A new family of positively based algebras ${\mathcal{H}}_n$

1.   Introduction

An algebra $A$ is called a positively based algebra if it has a positive basis. By the latter, we mean that a set of bases with all the structure constants with respect to this basis are non-negative real numbers. The concept of a positive basis can be traced back to the work of Schur on the centralizer algebra of a transitive permutation representation of a finite group in ^[16,17]. Since then, variations of the positively based algebras have been discovered in many different fields (see ^[1,3,15]). Examples of positively based algebras include the group algebras, semigroup algebras, table algebras, and Hecke algebras corresponding to Coxeter groups with respect to the Kazhdan-Lusztig basis. As an interesting infinite dimensional example, Thurston proved that the Kauffman bracket skein algebra for a compact oriented surface has a natural positive basis in ^[19], which generalized the positivity conjecture in cluster algebra. In ^[14], Mazorchuk et al. defined the cell 2-representations of finitary 2-categories. It turns out that the cell $2$-representation is a based module over some finite-dimensional positively based algebras on the level of the Grothendieck group. In addition, in the perspective of representation theory, Green algebra of a bialgebra is a central instance of positively based algebras. For example, Cao et al. ^[4] studied the cell modules of the Green algebra of the generalized Taft Hopf algebras, in which the approach depends on the Green ring.

In ^[10], Lin and Yang introduced a class of positively based algebras $A_{n, d}$ over the subring $\mathbb{K}$ of the complex field $\mathbb{C}$; all indecomposable modules and cell modules of $A_{n, d}$ are classified and constructed. It is noted that $A_{n, d}$ can be regarded as the Green algebra of weak Hopf algebras ${\mathfrak{w}}_{n, d}^1$ constructed from the generalized Taft algebra (see ^[18]). In the present paper, we introduce the $\mathbb{K}$-algebra ${\mathcal{H}}_n$, which is generated by $x, y, z$ with the relations $(1)$ and $(2)$; since we can construct a set of positive bases of ${\mathcal{H}}_n$, we can conclude that ${\mathcal{H}}_n$ is a positively based algebra. The algebra ${\mathcal{H}}_n$ can be viewed as the extension of the Green ring of the weak Hopf algebra $\mathfrak{w} H_{4n}$, which is constructed from neither the pointed nor the semisimple Hopf algebra $H_{4n}$ (see ^[5]). Recall that $A_{n, d}$ in [10,Definition 2.1] is generated by $x, y, z$ with the following relations:

We see that the generating relations of $A_{n, d}$ and ${\mathcal{H}}_n$ are strongly different. For instance, the representation type of $A_{n, d}$ depends on $d$, and it is of finite representation type if $d\leq 4$, of tame type if $d = 5$, and of wild type if $d\ge 6.$ However, the algebra ${\mathcal{H}}_n$ is only of finite representation type. Up to now, there is a limited number of works that classifies the representations of this family of positively based algebras. This provides a good chance to understand the representation theory of ${\mathcal{H}}_n$. In this paper, as the analog of the method in ^[10], we first construct a set of positive basis of ${\mathcal{H}}_n$ and show that ${\mathcal{H}}_n$ is a positively based algebra. Then, we make efforts to classify all the indecomposable modules of ${\mathcal{H}}_n$ by utilizing the representation theory of a quiver. It is easily seen that the algebra ${\mathcal{H}}_n$ is of finite representation type, and thus all the indecomposable modules of ${\mathcal{H}}_n$ are constructed. Finally, all the left (right) cells and left cell (right) modules of ${\mathcal{H}}_n$ are constructed. Moreover, the decompositions of the left cell modules are provided.

Let us describe the arrangement of this paper. In Section 2, we introduce the algebra ${\mathcal{H}}_n$ by generators and generating relations, and show that ${\mathcal{H}}_n$ is a positively based algebra. In Section 3, we focus on constructing all indecomposable modules of ${\mathcal{H}}_n$ and we see that ${\mathcal{H}}_n$ is of finite representation type. In Section 4, we classify all the left (right) cells and left (right) cell modules of ${\mathcal{H}}_n$.

2.   Preliminaries

Throughout the paper and unless otherwise stated, $\mathbb{C}$, $\mathbb{R}$, $\mathbb{Z}$ and $\mathbb{N}$, stand for the field of complex numbers, the field of real numbers, the ring of integers, and the set of natural numbers, respectively. Fixing an integer $n\geq 1$, we always suppose that $\mathbb{K}$ is an unital subring of $\mathbb{C}$ containing the primitive $2n$-th root of the unity:

Let $A$ be a $\mathbb{K}$-algebra of finite free rank $n$ with a basis $\mathfrak{B} = \{a_{i}\; |\; 1\leq i\leq n\}$. For any $1\leq i, j, k\leq n$, if

where $\gamma_{i, j}^{(k)}\in\mathbb{R}\geq 0$, then $\mathfrak{B}$ is said to be a positive basis of $A$, and $A$ is called a positively based algebra.

Now, we define the algebra ${\mathcal{H}}_n$ and describe its basic properties.

Definition 2.1. The $\mathbb{K}$-algebra ${\mathcal{H}}_n$ is generated by $x, y, z$ with the following relations$:$

(1) $x^{2n} = 1$, $y^2 = xy+y$, $xy = yx;$

(2) $z^2 = z,$ $xz = zx = z,$ $zy = 2z.$

It is easy to see that ${\mathcal{H}}_n$ is noncommutative, and it is noted that when $\mathbb{K} = {\mathbb C}$, the algebra ${\mathcal{H}}_n$ is just the Green algebra of the weak Hopf algebra $\mathfrak{w} H_{4n}$ (see ^[5]).

In the following, we construct a set of positive bases of ${\mathcal{H}}_n$ to show that ${\mathcal{H}}_n$ is a positively based algebra. For this purpose, we list one lemma as follows.

Lemma 2.2. (1) The $\mathbb{K}$-space $\mathcal{W}_{i}$ $(0\leq i\leq 2n-1)$ with a basis $\left\{\nu_{i}\right\}$ is an ${\mathcal{H}}_n$-module with the action of ${\mathcal{H}}_n:$

(2) The $\mathbb{K}$-space $\mathcal{V}_{1}$ with a basis $\left\{\theta_1, \theta_2\right\}$ is an ${\mathcal{H}}_n$-module with the action of ${\mathcal{H}}_n:$

(3) The $\mathbb{K}$-space $\mathcal{V}_{2}$ with a basis $\left\{\xi_1, \xi_2\right\}$ is an ${\mathcal{H}}_n$-module with the action of ${\mathcal{H}}_n:$

(4) The $\mathbb{K}$-space $\mathcal{V}_{3}$ with a basis $\left\{\mu_{j} \mid 0\leq j\leq 2n-1\right\}$ is an ${\mathcal{H}}_n$-module with the action of ${\mathcal{H}}_n:$

Proof. In the following, we only prove the statement (3); the proofs of the other statements are similar.

First, it is easy to see that

Then, the straightforward verification shows that

and

Moreover, it is obvious that

and

Finally, we see that

It follows that the generators $x, y$ and $z$ acting on $\mathcal{V}_{2}$ keep the defining relations.

Hence, the actions of $x, y$ and $z$ on $\mathcal{V}_{2}$ define an ${\mathcal{H}}_n$-module. □

Proposition 2.3. The set $\left\{x^{i}y^{j}, y^{k}z\; |\; 0\leq i\leq 2n-1, j, k = 0, 1\right\}$ forms a basis of ${\mathcal{H}}_n$.

Proof. By the defining relations of ${\mathcal{H}}_n$, one can see that ${\mathcal{H}}_n$ are spanned by the following:

Therefore, it is sufficient to prove that the set

consists of linearly independent elements.

Now, we assume the following:

Acting on $\mu_0$ by the both sides of $({\rm{I}})$, we have $\sum\limits_{i = 0}^{2n-1} a_{i}\mu_{i} = 0$ and

It yields that

Acting on $\left\{\nu_0, \cdots, \nu_{n-1}, \theta_2, \nu_{n+1}, \cdots\nu_{2n-1}\right\}$ by the both sides of $({\rm{II}})$, we have

and

Hence, obtain the following:

since $1+\eta^{i}\neq0$.

Thus,

It is easy to see that the determinant of the coefficient matrix is nonzero; thus, we have the following:

Moreover, it yields that

Acting on $\left\{\xi_1\right\}$ by the both sides of $({\rm{III}})$, we have

and $c_0 = c_1 = 0.$

In summary,

is a linearly independent set. Hence, it is a basis of ${\mathcal{H}}_n$.

The proof is finished. □

Set $L_{i} = x^{i}$, $M_{i} = x^{i}y$ for $0\leq i\leq 2n-1$, $N_{0} = z$ and $N_{1} = yz$. Let

Then, $\mathcal{B}$ is a basis of ${\mathcal{H}}_n$.

Proposition 2.4. For $0\leq i, j\leq 2n-1$, we have the following$:$

(1) $L_i \cdot L_j = L_j \cdot L_i = L_{i+j(\mathrm{mod}2n)};$

(2) $L_i \cdot M_j = M_j \cdot L_i = M_{i+j(\mathrm{mod}2n)};$

(3) $M_i \cdot M_j = M_j \cdot M_i = M_{i+j(\mathrm{mod}2n)}+ M_{i+j+1(\mathrm{mod}2n)};$

(4) $N_0 \cdot N_0 = N_0 \cdot L_i = L_i\cdot N_0 = N_0;$

(5) $N_0 \cdot N_1 = N_0 \cdot M_i = 2N_0;$

(6) $N_1 \cdot N_0 = M_i \cdot N_0 = N_1\cdot L_i = L_i \cdot N_1 = N_1;$

(7) $N_1 \cdot N_1 = N_1 \cdot M_i = M_i \cdot N_1 = 2N_1.$

Proof. For $0\leq i, j\leq 2n-1$, by a straightforward verification, we have

and

since $x^{2n} = 1, xy = yx$.

Hence, $\rm (1)$ and $\rm (2)$ hold.

(3) By $y^2 = xy+y$, we have the following:

(4) By $z^2 = z$, we have

and

(5) By $zy = 2z$, we have

and

(6) First, we have

and

Moreover, we have

and

Therefore, $N_1 \cdot N_0 = M_i \cdot N_0 = N_1\cdot L_i = L_i \cdot N_1 = N_1.$ (7) First, we have

and

since $zy = 2z$.

Moreover, we have the following:

Hence, $N_1 \cdot N_1 = N_1 \cdot M_i = M_i \cdot N_1 = 2N_1.$ □

Theorem 2.5. The algebra ${\mathcal{H}}_n$ is a positively based algebra.

Proof. By Proposition 2.4, one can easily check that

is a set of positive bases of ${\mathcal{H}}_n$.

The theorem follows. □

3.   The representations of ${\mathcal{H}}_{n}$

From this section, we always assume that $\mathbb{K}$ is a subfield of $\mathbb{C}$ containing $\eta$.

The aim of section is to construct all indecomposable modules of ${\mathcal{H}}_n$ with the help of the representation theory of a quiver. For this purpose, we first consider an algebra $\mathcal{A}$ over the field $\mathbb{K}$, which is generated by $x, y$ with the following relations:

By a straightforward calculation, we see that the following system of equations

has $4n-1$ distinct solutions given by the following:

For each solution $(x_1, x_2)\in\Gamma$, we can define a simple $\mathcal{A}$-module by

Hence, we have the following Lemma.

Lemma 3.1. (1) For $0\leq i\leq 2n-1$, there are $2n$ non-isomorphic $1$-dimensional $\mathcal{A}$-modules $S_{i}$ with the basis $\{v_{i}\}$. The action of $\mathcal{A}$ is given by the following$:$

(2) For $0\leq i\leq 2n-1$ and $i\neq n$, there are $2n-1$ non-isomorphic $1$-dimensional $\mathcal{A}$-modules $W_{i}$ with the basis $\{v_{i}\}$. The action of $\mathcal{A}$ is given by the following$:$

Proof. It is obvious. □

In the following, we describe all indecomposable modules of $\mathcal{A}$.

Let $\mathcal{D}$ be a $2$-dimensional vector space with the basis $\left\{v_{1}, v_{2}\right\}$. We define an action of $\mathcal{A}$ on $\mathcal{D}$ as follows:

Then, $\mathcal{D}$ is an indecomposable $\mathcal{A}$-module.

Proposition 3.2. Any $d$-dimensional, non-simple, indecomposable $\mathcal{A}$-module $V$ is isomorphic to $\mathcal{D}$.

Proof. Assume that $V$ is a $d$-dimensional, non-simple, indecomposable module with $d\geq 2$. Then, End$V$ is local and ${\rm{End}}V/ {\rm{rad }} {\rm{End}}V\cong\mathbb{K}$. Accordingly, we can assume that the matrices of $x$ and $y$ acting on some suitable basis of $V$ are $X = \eta^{i}E_{d}$ and

respectively, since $x^{2n} = 1$ and $xy = yx$.

One can easily see that matrices $X$ and $Y$ satisfy the following:

It follows that $i = n$, $d = 2$ and $\lambda = 1+\eta^{i} = 0.$ Thus, we have $V\cong\mathcal{D}$. □

In the following, set

for $0\leq i\leq 2n-1$.

One can easily check that $\left\{e_{0}, e_{1}, ..., e_{2n-1}\right\}$ is a set of central idempotents of ${\mathcal{H}}_n$, and

Therefore, we have the following:

Now, we can deduce the simple modules and indecomposable modules of ${\mathcal{H}}_n$ by Lemma 3.1 and Proposition 3.2:

$(1)$ For $0\leq i\leq 2n-1$, let $S_{i}$ be a $1$-dimensional ${\mathcal{H}}_n$-module with the basis $\{v_{i}\}$; the action of ${\mathcal{H}}_n$ is given by

$(2)$ For $0\leq i\leq 2n-1$ and $i\neq n$, let $\mathcal{W}_{i}$ be a $1$-dimensional ${\mathcal{H}}_n$-module with the basis $\{v_{i}\}$; the action of ${\mathcal{H}}_n$ is given by

$(3)$ The indecomposable ${\mathcal{H}}_n$-module $\mathcal{V}_{1}$ is produced by extending $\mathcal{D}$, which is given in Lemma 2.2.

Corollary 3.3. $\mathcal{V}_{1}\cong{\mathcal{H}}_ne_{n}$ is the projective cover of $S_n.$

Proof. First, by a straightforward calculation, we have

It follows that $\left\{ye_{n}, e_{n}\right\}$ is a basis of ${\mathcal{H}}_n e_{n}$; the action of ${\mathcal{H}}_n$ is given by

Hence, it is easy to obtain $\mathcal{V}_{1}\cong{\mathcal{H}}_ne_{n}$ and $S_n = {\rm{top}}\mathcal{V}_{1}$ by Lemma 2.2.

The proof is finished. □

Lemma 3.4. ${\mathcal{H}}_n z\cong \mathcal{V}_{2}$ is a $2$-dimensional indecomposable projective ${\mathcal{H}}_n$-module.

Proof. First, it is easy to have ${\mathcal{H}}_n z$ is projective, since ${\mathcal{H}}_n = {\mathcal{H}}_nz\oplus {\mathcal{H}}_n(1-z)$. Moreover, we have

It follows that $\left\{z, yz\right\}$ is a basis of ${\mathcal{H}}_n z$; the action of ${\mathcal{H}}_n$ is given by

By Lemma 2.2, it is easy to obtain that ${\mathcal{H}}_n z\cong \mathcal{V}_{2}$.

Suppose that $\mathcal{V}_{2} = A_1\oplus A_2$, where $A_1$ and $A_2$ are nonzero. One has $0\ne \varphi = k_1z+k_2yz\in A_{1}$ for some $k_1\neq 0$ or $k_2\ne 0$. Then,

If $k_1+2k_2 = 0$, we get that $\varphi = k_2(2z-yz)\in A_1$ with $k_2\ne 0$ and $2z-yz\in A_1$. Accordingly, the decomposition $yz = (yz-2z)+2z$ implies that $2z\in A_2$ and $y\cdot z = yz\in A_2$. Hence, $A_2 = \mathcal{V}_{2}$, which is a contradiction. This implies that $k_1+2k_2\ne 0$, $yz\in A_1$ and $z\in A_1$. Thus, we have $A_1 = \mathcal{V}_{2}$, which is also a contradiction.

Therefore, ${\mathcal{H}}_n z\cong \mathcal{V}_{2}$ is an indecomposable projective module. □

Assume $S = \mathbb{K}v$ is a $1$-dimensional ${\mathcal{H}}_n$-module defined by

Reviewing the action of ${\mathcal{H}}_n$ on $\mathcal{V}_{2}$, we let $\xi'_{1} = 2\xi_1-\xi_2$, $\xi'_{2} = \xi_2$; then the action of ${\mathcal{H}}_n$ can be written as follows:

We get that $S = \hbox{top}\mathcal{V}_{2}$ and the exact sequence

It concludes the following theorem.

Theorem 3.5. The following is a complete set of all simple ${\mathcal{H}}_n$-modules up to isomorphism:

(1) One non-projective simple module $S_{n}$ with the projective cover $\mathcal{V}_{1}$$;$

(2) One non-projective simple module $S$ with the projective cover $\mathcal{V}_2$$;$

(3) $4n-2$ projective simple modules $S_{i}$ and $\mathcal{W}_{i}$, where $0\leq i\leq 2n-1$ and $i\neq n$.

We also need more preparations to list all the indecomposable ${\mathcal{H}}_n$-modules.

Lemma 3.6. ${\mathcal{H}}_n(e_{0}-z) = S_{0}\oplus \mathcal{W}_{0}$.

Proof. For $0\leq i\leq 2n-1$, we have

and

Set $v_{1} = 2e_{0}-2z-ye_{0}+yz$, $v_{2} = ye_{0}-yz$; then, $\left\{v_{1}, v_{2}\right\}$ is a basis of ${\mathcal{H}}_n(e_{0}-z)$ and

By Lemma 2.2, ${\mathcal{H}}_n(e_{0}-z) = S_{0}\oplus \mathcal{W}_{0}$.

The result follows. □

Lemma 3.7. For $1\leq i\leq 2n-1$, we have the following$:$

(1) ${{{{\rm{Hom}}}}}_{{\mathcal{H}}_n}({\mathcal{H}}_n e_{i}, {\mathcal{H}}_n e_{0}) = 0;$

(2) ${{{{\rm{Hom}}}}}_{{\mathcal{H}}_n}({\mathcal{H}}_n e_{0}, {\mathcal{H}}_n e_{i}) = 0.$

Proof. (1) By [2,Lemma 4.2], we have

since $e_{i}e_{0} = e_{0}e_{i} = 0$, and $xy = yx, xz = zx$ for $1\leq i\leq 2n-1.$

The proof of (2) is similar to (1).□

Lemma 3.8. (1) ${{{{\rm{Hom}}}}}_{{\mathcal{H}}_n}(\mathcal{V}_{2}, \mathcal{W}_{0}) = 0$, ${{{{\rm{Hom}}}}}_{{\mathcal{H}}_n}(\mathcal{V}_{2}, S_{n}) = 0$$;$

(2) ${{{{\rm{Hom}}}}}_{{\mathcal{H}}_n}(S_{0}, \mathcal{V}_{2})\neq0$, ${{{{\rm{Hom}}}}}_{{\mathcal{H}}_n}(\mathcal{W}_{0}, \mathcal{V}_{2}) = 0$.

Proof. $\rm (1)$ By Lemmas 3.4 and 3.8, we have

Similarly,

$\rm (2)$ By (3.2), $S_{0}$ is the socle of $\mathcal{V}_{2}$. Thus,

For $\mathcal{W}_{0}$, the result is obvious.□

According to Lemmas 3.6–3.8, one can easily see that $\mathcal{V}_{2}\oplus S_{0}$ is a block of ${\mathcal{H}}_n$. Meanwhile, the indecomposable modules of other blocks are either $1$-dimensional or $2$-dimensional. Therefore, we only need to consider the indecomposable modules of the block $\mathcal{C} = \mathcal{V}_{2}\oplus S_{0}$.

Proposition 3.9. The quiver of the block $\mathcal{C}$ is as follows$:$

Proof. On the one hand, $\mathcal{V}_{2}$ is the projective cover of $S$, and $S_{0}$ is projective by Lemma 3.4. Therefore,

On the other hand, we have the following extension of $S_{0}$ by $S$:

Assume that

is another extension of $S_{0}$ by $S$, where $Q$ is indecomposable of the basis $\{v_{1}, v_{2} \}$; the action of ${\mathcal{H}}_n$ is given by

with $k\neq 0$.

Now, let $h:\mathcal{V}_{2}\rightarrow Q$, which is given by $\xi'_{1}\mapsto k v_{1}$, $\xi'_{2}\mapsto v_{2}$. Then, $h$ is an isomorphism. It follows that

The result follows. □

Theorem 3.10. The algebra ${\mathcal{H}}_n$ has $4n+2$ pairwise, non-isomorphic, indecomposable modules$:$

Proof. By Proposition 3.9, the quiver of block $\mathcal{C}$ is as follows:

It is easy to see that this block has non-isomorphic, indecomposable modules $S_{0}$, $S$ and $\mathcal{V}_{2}$. By Theorem 3.5, we get $4n+2$ pairwise, non-isomorphic, indecomposable modules of ${\mathcal{H}}_n$ as the aforementioned list. □

4.   The left cell modules of ${\mathcal{H}}_{n}$

Let $A$ be a positively based algebra with a fixed positive basis $\mathcal{B} = \{a_{i}\; |\; i \in I\}$ with the identity $a_{1}$ of $A$. For $i, j\in I$, set

This defines an associative multi-valued operation on the set $I$ and turns the latter set into a finite multi-semigroup, see [9,Subsection 3.7]. If there is an $s\in I$ such that $j\in s\star i$, then we denoted it by $i\leq_{L}j$. For $i, j\in I$, if $i\leq_{L}j$ and $j\leq_{L}i$, we denote it by $i\thicksim_{L}j$, which is an equivalent relation. The associated equivalence class is referred to as a left cell. Moreover, we write $i < _{L}j$, provided that $i\leq_{l}j$ and $i\nsim_{L}j.$

Assume that $\mathcal{L}$ is a left cell in $I$, and let $\overline{\mathcal{L}}$ be the union of all left cells $\mathcal{L'}$ in $I$ such that $\mathcal{L'}\geq\mathcal{L}.$ Set $\overline{\underline{\mathcal{L}}} = \overline{\mathcal{L}}\backslash\mathcal{L}.$ Let $M_{\mathcal{L}}$ be the vector space spanned by $a_{j}$, where $j\in\overline{\mathcal{L}},$ and $N_{\mathcal{R}}$ be the vector space spanned by $a_{j}$ with $j\in\overline{\underline{\mathcal{L}}}.$ According to [8,Proposition 1], $M_{\mathcal{L}}$ and $N_{\mathcal{L}}$ are submodules of $_{A}A$ and $N_{\mathcal{L}}\subset M_{\mathcal{L}}$. It allows us to define the cell module $C_{\mathcal{R}} = M_{\mathcal{L}}/ N_{\mathcal{L}}$. Especially, if $\overline{\underline{\mathcal{L}}} = \emptyset$, denote it by $N_{\mathcal{L}} = 0$. For the study of cells and cell modules, the readers can refer to ^{[6,7,8,11,12,13]}.

Now, we investigate the left (right) cells and left (right) cell modules of ${\mathcal{H}}_n$. By Theorem 2.5,

is a positive basis of ${\mathcal{H}}_n$.

Proposition 4.1. The algebra ${\mathcal{H}}_n$ has three left cells $\mathcal{L}_{1}$, $\mathcal{L}_{2}$ and $\mathcal{L}_{3}$, as listed in the following$;$

(1) $\mathcal{L}_{1} = \left\{i\; |\; i~~ {\mathit{{is ~~ the ~~ index~~ of}}}~~ L_i, i\in \mathbb{Z}_{2n}\right\};$

(2) $\mathcal{L}_{2} = \left\{j\; |\; j~~ {\mathit{{is~~ the~~ index ~~ of}}}~~ M_j, j\in \mathbb{Z}_{2n}\right\};$

(3) $\mathcal{L}_{3} = \left\{k\; |\; k~~ {\mathit{{is ~~ the ~~ index ~~ of}}}~~ N_k, k = 0, 1\right\}.$

Proof. (1) Set $i, i'\in \mathbb{Z}_{2n}$ and $i < i'$. We obtain $L_{i'-i}\cdot L_{i} = L_{i'}$ by Proposition 2.4. It follows that $i'\in (i'-i)\star i$, which implies that $i\leq_{L} i'$. Similarly, we have $L_{2n-i'+i}\cdot L_{i'} = L_{i}$; then, $i\in (2n-i'+i)\star i'$ and $i'\leq_{L} i$. The equivalent relation $i\sim_{L}i'$ holds and $\mathcal{L}_{1}$ is a left cell.

(2) Set $j, j'\in \mathbb{Z}_{2n}$ and $j < j'$. We have $j'\in (j'-j)\star j$ and $j\leq_{L}j'$, since $M_{j'-j}\cdot M_{j} = M_{j'}+M_{j'+1}$. Similarly, we obtain that $j\in (2n+j-j')\star j'$ and $j'\leq_{L}j$. Hence, $\mathcal{L}_{2}$ is a left cell.

(3) By Proposition 2.4, it is obvious that $N_0\cdot N_1 = 2N_0$ and $N_1\cdot N_0 = N_1$. Thus, we have $1\leq_{L}0$ and $0\leq_{L} 1$. Therefore, $\mathcal{L}_{3}$ is a left cell.

The proof is completed. □

Corollary 4.2. For the left cells of ${\mathcal{H}}_n$, we have $\mathcal{L}_{1} < _{L}\mathcal{L}_{2} < _{L}\mathcal{L}_{3}$.

Proof. For $i, j\in {\mathbb Z}_{2n}$, we have $M_j\cdot L_{i} = M_{i+j(\mathrm{mod}2n)}$ by Proposition 2.4. It is clear that $\mathcal{L}_{1} < _{L}\mathcal{L}_{2}$.

Furthermore, by Proposition 2.4, it is easy to see that

Thus, $\mathcal{L}_{2} < _{L}\mathcal{L}_{3}$ holds.

Consequently, we have $\mathcal{L}_{1} < _{L}\mathcal{L}_{2} < _{L}\mathcal{L}_{3}$. □

Proposition 4.3. For the left cells $\mathcal{L}_{1}$, $\mathcal{L}_{2}$ and $\mathcal{L}_{3}$, the corresponding left cell modules $\mathcal{C}_{\mathcal{L}_{1}}$, $\mathcal{C}_{\mathcal{L}_{2}}$ and $\mathcal{C}_{\mathcal{L}_{3}}$ are given as follows$:$

(1) $\mathcal{C}_{\mathcal{L}_{1}} = {{{{{\rm{Span}}}}}}\left\{\overline{L_i}\; |\; i\in \mathbb{Z}_{2n}\right\}$, where $\overline{L_i} = {L_i+\mathcal{N}_{\mathcal{L}_{1}}}$, $\mathcal{N}_{\mathcal{L}_{1}} = {{{{{\rm{Span}}}}}}\left\{M_j, N_k\; |\; j\in \mathbb{Z}_{2n}, k = 0, 1\right\};$

(2) $\mathcal{C}_{\mathcal{L}_{2}} = {{{{{\rm{Span}}}}}}\left\{\overline{M_j}\; |\; j\in \mathbb{Z}_{2n}\right\}$, where $\overline{M_j} = {M_j+\mathcal{N}_{\mathcal{L}_{2}}}$, and

(3) $\mathcal{C}_{\mathcal{L}_{3}} = {{{{{\rm{Span}}}}}}\left\{N_k\; |\; k = 0, 1\right\}$, where $\mathcal{N}_{\mathcal{L}_{3}} = \{0\}.$

Proof. (1) As is shown in Corollary 4.2, we have

By Proposition 4.1, it follows that

and

Hence,

(2) By Corollary 4.2, we have

It follows that

and

Therefore,

(3) By Proposition 4.1 and Corollary 4.2, we have

It follows that

Hence,

The proof is finished. □

By Proposition 4.3, we see that $\mathcal{C}_{\mathcal{L}_{1}} = {{{{{\rm{Span}}}}}}\left\{\overline{L_i}\; |\; i\in \mathbb{Z}_{2n}\right\}$ and write $\mu_i = \overline{L_i}$ with $0\leq i\leq 2n-1$. Then, the action of ${\mathcal{H}}_n$ on $\mathcal{C}_{\mathcal{L}_{1}}$ is defined by

Now, set

where $0\leq i\leq 2n-1$.

A straightforward calculation shows that

Theorem 4.4. $\mathcal{C}_{\mathcal{L}_{1}}$ is decomposable and

Proof. By Lemma 2.2, the 1-dimensional ${\mathcal{H}}_n$-module $\mathbb{K}\chi_i = S_{i}$. Hence, $S_{i}$ is a submodule of $\mathcal{C}_{\mathcal{L}_{1}}$. We easily see that

and conclude

The proof is completed. □

For the left cell module $\mathcal{C}_{\mathcal{L}_{2}}$, we have $\mathcal{C}_{\mathcal{L}_{2}} = {{{{{\rm{Span}}}}}}\left\{\overline{M_j}\; |\; j\in \mathbb{Z}_{2n}\right\}$ and denote $\nu_j = \overline{M_j}$ with $0\leq j\leq\; 2n-1$. It is easy to see that

Set

for $0\leq j\leq 2n-1$.

Then, a straightforward calculation shows that

Theorem 4.5. $\mathcal{C}_{\mathcal{L}_{2}}$ is decomposable and

Proof. On the one hand, by Lemma 2.2, we see that the ${\mathcal{H}}_n$-module $\mathbb{K}\omega_j\cong\mathcal{W}_{j}$ when $j \neq n$, and $\mathbb{K}\omega_n\cong S_{n}$. Hence, $\mathcal{W}_{j}$ is a submodule of $\mathcal{C}_{\mathcal{L}_{2}}$ for $0\leq j\leq 2n-1$ and $j \neq n$.

On the other hand, it is easy to see that

Hence,

The proof is finished. □

Theorem 4.6. $\mathcal{C}_{\mathcal{L}_{3}}\cong \mathcal{V}_2$ is indecomposable.

Proof. By Proposition 4.3, for the left cell module $\mathcal{C}_{\mathcal{L}_{3}}$, we have $\mathcal{C}_{\mathcal{L}_{3}} = {{{{{\rm{Span}}}}}}\left\{N_k\; |\; k = 0, 1\right\}$, on which the action of ${\mathcal{H}}_n$ is given by

Hence, $\mathcal{C}_{\mathcal{L}_{3}}\cong \mathcal{V}_2$ by Lemma 3.4.

The result follows. □

Finally, we give some remarks for the right cell modules of ${\mathcal{H}}_n$. For the positive basis $\mathcal{B}$, if there is an $s\in I$ such that $j\in i\star s,$ then one can denote it by $i\leq_{R}j$. If $i\leq_{R}j$ and $j\leq_{R}i,$ then one can denote it by $i\thicksim_{R}j$, which is an equivalent relation. The equivalence class is called a right cell. The similar statement to Proposition 4.1 shows that ${\mathcal{H}}_n$ has the following four right cells:

$(1)$ $\mathcal{R}_{1} = \left\{i\; |\; i{\mbox{ is the index of }}S_i, i\in \mathbb{Z}_{2n}\right\};$

$(2)$ $\mathcal{R}_{2} = \left\{j\; |\; j{\mbox{ is the index of }}M_j, j\in \mathbb{Z}_{2n}\right\};$

$(3)$ $\mathcal{R}_{3} = \left\{0\; |\; 0{\mbox{ is the index of }}N_0\right\};$

$(4)$ $\mathcal{R}_{4} = \left\{1\; |\; 1{\mbox{ is the index of }}N_1\right\}.$

We see that $\mathcal{R}_{1} < _{R}\mathcal{R}_{2} < _{R}\mathcal{R}_{3}$ and $\mathcal{R}_{1} < _{R}\mathcal{R}_{2} < _{R}\mathcal{R}_{4}.$ Consequently, we get the following:

$(1)$ $\mathcal{C}_{\mathcal{R}_{1}} = {{{{{\rm{Span}}}}}}\left\{\mu_i\; |\; i\in \mathbb{Z}_{2n}\right\}$, on which the action of ${\mathcal{H}}_n$ is given by

$(2)$ $\mathcal{C}_{\mathcal{R}_{2}} = {{{{{\rm{Span}}}}}}\left\{\nu_j\; |\; j\in \mathbb{Z}_{2n}\right\}$, on which the action of ${\mathcal{H}}_n$ is given by

$(3)$ $\mathcal{C}_{\mathcal{R}_{3}} = {{{{{\rm{Span}}}}}}\left\{\xi_1\right\}$, on which the action of ${\mathcal{H}}_n$ is given by

$(4)$ $\mathcal{C}_{\mathcal{R}_{4}}\cong \mathcal{C}_{\mathcal{R}_{3}}$.

5.   Conclusions

In the paper, all indecomposable modules of ${\mathcal{H}}_n$, a family of positively based algebras, are constructed and classified. Also, their left cell modules are described. In our further study, we will focus on the family of positive based algebras associated to the Green algebras of the dual of the generalized Taft algebra. These results may help us to understand the general representation theory of a positive based algebra.

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

Conflict of interest

All authors declare no conflict of interest in this paper.