The representation ring of a non-pointed bialgebra

1.   Introduction

The concept of representation rings, also known as the Green ring, was first introduced by Green ^[1] in the 1960's while studying the representations of finite groups. Benson and Carlson further developed this work, and applied it to the representation theory of algebras (see ^[2,3]). Recently, the study of representations and representation rings of Hopf algebras and, more generally, bialgebras has received considerable attention. For example, Chen, Van Oystaeyen, and Zhang in ^[4] studied the representation rings of Taft algebras. Wang, Li, and Zhang in ^[5] determined the structure of the Green ring for finite-dimensional pointed rank one Hopf algebras of nilpotent type. The authors also provided a classification of all indecomposable modules and constructed their tensor product decomposition formulas. Subsequently, they studied the Green rings of finite-dimensional pointed rank one Hopf algebras of non-nilpotent type in ^[6]. Guo and Yang in ^[7] gave a complete list of all indecomposable modules and described all components of Auslander-Reiten quivers of a class of non-pointed tame Hopf algebras. The authors also established their projective class rings as well as the projective class rings of the category of their Yetter-Drinfeld modules. Su and Yang gave the Green rings of $\Delta$-associative algebras (see ^[8]). Lin and Yang constructed and classified all string representations of a class of tame type Hopf algebras and gave the explicit description of projective class rings (see ^[9]).

Yang and Zhang in ^[10] classified the bialgebra Ore extensions of automorphism type for the $4$-dimensional Sweedler Hopf algebra and obtained a series of non-isomorphic bialgebras and Hopf algebras. The most interesting one of them, denoted by ${{\mathcal{A}_{24m}}}$, which is noncocommutative and non-pointed with dimension $24m,$ arouses us to study their representations and describe its representation ring. The results may help us to understand the more general case of non-pointed bialgebras. We first classify all simple and all indecomposable modules of ${{\mathcal{A}_{24m}}}$ explicitly. Then, the decomposition formulas of the tensor products of them are established, and the description of the representation ring is given.

The outline of this paper is as follows: In Section $1,$ the definition of the $24m$-dimensional bialgebras ${{\mathcal{A}_{24m}}}$ is reviewed. It is shown that ${{\mathcal{A}_{24m}}}$ is a basic algebra, and its corresponding bound quiver is described. In Section 2, all indecomposable modules of ${{\mathcal{A}_{24m}}}$ are classified, and decomposition formulas of their tensor products are established. In Section 3, we focus on characterizing its representation ring of ${{\mathcal{A}_{24m}}}.$

Throughout, $\mathbb{K}$ is assumed to be an algebraically closed field containing a non-trivial $6m$-th primitive root $q$ of unity, and $\omega = q^m.$

2.   The bialgebras ${{\mathcal{A}_{24m}}}$

First of all, let us review the family of bialgebras ${{\mathcal{A}_{24m}}}$ discussed in the paper.

By definition, the algebra ${{\mathcal{A}_{24m}}}$ is generated by $x, g, z$ subjecting to the following relations:

If we define two $\mathbb{K}$-maps $\Delta: {{\mathcal{A}_{24m}}}\to {{\mathcal{A}_{24m}}}\otimes {{\mathcal{A}_{24m}}}$ and $\epsilon: \; {{\mathcal{A}_{24m}}}\to \mathbb{K}$ on generators by

where $e_{1} = \frac {1-g} {2},$ and extend them to the whole space ${{\mathcal{A}_{24m}}}$ in the natural way. Then, ${{\mathcal{A}_{24m}}}$ becomes a $24m$-dimensional bialgebra with a basis

It is easy to see that ${{\mathcal{A}_{24m}}}$ is non-commutative, non-cocommutative and non-pointed. For the detailed construction, the readers are referred to [10, Page 20, Case 2: (a)(ⅰ)].

Obviously, $x{{\mathcal{A}_{24m}}}$ of ${{\mathcal{A}_{24m}}}$ is a bi-ideal; the bialgebra ${{\mathcal{A}_{24m}}}/x{{\mathcal{A}_{24m}}}$ deduced from ${{\mathcal{A}_{24m}}}$ is the group algebra $\mathbb{K}(C_2\times C_{6m}),$ which comultiplication is given by

Indeed, it is straightforward to see that $\mathbb{K}(C_2\times C_{6m})$ is a bialgebra in this setting.

For $i\in\mathbb{Z}_{6m}$, we set

Lemma 2.1. The set

is a complete set of primitive orthogonal idempotents of ${{\mathcal{A}_{24m}}}$.

Proof. For $n = 0, 1, \ldots, 6m-1$, we have

Similarly, $e_{1}xz^{n}\epsilon_{i}e_{1} = 0$. It follows that

Hence, $e_{0}\epsilon_{i}e_{0}$, $e_{1}\epsilon_{i}e_{1}$ are idempotents for all $i\in {\mathbb Z}_{6m}$.

Furthermore, for any $i\ne j$, we have

In a similar way, for any $i\ne j$,

and

for all $i = 0, 1$ and $j\in{\mathbb Z}_{6m}$. In other words, dim $e_i\epsilon_j e_i {{\mathcal{A}_{24m}}} e_i\epsilon_j e_i = 1.$

Also, we have

Therefore, the set

is indeed a complete set of primitive orthogonal idempotents of ${{\mathcal{A}_{24m}}}$. □

Lemma 2.2. The algebra ${{\mathcal{A}_{24m}}}$ is a basic algebra.

Proof. Recall that an algebra $A$ is basic if and only if the algebra $A/J$ is isomorphic to $\mathbb K\times\mathbb K\times\cdots\times \mathbb K,$ where $J = {\rm rad}A$ is the radical of $A$. Therefore, the statement is obvious since ${{\mathcal{A}_{24m}}}/J = {{\mathcal{A}_{24m}}}/x{{\mathcal{A}_{24m}}} \cong \mathbb{K}(C_2\times C_{6m}),$ which is basic, by the previous remark.

Here, we give the detailed proof by constructing the explicit correspondence between primitive idempotents $f_{ij} = e_{i}\epsilon_{j}e_{i} (i \in {\mathbb Z}_{2}, j \in {\mathbb Z}_{6m})$ of ${{\mathcal{A}_{24m}}}$ and those of the $12m$-dimensional diagonal matrix algebra. It is convenient in describing the bound quiver of ${{\mathcal{A}_{24m}}}$ and indecomposable projective modules of ${{\mathcal{A}_{24m}}}$ in the later. It is remarked that $f_{ij} = e_{i}\epsilon_{j}e_{i},$ the liftings of those of $\mathbb{K}(C_2\times C_{6m})$, are not in $\mathbb{K}(C_2\times C_{6m})$.

Now, for $i\in\mathbb{Z}_{2}$ and $j\in\mathbb{Z}_{6m}$, we set

It is easy to see that

of dimension $12m$, and

The $12m$-dimensional algebra

can be identified with the $12m$-dimensional diagonal matrix algebra

Let

where $p = 1, \cdots, 12m$, and only the $p$-th position is $1$ and others are $0$. Let

Now, define the mapping

by

where $j\in{\mathbb Z}_{6m}.$

We must prove that $\varphi$ is an algebra isomorphism. Indeed, we have

In the same way, we can obtain

Because

we have

Furthermore,

where $E$ is the identity matrix of $12m\times 12m.$

Hence, $\varphi$ is an algebra isomorphism, and the algebra ${{\mathcal{A}_{24m}}}$ is a basic algebra. □

Furthermore, we see that

Hence,

and

where $i, j\in \mathbb{Z}_{6m}.$

We can describe the bound quiver $(Q, I)$ of ${{\mathcal{A}_{24m}}}$. Now, let $a_{i, j}$ be the points corresponding to the primitive idempotents $f_{k_i, l_j}$, where

and the arrows are endowed with $\alpha_{i, j+1}: a_{i, j+1} \to a_{i, j},$ for all $(i, j)\in {\mathbb Z}_m\times {\mathbb Z}_{12}.$

For example, for $m = 1$, the bound quiver $(Q_{\mathcal{A}_{24}}, I)$ is

with the admissible ideal

In general, for $m\ge 1$, it is observed that the bound quiver of $(Q_{{{\mathcal{A}_{24m}}}}, I)$ is

with the admissible ideal

3.   Indecomposable ${{\mathcal{A}_{24m}}}$-modules

In this section, we mainly construct all mutually non-isomorphic indecomposable modules of ${{\mathcal{A}_{24m}}}$. Then, we consider their tensor products, respectively. Finally, we conclude that the decomposition formula of their tensor products.

By the bound quiver of algebra $({{\mathcal{A}_{24m}}}, I)$, we see that algebra ${{\mathcal{A}_{24m}}}$ can be decomposed into a direct sum of $m$ blocks ${\mathcal{C}}_i, i\in{\mathbb Z}_m$: ${{\mathcal{A}_{24m}}} = \bigoplus_{i\in{\mathbb Z}_m}{\mathcal{C}}_i,$ where ${\mathcal{C}}_i = \sum_{j\in{\mathbb Z}_{12}}{{\mathcal{A}_{24m}}} f_{k_i, l_j},$ and $k_i, l_j$ are as in (2.1). Consequently, by [11, Chap. 5, Theorem 3.5], any indecomposable ${{\mathcal{A}_{24m}}}$-module is either a $1$-dimensional simple module or a $2$-dimensional projective ${{\mathcal{A}_{24m}}}$-module.

Let $S_{ij}$ be a $1$-dimensional vector space spanned by $v_{ij}, i\in\mathbb{Z}_{2}, j\in\mathbb{Z}_{6m}$, we define the action of ${{\mathcal{A}_{24m}}}$ on $S_{ij}$ as follows:

It is straightforward to see that $S_{ij}$ is a simple ${{\mathcal{A}_{24m}}}$-module, and $S_{ij} \cong S_{kl}$ if and only if $i = k, j = l.$

Lemma 3.1. Any simple ${{\mathcal{A}_{24m}}}$-module $S$ is isomorphic to $S_{ij}$, where $i\in\mathbb{Z}_{2}, j\in\mathbb{Z}_{6m}.$

Proof. Assume that $S$ is a simple ${{\mathcal{A}_{24m}}}$-module. By Lemma 2.2, the dimension of the simple ${{\mathcal{A}_{24m}}}$-module $S$ is $1$. Let $v$ be the basis of $S$. We set

where $a, b, c\in\mathbb{K}.$ We have

Hence, we have $a = 0, \; b = \pm 1,$ and $c^{6m} = 1.$ Consequently, we can set $c = q^{j}$ for some $j\in{\mathbb Z}_{6m}$ and

or

Therefore, there exist $i\in \mathbb{Z}_{2},$ $j\in \mathbb{Z}_{6m}$ such that $S = S_{ij}.$ □

Furthermore, the module action of the submodule ${{\mathcal{A}_{24m}}} f_{ij}$ of the left regular ${{\mathcal{A}_{24m}}}$-module is as follows:

Let $P_{ij}$ be the projective cover of simple ${{\mathcal{A}_{24m}}}$-module $S_{ij}$ for $(i, j)\in\mathbb{Z}_{2}\times\mathbb{Z}_{6m}.$

Lemma 3.2. We have $P_{ij}\cong {{\mathcal{A}_{24m}}} f_{ij}.$

Proof. Note that $\{f_{ij}|i\in\mathbb{Z}_{2}, j\in\mathbb{Z}_{6m}\}$ is the complete set of orthogonal idempotent elements of the algebra ${{\mathcal{A}_{24m}}}.$ So we have

Also, ${{\mathcal{A}_{24m}}} f_{ij }$ is an indecomposable projective module.

We define the map $\varphi$ as follows:

for each $(i, j)\in\mathbb{Z}_{2}\times\mathbb{Z}_{6m}.$

It is easy to see that $\varphi$ is a surjective linear map. By Lemma 3.1, we have

So the map $\varphi: {{\mathcal{A}_{24m}}} f_{ij}\rightarrow S_{ij}$ can define a module epimorphism, and

It is easy to see that $\mathrm{ker}\varphi = \text{rad}({{\mathcal{A}_{24m}}} f_{ij}) = \mathbb{K}\{xf_{ij}\},$ and

However,

since $P_{ij}$ is a projective cover of simple ${{\mathcal{A}_{24m}}}$-module $S_ {ij}.$ By the uniqueness of the projective cover of a module, we have

Now, we identify $P_{ij}$ with ${{\mathcal{A}_{24m}}} f_{ij}$ for $(i, j)\in\mathbb{Z}_{2}\times\mathbb{Z}_{6m}.$

So far, we have provided a classification of all indecomposable ${{\mathcal{A}_{24m}}}$-modules. In the following, we consider their tensor products.

Proposition 3.3. $S_ {ij}\otimes S _{kl}\cong S_{(i+k)(\mathrm{mod}\, 2), (j+l)(\mathrm{mod}\, 6m)}\cong S_ {kl}\otimes S_{ij},$ for $i\in\mathbb{Z}_{2}, j\in\mathbb{Z}_{6m}.$

Proof. For simple modules $S_{ij}, S_{kl}$, the tensor product $S_{ij}\otimes S_{kl}$ is also an ${{\mathcal{A}_{24m}}}$-module with a basis $v_{ij}\otimes v_{kl}.$ By Lemma 3.1, we have

Hence,

The result follows. □

Proposition 3.4. $S_{ij}\otimes P_{kl}\cong P_{(i+k)(\mathrm{mod}\, 2), (j+l)(\mathrm{mod}\, 6m)}\cong P_{kl}\otimes S_{ij},$ for $i\in\mathbb{Z}_{2}, j\in\mathbb{Z}_{6m}.$

Proof. We consider $P_{kl} = {{\mathcal{A}_{24m}}} f_{kl}, \; (k\in\mathbb{Z}_{2}, l\in\mathbb{Z}_{6m}),$ which is of basis $\{f_{kl}, \; xf_{kl}\}.$ The action of ${{\mathcal{A}_{24m}}}$ on $P_{kl}$ is the following:

The corresponding matrix representations of $x, g, z$ are

respectively.

On the other hand, $\{ {v_{ij}} \otimes {f_{kl}}, \; {v_{ij}} \otimes {xf_{kl }} \}$ form a basis of the tensor product $S_{ij}\otimes P_{kl}.$ The action of ${{\mathcal{A}_{24m}}}$ on $S_{ij}\otimes P_{kl}$ is the following:

The corresponding matrix representations of $x, g, z$ are

respectively.

Hence, we have

In a similar way, we can also prove that

Proposition 3.5.

for $i\in\mathbb{Z}_{2}, j\in\mathbb{Z}_{6m}.$

Proof. Considering the exact sequence

where $\psi$ is defined by $\psi(f_{ij}) = (-1)^{i}v_{ij}, \ \psi (xf_{ij}) = 0.$ We have $\mathrm{Ker}\psi = \mathbb{K}\{xf_{ij}\} = \Omega \left({{S_{ij}}} \right)$, and

On the other hand, for a Hopf algebra $H$, and any $H$-module $M$, if $P$ is a projective $H$-module, then $P{\otimes} M$ or $M{\otimes} P$ is also a projective $H$-module (see, for example, ^[12]). Therefore, we obtain the following splitting short exact sequence:

Hence, we obtain

Note that $q$ is the $6m$-th primitive root of the unit, and $\omega$ is the $6$-th primitive root of the unit; we have $q^{ 3m} = -1.$ Also, without loss of generality, we assume that $q^{m} = \omega$. Accordingly, the action of the algebra ${{\mathcal{A}_{24m}}}$ on $\Omega \left({{S_{ij}}} \right)$ is as follows:

Reviewing the construction of simple ${{\mathcal{A}_{24m}}}$-module $S_{ij}$, one sees that

Therefore,

By Proposition 3.4, we obtain

4.   The representation ring of ${{\mathcal{A}_{24m}}}$

In this section, we mainly focus on establishing the representation ring of the algebra ${{\mathcal{A}_{24m}}}$. Let us first review the definition of the projective class ring.

Assuming that $H$ is a Hopf algebra, we denote by $[V]$, the isomorphic class of a finite-dimensional $H$-module $V$. Let $F(H)$ be the free abelian group generated by $[V]$, we define a multiplication on $F(H)$ as $[M][N] = [M\otimes N].$ Then, $F(H)$ is an associative ring with the identity element $[\mathbb{K}].$ The quotient ring $r(H): = F(H)/I$ is called the Green ring of $H$, where $I$ is the ideal generated by the relation $[M\otimes N]-[M][N]$ and $[M\oplus N]-[M]-[N].$ It is clear that $\{[V]\thinspace|\thinspace V\in \mathrm{ind} (H)\}$ forms a $\mathbb{Z}$-basis for $r(H),$ where $\mathrm{ind} (H)$ is the set consisting of isomorphism classes of all finite-dimensional indecomposable $H$-modules.

The subring $\mathcal{P}(H)$ of $r(H)$ generated by the projective modules and simple modules of $H$ is called the projective class ring of $H.$ It is noted that the representation ring of ${{\mathcal{A}_{24m}}}$ is the same as its projective class ring of ${{\mathcal{A}_{24m}}}.$

Now, let us determine the representation ring $r({{\mathcal{A}_{24m}}})$ of the algebra ${{\mathcal{A}_{24m}}}$.

Lemma 4.1. Let $t = [S_{01}], f = [P_{01}], u = [S_{11}], h = [P_{11}]$, then the following equations hold:

(1) $t^{6m} = 1, \thinspace u^{6m} = 1.$

(2) $tf = ft, \thinspace tu = ut, \thinspace th = ht, \thinspace uh = hu, \thinspace uf = fu.$

(3) $t^{2} = u^{2}, \thinspace fh = u^{4m+1}h+th, \thinspace hf = t^{m+1}f+uf.$

(4) $f^{2} = u^{4m+1}f+tf, \thinspace h^{2} = t^{m+1}h+uh.$

Proof. (1) By Proposition 3.3, we obtain

Therefore, $t^{6m} = 1.$ Similarly, $u^{6m} = 1$.

(2) By Proposition 3.4, $S_{ij}\otimes P_{kl} = P_{kl}\otimes S_{ij}$, so, $tf = [S_{01}][P_{01}] = [S_{01}\otimes P_{01}] = [P_{01}\otimes S_{01}] = [P_{01}][S_{01}] = ft$, thus, we have $tf = ft.$ Similarly, we obtain

(3) By Proposition 3.3, we obtain

Thus, $t^{2} = u^{2}$. By Proposition 3.5, we obtain

Thus, $fh = u^{4m+1}h+th, \thinspace hf = t^{m+1}f+uf.$

(4) By Proposition 3.5, we obtain

The lemma follows.

Corollary 4.2. $\{t^{i}f^{j}; u t^{i_{1}}f^{j_{1}}; u^{k}h^{l}; tu^{k_{1}}h^{l_{1}}\}$ is a $\mathbb{Z}$-basis of $r({{\mathcal{A}_{24m}}})$, where $i, i_{1}, k, k_{1}\in \mathbb{Z}_{6m}, j, j_{1}, l, l_{1}\in \mathbb{Z}_{2},$ and $i, k$ are even, $i_{1}, k_{1}$ are odd.

Proof. Note that $r({{\mathcal{A}_{24m}}})$ has a $\mathbb{Z}$-basis $\left \{ [S_{ij}], [P_{ij}] \mid i\in \mathbb{Z}_{2}, j\in \mathbb{Z}_{6m}\right \}.$ Hence, the rank of $r({{\mathcal{A}_{24m}}})$ is $24m$. By Lemma 4.1, $r({{\mathcal{A}_{24m}}})$ is spanned by the set $\{t^{i}f^{j}; u t^{i_{1}}f^{j_{1}}; u^{k}h^{l}; tu^{k_{1}}h^{l_{1}}\},$ which consists of $24m$ elements. Hence, the set is a $\mathbb{Z}$-basis of $r({{\mathcal{A}_{24m}}}).$

Now, we can obtain the main result of the paper.

Theorem 4.3. The representation ring $r({{\mathcal{A}_{24m}}})$ is isomorphic to the quotient ring $\mathbb{Z}\langle\alpha, \beta, \gamma, \eta\rangle/\mathcal{I}$, where $\mathcal{I}$ is the ideal generated by the relations

Proof. Let

be the natural epimorphism. It is straightforward to check that $\mathbb{Z}\langle\alpha, \beta, \gamma, \eta\rangle/\mathcal{I}$ is spanned by

where $i, i_{1}, k, k_{1}\in \mathbb{Z}_{6m}, j, j_{1}, l, l_{1}\in \mathbb{Z}_{2},$ and $i, k$ are even, $i_{1}, k_{1}$ are odd. This means that the rank of $\mathbb{Z}\langle\alpha, \beta, \gamma, \eta\rangle/\mathcal{I}$ is equal or less than $24m.$

By Corollary 4.2, $r({{\mathcal{A}_{24m}}})$ is generated by $t, f, u,$ and $h.$ So, there is a uniqueness ring epimorphism

satisfying

By Lemma 4.1, we see that

It follows that

Hence, $\Phi(\mathcal{I}) = 0,$ and $\Phi$ induces an epimorphism

satisfying $\overline{\Phi}(\overline{v}) = \Phi(v),$ for any $v\in\mathbb{Z}\langle\alpha, \beta, \gamma, \eta\rangle$, where $\overline{v} = \pi(v) = v+\mathcal{I}$.

Comparing the ranks of $\mathbb{Z}\langle\alpha, \beta, \gamma, \eta\rangle/\mathcal{I}$ with $r({{\mathcal{A}_{24m}}})$, we get that $\overline{\Phi}$ is indeed a ring isomorphism.

The result is obtained.

5.   Conclusions

In this paper, all indecomposable modules of ${{\mathcal{A}_{24m}}}$ are constructed and classified. Furthermore, we obtain the decomposition formulas for the tensor products of indecomposable modules as well as the representation ring of the bialgebra. These results may help us to understand the general representation theory of a non-pointed bialgebra.

Author Contributions

Shilin Yang contributed the creative ideas and proof techniques for this paper, including the construction of the article structure and the revision of content; Huaqing Gong consulted the relevant background of the paper and wrote the article. All authors have read and agreed to the published version of the manuscript.

Use of Generative-AI tools declaration

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

Conflict of interest

All authors declare no conflicts of interest in this paper.