted numerous attentions [4, 6, 13, 18]. But most of them are considered in the case of finite dimensional Hopf algebras or quantum groups.

In this paper, we describe an infinite-dimensional pointed Hopf algebra $H_{\beta}$ for any $\beta\in({\bf K^{*}})^{3}$ and its Grothendieck ring, which differs from the Green ring. The Hopf algebra is constructed by adding group-like elements to the restricted quantum enveloping algebra of $\mathfrak{sl}_{2}$. The Hopf algebra $H_{\beta}$ has many interesting finite dimensional images, such as Gelaki’s Hopf algebra ${\mathcal{U}}_{(n,N,\nu,q,\alpha,\beta,\gamma)}$ and Radford’s Hopf algebra $U_{(N,\nu,\omega)}$. Here, all irreducible representations of $H_{\beta}$ and the decomposition of the tensor product of two irreducible $H_{\beta}$-modules are determined as in [2], but we consider the decomposition in the sense of Grothendieck group. Thus we obtain the Grothendieck ring of the infinite-dimensional Hopf algebras $H_{\beta}$. Meanwhile we obtain the Grothendieck rings of Gelaki’s Hopf algebra ${\mathcal{U}}_{(n,N,\nu,q,\alpha,\beta,\gamma)}$ and Radford’s Hopf algebra $U_{(N,\nu,\omega)}$.

This paper is organized as follows. In Section 2, we introduce the infinite dimensional noncommutative and noncocommutative Hopf algebra $H_{\beta}$, which is a pointed $AS$-Gorenstein Hopf algebra of injective dimension 3. And we prove that the homological integral of $H_{\beta}$ is isomorphic to ${\bf K}$ as $H_{\beta}$-bimodules. To construct irreducible representations of $H_{\beta}$, we define a finite-dimensional quotient Hopf algebra $H_{\alpha,\beta}$ of Hopf algebra $H_{\beta}$ in Section 3, where $\alpha=(n,m,n_{1},n_{2},n_{3})\in{\bf N}^{5}$, $\beta=(\beta_{1},\beta_{2},\beta_{3})\in{\bf K}^{3}$. We prove that the algebra $H_{\alpha,\beta}$ could be decomposed into direct sum of matrix rings over some commutative local rings. Based on this result, all irreducible representations of $H_{\beta}$ are illustrated in Section 4. Besides, we point out that the category of all finite-dimensional representations of $H_{\beta}$ is not semisimple. Meanwhile we obtain all irreducible representations of Radford’s Hopf algebra $U_{(N,\nu,\omega)}$ and Gelaki’s Hopf algebra ${\mathcal{U}}_{(n,N,\nu,q,\alpha,\beta,\gamma)}$. In Section 5, the Grothendieck ring $G_{0}(H_{\beta})$ of $H_{\beta}$ is constructed. We illustrate the structure of the Grothendieck ring $G_{0}(H_{\beta})$ of $H_{\beta}$ in several cases. In details, Theorem 5.2(iv) describes the case in which $\beta_{1}=\beta_{2}=\beta_{3}=0$. Theorems 5.5, 5.12 and 5.16 illustrate the cases that iff two the three equations $\beta_{1}=0$, $\beta_{2}=0$ and $\beta_{3}=0$ hold. In theorems 5.10,   5.18 and 5.20, we give the cases in which iff one of the conditions $\beta_{1}=0$, $\beta_{2}=0$ and $\beta_{3}=0$ holds. In additional, we also obtain the Grothendieck rings $G_{0}(U_{(N,\nu,\omega)})$ and $G_{0}({\mathcal{U}}_{(n,N,\nu,q,\alpha,\beta,\gamma)})$. We prove that there exists non-isomorphic Hopf algebras with isomorphic Grothendieck rings.

Throughout this paper, ${\bf K}$ denotes an algebraically closed field of characteristic zero, ${\bf K}^{*}$ the group of all nonzero elements of ${\bf K}$ with the product of ${\bf K}$ and $q\in{\bf K}$ a primitive $n$-th root of unity, $n\geq 2$. For any $z\in{\bf K}^{*}$, we denote $\langle z\rangle$ the subgroup of ${\bf K}^{*}$ generated by $z$, the quotient group ${\bf K}^{*}/N$ of ${\bf K}^{*}$ with respect to a normal subgroup $N$, and $\bar{z}$ the element $zN$. We simplify the tensor product $V\otimes_{\bf K}W$ of two vector spaces over ${\bf K}$ as $V\otimes W$. The ring of integers is denoted by ${\mathbb{Z}}$ and $(N_{1},\cdots,N_{k})$ denotes the greatest common divisor of $N_{1},\cdots,N_{k}$, for any $N_{1},\cdots,N_{k}\in\mathbb{Z}$.

2. Definition of Hopf algebra $H_{\beta}$

In this section, we give the definition of Hopf algebra $H_{\beta}$. Then we prove that $H_{\beta}$ is an Artin-Schelter Gorenstein Hopf algebra with injective dimension 3.

Definition 2.1**.**

Let $n_{1}$ be a positive integer such that $2\nmid(n,n_{1})$ and $1\leq n_{1}\leq n$. Assume that $q\in{\bf K}$ is a primitive $n$-th root of unity. Suppose that $H_{\beta}$ is an algebra generated by $a,b,c,x,y$ with the relations

[TABLE]

[TABLE]

for $\beta=(\beta_{1},\beta_{2},\beta_{3})\in{\bf K}^{3}$. The coproduct $\Delta$ and counit $\varepsilon$ of $H_{\beta}$ is determined by

[TABLE]

and

[TABLE]

respectively. An anti-automorphism $s$ of $H_{\beta}$ is determined by

[TABLE]

In the following theorem, we prove that $H_{\beta}$ is an infinite dimensional Hopf algebra with the above coproduct $\Delta,$ counit $\varepsilon$ and antipode $s$.

Theorem 2.2**.**

The algebra $(H_{\beta},\ \Delta,\ \varepsilon,\ s)$ is a pointed Hopf algebra with a basis

[TABLE]

Proof.

First, we prove that $\Delta$ can determine a homomorphism of algebras from $H_{\beta}$ to $H_{\beta}\otimes H_{\beta}$.

By the definition of $H_{\beta}$, we obtain that $\Delta(x)^{k}=\sum\limits^{k}_{l=0}\binom{k}{l}_{q^{n_{1}}}b^{l}x^{k-l}\otimes a^{(k-l)n_{1}}x^{l}$, where $\binom{k}{l}_{q^{n_{1}}}=\frac{(k)_{q^{n_{1}}}!}{(l)_{q^{n_{1}}}!(k-l)_{q^{n_{1}}}!}$ and $(p)_{q^{n_{1}}}!=(p)_{q^{n_{1}}}(p-1)_{q^{n_{1}}}\cdots(1)_{q^{n_{1}}}$ for $(p)_{q^{n_{1}}}=1+q^{n_{1}}+\cdots+q^{(p-1)n_{1}}$. Hence

[TABLE]

Similarly, we can prove $\Delta(y^{n})=\Delta(y)^{n}$. Moreover,

[TABLE]

It is easy to verify that

[TABLE]

[TABLE]

[TABLE]

Thus $\Delta$ induces a homomorphism of algebras. Second, it is easy to prove that $\varepsilon$ is a homomorphism from the algebra $H_{\beta}$ to ${\bf K}$ and

[TABLE]

for any $K\in\{a,b,c,x,y\}$. Thus $H_{\beta}$ is a bialgebra.

At last, we need to prove that $s$ is an antipode of the algebra $H_{\beta}$.

By $s(x)=-q^{-n_{1}}a^{-n_{1}}b^{-1}x,$ we get that

[TABLE]

It is not difficult to prove that $s(x)^{n}=s(x^{n})$. Indeed, if $n$ is odd then $(-1)^{n+1}q^{\frac{1}{2}n(n-1)n_{1}}=1$. If $n$ is even then we obtain that $n_{1}$ is odd by $2\nmid(n,n_{1})$ and

[TABLE]

Similarly, we have $s(y^{n})=s(y)^{n}$. Moreover,

[TABLE]

It is easy to verify that

[TABLE]

[TABLE]

[TABLE]

Thus $s$ is an anti-automorphism of $H_{\beta}$. Hence, we only need to verify the antipode axiom on the generators $x,y,a,b,c$. It is easy to verify this. Altogether, we have proved that $H_{\beta}$ is a Hopf algebra.

By [12, Lemma 1] or [7, Lemma 1.1], we obtain that the Hopf algebra $H_{\beta}$ is pointed and the set $\{a^{i}b^{j}c^{k}|a,b,c\in{\mathbb{Z}}\}$ is of all group-like elements of $H_{\beta}$. Similarly to [7, 12], we can prove that $\{a^{i}b^{j}c^{k}x^{u}y^{v}|i,j,j\in{\mathbb{Z}},0\leq u,v\leq n-1\}$ is a basis of $H_{\beta}$ by the Diamond Lemma [1]. ∎

Remark 2.3**.**

Notice that if the generators of $H_{\beta}$ satisfy the relations

[TABLE]

*then the Hopf algebra $H_{\beta}$ is a Gelaki’s Hopf algebra ${\mathcal{U}}_{(n,N,n_{1},q,\beta_{1},\beta_{2},\beta_{3})}$, which is defined in [7]. *

Remark 2.4** (S. Gelaki [7]).**

Note that if $\omega$ is a primitive $N$-th root of unity and $N\nmid n_{1}^{2}$, then

[TABLE]

as Hopf algebras for any $\gamma\in{\bf K}^{*}$. Especially,

[TABLE]

Here, $U_{(N,n_{1},\omega)}$ is called Radford’s Hopf algebra.

Let $\mathfrak{S}=\{a^{kn}b^{l}c^{t}|k,l,t$ are nonnegative integers$\}$. Then $\mathfrak{S}$ is a multiplicatively closed set of ${\bf K}[a^{n},b,c]$. Since ${\bf K}[a^{\pm n},b^{\pm 1},c^{\pm 1}]$ is the localization of ${\bf K}[a^{n},b,c]$ with respect to $\mathfrak{S}$, we get that the Gelfand-Kirillov dimensions of ${\bf K}[a^{\pm n},b^{\pm 1},c^{\pm 1}]$ and ${\bf K}[a^{n},b,c]$ are equal by [11, Proposition 8.2.13]. For simplification, we write the Gelfand-Kirillov dimension as GK dim in this section.

Recall that a Hopf algebra $A$ over field ${\bf K}$ is Artin-Schelter Gorenstein (simply AS-Gorenstein) defined in [14, Definition 3.1] if

(AS1) injdim ${}_{A}A=d<\infty,$

(AS2) dim${}_{\bf K}Ext^{d}_{A}(_{A}{\bf K},_{A}A)=1,$ $Ext^{i}_{A}(_{A}{\bf K},_{A}A)=0$ for all $i\neq d$,

(AS3) the right $A$-module versions of the conditions (AS1,AS2) hold.

Theorem 2.5**.**

$H_{\beta}$* is an AS-Gorenstein Hopf algebra of injective dimension 3.*

Proof.

Let $T={\bf K}[a^{\pm n},b^{\pm 1},c^{\pm 1}]$. Then $H_{\beta}$ is finitely generated module over the Noetherian ring $T$. So $H_{\beta}$ is both right and left Noetherian. By [11, Corollary 13.1.13], $H_{\beta}$ is a $PI$ Hopf algebra. Moreover, by [11,  Propositions  8.2.9, 8.2.13 and 8.1.15] we obtain that

[TABLE]

By [14, Theorem 0.1], $H_{\beta}$ is an AS-Gorenstein Hopf algebra of injective dimension $3$. ∎

We recall the homological integral of $H$ from [10]. Let $H$ be an $AS$-Gorenstein Hopf algebra of injective dimension $d$ over field ${\bf K}$. The vector space of left homological integrals of $H$ is defined as $\int_{H}^{l}:=Ext_{H}^{d}(_{H}{\bf K},_{H}H)$ and the right homological integrals of $H$ as $\int_{H}^{r}:=Ext_{H}^{d}({\bf K}_{H},H_{H})$. Besides, homological integrals agree with the classical integrals for finite dimensional Hopf algebra in [10, section 1]. Then we get the following corollary.

Corollary 2.6**.**

$Ext^{3}_{H_{\beta}^{op}}({\bf K}_{H_{\beta}},{H_{\beta}}_{H_{\beta}})\simeq Ext^{3}_{H_{\beta}}(_{H_{\beta}}{\bf K},_{H_{\beta}}H_{\beta})\simeq{\bf K}$* as two sided $H_{\beta}$-modules.*

Proof.

Let $H^{\prime}=H_{\beta}/(b-1)$,$H^{\prime\prime}=H^{\prime}/(c-1)$ and $H^{\prime\prime\prime}=H^{\prime\prime}/(a^{n(n-1)}-1)$. Then $H^{\prime\prime\prime}$ is isomorphic to a Gelaki’s Hopf algebra ${\mathcal{U}}_{(n,n(n-1),n_{1},q,\beta_{1},\beta_{2},\beta_{3})}$. Since $a^{(n-1)n}-1,b-1,$ $c-1$ are in the center of $H_{\beta}$, we get $\int^{l}_{H_{\beta}}=\int^{l}_{H^{\prime}}=\int^{l}_{H^{\prime\prime}}=\int^{l}_{H^{\prime\prime\prime}}$ by [10, Lemma 2.6]. Since $H^{\prime\prime\prime}$ is finite dimensional, by [7, Proposition 3.2] we obtain that $\int^{l}_{H^{\prime\prime\prime}}={\bf K}\lambda$, where $\lambda=\frac{1}{n(n-1)}(\sum\limits_{i=0}^{n(n-1)-1}a^{i})x^{n-1}y^{n-1}$. It follows that ${\bf K}\lambda\simeq{\bf K}$ as $H_{\beta}$-bimodules, since $k\cdot\lambda=\varepsilon(k)\lambda$ for any $k\in H_{\beta}$ the same action as on $1\in{\bf K}$. ∎

3. Properties of $H_{\alpha,\beta}$

In this section, we construct and illustrate some properties of the Hopf algebra $H_{\alpha,\beta}$, which is a quotient of $H_{\beta}$. It is necessary to determine all irreducible representations of $H_{\beta}$ in the next section.

Definition 3.1**.**

Assume that $N=mn(n-1)$, where $m\geq 1$. Let $\alpha=(n,m,n_{1},n_{2},n_{3})\in{\bf N}^{5}$, where $1\leq n_{1}<m(n-1)$, $0\leq n_{2},n_{3}<n-1$. Let $I$ be an ideal of $H_{\beta}$ generated by $a^{N}-1,$ $b-a^{mnn_{2}}$ and $c-a^{mnn_{3}}$. Let

[TABLE]

Then $H_{\alpha,\beta}$ is a Hopf algebra generated by $a,x,y$ satisfying

[TABLE]

[TABLE]

The coalgebra structure of $H_{\alpha,\beta}$ is determined by

[TABLE]

[TABLE]

The antipode of $H_{\alpha,\beta}$ is determined by

[TABLE]

Besides, we can prove that $\lambda=\sum\limits_{i=0}^{N-1}\frac{a^{i}}{N}x^{n-1}y^{n-1}$ is a two-sided integral of $H_{\alpha,\beta}$. Indeed, by $yx-q^{-n_{1}}xy=\beta_{3}(a^{2n_{1}}-a^{mn(n_{2}+n_{3})})$ we get that

[TABLE]

where $u_{k}=q^{-(k-1)n_{1}}+\cdots+q^{-n_{1}}+1.$

Let $\alpha=(n,m,n_{1},n_{2},n_{3}),\alpha^{\prime}=(n^{\prime},m^{\prime},n_{1}^{\prime},n_{2}^{\prime},n_{3}^{\prime})$, $\beta=(1,1,\beta_{3})$ and $\beta^{\prime}=(1,1,\beta_{3}^{\prime})$. Suppose that $\beta_{1}\beta_{2}\neq 0$. Then we can assume that $\beta_{1}=\beta_{2}=1$. Moreover, we assume that $x^{n}\neq 0$, $y^{n}\neq 0$ and $yx\neq q^{-n_{1}}xy$. Under these assumptions, we have the following proposition.

Proposition 3.2**.**

$H_{\alpha,\beta}\simeq H_{\alpha^{\prime},\beta^{\prime}}$* if and only if $\alpha=\alpha^{\prime}$ and $\beta_{3}=c\beta_{3}^{\prime}$, where $c^{n}=1$.*

Proof.

Let $f$ be an isomorphism from $H_{\alpha,\beta}$ to $H_{\alpha^{\prime},\beta^{\prime}}$ and $H_{\alpha^{\prime},\beta^{\prime}}$ be a Hopf algebra generated by $g=a+I^{\prime}$, $z=x+I^{\prime}$ and $w=y+I^{\prime}$, where $I^{\prime}$ is the ideal generated by $a^{n^{\prime}(n^{\prime}-1)m^{\prime}}-1,$ $b-a^{n^{\prime}m^{\prime}n^{\prime}_{2}}$ and $c-a^{m^{\prime}n^{\prime}n^{\prime}_{3}}$. Notice that the subset $G(H_{\alpha,\beta})$ of all group-like elements of $H_{\alpha,\beta}$ is equal to $\{a^{t}|0\leq t\leq n(n-1)m-1\}$, and the subset $G(H_{\alpha^{\prime},\beta^{\prime}})$ of all group-like elements of $H_{\alpha^{\prime},\beta^{\prime}}$ is equal to $\{g^{t}|0\leq t\leq n^{\prime}(n^{\prime}-1)m^{\prime}-1\}$. Since $f$ induces an isomorphism from $G(H_{\alpha,\beta})$ to $G(H_{\alpha^{\prime},\beta^{\prime}})$ and $mn(n-1)=m^{\prime}n^{\prime}(n^{\prime}-1)$. Moreover, we can assume that $f(a)=g$. Since $f(a)f(x)=qf(x)f(a)$, $f(x)=\sum\limits_{i=0}^{N-1}\sum\limits_{j=1}^{n^{\prime}-1}x_{ij}g^{i}z^{j}w^{j-1}$ for some $x_{ij}\in{\bf K}$. By $\Delta(f(x))=(f\otimes f)\Delta(x)$, we get $f(x)=x_{01}z$. It is obvious that $x_{01}\neq 0$. Similarly, we can prove that $f(y)=uy$ for some nonzero $u\in{\bf K}$. Since $f(s(x))=s(f(x))$, we have $q^{-n_{1}}a^{-mnn_{2}-n_{1}}x=q^{-n^{\prime}_{1}}a^{-m^{\prime}n^{\prime}n^{\prime}_{2}-n^{\prime}_{1}}x$. Then $mnn_{2}=m^{\prime}n^{\prime}n^{\prime}_{2}$ and $n_{1}=n^{\prime}_{1}$. Since $x^{n}=a^{nn_{1}}-a^{mnn_{2}}$, we have

[TABLE]

Since $g^{nn_{1}-mnn_{2}}\neq 0$, we have $x_{01}^{n}=1$ and $n=n^{\prime}$. Thus $m=m^{\prime}$ and $n_{2}=n^{\prime}_{2}$. Similarly, we can prove that $u^{n}(g^{n^{\prime}n_{1}^{\prime}}-g^{m^{\prime}n^{\prime}n^{\prime}_{3}})=g^{nn_{1}}-g^{mnn_{3}}$. Hence $n_{3}=n_{3}^{\prime}$ and $u^{n}=1$. Notice that by applying the function $f$ to the equation $yx-q^{-n_{1}}xy=\beta_{3}(a^{2n_{1}}-a^{mn(n_{2}+n_{3})})$, we obtain the equation $\beta_{3}=ux_{01}\beta_{3}^{\prime}.$ ∎

Remark 3.3**.**

The Hopf algebra $H_{\alpha,\beta}$ is finite dimensional and by Theorem 2.2 we get

[TABLE]

Especially, if $n_{2}=n_{3}=0$ then $H_{\alpha,\beta}$ is Gelaki’s Hopf algebra by Remark 2.3.

Next, we determine the algebraic structure of $H_{\alpha,\beta}$. For this purpose, we construct some idempotent elements of $H_{\alpha,\beta}$. Suppose that $\omega_{0}$ is a primitive $m(n-1)$-th root of unity and

[TABLE]

Note that $\sum\limits_{j=0}^{m(n-1)-1}(\omega_{0}^{i})^{j}=0$ for any $i\neq 0$. Thus we have the following claim.

Lemma 3.4**.**

$1=e_{0}+e_{1}+\cdots+e_{m(n-1)-1}$* is a sum of central idempotent elements.*

Proof.

Since $a^{n}x=xa^{n},a^{n}y=ya^{n}$, $e_{i}$ are in the center of $H_{\alpha,\beta}$. Hence

[TABLE]

and

[TABLE]

where $\delta_{ij}$ is the Kronecker symbol. The claim is true.∎

Let $A_{i}=e_{i}H_{\alpha,\beta}$, where $i\in\{0,1,\cdots,m(n-1)-1\}$. Since $e_{i}a^{n}=\omega_{0}^{-i}e_{i}$, we obtain that the set $\{e_{i}a^{j}x^{l}y^{t}|0\leq j,l,t\leq n-1\}$ is a basis of $A_{i}$ and $dim_{\bf{K}}A_{i}=n^{3}$. It follows that $H_{\alpha,\beta}$ is a direct sum of algebras $A_{i}$ by Lemma 3.4.

Lemma 3.5**.**

Let $\omega$ be a primitive $N$-th root of unity such that $\omega^{n}=\omega_{0}$. Then $A_{i}$ is generated by $g=\omega^{i}e_{i}a,x^{\prime},y^{\prime}$, satisfying the following relations

[TABLE]

and

[TABLE]

where \beta_{1}^{\prime}=\beta_{1}(\omega_{0}^{-n_{1}i}-\omega_{0}^{-mn_{2}i}),\ \beta_{2}^{\prime}=\left\{\begin{array}[]{ll}\beta_{2}(\omega_{0}^{-n_{1}i}-\omega_{0}^{-mn_{3}i})&\beta_{3}=0\\ \beta_{2}\beta_{3}^{-n}(\omega_{0}^{n_{1}i}-\omega_{0}^{(2n_{1}-mn_{3})i})&\beta_{3}\neq 0\end{array}\right..

Proof.

Let $x^{\prime}=xe_{i}$ and y^{\prime}=\left\{\begin{array}[]{ll}ye_{i}&\beta_{3}=0\\ \beta_{3}^{-1}\omega^{2n_{1}i}ye_{i}&\beta_{3}\neq 0\end{array}\right.. It is obvious that $A_{i}$ is generated by $g,x^{\prime},y^{\prime}$. It is easy to check the relations on these generators $g,x^{\prime},y^{\prime}$ in this lemma. ∎

For the rest of this section, we illustrate the properties of $A_{i}$. For simplification, we denote $A$ as an algebra generated by $g,x,y$, satisfying

[TABLE]

and

[TABLE]

Let $f_{i}=\frac{1}{n}\sum\limits_{j=0}^{n-1}(q^{i}g)^{j},i=0,1,\cdots,n-1$. Then

[TABLE]

and $f_{i}f_{j}=\delta_{ij}$. Thus

[TABLE]

as a direct sum of left ideals.

Lemma 3.6**.**

Suppose that either $\beta_{1}^{\prime}\neq 0$ or $\beta_{2}^{\prime}\neq 0$. Then A is isomorphic to

[TABLE]

where $R_{i}$ are commutative local rings.

Proof.

It is easy to verify that $f_{i}x=xf_{i-1}$ and $f_{i}y=yf_{i+1}$. If $\beta_{1}^{\prime}\neq 0$, then $Af_{i}$ is isomorphic to $Af_{i-1}$ as left $A$-modules by multiplying $x$ from the right. If $\beta_{2}^{\prime}\neq 0$, then $Af_{i}$ is isomorphic to $Af_{i+1}$ as left $A$-modules by multiplying $y$ from the right. Thus $A$ is isomorphic to $Hom_{A}(A,A)\simeq M_{n}(End_{A}(Af_{i}))$. Since

[TABLE]

$End_{A}(Af_{i})=f_{i}Af_{i}=span\{f_{i}x^{t}y^{t}f_{i}|t=0,1,\cdots,n-1\}$.

Notice that $f_{l}yxf_{l}-q^{-n_{1}}f_{l}xyf_{l}=(q^{-2n_{1}l}-\omega_{0}^{2n_{1}i-m(n_{2}+n_{3})i})f_{l}$ when $\beta_{3}\neq 0$. Let $\gamma_{l}=0$ if $\beta_{3}=0$ and $\gamma_{l}=q^{-2n_{1}l}-\omega_{0}^{2n_{1}i-m(n_{2}+n_{3})i}$ if $\beta_{3}\neq 0$. Then

[TABLE]

Similarly, we can prove that $(f_{i}xyf_{i})^{k}=q^{-(k-1)n_{1}}f_{i}x^{k}y^{k}f_{i}+b_{k}f_{i}x^{k-1}y^{k-1}f_{i}+\cdots+b_{1}$ for $k\geq 2$. Hence $R=End_{A}(Af_{i})$ is generated by $f_{i}xyf_{i}$ over the field ${\bf K}$ as an algebra. Since $dim_{\bf{K}}R=n$, there exists a minimal polynomial $p(x)$ such that $p(f_{i}xyf_{i})=0$. Therefore, $R=R_{1}\oplus\cdots\oplus R_{t}$ is a direct sum of local rings.∎

Lemma 3.7**.**

Suppose that $\beta_{1}^{\prime}=\beta_{2}^{\prime}=\beta_{3}=0$. Then the Jacobson radical $J(A)$ of $A$ is spanned by $\{g^{i}x^{j}y^{k}|0\leq i,j,k\leq n-1,j+k>0\}$ and $A/J(A)\cong{\bf K}[{\mathbb{Z}}_{n}]$.

Proof.

Since the ideal $I=\langle x,y\rangle$ is nilpotent, we have $I\subseteq J(A).$ Besides, there is a $\bf{K}$ algebra isomorphism $A/I={\bf{K}}[g]/\langle g^{n}-1\rangle\cong{\bf{K}}[{\mathbb{Z}}_{n}]$. It follows that $J(A)=I.$ ∎

Lemma 3.8**.**

Suppose that $\beta_{1}^{\prime}=\beta_{2}^{\prime}=0$ and $\beta_{3}\not=0$. Then $A$ is generated by $K,E,F$ with relations

[TABLE]

[TABLE]

where $\rho=\omega_{0}^{2n_{1}i-mn(n_{2}+n_{3})i}$.

Proof.

Let $E=\frac{1}{q^{n_{1}}-1}yg^{n-n_{1}},F=\frac{1}{q^{n_{1}}+1}x$, $K=g$, and $K^{-1}=g^{n-1}$. It is obvious that $A$ is generated by $E,F,K$. Thus we have $EF-FE=K^{2n_{1}}\frac{K^{n_{1}}-\rho K^{-n_{1}}}{q^{n_{1}}-q^{-n_{1}}}$ from $yx-q^{-n_{1}}xy=g^{2n_{1}}-\omega^{2n_{1}i-mn(n_{2}+n_{3})i}.$ By $E=\frac{1}{q^{n_{1}}-1}yg^{n-n_{1}}$, then $E^{n}=(\frac{1}{q^{n_{1}}-1}yg^{n-n_{1}})^{n}=\frac{1}{(q^{n_{1}}-1)^{n}}q^{\frac{n(n-1)(n-n_{1})}{2}}y^{n}g^{(n-n_{1})n}$. Hence $E^{n}=0$ by the relation (2) $y^{n}=\beta_{2}^{\prime}=0$ of $A$. The other relations hold trivially. ∎

Notice that a weak Hopf algebra related to the algebra $A_{i}$ in Lemma 3.8 has been studied in [17]. Especially, if $N|[2n_{1}i-mn(n_{2}+n_{3})i]$ then the algebra $A$ is isomorphic to Radford’s Hopf algebra $U_{(n,n_{1},\omega)}$.

4. Irreducible representations of $H_{\beta}$

In this section, we determine all irreducible representations of $H_{\beta}$. The following key Lemma is necessary.

Lemma 4.1**.**

Every irreducible representation of $H_{\beta}$ is finite-dimensional.

Proof.

The proof is similar to that of [15, Corollary 1.2]. For the sake of completeness, we give a sketch. Let $M$ be a simple module over $H_{\beta}$ and $T:={\bf K}[a^{\pm n},b^{\pm 1},c^{\pm 1}]$. Since $T$ is contained in the center of $H_{\beta}$, every element $t$ of $T$ induces an endomorphism of the $H_{\beta}$-module $M$. We denote $\varphi(t)$ as the endomorphism of $M$ induced by $t\in T$. Suppose that ${\mathcal{K}}=\{\varphi(t)|t\in T\}$. Then $kM=M$ for any nonzero $k\in{\mathcal{K}}$. Since $H_{\beta}$ is finitely generated over $T$, $M$ is a finitely generated ${\mathcal{K}}$-module. Let $m_{1},m_{2},\cdots,m_{r}$ be the generators of $M$. By $kM=M$, we obtain

[TABLE]

for some $a_{ij}\in{\mathcal{K}}$. Then $det(I-k(a_{ij}))=0$, where $I$ is the identity matrix with order $r$. Thus, there exists $h\in{\mathcal{K}}$ such that $kh=1$. Hence ${\mathcal{K}}$ is a field. Notice that ${\mathcal{K}}={\bf K}[\varphi(a^{\pm n}),\varphi(b^{\pm 1}),\varphi(c^{\pm 1})]$. It is an algebraic extension of ${\bf K}$. Since ${\bf K}$ is an algebraically closed field, ${\bf K}={\mathcal{K}}$. Consequently, $M$ is finite-dimensional. ∎

From the proof of Lemma 4.1, we can construct an algebra homomorphism $\lambda_{M}\in Hom_{\bf K}(T,{\bf K})$ for any simple module $M$. Let $\lambda_{M}(a^{n})=\gamma_{1},\lambda_{M}(b)=\gamma_{2}$ and $\lambda_{M}(c)=\gamma_{3}$. Since $a,b,c$ are invertible in $T$, $\gamma_{i}\neq 0$ for $i=1,2,3$. Then $M$ can be viewed as a module over $H_{M}:=H_{\beta}/(a^{n}-\gamma_{1},b-\gamma_{2},c-\gamma_{3})$. Let $a^{\prime}=\frac{1}{\sqrt[n]{\gamma_{1}}}a$, $b^{\prime}=\frac{1}{\sqrt[n]{\gamma_{1}^{n_{1}}}}b$, $c^{\prime}=\frac{1}{\sqrt[n]{\gamma_{1}}^{n_{1}}}c$, $x^{\prime}=x$ and y^{\prime}=\left\{\begin{array}[]{ll}y&\beta_{3}=0\\ \beta_{3}^{-1}\gamma_{1}^{-\frac{2n_{1}}{n}}y&\beta_{3}\neq 0\end{array}\right.. Then $H_{\beta}$ is generated by $a^{\prime},b^{\prime},c^{\prime},x^{\prime},y^{\prime}$ with the relations

[TABLE]

[TABLE]

[TABLE]

where $\beta_{1}^{\prime}=\gamma_{1}^{n_{1}}\beta_{1},$ \beta_{2}^{\prime}=\left\{\begin{array}[]{lll}\gamma_{1}^{n_{1}}\beta_{2}&&\beta_{3}=0\\ \beta_{3}^{-n}\gamma_{1}^{-n_{1}}\beta_{2}&&\beta_{3}\neq 0\end{array}\right.. Thus the generators $a^{\prime},x^{\prime},y^{\prime}$ in $H_{M}$ satisfy

[TABLE]

[TABLE]

In the following we illustrate all irreducible representations of $H_{\beta}$.

Lemma 4.2**.**

Let $\beta_{1}^{\prime\prime}=\beta_{1}^{\prime}(1-\frac{\gamma_{2}^{n}}{\gamma_{1}^{n_{1}}}),\ \beta_{2}^{\prime\prime}=\beta_{2}^{\prime}(1-\frac{\gamma_{3}^{n}}{\gamma_{1}^{n_{1}}}),$ and

[TABLE]

• (I)

Suppose that $\beta_{1}^{\prime\prime}\neq 0$. Then the irreducible $H_{\beta}$-module $M$ has a basis $\{m_{0},m_{1},\cdots,m_{n-1}\}$ such that the actions of $a,b,c,x,y$ on $M$ with respect to this basis are given by

[TABLE]

[TABLE]

[TABLE]

where $k_{1},\cdots,k_{n}$ are determined by

[TABLE]

for $2\leq l\leq n$ and $k_{1}k_{2}\cdots k_{n}=(\gamma_{1}^{n_{1}}-\gamma_{3}^{n})\beta_{2}$. We denote this irreducible module $M$ by $V_{I}(\gamma_{1},\gamma_{2},\gamma_{3};i)$.

• (II)

Suppose that $\beta_{1}^{\prime\prime}=0$ and $\beta_{2}^{\prime\prime}\neq 0$. Then the irreducible $H_{\beta}$-module $M$ has a basis

[TABLE]

such that the actions of $a,b,c,x,y$ on $M$ with respect to this basis satisfying

[TABLE]

[TABLE]

[TABLE]

where $k_{1},\cdots,k_{n}$ are determined by

[TABLE]

for $1\leq l\leq n-1$ and $k_{1}k_{2}\cdots k_{n}=(\gamma_{1}^{n_{1}}-\gamma_{2}^{n})\beta_{1}$. We denote this irreducible module $M$ by $V_{II}(\gamma_{1},\gamma_{2},\gamma_{3};i)$.

Proof.

The proof of (I) is similar to that of (II). We only give the proof of (II). Similarly to the proof of Lemma 3.6, we can prove that $H_{M}$ is a direct sum of $M_{n}(R_{t})$, where $R_{t}$ are commutative local rings. Hence every irreducible module over $H_{M}$ is of dimension $n$. Since $H_{\beta}$ module $M$ can be viewed as a $H_{M}$ module, $dim(M)=n$. Let $m_{0}$ be an eigenvector of $a^{\prime}$ with eignvalue $q^{i}$ for some $0\leq i\leq n-1$. Since $y^{\prime n}m_{0}=\beta_{2}^{\prime\prime}m_{0}\neq 0$, $\{m_{0},y^{\prime}m_{0},\cdots,y^{\prime n-1}m_{0}\}$ is a basis of $M$. Moreover $a^{\prime}x^{\prime}m_{0}=q^{-1}x^{\prime}a^{\prime}m_{0}=q^{i-1}x^{\prime}m_{0}$. Therefore $x^{\prime}m_{0}=k^{\prime}_{n}y^{\prime n-1}m_{0}$ for some $k^{\prime}_{n}\in{\bf K}$. Thus $M$ is a $H_{\beta}$ module with a basis $\{m_{0},ym_{0},\cdots,y^{n-1}m_{0}\}$ and the action of $a$ and $y$ on $M$ with respect to this basis is given by (6) and (7) respectively. Meanwhile, $xm_{0}=k_{n}y^{n-1}m_{0}$ for some $k_{n}\in{\bf K}$. Since $xa=qax$ and $y^{l}x-q^{-ln_{1}}xy^{l}=\beta_{3}\sum^{l-1}_{j=0}q^{-jn_{1}}(q^{-(l-1)n_{1}}a^{2n_{1}}-bc)y^{l-1}$, the action of $x$ can be realized by

[TABLE]

where

[TABLE]

Since ${\mathcal{X}}^{n}=k_{1}k_{2}\cdots k_{n}I_{n}=\beta_{1}(\gamma_{1}^{n_{1}}-\gamma_{2}^{n})I_{n}$, $k_{n}$ is a root of the equation

[TABLE]

Hence, the claim is true. ∎

Lemma 4.3**.**

Suppose that $\beta_{1}^{\prime\prime}=\beta_{2}^{\prime\prime}=\beta_{3}^{\prime\prime}(i)=0$ for some $0\leq i\leq n-1$. Then the irreducible $H_{\beta}$ module $M={\bf K}$ is determined by $a\cdot 1=\sqrt[n]{\gamma_{1}}q^{i},\ b\cdot 1=\gamma_{2},\ c\cdot 1=\gamma_{3}$ and $x\cdot 1=y\cdot 1=0$ for some $0\leq i\leq n-1$. We denote this irreducible module $M$ by $V_{0}(\gamma_{1},\gamma_{2},\gamma_{3};i)$ in the sequel.

Proof.

We first consider $M$ as the irreducible $H_{M}$ module. Suppose that $m_{0}$ is an eigenvector of $a^{\prime}$. Since $x^{\prime n}m_{0}=0$, we can assume that $x^{\prime}m_{0}=0$. If $y^{\prime}m_{0}\neq 0$, then there exist $j\in\{1,\cdots,n-1\}$ such that $y^{\prime j}m_{0}\neq 0$ and $y^{\prime j+1}m_{0}=0$. Hence $\{m_{0},y^{\prime}m_{0},y^{\prime 2}m_{0},\cdots,y^{\prime j}m_{0}\}$ is a basis of the module $M$. $<y^{\prime j}m_{0}>$ is the submodule of $M$ by $x^{\prime}y^{\prime j}m_{0}=q^{jn_{1}}y^{\prime j}x^{\prime}m_{0}=0$. This gives a contradiction. Thus $x\cdot m_{0}=y\cdot m_{0}=0$ and $M={\bf K}m_{0}\cong{\bf K}$. ∎

Suppose that $\beta_{3}(i)^{\prime\prime}=\beta_{3}(q^{2n_{1}i}-\gamma_{1}^{-\frac{2n_{1}}{n}}\gamma_{2}\gamma_{3})\neq 0$. If there is an integer $v$ such that

[TABLE]

then there is a minimal positive integer $r$ such that $q^{(2i-r+1)n_{1}}=\gamma_{1}^{-\frac{2n_{1}}{n}}\gamma_{2}\gamma_{3}$. As $\beta_{3}(i)^{\prime\prime}\neq 0$, we have $q^{(2i-r+1)n_{1}}\neq q^{2n_{1}i}$ and $q^{(1-r)n_{1}}\neq 1$. Thus $2\leq r\leq t,$ where $t=|q^{n_{1}}|$ the order of $q^{n_{1}}$. If there exists no integer $v$ such that $q^{(2i-v+1)n_{1}}=\gamma_{1}^{-\frac{2n_{1}}{n}}\gamma_{2}\gamma_{3}$, then we define $r:=t$.

Lemma 4.4**.**

Suppose that $\beta_{1}^{\prime\prime}=\beta_{2}^{\prime\prime}=0$, but $\beta_{3}^{\prime\prime}(i)\not=0$ for some $0\leq i\leq n-1$. Then the irreducible $H_{\beta}$ module $M$ has a basis $\{m_{i}|0\leq i\leq r-1\}$ such that the actions of $a,b,c,x,y$ are given by

[TABLE]

[TABLE]

[TABLE]

where $k_{1},\cdots,k_{r}$ are determined by

[TABLE]

for $1\leq l\leq r-1.$

Proof.

We first consider $M$ as the irreducible $H_{M}$ module. Suppose that $m_{0}$ is an eigenvector of $a^{\prime}$. Since $y^{\prime n}m_{0}=0$, we can assume that $y^{\prime}m_{0}=0$. Moreover, we can assume that $x^{\prime j}m_{0}\neq 0$ for $0\leq j\leq r^{\prime}-1$ and $x^{\prime r^{\prime}}m_{0}=0$. It is obvious that $r^{\prime}\leq n$ and $a^{\prime}m_{0}=q^{i}m_{0}$ for some $1\leq i\leq n$. It is easy to check that

[TABLE]

Since

[TABLE]

either $q^{r^{\prime}n_{1}}=1$, or $q^{(2i-r^{\prime}+1)n_{1}}=\gamma_{1}^{-\frac{2n_{1}}{n}}\gamma_{2}\gamma_{3}$. If $r^{\prime}>t$, then $span\{x^{\prime t}m_{0},\cdots,x^{\prime r^{\prime}-1}m_{0}\}$ is a nonzero submodule of the simple module $M$. This is impossible. Thus $r^{\prime}\leq t$. Similarly, we have $r\leq r^{\prime}$. Hence $r^{\prime}=min\{t,r\}.$ Since $y^{\prime}x^{\prime j}m_{0}=(\sum\limits_{v=0}^{j-1}q^{(2i-v)n_{1}}-\sum\limits_{v=0}^{j-1}q^{vn_{1}}\gamma_{1}^{-\frac{2n_{1}}{n}}\gamma_{2}\gamma_{3})x^{\prime j-1}m_{0}$ for all $j\geq 1$, $y^{\prime}(x^{\prime l}m_{0})=k^{\prime}_{l}(x^{\prime l-1}m_{0})$ for $1\leq l\leq r-1$, where $k^{\prime}_{l}=\sum^{l-1}_{j=0}q^{-jn_{1}}(q^{(2i-l+1)n_{1}}-\gamma^{-\frac{2n_{1}}{n}}_{1}\gamma_{2}\gamma_{3}).$ Finally, we prove that $M=span\{x^{\prime v}m_{0}|0\leq v\leq r-1\}$ is a simple $H_{M}$-module. Let $V^{\prime}$ be a nonzero module of $M$. Then there exists $x^{\prime i-s}m_{0}\in V^{\prime}$ for some $0\leq s\leq r-1$. If $s=i$, then $m_{0}\in V^{\prime}$. If $s\neq i$, then $y^{\prime i-s}x^{\prime i-s}m_{0}=k^{\prime}_{1}k^{\prime}_{2}\cdots k^{\prime}_{i-s}m_{0}\in V^{\prime}$. Thus $m_{0}\in V^{\prime}$ and $V^{\prime}=M$. Since $a^{\prime}=\frac{1}{\sqrt[n]{\gamma_{1}}}a$, $b^{\prime}=\frac{1}{\sqrt[n]{\gamma_{1}^{n_{1}}}}b$, $c^{\prime}=\frac{1}{\sqrt[n]{\gamma_{1}}^{n_{1}}}c$, $x^{\prime}=x$, and $y^{\prime}=\beta_{3}^{-1}\gamma_{1}^{-\frac{2n_{1}}{n}}y$, the action of $a,b,c,x,y$ on $M$ is given by (9)(10)(11). ∎

Recall the modules $V_{I}(\gamma_{1},\gamma_{2},\gamma_{3};i)$, $V_{II}(\gamma_{1},\gamma_{2},\gamma_{3};i)$ and $V_{0}(\gamma_{1},\gamma_{2},\gamma_{3};i)$ defined in Lemma 4.2 and Lemma 4.3. We denote the irreducible module described in Lemma 4.4 by $V_{r}(\gamma_{1},\gamma_{2},\gamma_{3};i)$. By all the above lemmas, we obtain the following classification theorem.

Theorem 4.5**.**

Let $M$ be an irreducible representation of $H_{\beta}$. Then $M$ is isomorphic to one of the following types: $V_{I}(\gamma_{1},\gamma_{2},\gamma_{3};i)$, $V_{II}(\gamma_{1},\gamma_{2},\gamma_{3};i)$, $V_{0}(\gamma_{1},\gamma_{2},\gamma_{3};i)$, and $V_{r}(\gamma_{1},\gamma_{2},\gamma_{3};i)$.

Proof.

It is easy to verify that $V_{I}(\gamma_{1},\gamma_{2},\gamma_{3};i)$, $V_{II}(\gamma_{1},\gamma_{2},\gamma_{3};i)$, $V_{0}(\gamma_{1},\gamma_{2},\gamma_{3};i)$, and $V_{r}(\gamma_{1},\gamma_{2},\gamma_{3};i)$ are irreducible representations of $H_{\beta}$. Let $M$ be an irreducible representation of $H_{\beta}$. Then there exists an algebra homomorphism $\lambda\in Hom_{\bf{K}}(T,{\bf K})$ such that $\lambda(a^{n})=\gamma_{1}\neq 0$, $\lambda(b)=\gamma_{2}\neq 0$ and $\lambda(c)=\gamma_{3}\neq 0$. Since $\gamma_{1},\gamma_{2},\gamma_{3}$ lie in one of the above four cases, $M$ can be viewed as a representation of $H_{M}$ described by Lemma 4.2, Lemma 4.3 and Lemma 4.4. Being careful with the change of the generators, $M$ is isomorphic to one of the above four kinds of irreducible representations. ∎

Next proposition shows that the category of all $H_{\beta}$ modules is not semisimple.

Proposition 4.6**.**

Suppose that $\gamma_{1}^{-\frac{2n_{1}}{n}}\gamma_{2}\gamma_{3}\neq q^{vn_{1}}$ for any $v$ and $(\gamma_{1}^{n_{1}}-\gamma_{2}^{n})\beta_{1}=(\gamma_{1}^{n_{1}}-\gamma_{3}^{n})\beta_{2}=0$. Then $Ext_{H_{\beta}}(V_{t}(\gamma_{1},\gamma_{2},\gamma_{3};i-t+n),V_{t}(\gamma_{1},\gamma_{2},\gamma_{3};i))\neq 0$ for all $1\leq i\leq n-1$, where $t=|q^{n_{1}}|$.

Proof.

Let $k_{u}(i)=\frac{1-q^{-un_{1}}}{1-q^{-n_{1}}}(q^{(2i-u+1)n_{1}}-\gamma_{1}^{-\frac{2n_{1}}{n}}\gamma_{2}\gamma_{3})$ and $L$ be a vector space with a basis $\{v_{1},\cdots,v_{2t}\}$. Set

[TABLE]

Then $span\{v_{t+1},\cdots,v_{2t}\}$ is isomorphic to $V_{t}(\gamma_{1},\gamma_{2},\gamma_{3};i-t+n)$ and

[TABLE]

Obviously, the sequence $0\rightarrow V_{t}(\gamma_{1},\gamma_{2},\gamma_{3};i-t+n)\rightarrow L\rightarrow V_{t}(\gamma_{1},\gamma_{2},\gamma_{3};i)\rightarrow 0$ is not splitting. Hence $Ext_{H_{\beta}}(V_{t}(\gamma_{1},\gamma_{2},\gamma_{3};i-t+n),V_{t}(\gamma_{1},\gamma_{2},\gamma_{3};i))\neq 0$.∎

Remark 4.7**.**

Since $H_{\beta}$ is a Hopf algebra, the dual of any $H_{\beta}$ module is still a $H_{\beta}$ module. It is easy to prove that the dual of any irreducible representation is irreducible. Therefore the dual module $V_{I}(\gamma_{1},\gamma_{2},\gamma_{3};i)^{*}$, $V_{II}(\gamma_{1},\gamma_{2},\gamma_{3};i)^{*}$, $V_{0}(\gamma_{1},\gamma_{2},\gamma_{3};i)^{*}$, and $V_{r}(\gamma_{1},\gamma_{2},\gamma_{3};i)^{*}$ are also irreducible representations of $H_{\beta}$.

5. Grothendieck ring $G_{0}(H_{\beta})$

We recall some notations. Let $R$ be an algebra over field K. Recall that the Grothendieck group $G_{0}(R)$ is the abelian group generated by the set $\{[M]|$ $M$ is a left $R$-module$\}$ of isomorphic classes of finite-dimensional $R$-modules with relations $[B]=[A]+[C]$ for every short exact sequence $0\to A\to B\to C\to 0$ of R-modules. Unlike ordinary algebras, the Grothendieck group $G_{0}(H)$ of a Hopf algebra $H$ has a product given by $[M][N]=[M\otimes N]$. With this product, Grothendieck group $G_{0}(H)$ becomes a ring, which is called the Grothendieck ring of the Hopf algebra $H$.

Assume that $H_{\beta}$ is the Hopf algebra defined in Section 2 and the modules

[TABLE]

are defined in Lemma 4.2, Lemma 4.3 and Lemma 4.4. In this section, we determine the Grothendieck ring $G_{0}(H_{\beta})$ of the Hopf algebra $H_{\beta}$ in several cases, which is generated by

[TABLE]

where $0\leq i\leq n-1$, $2\leq r\leq t$, $(\gamma_{1},\gamma_{2},\gamma_{3})\in(\bf{K}^{*})^{3}$ and $t=|q^{n_{1}}|$ is the order of $q^{n_{1}}$. The classification results of all cases are illustrated in several lemmas and theorems. In details, the Grothendieck rings in cases that iff two the three equations $\beta_{1}=0$, $\beta_{2}=0$ and $\beta_{3}=0$ hold are stated in theorems 5.5, 5.12 and 5.16. To do that, lemmas 5.4,  5.7 and 5.14 are necessary. In theorems 5.10,   5.18 and 5.20, we give the cases in which iff one of the conditions $\beta_{1}=0$, $\beta_{2}=0$ and $\beta_{3}=0$ holds. In additional, we also obtain properties of the Grothendieck rings $G_{0}(U_{(N,\nu,\omega)})$ and $G_{0}({\mathcal{U}}_{(n,N,\nu,q,\alpha,\beta,\gamma)})$ in several corollaries. We also observe that there exist non-isomorphic Hopf algebras with isomorphic Grothendieck rings.

In order to discuss $G_{0}(H_{\beta})$ in which $\beta_{1}=\beta_{2}=\beta_{3}=0$, the analyze to the subring of $G_{0}(H_{\beta})$ generated by all $[V_{0}(\gamma_{1},\gamma_{2},\gamma_{3};i)]$ for all nonzero $\gamma_{i}\in{\bf K}$ and $i\in\mathbb{Z}_{n}$ is crucial. We present the results of this case in Theorem 5.2 and the following Lemma 5.1 is necessary.

Lemma 5.1**.**

The product of the Grothendieck ring $G_{0}(H_{\beta})$ satisfies

[TABLE]

for any $i,j\in\mathbb{Z}_{n}$ and $\gamma_{k},\gamma_{k}^{\prime}\in{\bf K}^{*}$, $k\in\{1,2,3\}$.

Proof.

By the definition of $V_{0}(\gamma_{1},\gamma_{2},\gamma_{3};i)$ in Lemma 4.4, we obtain that the product

[TABLE]

is generated by $1\otimes 1$. Since

[TABLE]

and

[TABLE]

we have

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Therefore, $V_{0}(\gamma_{1},\gamma_{2},\gamma_{3};i)\otimes V_{0}(\gamma^{\prime}_{1},\gamma^{\prime}_{2},\gamma^{\prime}_{3};j)\cong V_{0}(\gamma_{1}\gamma^{\prime}_{1},\gamma_{2}\gamma^{\prime}_{2},\gamma_{3}\gamma^{\prime}_{3};i+j).$ ∎

By Lemma 5.1, we can prove the following result.

Theorem 5.2**.**

Let $R_{1}$ be the subring of the Grothendieck ring $G_{0}(H_{\beta})$ generated by all $[V_{0}(\gamma_{1},\gamma_{2},\gamma_{3};i)]$ for all nonzero $\gamma_{i}\in{\bf K}$ and $i\in\mathbb{Z}_{n}$. Then we have the following.

• (i)

If at most one of $\beta_{1},\beta_{2},\beta_{3}$ equals to zero, then $R_{1}$ is isomorphic to $\mathbb{Z}[\mathbb{Z}_{n_{1}}\times\mathbb{Z}_{n}^{2}\times{\bf K^{*}}]$;

• (ii)

If only $\beta_{1}\neq 0$ or $\beta_{2}\neq 0$, then $R_{1}$ is isomorphic to $\mathbb{Z}[\mathbb{Z}_{n_{1}}\times\mathbb{Z}_{n}\times{(\bf K^{*})}^{2}]$;

• (iii)

If only $\beta_{3}\neq 0$, then $R_{1}$ is isomorphic to $\mathbb{Z}[\mathbb{Z}_{2n_{1}}\times\mathbb{Z}_{n}\times{(\bf K^{*})}^{2}]$;

• (iv)

If $\beta_{1}=\beta_{2}=\beta_{3}=0$, then $R_{1}$ is isomorphic to $\mathbb{Z}[\mathbb{Z}_{n}\times{\bf K^{*}}\times{\bf K^{*}}\times{\bf K^{*}}]$.

Thus $G_{0}(H_{\beta})\cong\mathbb{Z}[\mathbb{Z}_{n}\times{\bf K^{*}}\times{\bf K^{*}}\times{\bf K^{*}}]$ if $\beta=(\beta_{1},\beta_{2},\beta_{3})=0$.

Proof.

We prove this claim by several cases.

• (i)

First, we consider the case where $\beta_{1}\beta_{2}\beta_{3}\neq 0$. By $\beta^{\prime\prime}_{1}=0$ and $\beta^{\prime\prime}_{2}=0$, we obtain $\beta_{1}^{\prime\prime}=\beta_{1}^{\prime}(1-\frac{\gamma_{2}^{n}}{\gamma_{1}^{n_{1}}})=0$ and $\beta_{2}^{\prime\prime}=\beta_{2}^{\prime}(1-\frac{\gamma_{3}^{n}}{\gamma_{1}^{n_{1}}})=0$. Hence $\gamma_{1}=\gamma_{2}^{\frac{n}{n_{1}}}\bar{\omega}^{p}=\gamma_{3}^{\frac{n}{n_{1}}}\bar{\omega}^{l}$ and $\gamma_{3}=q^{u}\gamma_{2}$ for some intergers $p,l,u$, where $\bar{\omega}$ is the primitive $n_{1}$-th unitary root. Thus

[TABLE]

for some nonzero $\gamma_{2}$ and $i\in\mathbb{Z}_{n}$. Then $R_{1}$ is isomorphic to the group algebra $\mathbb{Z}[\mathbb{Z}_{n_{1}}\times\mathbb{Z}^{2}_{n}\times{\bf K^{*}}]$, where ${\bf K^{*}}$ is the multiplicative group of all nonzero elements of ${\bf K}$.

If $\beta_{2}\beta_{3}\neq 0$ and $\beta_{1}=0$, then $\gamma_{1}=\gamma_{3}^{\frac{n}{n_{1}}}\bar{\omega}^{p}$ for some integer $p$. Since $\gamma_{1}^{-\frac{2n_{1}}{n}}\gamma_{2}\gamma_{3}=q^{2n_{1}i}$, $\gamma_{2}=q^{2n_{1}i+u}\gamma_{3}$ for some integer $u$. Thus

[TABLE]

Hence $R_{1}$ is isomorphic to the group algebra $\mathbb{Z}[\mathbb{Z}_{n_{1}}\times\mathbb{Z}_{n}^{2}\times{\bf K^{*}}]$. Similarly, we can prove that $R_{1}$ is also isomorphic to $\mathbb{Z}[\mathbb{Z}_{n_{1}}\times\mathbb{Z}_{n}^{2}\times{\bf K}^{*}]$ if $\beta_{1}\beta_{3}\neq 0$ and $\beta_{2}=0$.

If $\beta_{1}\beta_{2}\neq 0$ and $\beta_{3}=0$, then $\gamma_{1}=\gamma_{3}^{\frac{n}{n_{1}}}\bar{\omega}^{p}$ and $\gamma_{2}=\gamma_{3}q^{u}$ for some integers $p,u$. Thus

[TABLE]

Hence $R_{1}$ is isomorphic to $\mathbb{Z}[\mathbb{Z}_{n_{1}}\times\mathbb{Z}^{2}_{n}\times{\bf K}^{*}]$.

• (ii)

If $\beta_{1}\neq 0$ and $\beta_{2}=\beta_{3}=0$, then $\gamma_{1}=\gamma_{2}^{\frac{n}{n_{1}}}\bar{\omega}^{p}$ for some integer $p$. Thus

[TABLE]

Hence $R_{1}$ is isomorphic to $\mathbb{Z}[\mathbb{Z}_{n_{1}}\times\mathbb{Z}_{n}\times{({\bf K}^{*})}^{2}]$. Similarly, we can prove that $R_{1}$ is isomorphic to $\mathbb{Z}[\mathbb{Z}_{n_{1}}\times\mathbb{Z}_{n}\times{({\bf K}^{*})}^{2}]$ if $\beta_{2}\neq 0$ and $\beta_{1}=\beta_{3}=0$.

• (iii)

If $\beta_{1}=\beta_{2}=0$ and $\beta_{3}\neq 0$, then $\gamma_{1}=(\gamma_{2}\gamma_{3})^{\frac{n}{2n_{1}}}\bar{\omega}^{\frac{p}{2}}$ for some integer $p$. Thus

[TABLE]

Hence $R_{1}$ is isomorphic to $\mathbb{Z}[\mathbb{Z}_{2n_{1}}\times\mathbb{Z}_{n}\times{({\bf K}^{*})}^{2}]$.

• (iv)

Finally, if $\beta=0$, then \begin{array}[]{lll}V_{0}(\gamma_{1},\gamma_{2},\gamma_{3},i)&\cong&V_{0}(1,1,1;i)\otimes V_{0}(\gamma_{1},\gamma_{2},\gamma_{3};0).\end{array} Hence $R_{1}$ is isomorphic to $\mathbb{Z}[\mathbb{Z}_{n}\times{\bf K}^{*}\times{\bf K}^{*}\times{\bf K}^{*}]$.

∎

Since $H_{\beta}/(a^{N}-1,b-1,c-1)$ is isomorphic to Gelaki’s Hopf algebra ${\mathcal{U}}_{(n,N,n_{1},q,\beta_{1},\beta_{2},\beta_{3})}$, where $n|N$, we obtain the following properties of the Gelaki’s Hopf alegbra.

Corollary 5.3**.**

Suppose that $n|N$. Let $\mathfrak{q}$ be a primitive $N$-th root of unity and $S_{1}$ be the subring of $R_{1}$ generated by $[V_{0}(\gamma,1,1;i)]$, where $\gamma^{\frac{N}{n}}=1$. Then we have the following result.

• (1)

If either $\beta_{1}\beta_{2}\neq 0$ and $\beta_{3}=0$, or $\beta_{1}\neq 0$ and $\beta_{2}=\beta_{3}=0$, or $\beta_{2}\neq 0$ and $\beta_{1}=\beta_{3}=0$, then $S_{1}=\mathbb{Z}[g,h]$ for $g=[V_{0}(1,1,1;1)]$ and $h=[V_{0}(\mathfrak{q}^{\frac{N}{(N/n,n_{1})}},1,1;0)]$. Moreover,

[TABLE]

• (2)

If $\beta_{3}\neq 0$ and $\beta_{1}=\beta_{2}=0$, then $S_{1}=\mathbb{Z}[g,h_{1}]$ for $g=[V_{0}(1,1,1;1)]$ and

[TABLE]

Moreover, $g^{n}=h_{1}^{\mu}=1$ and $\mathbb{Z}[g,h_{1}]\simeq\mathbb{Z}[\mathbb{Z}_{\mu}\times\mathbb{Z}_{n}]$, where $\mu=\frac{(N,2n_{1})}{(n,2n_{1})}$.

• (3)

If either $\beta_{1}\beta_{3}\neq 0$ and $\beta_{2}=0$, or $\beta_{2}\beta_{3}\neq 0$ and $\beta_{1}=0$, or $\beta_{1}\beta_{2}\beta_{3}\neq 0$, then $S_{1}=\mathbb{Z}[g,h_{2}]$ for $g=[V_{0}(1,1,1;1)]$ and $h_{2}=[V_{0}(\mathfrak{q}^{n\nu},1,1;0)]$, where $\nu=\frac{N}{(N,2n_{1},nn_{1})}$. Moreover, $g^{n}=h_{2}^{\mu^{\prime}}=1$ and $\mathbb{Z}[g,h_{2}]\simeq\mathbb{Z}[\mathbb{Z}_{\mu^{\prime}}\times\mathbb{Z}_{n}]$, where $\mu^{\prime}=\frac{(N,2n_{1},nn_{1})}{(n,2n_{1})}$.

• (4)

If either $\beta_{1}=\beta_{2}=\beta_{3}=0$, then $S_{1}=\mathbb{Z}[g,h_{3}]$ for $g=[V_{0}(1,1,1;1)]$ and $h_{3}=[V_{0}(\mathfrak{q}^{n},1,1;0)]$. Moreover, $g^{n}=h_{3}^{\frac{N}{n}}=1$ and $\mathbb{Z}[g,h_{3}]\simeq\mathbb{Z}[\mathbb{Z}_{N/n}\times\mathbb{Z}_{n}].$

• (5)

If $\beta_{1}=\beta_{2}=\beta_{3}=0$, then the Grothendieck ring $G_{0}({\mathcal{U}}_{(n,N,n_{1},q,\beta_{1},\beta_{2},\beta_{3})})$ of the Gelaki’s Hopf algebra ${\mathcal{U}}_{(n,N,n_{1},q,0,0,0)}$ is equal to $\mathbb{Z}[g,h_{3}]$.

Proof.

If $\beta_{1}\beta_{2}\neq 0$ and $\beta_{3}=0$, then $\gamma=\bar{\omega}^{p}$ satisfying $\gamma^{\frac{N}{n}}=1$, where $\bar{\omega}$ is a primitive $n_{1}$-th root of unity. Thus $n_{1}|\frac{N}{n}p$ and $\frac{n_{1}}{(n_{1},\frac{N}{n})}|p$. Therefore $\gamma\in\langle\bar{\omega}^{\frac{n_{1}}{(N/n,n_{1})}}\rangle=\langle\mathfrak{q}^{\frac{N}{(N/n,n_{1})}}\rangle$, where $\mathfrak{q}$ is a primitive $N$-th root of unity. Hence $S_{1}=\mathbb{Z}[g,h]\simeq\mathbb{Z}[\mathbb{Z}_{{(N/n,n_{1})}}\times\mathbb{Z}_{n}]$ by (12), where $g=[V_{0}(1,1,1;1)]$, $h=[V_{0}(\mathfrak{q}^{\frac{N}{(N,n_{1})}},1,1;0)]$. It is obvious that $g^{n}=h^{{(N/n,n_{1})}}=1$.

If $\beta_{3}\neq 0$ and $\beta_{1}=\beta_{2}=0$, then $\gamma^{-\frac{2n_{1}}{n}}=1$ for $V_{0}(\gamma,1,1;0)$. Since $\gamma^{\frac{N}{n}}=1$, $\gamma=\mathfrak{q}^{nr}$ for some integer $r$, where $\mathfrak{q}$ is a primitive $N$-th root of unity. Then $\mathfrak{q}^{-2rn_{1}}=1$ and $N|2rn_{1}$. Thus $r=\frac{N}{(N,2n_{1})}p$ for any integer $p$. Hence $\gamma\in\langle\mathfrak{q}^{\frac{Nn}{(N,2n_{1})}}\rangle$. Thus $V_{0}(\gamma,1,1;i)\cong V_{0}(\mathfrak{q}^{\frac{Nn}{(N,2n_{1})}},1,1;0)^{\otimes p}\otimes V_{0}(1,1,1;1)^{\otimes i}.$ Hence $S_{1}=\mathbb{Z}[g,h_{1}]$ for $h_{1}=[V_{0}(\mathfrak{q}^{\frac{Nn}{(N,2n_{1})}},1,1;0)].$ It is obvious that $S_{1}$ is isomorphic to the group algebra $\mathbb{Z}[\mathbb{Z}_{\mu}\times\mathbb{Z}_{n}]$, where $\mu=\frac{(N,2n_{1})}{(n,2n_{1})}$.

If either $\beta_{1}\beta_{3}\neq 0$, or $\beta_{2}\beta_{3}\neq 0$, or $\beta_{1}\beta_{2}\beta_{3}\neq 0$, then $\gamma^{\frac{N}{n}}=\gamma^{n_{1}}=\gamma^{\frac{2n_{1}}{n}}=1$ for $V_{0}(\gamma,1,1;0)$. Thus $\gamma=\mathfrak{q}^{nr}$ for some integer $r$ and $\mathfrak{q}^{nrn_{1}}=\mathfrak{q}^{2n_{1}r}=1$. Hence $r=\frac{N}{(N,nn_{1})}p=\frac{N}{(N,2n_{1})}p^{\prime}$ for some integers $p,p^{\prime}$. Let $\bar{d}=(\frac{N}{(N,nn_{1})},\frac{N}{(N,2n_{1})})$ and $\nu=\frac{N^{2}}{\bar{d}(N,nn_{1})(N,2n_{1})}=\frac{N}{(N,2n_{1},nn_{1})}$. Then $\gamma\in\langle\mathfrak{q}^{n\nu}\rangle$. Thus $S_{1}=\mathbb{Z}[g,h_{2}]$ for $h_{2}=[V_{0}(\mathfrak{q}^{n\nu},1,1;0)].$ It is obvious that $S_{1}$ is isomorphic to the group algebra $\mathbb{Z}[\mathbb{Z}_{\mu^{\prime}}\times\mathbb{Z}_{n}]$, where $\mu^{\prime}=|\mathfrak{q}^{n\nu}|=\frac{(N,2n_{1},nn_{1})}{(n,2n_{1})}$.

If $\beta_{1}\neq 0$ and $\beta_{2}=\beta_{3}=0$, then $\gamma=\bar{\omega}^{p}$ for some integer $p$. Thus $S_{1}=\mathbb{Z}[g,h]$, which is isomorphic to $\mathbb{Z}[\mathbb{Z}_{(N/n,n_{1})}\times\mathbb{Z}_{n}]$. Similarly, we can prove that $S_{1}$ is isomorphic to $\mathbb{Z}[\mathbb{Z}_{(N/n,n_{1})}\times\mathbb{Z}_{n}]$ if $\beta_{2}\neq 0$ and $\beta_{1}=\beta_{3}=0$.

Finally, if $\beta=0$, then \begin{array}[]{lll}V_{0}(\gamma,1,1,i)&\cong&V_{0}(1,1,1;i)\otimes V_{0}(\gamma,1,1;0).\end{array} Hence $S_{1}=\mathbb{Z}[g,h_{3}]$, where $h_{3}=[V_{0}(\mathfrak{q}^{n},1,1;0)].$ Thus $S_{1}$ is isomorphic to $\mathbb{Z}[\mathbb{Z}_{N/n}\times\mathbb{Z}_{n}]$. ∎

Let $\mathfrak{q}$ is a primitive $N$-th root of unity. Set $g:=[V_{0}(1,1,1;1)],\;\mathfrak{s}:=\sum\limits_{p=0}^{n-1}g^{p},\;\mathfrak{s}^{\prime}:=\sum\limits_{k=1}^{u}g^{kt},\\ \mathfrak{s}^{\prime\prime}:=\sum\limits_{k=0}^{t-1}g^{n-k},\;h:=[V_{0}(\mathfrak{q}^{\frac{N}{(N,n_{1})}},1,1;0)],\;h_{1}:=[V_{0}(\mathfrak{q}^{\frac{Nn}{(N,2n_{1})}},1,1;0)],\;h_{2}:=[V_{0}(\mathfrak{q}^{n\nu},1,1;0)],\\ h_{3}:=[V_{0}(\mathfrak{q}^{n},1,1;0)],\;$ and $g_{\gamma_{1},\gamma_{2},\gamma_{3}}:=[V_{0}(\gamma_{1},\gamma_{2},\gamma_{3};0)]\in G_{0}(H_{\beta}).$

To determine the Grothendieck ring $G_{0}(H_{\beta})$ for $\beta_{3}\neq 0$, the following lemma is necessary.

Lemma 5.4**.**

Suppose that $\beta_{3}(q^{2n_{1}i}-\gamma_{1}^{-\frac{2n_{1}}{n}}\gamma_{2}\gamma_{3})\neq 0$, $\beta_{1}(\gamma_{1}^{\prime n_{1}}-\gamma_{3}^{\prime n})=\beta_{2}(\gamma_{2}^{\prime n}-\gamma_{3}^{\prime n})=0$ and $\beta_{3}(\gamma_{1}^{\prime-\frac{2n_{1}}{n}}\gamma_{2}^{\prime}\gamma_{3}^{\prime}-q^{2n_{1}j})=0$. Then

[TABLE]

Proof.

As $\beta_{3}(\gamma_{1}^{\frac{2n_{1}}{n}}q^{2n_{1}i}-\gamma_{2}\gamma_{3})\neq 0$, we have $\beta_{3}\neq 0$. Moreover,

[TABLE]

Since $\beta_{1}(\gamma_{1}^{n_{1}}-\gamma_{2}^{n})=\beta_{2}(\gamma_{1}^{n_{1}}-\gamma_{3}^{n})=0$ and $\beta_{1}(\gamma_{1}^{\prime n_{1}}-\gamma_{2}^{\prime n})=\beta_{2}(\gamma_{1}^{\prime n_{1}}-\gamma_{3}^{\prime n})=0$, we have

[TABLE]

If $\gamma_{1}^{-\frac{2n_{1}}{n}}\gamma_{2}\gamma_{3}=q^{n_{1}(2i-r+1)}$, then $(\gamma_{1}\gamma^{\prime}_{1})^{-\frac{2n_{1}}{n}}(\gamma_{2}\gamma_{2}^{\prime})(\gamma_{3}\gamma_{3})=q^{n_{1}(2(i+j)-r+1)}$. If $\gamma_{1}^{\frac{-2n_{1}}{n}}\gamma_{2}\gamma_{3}\neq q^{n_{1}v}$ for any integer $v$, then $(\gamma_{1}\gamma^{\prime}_{1})^{-\frac{2n_{1}}{n}}(\gamma_{2}\gamma_{2}^{\prime})(\gamma_{3}\gamma_{3})\neq q^{n_{1}v}$ for any integer $v$. Let $\{m_{0},m_{1},\cdots,m_{r-1}\}$ be the basis of $V_{r}(\gamma_{1},\gamma_{2},\gamma_{3};i)$ with the action given by (9)-(11). Then $\{m_{0}\otimes 1,m_{1}\otimes 1,\cdots,m_{r-1}\otimes 1\}$ is a basis of $V_{r}(\gamma_{1},\gamma_{2},\gamma_{3};i)\otimes V_{0}(\gamma^{\prime}_{1},\gamma^{\prime}_{2},\gamma^{\prime}_{3};j)$ and the action on this basis is as follows:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Let $u_{l}:=\gamma^{\prime\frac{n_{1}l}{n}}_{1}q^{jn_{1}l}m_{l}\otimes 1$, then $\{u_{0},\cdots,u_{r-1}\}$ is a new basis of $V_{r}(\gamma_{1},\gamma_{2},\gamma_{3};i)\otimes V_{0}(\gamma^{\prime}_{1},\gamma^{\prime}_{2},\gamma^{\prime}_{3};j)$. Under this basis,

[TABLE]

[TABLE]

[TABLE]

Hence $V_{r}(\gamma_{1},\gamma_{2},\gamma_{3};i)\otimes V_{0}(\gamma^{\prime}_{1},\gamma^{\prime}_{2},\gamma^{\prime}_{3};j)\cong V_{r}(\gamma_{1}\gamma^{\prime}_{1},\gamma_{2}\gamma^{\prime}_{2},\gamma_{3}\gamma^{\prime}_{3};i+j)$. Similarly, we get

[TABLE]

∎

Next, we discuss the case in which there is an integer $v$ such that $q^{(2i-v+1)n_{1}}=\gamma_{1}^{-\frac{2n_{1}}{n}}\gamma_{2}\gamma_{3}$. Let $r$ be the minimal positive integer such that $q^{(2i-r+1)n_{1}}=\gamma_{1}^{-\frac{2n_{1}}{n}}\gamma_{2}\gamma_{3}$. From Lemma 5.4, we obtain

[TABLE]

By Equations (10) and (11), the action of $H_{\beta}$ on $V_{r}(1,1,1;0)$ is given by $a\cdot m_{j}=q^{\frac{1}{2}(r-1)-j}m_{j}$, $b\cdot m_{j}=c\cdot m_{j}=m_{j}$, where $k_{1},\cdots,k_{r}$ are determined by $k_{l}=\sum\limits^{l-1}_{j=0}q^{-jn_{1}}\beta_{3}(q^{(r-l)n_{1}}-1)$ for $1\leq l\leq r-1.$

In the case that $\gamma_{1}^{\frac{2n_{1}}{n}}q^{vn_{1}}\neq\gamma_{2}\gamma_{3}$ for any integer $v$, we have

[TABLE]

where $\gamma_{1}^{\prime}=(\gamma_{3}\gamma_{2})^{-\frac{n}{2n_{1}}}\gamma_{1}\neq\bar{\omega}^{\frac{p(n,n_{1})}{2}}$ for any integer $p$.

Let $R_{2}$ be the subring of the Grothendieck ring $G_{0}(H_{\beta})$ for $\beta_{3}\neq 0$ generated by $R_{1}$ and $V_{r}(\gamma_{1},\gamma_{2},\gamma_{3};i)$. Then $R_{2}$ is isomorphic to a quotient ring of $R_{1}[z_{r},z_{\xi}^{\prime}|2\leq r\leq t,\xi\in\widetilde{{\bf K}_{0}}]$, where $\widetilde{{\bf K}_{0}}:=({\bf K}^{*}/\{\bar{\omega}^{\frac{p(n,n_{1})}{2}}\mid p\in\mathbb{Z}\})\setminus\{\overline{1}\}$, $z_{r}=[V_{r}(1,1,1;0)]$ for $2\leq r\leq t$ and $z_{\xi}^{\prime}=[V_{t}(\xi,1,1;0)]$ for $\xi\in\widetilde{{\bf K}_{0}}$. Moreover, if $\beta_{3}\neq 0$ and $\beta_{1}=\beta_{2}=0$, then $R_{2}=G_{0}(H_{\beta})$.

Let $r_{p}=(r-2p)\ mod(t)$ and $1\leq r_{p}\leq t$, where $r$ is the minimal positive integer such that $\xi\xi^{\prime}=q^{(-r+1)n_{1}}$. Then we obtain the following theorem about the Grothendieck ring $G_{0}(H_{\beta})$.

Theorem 5.5**.**

Suppose that $\beta_{1}=\beta_{2}=0$, $\beta_{3}\neq 0$. Then the Grothendieck ring $G_{0}(H_{\beta})$ is isomorphic to $R_{2}:=R_{1}[z_{2},z_{\xi}^{\prime}|\xi\in\widetilde{{\bf K}_{0}}]$ with relations

[TABLE]

[TABLE]

and

[TABLE]

where $z_{r}=\sum\limits_{v=0}^{[\frac{r-1}{2}]}(-1)^{v}\binom{r-1-v}{v}g^{(n-1)v}z_{2}^{r-1-2v}$ for $3\leq r\leq t$.

Proof.

(1) Suppose that $\{m_{0},\cdots,m_{r-1}\}$ and $\{m_{0}^{\prime},m_{1}^{\prime}\}$ are the basis of $V_{r}(1,1,1;0)$ and $V_{2}(1,1,1;0)$, respectively. Then $\{m_{l}\otimes m^{\prime}_{k},\;0\leq l\leq r-1,0\leq k\leq 1\}$ is a basis of $V_{r}(1,1,1;0)\otimes V_{2}(1,1,1;0)$. With this basis, we have

[TABLE]

[TABLE]

[TABLE]

Since $\Delta(x^{k})=\sum\limits^{k}_{l=0}\binom{k}{l}_{q^{n_{1}}}b^{l}x^{k-l}\otimes a^{(k-l)n_{1}}x^{l}$ in $H_{\beta}$, where $\binom{k}{l}_{q^{n_{1}}}=\frac{(k)_{q^{n_{1}}}!}{(l)_{q^{n_{1}}}!(k-l)_{q^{n_{1}}}!}$ and $(p)_{q^{n_{1}}}!=(p)_{q^{n_{1}}}(p-1)_{q^{n_{1}}}\cdots(1)_{q^{n_{1}}}$ for $(p)_{q^{n_{1}}}=1+q^{n_{1}}+\cdots+q^{(p-1)n_{1}}$, we have

[TABLE]

Thus $x^{k^{\prime}}(m_{0}\otimes m^{\prime}_{0})=0$ only if $k^{\prime}\geq l_{0}:=min\{t,r+1\}$.

If $2\leq r\leq t-1$, then $x^{r+1}(m_{0}\otimes m_{0}^{\prime})=0$. In addition,

[TABLE]

Let $u_{l}=x^{l}(m_{0}\otimes m_{0}^{\prime})$. Then $\{u_{0},\cdots,u_{r}\}$ is a basis of $H_{\beta}(m_{0}\otimes m^{\prime}_{0})$, which is isomorphic to $V_{r+1}(1,1,1;0)$ for $r\leq t-1$. Let $A_{1}:=H_{\beta}(m_{0}\otimes m^{\prime}_{1})/H_{\beta}(m_{0}\otimes m^{\prime}_{0})$. Then

[TABLE]

It is easy to prove that $x^{p}(m_{0}\otimes m_{1}^{\prime})=q^{\frac{pn_{1}}{2}}m_{p}\otimes m_{1}^{\prime}\notin H_{\beta}(m_{0}\otimes m_{0}^{\prime})$ for $0\leq p\leq r-2$ and $x^{r-1}(m_{0}\otimes m_{1}^{\prime})\in H_{\beta}(m_{0}\otimes m_{0}^{\prime})$. Let $u_{l}=x^{l}(\overline{m_{0}\otimes m_{1}^{\prime}})$. Then $\{u_{0},\cdots,u_{r-2}\}$ is a basis of $H_{\beta}(\overline{m_{0}\otimes m^{\prime}_{1}})$. Since $a(m_{0}\otimes m_{1}^{\prime})=q^{\frac{1}{2}(r-2)}m_{0}\otimes m_{1}^{\prime},$ $H_{\beta}(\overline{m_{0}\otimes m_{1}^{\prime}})\cong V_{r-1}(1,1,1;0)\otimes V_{0}(1,1,1,n-1)$. Thus $z_{r}z_{2}=z_{r+1}+g^{n-1}z_{r-1}$ for $2\leq r\leq t-1$. In particular, $z_{2}^{2}=z_{3}+g^{n-1}.$ So $z_{r+1}=z_{r}z_{2}-g^{n-1}z_{r-1}$ for $2\leq r\leq t-1$.

If $r=t$, then $yx(m_{0}\otimes m_{0}^{\prime})=0$ by (17). Since $ax(m_{0}\otimes m_{0}^{\prime})=q^{\frac{1}{2}(t-2)}x(m_{0}\otimes m_{0}^{\prime})$, $H_{\beta}x(m_{0}\otimes m_{0}^{\prime})\cong V_{t-1}(1,1,1;0)\otimes V_{0}(1,1,1;n-1)$. Thus $H_{\beta}(m_{0}\otimes m^{\prime}_{0})/H_{\beta}x(m_{0}\otimes m_{0}^{\prime})\cong V_{0}(1,1,1;0)$ and $[H_{\beta}(m_{0}\otimes m_{0}^{\prime})]=g^{n-1}z_{t-1}+1$, where $g=[V_{0}(1,1,1;1)]$. Let $A_{1}:=H_{\beta}(m_{0}\otimes m^{\prime}_{1})/H_{\beta}(m_{0}\otimes m^{\prime}_{0})$. Then $y(\overline{m_{0}\otimes m^{\prime}_{1}})=\overline{0}$ and

[TABLE]

only if $k^{\prime}\geq t$. Since $yx^{t-1}(m_{0}\otimes m_{1}^{\prime})=0$. We have $H_{\beta}x^{t-1}(\overline{m_{0}\otimes m_{1}^{\prime}})\cong V_{0}(1,1,1;n-t)$ and $H_{\beta}(\overline{m_{0}\otimes m^{\prime}_{1}})/H_{\beta}x^{t-1}(\overline{m_{0}\otimes m_{1}^{\prime}})\cong V_{t-1}(1,1,1;n-1)\cong V_{t-1}(1,1,1;0)\otimes V_{0}(1,1,1;n-1)$. Hence $z_{t}z_{2}=2g^{n-1}z_{t-1}+g^{n-t}+1$. Consequently, we have $z_{r}=\sum\limits_{v=0}^{[\frac{r-1}{2}]}(-1)^{v}\binom{r-1-v}{v}g^{(n-1)v}z_{2}^{r-1-2v}$ for $3\leq r\leq t$ and $\sum\limits_{v=0}^{[\frac{t}{2}]}(-1)^{v}\frac{t}{t-v}\binom{r-v}{v}g^{(n-1)v}z_{2}^{t-2v}-g^{n-t}-1=0$.

(2) Now we consider the product $z_{2}z_{\xi}^{\prime}$. Similar to (1), we get $H_{\beta}(m_{0}\otimes m_{0}^{\prime})\cong V_{t}(\xi q^{\frac{n}{2}},1,1;0)$ and $H_{\beta}(\overline{m_{0}\otimes m^{\prime}_{1}})\cong V_{t}(\xi q^{\frac{n}{2}},1,1;n-1)\cong V_{t}(\xi q^{\frac{n}{2}},1,1;0)\otimes V_{0}(1,1,1;n-1)$. Thus $z_{2}z_{\xi}^{\prime}=z_{\xi q^{\frac{n}{2}}}^{\prime}+g^{n-1}z_{\xi q^{\frac{n}{2}}}^{\prime}$.

(3) Here, we determine the product $z_{\xi}^{\prime}z_{\xi^{\prime}}^{\prime}$. Similar to the proof of (1), we get $l_{p}=t$ and $V_{t}(\xi,1,1;0)\otimes V_{t}(\xi^{\prime},1,1;0)=\sum\limits_{p=0}^{t-1}H_{\beta}(m_{0}\otimes m_{p}^{\prime})$. Suppose that $\xi\xi^{\prime}\in\widetilde{\bf K}_{0}$. Hence we get that

[TABLE]

Thus $z_{\xi}^{\prime}z_{\xi^{\prime}}^{\prime}=\sum\limits_{p=0}^{t-1}g^{n-p}z_{\xi\xi^{\prime}}^{\prime}$, where $g=[V_{0}(1,1,1;1)]\in R_{1}$.

(4) Finally, we assume that $\xi\xi^{\prime}\in\{\bar{\omega}^{\frac{p(n,n_{1})}{2}}\mid p\in\mathbb{Z}\}$. Let $r$ be the minimal positive integer such that $(\xi\xi^{\prime})^{\frac{2n_{1}}{n}}q^{(-r+1)n_{1}}=1$. Then $(\xi\xi^{\prime})^{\frac{2n_{1}}{n}}q^{(-2p-r_{p}+1)n_{1}}=1$ for $1\leq r_{p}\leq t$ satisfying $r_{p}\equiv(r-2p)\ mod(t)$. Hence $H_{\beta}(\overline{m_{0}\otimes m_{p}^{\prime}})\cong V_{t}(\xi\xi^{\prime},1,1;n-p)\cong V_{t}(\xi\xi^{\prime},1,1;0)\otimes V_{0}(1,1,1;n-p)$ if $r_{p}=t$. Otherwise, we have $H_{\beta}x^{r_{p}}(\overline{m_{0}\otimes m_{p}^{\prime}})\cong V_{t-r_{p}}(\xi\xi^{\prime},1,1;n-p-r_{p})\cong V_{t-r_{p}}(\xi\xi^{\prime},1,1;0)\otimes V_{0}(1,1,1;n-p-r_{p})$ and $H_{\beta}(\overline{m_{0}\otimes m^{\prime}_{p}})/H_{\beta}x^{r_{p}}(\overline{m_{0}\otimes m_{p}^{\prime}})\cong V_{r_{p}}(\xi\xi^{\prime},1,1;n-p)\cong V_{r_{p}}(\xi\xi^{\prime},1,1;0)\otimes V_{0}(1,1,1;n-p)$. Thus $z_{\xi}^{\prime}z_{\xi^{\prime}}^{\prime}=\sum\limits^{t-1}_{p=0}W^{\prime}(p),$ where $W^{\prime}(p)=g^{n-p}z_{t}$ if $r_{p}=t$ and $W^{\prime}(p)=g^{n-p-r_{p}}z_{t-r_{p}}+g^{n-p}z_{r_{p}}$ if $r_{p}<t$. The proof is completed. ∎

Especially, we obtain the structure of the Grothendieck ring of the Gelaki’s Hopf alegbra where $\beta_{3}\neq 0$.

Corollary 5.6**.**

Suppose that $\beta_{3}\neq 0$.

• (1)

If $N\nmid 2n_{1}t$, then one of the following satisfies.

• (1.1)

If either $2n\mid N$, or $2n\nmid N$ and $2(n,n_{1})\nmid n$, then the Grothendieck ring $G_{0}(\mathcal{U}_{(n,N,n_{1},q,0,0,\beta_{3})})$ is isomorphic to $S_{2}:=\mathbb{Z}[g,h_{1},z_{2},z^{\prime}]$ with relations (14), $z_{2}z^{\prime}=(1+g^{n-1})z^{\prime}$ and

[TABLE]

where $z_{0}=0$, $z_{1}=1$, $z_{p}=\sum\limits_{v=0}^{[\frac{p-1}{2}]}(-1)^{v}\binom{p-1-v}{v}g^{(n-1)v}z_{2}^{p-1-2v}$ for $3\leq p\leq t$ and $v_{0}=\frac{N}{(2n_{1}t,N)}$.

• (1.2)

If $2n\nmid N$ and $2(n,n_{1})\mid n$, then the Grothendieck ring $G_{0}(\mathcal{U}_{(n,N,n_{1},q,0,0,\beta_{3})})$ is isomorphic to $S_{2}:=\mathbb{Z}[g,h_{1},z_{3},z^{\prime}]$ with relations (18), $z_{3}z^{\prime}=(1+g^{n-1}+g^{n-2})z^{\prime}$ and

[TABLE]

where $z_{0}=0$, $z_{1}=1$, $v_{0}=\frac{N}{(2n_{1}t,N)}$ and $z_{2l+1}=\sum\limits_{v=0}^{[\frac{l-1}{2}]}(-1)^{v}\binom{l-1-v}{v}g^{(n-2)v}(z_{3}-g^{n-1})^{l-1-2v}$ for $3\leq 2l+1\leq t-1$.

• (2)

If $N\mid 2n_{1}t$, then the Grothendieck ring $G_{0}(\mathcal{U}_{(n,N,n_{1},q,0,0,\beta_{3})})$ is isomorphic to

[TABLE]

which is a subring of $S_{2}$.

Proof.

Since $\gamma^{\frac{N}{n}}=1$, $\gamma=\mathfrak{q}^{np}$ for some integer $p$. If $\gamma^{\frac{2n_{1}}{n}}=q^{vn_{1}}$ for some integer $v$, then $\mathfrak{q}^{2n_{1}p}=q^{vn_{1}}$. Hence $\mathfrak{q}^{2n_{1}pt}=1$, i.e. $N\mid 2n_{1}pt$. Thus $\frac{N}{(N,2n_{1}t)}\mid p$. Then $z^{\prime}_{\gamma}\in S_{2}$ only if $\gamma\in\langle\mathfrak{q^{n}}\rangle$ and $\gamma\notin\langle\mathfrak{q}^{nv_{0}}\rangle$ for $v_{0}=\frac{N}{(2n_{1}t,N)}$. If either $2n\mid N$, or $2n\nmid N$ and $2(n,n_{1})\nmid n$, then we have generator $z_{2}=[V_{0}(1,1,1;0)]$. If $v_{0}=1$, that is $N(n,n_{1})|2nn_{1}$, then $S_{2}=S_{1}[z_{2}]$. Otherwise, $S_{2}=S_{1}[z_{2},z^{\prime}]$ for $z^{\prime}=[V_{t}(\mathfrak{q}^{n},1,1;0)]$. If $2n\nmid N$ and $2(n,n_{1})\mid n$, then $[V_{r}(\gamma,1,1;0)]\in\mathcal{U}_{(n,N,n_{1},q,0,0,\beta_{3})}$ for $\gamma^{\frac{2n_{1}}{n}}\in\langle q^{n_{1}}\rangle$ implies that $r$ is an odd integer. Let $r_{p}\equiv(r-2p)\ mod(t)$ for $1\leq p\leq t$ and $z^{\prime}=[V_{t}(\mathfrak{q}^{n},1,1;0)]$. We further assume that $1\leq r_{p}\leq t$. Then $(z^{\prime})^{v_{0}}=z^{\prime}(z^{\prime})^{v_{0}-1}={\mathfrak{s}^{\prime\prime}}^{v_{0}-2}z^{\prime}z^{\prime}_{\mathfrak{q}^{v_{0}-1}}$ by Theorem 5.5. Since

[TABLE]

for some $2\leq r\leq t$, $z^{\prime}z_{\mathfrak{q}^{v_{0}-1}}^{\prime}=\sum\limits^{t-1}_{p=0,r_{p}<t}(g^{n-p-r_{p}}z_{t-r_{p}}+g^{n-p}z_{r_{p}})+\sum\limits_{i=0,r_{p}=t}^{t-1}g^{n-p}z_{t},$ where

[TABLE]

for $3\leq p\leq t$. Since $\mathfrak{q}^{n}q^{\frac{n}{2}}=\mathfrak{q}^{n}$, $z_{2}z^{\prime}=(g^{n-1}+1)z^{\prime}$ by Theorem 5.5. Similarly to the proof of Theorem 5.5, we have $z_{3}z^{\prime}=(g^{n-2}+g^{n-1}+1)z^{\prime}$,

[TABLE]

and $z_{2l+1}=\sum\limits_{v=0}^{[\frac{l-1}{2}]}(-1)^{v}\binom{l-1-v}{v}g^{(n-2)v}(z_{3}-g^{n-1})^{l-1-2v}$ for $3\leq 2l+1\leq t-1$. ∎

We always assume that $z_{0}=0$, $z_{1}=1$, $z_{2}=[V_{2}(1,1,1;0)]$ and

[TABLE]

To determine the Grothendieck ring $G_{0}(H_{\beta})$ for $\beta_{1}\neq 0$, we need the following lemma.

Lemma 5.7**.**

Suppose that $\beta_{1}(\gamma_{1}^{n_{1}}-\gamma_{2}^{n})\neq 0$. Then

[TABLE]

Proof.

Since $(\gamma_{1}^{n_{1}}-\gamma_{2}^{n})\beta_{1}\neq 0$, $\beta_{1}\neq 0$. Thus $\gamma_{1}^{\prime n_{1}}=\gamma_{2}^{\prime n}$ and $\gamma_{2}^{\prime}=\gamma_{1}^{\prime\frac{n_{1}}{n}}q^{s}$ for some $0\leq s\leq n-1$. Hence $((\gamma_{1}\gamma_{1}^{\prime})^{n_{1}}-(\gamma_{2}\gamma_{2}^{\prime})^{n})\beta_{1}=(\gamma_{1}^{\prime})^{n_{1}}(\gamma_{1}^{n_{1}}-\gamma_{2}^{n})\beta_{1}\neq 0$. Let $\{m_{0},m_{1},\cdots,m_{n-1}\}$ be the basis of $V_{I}(\gamma_{1},\gamma_{2},\gamma_{3};i)$ with the action given by Equations (3)-(5). Then $\{m_{0}\otimes 1,m_{1}\otimes 1,\cdots,m_{n-1}\otimes 1\}$ is a basis of $V_{I}(\gamma_{1},\gamma_{2},\gamma_{3};i)\otimes V_{0}(\gamma_{1}^{\prime},\gamma_{2}^{\prime},\gamma_{3}^{\prime};j)$. With this basis, we have

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Let $u_{l}=\gamma^{\prime\frac{n_{1}l}{n}}_{1}q^{jn_{1}l}m_{l}\otimes 1$. Then $\{u_{0},\cdots,u_{n-1}\}$ is a new basis of $V_{I}(\gamma_{1},\gamma_{2},\gamma_{3};i)\otimes V_{0}(\gamma_{1}^{\prime},\gamma_{2}^{\prime},\gamma_{3}^{\prime};j)$. Under this basis, we have

[TABLE]

[TABLE]

[TABLE]

Thus $V_{I}(\gamma_{1},\gamma_{2},\gamma_{3};i)\otimes V_{0}(\gamma^{\prime}_{1},\gamma^{\prime}_{2},\gamma^{\prime}_{3};j)\cong V_{I}(\gamma_{1}\gamma^{\prime}_{1},\gamma_{2}\gamma^{\prime}_{2},\gamma_{3}\gamma^{\prime}_{3};i+j)$. Similarly, we get

[TABLE]

∎

Remark 5.8**.**

Lemma 5.7 implies that

[TABLE]

where $\gamma_{1}^{\prime}=\gamma_{1}\gamma_{2}^{-\frac{n}{n_{1}}}$ satisfying $\gamma_{1}^{\prime n_{1}}\neq 1$, $\gamma_{3}^{\prime}=\gamma_{3}\gamma_{2}^{-1}q^{-2n_{1}i}$. If $\beta_{2}=\beta_{3}=0$, then $V_{I}(\gamma_{1}^{\prime},1,\gamma_{3}^{\prime};0)\cong V_{I}(\gamma_{1}^{\prime},1,1;0)\otimes V_{0}(1,1,\gamma_{3}^{\prime};0)$. If $\beta_{2}\neq 0$, $\beta_{3}=0$ and $\gamma_{3}^{\prime}=q^{u}$ for some integer $u$, then $V_{I}(\gamma_{1}^{\prime},1,\gamma_{3}^{\prime};0)\cong V_{I}(\gamma_{1}^{\prime},1,1;0)\otimes V_{0}(1,1,\gamma_{3}^{\prime};0)$. If $\beta_{2}=0$, $\beta_{3}\neq 0$ and $(\gamma_{3}^{\prime})^{\frac{n}{2}}=1$, then $V_{I}(\gamma_{1}^{\prime},1,\gamma_{3}^{\prime};0)\cong V_{I}(\gamma_{1}^{\prime}(\gamma_{3}^{\prime})^{-\frac{n}{2n_{1}}},$ $1,1;0)\otimes V_{0}((\gamma_{3}^{\prime})^{\frac{n}{2n_{1}}},1,\gamma_{3}^{\prime};0)$. If $\beta_{2}\beta_{3}\neq 0$ and $(\gamma_{3}^{\prime})^{\frac{n}{2}}=\gamma_{3}^{\prime n}=1$, then $V_{I}(\gamma_{1}^{\prime},1,\gamma_{3}^{\prime};0)\cong V_{I}(\gamma_{1}^{\prime}(\gamma_{3}^{\prime})^{-\frac{n}{2n_{1}}},$ $1,1;0)\otimes V_{0}((\gamma_{3}^{\prime})^{\frac{n}{2n_{1}}},1,\gamma_{3}^{\prime};0)$.

Remark 5.9**.**

Let $R_{3}$ be the subring of the Grothendieck ring $G_{0}(H_{\beta})$ for $\beta_{1}\neq 0$ generated by $R_{2}$ and $\{[V_{I}(\gamma_{1},\gamma_{2},\gamma_{3};i)]|\gamma_{1}^{n_{1}}\neq\gamma_{2}^{n},i\in\mathbb{Z}_{n}\}$. Then $R_{3}$ is isomorphic to a quotient ring of $R_{2}[x_{\zeta_{1},\zeta_{2}}|\zeta_{2}\in{\bf K}^{*},\zeta_{1}\in\widehat{{\bf K}_{0}}]$, where $\widehat{{\bf K}_{0}}=({\bf K}^{*}/\{\bar{\omega}^{r}|r\in\mathbb{Z}\})\setminus\{\overline{1}\}$ and $x_{\zeta_{1},\zeta_{2}}=[V_{I}(\zeta_{1},1,\zeta_{2};0)]$. Moreover, if $\beta_{2}=0$ and $\beta_{1}\beta_{3}\neq 0$, then $R_{3}$ is the Grothendieck ring of $G_{0}(H_{\beta})$. Suppose that $\beta_{2}=\beta_{3}=0$ and $\beta_{1}\neq 0$. Then the Grothendieck ring $G_{0}(H)$ is isomorphic to a quotient ring of the algebra $R_{2}^{\prime}=R_{1}[x_{\zeta,1}|\zeta\in\widehat{{\bf K}_{0}}]$.

Let $r_{p}=(r-2p)\ mod(t)$ and $1\leq r_{p}\leq t$, where $r$ is the minimal positive integer such that $\zeta_{2}\zeta_{2}^{\prime}(\zeta_{1}\zeta_{1}^{\prime})^{-\frac{2n_{1}}{n}}=q^{(-r+1)n_{1}}$. Then $r_{p}$ is the minimal positive integer such that

[TABLE]

Theorem 5.10**.**

Suppose that $\beta_{1}\beta_{3}\neq 0$, $\beta_{2}=0$ and ${\bf K}^{**}=({\bf K}^{*}/\{\zeta\in{\bf K}|\zeta^{\frac{n}{2}}=1\})\setminus\{\bar{1}\}$. Let $z_{\xi}^{\prime\prime}=[V_{t}(1,1,\xi;0)]$ for $\xi\in\overline{{\bf K}^{*}}:=({\bf K}^{*}/\langle q^{n_{1}}\rangle)\setminus\{\bar{1}\}$. Then the Grothendieck ring $R_{3}:=G_{0}(H_{\beta})$ is isomorphic to the commutative ring $R_{1}[z_{2},x_{\zeta_{1},\zeta_{2}},z_{\xi}^{\prime\prime}|\zeta_{1}\in\widehat{{\bf K}_{0}},\zeta_{2}\in{\bf K}^{**},\xi\in\overline{{\bf K}^{*}}]$ with relations (14),

[TABLE]

[TABLE]

[TABLE]

for $r^{\prime}$ is the minimal positive integer such that $\xi\xi^{\prime}=q^{r^{\prime}n_{1}}$, and

[TABLE]

where $\mathfrak{s}^{\prime}=\sum\limits_{k=1}^{u}g^{kt}\in R_{1}$, $u=\frac{n}{t}$, $z_{r}=\sum\limits_{v=0}^{[\frac{r-1}{2}]}(-1)^{v}\binom{r-1-v}{v}g^{(n-1)v}z_{2}^{r-1-2v}$ for $3\leq r\leq t$ and $\eta=[V_{0}(q^{\frac{n}{2}},1,q^{n_{1}};0)]$.

Proof.

Since $\beta_{2}=0$, $G_{0}(H_{\beta})$ is generated by $[V_{I}(\gamma_{1},\gamma_{2},\gamma_{3};i)],[V_{0}(\gamma_{1},\gamma_{2},\gamma_{3};i)]$ and $[V_{r}(\gamma_{1},\gamma_{2},\gamma_{3};i)]$ for $r\leq t$. Suppose that $\gamma_{1}^{n_{1}}=\gamma_{2}^{n}$ and $\gamma_{1}^{-\frac{2n_{1}}{n}}\gamma_{2}\gamma_{3}\neq q^{n_{1}v}$ for any integer $v$. Then

[TABLE]

where $\gamma_{2}^{-1}\gamma_{3}\neq q^{n_{1}v}$ for any integer $v$. In addition, we have

[TABLE]

Therefore $G_{0}(H_{\beta})$ is generated by $R_{1}$, $z_{2}$, $x_{\zeta_{1},\zeta_{2}}$ and $z_{\zeta}^{\prime\prime}=[V_{t}(1,1,\xi;0)]$ for $\xi\in({\bf K}^{*}/\langle q^{n_{1}}\rangle)\setminus\{\bar{1}\}$.

In the following, we determine the relations of the generators in several subcases.

(1) Similar to the proof of Theorem 5.5, we have Equation (14),

[TABLE]

and

[TABLE]

for $\xi\xi^{\prime}=q^{r^{\prime}n_{1}}$, where $z_{r}=\sum\limits_{v=0}^{[\frac{r-1}{2}]}(-1)^{v}\binom{r-1-v}{v}g^{(n-1)v}z_{2}^{r-1-2v}$ for $3\leq r\leq t$, $\eta=[V_{0}(q^{\frac{n}{2}},1,q^{n_{1}};0)]$, and $r_{p}^{\prime}=(r^{\prime}-2p)(mod\ t)$ for $1\leq p\leq t$, $r^{\prime}$ is the minimal positive integer such that $\xi\xi^{\prime}=q^{r^{\prime}n_{1}}$.

(2) Let $\{m_{0},\cdots,m_{n-1}\}$ (resp. $\{m_{0}^{\prime},m_{1}^{\prime}\}$) be the basis of $V_{I}(\zeta_{1},1,\zeta_{2};0)$ (resp. $V_{2}(1,1,1;0)$) with the actions given by Equations (3)-(5)(resp. Equations (9)-(11), where $k_{i}$ is replaced by $k_{i}^{\prime}$). Then $\{m_{l}\otimes m^{\prime}_{v},\;0\leq l\leq n-1,0\leq v\leq 1\}$ is a basis of $V_{I}(\zeta_{1},1,\zeta_{2};0)\otimes V_{2}(1,1,1;0)$. With this basis, we have

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

If $(n_{1},n)=u$, then $|q^{n_{1}}|=|q^{u}|=\frac{n}{u}$. Hence $n=ut$ and

[TABLE]

Let $v_{0}=m_{0}\otimes m^{\prime}_{0}+\alpha_{1}m_{n-1}\otimes m^{\prime}_{1}$ , $v_{1}=m_{0}\otimes m^{\prime}_{1}$ and $yv_{0}=\theta x^{n-1}v_{0}$. Then

[TABLE]

and

[TABLE]

Thus

[TABLE]

Hence we can determine the $\alpha_{1}$. In addition, $H_{\beta}v_{0}\cong V_{I}(\zeta_{1}q^{\frac{n}{2}},1,\zeta_{2};0)$ is an irreducible submodule of $V_{I}(\zeta_{1},1,\zeta_{2};0)\otimes V_{2}(1,1,1;0)$ by the proof of Lemma 4.2.

By the same method, we prove that

[TABLE]

is an irreducible submodule of $V_{I}(\zeta_{1},1,\zeta_{2};0)\otimes V_{2}(1,1,1;0)/H_{\beta}v_{0}$. Hence

[TABLE]

where $g=[V_{0}(1,1,1;1)]$. Likewise, we have $z_{2}x_{\zeta_{1},\zeta_{2}}=x_{\zeta_{1}q^{\frac{n}{2}},\zeta_{2}}+g^{n-1}x_{\zeta_{1}q^{\frac{n}{2}},\zeta_{2}}.$

(3) Similarly, we obtain

[TABLE]

Hence $x_{\zeta_{1},\zeta_{2}}z_{\xi}^{\prime\prime}=\sum\limits^{t-1}_{p=0}g^{n-p}x_{\zeta_{1},\zeta_{2}\xi}$.

(4) Let $\{m_{0},\cdots,m_{n-1}\}$ (resp. $\{m_{0}^{\prime},\cdots,m_{n-1}^{\prime}\}$) be a basis of $V_{I}(\zeta_{1},1,\zeta_{2};0)$ (resp. $V_{I}(\zeta_{1}^{\prime},1,\zeta_{2}^{\prime};0)$) with the action given by (3)-(5) (resp. (3)-(5) with $k_{i}$ replaced by $k_{i}^{\prime}$). Then $\{m_{l}\otimes m^{\prime}_{v},\;0\leq l,v\leq n-1\}$ is a basis of $V_{I}(\zeta_{1},1,\zeta_{2};0)\otimes V_{I}(\zeta_{1}^{\prime},1,\zeta_{2}^{\prime};0)$ and the action on this basis is given by

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

(4.1) If $((\zeta_{1}\zeta_{1}^{\prime})^{n_{1}}-1)\beta_{1}\neq 0$, let $v_{p}=m_{0}\otimes m^{\prime}_{p}+\sum\limits^{n-1}_{l=p+1}\alpha_{l}m_{n+p-l}\otimes m^{\prime}_{l}$ for $0\leq p\leq n-1$. Assume that $yv_{0}=\theta x^{n-1}v_{0}$, similar to (1), we can determine $\alpha_{l},1\leq l\leq n-1$. Hence, $A_{0}:=H_{\beta}v_{0}\cong V_{I}(\zeta_{1}\zeta_{1}^{\prime},1,\zeta_{2}\zeta_{2}^{\prime};0)$ is an irreducible submodule of $V_{I}(\zeta_{1},1,\zeta_{2};0)\otimes V_{I}(\zeta_{1}^{\prime},1,\zeta_{2}^{\prime};0)$.

By using the same method, we can determine that

[TABLE]

is an irreducible submodule of $V_{I}(\zeta_{1},1,\zeta_{2};0)\otimes V_{I}(\zeta_{1}^{\prime},1,\zeta_{2}^{\prime};0)/\sum^{p-1}_{l=0}H_{\beta}v_{l}$ for $1\leq p\leq n-1$. Hence $[V_{I}(\zeta_{1},1,\zeta_{2};0)][V_{I}(\zeta_{1}^{\prime},1,\zeta_{2}^{\prime};0)]=\sum\limits^{n-1}_{p=0}[V_{I}(\zeta_{1}\zeta_{1}^{\prime},1,\zeta_{2}\zeta_{2}^{\prime};n-p)]=\sum\limits^{n-1}_{p=0}[V_{I}(\zeta_{1}\zeta_{1}^{\prime},1,\zeta_{2}\zeta_{2}^{\prime};0)][V_{0}(1,1,1;n-p)]$. Consequently, $x_{\zeta_{1},\zeta_{2}}x_{\zeta_{1}^{\prime},\zeta_{2}^{\prime}}=\sum\limits^{n-1}_{p=0}g^{n-p}x_{\zeta_{1}\zeta_{1}^{\prime},\zeta_{2}\zeta_{2}^{\prime}}=\sum\limits^{n-1}_{p=0}g^{p}x_{\zeta_{1}\zeta_{1}^{\prime},\zeta_{2}\zeta_{2}^{\prime}}$.

(4.2) We discuss the case in which $((\zeta_{1}\zeta_{1}^{\prime})^{n_{1}}-1)\beta_{1}=0$. Since $\beta_{2}=0$, we have $y^{n}=0$. There exists an element $v^{\prime}\in V_{I}(\zeta_{1},1,\zeta_{2};0)\otimes V_{I}(\zeta_{1}^{\prime},1,\zeta_{2}^{\prime};0)$ such that $yv^{\prime}=0$. We assume that $v^{\prime}=\sum\limits^{d}_{l=0}\alpha_{l}m_{d-l}\otimes m^{\prime}_{l}+\sum\limits^{n-1}_{l=d+1}\alpha_{l}m_{n+d-l}\otimes m^{\prime}_{l}$ for $ay=qya$. Let

[TABLE]

then $det(C)=0$. Let $v_{p}=\sum\limits^{p}_{l=0}\alpha_{l}m_{p-l}\otimes m^{\prime}_{l}+\sum\limits^{n-1}_{l=p+1}\alpha_{l}m_{n+p-l}\otimes m^{\prime}_{l}$ for $0\leq p\leq n-1$,

[TABLE]

[TABLE]

where

[TABLE]

[TABLE]

Since $det(D_{p})=det(C)=0$ and $D_{p}-\lambda B_{p}\neq 0$ for any $\lambda$, there exists $\alpha_{l}$ for $0\leq l\leq n-1$ such that $yv_{p}=0$ and $x^{n-1}v_{p}\neq 0$. In fact, $Hv_{i}\cap Hv_{j}=0$ for any $0\leq i<j\leq n-1$. If there is a $\lambda\in\bf{K}$ such that $v_{j}=\lambda x^{j-i}v_{i}$. Then $x^{n-1}v_{1}=\lambda x^{n-1+j-i}v_{0}=\lambda\beta_{1}((\zeta_{1}\zeta_{1}^{\prime})^{n_{1}}-1)x^{j-i-1}v_{i}=0$, which is a contradiction. Thus, $V_{I}(\zeta_{1},1,\zeta_{2};0)\otimes V_{I}(\zeta_{1}^{\prime},1,\zeta_{2}^{\prime};0)=\bigoplus\limits^{n-1}_{p=0}H_{\beta}v_{p}$. Notice that

[TABLE]

(4.2.1) In the case that there exists an integer $l$ such that $\zeta_{2}\zeta_{2}^{\prime}-(\zeta_{1}\zeta_{1}^{\prime})^{\frac{2n_{1}}{n}}q^{(-2p-l+1)n_{1}}=0$. Let $r_{p}$ be the minimal positive integer such that $\zeta_{2}\zeta_{2}^{\prime}(\zeta_{1}\zeta_{1}^{\prime})^{-\frac{2n_{1}}{n}}=q^{(-2p-r_{p}+1)n_{1}}$. Then $1\leq r_{p}\leq t$ and $r_{p}=(r-2p)\ mod(t)$, where $r$ is the minimal positive integer such that $\zeta_{2}\zeta_{2}^{\prime}(\zeta_{1}\zeta_{1}^{\prime})^{-\frac{2n_{1}}{n}}=q^{(-r+1)n_{1}}$. So $r_{p}$, which satisfies $1\leq r_{p}\leq t$, is the minimal positive integer such that $yx^{r_{p}}v_{p}=0$. Thus $yx^{kt+r_{p}}v_{p}=0$ for $0\leq k\leq u-1$, where $u=\frac{n}{t}$. Notice that $yx^{kt}v_{p}=0$ for $0\leq k\leq u-1$. Hence $A_{p,2k+1}:=H_{\beta}x^{(u-k-1)t+r_{p}}v_{p}/H_{\beta}x^{(u-k)t}v_{p}\cong V_{t-r_{p}}(\zeta_{1}\zeta_{1}^{\prime},1,\zeta_{2}\zeta_{2}^{\prime};n-p+(k+1)t-r_{p})\cong V_{t-r_{p}}(1,1,1;0)\otimes V_{0}(\zeta_{1}\zeta_{1}^{\prime},1,\zeta_{2}\zeta_{2}^{\prime};n-p+(k+1)t-r_{p})\cong V_{t-r_{p}}(1,1,1;0)\otimes V_{0}(\zeta_{1}\zeta_{1}^{\prime},1,\zeta_{2}\zeta_{2}^{\prime};0)\otimes V_{0}(1,1,1;1)^{\otimes(n-p+(k+1)t-r_{p})}$, $A_{p,2(k+1)}:=H_{\beta}x^{(u-k-1)t}v_{p}/H_{\beta}x^{(u-k-1)t+r_{p}}v_{p}\cong V_{r_{p}}(\zeta_{1}\zeta_{1}^{\prime},1,\zeta_{2}\zeta_{2}^{\prime};n-p+(k+1)t)\cong V_{r_{p}}(1,1,1;0)\otimes V_{0}(\zeta_{1}\zeta_{1}^{\prime},1,\zeta_{2}\zeta_{2}^{\prime};n-p+(k+1)t)\cong V_{r_{p}}(1,1,1;0)\otimes V_{0}(\zeta_{1}\zeta_{1}^{\prime},1,\zeta_{2}\zeta_{2}^{\prime};0)\otimes V_{0}(1,1,1;1)^{\otimes(n-p+(k+1)t)}$ for $0\leq k\leq u-1$. Therefore $[A_{p,2k+1}]=z_{t-r_{p}}g^{(k+1)t-p-r_{p}}g_{\zeta_{1}\zeta_{1}^{\prime},1,\zeta_{2}\zeta_{2}^{\prime}}$ and $[A_{p,2(k+1)}]=z_{r_{p}}g^{(k+1)t-p}g_{\zeta_{1}\zeta_{1}^{\prime},1,\zeta_{2}\zeta_{2}^{\prime}}$. Consequently,

[TABLE]

where $\mathfrak{s}^{\prime}=\sum\limits_{k=1}^{u}g^{kt}$.

(4.2.2) If $\zeta_{2}\zeta_{2}^{\prime}\neq(\zeta_{1}\zeta_{1}^{\prime})^{\frac{2n_{1}}{n}}q^{vn_{1}}$ for any integer $v$, then $yx^{kt}v_{p}=0$ for $0\leq k\leq u-1$, where $u=\frac{n}{t}$. Thus $A_{p,v}:=H_{\beta}x^{(u-v)t}v_{p}/H_{\beta}x^{(u-v+1)t}v_{p}\cong V_{t}(\zeta_{1}\zeta_{1}^{\prime},1,\zeta_{2}\zeta_{2}^{\prime};n-p+vt)\cong V_{0}(\zeta_{1}\zeta_{1}^{\prime},1,1;n-p+vt)\otimes V_{t}(1,1,\zeta_{2}\zeta_{2}^{\prime};0)$ for $1\leq v\leq u$. Hence $[A_{p,v}]=g_{\zeta_{1}\zeta_{1}^{\prime},1,1}g^{n-p+vt}z_{\zeta_{2}\zeta_{2}^{\prime}}^{\prime\prime}$. Consequently, $x_{\zeta_{1},\zeta_{2}}x_{\zeta_{1}^{\prime},\zeta_{2}^{\prime}}=\sum\limits^{n-1}_{p=0}\sum\limits^{u}_{v=1}[A_{p,v}]=u\mathfrak{s}g_{\zeta_{1}\zeta_{1}^{\prime},1,1}z_{\zeta_{2}\zeta_{2}^{\prime}}^{\prime\prime}$, where $u=\frac{n}{t}$. ∎

Especially, we obtain the structure of the Grothendieck ring of the Gelaki’s Hopf alegbra where $\beta_{1}\beta_{3}\neq 0$.

Corollary 5.11**.**

(1)* Suppose that $\beta_{1}\beta_{3}\neq 0$ and $(N,nn_{1},2n_{1}t)<(nn_{1},N)$.*

(1.1)* If either $2n\mid N$, or $2n\nmid N$ and $2(n,n_{1})\nmid n$, then the Grothendieck ring*

[TABLE]

is isomorphic to the commutative ring $S_{2}[x^{*}]=\mathbb{Z}[g,h_{2},z_{2},z^{\prime\prime},x^{*}]$ with relations Equation (14),

[TABLE]

$z_{2}z^{\prime\prime}=(1+g^{n-1})z^{\prime\prime},$* $z_{2}x^{*}=(1+g^{n-1})x^{*}$ and ${x^{*}}^{\frac{N}{(N/n,n_{1})}}=u\mathfrak{s}^{\frac{N}{(N/n,n_{1})}-1}z^{\prime\prime}$.*

(1.2)* If $2n\nmid N$ and $2(n,n_{1})\mid n$, then the Grothendieck ring $S_{3}:=G_{0}(\mathcal{U}_{(n,N,n_{1},q,\beta_{1},0,\beta_{3})})$ is isomorphic to the commutative ring $S_{2}[x^{*}]=\mathbb{Z}[g,h_{2},z_{3},z^{\prime\prime},x^{*}]$ with relations (19), (21), $z_{3}z^{\prime\prime}=(1+g^{n-1}+g^{n-2})z^{\prime\prime},$ $z_{3}x^{*}=(1+g^{n-1}+g^{n-2})x^{*}$ and ${x^{*}}^{\frac{N}{(N/n,n_{1})}}=u\mathfrak{s}^{\frac{N}{(N/n,n_{1})}-1}z^{\prime\prime}$.*

(2)* Suppose that $\beta_{1}\beta_{3}\neq 0$ and $(N,nn_{1},2n_{1}t)=(nn_{1},N)$. Then the Grothendieck ring*

[TABLE]

is a subring of $S_{3}$.

Proof.

Suppose that $V_{I}(\gamma,1,1;0)$ is an irreducible representation of $\mathcal{U}_{(n,N,n_{1},q,\beta_{1},0,\beta_{3})}$. Then $\gamma^{\frac{N}{n}}=1$ and $\gamma^{n_{1}}\neq 1$. Assume that $\gamma^{\frac{N}{n}}=\gamma^{n_{1}}=1$. Then $\gamma=\mathfrak{q}^{nr}$ for some integer $r$ and $N|nn_{1}r$. Thus $\gamma\in\langle\mathfrak{q}^{\frac{Nn}{(N,nn_{1})}}\rangle$. We further assume that $\gamma^{\frac{2n_{1}}{n}}\in\langle q^{n_{1}}\rangle$, then $\gamma\in\langle\mathfrak{q}^{\frac{Nn}{(N,nn_{1})}}\rangle\cap\langle\mathfrak{q}^{\frac{Nn}{(N,2n_{1}t)}}\rangle=\langle\mathfrak{q}^{\frac{Nn}{(N,nn_{1},2n_{1}t)}}\rangle$. Let $v_{1}=\frac{(nn_{1},N)}{(nn_{1},2n_{1}t,N)}$ and $\mathfrak{q}^{\frac{2n_{1}N}{(N,nn_{1},2n_{1}t)}}=q^{(r-1)n_{1}}$ for some $2\leq r\leq t$. Then

[TABLE]

$z_{2}z^{\prime\prime}=(1+g^{n-1})z^{\prime\prime}$ and $z_{3}z^{\prime\prime}=(1+g^{n-1}+g^{n-2})z^{\prime\prime},$ where $z^{\prime\prime}=[V_{t}(q^{\frac{Nn}{(N,nn_{1})}},1,1;0)]$. Let $x^{*}=[V_{I}(\mathfrak{q}^{n},1,1;0)]$. Then $z_{2}x^{*}=(1+g^{n-1})x^{*}$, $z_{3}x^{*}=(1+g^{n-1}+g^{n-2})x^{*}$, and ${x^{*}}^{\frac{N}{(N/n,n_{1})}}=u\mathfrak{s}^{\frac{N}{(N/n,n_{1})}-1}z^{\prime\prime}$ by Theorem 5.10. ∎

Theorem 5.12**.**

Suppose that $\beta_{1}\neq 0$ and $\beta_{2}=\beta_{3}=0$. Then the Grothendieck ring $G_{0}(H_{\beta})=R_{2}^{\prime}:=R_{1}[x_{\zeta,1}|\zeta\in\widehat{{\bf K}_{0}}]$ with relations

[TABLE]

Proof.

In the case that $(\zeta_{1}\zeta_{1}^{\prime})^{n_{1}}-1\neq 0$, according to the proof of Theorem 5.10, we have

[TABLE]

Hence, $x_{\zeta,1}x_{\zeta^{\prime},1}=\sum\limits^{n-1}_{p=0}g^{n-p}x_{\zeta\zeta^{\prime},1}=\mathfrak{s}x_{\zeta\zeta^{\prime},1}$.

Next we discuss the case that $(\zeta_{1}\zeta_{1}^{\prime})^{n_{1}}=1$. Let $\{m_{0},\cdots,m_{n-1}\}$ and $\{m_{0}^{\prime},\cdots,m_{n-1}^{\prime}\}$ be the basis of $V_{I}(\zeta_{1},1,1;0)$ and $V_{I}(\zeta_{1}^{\prime},1,1;0)$ with the actions given by Equations (3)-(5) respectively. Then $\{m_{l}\otimes m^{\prime}_{k},\;0\leq l,k\leq n-1\}$ is a basis of $V_{I}(\zeta_{1},1,1;0)\otimes V_{I}(\zeta_{1}^{\prime},1,1;0)$. From the proof of Theorem 5.10, we have $V_{I}(\zeta_{1},1,1;0)\otimes V_{I}(\zeta_{1}^{\prime},1,1;0)=\bigoplus\limits^{n-1}_{p=0}H_{\beta}v_{p}$, where $v_{p}=\sum\limits^{p}_{l=0}\alpha_{l}m_{p-l}\otimes m^{\prime}_{l}+\sum\limits^{n-1}_{l=p+1}\alpha_{l}m_{n+p-l}\otimes m^{\prime}_{l}$ in the proof of Theorem 5.10 for $0\leq p\leq n-1$. Since $\beta_{3}=0$, we obtain that $yxv_{p}=0$ for $0\leq p\leq n-1$. Hence

[TABLE]

Since $g^{n}=1$, $\sum\limits_{p=0}^{n-1}\sum\limits_{v=1}^{n}g^{n-v-p}=\sum\limits_{p=0}^{n-1}g^{n-p}\sum\limits_{v=1}^{n}g^{n-v}=n\sum\limits_{p=0}^{n-1}g^{p}.$ Thus

[TABLE]

∎

Especially, we obtain the structure of the Grothendieck ring of the Gelaki’s Hopf alegbra where $\beta_{1}\neq 0$.

Corollary 5.13**.**

Suppose that $\beta_{1}\neq 0$. Then the Grothendieck ring $G_{0}(\mathcal{U}_{(n,N,n_{1},q,\beta_{1},0,0)}=S_{1}[x^{*}]=\mathbb{Z}[g,h,x^{*}]$ with relations

[TABLE]

Proof.

Suppose that $V_{I}(\gamma,1,1;0)$ is an irreducible representation of $\mathcal{U}_{(n,N,n_{1},q,\beta_{1},0,0)}$. Then $\gamma^{\frac{N}{n}}=1$ and $\gamma^{n_{1}}\neq 1$. Thus $\gamma\in\langle\mathfrak{q}^{n}\rangle\setminus\langle\mathfrak{q}^{\frac{N}{(N/n,n_{1})}}\rangle$ by the proof of Corollary 5.11. Let $x^{*}=[V_{I}(\mathfrak{q}^{n},1,1;0)]$. Then $G_{0}(\mathcal{U}_{(n,N,n_{1},q,\beta_{1}2,0,0)})=\mathbb{Z}[g,h,x^{*}]$. By $\mathfrak{q}^{\frac{Nn_{1}}{(N/n,n_{1})}}=1$, by Theorem 5.12 we get

[TABLE]

∎

To determine the Grothendieck ring $G_{0}(H_{\beta})$ for $\beta_{2}\neq 0$, we need the following lemma.

Lemma 5.14**.**

Suppose that $\beta_{1}(\gamma_{1}^{n_{1}}-\gamma_{2}^{n})=0$, $\beta_{2}(\gamma_{1}^{n_{1}}-\gamma_{3}^{n})\neq 0$ and $\beta_{1}(\gamma_{1}^{\prime n_{1}}-\gamma_{2}^{\prime n})=\beta_{2}(\gamma_{1}^{\prime n_{1}}-\gamma_{3}^{\prime n})=\beta_{3}({\gamma^{\prime}_{1}}^{-\frac{2n_{1}}{n}}\gamma_{2}^{\prime}\gamma_{3}^{\prime}-q^{2n_{1}j})=0$. Then

[TABLE]

Proof.

Since $\beta_{2}(\gamma_{1}^{n_{1}}-\gamma_{3}^{n})\neq 0$, $\beta_{2}\neq 0$. Thus $\gamma_{1}^{\prime n_{1}}=\gamma_{3}^{\prime n}$ and $\gamma_{3}^{\prime}=\gamma_{1}^{\prime\frac{n_{1}}{n}}q^{s}$ for some integer $1\leq s\leq n-1$. Hence $\beta_{2}((\gamma_{1}\gamma_{1}^{\prime})^{n_{1}}-(\gamma_{3}\gamma_{3}^{\prime})^{n})=\beta_{2}(\gamma_{1}^{n_{1}}-\gamma_{3}^{n})(\gamma^{\prime}_{3})^{n}\neq 0$. Since $\beta_{1}(\gamma_{1}^{n_{1}}-\gamma_{2}^{n})=0$ and $\beta_{1}(\gamma_{1}^{\prime n_{1}}-\gamma_{2}^{\prime n})=0$, $\beta_{1}((\gamma_{1}\gamma^{\prime}_{1})^{n_{1}}-(\gamma_{2}\gamma^{\prime}_{2})^{n})=0$. Let $\{m_{0},m_{1},\cdots,m_{n-1}\}$ be the basis of $V_{II}(\gamma_{1},\gamma_{2},\gamma_{3};i)$ with the actions given by Equations (6)-(8). Then $\{m_{0}\otimes 1,m_{1}\otimes 1,\cdots,m_{n-1}\otimes 1\}$ is a basis of $V_{II}(\gamma_{1},\gamma_{2},\gamma_{3};i)\otimes V_{0}(\gamma^{\prime}_{1},\gamma^{\prime}_{2},\gamma^{\prime}_{3};j)$. With this basis, we have

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Let $u_{l}=\gamma^{\prime\frac{n_{1}l}{n}}_{1}q^{jn_{1}l}m_{l}\otimes 1$. Then $\{u_{0},\cdots,u_{n-1}\}$ is also a basis of $V_{II}(\gamma_{1},\gamma_{2},\gamma_{3};i)\otimes V_{0}(\gamma^{\prime}_{1},\gamma^{\prime}_{2},\gamma^{\prime}_{3};j)$. Then we have

[TABLE]

[TABLE]

[TABLE]

Thus

[TABLE]

Similarly, we get

[TABLE]

∎

Remark 5.15**.**

By Lemma 5.14, we get

[TABLE]

where $\gamma_{1}^{\prime}=\gamma_{1}\gamma_{3}^{-\frac{n}{n_{1}}}$ satisfying $\gamma_{1}^{\prime n_{1}}\neq 1$, $\gamma_{2}^{\prime}=\gamma_{2}\gamma_{3}^{-1}q^{-2n_{1}i}$. If $\beta_{1}=\beta_{3}=0$, then $V_{II}(\gamma_{1},\gamma_{2},1;0)\cong V_{II}(\gamma_{1},1,1;0)\otimes V_{0}(1,\gamma_{2},1;0)$. If $\beta_{1}\neq 0$, then $V_{II}(\gamma_{1},\gamma_{2},1;0)=V_{II}(\gamma_{1},\gamma_{1}^{\frac{n_{1}}{n}}q^{u},1;0)=V_{II}(\gamma_{1},\gamma_{1}^{\frac{n_{1}}{n}},1;0)\otimes V_{0}(1,q^{u},1;0)$ for some integer $u$. If $\beta_{1}=0$, $\beta_{3}\neq 0$ and $\gamma_{2}=q^{u}$ for some integer $u$, then $V_{II}(\gamma_{1},\gamma_{2},1;0)\cong V_{II}(\gamma_{1},1,1;0)\otimes V_{0}(1,\gamma_{2},1;0)$.

Let $y_{\epsilon_{1},\epsilon_{2}}=[V_{II}(\epsilon_{1},\epsilon_{2},1;0)]$ for $\epsilon_{1}\in\widehat{{\bf K}_{0}}:=({\bf K}^{*}/\{\bar{\omega}^{v}\mid v\in\mathbb{Z}\})\setminus\{\overline{1}\}$, $\epsilon_{2}\in{\bf K}^{*}$. If $\beta_{2}\neq 0$ and $\beta_{1}=\beta_{3}=0$, then the Grothendieck ring of $G_{0}(H_{\beta})$ is isomorphic to a quotient ring of $R_{2}^{\prime\prime}=R_{1}[y_{\epsilon,1}|\epsilon\in\widehat{{\bf K}_{0}}]$. Similar to the proof of Theorem 5.12, we have the following theorem.

Theorem 5.16**.**

Suppose that $\beta_{2}\neq 0$ and $\beta_{1}=\beta_{3}=0$. Then the Grothendieck ring $G_{0}(H_{\beta})=R_{2}^{\prime\prime}:=R_{1}[y_{\epsilon,1}|\epsilon\in\widehat{{\bf K}_{0}}]$ with the relations

[TABLE]

Especially, we obtain the structure of the Grothendieck ring of the Gelaki’s Hopf alegbra where $\beta_{2}\neq 0$.

Corollary 5.17**.**

Suppose that $\beta_{2}\neq 0$. Then the Grothendieck ring $G_{0}(\mathcal{U}_{(n,N,n_{1},q,0,\beta_{2},0)})=S_{2}^{\prime\prime}:=S_{1}[y^{*}]$ with the relation

[TABLE]

Proof.

Similar to the proof of Corollary 5.13.∎

Theorem 5.18**.**

Suppose that $\beta_{1}\beta_{2}\neq 0$ and $\beta_{3}=0$. Then the Grothendieck ring $R_{3}^{\prime}$ of $H_{\beta}$ is equal to $R_{1}[x_{\zeta_{1},\zeta_{2}},y_{\epsilon_{1},\epsilon_{2}}|(\zeta_{1},\zeta_{2})\in{\widehat{\bf K_{0}}\times\bf{K}^{*}},\epsilon_{1}\in{\widehat{\bf K_{0}}},\epsilon_{2}=\epsilon_{1}^{\frac{n_{1}}{n}}]$ with relations

[TABLE]

[TABLE]

[TABLE]

where $\zeta=\zeta_{2}\zeta_{2}^{\prime}$.

Proof.

By the proof of Theorem 5.12 and Theorem 5.16, we have the Equation (27) except for the case of $(\zeta_{1}\zeta_{1}^{\prime})^{n_{1}}-1=0,(\zeta_{1}\zeta_{1}^{\prime})^{n_{1}}-(\zeta_{2}\zeta_{2}^{\prime})^{n}\neq 0$ and Equation (28).

(1) Here, we discuss the case in which $((\zeta_{1}\zeta_{1}^{\prime})^{n_{1}}-1)\beta_{1}=0$ and $((\zeta_{1}\zeta_{1}^{\prime})^{n_{1}}-(\zeta_{2}\zeta_{2}^{\prime})^{n})\beta_{2}\neq 0$. Then $(\zeta_{1}\zeta_{1}^{\prime})^{n_{1}}=1$ and $(\zeta_{2}\zeta_{2}^{\prime})^{n}\neq(\zeta_{1}\zeta_{1}^{\prime})^{n_{1}}$. Let $\{m_{0},\cdots,m_{n-1}\}$ and $\{m_{0}^{\prime},\cdots,m_{n-1}^{\prime}\}$ be the basis of $V_{I}(\zeta_{1},1,\zeta_{2};0)$ and $V_{I}(\zeta_{1}^{\prime},1,\zeta_{2}^{\prime};0)$ with the actions given by Equations (3)-(5) respectively. Then $\{m_{l}\otimes m^{\prime}_{k},\;0\leq l,k\leq n-1\}$ is a basis of $V_{I}(\zeta_{1},1,\zeta_{2};0)\otimes V_{I}(\zeta_{1}^{\prime},1,\zeta_{2}^{\prime};0)$. Let $v_{p}=m_{0}\otimes m^{\prime}_{p}+\sum\limits^{n-1}_{l=p+1}\alpha_{l}m_{n+p-l}\otimes m^{\prime}_{l}$ for $0\leq p\leq n-1$. Assume that $xv_{0}=\theta y^{n-1}v_{0}$. Similar to the proof of Theorem 5.12, we can determine $\alpha_{l},1\leq l\leq n-1$. Hence, $H_{\beta}v_{0}\cong V_{II}(\zeta_{1}\zeta_{1}^{\prime},1,\zeta_{2}\zeta_{2}^{\prime};0)\cong V_{0}((\zeta_{2}\zeta_{2}^{\prime})^{\frac{n}{n_{1}}},\zeta_{2}\zeta_{2}^{\prime},\zeta_{2}\zeta_{2}^{\prime};0)\otimes V_{II}(\zeta_{1}\zeta_{1}^{\prime}(\zeta_{2}\zeta_{2}^{\prime})^{-\frac{n}{n_{1}}},(\zeta_{2}\zeta_{2}^{\prime})^{-1},1;0)$ and $H_{\beta}v_{0}$ is an irreducible submodule of $V_{I}(\zeta_{1},1,\zeta_{2};0)\otimes V_{I}(\zeta_{1}^{\prime},1,\zeta_{2}^{\prime};0)$.

Similarly, we can show that

[TABLE]

and $(H_{\beta}v_{p}+\sum^{p-1}_{l=0}H_{\beta}v_{l})/\sum\limits^{p-1}_{l=0}H_{\beta}v_{l}$ is an irreducible submodule of $V_{I}(\zeta_{1},1,\zeta_{2};0)\otimes V_{I}(\zeta_{1}^{\prime},1,\zeta_{2}^{\prime};0)/\sum\limits^{p-1}_{l=0}H_{\beta}v_{l}$ for $1\leq p\leq n-1$. Hence $[V_{I}(\zeta_{1},1,\zeta_{2};0)][V_{I}(\zeta_{1}^{\prime},1,\zeta_{2}^{\prime};0)]=\sum\limits^{n-1}_{p=0}[V_{II}(\zeta_{1}\zeta_{1}^{\prime},1,\zeta_{2}\zeta_{2}^{\prime};n-p)]$. Consequently,

[TABLE]

where $c_{2}=\sum\limits^{n-1}_{p=0}[V_{0}((\zeta_{2}\zeta_{2}^{\prime})^{\frac{n}{n_{1}}},\zeta_{2}\zeta_{2}^{\prime},\zeta_{2}\zeta_{2}^{\prime};n-p)]=\sum\limits^{n-1}_{p=0}g^{p}[V_{0}((\zeta_{2}\zeta_{2}^{\prime})^{\frac{n}{n_{1}}},\zeta_{2}\zeta_{2}^{\prime},\zeta_{2}\zeta_{2}^{\prime};0)]$.

(2) Let $\{m_{0},\cdots,m_{n-1}\}$ and $\{m_{0}^{\prime\prime},\cdots,m_{n-1}^{\prime\prime}\}$ be the basis of $V_{I}(\zeta_{1},1,\zeta_{2};0)$ and $V_{II}(\epsilon_{1},\epsilon_{2},1;0)$ with the actions given by Equations(3)-(5) and (6)-(8) respectively. Then $\{m_{l}\otimes m_{k}^{\prime\prime},\;0\leq l,k\leq n-1\}$ is a basis of $V_{I}(\zeta_{1},1,\zeta_{2};0)\otimes V_{II}(\epsilon_{1},\epsilon_{2},1;0)$. Let $v_{p}^{\prime}=\sum\limits^{n-p-1}_{v=0}\alpha_{v}m_{p+v}\otimes m_{v}^{\prime\prime}$ for $\alpha_{0}=1$. Similar to the proof of Theorem 5.10, we can prove

[TABLE]

and $\sum^{p}_{l=0}H_{\beta}v_{l}^{\prime\prime}/\sum^{p-1}_{l=0}H_{\beta}v_{l}^{\prime\prime}$ is an irreducible submodule of $V_{I}(\zeta_{1},1,\zeta_{2};0)\otimes V_{II}(\epsilon_{1},\epsilon_{2},1;0)/\sum\limits^{p-1}_{l=0}H_{\beta}v_{l}^{\prime\prime}$. Hence

[TABLE]

Similarly, we have $V_{II}(\epsilon_{1},\epsilon_{2},1;0)\otimes V_{I}(\zeta_{1},1,\zeta_{2};0)=\sum\limits^{n-1}_{p=0}V_{I}(\zeta_{1},1,\zeta_{2}\epsilon_{2}^{-1};0)\otimes V_{0}(\epsilon_{1},\epsilon_{2},\epsilon_{2};n-p)$. Consequently, we obtain $x_{\zeta_{1},\zeta_{2}}y_{\epsilon_{1},\epsilon_{2}}=c_{3}x_{\zeta_{1},\zeta_{2}\epsilon_{2}^{-1}}=y_{\epsilon_{1},\epsilon_{2}}x_{\zeta_{1},\zeta_{2}}$, where $c_{3}=\sum\limits^{n-1}_{p=0}[V_{0}(\epsilon_{1},\epsilon_{2},\epsilon_{2};n-p)]$. ∎

Especially, we obtain the structure of the Grothendieck ring of the Gelaki’s Hopf alegbra where $\beta_{1}\beta_{2}\neq 0$.

Corollary 5.19**.**

Suppose that $\beta_{1}\beta_{2}\neq 0$. Then the Grothendieck ring $G_{0}(\mathcal{U}_{(n,N,n_{1},q,\beta_{1},\beta_{2},0)})=S_{3}^{\prime}:=S_{1}[x^{*}]=\mathbb{Z}[g,h,x^{*}]$ with relation

[TABLE]

Proof.

Since $\beta_{1}^{\prime\prime}=0$ implies $\gamma^{n_{1}}=1$, $\mathcal{U}_{(n,N,n_{1},q,\beta_{1},\beta_{2},0)}$ has no irreducible representation $V_{II}(\gamma,1,1;0)$. Thus $S_{3}^{\prime}=S_{1}[x^{*}]$. The rest of the proof is similar to that of Corollary 5.13.∎

Let $\{m_{0}^{\prime},\cdots,m_{r-1}^{\prime}\}$ be the basis of $V_{r}(\gamma^{\prime}_{1},\gamma^{\prime}_{2},\gamma^{\prime}_{3};j)$ with the actions given by Equations (9)-(11). Since $k^{\prime}_{l}\neq 0$ for $1\leq l\leq r-1$, $\{m^{\prime\prime}_{l}:=\prod^{r-1}_{v=r-l}k^{\prime}_{v}m^{\prime}_{r-l-1}|0\leq l\leq r-1\}$ is also a basis of $V_{r}(\gamma^{\prime}_{1},\gamma^{\prime}_{2},\gamma^{\prime}_{3};j)$. The action of $H_{\beta}$ on this basis is given by

$am^{\prime\prime}_{l}=\prod\limits^{r-1}_{v=r-l}k^{\prime}_{v}am^{\prime}_{r-l-1}=(\gamma^{\prime}_{1})^{\frac{1}{n}}q^{j-r+1+l}m^{\prime\prime}_{l},$  for $0\leq l\leq r-1$;

$xm^{\prime\prime}_{l}=k^{\prime}_{r-l}m^{\prime\prime}_{l-1}$,  for $1\leq l\leq r-1$;     $xm^{\prime\prime}_{0}=0$;

$ym^{\prime\prime}_{l}=m^{\prime\prime}_{l+1}$,    for $0\leq l\leq r-2$;     $ym^{\prime\prime}_{r-1}=0$.

Let $\overline{\bf{K_{0}}}:=({\bf{K}^{*}}/\langle q\rangle)\setminus\{\bar{1}\}$. By the same method as Theorem 5.10, we can get the following theorem.

Theorem 5.20**.**

Suppose that $\beta_{2}\beta_{3}\neq 0$ and $\beta_{1}=0$. Let $\widetilde{z}_{\xi}=[V_{t}(1,\xi,1;0)]$, where $\xi\in\overline{{\bf K}^{*}}$. Then the Grothendieck ring $R_{3}^{\prime\prime}$ of $H_{\beta}$ is isomorphic to the commutative ring $R_{1}[z_{2},\tilde{z}_{\xi},y_{\epsilon_{1},\epsilon_{2}}|(\epsilon_{1},\epsilon_{2})\in\widehat{{\bf K}_{0}}\times\overline{{\bf K}_{0}},\xi\in\overline{{\bf K}^{*}}]$ with relations

[TABLE]

[TABLE]

[TABLE]

for $\xi\xi^{\prime}=q^{rn_{1}},$

[TABLE]

where $\eta^{\prime}=[V_{0}(q^{\frac{n}{2}},q^{n_{1}},1;0)]$ and $r_{p}=(r-2p)\ mod(t)$, $r$ is the minimal positive integer such that $\epsilon_{1}\epsilon_{2}=q^{rn_{1}}$.

Especially, we obtain the structure of the Grothendieck ring of the Gelaki’s Hopf alegbra where $\beta_{2}\beta_{3}\neq 0$.

Corollary 5.21**.**

Suppose that $\beta_{2}\beta_{3}\neq 0$.

• (1)

Suppose $(N,nn_{1},2n_{1}t)<(nn_{1},N)$. If either $2n\mid N$, or $2n\nmid N$ and $2(n,n_{1})\nmid n$, then the Grothendieck ring $S_{3}^{\prime}=G_{0}(\mathcal{U}_{(n,N,n_{1},q,0,\beta_{2},\beta_{3})})$ is isomorphic to the commutative ring $S_{2}[y^{*}]=\mathbb{Z}[g,h_{2},z_{2},z^{\prime\prime},y^{*}]$ with relations (14), (21), $z_{2}z^{\prime\prime}=(1+g^{n-1})z^{\prime\prime}$, $z_{2}y^{*}=(1+g^{n-1})y^{*}$ and ${y^{*}}^{\frac{N}{(N/n,n_{1})}}=u\mathfrak{s}^{\frac{N}{(N/n,n_{1})}-1}z^{\prime\prime}$. If $2n\nmid N$ and $2(n,n_{1})\mid n$, then the Grothendieck ring $S_{3}^{\prime}:=G_{0}(\mathcal{U}_{(n,N,n_{1},q,0,\beta_{2},\beta_{3})})$ is isomorphic to the commutative ring $S_{2}[y^{*}]=\mathbb{Z}[g,h_{2},z_{3},z^{\prime\prime},y^{*}]$ with relations (19), (21), $z_{3}z^{\prime\prime}=(1+g^{n-1}+g^{n-2})z^{\prime\prime}$, $z_{3}y^{*}=(1+g^{n-1}+g^{n-2})y^{*}$ and ${y^{*}}^{\frac{N}{(N/n,n_{1})}}=u\mathfrak{s}^{\frac{N}{(N/n,n_{1})}-1}z^{\prime\prime}$.

• (2)

Suppose that $\beta_{2}\beta_{3}\neq 0$ and $(N,nn_{1},2n_{1}t)=(nn_{1},N)$. Then the Grothendieck ring

[TABLE]

is a subring of $S_{3}^{\prime}$.

Similar to the proof of Theorem 5.5-Theorem 5.20, we have the following theorem.

Theorem 5.22**.**

Suppose that $\beta_{1}\beta_{2}\beta_{3}\neq 0$. Then the Grothendieck ring $R_{4}$ of $H_{\beta}$ is equal to $R_{1}[z_{2},z_{\xi}^{\prime},x_{\zeta_{1},\zeta_{2}},$ $y_{\epsilon_{1},\epsilon_{1}}|\xi\in\widetilde{\bf K_{0}},(\zeta_{1},\zeta_{2})\in\overline{\bf K_{0}}\otimes{\bf K}^{*},(\epsilon_{1},\epsilon_{2})\in{\widehat{{\bf K}_{0}}}\times\overline{{\bf K}_{0}},\epsilon_{1}^{n_{1}}=\epsilon_{2}^{n}]$ with relations (14), (30),

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

where $\zeta=(\zeta_{2}\zeta_{2}^{\prime})^{\frac{n}{n_{1}}}$, $\zeta^{*}=\zeta_{2}\zeta_{2}^{\prime}$, $\zeta^{\prime\prime}=\zeta_{1}\zeta_{1}^{\prime}$, $r_{p}=(r-2p)\ mod(t)$ and $1\leq r_{p}\leq t$, $r$ is the minimal positive integer such that $\zeta_{2}\zeta_{2}^{\prime}=(\zeta_{1}\zeta_{1}^{\prime})^{\frac{2n_{1}}{n}}q^{(-r+1)n_{1}}$.

Especially, we obtain the structure of the Grothendieck ring of the Gelaki’s Hopf alegbra where $\beta_{1}\beta_{2}\beta_{3}\neq 0$.

Corollary 5.23**.**

Suppose that $\beta_{1}\beta_{2}\beta_{3}\neq 0$.

• (1)

Suppose $(N,nn_{1},2n_{1}t)<(nn_{1},N)$. If either $2n\mid N$, or $2n\nmid N$ and $2(n,n_{1})\nmid n$, then the Grothendieck ring $S_{3}^{\prime\prime}=G_{0}(\mathcal{U}_{(n,N,n_{1},q,\beta_{1},\beta_{2},\beta_{3})})$ is isomorphic to the commutative ring $S_{2}[x^{*}]=\mathbb{Z}[g,h_{2},z_{2},z^{\prime\prime},x^{*}]$ with relations in Corollary 5.11(1)(I). If $2n\nmid N$ and $2(n,n_{1})\mid n$, then the Grothendieck ring $S_{3}^{\prime\prime}:=G_{0}(\mathcal{U}_{(n,N,n_{1},q,\beta_{1},\beta_{2},\beta_{3})})$ is isomorphic to the commutative ring $S_{2}[x^{*}]=\mathbb{Z}[g,h_{2},z_{3},z^{\prime\prime},x^{*}]$ with relations in Corollary 5.11(1)(II).

• (2)

Suppose that $\beta_{1}\beta_{2}\beta_{3}\neq 0$ and $(N,nn_{1},2n_{1}t)=(nn_{1},N)$. Then the Grothendieck ring

[TABLE]

is a subring of $S_{3}^{\prime\prime}$.

Let $\omega$ be a primitive $N$-th root of unity. If $N\nmid n_{1}^{2}$, then $U_{(N,n_{1},\omega)}={\mathcal{U}}_{(N/(N,n_{1}),N,n_{1},\omega^{n_{1}},0,0,1)}$ and ${\mathcal{U}}_{(N/(N,n_{1}),N,n_{1},\omega^{n_{1}},0,0,\gamma)}\simeq U_{(N,n_{1},\omega)}$ as Hopf algebras for any $\gamma\in{\bf K}^{*}$. Hence $V_{0}(\gamma_{1},1,1,i)$ and $V_{r}(\gamma_{1},1,1,i)$ are irreducible representations of $U_{(N,n_{1},\omega)}$, where $\gamma_{1}^{(N,n_{1})}=1$. Thus, the Grotendieck ring of $U_{(N,n_{1},\omega)}$ is the same as the Grothendieck ring $G_{0}({\mathcal{U}}_{(n,N,n_{1},q,0,0,\beta_{3})})$ with $\beta_{3}\neq 0$ in Corollary 5.6.

Corollary 5.24**.**

Let $h^{\prime}=[V_{0}(\omega^{n^{\prime}},1,1;0)]$ and $z^{\prime}=[V_{t}(\omega^{n},1,1;0)]$. Then $g^{n}=h^{\prime\frac{N}{(N,n^{\prime})}}=1$, where $n=\frac{N}{(N,\nu)}$ and $n^{\prime}=\frac{N^{2}}{(N^{2},N\nu,2\nu^{2})}$.

• (1)

Suppose that $(\nu^{2},N)\nmid 2\nu$. If either $2\mid(N,\nu)$, or $2\nmid(N,\nu)$ and $2(N,\nu^{2})\nmid N$, then the Grothendieck ring $S=G_{0}(U_{(N,\nu,\omega)})\cong\mathbb{Z}[g,h^{\prime},z_{2},z^{\prime}]$ with relations (14), (18) and $z_{2}z^{\prime}=(1+g^{n-1})z^{\prime}$. If $2\nmid(N,\nu)$ and $2(N,\nu^{2})\mid N$, then the Grothendieck ring $S=G_{0}(U_{(N,\nu,\omega)})\cong\mathbb{Z}[g,h^{\prime},z_{3},z^{\prime}]$ with relations (18), (19) and $z_{3}z^{\prime}=(1+g^{n-1}+g^{n-2})z^{\prime}$.

• (2)

Suppose that $2\nu=(\nu^{2},N)$. Then the Grothendieck ring

[TABLE]

is a subring of $S$.

Remark 5.25**.**

By Corollary 5.6-Corollary 5.23, we have

[TABLE]

and

[TABLE]

But ${\mathcal{U}}_{(n,N,n_{1},q,\beta_{1},\beta_{2},\beta_{3})}\ncong{\mathcal{U}}_{(n,N,n_{1},q,\beta_{1},0,\beta_{3})}$ by [7, Proposition 3.2].