Ore Extensions for the Sweedler’s Hopf Algebra H₄

Abstract

The aim of this paper is to classify all Hopf algebra structures on the quotient of Ore extensions ${\mathbb{H}}_4[z;\sigma ]$ of automorphism type for the Sweedler${}^{\prime }$s 4-dimensional Hopf algebra ${\mathbb{H}}_4$. Firstly, we calculate all equivalent classes of twisted homomorphisms $(\sigma ,J)$ for ${\mathbb{H}}_4$. Then we give the classification of all bialgebra (Hopf algebra) structures on the quotients of ${\mathbb{H}}_4[z;\sigma ]$ up to isomorphism.

1. Introduction

Ore extensions play a key role in classifying pointed Hopf algebras, see for example [1,2,3,4,5]. It also can provide some neither pointed nor semisimple Hopf algebras, see for example ([6], Examples 3.1 and 3.3).

Panov [7] introduced the concept of Hopf-Ore extension $R[x;\sigma ,\delta ]$ of which the variable x is restricted to a skew primitive element and gave some equivalent descriptions. Later the Hopf-Ore extensions for some special Hopf algebras were obtained, such as quasitriangular Hopf algebras and multiplier Hopf algebras, see for example [8,9,10,11]. The authors [12] gave the realization of PBW-deformations of an quantum group via iterated Ore extensions. As is well-known, a Drinfeld’s twist $\mathcal{J}$ for a Hopf algebra R gives rise to a new Hopf algebra $R^{\mathcal{J}}$ with the same underlying algebra and the coalgebra structure, which is twisted from $\Delta$ by $\mathcal{J}$. Moreover, if $(R,\mathcal{R})$ is a quasitriangular Hopf algebra, so is $(R^{\mathcal{J}},{\mathcal{R}}^{\mathcal{J}})$, where ${\mathcal{R}}^{\mathcal{J}}={\mathcal{J}}_{21}\mathcal{R}{\mathcal{J}}^{-1}$ (see for example [13,14,15,16,17]). Now, let R be a bialgebra or Hopf algebra and $\sigma$ an automorphism of R. Recently, Yang and Zhang [6] described Hopf algebra structures on the localization of skew polynomial ring $R[z;\sigma ]$ and the quotients of $R[z;\sigma ]/I$, for a certain Hopf ideal I of $R[z;\sigma ]$, where $R[z;\sigma ]$ is a Ore extension of automorphism type. Recall that the Sweedler${}^{\prime }$s Hopf Algebra ${\mathbb{H}}_4$ is a noncommutative and noncocommutative quasitriangular Hopf algebra of the smallest dimension. It is one of the few examples discovered in the early stage of the exhibition, and now it still plays an important role in the theoretical development of Hopf algebra [18]. In this paper, we study Ore extensions of automorphism type for the Hopf algebra ${\mathbb{H}}_4$. Consequently, Ore extensions of ${\mathbb{H}}_4$ that are of bialgebras are classified. Some new examples of Hopf algebras of dimension $4n$, consisting some neither pointed nor semisimple Hopf algebras are given.

The paper is organized as follows.

In Section 1, some basic notions of Ore extensions, Hopf algebras, Drinfeld${}^{\prime }$s twists and twisted homomorphisms are reviewed. One fundamental result (Theorem 1) about the Ore extensions of automorphism type for Hopf algebras is established. In Section 2, we firstly compute the equivalent classes of twisted homomorphisms for ${\mathbb{H}}_4$. Equivalently, the isomorphism classes of Ore extensions of automorphism type for ${\mathbb{H}}_4$ are described. In Section 3, up to bialgebra isomorphisms, the classes of the Ore extensions of automorphism type for ${\mathbb{H}}_4$ are determined completely (Theorem 3). All Hopf algebras structures on the quotients of Ore extensions of automorphism type for ${\mathbb{H}}_4$ are also classified (Theorem 4).

2. Preliminaries

Throughout the paper, we work over the fixed field $\mathbb{k}$ containing some primitive root $\omega$ of unity. All algebras, modules, homomorphisms and tensor products are defined over the field $\mathbb{k}$.

The group of automorphisms of an algebra R is denoted by $\mathrm{Aut}\left(R\right)$ and $\sigma \in \mathrm{Aut}\left(R\right)$ unless otherwise stated.

Let us recall some basic notions and results about Ore extensions and Hopf algebras. For more details, the readers can refer to [18,19].

Suppose that R is a ring, $\sigma :R\to R$ a ring homomorphism, and $\delta :R\to R$ a $\sigma$-derivation of R, which means that $\delta$ is a homomorphism of abelian groups satisfying

$$\delta \left(r_1{r}_2\right)=\sigma \left(r_1\right)\delta \left(r_2\right)+\delta \left(r_1\right)r_2.$$

Then an Ore extension $R[x;\sigma ,\delta ]$, is defined by a noncommutative ring obtained by giving the ring of polynomials $R\left[x\right]$ a new multiplication, subject to the identity

$$xr=\sigma \left(r\right)x+\delta \left(r\right).$$

If $\delta =0$, the Ore extension is denoted by $R[x;\sigma ]$, and it is called an Ore extension of automorphism type for R.

A bialgebra over the the field $\mathbb{k}$ is a vector space which is both a unital associative algebra and a coalgebra. The algebraic and coalgebraic structures are compatible with a few more axioms: the comultiplication and the counit are both unital algebra homomorphisms, or equivalently, the multiplication and the unit of the algebra both are coalgebra morphisms.

Let R be a bialgebra with the comultiplication $\Delta :R\to R\otimes R$, the counit $ϵ:R\to \mathbb{k}$. If there exist a $\mathbb{k}$-map $S:R\to R$ such that

$$\sum _{\left(h\right)}S\left(h_1\right)h_2=\sum _{\left(h\right)}{h}_{1}S\left(h_2\right)=ϵ\left(h\right)$$

for all $h\in R$, then R is called a Hopf algebra, where we use the sigma notations

$$\Delta \left(h\right)=\sum _{\left(h\right)}{h}_1\otimes h_2.$$

Example 1.

The Sweedler${}^{\prime }$s 4-dimensional Hopf algebra ${\mathbb{H}}_4$ is defined by

$$\begin{array}{c}\mathit{as}\mathit{an}\mathit{algebra}:{\mathbb{H}}_4=\mathbb{k}〈1,g,x,gx|g^2=1,x^2=0,xg=-gx〉;\hfill \\ \mathit{the}\mathit{coalgebra}:\Delta \left(g\right)=g\otimes g,\Delta \left(x\right)=x\otimes 1+g\otimes x,\hfill \\ ϵ\left(g\right)=1,ϵ\left(x\right)=0,\hfill \\ S\left(g\right)=g,S\left(x\right)=-gx.\hfill \end{array}$$

${\mathbb{H}}_4$ is a unique and non-commutative quasi-triangular Hopf algebra of dimension 4. It is one of the few examples discovered in the early stage of the exhibition, and now it still plays an important role in the theoretical development of Hopf algebra.

It is easy to see that the set of grouplikes of ${\mathbb{H}}_4$ is $G\left({\mathbb{H}}_4\right)=\{1,g\}$ and $x\in P_{1,g}\left({\mathbb{H}}_4\right)$, where

$$P_{g,h}\left(H\right)=\{a\in H|\Delta \left(a\right)=a\otimes g+h\otimes a\}.$$

Let

$$e_0=\frac{1+g}{2},e_1=\frac{1-g}{2}.$$

Then the $\{e_0,e_1,e_{0}x,e_{1}x\}$ forms another basis of ${\mathbb{H}}_4$ and

$$e_0{e}_1=e_1{e}_0=0,e_0^2=e_1^2=1,$$

$$e_{0}g=g{e}_0=e_0,e_{1}g=g{e}_1=-e_1,$$

$$e_{0}x=x{e}_1,e_{1}x=x{e}_0.$$

It is straightforward to see that if $\sigma \in \mathrm{Aut}\left({\mathbb{H}}_4\right)$, then

$$\sigma \left(x\right)=px+qgx=p(e_0+e_1)x+q(e_0-e_1)x=(p+q)e_{0}x+(p-q)e_{1}x$$

and

$$\sigma \left(g\right)=\nu g+mx+ngx=\nu g+(m+n)e_{0}x+(m-n)e_{1}x,$$

for some $p,q,m,n\in \mathbb{k}$ and $\nu =\pm 1$ with some relations.

We denote a copy of the Sweedler’s Hopf algebra ${\mathbb{H}}_4$ by ${\mathbb{H}}_4^{\prime }$, which generators are replaced by $g^{\prime },x^{\prime }$ satisfying the same relations. The analogous notations are allowed for twisted homomorphisms for ${\mathbb{H}}_4^{\prime }$.

Furthermore, if $\mathsf{\Phi }:{\mathbb{H}}_4\to {\mathbb{H}}_4^{\prime }$ is the bialgebra isomorphism, then

$$\mathsf{\Phi }\left(g\right)=g^{\prime },\mathsf{\Phi }\left(x\right)=\mu x^{\prime },\mathrm{for}\mathrm{some}\mu \ne 0$$

(see [20]).

Definition 1. 

([21]) Let R be a bialgebra or Hopf algebra, the invertible element $J\in R\otimes R$, and $\sigma \in \mathrm{Aut}\left(R\right)$. The pair $(\sigma ,J)$ is said to be a twisted homomorphism for R if

$$(\Delta \otimes id)\left(J\right)(J\otimes 1)=(id\otimes \Delta )\left(J\right)(1\otimes J),$$

(1)

$$(id\otimes ϵ)\left(J\right)=(ϵ\otimes id)\left(J\right)=1,$$

(2)

$$\begin{array}{c}J(\sigma \otimes \sigma )\Delta \left(h\right)=\Delta \left(\sigma \right(h\left)\right)J,\hfill \end{array}$$

(3)

$$\begin{array}{c}ϵ\circ \sigma =ϵ\hfill \end{array}$$

(4)

for all $h\in R$.

Assume that R is a Hopf algebra. By ([22], Theorem 2.4), $R[z;\sigma ]$ is a bialgebra with $R\subset R[z;\sigma ]$ defined by

$$\Delta \left(z\right)=J(z\otimes z),ϵ\left(z\right)=1_{\mathbb{k}}$$

(5)

if and only if $(\sigma ,J)$ is a twisted homomorphism for R. In this case, $R[z;\sigma ]$ is called a bialgebra Ore extension of automorphism type (simply, BOEA ) for Hopf algebra R.

Let R be a Hopf algebra with the antipode S, $\sigma \in \mathrm{Aut}\left(R\right)$, and $(\sigma ,J)$ a twisted homomorphism for R. Let

$$\vartheta =\prod _{i=0}^{n-1}({\sigma }^i\otimes {\sigma }^i)\left(J\right)=\prod _{i=0}^{n-1}\left(\sum _J{\sigma }^i\left(J^1\right)\otimes {\sigma }^i\left(J^2\right)\right)=\sum _{\vartheta }{\vartheta }^1\otimes {\vartheta }^2.$$

We denote $S^{\sigma }=\sigma \circ S\circ {\sigma }^{-1}$ and

$$\theta :={\theta }_l=\sum _{J}S\left(J^1\right)J^2,{\theta }_r=\sum _J{J}^1{S}^{\sigma }\left(J^2\right),$$

and

$$\varrho :={\varrho }_l=\sum _{\vartheta }S\left({\vartheta }^1\right){\vartheta }^2,{\varrho }_r=\sum _{\vartheta }{\vartheta }^{1}S\left({\vartheta }^2\right),$$

where $J=\sum J^1\otimes J^2\in R\otimes R$.

Theorem 1. 

([6], Theorem 2.6)Let R be a Hopf algebra with the antipode S, $(\sigma ,J)$ a twisted homomorphism for R, and $R[z;\sigma ]$ aBOEAfor R. Suppose that there exists a nonzero $t\in R$, with

$$\Delta \left(t\right)=\prod _{i=0}^{n-1}({\sigma }^i\otimes {\sigma }^i)\left(J\right)(t\otimes t),th={\sigma }^n\left(h\right)t$$

for all $h\in R$. Then $H=R[z;\sigma ]/〈z^n-t〉$ is a Hopf algebra with the antipode S such that $S\left(z\right)=z^{-1}{\theta }^{-1}$ if and only if

• ${\theta }_r=\theta$ and ${\sigma }^n\left({\varrho }_r\right)=\varrho ;$
• $S^{\sigma }\left(h\right)={\theta }^{-1}S\left(h\right)\theta$ for all $h\in R$;
• ${\displaystyle \prod _{i=0}^{n-1}}{\sigma }^i\left(\theta \right)=\varrho .$

The authors in [6] gave some nontrivial examples on Theorem 1. Here we give more example as follows.

Example 2.

For the Hopf algebra ${\mathbb{H}}_4$, $\omega \in \mathbb{k}$ a primitive $2n$-th root of unity. Let

$$\sigma \left(g\right)=g,\sigma \left(x\right)=\omega x,t=g$$

$$J=1\otimes 1+dxg\otimes x.$$

for any $d\in \mathbb{k}$. Then $(\sigma ,J)$ is a twisted homomorphism for ${\mathbb{H}}_4$ and satisfies the condtions of Theorem 1. Thus up to isomorphism, we get a Hopf algebra ${\mathbb{H}}_{4n}$, generated by $x,z$ with the relations

$$\begin{array}{c}{z}^{2n}=1,x^2=0,zx=\omega xz.\hfill \end{array}$$

The coalgebra is

$$\begin{array}{c}\Delta \left(x\right)=x\otimes 1+z^n\otimes x,\Delta \left(z\right)=z\otimes z+dx{z}^{n+1}\otimes xz,\hfill \\ \epsilon \left(z\right)=1,\epsilon \left(x\right)=0,\hfill \\ S\left(z\right)=z^{-1},S\left(x\right)=-z^{n}x.\hfill \end{array}$$

${\mathbb{H}}_{4n}$ is a neither pointed nor semisimple quasitrangular Hopf algebra of dimension $4n$ extended by ${\mathbb{H}}_4$. In the present paper, we shall investigate the bialgebra (Hopf algebra) structures on the quotients ${\mathbb{H}}_4[z;\sigma ]/〈z^n-t〉$ in general.

3. Classification of Twisted Homomorphisms for ${\mathbb{H}}_4$

In this section, we give the classification of twisted homomorphisms for ${\mathbb{H}}_4$.

Let $(\sigma ,J)$ and $({\sigma }^{\prime },J^{\prime })$ be twisted homomorphisms for ${\mathbb{H}}_4$ and ${\mathbb{H}}_4^{\prime }$ respectively, and ${\mathbb{H}}_4[z,\sigma ]$ and ${\mathbb{H}}_4^{\prime }[z^{\prime },{\sigma }^{\prime }]$ the corresponding Ore extensions of automorphism type. The datum $(\sigma ,J)$ is said to be equivalent to $({\sigma }^{\prime },J^{\prime })$, denoted by $(\sigma ,J)\approx ({\sigma }^{\prime },J^{\prime })$, if there is a bi-algebraic isomorphism $\mathsf{\Phi }:{\mathbb{H}}_4[z,\sigma ]\to {\mathbb{H}}_4^{\prime }[z^{\prime },{\sigma }^{\prime }]$ such that $\mathsf{\Phi }\left({\mathbb{H}}_4\right)={\mathbb{H}}_4^{\prime }$ as bialgebras.

Therefore, if $\mathsf{\Phi }:{\mathbb{H}}_4[z;\sigma ]\cong {\mathbb{H}}_4^{\prime }[z^{\prime };{\sigma }^{\prime }]$ as bialgebras, then

$$\begin{array}{c}(\mathsf{\Phi }\otimes \mathsf{\Phi })\left(J\right)(\mathsf{\Phi }\left(z\right)\otimes \mathsf{\Phi }\left(z\right))={\Delta }^{\prime }\left(\mathsf{\Phi }\left(z\right)\right)\hfill \end{array}$$

(6)

$$\begin{array}{c}\mathsf{\Phi }\left(z\right)\mathsf{\Phi }\left(h\right)=\mathsf{\Phi }\left(\sigma \right(h\left)\right)\mathsf{\Phi }\left(z\right),\hfill \end{array}$$

(7)

$$\begin{array}{c}ϵ\left(\mathsf{\Phi }\right(z\left)\right)=1\hfill \end{array}$$

(8)

for all $h\in R$.

We have the following main result.

Theorem 2.

Any twisted homomorphism $(\sigma ,J)$ for ${\mathbb{H}}_4$ is equivalent to one of the following lists.

(a)

the pair $({\sigma }_{1,s},J_{1,d}):$

$$\begin{array}{cc}\hfill & {\sigma }_{1,s}\left(g\right)=g,{\sigma }_{1,s}\left(x\right)=sx,\hfill \\ J_{1,d}=1\otimes 1+d\left(gx\otimes x\right).\hfill \end{array}$$

for any $d\in \mathbb{k}$ and $0\ne s\in \mathbb{k}$.

(b)

the pair $({\sigma }_{2,s},J_{2,d}):$

$$\begin{array}{c}{\sigma }_{2,s}\left(g\right)=g+2gx,{\sigma }_{2,s}\left(x\right)=sx,\hfill \\ J_{2,d}=1\otimes 1+2e_1\otimes x+d(gx\otimes x).\hfill \end{array}$$

for any $d\in \mathbb{k}$ and $0\ne s\in \mathbb{k}$.

(c)

the pair $({\sigma }_{3,s},J_3):$

$$\begin{array}{c}{\sigma }_{3,s}\left(g\right)=g+2x,{\sigma }_{3,s}\left(x\right)=sgx,\hfill \\ J_3=1\otimes 1-2e_1\otimes e_1+2e_1\otimes x.\hfill \end{array}$$

for any $0\ne s\in \mathbb{k}$.

Proof. 

Let $(\sigma ,J)$ be a twisted homomorphism for ${\mathbb{H}}_4$. The proof is given in three steps as follows.

Step 1: Firstly, we assume that

$$\begin{array}{c}\hfill J=\sum _{i,j=0}^1(a_{ij}{g}^i\otimes g^j+b_{ij}{g}^i\otimes g^{j}x+c_{ij}{g}^{i}x\otimes g^j+d_{ij}{g}^{i}x\otimes g^{j}x)\in {\mathbb{H}}_4\otimes {\mathbb{H}}_4\end{array}$$

satisfies Equations (1) and (2). Tedious computations and comparing the coefficients of terms by Equations (1) and (2) show that the Drinfeld twist J for ${\mathbb{H}}_4$ must be one of the following

(1)

$$J_1=1\otimes 1+4(a-1)e_1\otimes e_1+2cgx\otimes e_1+d\left(gx\otimes x+4(a-1)e_{0}x\otimes e_{0}x\right)$$

for any $a,c,d$.

(2)

$$J_2=1\otimes 1+2b{e}_1\otimes x+d(gx\otimes x),$$

for any $b\ne 0$ and d.

(3)

$$J_3=1\otimes 1+4(a-1)e_1\otimes e_1+2b{e}_1\otimes x+2cgx\otimes e_1+d^{\prime }(gx\otimes x)$$

for any $a\ne 1,b\ne 0$ and c, where $d^{\prime }=\frac{bc}{a-1}$.

Step 2: Secondly, we note that

$$\begin{array}{c}{J}_1(\sigma \otimes \sigma )\Delta \left(h\right)=\Delta \left(\sigma \left(h\right)\right)J_1,\hfill \end{array}$$

(9)

$$\begin{array}{c}ϵ\circ \sigma =ϵ.\hfill \end{array}$$

(10)

The Equation (10) and $\sigma \in \mathrm{Aut}\left({\mathbb{H}}_4\right)$ imply that

$$\sigma \left(g\right)=g+mx+ngx,\sigma \left(x\right)=px+qgx.$$

for some $p,q,m,n\in \mathbb{k}$. Recall that $J_1$ can be written as

$$\begin{array}{cc}{J}_1=\hfill & a(1\otimes 1)+(1-a)(g\otimes 1)+(1-a)(1\otimes g)+(a-1)(g\otimes g)+c(gx\otimes 1-gx\otimes g)\hfill \\ & +d\left(\right(a-1)1\otimes 1+ag\otimes 1+(a-1)1\otimes g+(a-1)g\otimes g)(x\otimes x)\hfill \end{array}$$

for any $a,c,d$. Equation (9) imply that

$$J_1(\sigma \otimes \sigma )\Delta \left(g\right)=\Delta \left(\sigma \left(g\right)\right)J_1,J_1(\sigma \otimes \sigma )\Delta \left(x\right)=\Delta \left(\sigma \left(x\right)\right)J_1.$$

The above two equations show that

$$\begin{array}{c}2(a-1)p=(2a-1)q,\hfill \end{array}$$

(11)

$$\begin{array}{c}2(a-1)m=(2a-1)n,\hfill \end{array}$$

(12)

$$\begin{array}{c}2(a-1)n+2c=(2a-1)m,\hfill \end{array}$$

(13)

$$\begin{array}{c}a{m}^2+2(1-a)mn+(a-1)n^2=2cm-cn,\hfill \end{array}$$

(14)

$$\begin{array}{c}amn+(1-a)n^2+(1-a)m^2+(a-1)mn=cn,\hfill \end{array}$$

(15)

$$\begin{array}{c}a{n}^2+2(1-a)mn+(a-1)m^2=cn,\hfill \end{array}$$

(16)

$$\begin{array}{c}amp+(1-a)np+(1-a)mq+(a-1)nq=2cp-cq,\hfill \end{array}$$

(17)

$$\begin{array}{c}amq+(1-a)nq+(1-a)mp+(a-1)np=cq,\hfill \end{array}$$

(18)

$$\begin{array}{c}anp+(1-a)mp+(1-a)nq+(a-1)mq=cq,\hfill \end{array}$$

(19)

$$\begin{array}{c}anq+(1-a)mq+(1-a)np+(a-1)mp=cq.\hfill \end{array}$$

(20)

If $a=\frac{3}{4}$, then $c=0$, $m+n=0,p+q=0$ by Equations (11)–(13) and Equations (14)–(20) hold automatically. However, $\sigma \notin \mathrm{Aut}\left({\mathbb{H}}_4\right)$. Therefore, $a\ne \frac{3}{4}$. Now we set $\kappa =\frac{1}{4a-3}$, then

$$m=(1+\kappa )c,n=(1-\kappa )c$$

by Equations (12) and (13) and we can rewrite

$$p=\frac{1}{2}(1+\kappa )s,q=\frac{1}{2}(1-\kappa )s$$

by Equation (11). Also, such $m,n$ and $p,q$ enjoy Equations (14)–(20) automatically and ${\sigma }_1\in \mathrm{Aut}\left({\mathbb{H}}_4\right)$.

Therefore, we get a twisted homomorphism $({\sigma }_1^{\kappa ,s},J_1^{c,d}),$ for ${\mathbb{H}}_4$, where

$$\begin{array}{c}{\sigma }_1^{\kappa ,s}\left(g\right)=g+2c\left(e_0+\kappa e_1\right)x,{\sigma }_1^{\kappa ,s}\left(x\right)=s(e_0+\kappa e_1)x(s\ne 0),\hfill \\ J_1^{c,d}=1\otimes 1+\left(\frac{1}{\kappa }-1\right)1\otimes e_1+2cgx\otimes e_1+d\left(gx\otimes x+\left(\frac{1}{\kappa }-1\right)e_{0}x\otimes e_{0}x\right)\hfill \end{array}$$

for any $\kappa \ne 0,c,d$.

Similarly, we get the twisted homomorphism $({\sigma }_2^s,J_2^{b,d})$, where

$$\begin{array}{c}{\sigma }_2^s\left(g\right)=g+2bgx,{\sigma }_2^s\left(x\right)=sx(s\ne 0),\hfill \\ J_2^{b,d}=1\otimes 1+2b(e_1\otimes x)+d(gx\otimes x)\hfill \end{array}$$

for any $b\ne 0$ and d, and the twisted homomorphism $({\sigma }_3^{\kappa ,s},J_3^{b,c})$, where

$$\begin{array}{c}{\sigma }_3^{\kappa ,s}\left(g\right)=g+2\left((b+c)e_0+\kappa (c-b)e_1)\right)x,{\sigma }_3^{\kappa ,s}\left(x\right)=s(e_0+\kappa e_1)x(s\ne 0),\hfill \\ J_3^{b,c}=1\otimes 1+\left(\frac{1}{\kappa }-1\right)e_1\otimes e_1+2b{e}_1\otimes x+2cgx\otimes e_1+\frac{4\kappa bc}{1-\kappa }(gx\otimes x),\hfill \end{array}$$

for any $\kappa \ne 0,1,b\ne 0$, c.

Step 3: Now, assume that $\mathsf{\Phi }:H={\mathbb{H}}_4[z,\sigma ]\to H^{\prime }={\mathbb{H}}_4^{\prime }[z^{\prime },{\sigma }^{\prime }]$ is a bialgebra isomorphism. Define $degr=0$ for all $r\in {\mathbb{H}}_4$ and $degz=1$. Then we can extend this to define the lexicographic order on ${\mathbb{H}}_4[z;\sigma ]$. Since $\mathsf{\Phi }$ is bialgebra isomorphism from H to $H^{\prime }$, we have $\mathsf{\Phi }\left(z\right)=A^{\prime }{z}^{\prime }+B^{\prime }$ by considering the expression for $\mathsf{\Phi }\left(z\right)$ as a polynomial z with coefficients in ${\mathbb{H}}_4^{\prime }$, where $A^{\prime },B^{\prime }\in {\mathbb{H}}_4^{\prime }$. It is easy to see that $A^{\prime }$ is invertible in ${\mathbb{H}}_4^{\prime }$ since $\mathsf{\Phi }$ is an isomorphism.

By Equation (6), we have

$$\begin{array}{cc}& {\Delta }^{\prime }\left(A^{\prime }\right)J^{\prime }(z^{\prime }\otimes z^{\prime })+{\Delta }^{\prime }\left(B^{\prime }\right)\hfill \\ =& (\mathsf{\Phi }\otimes \mathsf{\Phi })\left(J\right)((A^{\prime }\otimes A^{\prime })(z^{\prime }\otimes z^{\prime })+(A^{\prime }\otimes B^{\prime })(z^{\prime }\otimes 1)+(B^{\prime }\otimes A^{\prime })(1\otimes z^{\prime })+B^{\prime }\otimes B^{\prime }).\hfill \end{array}$$

Comparing the coefficients of $z^{\prime }\otimes 1$ and $1\otimes z^{\prime }$, we get $B^{\prime }=0$. Hence we have $\mathsf{\Phi }\left(z\right)=A^{\prime }{z}^{\prime }$. Equation (6) holds if and only if

$${\Delta }^{\prime }\left(A^{\prime }\right)J^{\prime }=(\mathsf{\Phi }\otimes \mathsf{\Phi })\left(J\right)(A^{\prime }\otimes A^{\prime }).$$

(21)

Now, by Equation (8), it follows that

$$A^{\prime }=e_0^{\prime }+b_0{e}_1^{\prime }+c_0{e}_0^{\prime }{x}^{\prime }+d_0{e}_1^{\prime }{x}^{\prime }.$$

Let us investigate them case by case.

(1) For the twisted homomorphism $({\sigma }_1^{\kappa ,s},J_1^{c,d}),$ where

$$\begin{array}{c}{\sigma }_1^{\kappa ,s}\left(g\right)=g+2c(e_0+\kappa e_1)x,{\sigma }_1^{\kappa ,s}\left(x\right)=s(e_0+\kappa e_1)x(s\ne 0),\hfill \\ J_1^{c,d}=1\otimes 1+({\kappa }^{-1}-1)e_1\otimes e_1+2cgx\otimes e_1+d(({\kappa }^{-1}-1)e_0\otimes e_0+g\otimes 1)\left(x\otimes x\right).\hfill \end{array}$$

Assume that

$$({\sigma }_1^{\kappa ,s},J_1^{c,d})\approx ({\sigma }_1^{{\kappa }^{\prime },s^{\prime }},{J_1}^{c^{\prime },d^{\prime }}),$$

where

$$\begin{array}{c}{\sigma }_1^{{\kappa }^{\prime },s^{\prime }}\left(g^{\prime }\right)=g^{\prime }+2c^{\prime }(e_0^{\prime }+{\kappa }^{\prime }{e}_1^{\prime })x^{\prime },{\sigma }_1^{{\kappa }^{\prime },s^{\prime }}\left(x^{\prime }\right)=s^{\prime }(e_0^{\prime }+{\kappa }^{\prime }{e}_1^{\prime })x^{\prime }(s^{\prime }\ne 0),\hfill \\ J_1^{c^{\prime },d^{\prime }}=1\otimes 1+({{\kappa }^{\prime }}^{-1}-1)e_1^{\prime }\otimes e_1^{\prime }+2c^{\prime }{g}^{\prime }{x}^{\prime }\otimes e_1^{\prime }+d^{\prime }(({{\kappa }^{\prime }}^{-1}-1)e_0^{\prime }\otimes e_0^{\prime }+g^{\prime }\otimes 1)\left(x^{\prime }\otimes x^{\prime }\right)\hfill \end{array}$$

in ${\mathbb{H}}_4^{\prime }$. By Equation (6), we have

$$\mathsf{\Phi }\left(z\right)\mathsf{\Phi }\left(x\right)=\mathsf{\Phi }\left(\sigma \right(x\left)\right)\mathsf{\Phi }\left(z\right),\mathsf{\Phi }\left(z\right)\mathsf{\Phi }\left(g\right)=\mathsf{\Phi }\left(\sigma \right(g\left)\right)\mathsf{\Phi }\left(z\right).$$

Therefore, we have

$$\begin{array}{cc}\hfill & s{b}_0=s^{\prime },s\kappa =b_0{s}^{\prime }{\kappa }^{\prime },c_0=c^{\prime }-c\mu b_0,d_0=c\mu \kappa -c^{\prime }{b}_0{\kappa }^{\prime }.\hfill \end{array}$$

(22)

It is noted that $b_0=\frac{s^{\prime }}{s},b_0^2=\frac{\kappa }{{\kappa }^{\prime }}.$

On the other hand, it is straightforward to see that by Equation (21),

$$(\mathsf{\Phi }\otimes \mathsf{\Phi })\left(J_1^{c,d}\right)\left(A^{\prime }\otimes A^{\prime }\right)={\Delta }^{\prime }\left(A^{\prime }\right)J_1^{c^{\prime },d^{\prime }}.$$

Comparing coefficients of all terms and noting that Equation (22), one get

$$d^{\prime }=d{\mu }^2+2c\mu d_0-{\kappa }^{-1}{d}_0^2.$$

Hence

$$\begin{array}{c}{a}_0=1,b_0=\frac{s^{\prime }}{s}=\epsilon \sqrt{\frac{\kappa }{{\kappa }^{\prime }}},\hfill \\ c_0=c^{\prime }-c\mu \sqrt{\frac{\kappa }{{\kappa }^{\prime }}}\epsilon ,d_0=-\sqrt{\kappa {\kappa }^{\prime }}(c^{\prime }-c\mu \sqrt{\frac{\kappa }{{\kappa }^{\prime }}}\epsilon ),\hfill \\ d^{\prime }=d{\mu }^2+\kappa c^2{\mu }^2-{\kappa }^{\prime }{c}^{\prime 2}\hfill \end{array}$$

for any $c,d,c^{\prime },d^{\prime }$ and $s\kappa \ne 0,s^{\prime }{\kappa }^{\prime }\mu \ne 0$.

In particular, for any $c,d$ and $s\kappa \ne 0$, we can choose suitable triples $(b_0,c_0,d_0)$, such that

$$c^{\prime }=0,{\kappa }^{\prime }=1,s^{\prime }=\epsilon \sqrt{\kappa }s,d^{\prime }=d{\mu }^2+\kappa c^2{\mu }^2.$$

Therefore, we get

$$({\sigma }_1^{\kappa ,s},J_1^{c,d})\approx ({\sigma }_{1,s^{\prime }},J_{1,d^{\prime }}),$$

where

$$\begin{array}{cc}\hfill & {\sigma }_{1,s^{\prime }}\left(g\right)=g,{\sigma }_{1,s^{\prime }}\left(x\right)=s^{\prime }x(s^{\prime }\ne 0),J_{1,d^{\prime }}=1\otimes 1+d^{\prime }\left(gx\otimes x\right),\hfill \end{array}$$

for any $d^{\prime }\in \mathbb{k}$ and $0\ne s^{\prime }\in \mathbb{k}$.

(2) For the twisted homomorphism $({\sigma }_2^s,J_2^{b,d}),$ where

$$\begin{array}{c}\hfill {\sigma }_2^s\left(g\right)=g+2bgx,{\sigma }_2^s\left(x\right)=sx(s\ne 0),J_2^{b,d}=1\otimes 1+2b(e_1\otimes x)+d(gx\otimes x)\end{array}$$

for any d and $b\ne 0$.

Assume that

$$({\sigma }_2^s,J_2^{b,d})\approx ({\sigma }_2^{s^{\prime }},{J_2}^{b^{\prime },d^{\prime }}),$$

where

$$\begin{array}{c}\hfill {\sigma }_2^{s^{\prime }}\left(g^{\prime }\right)=g^{\prime }+2b^{\prime }{g}^{\prime }{x}^{\prime },{\sigma }_2^{s^{\prime }}\left(x^{\prime }\right)=s^{\prime }{x}^{\prime }(s^{\prime }\ne 0),J_2^{b^{\prime },d^{\prime }}=1\otimes 1+2b^{\prime }(e_1^{\prime }\otimes x^{\prime })+d^{\prime }(g^{\prime }{x}^{\prime }\otimes x^{\prime })\end{array}$$

in ${\mathbb{H}}_4^{\prime }$. By Equation (8), we have

$$\mathsf{\Phi }\left(z\right)\mathsf{\Phi }\left(x\right)=\mathsf{\Phi }\left(\sigma \right(x\left)\right)\mathsf{\Phi }\left(z\right),\mathsf{\Phi }\left(z\right)\mathsf{\Phi }\left(g\right)=\mathsf{\Phi }\left(\sigma \right(g\left)\right)\mathsf{\Phi }\left(z\right).$$

Therefore, we have

$$\begin{array}{cc}\hfill & s{b}_0=s^{\prime },s=b_0{s}^{\prime },c_0=b^{\prime }-b\mu b_0,d_0=b_0{b}^{\prime }-b\mu \kappa .\hfill \end{array}$$

(23)

It is noted that $b_0=\frac{s^{\prime }}{s}$ and get that $b_0^2=1.$

Computing

$$(\mathsf{\Phi }\otimes \mathsf{\Phi })\left(J_2^{b,d}\right)\left(A^{\prime }\otimes A^{\prime }\right)\mathrm{and}{\Delta }^{\prime }\left(A^{\prime }\right)J_2^{b^{\prime },d^{\prime }}$$

as in the case (1), and comparing coefficients of all terms and noticing that Equation (23), one get that

$$d^{\prime }=d{\mu }^2-2b\mu d_0-d_0^2.$$

Hence

$$\begin{array}{c}{a}_0=1,b_0=\frac{s^{\prime }}{s}=\epsilon ,\hfill \\ c_0=b^{\prime }-b\mu \epsilon ,d_0=\epsilon b^{\prime }-b\mu ,\hfill \\ d^{\prime }=d{\mu }^2-b^{\prime 2}+b^2{\mu }^2\hfill \end{array}$$

for any $d,d^{\prime },b\ne 0,b^{\prime }\ne 0$ and $s\ne 0,s^{\prime }\mu \ne 0$.

In particular, for any $d,b\ne 0$ and $s\ne 0$, we can choose suitable triples $(b_0,c_0,d_0)$, such that

$$b^{\prime }=1,s^{\prime }=\epsilon s,d^{\prime }=d{\mu }^2+b^2{\mu }^2-1.$$

This means that

$$({\sigma }_2^s,J_2^{b,d})\approx ({\sigma }_{2,s^{\prime }},J_{2,d^{\prime }}),$$

where

$$\begin{array}{c}{\sigma }_{2,s^{\prime }}\left(g\right)=g+2gx,{\sigma }_{2,s^{\prime }}\left(x\right)=s^{\prime }x(s^{\prime }\ne 0),\hfill \\ J_{2,d^{\prime }}=1\otimes 1+2(e_1\otimes x)+d^{\prime }\left(gx\otimes x\right)\hfill \end{array}$$

for any $d^{\prime }\in \mathbb{k}$ and $0\ne s^{\prime }\in \mathbb{k}$.

(3) For the twisted homomorphism pair $({\sigma }_3^{\kappa ,s},J_3^{b,c}),$ where

$$\begin{array}{c}{\sigma }_3^{\kappa ,s}\left(g\right)=2\left((b+c)e_0+\kappa (c-b)e_1)\right)x,{\sigma }_3^{\kappa ,s}\left(x\right)=s(e_0+\kappa e_1)x(s\ne 0),\hfill \\ J_3^{b,c}=1\otimes 1+({\kappa }^{-1}-1)e_1\otimes e_1+2b{e}_1\otimes x+2cgx\otimes e_1+d^{\prime }(gx\otimes x)\hfill \end{array}$$

for any $b\ne 0$ and c, where $d^{\prime }=\frac{4\kappa bc}{1-\kappa }$ with $\kappa \ne 0,1$.

Now, we assume that

$$({\sigma }_3^{\kappa ,s},J_3^{b,c})\approx ({\sigma }_3^{{\kappa }^{\prime },s^{\prime }},{J_3}^{b^{\prime },c^{\prime }}),$$

where

$$\begin{array}{c}{\sigma }_3^{{\kappa }^{\prime },s^{\prime }}\left(g^{\prime }\right)=g^{\prime }+2\left((b^{\prime }+c^{\prime })e_0^{\prime }+{\kappa }^{\prime }(c^{\prime }-b^{\prime })e_1^{\prime })\right)x^{\prime },{\sigma }_3^{{\kappa }^{\prime },s^{\prime }}\left(x^{\prime }\right)=s^{\prime }(e_0^{\prime }+{\kappa }^{\prime }{e}_1^{\prime })x^{\prime }(s^{\prime }\ne 0),\hfill \\ J_3^{b^{\prime },c^{\prime }}=1\otimes 1+({\kappa }^{\prime -1}-1)e_1^{\prime }\otimes e_1^{\prime }+2b^{\prime }{e}_1^{\prime }\otimes x^{\prime }+2c^{\prime }{g}^{\prime }{x}^{\prime }\otimes e_1^{\prime }+d^{\prime \prime }(g^{\prime }{x}^{\prime }\otimes x^{\prime })\hfill \end{array}$$

in ${\mathbb{H}}_4^{\prime }$. By Equation (7), we have

$$\mathsf{\Phi }\left(z\right)\mathsf{\Phi }\left(x\right)=\mathsf{\Phi }\left(\sigma \right(x\left)\right)\mathsf{\Phi }\left(z\right),\mathsf{\Phi }\left(z\right)\mathsf{\Phi }\left(g\right)=\mathsf{\Phi }\left(\sigma \right(g\left)\right)\mathsf{\Phi }\left(z\right).$$

From the above equations, we easily get that

$$\begin{array}{cc}\hfill & s{b}_0=s^{\prime },s\kappa =b_0{s}^{\prime }{\kappa }^{\prime },c_0=b^{\prime }+c^{\prime }-\mu b_0(b+c),d_0=\mu \kappa (c-b)-b_0{\kappa }^{\prime }(c^{\prime }-b^{\prime }).\hfill \end{array}$$

(24)

It is noted that $b_0=\frac{s^{\prime }}{s}$ and $b_0^2=\frac{\kappa }{{\kappa }^{\prime }}.$

Furthermore, tediously computing

$$(\mathsf{\Phi }\otimes \mathsf{\Phi })\left(J_3^{b,c}\right)\left(A^{\prime }\otimes A^{\prime }\right)\mathrm{and}{\Delta }^{\prime }\left(A^{\prime }\right)J_3^{b^{\prime },c^{\prime }}$$

and comparing coefficients of all terms of them, one get that

$$d^{\prime \prime }=d^{\prime }{\mu }^2+2c\mu d_0-2b\mu d_0-{\kappa }^{-1}{d}_0^2.$$

Hence

$$\begin{array}{c}{a}_0=1,b_0=\frac{s^{\prime }}{s}=\epsilon \sqrt{\frac{\kappa }{{\kappa }^{\prime }}},\hfill \\ c_0=(b^{\prime }+c^{\prime })-(b+c)\mu \sqrt{\frac{\kappa }{{\kappa }^{\prime }}}\epsilon ,d_0=\mu \kappa (c-b)-\sqrt{\kappa {\kappa }^{\prime }}(c^{\prime }-b^{\prime }),\hfill \\ d^{\prime \prime }=d^{\prime }{\mu }^2+\kappa {(c-b)}^2{\mu }^2-{\kappa }^{\prime }{(c^{\prime }-b^{\prime })}^2\hfill \end{array}$$

for any $c,c^{\prime },b,b^{\prime }\ne 0$ and $s\kappa \ne 0,s^{\prime }{\kappa }^{\prime }\mu \ne 0$.

In particular, for any c and $s\kappa \ne 0$, we can choose suitable triples $(b_0,c_0,d_0)$, such that

$$b^{\prime }=1,c^{\prime }=0,{\kappa }^{\prime }=-1,s^{\prime }=\epsilon \sqrt{-\kappa }s,d^{\prime \prime }=0$$

and we get

$$({\sigma }_3^{\kappa ,s},J_3^{b,c})\approx ({\sigma }_{3,s^{\prime }},J_3),$$

where

$$\begin{array}{cc}\hfill & {\sigma }_{3,s^{\prime }}\left(g\right)=g+2x,{\sigma }_{3,s^{\prime }}\left(x\right)=s^{\prime }gx(s^{\prime }\ne 0),J_3=1\otimes 1-2e_1\otimes e_1+2e_1\otimes x\hfill \end{array}$$

for any $0\ne s^{\prime }\in \mathbb{k}$.

The proof is completed. □

Using Theorem 2 and ([22], Theorem 2.4), we deduce the following result.

Corollary 1.

Assume that $H={\mathbb{H}}_4[z;\sigma ]$, aBOEAfor ${\mathbb{H}}_4$. Then H is one of the following lists up to isomorphism.

(1)

$H_1^{s,d}$: $H_1^{s,d}$ is generated by $g,x,z$ subjecting to the relations

$$\begin{array}{c}{g}^2=1,x^2=0,xg=-gx,zg=gz,zx=sxz(s\ne 0).\hfill \end{array}$$

The coalgebra is defined by

$$\begin{array}{c}\Delta \left(g\right)=g\otimes g,\Delta \left(x\right)=x\otimes 1+g\otimes x,\hfill \\ \Delta \left(z\right)=z\otimes z+dgxz\otimes xz,\hfill \\ ϵ\left(g\right)=1,ϵ\left(z\right)=1,ϵ\left(x\right)=0\hfill \end{array}$$

for any d.

(2)

$H_2^{s,d}$: $H_2^{s,d}$ is generated by $g,x,z$ subjecting to the relations

$$g^2=1,x^2=0,xg=-gx,zx=sxz,zg=gz+2gxz(s\ne 0).$$

The coalgebra is defined by

$$\begin{array}{c}\Delta \left(g\right)=g\otimes g,\Delta \left(x\right)=x\otimes 1+g\otimes x,\hfill \\ \Delta \left(z\right)=z\otimes z+2e_{1}z\otimes xz+dgxz\otimes xz\hfill \\ ϵ\left(g\right)=1,ϵ\left(z\right)=1,ϵ\left(x\right)=0\hfill \end{array}$$

for any d.

(3)

$H_3^s$: $H_3^s$ is generated by $g,x,z$ subjecting to the relations

$$g^2=1,x^2=0,xg=-gx,zx=sgxz(s\ne 0),zg=gz+2xz.$$

The coalgebra is defined by

$$\begin{array}{c}\Delta \left(g\right)=g\otimes g,\Delta \left(x\right)=x\otimes 1+g\otimes x,\hfill \\ \Delta \left(z\right)=z\otimes z-2e_{1}z\otimes e_{1}z+2e_{1}z\otimes xz,\hfill \\ ϵ\left(g\right)=1,ϵ\left(z\right)=1,ϵ\left(x\right)=0.\hfill \end{array}$$

In the sequel, we always suppose that $H={\mathbb{H}}_4[z;\sigma ]$, a BOEA for ${\mathbb{H}}_4$.

4. The Quotients of the BOEA for ${\mathbb{H}}_4$

Let R be a Hopf algebra, and $R[z;\sigma ]$ a BOEA for R. Suppose that there exists $0\ne t\in R$ such that

$$\Delta \left(t\right)=\prod _{i=0}^{n-1}({\sigma }^i\otimes {\sigma }^i)\left(J\right)(t\otimes t)\mathrm{and}th={\sigma }^n\left(h\right)t(*)$$

for all $h\in R$. we get by ([6], Lemma 2.5) that $R[z;\sigma ]/\langle z^n-t\rangle$ is a bialgebra.

The aim of this section is to investigate all bialgebra structures on the quotients ${\mathbb{H}}_4[z;\sigma ]/〈z^n-t〉,$ where $t\in {\mathbb{H}}_4$ satisfying (*). Firstly, up to equivalence, $(\sigma ,J)$ should be one of the twisted homomorphisms given in Theorem 2.

Let us determine all $t\in {\mathbb{H}}_4$ satisfying (*).

(a) For the twisted homomorphism $({\sigma }_{1,s},J_{1,d})$, where

$$\begin{array}{c}{\sigma }_{1,s}\left(g\right)=g,{\sigma }_{1,s}\left(x\right)=sx,\hfill \\ J_{1,d}=1\otimes 1+d\left(gx\otimes x\right)\hfill \end{array}$$

for any $d\in \mathbb{k}$ and $s\ne 0$.

Now we assume that

$$t=a_1{e}_0+b_1{e}_1+c_1{e}_{0}x+d_1{e}_{1}x.$$

It is easy to see that $a_1=1$ since $ϵ\left(t\right)=1$. On the other hand, since

$$tx={\sigma }_{1,s}^n\left(x\right)t,tg={\sigma }_{1,s}^n\left(g\right)t,$$

It follows that

$$\begin{array}{c}{b}_1=s^n,b_1{s}^n=1,\hfill \\ c_1=d_1=0.\hfill \end{array}$$

Hence $t=e_0+\epsilon e_1,$ where $b_1=s^n=\epsilon =\pm 1.$ For simplicity of discussion, we denote t by $t_{\epsilon }$ and ${\sigma }_{1,s}$ by $\sigma$. It is easy to see that

$$\Delta \left(t_{\epsilon }\right)=t_{\epsilon }\otimes t_{\epsilon }.$$

We have the following lemma for the case (a).

Lemma 1.

The element $t_{\epsilon }$ satisfies the following.

(1)

If $s^2=1$, then $\Delta \left(t_{\epsilon }\right)=\prod _{i=0}^{n-1}\left({\sigma }^i\otimes {\sigma }^i\right)\left(J_{1,d}\right)\left(t_{\epsilon }\otimes t_{\epsilon }\right)$ if and only if $d=0$.

(2)

If $s^2\ne 1$, then $\Delta \left(t_{\epsilon }\right)=\prod _{i=0}^{n-1}\left({\sigma }^i\otimes {\sigma }^i\right)\left(J_{1,d}\right)(t_{\epsilon }\otimes t_{\epsilon })$.

Proof. 

Noting that

$$(\sigma \otimes \sigma )\left(J_{1,d}\right)=(\sigma \otimes \sigma )(1\otimes 1+dgx\otimes x)=1\otimes 1+d{s}^{2}gx\otimes x,$$

we have

$$({\sigma }^i\otimes {\sigma }^i)\left(J_{1,d}\right)=1\otimes 1+d{s}^{2i}gx\otimes x$$

and

$$\prod _{i=0}^{\ell -1}\left({\sigma }^i\otimes {\sigma }^i\right)\left(J_{1,d}\right)=\prod _{i=0}^{\ell -1}\left(1\otimes 1+d{s}^{2i}gx\otimes x)\right)=1\otimes 1+d\left(\sum _{i=0}^{\ell -1}{s}^{2i}\right)gx\otimes x.$$

(i)

If $s^2=1$, then we have

$$\sum _{i=0}^{n-1}{s}^{2i}=\underset{n\mathrm{times}}{\underbrace{1+1+\cdots +1}}=n,$$

$$\prod _{i=0}^{n-1}\left({\sigma }^i\otimes {\sigma }^i\right)\left(J_{1,d}\right)=1\otimes 1+ndgx\otimes x.$$

One sees that

$$\begin{array}{c}\hfill \prod _{i=0}^{n-1}\left({\sigma }^i\otimes {\sigma }^i\right)\left(J_{1,d}\right)\left(t_{\epsilon }\otimes t_{\epsilon }\right)=t_{\epsilon }\otimes t_{\epsilon }+nd{t}_{-\epsilon }x\otimes t_{\epsilon }x.\end{array}$$

Hence if $s^2=1$, then $\Delta \left(t_{\epsilon }\right)=\prod _{i=0}^{n-1}\left({\sigma }^i\otimes {\sigma }^i\right)\left(J_{1,d}\right)\left(t_{\epsilon }\otimes t_{\epsilon }\right)$ if and only if $d=0$.

(ii)

If $s^2\ne 1$, then

$$\prod _{i=0}^{\ell -1}\left({\sigma }^i\otimes {\sigma }^i\right)\left(J_{1,d}\right)=1\otimes 1+d\frac{s^{2\ell }-1}{s^2-1}gx\otimes x.$$

Noting that

$$\sum _{i=0}^{n-1}{s}^{2i}=\frac{s^{2n}-1}{s^2-1}=0,$$

we have

$$\prod _{i=0}^{n-1}\left({\sigma }^i\otimes {\sigma }^i\right)\left(J_{1,d}\right)=1\otimes 1$$

and

$$\Delta \left(t_{\epsilon }\right)=\prod _{i=0}^{n-1}\left({\sigma }^i\otimes {\sigma }^i\right)\left(J_{1,d}\right)(t_{\epsilon }\otimes t_{\epsilon })$$

for any d.

The proof is completed. □

For suitable elements $d\in \mathbb{k}$ and $s=\omega \in \mathbb{k}$, if $t_{\epsilon }$ satisfies the hypothesis of Lemma 1, then

$$\overline{H}={\mathbb{H}}_4[z;\sigma ]/\langle z^n-t_{\epsilon }\rangle$$

is a bialgebra.

In this case, we have ${\omega }^{2n}=1$ and the bialgebra $\overline{H}$ is one of the following lists.

Case 1: If $\omega$ is 2-th primitive root of unity: ${\omega }^2=1$.

(i)

if $n\ge 2$ is even, for example $n=2m(m\ge 1)$, then ${\omega }^{2m}=1$ and $t_{\epsilon }=e_0+{\omega }^{2m}{e}_1=1.$ Thus we get the bialgebra $H_{8m}^1(m\ge 1)$ generated by $g,x,z$ with the relations

$$g^2=1,xg=-gx,x^2=0,zg=gz,zx=-xz,z^{2m}=1.$$

The coalgebra is

$$\begin{array}{c}\Delta \left(g\right)=g\otimes g,\Delta \left(x\right)=x\otimes 1+g\otimes x,\Delta \left(z\right)=z\otimes z;\hfill \\ ϵ\left(g\right)=1,ϵ\left(z\right)=1,ϵ\left(x\right)=0.\hfill \end{array}$$

(ii)

if $n>2$ is odd, for example $n=2m+1(m\ge 1)$, then ${\omega }^{2m+1}=-1$ and $t_{\epsilon }=e_0+{\omega }^{2m+1}{e}_1=g.$ Then we get the bialgebra $H_{4(2m+1)}^2(m\ge 1)$ generated by $g,x,z$ with the relations

$$g^2=1,xg=-gx,x^2=0,zg=gz,zx=-xz,z^{2m+1}=g.$$

The coalgebra is

$$\Delta \left(g\right)=g\otimes g,\Delta \left(x\right)=x\otimes 1+g\otimes x,\Delta \left(z\right)=z\otimes z;$$

$$ϵ\left(g\right)=1,ϵ\left(z\right)=1,ϵ\left(x\right)=0.$$

In fact, up to isomorphism, $H_{4(2m+1)}^2(m\ge 1)$ is generated by $z,x$ with the relations

$$z^{4m+2}=1,x^2=0,zx=-xz.$$

The coalgebra is

$$\Delta \left(x\right)=x\otimes 1+z^{2m+1}\otimes x,\Delta \left(z\right)=z\otimes z;$$

$$ϵ\left(z\right)=1,ϵ\left(x\right)=0.$$

Case 2: Assume that $n\ge 2$ and let $\omega$ be the $2r$-th primitive root of unity with $r>1$. Since ${\omega }^{2n}=1$, we have $r|n$. Let $n=\ell r$. Then ${\omega }^n={(-1)}^{\ell }$.

(i)

If ℓ is even, for example $\ell =2m(m\ge 1)$, then ${\omega }^n=1$ and $t_{\epsilon }=e_0+{\omega }^n{e}_1=1.$ Thus we get the bialgebra $H_{8mr}^1(2m,r,d)(m\ge 1)$ generated by $g,x,z$ with the relations

$$g^2=1,xg=-gx,x^2=0,zg=gz,zx=\omega xz,z^{2mr}=1.$$

The coalgebra is

$$\Delta \left(g\right)=g\otimes g,\Delta \left(x\right)=x\otimes 1+g\otimes x,\Delta \left(z\right)=z\otimes z-dxgz\otimes xz;$$

$$ϵ\left(g\right)=1,ϵ\left(z\right)=1,ϵ\left(x\right)=0.$$

(ii)

If ℓ is odd, for example $\ell =2m+1(m\ge 0)$, then ${\omega }^n=-1$ and $t_{\epsilon }=e_0+{\omega }^n{e}_1=g.$ Thus we get the bialgebra $H_{4(2m+1)r}^2(2m+1,r,d)(m\ge 0)$ generated by $g,x,z$ with the relations

$$g^2=1,xg=-gx,x^2=0,zg=gz,zx=\omega xz,z^{(2m+1)r}=g.$$

The coalgebra is

$$\Delta \left(g\right)=g\otimes g,\Delta \left(x\right)=x\otimes 1+g\otimes x,\Delta \left(z\right)=z\otimes z-dxgz\otimes xz;$$

$$ϵ\left(g\right)=1,ϵ\left(z\right)=1,ϵ\left(x\right)=0.$$

In fact, up to isomorphism, $H_{4(2m+1)}^2(2m+1,r,d)(m\ge 0)$ is generated by $z,x$ with the relations

$$z^{2(2m+1)r}=1,x^2=0,zx=\omega xz.$$

The coalgebra is

$$\Delta \left(x\right)=x\otimes 1+z^{(2m+1)r}\otimes x,\Delta \left(z\right)=z\otimes z-dx{z}^{(2m+1)r+1}\otimes xz;$$

$$ϵ\left(z\right)=1,ϵ\left(x\right)=0.$$

In particular, if $r=n$, then $\ell =1$ and the bialgebra $H_{4n}(1,n,d)$ is generated by $x,z$ with the relations

$$z^{2n}=1,x^2=0,zx=\omega xz.$$

The coalgebra is

$$\Delta \left(z\right)=z\otimes z-dx{z}^{n+1}\otimes xz,\Delta \left(x\right)=x\otimes 1+z^n\otimes x;$$

$$ϵ\left(g\right)=1,ϵ\left(z\right)=1,ϵ\left(x\right)=0.$$

Case 3: Let ℓ be odd and $\omega$ be ℓ-th primitive root of unity with $\ell >1$. Setting ${\omega }^{\prime }=-\omega$, then ${\omega }^{\prime }$ is the $2\ell$-th primitive root of unity with order $2\ell$ and ${\left({\omega }^{\prime }\right)}^{2n}={\omega }^{2n}=1$. This case turns into Case 2.

(b) For the twisted homomorphism $({\sigma }_{2,s},J_{2,d})$, where

$$\begin{array}{cc}\hfill & {\sigma }_{2,s}\left(g\right)=g+2gx,{\sigma }_{2,s}\left(x\right)=sx,\hfill \\ J_{2,d}=1\otimes 1+2e_1\otimes x+d\left(gx\otimes x\right)\hfill \end{array}$$

for any $d\in \mathbb{k}$ and $s\ne 0$.

It is easy to see that if $s=1$, then

$${\sigma }_{2,s}^n\left(x\right)=x,{\sigma }_{2,s}^n\left(g\right)=g+2ngx$$

(25)

and if $s\ne 1$, then

$${\sigma }_{2,s}^n\left(x\right)=s^{n}x,{\sigma }_{2,s}^n\left(g\right)=g+\frac{2(1-s^n)}{1-s}gx.$$

(26)

Now we assume that

$$t=a_2{e}_0+b_2{e}_1+c_2{e}_{0}x+d_2{e}_{1}x.$$

One see that $a_2=1$ since $ϵ\left(t\right)=1$. On the other hand,

$$tx={\sigma }_{2,s}^n\left(x\right)t,tg={\sigma }_{2,s}^n\left(g\right)t.$$

(i)

If $s=1$, then

$$\begin{array}{c}{e}_{0}x+b_2{e}_{1}x=e_{1}x+b_2{e}_{0}x,\hfill \\ g-c_2{e}_{0}x+d_2{e}_{1}x=g+(c_1+2n)e_{0}x-(d_2+2n)e_{1}x.\hfill \end{array}$$

Therefore, we have $b_2=1,c_2=d_2=-n,$ and we get $t_0=1-nx\in {\mathbb{H}}_4$ satisfying (*) and

$$\Delta \left(t_0\right)=1\otimes 1-nx\otimes 1-ng\otimes x.$$

(ii)

If $s\ne 1$, then

$$\begin{array}{c}{e}_{0}x+b_1{e}_{1}x=s^n{e}_{1}x+s^n{b}_2{e}_{0}x,\hfill \\ -c_2{e}_{0}x+d_2{e}_{1}x=\left(c_2+b_2\frac{2(1-s^n)}{1-s}\right)e_{0}x-\left(d_2+\frac{2(1-s^n)}{1-s}\right)e_{1}x.\hfill \end{array}$$

It follows that

$$\begin{array}{c}{b}_2=s^n,b_2{s}^n=1,\hfill \\ c_2=b_2\frac{s^n-1}{1-s},d_2=\frac{s^n-1}{1-s}.\hfill \end{array}$$

Hence

$$t_{\epsilon }=e_0+\epsilon e_1+\frac{1-\epsilon }{1-s}{e}_{0}x+\frac{\epsilon -1}{1-s}{e}_{1}x=e_0+\epsilon e_1+\frac{1-\epsilon }{1-s}gx,$$

where $b_1=s^n=\epsilon =\pm 1.$ We denote t by $t_{\epsilon }$ and ${\sigma }_{2,s}$ by $\sigma$ in discussion.

We have the following in the case (b).

Lemma 2.

The element $t_{\epsilon }$ satisfies the following condition:

(1)

If $s^2=1$, then $\Delta \left(t_{\epsilon }\right)=\prod _{i=0}^{n-1}\left({\sigma }^i\otimes {\sigma }^i\right)\left(J_{2,d}\right)\left(t_{\epsilon }\otimes t_{\epsilon }\right)$ if and only if $d=-1$.

(2)

If $s^2\ne 1$, then $\Delta \left(t_{\epsilon }\right)=\prod _{i=0}^{n-1}\left({\sigma }^i\otimes {\sigma }^i\right)\left(J_{2,d}\right)(t_{\epsilon }\otimes t_{\epsilon })$.

Proof. 

Analogous argument to the proof of Lemma 1.

For suitable elements $d\in \mathbb{k}$ and $s=\omega \in \mathbb{k}$, if $t_{\epsilon }$ satisfies the hypothesis of Lemma 2, then $\overline{B}={\mathbb{H}}_4[z;\sigma ]/\langle z^n-t_{\epsilon }\rangle$ is a bialgebra.

In this case, ${\omega }^{2n}=1$ and $\epsilon ={\omega }^n$. The bialgebra $\overline{B}$ is one of the following lists.

Case 1: If $\omega =1$, of course ${\omega }^2={\omega }^n={\omega }^{2n}=1$. Thus we get the bialgebra $B_{4n}^1$ generated by $g,x,z$ with the relations

$$g^2=1,xg=-gx,x^2=0,zg=gz+2gxz,zx=xz,z^n=1-nx.$$

The coalgebra is

$$\Delta \left(g\right)=g\otimes g,\Delta \left(x\right)=x\otimes 1+g\otimes x,\Delta \left(z\right)=z\otimes z+2e_{1}z\otimes xz-gxz\otimes xz;$$

$$ϵ\left(g\right)=1,ϵ\left(z\right)=1,ϵ\left(x\right)=0.$$

Case 2: If $\omega =-1$, then ${\omega }^2=1$.

(i)

Assume that $n\ge 2$ is even, for example $n=2m(m\ge 1)$, then ${\omega }^{2m}=1$. Thus we get the bialgebra $B_{8m}^2(m\ge 1)$ generated by $g,x,z$ with the relations

$$g^2=1,xg=-gx,x^2=0,zg=gz+2gxz,zx=-xz,z^{2m}=1.$$

The coalgebra is

$$\Delta \left(g\right)=g\otimes g,\Delta \left(x\right)=x\otimes 1+g\otimes x,\Delta \left(z\right)=z\otimes z+2e_{1}z\otimes xz-gxz\otimes xz;$$

$$ϵ\left(g\right)=1,ϵ\left(z\right)=1,ϵ\left(x\right)=0.$$

(ii)

if $n>2$ is odd, for example $n=2m+1(m\ge 1)$, then ${\omega }^n={\omega }^{2m+1}=-1$. Thus we get the bialgebra $B_{4(2m+1)}^3(m\ge 1)$ generated by $g,x,z$ with the relations

$$g^2=1,xg=-gx,x^2=0,zg=gz+2gxz,zx=-xz,z^{2m+1}=g+gx.$$

The coalgebra is

$$\Delta \left(g\right)=g\otimes g,\Delta \left(x\right)=x\otimes 1+g\otimes x,\Delta \left(z\right)=z\otimes z+2e_{1}z\otimes xz-gxz\otimes xz;$$

$$ϵ\left(g\right)=1,ϵ\left(z\right)=1,ϵ\left(x\right)=0.$$

Case 3: Let $\omega$ be $2r$-th primitive root of unity with $r>1$. Since ${\omega }^{2n}=1$, we have $r|n$. Let $n=\ell r$. Therefore, ${\omega }^n={(-1)}^{\ell }$.

(i)

If ℓ is even, for example $\ell =2m$, we get the bialgebra $B_{8mr}^1(2m,r,d)$ generated by $g,x,z$ with the relations

$$g^2=1,xg=-gx,x^2=0,zg=gz+2gxz,zx=\omega xz,z^{2mr}=1.$$

The coalgebra is

$$\Delta \left(g\right)=g\otimes g,\Delta \left(x\right)=x\otimes 1+g\otimes x,\Delta \left(z\right)=z\otimes z+2e_{1}z\otimes xz+dgxz\otimes xz;$$

$$ϵ\left(g\right)=1,ϵ\left(z\right)=1,ϵ\left(x\right)=0.$$

(ii)

If ℓ is odd with $\ell =2m+1$, we get a bialgebra $B_{4(2m+1)r}^2(2m+1,r,d)$ generated by $g,x,z$ with the relations

$$g^2=1,xg=-gx,x^2=0,zg=gz+2gxz,zx=\omega xz,z^{(2m+1)r}=g+\frac{2}{1-\omega }gx.$$

The coalgebra is

$$\Delta \left(g\right)=g\otimes g,\Delta \left(x\right)=x\otimes 1+g\otimes x,\Delta \left(z\right)=z\otimes z+2e_{1}z\otimes xz+dgxz\otimes xz;$$

$$ϵ\left(g\right)=1,ϵ\left(z\right)=1,ϵ\left(x\right)=0.$$

Case 4: Let ℓ be odd and $\omega$ be ℓ-th primitive root of unity with $\ell >1$. Setting ${\omega }^{\prime }=-\omega$, then ${\omega }^{\prime }$ is the $2\ell$-th primitive root of unity with order $2\ell$ and ${\left({\omega }^{\prime }\right)}^{2n}={\omega }^{2n}=1$. This case turns into Case 3.

(c) For the twisted homomorphism $({\sigma }_{3,s},J_3)$, we also denote ${\sigma }_{3,s}$ by $\sigma$ in the case. It is easy to see that

$${\sigma }^i\left(e_0\right)=e_0+f_i\left(s\right)e_{0}x+f_i(-s)e_{1}x,{\sigma }^i\left(e_1\right)=e_1-f_i\left(s\right)e_{0}x-f_i(-s)e_{1}x.$$

$${\sigma }^i\left(e_{0}x\right)=s^i{e}_{0}x,{\sigma }^i\left(e_{1}x\right)={(-s)}^i{e}_{1}x,$$

where $f_0\left(x\right)=0$ and for $i>0$

$$f_i\left(x\right)=1+x+\cdots +x^{i-1}=\left\{\begin{array}{c}i,\mathrm{for}x=1,\\ \frac{x^i-1}{x-1},\mathrm{for}x\ne 1.\end{array}\right.$$

We see that

$$f_i\left(x\right)+x^i=f_{i+1}\left(x\right)$$

and the following equations hold by induction.

$$\begin{array}{c}2\sum _{i=0}^k{(-1)}^{k-i}{f}_i\left(x\right)-f_{k+1}\left(x\right)={(-1)}^{k+1}{f}_{k+1}(-x),\hfill \end{array}$$

(27)

$$\begin{array}{c}2\sum _{i=0}^k{(-1)}^{k-i}{f}_{i+1}\left(x\right)-f_{k+2}\left(x\right)={(-1)}^{k+2}{f}_{k+2}(-x).\hfill \end{array}$$

(28)

We also see that

$$\begin{array}{cc}& \left({\sigma }^i\otimes {\sigma }^i\right)\left(J_3\right)\hfill \\ =& \left({\sigma }^i\otimes {\sigma }^i\right)(1\otimes 1-2e_1\otimes e_1+2e_1\otimes e_{0}x+2e_1\otimes e_{1}x)\hfill \\ =& \mathcal{A}-2f_i\left(s\right)f_{i+1}\left(s\right)e_{0}x\otimes e_{0}x-2f_i(-s)f_{i+1}(-s)e_{1}x\otimes e_{1}x\hfill \\ & +2f_{i+1}\left(s\right)e_1\otimes e_{0}x+2f_{i+1}(-s)e_1\otimes e_{1}x+2f_i\left(s\right)e_{0}x\otimes e_1\hfill \\ & +2f_i(-s)e_{1}x\otimes e_1-2f_i\left(s\right)f_{i+1}(-s)e_{0}x\otimes e_{1}x-2f_i(-s)f_{i+1}\left(s\right)e_{1}x\otimes e_{0}x\hfill \end{array}$$

where $\mathcal{A}=1\otimes 1-2e_1\otimes e_1$. It is noted that

$${\mathcal{A}}^l=\left\{\begin{array}{c}1\otimes 1,\mathrm{if}l\mathrm{is}\mathrm{even};\\ \mathcal{A},\mathrm{if}l\mathrm{is}\mathrm{odd}.\end{array}\right.$$

Lemma 3.

We have

$$\begin{array}{c}\prod _{i=0}^l\left({\sigma }^i\otimes {\sigma }^i\right)\left(J_3\right)={\mathcal{A}}^{l+1}\hfill \\ +2\left(\sum _{i=0}^l{(-1)}^{l-i}{f}_{i+1}\left(s\right)\right)e_1\otimes e_{0}x+2\left(\sum _{i=0}^l{(-1)}^i{f}_{i+1}(-s)\right)e_1\otimes e_{1}x\hfill \\ +2\left(\sum _{i=0}^l{(-1)}^{l-i}{f}_i\left(s\right)\right)e_{0}x\otimes e_1+2\left(\sum _{i=0}^l{(-1)}^i{f}_i(-s)\right)e_{1}x\otimes e_1\hfill \\ +2\left(\sum _{i=0}^l{(-1)}^{l-i+1}{f}_i\left(s\right)f_{i+1}\left(s\right)\right)e_{0}x\otimes e_{0}x+2\left(\sum _{i=0}^l{(-1)}^{i+1}{f}_i(-s)f_{i+1}(-s)\right)e_{1}x\otimes e_{1}x\hfill \\ +2\left(\sum _{i=0}^l{(-1)}^i{f}_i(-s)f_{i+1}(-s)\right)e_{0}x\otimes e_{1}x+2\left(\sum _{i=0}^l{(-1)}^{i+1}{f}_i(-s)f_{i+1}(-s)\right)e_{1}x\otimes e_{0}x.\hfill \end{array}$$

Proof. 

The equation is trivial if $l=0$. Applying Equations (27) and (28), one can get the result by induction.

The proof is finished. □

Assume that

$$tx={\sigma }_{3,s}^n\left(x\right)t,tg={\sigma }_{3,s}^n\left(g\right)t,$$

and

$$t=a_3{e}_0+b_3{e}_1+c_3{e}_{0}x+d_3{e}_{1}x.$$

It is easy to see that $a_3=1$ since $ϵ\left(t\right)=1$ and

$${\sigma }^i\left(e_{0}x\right)=s^i{e}_{0}x,{\sigma }^i\left(e_{1}x\right)={(-s)}^i{e}_{1}x,$$

$$\begin{array}{c}{e}_{0}x+b_3{e}_{1}x=b_3{s}^n{e}_{0}x+{(-s)}^n{e}_{1}x,\hfill \\ e_0-b_1{e}_1-c_1{e}_{0}x+d_1{e}_{1}x=e_0-b_1{e}_1+(c_1+2b_1{f}_n\left(s\right))e_{0}x+(2f_n(-s)-d_1)e_{1}x.\hfill \end{array}$$

It follows that

$$\begin{array}{c}{b}_3=s^{-n}={(-s)}^n,c_3=-b_3{f}_n\left(s\right),d_3=f_n(-s).\hfill \end{array}$$

Therefore, we have

$$t=e_0+b_3{e}_1-b_3{f}_n\left(s\right)e_{0}x+f_n(-s)e_{1}x,$$

where ${(-s^2)}^n=1$ and $b_3={(-1)}^n{s}^n$. We also have $b_3^2=s^{2n}={(-1)}^n$.

We denote t by $t_{\nu }$ where $\nu ={(-1)}^n{s}^n$ with ${(-s^2)}^n=1$. In this case

$$t_{\nu }=e_0+\nu e_1-\nu f_n\left(s\right)e_{0}x+f_n(-s)e_{1}x.$$

It is easy to see that

$$\begin{array}{ccc}\hfill \Delta \left(t_{\nu }\right)& =& e_0\otimes e_0+e_1\otimes e_1+b_3{e}_1\otimes e_0+b_3{e}_0\otimes e_1+c_3{e}_{0}x\otimes e_0+c_3{e}_0\otimes e_{0}x\hfill \\ & & +c_3{e}_{1}x\otimes e_1-c_3{e}_1\otimes e_{1}x+d_3{e}_{1}x\otimes e_0-d_3{e}_1\otimes e_{0}x+d_3{e}_{0}x\otimes e_1+d_3{e}_0\otimes e_{1}x.\hfill \end{array}$$

Lemma 4.

The following condition holds

$$\Delta \left(t_{\nu }\right)=\prod _{i=0}^{n-1}\left({\sigma }^i\otimes {\sigma }^i\right)\left(J_3\right)(t_{\nu }\otimes t_{\nu })$$

if and only if

$$\sum _{i=0}^{n-1}{(-s^2)}^i=0.$$

Proof. 

(Sketch) By Lemma 3, it is straightforward to see

$$\begin{array}{cc}& \prod _{i=0}^{n-1}\left({\sigma }^i\otimes {\sigma }^i\right)\left(J_3\right)(t_{\nu }\otimes t_{\nu })\hfill \\ =& \Delta \left(t_{\nu }\right)+\left(\sum _{i=0}^{n-1}{(-1)}^{i+1}{s}^{2i}\right)e_{0}x\otimes e_{0}x+\nu \left(\sum _{i=0}^{n-1}{(-1)}^{i+1}{s}^{2i}\right)e_{0}x\otimes e_{1}x\hfill \\ & +\nu \left(\sum _{i=0}^{n-1}{(-1)}^i{s}^{2i}\right)e_{1}x\otimes e_{0}x+\left(\sum _{i=0}^{n-1}{(-1)}^{n-i}{s}^{2i}\right)e_{1}x\otimes e_{1}x.\hfill \end{array}$$

Comparing the coefficients of each term of two-hand side of $\Delta \left(t_{\nu }\right)$, we have

$$\Delta \left(t_{\nu }\right)=\prod _{i=0}^{n-1}\left({\sigma }^i\otimes {\sigma }^i\right)\left(J_3\right)(t_{\nu }\otimes t_{\nu })$$

if and only if

$$\sum _{i=0}^{n-1}{(-s^2)}^i=0.$$

The proof is completed.  □

Now we assume that $n\ge 2$, $s=\omega \in \mathbb{k}$, and

$$t_{\nu }=e_0+\nu e_1-\nu f_n\left(s\right)e_{0}x+f_n(-s)e_{1}x,$$

where

$$\nu ={(-1)}^n{\omega }^n,{(-{\omega }^2)}^n=1,\sum _{i=0}^{n-1}{(-{\omega }^2)}^i=0.$$

Thus, we get that

$$\overline{C}={\mathbb{H}}_4[z;\sigma ]/\langle z^n-t_{\nu }\rangle$$

is a bialgebra.

In this case, ${(-{\omega }^2)}^n=1(n\ge 2)$, ${\omega }^2\ne -1$, and $\nu ={(-1)}^n{\omega }^n$. The bialgebra $\overline{C}$ is one of the following lists.

Case 1: If $-{\omega }^2=-1$, then $\omega =\pm 1$. Hence n must be an even and set $n=2m$. Then

$$\sum _{i=0}^{n-1}{(-{\omega }^2)}^i=0.$$

(i)

If $\omega =1$, then $\nu ={(-1)}^n{\omega }^n=1$ and

$$t_1=1-2m{e}_{0}x.$$

We get the bialgebra $C_{8m}^1(m\ge 1$) generated by $g,x,z$ with the relations

$$g^2=1,xg=-gx,x^2=0,zg=gz+2xz,zx=gxz,z^{2m}=1-2m{e}_{0}x.$$

The coalgebra is

$$\Delta \left(g\right)=g\otimes g,\Delta \left(x\right)=x\otimes 1+g\otimes x,\Delta \left(z\right)=z\otimes z-2e_{1}z\otimes e_{1}z+2e_{1}z\otimes xz;$$

$$ϵ\left(g\right)=1,ϵ\left(z\right)=1,ϵ\left(x\right)=0.$$

(ii)

If $\omega =-1$, then $\nu ={(-1)}^n{\omega }^n=1$ and

$$t_1=1+2m{e}_{1}x.$$

We get the bialgebra $C_{8m}^2(m\ge 1)$ generated by $g,x,z$ with the relations

$$g^2=1,xg=-gx,x^2=0,zg=gz+2xz,zx=-gxz,z^{2m}=1+2m{e}_{1}x.$$

The coalgebra is

$$\Delta \left(g\right)=g\otimes g,\Delta \left(x\right)=x\otimes 1+g\otimes x,\Delta \left(z\right)=z\otimes z-2e_{1}z\otimes e_{1}z+2e_{1}z\otimes xz;$$

$$ϵ\left(g\right)=1,ϵ\left(z\right)=1,ϵ\left(x\right)=0.$$

Case 2: Assume that ${\omega }^2\ne \pm 1$ and $n\ge 4$ is even, we have ${\omega }^{2n}=1$ and

$$\sum _{i=0}^{n-1}{(-{\omega }^2)}^i=0$$

always holds.

(a)

Let $\omega$ be a $2r$-th primitive root of unity with $r>2$. Then we also have $r|n$. Let $n=\ell r$. Then $\nu ={(-\omega )}^n={\omega }^n={(-1)}^{\ell }$.

(i)

If ℓ is even with $\ell =2m$, then $\nu =1$ and

$$f_n\left(\omega \right)=\frac{{\omega }^n-1}{\omega -1}=0,f_n(-\omega )=\frac{{(-\omega )}^n-1}{-\omega -1}=0.$$

Hence $t_1=e_0+e_1=1$ and we get a bialgebra $C_{8mr}^1(2m,r)(m\ge 1)$ generated by $g,x,z$ with the relations

$$g^2=1,xg=-gx,x^2=0,zg=gz+2xz,zx=\omega gxz,z^{2mr}=1.$$

The coalgebra is

$$\Delta \left(g\right)=g\otimes g,\Delta \left(x\right)=x\otimes 1+g\otimes x,\Delta \left(z\right)=z\otimes z-2e_{1}z\otimes e_{1}z+2e_{1}z\otimes xz;$$

$$ϵ\left(g\right)=1,ϵ\left(z\right)=1,ϵ\left(x\right)=0.$$

(ii)

If ℓ is odd with $\ell =2m+1(m\ge 1)$, then $\nu =-1$ and

$$f_n\left(\omega \right)=\frac{{\omega }^n-1}{\omega -1}=\frac{2}{1-\omega },f_n(-\omega )=\frac{{(-\omega )}^n-1}{-\omega -1}=\frac{2}{1+\omega }.$$

Hence

$$t_{-1}=g+\frac{2}{1-\omega }{e}_{0}x+\frac{2}{1+\omega }{e}_{1}x.$$

We get a bialgebra $C_{4(2m+1)r}^2(2m+1,r)(m\ge 1,r$ is even) generated by $g,x,z$ with the relations

$$g^2=1,xg=-gx,x^2=0,zg=gz+2xz,zx=\omega gxz,$$

$$z^{(2m+1)r}=g+\frac{2}{1-\omega }{e}_{0}x+\frac{2}{1+\omega }{e}_{1}x.$$

The coalgebra is

$$\Delta \left(g\right)=g\otimes g,\Delta \left(x\right)=x\otimes 1+g\otimes x,\Delta \left(z\right)=z\otimes z-2e_{1}z\otimes e_{1}z+2e_{1}z\otimes xz;$$

$$ϵ\left(g\right)=1,ϵ\left(z\right)=1,ϵ\left(x\right)=0.$$

(b)

Let ℓ be odd and $\omega$ be ℓ-th primitive root of unity with $\ell >1$. We can replace $\omega$ by $-\omega$. This case turns into the case (a) above.

Case 3: Assume that ${\omega }^2\ne \pm 1$ and $n=2m+1(m\ge 1)$, we have ${\omega }^n=\pm \mathbf{i}$ and

$$\sum _{i=0}^{n-1}{(-{\omega }^2)}^i=0.$$

It is easy to see that

$$\nu ={(-\omega )}^n=-{\omega }^n,f_n\left(\omega \right)=\frac{{\omega }^n-1}{\omega -1},f_n(-\omega )=\frac{{\omega }^n+1}{\omega +1}.$$

Hence

$$t_{\nu }=e_0-{\omega }^n{e}_1+\frac{{\omega }^n+1}{1-\omega }{e}_{0}x+\frac{{\omega }^n+1}{1+\omega }{e}_{1}x.$$

We get the bialgebra $C_{4(2m+1)}^{\pm }$ ($m\ge 1$) generated by $g,x,z$ with the relations

$$g^2=1,xg=-gx,x^2=0,zg=gz+2xz,zx=\omega gxz,$$

$$z^{2m+1}=e_0\mp \mathbf{i}{e}_1+\frac{\pm \mathbf{i}+1}{1-\omega }{e}_{0}x+\frac{\pm \mathbf{i}+1}{\omega +1}{e}_{1}x,$$

where ${\omega }^{2m+1}=\pm \mathbf{i}$.

The coalgebra is

$$\Delta \left(g\right)=g\otimes g,\Delta \left(x\right)=x\otimes 1+g\otimes x,\Delta \left(z\right)=z\otimes z-2e_{1}z\otimes e_{1}z+2e_{1}z\otimes xz;$$

$$ϵ\left(g\right)=1,ϵ\left(z\right)=1,ϵ\left(x\right)=0.$$

In summary we get the following result.

Theorem 3.

Let $H={\mathbb{H}}_4[z;\sigma ]$ be theBOEAfor ${\mathbb{H}}_4$ and $\overline{H}:=H/〈z^n-t〉$, where $\sigma \in \mathrm{Aut}\left({\mathbb{H}}_4\right)$ and $t\in {\mathbb{H}}_4$ satisfy

$$\Delta \left(t\right)=\prod _{i=0}^{n-1}({\sigma }^i\otimes {\sigma }^i)\left(J\right)(t\otimes t)$$

and

$$th={\sigma }^n\left(h\right)t,\mathrm{for}\mathrm{all}h\in {\mathbb{H}}_4.$$

Then $\overline{H}$ is a bialgebra and up to isomorphism, it is one of the following lists.

(a)

$H_{8m}^1(m\ge 1),$$H_{4(2m+1)}^2(m\ge 1),$

$H_{8mr}^1(2m,r,d)(m\ge 1),H_{4(2m+1)r}^2(2m+1,r,d)(m\ge 0),$ where ω is $2r$-th root of unity with $r>1$;

(b)

$B_{4m}^1(m\ge 1)$, $B_{8m}^2(m\ge 1)$,

$B_{4(2m+1)}^3(m\ge 1)$, $B_{8mr}^1(2m,r,d)(m\ge 1)$, $B_{4(2m+1)r}^2(2m+1,r,d)(m\ge 0)$, where ω is $2r$-th root of unity with $r>1$;

(c)

$C_{8m}^1(m\ge 1)$, $C_{8m}^2(m\ge 1)$,

$C_{8mr}^3(2m,r)(m\ge 1)$, $C_{4(2m+1)r}^4(2m+1,r)(m\ge 1,r\mathrm{is}\mathrm{even})$, where ω is an $2r$-th root of unity with $r>1$;

(d)

$C_{4(2m+1)}^{\pm }(m\ge 1)$, where ${\omega }^{2m+1}=\pm \mathbf{i}$.

5. Hopf Algebra Structures for ${\mathbb{H}}_4$-Ore Extension of Automorphism Type

Let $H={\mathbb{H}}_4[z;\sigma ]$ be the BOEA for ${\mathbb{H}}_4$, and $H_z$ an algebra obtained from H by adding a new generator $z^{-1}$ such that

$$z{z}^{-1}=z^{-1}z=1.$$

Theorem 4.

Keeping notations as above. Then up to isomorphism, $H_z$ is a Hopf algebra if and only if $H_z$ is generated by $g,x,z$ subjecting to relations

$$\begin{array}{c}{g}^2=1,x^2=0,z{z}^{-1}=z^{-1}z=1,\hfill \\ xg=-gx,zg=gz,zx=sxz(s\ne 0).\hfill \end{array}$$

The coalgebra is defined by

$$\begin{array}{c}\Delta \left(g\right)=g\otimes g,\Delta \left(x\right)=x\otimes 1+g\otimes x,\hfill \\ \Delta \left(z\right)=z\otimes z+dgxz\otimes xz,\hfill \\ ϵ\left(g\right)=1,ϵ\left(z\right)=1,ϵ\left(x\right)=0,\hfill \\ S\left(g\right)=g,S\left(z\right)=z^{-1},S\left(x\right)=-gx\hfill \end{array}$$

for any d.

Proof. 

Up to equivalence we have yielded the twisted homomorphism $(\sigma ,J)$ for ${\mathbb{H}}_4$ listed in Theroem 2.

(a)

For the twisted homomorphism $({\sigma }_{1,s},J_{1,d}),$ where

$$\begin{array}{c}{\sigma }_{1,s}\left(g\right)=g,{\sigma }_{1,s}\left(x\right)=sx,\hfill \\ J_{1,d}=1\otimes 1+d\left(gx\otimes x\right)\hfill \end{array}$$

for any $d\in \mathbb{k}$ and $s\ne 0$. One see that

$$S^{{\sigma }_{1,s}}\left(g\right)=g=S\left(g\right),S^{{\sigma }_{1,s}}\left(x\right)={\sigma }_{1,s}S\left(s^{-1}x\right)=-s^{-1}\sigma \left(gx\right)=-gx=S\left(x\right).$$

It follows that ${\theta }_l={\theta }_r=1$ and

$$S^{{\sigma }_{1,s}}\left(h\right)=S\left(h\right)$$

for all $h\in {\mathbb{H}}_4$. Hence $H_z$ is a Hopf algebra with $S\left(z\right)=z^{-1}$.

(b)

For the twisted homomorphism $({\sigma }_{2,s},J_{2,d})$, where

$$\begin{array}{c}{\sigma }_{2,s}\left(g\right)=g+2gx,{\sigma }_{2,s}\left(x\right)=sx,\hfill \\ J_{2,d}=1\otimes 1+2e_1\otimes x+d(gx\otimes x)\hfill \end{array}$$

for any $d\in k$ and $s\ne 0$. one see that

$${\sigma }_{2,s}^{-1}\left(g\right)=g-2s^{-1}gx,{\sigma }_{2,s}^{-1}\left(x\right)=s^{-1}x.$$

Hence we have

$$S^{{\sigma }_{2,s}}\left(g\right)=g+4e_{0}x,S^{{\sigma }_{2,s}}\left(x\right)=gx.$$

In this case, we have $\theta :={\theta }_l=1+2e_{1}x={\theta }_r.$ But

$$S^{{\sigma }_{2,s}}\left(g\right)=g+4e_{0}x,{\theta }^{-1}S\left(g\right)\theta =(1-2e_{1}x)g(1+2e_{1}x)=g+4e_{1}x.$$

It follows that $S^{{\sigma }_{2,s}}\left(g\right)\ne {\theta }^{-1}S\left(g\right)\theta$ and $H_z$ is not a Hopf algebra by Theorem 1.

(c)

For the twisted homomorphism $({\sigma }_{3,s},J_3)$, where

$$\begin{array}{c}{\sigma }_{3,s}\left(g\right)=g+2x,{\sigma }_{3,s}\left(x\right)=sgx(s\ne 0),\hfill \\ J_3=1\otimes 1-2e_1\otimes e_1+2e_1\otimes x.\hfill \end{array}$$

Similarly, $H_z$ also is not a Hopf algebra.

The proof is completed. □

Now, we consider Hopf algebra structure on the quotient

$${\mathbb{H}}_4[z;\sigma ]/〈z^n-t〉,$$

where ${\mathbb{H}}_4[z;\sigma ]$ is the BOEA for ${\mathbb{H}}_4$ satisfying

$$\begin{array}{c}\hfill \Delta \left(t\right)=\prod _{i=0}^{n-1}({\sigma }^i\otimes {\sigma }^i)\left(J\right)(t\otimes t)\end{array}$$

and

$$\begin{array}{c}\hfill th={\sigma }^n\left(h\right)t,\mathrm{for}\mathrm{all}h\in {\mathbb{H}}_4.\end{array}$$

The following result is one of the main results.

Theorem 5.

Let $H={\mathbb{H}}_4[z;\sigma ]$ be theBOEAfor ${\mathbb{H}}_4$ and $\overline{H}:=H/〈z^n-t〉$, where $\sigma \in \mathrm{Aut}\left({\mathbb{H}}_4\right)$ and $t\in {\mathbb{H}}_4$ satisfy

$$\Delta \left(t\right)=\prod _{i=0}^{n-1}({\sigma }^i\otimes {\sigma }^i)\left(J\right)(t\otimes t)$$

and

$$th={\sigma }^n\left(h\right)t,\mathrm{for}\mathrm{all}h\in {\mathbb{H}}_4.$$

If $\overline{H}$ is a Hopf algebra, then it is one of the following lists up to isomorphism.

(a)

$H_{8m}^1(m\ge 1)$, $H_{4(2m+1)}^2(m\ge 1);$

(b)

$H_{8mr}^1(2m,r,d)(m\ge 1)$, $H_{4(2m+1)r}^2(2m+1,r,d)(m\ge 0)$, where ω is $2r$-th root of unity with $r>1$.

Proof. 

By the proof of Theorem 4, we see that only for the twisted homomorphism $({\sigma }_{1,s},J_{1,d})$, ${\theta }_l$ and ${\theta }_r$ enjoy the conditions (1) and (2) in Theorem 1.

Now, we assume that $d=0$ if ${\omega }^2=1$, and d is arbitrary if ${\omega }^2\ne 1$ and ${\omega }^{2n}=1$. Note that

$$\prod _{i=0}^{n-1}\left({\sigma }^i\otimes {\sigma }^i\right)\left(J_{1,d}\right)=\prod _{i=0}^{n-1}\left(1\otimes 1+d{\omega }^{2i}gx\otimes x)\right)=1\otimes 1+d\left(\sum _{i=0}^{n-1}{\omega }^{2i}\right)gx\otimes x.$$

Hence

$$\prod _{i=0}^{n-1}\left({\sigma }^i\otimes {\sigma }^i\right)\left(J_{1,d}\right)=1\otimes 1.$$

Therefore, the remaining conditions in Theorem 1 also hold. By Theorem 1, we get that $H_{8m}^1(m\ge 1)$, $H_{4(2m+1)}^2(m\ge 1);$ $H_{8mr}^1(2m,r,d)(m\ge 1)$, $H_{4(2m+1)r}^2(2m+1,r,d)(m\ge 0)$, where $\omega$ is $2r$-th root of unity with $r>1$, are all Hopf algebras. The antipodes S can be easily given by Theorem 1.

The remaining two cases are referred to Theorem 3 and the proof of Theorem 4.

This completes the proof. □