Representation rings of extensions of Hopf algebra of Kac-Paljutkin type

1.   Introduction

As an invariant of a finite tensor category, the representation ring plays a very important role, having gained great attention in recent years. For example, Chen et al. ^[1] constructed the representation rings of Taft algebras $H_{n}(q)$. Wang et al. studied the Green rings of finite-dimensional pointed rank-one Hopf algebras of nilpotent and non-nilpotent types in ^[2,3]. It should be noted that the small quantum group ^[4,5] and the Drinfeld double of a Taft algebra ^[6,7] are not of finite representation type. Their representation rings are not finitely generated, as described by ^[8] and ^[9,10], respectively. In ^[11,12], the authors considered the Grothendieck rings of the quotient algebras of Wu-Liu-Ding algebras (see ^[13,14] for definitions), and provided their Casimir numbers, which are a class of non-pointed semi-simple Hopf algebras. The representations and Grothendieck rings of the Hopf algebra $H_{2n^{2}}$ are described in ^[15]. Furthermore, Guo and Yang ^[16] explicitly described the Grothendieck rings of the category of Yetter-Drinfeld modules over $H_{2n^{2}}$ by generators and relations. Sun et al. ^[17] described the structure of the representation ring of the small quasi-quantum group. For more results on representations and representation rings, one can refer to ^{[18,19,20,21,22,23,24,25]}.

In order to understand and extend the concept of representation rings, we can weaken the definition of weak Hopf algebras to more general cases. The most interesting one is the so-called $\Delta$-associative algebra, which was introduced in ^[26], where the representation ring of $\Delta$-associative algebra was defined and described for Kac-Paljutkin Hopf algebra $K_8 ($i.e., $H_8)$. One consideration is that the representation rings of $\Delta$-associative algebras may not be commutative and may not even contain the identity (see ^[26]). In this paper, we provide two classes of $\Delta$-associative algebras to understand them. Roughly speaking, based on Kac-Paljutkin-type Hopf algebras $H_{2n^2}$ (see ^[27] or ^[15]), which are generalizations of the $8$-dimensional Kac-Paljutkin Hopf algebra, we weaken the definition of Hopf algebra $H_{2n^2}$ to obtain two classes of $\Delta$-associative algebras, $\overline{H}_{2n^2}$ and $\widehat{H}_{2n^2}$. One can see that $\overline{H}_{2n^2}$ is still a weak Hopf algebra but $\widehat{H}_{2n^2}$ is a non-trivial $\Delta$-associative algebra. Following that, we describe their representations as well as their representation rings. We show that the representation ring $r(\overline{H}_{2n^2})$ is a commutative ring, but the representation ring $r(\widehat{H}_{2n^2})$ is noncommutative with no identity. However, we can embed the ring $r(\widehat{H}_{2n^2})$ into a ring $r^{*}(\widehat{H}_{2n^2})$ with an identity in the natural way. The rings $r(\overline{H}_{2n^2})$ and $r^{*}(\widehat{H}_{2n^2})$ are described by generators with suitable relations.

The paper is organized as follows. In Section 2, we recall the definition of $\Delta$-associative algebra, and review the definition and representations of the semi-simple Hopf algebra $H_{2n^{2}}$. In Section 3, by weakening the definition of Hopf algebra $H_{2n^2}$, we obtain two classes of $\Delta$-associative algebras $\overline{H}_{2n^2}$ and $\widehat{H}_{2n^2}$, where $\overline{H}_{2n^2}$ is a weak Hopf algebra, and $\widehat{H}_{2n^2}$ is just a non-trivial $\Delta$-associative algebra. Some properties of $\overline{H}_{2n^2}$ and $\widehat{H}_{2n^2}$ are discussed. In Section 4, all irreducible modules of $\overline{H}_{2n^2}$ are listed, and there are $(n^{2}+7n+2)/2$ non-isomorphic finite dimensional irreducible modules for $\overline{H}_{2n^2}$. The decomposition formulas of the tensor product of two arbitrary irreducible $\overline{H}_{2n^2}$-modules are also established. The representation ring $r(\overline{H}_{2n^2})$, described by generators and relations, shows that $r(\overline{H}_{2n^2})$ is a commutative ring. In Section 5, the representations and representation ring of $\widehat{H}_{2n^2}$ are studied. We found that the representation ring $r(\widehat{H}_{2n^2})$ is a noncommutative ring with no identity, which can be embedded into a ring $r^{*}(\widehat{H}_{2n^2})$ with an identity.

2.   Preliminaries

Throughout this research, we work over a fixed algebraically closed field ${\mathbb{k}}$ of characteristic 0, unless otherwise stated. All algebras, Hopf algebras, and modules are defined over ${\mathbb{k}}$; all modules are left modules and finite dimensional; all maps are ${\mathbb{k}}$-linearity; $\dim, \otimes$, and $\hom$ stand for $\dim_{\mathbb{k}}$, $\otimes_{\mathbb{k}},$ and $\hom_{\mathbb{k}}$, respectively.

The definition of weak Hopf algebra was introduced by Li (see ^[28]). We recall that a ${\mathbb{k}}$-bialgebra $(H, \mu, \eta, \Delta, \epsilon)$ is called a weak Hopf algebra if there exists a map $T\in \mathrm{hom}(H, H)$ such that $T\ast \mathrm{id} \ast T = T$ and $\mathrm{id}\ast T\ast \mathrm{id} = \mathrm{id}$, where $\ast$ is the convolution map in $\mathrm{hom}(H, H)$. By weakening the definition of weak Hopf algebras, the following definition is established.

Definition 2.1. An associative ${\mathbb{k}}$-algebra $A$ with an identity is called a $\Delta$-associative algebra if there exists an algebra homomorphism ${\Delta}: A\to A{\otimes} A$ such that $\left({{\Delta}{\otimes} {{\rm{id}}}}\right){\Delta} = \left({{{\rm{id}}}{\otimes}{\Delta}}\right){\Delta}.$

All Hopf algebras, bialgebras, and weak Hopf algebras are $\Delta$-associative algebras. If the $\Delta$-associative algebra $A$ is not a coalgebra, $A$ is said to be non-trivial.

In the sequel, we always assume that $A$ is a $\Delta$-associative algebra, and the Sweedler's notations ^[29] are used. For example, for $a\in A$, we denote

Now, suppose that $M$ and $N$ are two $A$-modules, then $M{\otimes} N$ is also an $A$-module defined as follows:

for all $m\in M, n\in N$.

We denote by $[M]$ the isomorphism class of an $A$-module $M$ and

Let $R(A)$ be a free abelian group spanned by ${\mathscr{P}}$. For all $[M], [N]\in {\mathscr{P}}$, we define $[M][N] = [M{\otimes} N]$, and it is easy to see that $R(A)$ is a ring. Let

where $\langle\, {[M\oplus N]-[M]-[N]}\rangle$ is the ideal of $R(A)$ generated by $[M\oplus N]-[M]-[N]$ for all $[M], [N]\in{\mathscr{P}}$, then $r(A)$ is also a ring and is called the representation ring or the Green ring of $A$. If $A$ is a non-trivial $\Delta$-associative algebra, then $r(A)$ may have no identity $[{\mathbb{k}}_{\varepsilon}]$, the trivial $1$-dimensional $A$-module. In this case, let

For $(k, \alpha)$, $(k', \alpha')\in r^{*}(A)$, we define the addition and multiplication in $r^{*}(A)$ as the following

Then, $r^{*}(A)$ is a ring with the identity $(1, 0)$, and $r(A)$ is embedded into $r^*(A)$ naturally.

In the sequel, we always assume that $n > 1$ and $q$ is a primitive $n$-th root of unity.

The Hopf algebra $H_{2n^2}$ can be found in ^[15,27], which is a generalization of Kac-Paljutkin Hopf algebra.

Definition 2.2. The Hopf algebra $H_{2n^2}$ is an associative algebra generated by $x, y$ and $z$, with the following relations

The comultiplication, counit, and antipode are as follows

The Hopf algebra $H_{2n^2}$ is a $2n^2$-dimensional semi-simple Hopf algebra. It is of a basis $\{x^iy^j, \ x^iy^j z | 0\leq i, j\leq n-1\}$. Indeed,

is the left and right integral of $H_{2n^2}$ and $\epsilon(\ell) = 2n^2\ne 0.$ It is also a quasi-triangular Hopf algebra with a universal $R$-matrix

Therefore, the representation ring of $H_{2n^2}$ is semi-simple and commutative. The representations and representation ring of the Hopf algebra $H_{2n^{2}}$ are described in ^[15]. We list them as follows, where $q$ is any primitive $n$-th root of unity, but $q = e^{2p\pi i/n}$ with an even number $p\in \mathbb{Z}$ if $n$ is odd for convenience.

Set

(a) 1-dimensional irreducible $H_{2n^2}$-module $S_{m}, m\in \mathbb{Z}_{2n}$: it is of basis $v^{m}$, and the actions of $H_{2n^2}$ on $S_{m}$ are

(b) 2-dimensional irreducible $H_{2n^2}$-module $S_{i, j}, 0\leq i < j\leq n-1$: it is of basis $v_{1}^{ij}$ and $v_{2}^{ij}$, and the actions of $H_{2n^2}$ on $S_{i, j}$ are

The set

forms a complete list of non-isomorphic irreducible $H_{2n^2}$-modules.

Lemma 2.3. [15,Theorem 1] The decomposition formulas of tensor product of two $H_{2n^2}$-modules in $\mathscr{S}$ are as follows.

(1) (a) For all $0\leq m, m'\leq n-1$, we have

(b) for all $0\leq m\leq n-1, n\leq m'\leq 2n-1$ and $n\leq m+m'\leq 2n-1$, or $0\leq m'\leq n-1, n\leq m\leq 2n-1$ and $n\leq m'+m\leq 2n-1$, we have

(c) for all $0\leq m\leq n-1, n\leq m'\leq 2n-1$ and $2n\leq m+m'\leq 3n-1$, or $0\leq m'\leq n-1, n\leq m\leq 2n-1$ and $2n\leq m'+m\leq 3n-1$, we have

(d) for all $n\leq m, m'\leq 2n-1$ and $2n\leq m+m'\leq 3n-1$, we have

(e) for all $n\leq m, m'\leq 2n-1$ and $3n\leq m+m'\leq 4n-1$, we have

(2) For all $m\in \mathbb{Z}_{2n}, 0\leq i < j\leq n-1$, we have

(3) Set $I_{1} = \{0\leq i < j, k < l < n \mid i+k\equiv j+l(\mathrm{mod}\; n)\}$ and $I_{2} = \{0\leq i < j, k < l < n \mid i+l\equiv j+k(\mathrm{mod}\; n)\},$ then

(a) if $i, j, k, l\notin I_{1}\cup I_{2}$, we have

(b) if $i, j, k, l\in I_{1} \backslash I_{2}$, we have

(c) if $i, j, k, l\in I_{2} \backslash I_{1}$, we have

(d) if $n$ is even, and $i, j, k, l\in I_{1}\cap I_{2}$, we have

Let

and

Then, we have

[15,Theorem 2,Theorem 3] can be stated as follows.

Lemma 2.4. Assume that $n\geq 2$ is an integer, we have

(1) if $n\geq 3$ is odd and $m = \frac{n-1}{2}$, then

(2) if $n = 2$, then

(3) if $n > 2$ is even with $m = \frac{n}{2}$, then

3.   Extensions of Hopf algebra of Kac-Paljutkin type

In this section, we establish two classes of extensions of Hopf algebra $H_{2n^2}$ of Kac-Paljutkin type, which are denoted by $\overline{H}_{2n^2}$ and $\widehat{H}_{2n^2}$, where $\overline{H}_{2n^2}$ is a weak Hopf algebra in the sense of Li in ^[28], but $\widehat{H}_{2n^2}$ is a non-trivial $\Delta$-associative algebra.

Definition 3.1. Let $\overline{H}_{2n^{2}}$ be an associative algebra generated by $x, y,$ and $z$, with the following relations

For convenience, we set

for $i = 1, 2, \cdots, n,$ then

for $s, t = 1, 2, \cdots, n,$ and

We also have

Let $J_1 = x^n, J_2 = y^n$ and $J = J_1J_2$.

The following lemma helps us to understand the deep relations between the algebra ${\overline{H}_{2n^{2}}}$ and the algebra $H_{2n^2}$.

Lemma 3.2. $z = yzx^{n-1}, \ z = xzy^{n-1}$ if and only if $zJ = Jz = z, \ zx = yz, \ zy = xz.$

Proof. Indeed, assume that $z = yzx^{n-1}, \ z = xzy^{n-1},$ then

It follows that

and $zJ = Jz = z.$ Also,

Conversely, if $zJ = Jz = z$ and $xz = zy, \ yz = zx$, then

for $i = 1, 2$. So we have

The result follows. □

It is easy to see that

is a set of orthogonal central idempotents of $\overline{H}_{2n^{2}}$.

Now, we define three maps ${\Delta}: {\overline{H}_{2n^{2}}}\to {\overline{H}_{2n^{2}}}{\otimes} {\overline{H}_{2n^{2}}}$, ${\varepsilon}: {\overline{H}_{2n^{2}}}\to {\mathbb{k}}$, and $T: {\overline{H}_{2n^{2}}} \to {\overline{H}_{2n^{2}}}$ as follows

We get the first main result as follows.

Theorem 3.3. ${\overline{H}_{2n^{2}}}$ is a weak Hopf algebra with the weak antipode $T$.

Proof. We will prove this theorem in three steps.

a) The ${\mathbb{k}}$-map $\Delta$ keeps the defining relations, hence it can be extended into the whole algebra ${\overline{H}_{2n^{2}}}$ such that ${\Delta}$ is an algebra homomorphism. Indeed, we note that

So it is easy to see that

Hence, ${\Delta}$ can define a homomorphism of algebra. Similarly, $\varepsilon$ can also define a homomorphism of algebra.

b) $({\overline{H}_{2n^{2}}}, \Delta, \varepsilon)$ is a coalgebra. Indeed, the ${\mathbb{k}}$-map $\Delta$ is coassociative. To see this fact, we have

and

Hence,

for $a\in \{x, y, z\}$. Then, the map $\Delta$ is coassociative in $\overline{H}_{2n^{2}}$ by the statement a). Furthermore, ${\varepsilon}$ also satisfies the counit axiom.

By the statements a) and b), we see that ${\overline{H}_{2n^{2}}}$ is a bialgebra.

c) The map $T$ can define a weak antipode of $\overline{H}_{2n^2}$ in a natural way. To see this fact, we note that

Thus, the map $T$ can be extended into an anti-algebra homomorphism $T:\overline{H}_{2n^2}\to \overline{H}_{2n^2}$.

Furthermore, it is easy to see that

In addition, we have

The elements $J_1, J_2$ and $J$ are all the center of ${\overline{H}_{2n^{2}}}$ by Lemma 3.2. Hence,

the center of ${\overline{H}_{2n^{2}}}$, for all $a\in{\overline{H}_{2n^{2}}}$.

The above facts show that

for all $a\in \overline{H}_{2n^2}$ by induction.

This means that $T$ can indeed define a weak antipode of $\overline{H}_{2n^2}$, and hence $\overline{H}_{2n^2}$ is a weak Hopf algebra. Here, we point that ${\overline{H}_{2n^{2}}}$ is a noncommutative and noncocommutative weak Hopf algebra. □

Now, we assume that

One can get the following result.

Proposition 3.4. As algebras, we have

(1) $\overline{H}_{2n^{2}} = \overline{A}_0 \oplus\overline{A}_1\oplus\overline{A}_2\oplus\overline{A}_3;$

(2) $\overline{A}_0\cong H_{2n^{2}};$

(3) $\overline{A}_1\cong \overline{A}_2\cong{\mathbb{k}}[g]/(g^{n}-1);$

(4) $\overline{A}_3\cong{\mathbb{k}}.$

Proof. (1) The first statement follows from the fact that $\{J, J_1-J, J_2-J, 1+J-J_1-J_2\}$ is a set of orthogonal central idempotents.

(2) For the statement, we note that $\overline{A}_0$ can be viewed as a subalgebra of $\overline{H}_{2n^{2}}$ with the identity element $J$, generated by $Jx$, $Jy,$ and $Jz$ with the relations

Let $\rho_0: H_{2n^{2}}\rightarrow \overline{A}_{0}$ be the map defined by

it is straightforward to see that $\rho_{0}$ is a well-defined surjective algebra homomorphism. Let $\phi: \overline{H}_{2n^{2}}\rightarrow H_{2n^{2}}$ be the map given by

The map $\phi$ is also a well-defined algebra epimorphism. Considering the restriction ${\phi}|_{\overline{A}_{0}}$ of $\phi$ on $\overline{A}_{0}$, we have ${\phi}|_{\overline{A}_{0}}\circ \rho_0 = {{\rm{id}}}_{H_{2n^{2}}}$. Hence, $\rho_0$ is injective and $\overline{A}_0\cong H_{2n^{2}}$ as algebras.

(3) We note that $\overline{A}_{1}$ and $\overline{A}_{2}$ are the subalgebras of $\overline{H}_{2n^{2}}$ with the unit $J_1-J$ and $J_2-J$, respectively. In $\overline{A}_{1}$, we have

It follows that $\overline{A}_{1}$ is generated by $(J_1-J)x$, with the relation

Similarly, the subalgebra $\overline{A}_{2}$ is generated by $(J_2-J)y$, with the relations

and $(J_2-J)x = J_2x-J_2x = 0, \; (J_2-J)z = z-z = 0$. Finally, it is easy to see that

(4) It is obvious that $\overline{A}_{3}$ is a one-dimensional subalgebra of $\overline{H}_{2n^{2}}$ with the unit $1+J-J_1-J_2$. Indeed,

So, $\overline{A}_{3} = {\mathbb{k}}(1+J-J_1-J_2)$.

The proof is completed. □

By Proposition 3.4, ${\overline{H}_{2n^{2}}}$ is a semi-simple algebra of dimension $2n^2+2n+1$ with a ${\mathbb{k}}$-basis:

Now, we consider the second class of extensions of $H_{2n^2}$. The definition is given as follows.

Definition 3.5. The associative algebra ${\widehat{H}_{2n^{2}}}$ is generated by $x, y$, and $z$, with the relations

We define ${\Delta}: {\widehat{H}_{2n^{2}}}\to {\widehat{H}_{2n^{2}}}{\otimes}{\widehat{H}_{2n^{2}}},$ as follows

and extend it to ${\widehat{H}_{2n^{2}}}$ in the natural way. As previously, let $J_1 = x^n, J_2 = y^n$ and $J = J_1J_2$.

Theorem 3.6. ${\widehat{H}_{2n^{2}}}$ is a $\Delta$-associative algebra.

Proof. We should show that the $\Delta$ is an algebra homomorphism of ${\widehat{H}_{2n^{2}}}$. In fact,

Furthermore, we have that

by the proof of Theorem 3.3 a). Hence, the map ${\Delta}$ can be extended to a homomorphism of algebra in $\widehat{H}_{2n^2}$. Also, $\Delta$ satisfies the coassociative axiom by the proof of Theorem 3.3 b). Hence, ${\widehat{H}_{2n^{2}}}$ is a $\Delta$-associative algebra. □

Remark 3.7. The ${\Delta}$-associative algebra ${\widehat{H}_{2n^{2}}}$ may be non-trivial. Indeed, if there exists a ${\mathbb{k}}$-map ${\varepsilon}: {\widehat{H}_{2n^{2}}}\to {\mathbb{k}}:$

such that ${\varepsilon}$ is a counit of ${\widehat{H}_{2n^{2}}}$. Then, we have

we get that $a = b = 1$ and

where $m$ is the multiplication of ${\widehat{H}_{2n^{2}}}$.

If there exists a $c\in{\mathbb{k}}$ such that $cJ_1z = cJ_2z = z$, then ${\widehat{H}_{2n^{2}}}$ is trivial. Otherwise, ${\widehat{H}_{2n^{2}}}$ is non-trivial.

Here, we always assume that $x, \ y,$ and $z$ are freely without any additional conditions in the definition of ${\widehat{H}_{2n^{2}}}.$ Consequently, in this case, ${\widehat{H}_{2n^{2}}}$ is a non-trivial ${\Delta}$-associative algebra.

Recall that in $\widehat{H}_{2n^{2}}$, the set

is a set of orthogonal central idempotents. Set

Proposition 3.8. As an algebra, we have

(1) $\widehat{H}_{2n^{2}} = \widehat{A}_0 \oplus\widehat{A}_1\oplus\widehat{A}_2.$

(2) $\widehat{A}_0\cong H_{2n^{2}}.$

(3) $\widehat{A}_1\cong\underbrace{H_2 \oplus H_2 \oplus\cdots \oplus H_2}_{n\; \mathrm{copies}}$, where $H_2$ is the Sweedler's algebra.

(4) $\widehat{A}_2 \cong{\mathbb{k}}[h]/(h^{2}).$

Proof. (1) It is obvious.

(2) The proof is similar to Proposition 3.4 (2).

(3) Let

Note that $\widehat{A}_{1}$ is a subalgebra of $\widehat{H}_{2n^{2}}$ with the identity $J_1+J_2-2J$, which is generated by $x_{1}$, $y_{1}$, and $z_{1}$ with the relations

Let

for $s = 1, 2, \cdots, n$. One can see that

is a complete set of primitive orthogonal idempotents of $\widehat{A}_1$, and

Thus, the bounded quiver of $\widehat{A}_{1}$ is as follows

with the admissible ideal

Therefore, the set

forms a complete set of orthogonal central idempotents of $\widehat{A}_{1}$ and

In addition, each ideal $(\lambda_{l}+\mu_{l})\widehat{A}_1$ of $\widehat{A}_1$ is a 4-dimensional subalgebra of $\widehat{A}_{1}$ with the identity $\lambda_{l}+\mu_{l}$. Also, it has the ${\mathbb{k}}$-basis $\{\lambda_{l}, \mu_{l}, \lambda_{l}z, \mu_{l}z\}$ for $l\in\{1, 2, \cdots, n\}$.

Let $H_2$ be the 4-dimensional Sweedler's algebra generated by $\eta, \chi$ subjecting to the relations

For $l = 1, 2, \cdots, n$, let $\varrho_{l}: (\lambda_{l}+\mu_{l})\widehat{A}_1 \rightarrow H_2$ be the map given by

Obviously, $\varrho_{l}$ is an algebra isomorphism.

(4) The ideal $\widehat{A}_{2}$ is a subalgebra of $\widehat{H}_{2n^{2}}$ with the identity $1+J-J_1-J_2$. Since

it can be viewed as an algebra generated by $(1+J-J_1-J_2)z$ with the relation

This shows that $\widehat{A}_{2}$ is isomorphic to ${\mathbb{k}}[h]/(h^{2})$.

The proof is completed. □

By Proposition 3.8, we see that $\widehat{H}_{2n^{2}}$ is not a semi-simple algebra with the dimension $2n^2+4n+2$. Its radical has a ${\mathbb{k}}$-basis as follows

4.   The representations of $\overline{H}_{2n^2}$

In this section, the aim is to classify all indecomposable representations of the weak Hopf algebra $\overline{H}_{2n^{2}}$, and then to characterize its representation ring using generators with generating relations.

The representation theory of semi-simple algebra $H_{2n^2}$ and ${\mathbb{k}}[g]/(g^{n}-1)$ was studied by many authors; see, for example, ^[15,30,31]. Let ind-$R$ denote the set of all indecomposable representations of some ring $R$. By Proposition 3.4, we have

as algebras, therefore,

where ind-$\overline{A}_i$ can be viewed as the subset of ind-${\overline{H}_{2n^{2}}}$ in the natural way. We note that $\overline{A}_0\cong H_{2n^2}$, $\overline{A}_1\cong \overline{A}_2\cong {\mathbb{k}}[g]/(g^{n}-1)$ and $\overline{A}_3\cong {\mathbb{k}}$,

Now, we list all irreducible representations of $\overline{A}_i$, $i = 1, 2, 3.$

(1) Let $M_{i}, i\in \mathbb{Z}_{n}$. The 1-dimensional irreducible $\overline{H}_{2n^2}$-module with basis $v^{(i)}$, the actions of $\overline{H}_{2n^2}$ on $M_{i}$ are

In fact,

(2) Let $N_{i}, i\in \mathbb{Z}_{n}$. The 1-dimensional irreducible $\overline{H}_{2n^2}$-module with basis $w^{(i)}$, the actions of $\overline{H}_{2n^2}$ on $N_{i}$ are

In fact,

(3) Let $L$ be a 1-dimensional irreducible $\overline{H}_{2n^2}$-module with basis $u$, the actions of $\overline{H}_{2n^2}$ on $L$ are

In fact,

Therefore, we have the following results.

Proposition 4.1. The set

forms a complete list of non-isomorphic irreducible representations of $\overline{H}_{2n^2}$.

For the decomposition formulas of the tensor product of two irreducible representations of $\overline{H}_{2n^2}$, we have the following several lemmas.

Lemma 4.2. For all $i\in \mathbb{Z}_{n}$, $m\in \mathbb{Z}_{2n}$ and $0\leq s < t\leq n-1$, we have

(1) $M_{i}\otimes S_{m}\cong S_{m}\otimes M_{i}\cong M_{m+i(\mathrm{mod}\; n)};$

(2) $M_{i}\otimes S_{s, t}\cong S_{s, t}\otimes M_{i}\cong M_{i+s(\mathrm{mod}\; n)}\oplus M_{i+t(\mathrm{mod}\; n)};$

(3) $N_{i}\otimes S_{m}\cong S_{m}\otimes N_{i}\cong N_{m+i(\mathrm{mod}\; n)};$

(4) $N_{i}\otimes S_{s, t}\cong S_{s, t}\otimes N_{i}\cong N_{i+s(\mathrm{mod}\; n)}\oplus N_{i+t(\mathrm{mod}\; n)};$

(5) $L\otimes S_{m}\cong S_{m}\otimes L\cong L;$

(6) $L\otimes S_{s, t}\cong S_{s, t}\otimes L\cong L\oplus L.$

Proof. Suppose that $S_{m}$ and $S_{s, t}$ are irreducible $\overline{H}_{2n^2}$-modules with the basis $\{ v^{m} \}$ and $\{ v_{1}^{st}$, $v_{2}^{st} \}$, respectively, for all $m\in \mathbb{Z}_{2n}$ and $0\leq s < t\leq n-1$.

(1) Considering the decomposition formulas of tensor product $M_{i}\otimes S_{m}$ for $i\in \mathbb{Z}_{n}$. We have

Hence, $M_{i}\otimes S_{m}\cong M_{i+m(\mathrm{mod}\; n)}$. Similarly, one can show that $S_{m}\otimes M_{i}\cong M_{i+m(\mathrm{mod}\; n)}$.

(2) Considering the decomposition formulas of tensor product $M_{i}\otimes S_{s, t}$ for all $i\in \mathbb{Z}_{n}$. We have

So, $M_{i}\otimes S_{s, t}\cong M_{i+s(\mathrm{mod}\; n)}\oplus M_{i+t(\mathrm{mod}\; n)}$. Similarly, $S_{s, t}\otimes M_{i}\cong M_{i+s(\mathrm{mod}\; n)}\oplus M_{i+t(\mathrm{mod}\; n)}$ can be proven.

The remaining cases can be proven in a similar way. □

Lemma 4.3. For all $i, i'\in \mathbb{Z}_{n}$, we have

(1) $M_{i}\otimes M_{i'}\cong M_{i+i'(\mathrm{mod}\; n)};$

(2) $M_{i}\otimes N_{i'}\cong N_{i'}\otimes M_{i}\cong L;$

(3) $M_{i}\otimes L\cong L\otimes M_{i}\cong L;$

(4) $N_{i}\otimes N_{i'}\cong N_{i+i'(\mathrm{mod}\; n)};$

(5) $N_{i}\otimes L\cong L\otimes N_{i}\cong L;$

(6) $L\otimes L\cong L.$

Proof. It is similar to the proof of Lemma 4.2. □

Let $a = [S_1], \ b = [S_{n+1}], \ c = [S_{0, 1}], \ d = [M_0], \; e = [N_0]$ in $r(\overline{H}_{2n^{2}})$, we have the following lemma.

Lemma 4.4. For all $i, i'\in \mathbb{Z}_{n}$, the following statements hold in $r(\overline{H}_{2n^{2}})$.

(1) If $n$ is odd, then

(a) $[M_{i}] = b^{i}d,$ and $b^{n}d = d$;

(b) $[N_{i}] = b^{i}e,$ and $b^{n}e = e$;

(c) $[L] = ed$;

(d) $cd = d+bd,$ $ce = e+be,$ $d^{2} = d$ and $e^{2} = e$.

(2) If $n$ is even, then

(a) $[M_{i}] = a^{i}d,$ and $a^{n}d = d$;

(b) $[N_{i}] = a^{i}e,$ and $a^{n}e = e$;

(c) $[L] = ed$;

(d) $bd = ad,$ $be = ae,$ $cd = d+ad,$ $ce = e+ae,$ $d^{2} = d$ and $e^{2} = e$.

Proof. (1) If $n$ is odd, it can be concluded that $b^{i} = S_{i+\frac{(-1)^{i+1}+1}{2}n}$, for $i\in \mathbb{Z}_{n}$ and $b^{n} = S_{n}$ by [15, Lemma 3]. Hence, we get that

by Lemma 4.2. One sees that $[M_{i}] = b^{i}d,$ $b^{n}d = d$, $[N_{i}] = b^{i}e$ and $b^{n}e = e$. The statements (a) and (b) of (1) are followed.

Moreover, based on statements (a) and (b), one can get that $cd = d+bd,$ $ce = e+be.$ Also,

by Lemma 4.3, then $[L] = ed$, $d^{2} = d$ and $e^{2} = e$.

The first claim follows.

(2) If $n$ is even, by [15, Lemma 4], it can be concluded that

The remaining statements can be proven in a similar way of claim (1). □

Corollary 4.5. For the representation ring $r(\overline{H}_{2n^{2}})$, we have

(1) if $n$ is odd, the set

forms a $\mathbb{Z}$-basis of $r(\overline{H}_{2n^{2}})$;

(2) if $n$ is even, the set

forms a $\mathbb{Z}$-basis of $r(\overline{H}_{2n^{2}})$.

Proof. (1) If $n$ is odd, the set

corresponds one-to-one to the set of irreducible $\overline{H}_{2n^{2}}$-modules

by [15, Corollary 2 (1)]. We have

by Lemma 4.4 (1).

Similarly, the set $\left\{b^{l}d, b^{l}e, de| 0\leq l\leq n-1\right\}$ corresponds one-to-one to the set of irreducible $\overline{H}_{2n^{2}}$-modules $\left\{[M_l], [N_l], [L]| 0\leq l\leq n-1\right\}$.

(2) If $n$ is even, the set

corresponds one-to-one to the set of irreducible $\overline{H}_{2n^{2}}$-modules

by [15, Corollary 2 (2)].

On the other hand, we have

by Lemma 4.4 (2). Hence, the set $\left\{a^{l}d, a^{l}e, de| 0\leq l\leq n-1\right\}$ corresponds one-to-one to the set of irreducible $\overline{H}_{2n^{2}}$-modules $\left\{[M_l], [N_l], [L]| 0\leq l\leq n-1\right\}$.

Accordingly, the proof is finished. □

Now, we can prove the main result of this section.

Theorem 4.6. Assume that $n\in \mathbb{N}$ and $n\ge 2$, then the representation ring $r(\overline{H}_{2n^{2}})$ is a commutative ring, which can be characterized by generators and relations as follows:

(1) if $n$ is odd, then $r(\overline{H}_{2n^{2}})\cong\mathbb{Z}[y, z, \alpha, \beta]/\mathscr{I}_{1}$, where $\mathscr{I}_{1}$ is the ideal generated by the set

(2) if $n = 2$, then $r(\overline{H}_{2n^{2}})\cong\mathbb{Z}[y, z, \alpha, \beta]/\mathscr{I}_{2}$, where $\mathscr{I}_{2}$ is the ideal generated by the set

(3) if $n > 2$ is even, then $r(\overline{H}_{2n^{2}})\cong\mathbb{Z}[y, z, \alpha, \beta]/\mathscr{I}_{3}$, where $\mathscr{I}_{3}$ is the ideal generated by the set

Proof. It is easy to see that $r(\overline{H}_{2n^{2}})$ is commutative.

(1) By Corollary 4.5, if $n$ is odd, then $r(\overline{H}_{2n^{2}})$ is generated by $b, c, d$ and $e$. Hence, there is a unique ring epimorphism

such that

We note that

by Lemma 2.4, and Lemma 4.4, we have

where $m: = \frac{n-1}{2}$. It follows that $\Phi(\mathscr{I}_{1}) = 0$, and $\Phi$ induces a ring epimorphism

such that

where $\overline{\nu} = \pi(\nu)$, and $\pi$ is a natural epimorphism $\mathbb{Z}[y, z, \alpha, \beta]\rightarrow \mathbb{Z}[y, z, \alpha, \beta]/\mathscr{I}_{1}.$

We note that the ring $r(\overline{H}_{2n^{2}})$ is the free $\mathbb{Z}$-module of rank $4n+1+\frac{n(n-1)}{2}$, with the $\mathbb{Z}$-basis

So we can define a $\mathbb{Z}$-module homomorphism:

where $0\leq k\leq2n-1, 1\leq i\leq\frac{n-1}{2}, 0\leq j\leq n-1$ and $0\leq l\leq n-1$.

On the other hand, as the $\mathbb{Z}$-module, $\mathbb{Z}[y, z, \alpha, \beta]/\mathscr{I}_{1}$ is generated by elements

Let

it is straightforward to check that $\Psi\overline{\Phi}(\overline{a}) = \overline{a}$. Hence $\Psi\overline{\Phi} = {{\rm{id}}}$, which implies that $\overline{\Phi}$ is a monomorphism, and hence $\overline{\Phi}$ is an isomorphism.

The proofs of the remaining statements of the theorem are similar to the above. □

5.   The representations of $\widehat{H}_{2n^2}$

In this section, the representations and representation ring of the $\Delta$-associative algebra $\widehat{H}_{2n^{2}}$ are given. Due to the similar research methods used as in Section 4, we directly provide relevant conclusions and omit their proofs.

By Proposition 3.8, it follows that $\widehat{H}_{2n^2} = \widehat{A}_{0}\oplus \widehat{A}_{1}\oplus \widehat{A}_{2}$ as algebras, then

where ind-$\widehat{A}_i$ can be viewed as the subset of ind-${\widehat{H}_{2n^{2}}}$ in the natural way. As an algebra, we note that

where $H_2$ is the 4-dimensional Sweedler's algebra.

It is noted that the algebra $H_{2}$ and ${\mathbb{k}}[y]/(y^{d})$ are Nakayama algebras, and their representations are also easy to construct; for examples, see ^[21,30,31]. Therefore, we can give a complete list of finite dimensional indecomposable representations in ind-$\widehat{A}_1$ and ind-$\widehat{A}_2$ as follows.

(1) Let $\widehat{M}_{i}, i\in \mathbb{Z}_{n}$. The 1-dimensional irreducible ${\widehat{H}_{2n^{2}}}$-module with basis $\widehat{v}^{(i)}$, the actions of ${\widehat{H}_{2n^{2}}}$ on $\widehat{M}_{i}$ are

(2) Let $\widehat{N}_{i}, i\in \mathbb{Z}_{n}$. The 1-dimensional irreducible ${\widehat{H}_{2n^{2}}}$-module with basis $\widehat{w}^{(i)}$, the actions of ${\widehat{H}_{2n^{2}}}$ on $\widehat{N}_{i}$ are

(3) Let $L_{i}, i\in \mathbb{Z}_{n}$. The 2-dimensional indecomposable ${\widehat{H}_{2n^{2}}}$-module with basis $\widehat{v}_{1}^{(i)}, \widehat{v}_{2}^{(i)}$, the actions of ${\widehat{H}_{2n^{2}}}$ on $L_{i}$ are

(4) Let $P_{i}, i\in \mathbb{Z}_{n}$. The 2-dimensional indecomposable ${\widehat{H}_{2n^{2}}}$-module with basis $\widehat{w}_{1}^{(i)}, \widehat{w}_{2}^{(i)}$, the actions of ${\widehat{H}_{2n^{2}}}$ on $P_{i}$ are

In fact,

(5) Let $Q$ be a 1-dimensional irreducible ${\widehat{H}_{2n^{2}}}$-module with basis $\widehat{u}$, the actions of ${\widehat{H}_{2n^{2}}}$ on $Q$ are

(6) Let $R$ be the 2-dimensional indecomposable ${\widehat{H}_{2n^{2}}}$-module with basis $\widehat{u}_{1}, \widehat{u}_{2}$, the actions of ${\widehat{H}_{2n^{2}}}$ on $R$ are

In fact,

Therefore, we have

Proposition 5.1. The set

forms a complete list of non-isomorphic indecomposable ${\widehat{H}_{2n^{2}}}$-modules.

Also, we have several lemmas as follows. As an example, we only give the proof of the first one.

Lemma 5.2. For all $i, i'\in \mathbb{Z}_{n}$, $m\in \mathbb{Z}_{2n}$ and $0\leq s < t\leq n-1$, we have

(1) $L_{i}\otimes S_{m}\cong \widehat{M}_{m+i(\mathrm{mod}\; n)}\oplus\widehat{N}_{m+i(\mathrm{mod}\; n)};$

(2) $S_{m}\otimes L_{i} \cong L_{m+i(\mathrm{mod}\; n)};$

(3) $L_{i}\otimes S_{s, t}\cong \widehat{M}_{i+s(\mathrm{mod}\; n)}\oplus \widehat{M}_{i+t(\mathrm{mod}\; n)}\oplus\widehat{N}_{i+s(\mathrm{mod}\; n)}\oplus \widehat{N}_{i+t(\mathrm{mod}\; n)};$

(4) $S_{s, t}\otimes L_{i}\cong L_{i+s(\mathrm{mod}\; n)}\oplus L_{i+t(\mathrm{mod}\; n)};$

(5) $L_{i}\otimes L_{i'}\cong \widehat{M}_{i+i'(\mathrm{mod}\; n)}\oplus\widehat{N}_{i+i'(\mathrm{mod}\; n)}\oplus Q\oplus Q;$

(6) $L_{i}\otimes P_{i'}\cong \widehat{M}_{i+i'(\mathrm{mod}\; n)}\oplus \widehat{N}_{i+i'(\mathrm{mod}\; n)}\oplus Q\oplus Q;$

(7) $P_{i'}\otimes L_{i}\cong R\oplus \widehat{M}_{i+i'(\mathrm{mod}\; n)}\oplus \widehat{N}_{i+i'(\mathrm{mod}\; n)};$

(8) $L_{i}\otimes Q\cong Q\otimes L_{i}\cong Q\oplus Q;$

(9) $L_{i}\otimes R\cong R\otimes L_{i}\cong Q\oplus Q\oplus Q\oplus Q.$

Proof. (1) In $\mathscr{G}$, considering the tensor product $L_{i}\otimes S_{m}$ for $i\in \mathbb{Z}_{n}$, $m\in \mathbb{Z}_{2n}$, we have

Hence, $L_{i}\otimes S_{m}\cong \widehat{N}_{i+m(\mathrm{mod}\; n)}\oplus\widehat{M}_{i+m(\mathrm{mod}\; n)}$.

(2) Considering the tensor product $S_{m}\otimes L_{i}$ for $i\in \mathbb{Z}_{n}$, $m\in \mathbb{Z}_{2n}$, we have

Let $\pi_{1} = v^{m}\otimes\widehat{v}_{1}^{(i)}, \quad \pi_{2} = \frac{1}{n}\sigma(m)q^{\frac{m^{2}}{2}}\sum\limits_{i', j' = 1}^{n} q^{-i'j'+mi'+ij'}v^{m}\otimes\widehat{v}_{2}^{(i)},$ then

Thus, $S_{m}\otimes L_{i}\cong L_{i+m(\mathrm{mod}\; n)}.$

(3) Considering the tensor product $L_{i}\otimes S_{s, t}$ for $i\in \mathbb{Z}_{n}$ and $0\leq s < t\leq n-1$, we have

Hence, $L_{i}\otimes S_{s, t}\cong \widehat{M}_{i+s(\mathrm{mod}\; n)}\oplus\widehat{M}_{i+t(\mathrm{mod}\; n)}\oplus \widehat{N}_{i+s(\mathrm{mod}\; n)}\oplus\widehat{N}_{i+t(\mathrm{mod}\; n)}$.

(4) Considering the tensor product $S_{s, t} \otimes L_{i}$ for all $i\in \mathbb{Z}_{n}$ and $0\leq s < t\leq n-1$, we have

Let

then

it follows that $S_{s, t}\otimes L_{i}\cong L_{i+s(\mathrm{mod}\; n)}\oplus L_{i+t(\mathrm{mod}\; n)}$.

The remaining statements can be proven similarly. □

Lemma 5.3. For $i, i'\in \mathbb{Z}_{n}$, $m\in \mathbb{Z}_{2n}$ and $0\leq s < t\leq n-1$, we have

(1) $\widehat{M}_{i}\otimes S_{m}\cong S_{m}\otimes \widehat{M}_{i}\cong \widehat{M}_{m+i(\mathrm{mod}\; n)};$

(2) $\widehat{M}_{i}\otimes S_{s, t}\cong S_{s, t}\otimes \widehat{M}_{i}\cong \widehat{M}_{i+s(\mathrm{mod}\; n)}\oplus \widehat{M}_{i+t(\mathrm{mod}\; n)};$

(3) $\widehat{M}_{i}\otimes \widehat{M}_{i'}\cong \widehat{M}_{i+i'(\mathrm{mod}\; n)};$

(4) $\widehat{M}_{i}\otimes N_{i'}\cong N_{i'}\otimes \widehat{M}_{i}\cong Q;$

(5) $\widehat{M}_{i}\otimes L_{i'}\cong L_{i'}\otimes \widehat{M}_{i}\cong Q\oplus \widehat{M}_{i+i'(\mathrm{mod}\; n)};$

(6) $\widehat{M}_{i}\otimes P_{i'}\cong P_{i'}\otimes \widehat{M}_{i}\cong Q\oplus \widehat{M}_{i+i'(\mathrm{mod}\; n)};$

(7) $\widehat{M}_{i}\otimes Q\cong Q\otimes \widehat{M}_{i}\cong Q;$

(8) $\widehat{M}_{i}\otimes R\cong R\otimes \widehat{M}_{i}\cong Q\oplus Q.$

Lemma 5.4. For all $i, i'\in \mathbb{Z}_{n}$, $m\in \mathbb{Z}_{2n}$ and $0\leq s < t\leq n-1$, we have

(1) $\widehat{N}_{i}\otimes S_{m}\cong S_{m}\otimes \widehat{N}_{i}\cong \widehat{N}_{m+i(\mathrm{mod}\; n)};$

(2) $\widehat{N}_{i}\otimes S_{s, t}\cong S_{s, t}\otimes \widehat{N}_{i}\cong \widehat{N}_{i+s(\mathrm{mod}\; n)}\oplus \widehat{N}_{i+t(\mathrm{mod}\; n)};$

(3) $\widehat{N}_{i}\otimes \widehat{N}_{i'}\cong \widehat{N}_{i+i'(\mathrm{mod}\; n)};$

(4) $\widehat{N}_{i}\otimes L_{i'}\cong L_{i'}\otimes \widehat{N}_{i}\cong\widehat{N}_{i+i'(\mathrm{mod}\; n)}\oplus Q;$

(5) $\widehat{N}_{i}\otimes P_{i'}\cong P_{i'}\otimes \widehat{N}_{i}\cong Q\oplus \widehat{N}_{i+i'(\mathrm{mod}\; n)};$

(6) $\widehat{N}_{i}\otimes Q\cong Q\otimes \widehat{N}_{i}\cong Q;$

(7) $\widehat{N}_{i}\otimes R\cong R\otimes \widehat{N}_{i}\cong Q\oplus Q.$

Lemma 5.5. For all $i, i'\in \mathbb{Z}_{n}$, $m\in \mathbb{Z}_{2n}$ and $0\leq s < t\leq n-1$, we have

(1) $P_{i}\otimes S_{m}\cong P_{m+i(\mathrm{mod}\; n)};$

(2) $S_{m}\otimes P_{i} \cong \widehat{M}_{m+i(\mathrm{mod}\; n)}\oplus \widehat{N}_{m+i(\mathrm{mod}\; n)};$

(3) $P_{i}\otimes S_{s, t}\cong \widehat{P}_{i+s(\mathrm{mod}\; n)}\oplus \widehat{P}_{i+t(\mathrm{mod}\; n)};$

(4) $S_{s, t}\otimes P_{i}\cong M_{i+s(\mathrm{mod}\; n)}\oplus M_{i+t(\mathrm{mod}\; n)}\oplus N_{i+s(\mathrm{mod}\; n)}\oplus N_{i+t(\mathrm{mod}\; n)};$

(5) $P_{i}\otimes P_{i'}\cong \widehat{M}_{i+i'(\mathrm{mod}\; n)}\oplus\widehat{N}_{i+i'(\mathrm{mod}\; n)}\oplus Q\oplus Q;$

(6) $P_{i}\otimes Q\cong Q\otimes P_{i}\cong Q\oplus Q;$

(7) $P_{i}\otimes R\cong R\otimes P_{i}\cong Q\oplus Q\oplus Q\oplus Q.$

Lemma 5.6. For all $i\in \mathbb{Z}_{n}$, $m\in \mathbb{Z}_{2n}$ and $0\leq s < t\leq n-1$, we have

(1) $Q\otimes S_{m}\cong S_{m}\otimes Q\cong Q;$

(2) $Q\otimes S_{s, t}\cong S_{s, t}\otimes Q\cong Q\oplus Q;$

(3) $Q\otimes Q\cong Q;$

(4) $Q\otimes R\cong R\otimes Q\cong Q\oplus Q;$

(5) $R\otimes S_{m}\cong S_{m}\otimes R\cong Q\oplus Q;$

(6) $R\otimes S_{ij}\cong S_{ij}\otimes R\cong Q\oplus Q\oplus Q\oplus Q;$

(7) $R\otimes R\cong Q\oplus Q\oplus Q\oplus Q.$

Let

Then, we have the following result.

Lemma 5.7. For all $i, i'\in \{1, 2, \cdots, n-1\}$, the following statements hold in $r(\widehat{H}_{2n^{2}})$.

(1) If $n$ is odd, then

(a) $[\widehat{M}_{i}] = b^{i}d' = d'b^{i}$ and $[\widehat{M}_{0}] = b^{n}d' = d'b^{n} = d'$;

(b) $[\widehat{N}_{i}] = b^{i}e' = e'b^{i}$ and $[\widehat{N}_{0}] = b^{n}e' = e'b^{n} = e'$;

(c) $[L_{i}] = b^{i}f',$ and $[L_{0}] = b^{n}f' = f'$;

(d) $[P_{i}] = g'b^{i},$ and $[P_{0}] = g'b^{n} = g'$;

(e) $[Q] = e'd' = d'e'$;

(f) $[R] = g'f'-d'-e'$;

(g) $f'b = bg' = b(d'+e')$;

(h) $cd' = d'c = d'+bd',$ $ce' = e'c = e'+be',$ $cf' = f'+bf',$ $f'c = cg' = d'+e'+bd'+be'$ and $g'c = g'+g'b$;

(i) $d'^{2} = d'$, $e'^{2} = e'$, $f'g' = g'^{2} = f'^{2} = d'+e'+2e'd'$, $d'f' = f'd' = d'g' = g'd' = d'+e'd'$, and $e'f' = f'e' = e'g' = g'e' = e'+e'd'$.

(2) If $n$ is even, then

(a) $[\widehat{M}_{i}] = a^{i}d' = d'a^{i}$ and $[\widehat{M}_{0}] = a^{n}d' = d'a^{n} = d'$;

(b) $[\widehat{N}_{i}] = a^{i}e' = e'a^{i}$ and $[\widehat{N}_{0}] = a^{n}e' = e'a^{n} = e'$;

(c) $[L_{i}] = a^{i}f'$ and $[L_{0}] = a^{n}f' = f'$;

(d) $[P_{i}] = g'a^{i}$ and $[P_{0}] = g'a^{n} = g'$;

(e) $[Q] = e'd' = d'e'$;

(f) $[R] = g'f'-d'-e'$;

(g) $f'a = ag' = a(d'+e')$;

(h) $bd' = d'b = ad'$, $be' = e'b = ae'$, $bf' = af'$, $bg' = f'b = f'a$ and $g'b = g'a$;

(i) $cd' = d'c = d'+ad'$, $ce' = e'c = e'+ae'$, $cf' = f'+af'$, $f'c = cg' = d'+e'+ad'+ae'$ and $g'c = g'+g'a$;

(j) $d'^{2} = d'$, $e'^{2} = e'$, $f'g' = g'^{2} = f'^{2} = d'+e'+2e'd'$, $d'f' = f'd' = d'g' = g'd' = d'+e'd'$ and $e'f' = f'e' = e'g' = g'e' = e'+e'd'$.

Corollary 5.8. (1) If $n$ is odd, the set

forms a $\mathbb{Z}$-basis of $r(\widehat{H}_{2n^{2}})$.

(2) If $n$ is even, the set

forms a $\mathbb{Z}$-basis of $r(\widehat{H}_{2n^{2}})$.

Proof. By Lemmas 5.3–5.7, we see that $[S_0]$ is no longer the identity of $r(\widehat{H}_{2n^{2}})$. By Lemma 2.3, $[S_0] = [S_{2n}]$, and if $n$ is odd, the set

corresponds one-to-one to the set of irreducible $\widehat{H}_{2n^{2}}$-modules

if $n$ is even, the set

corresponds one-to-one to the set of irreducible $\widehat{H}_{2n^{2}}$-modules

On the other hand, if $n$ is odd, we have

by Lemma 5.7, where $1\leq s\leq n-1$. Hence, the set

corresponds one-to-one to the set of indecomposable $\widehat{H}_{2n^{2}}$-modules

The remaining can be proven in a similar ways. □

By Lemma 5.3–5.7, we see that $r(\widehat{H}_{2n^{2}})$ is a noncommutative ring without an identity. Let $r^{*}(\widehat{H}_{2n^{2}})$ be the ring with the identity extended from the ring $r(\widehat{H}_{2n^{2}})$ in the natural way. For the general definition one can refer to Section 2. Therefore, $r^{*}(\widehat{H}_{2n^{2}})$ is a ring with the identity $(1, 0)$ and

Now, we can explicitly describe $r^{*}(\widehat{H}_{2n^{2}})$ by generators with relations.

Theorem 5.9. Assume that $n\in \mathbb{N}$ and $n\geq 2$, we have the following statements:

(1) if $n$ is odd, then $r^{*}(\widehat{H}_{2n^{2}})\cong\mathbb{Z}\langle y, z, \alpha, \beta, \gamma, \delta\rangle/\mathscr{J}_{1}$, where $\mathscr{I}_{1}$ is the ideal generated by the following set

(2) if $n = 2$, then $r^{*}(\widehat{H}_{2n^{2}})\cong\mathbb{Z}\langle x, y, z, \alpha, \beta, \gamma, \delta\rangle/\mathscr{J}_{2}$, where $\mathscr{I}_{2}$ is the ideal generated by the following set

(3) if $n > 2$ is even, then $r^{*}(\widehat{H}_{2n^{2}})\cong\mathbb{Z}\langle x, y, z, \alpha, \beta, \gamma, \delta\rangle/\mathscr{J}_{3}$, where $\mathscr{I}_{3}$ is the ideal generated by the following set

Proof. It is easy to see that $r^{*}(\widehat{H}_{2n^{2}})$ is a noncommutative ring.

(1) Let $\pi:\mathbb{Z}\langle y, z, \alpha, \beta, \gamma, \delta\rangle\rightarrow \mathbb{Z}\langle y, z, \alpha, \beta, \gamma, \delta\rangle/\mathscr{J}_1$ be the natural epimorphisms and $\overline{a} = \pi(a)$ for any $a\in \mathbb{Z}\langle y, z, \alpha, \beta, \gamma, \delta\rangle$.

By Lemma 5.7(1) and Corollary 5.8(1), if $n$ is odd, $r(\widehat{H}_{2n^{2}})$ is generated as a ring by $b, c, d', e', f', g'$, and any element of $r(\widehat{H}_{2n^{2}})$ can be written as

where $\varrho_{k_{1}k_{2}\cdots k_{6}}, k_{1}, k_{2}, \cdots, k_{6}\in \mathbb{Z}$. Therefore,

We defined a $\mathbb{Z}$-map

as

It is easy to get that $\Phi$ can be extended to a ring epimorphism in the natural way. In fact, any $\upsilon \in \mathbb{Z}\langle y, z, \alpha, \beta, \gamma, \delta\rangle$ can be written as

where $\varrho_{k_{1}k_{2}\cdots k_{6}}, k_{1}, k_{2}, \cdots, k_{6}, k\in \mathbb{Z}$. Thus,

By Lemma 5.7(1), one sees that $\Phi(\omega) = (0, 0)$ for all $\omega\in T_1$. Hence, $\Phi(\mathscr{J}_1) = (0, 0)$ and $\Phi$ induces a unique ring epimorphism

with $\overline{\Phi}(\overline{\nu}) = \Phi(\nu)$ for all $\nu\in \mathbb{Z}\langle y, z, \alpha, \beta, \gamma, \delta\rangle.$

We note that the ring $r^{*}(\widehat{H}_{2n^{2}})$ is the free $\mathbb{Z}$-module of rank $6n+3+\frac{n(n-1)}{2}$, with the $\mathbb{Z}$-basis

So we can define a $\mathbb{Z}$-module homomorphism:

by

where $1\leq k\leq2n, 1\leq i\leq\frac{n-1}{2}, 1\leq j\leq n$ and $1\leq s\leq n$. Obviously, as the $\mathbb{Z}$-module, $\mathbb{Z}\langle y, z, \alpha, \beta, \gamma, \delta\rangle/\mathscr{J}_1$ is generated by the set

For any $\overline{a}\in S,$ we see that $\Psi\overline{\Phi}(\overline{a}) = \overline{a}$. Hence $\Psi\overline{\Phi} = {{\rm{id}}}$, which implies that $\overline{\Phi}$ is a monomorphism, and hence $\overline{\Phi}$ is an isomorphism.

The proofs of the remaining statements are similar. □

Remark 5.10. By Theorems 4.6 and 5.9, as the rings, it is easy to see that

(1) if $n$ is odd, then

(2) if $n$ is even, then

6.   Conclusions

We have described the representation rings of two classes of $\Delta$-associative algebras, $\overline{H}_{2n^{2}}$ and $\widehat{H}_{2n^{2}}$, extended from Hopf algebra $H_{2n^2}$ of Kac-Paljutkin type. It may be interesting to consider the category of representations of these representation rings, as this could be helpful in understanding the invariants of the more general tensor categories. It is challenging to consider non-Hermitian linear systems over these rings, similar to non-Hermitian quaternion linear systems over the quaternion algebra. Readers are referred to related works, such as ^[32,33,34].

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

This work is supported by National Natural Science Foundation of China (Grant No. 12201187, 12471038) and Natural Science Foundation of Henan Province (Grant No. 222300420156, 242300421385)

Conflict of interest

The authors declare there is no conflicts of interest.