Weak Hopf Algebra Structures on Hybrid Numbers

Abstract

Let $\mathbb{K}$ be the algebra of hybrid numbers. In this paper, a weak Hopf algebra structure is endowed on $\mathbb{K}$. Additionally, the integrals in this weak Hopf algebra $\mathbb{K}$ are discussed.

1. Introduction

Weak Hopf algebras were introduced by Böhm, Nill and Szlachányi as a generalization of both ordinary Hopf algebras and groupoid algebras [1], providing powerful tools for studying partial or local symmetries in mathematics and physics. In the definition of a weak Hopf algebra, the conditions that $∆$, $\epsilon$, and the antipode S satisfy are relaxed; for instance, the comultiplication of a weak Hopf algebra is no longer required to preserve the unit. The axioms governing the weak Hopf algebra exhibit self-dual characteristics, which guarantee that for any finite dimensional weak Hopf algebra H, its dual vector space $H^{\ast }$ naturally acquires the structure of a weak Hopf algebra. The study of a weak Hopf algebra is motivated by its association with the theory of algebra extensions [2] and its important applications in providing a natural framework for the study of dynamical twists in Hopf algebras [3].

Hybrid numbers, a generalization of complex, hyperbolic, and dual numbers, were introduced in [4], unifying multiple number systems and their symmetries. A hybrid number is defined as any combination of complex, hyperbolic, and dual numbers that satisfies a certain relation. Hybrid numbers are mathematically intriguing as they reside at the intersection of multiple number systems, providing fertile ground for algebraic generalizations. Further research may reveal deeper connections with deformation quantization, gauge theory, or innovative computational techniques. In recent years, significant progress has been made in the study of hybrid numbers [5,6,7,8,9,10,11,12,13,14]. In [5], with the help of the Fibonacci divisor, Fibonacci divisor hybrid numbers were introduced and the miscellaneous algebraic properties of the Fibonacci divisor hybrid numbers were obtained. Tasci and Sevgi defined Mersenne hybrid numbers inspired by the definition of hybrid numbers and gave some algebraic properties of Mersenne, Jacobsthal, and Jacobsthal–Lucas hybrid numbers in [7]. In [12], the authors provided the Euler and De Moivre formulas for the 4 × 4 matrices associated with hybrid numbers by using trigonometric identities and gave the roots of the matrices of hybrid numbers.

A primary objective of this paper is to explore the weak Hopf algebra structures on the algebras of hybrid numbers, denoted by $\mathbb{K}$. By discussing the weak Hopf algebra structure on the algebras of hybrid numbers, a new direction for the study of hybrid numbers has been opened.

This paper is organized as follows: In Section 2, a comprehensive review of the fundamental definitions and properties concerning weak Hopf algebras and hybrid numbers is provided. In Section 3, we discuss weak Hopf algebra structures on $\mathbb{K}$ and compute the source and target algebras for such structures. In Section 4, we examine left and right integrals in such weak Hopf algebras.

2. Preliminaries

Throughout this paper, all vector spaces, linear maps, and tensor products are over $\mathbb{C}$. The terminology regarding algebra, coalgebra, and modules is taken from Montgomery’s book [15]; however, conventional summation indices and symbols are omitted.

2.1. Weak Hopf Algebras

A six-tuple $(H,m,1,∆,\epsilon ,S)$ is a weak Hopf algebra with the antipode S if $(H,m,1)$ is an algebra and $(H,∆,\epsilon )$ is a coalgebra, and the following conditions hold:

$$∆\left(kh\right)=∆\left(k\right)∆\left(h\right),\forall h,k\in H,$$

(1)

$$\epsilon \left(k{h}_{\left(1\right)}\right)\epsilon \left(h_{\left(2\right)}g\right)=\epsilon \left(khg\right)=\epsilon \left(k{h}_{\left(2\right)}\right)\epsilon \left(h_{\left(1\right)}g\right),\forall h,k,g\in H,$$

(2)

$$(1\otimes ∆\left(1\right))(∆\left(1\right)\otimes 1)={∆}^2\left(1\right)=(∆\left(1\right)\otimes 1)(1\otimes ∆\left(1\right)),$$

(3)

$$h_{\left(1\right)}S\left(h_{\left(2\right)}\right)={\epsilon }_t\left(h\right),S\left(h_{\left(1\right)}\right)h_{\left(2\right)}={\epsilon }_s\left(h\right),S\left(h_{\left(1\right)}\right)h_{\left(2\right)}S\left(h_{\left(3\right)}\right)=S\left(h\right),$$

(4)

where ${\epsilon }_t,{\epsilon }_s:H\to H$ are defined by ${\epsilon }_t\left(h\right)=\epsilon \left(1_{\left(1\right)}h\right)1_{\left(2\right)},{\epsilon }_s\left(h\right)=\epsilon \left(h{1}_{\left(2\right)}\right)1_{\left(1\right)}$. We denote $H_t={\epsilon }_t\left(H\right)$ and $H_s={\epsilon }_s\left(H\right)$.

Recall from [16] that an algebra A is separable if and only if there is a $q\in A\otimes A$ such that

$$(x\otimes 1)q=q(1\otimes x)$$

holds, for all $x\in A$ and, furthermore, $m_A\left(q\right)=1$. Such a q will be called a separable idempotent. Let H be a weak Hopf algebra. Based on Proposition 2.11 in [1], we know that $H_t\left(H_s\right)$ is a separable algebra with a separable idempotent given by

$$q=S\left(1_{\left(1\right)}\right)\otimes 1_{\left(2\right)}(q=1_{\left(1\right)}\otimes S\left(1_{\left(2\right)}\right)).$$

A left (right) integral in a weak Hopf algebra H is an element $l\in H(r\in H)$ satisfying

$$xl={\epsilon }_t\left(x\right)l(rx=r{\epsilon }_s\left(x\right)),$$

for all $x\in H$. The space of left (right) integrals in H is denoted by $I^L\left(H\right)$$\left(I^R\left(H\right)\right)$. $l\in I^L\left(H\right)$ is called normalized if ${\epsilon }_t\left(l\right)=1$. For a weak Hopf algebra H, based on Theorem 3.13 in [1], we know that H is semisimple if and only if there is a normalized left integral $l\in H$.

2.2. Hybrid Numbers

The algebras of hybrid numbers, denoted by $\mathbb{K}$, are defined as

$$\mathbb{K}=\{a+bg+c\mu +d\nu |a,b,c,d\in \mathbb{C},g^2=-1,{\mu }^2=0,{\nu }^2=1,g\nu =-\nu g=\mu +g\}.$$

Using these equalities

$$g^2=-1,{\mu }^2=0,{\nu }^2=1,g\nu =-\nu g=\mu +g,$$

we can derive the following multiplication table:

	1	g	μ	v
1	1	g	μ	v
g	g	−1	1 − v	μ + g
μ	μ	v + 1	0	−μ
v	v	−μ − g	μ	1

As demonstrated in the above table, the multiplication operation in the hybrid numbers is not commutative; however, it does possess the property of associativity.

3. Weak Hopf Algebra Structures on $\mathbb{K}$

In this section, we discuss when $\mathbb{K}$ becomes a weak Hopf algebra. First, we address the issue of the coalgebra structure on $\mathbb{K}$.

Lemma 1.

For a fixed $b\in \mathbb{C}$, $\mathbb{K}$ is endowed with the coalgebra structure as follows:

$$\left\{\begin{array}{cc}∆\left(1\right)={\displaystyle \frac{1}{2}}1\otimes 1-{\displaystyle \frac{1}{2}}\mu \otimes \mu -{\displaystyle \frac{i}{2}}\nu \otimes \mu -{\displaystyle \frac{i}{2}}\mu \otimes \nu +{\displaystyle \frac{1}{2}}\nu \otimes \nu ,\hfill & \\ ∆\left(g\right)=-{\displaystyle \frac{1}{2}}\mu \otimes 1-{\displaystyle \frac{i}{2}}\nu \otimes 1+bg\otimes g+b\mu \otimes g+ib\nu \otimes g-{\displaystyle \frac{1}{2}}1\otimes \mu +bg\otimes \mu +b\mu \otimes \mu \hfill \\ +ib\nu \otimes \mu -{\displaystyle \frac{i}{2}}1\otimes \nu +ibg\otimes \nu +ib\mu \otimes \nu -b\nu \otimes \nu ,\hfill & \\ ∆\left(\mu \right)={\displaystyle \frac{1}{2b}}\mu \otimes \mu ,\hfill & \\ ∆\left(\nu \right)=-{\displaystyle \frac{i}{2}}\mu \otimes 1+{\displaystyle \frac{1}{2}}\nu \otimes 1-{\displaystyle \frac{i}{2}}1\otimes \mu +{\displaystyle \frac{i}{2b}}\mu \otimes \mu +{\displaystyle \frac{1}{2}}1\otimes \nu ,\hfill & \\ \epsilon \left(1\right)=2,\epsilon \left(g\right)={\displaystyle \frac{1}{b}},\epsilon \left(\mu \right)=2b,\epsilon \left(\nu \right)=2ib.\hfill \end{array}\right.$$

Proof. 

First, we verify that, for all $x\in \{1,g,\mu ,\nu \}$, the following equality holds:

$$(∆\otimes id)∆\left(x\right)=(id\otimes ∆)∆\left(x\right).$$

For $x=\nu$, on one hand, we have

$$\begin{array}{ccc}\hfill (∆\otimes id)∆\left(\nu \right)& =& (∆\otimes id)(-{\displaystyle \frac{i}{2}}\mu \otimes 1+{\displaystyle \frac{1}{2}}\nu \otimes 1-{\displaystyle \frac{i}{2}}1\otimes \mu +{\displaystyle \frac{i}{2b}}\mu \otimes \mu +{\displaystyle \frac{1}{2}}1\otimes \nu )\hfill \\ & =& -{\displaystyle \frac{i}{4}}\mu \otimes 1\otimes 1+{\displaystyle \frac{1}{4}}\nu \otimes 1\otimes 1-{\displaystyle \frac{i}{4}}1\otimes \mu \otimes 1+{\displaystyle \frac{1}{4}}1\otimes \nu \otimes 1\hfill \\ & & -{\displaystyle \frac{i}{4}}1\otimes 1\otimes \mu +{\displaystyle \frac{i{b}^2+i}{4b^2}}\mu \otimes \mu \otimes \mu -{\displaystyle \frac{1}{4}}\nu \otimes \mu \otimes \mu \hfill \\ & & -{\displaystyle \frac{1}{4}}\mu \otimes \nu \otimes \mu -{\displaystyle \frac{i}{4}}\nu \otimes \nu \otimes \mu +{\displaystyle \frac{1}{4}}1\otimes 1\otimes \nu \hfill \\ & & -{\displaystyle \frac{1}{4}}\mu \otimes \mu \otimes \nu -{\displaystyle \frac{i}{4}}\nu \otimes \mu \otimes \nu -{\displaystyle \frac{i}{4}}\mu \otimes \nu \otimes \nu +{\displaystyle \frac{1}{4}}\nu \otimes \nu \otimes \nu .\hfill \end{array}$$

On the other hand, we obtain

$$\begin{array}{ccc}\hfill (id\otimes ∆)∆\left(\nu \right)& =& (id\otimes ∆)(-{\displaystyle \frac{i}{2}}\mu \otimes 1+{\displaystyle \frac{1}{2}}\nu \otimes 1-{\displaystyle \frac{i}{2}}1\otimes \mu +{\displaystyle \frac{i}{2b}}\mu \otimes \mu +{\displaystyle \frac{1}{2}}1\otimes \nu )\hfill \\ & =& -{\displaystyle \frac{i}{4}}\mu \otimes 1\otimes 1+{\displaystyle \frac{i{b}^2+i}{4b^2}}\mu \otimes \mu \otimes \mu -{\displaystyle \frac{1}{4}}\mu \otimes \nu \otimes \mu -{\displaystyle \frac{1}{4}}\mu \otimes \mu \otimes \nu \hfill \\ & & -{\displaystyle \frac{i}{4}}\mu \otimes \nu \otimes \nu +{\displaystyle \frac{1}{4}}\nu \otimes 1\otimes 1-{\displaystyle \frac{1}{4}}\nu \otimes \mu \otimes \mu \hfill \\ & & -{\displaystyle \frac{i}{4}}\nu \otimes \nu \otimes \mu -{\displaystyle \frac{i}{4}}\nu \otimes \mu \otimes \nu +{\displaystyle \frac{1}{4}}\nu \otimes \nu \otimes \nu \hfill \\ & & -{\displaystyle \frac{i}{4}}1\otimes \mu \otimes 1+{\displaystyle \frac{1}{4}}1\otimes \nu \otimes 1-{\displaystyle \frac{i}{4}}1\otimes 1\otimes \mu +{\displaystyle \frac{1}{4}}1\otimes 1\otimes \nu .\hfill \end{array}$$

Comparing the above two identities, we have

$$(∆\otimes id)∆\left(\nu \right)=(id\otimes ∆)∆\left(\nu \right).$$

Since

$$\begin{array}{ccc}& & (id\otimes \epsilon )∆\left(g\right)\hfill \\ & =& (id\otimes \epsilon )(-{\displaystyle \frac{1}{2}}\mu \otimes 1-{\displaystyle \frac{i}{2}}\nu \otimes 1+bg\otimes g+b\mu \otimes g+ib\nu \otimes g\hfill \\ & & -{\displaystyle \frac{1}{2}}1\otimes \mu +bg\otimes \mu +b\mu \otimes \mu +ib\nu \otimes \mu -{\displaystyle \frac{i}{2}}1\otimes \nu \hfill \\ & & +ibg\otimes \nu +ib\mu \otimes \nu -b\nu \otimes \nu )\hfill \\ & =& -\mu -i\nu +g+\mu +i\nu -b1+2b^{2}g+2b^2\mu +2i{b}^2\nu +b1-2b^{2}g-2b^2\mu -2i{b}^2\nu \hfill \\ & =& g\hfill \end{array}$$

and

$$\begin{array}{ccc}& & (\epsilon \otimes id)∆\left(g\right)\hfill \\ & =& (\epsilon \otimes id)(-{\displaystyle \frac{1}{2}}\mu \otimes 1-{\displaystyle \frac{i}{2}}\nu \otimes 1+bg\otimes g+b\mu \otimes g+ib\nu \otimes g\hfill \\ & & -{\displaystyle \frac{1}{2}}1\otimes \mu +bg\otimes \mu +b\mu \otimes \mu +ib\nu \otimes \mu -{\displaystyle \frac{i}{2}}1\otimes \nu \hfill \\ & & +ibg\otimes \nu +ib\mu \otimes \nu -b\nu \otimes \nu )\hfill \\ & =& -b1+b1+g+2b^{2}g-2b^{2}g-\mu +\mu +2b^2\mu -2b^2\mu -i\nu +i\nu +2i{b}^2\nu -2i{b}^2\nu \hfill \\ & =& g,\hfill \end{array}$$

it follows that

$$(id\otimes \epsilon )∆\left(g\right)=g=(\epsilon \otimes id)∆\left(g\right).$$

Similarly, we have

$$(id\otimes \epsilon )∆\left(x\right)=x=(\epsilon \otimes id)∆\left(x\right),x\in \{1,\mu ,\nu \}.$$

Thus, we obtain

$$(id\otimes \epsilon )∆=id=(\epsilon \otimes id)∆.$$

 □

Lemma 2.

The comultiplication Δ is given in Lemma 1. Then, Δ satisfies the conditions in (1) and (3).

Proof. 

Since

$$\begin{array}{ccc}& & (1\otimes ∆(1))(∆(1)\otimes 1)\hfill \\ & =& (1\otimes ({\displaystyle \frac{1}{2}}1\otimes 1-{\displaystyle \frac{1}{2}}\mu \otimes \mu -{\displaystyle \frac{i}{2}}\nu \otimes \mu -{\displaystyle \frac{i}{2}}\mu \otimes \nu +{\displaystyle \frac{1}{2}}\nu \otimes \nu ))\hfill \\ & & (({\displaystyle \frac{1}{2}}1\otimes 1-{\displaystyle \frac{1}{2}}\mu \otimes \mu -{\displaystyle \frac{i}{2}}\nu \otimes \mu -{\displaystyle \frac{i}{2}}\mu \otimes \nu +{\displaystyle \frac{1}{2}}\nu \otimes \nu )\otimes 1)\hfill \\ & =& {\displaystyle \frac{1}{4}}1\otimes 1\otimes 1-{\displaystyle \frac{1}{4}}\mu \otimes \mu \otimes 1-{\displaystyle \frac{i}{4}}\nu \otimes \mu \otimes 1-{\displaystyle \frac{i}{4}}\mu \otimes \nu \otimes 1+{\displaystyle \frac{1}{4}}\nu \otimes \nu \otimes 1\hfill \\ & & -{\displaystyle \frac{1}{4}}1\otimes \mu \otimes \mu -{\displaystyle \frac{i}{4}}1\otimes \nu \otimes \mu -{\displaystyle \frac{i}{4}}1\otimes \mu \otimes \nu +{\displaystyle \frac{1}{4}}1\otimes \nu \otimes \nu \hfill \\ & & -{\displaystyle \frac{1}{4}}\mu \otimes 1\otimes \mu -{\displaystyle \frac{i}{4}}\nu \otimes 1\otimes \mu -{\displaystyle \frac{i}{4}}\mu \otimes 1\otimes \nu +{\displaystyle \frac{1}{4}}\nu \otimes 1\otimes \nu \hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill {∆}^2\left(1\right)& =& (id\otimes ∆)({\displaystyle \frac{1}{2}}1\otimes 1-{\displaystyle \frac{1}{2}}\mu \otimes \mu -{\displaystyle \frac{i}{2}}\nu \otimes \mu -{\displaystyle \frac{i}{2}}\mu \otimes \nu +{\displaystyle \frac{1}{2}}\nu \otimes \nu )\hfill \\ & =& {\displaystyle \frac{1}{4}}1\otimes 1\otimes 1-{\displaystyle \frac{1}{4}}1\otimes \mu \otimes \mu -{\displaystyle \frac{i}{4}}1\otimes \nu \otimes \mu -{\displaystyle \frac{i}{4}}1\otimes \mu \otimes \nu \hfill \\ & & +{\displaystyle \frac{1}{4}}1\otimes \nu \otimes \nu -{\displaystyle \frac{1}{4}}\mu \otimes \mu \otimes 1-{\displaystyle \frac{i}{4}}\mu \otimes \nu \otimes 1\hfill \\ & & -{\displaystyle \frac{1}{4}}\mu \otimes 1\otimes \mu -{\displaystyle \frac{i}{4}}\mu \otimes 1\otimes \nu -{\displaystyle \frac{i}{4}}\nu \otimes \mu \otimes 1+{\displaystyle \frac{1}{4}}\nu \otimes \nu \otimes 1\hfill \\ & & -{\displaystyle \frac{i}{4}}\nu \otimes 1\otimes \mu +{\displaystyle \frac{1}{4}}\nu \otimes 1\otimes \nu ,\hfill \end{array}$$

it follows that

$$(1\otimes ∆\left(1\right))(∆\left(1\right)\otimes 1)={∆}^2\left(1\right).$$

Similarly, we can check that

$${∆}^2\left(1\right)=(∆\left(1\right)\otimes 1)(1\otimes ∆\left(1\right))$$

holds.

Now, we shall check that $∆$ preserves the multiplication. For example, we check the following:

$$\begin{array}{ccc}& & ∆\left(g\right)∆\left(\nu \right)\hfill \\ & =& (-{\displaystyle \frac{1}{2}}\mu \otimes 1-{\displaystyle \frac{i}{2}}\nu \otimes 1+bg\otimes g+b\mu \otimes g+ib\nu \otimes g\hfill \\ & & -{\displaystyle \frac{1}{2}}1\otimes \mu +bg\otimes \mu +b\mu \otimes \mu +ib\nu \otimes \mu \hfill \\ & & -{\displaystyle \frac{i}{2}}1\otimes \nu +ibg\otimes \nu +ib\mu \otimes \nu -b\nu \otimes \nu \left)\right(-{\displaystyle \frac{i}{2}}\mu \otimes 1+{\displaystyle \frac{1}{2}}\nu \otimes 1\hfill \\ & & -{\displaystyle \frac{i}{2}}1\otimes \mu +{\displaystyle \frac{i}{2b}}\mu \otimes \mu +{\displaystyle \frac{1}{2}}1\otimes \nu )\hfill \\ & =& -{\displaystyle \frac{1}{4}}\mu \nu \otimes 1+{\displaystyle \frac{i}{2}}\mu \otimes \mu -{\displaystyle \frac{1}{2}}\mu \otimes \nu -{\displaystyle \frac{1}{4}}\nu \mu \otimes 1-{\displaystyle \frac{i}{4}}{\nu }^2\otimes 1\hfill \\ & & -{\displaystyle \frac{1}{2}}\nu \otimes \mu +{\displaystyle \frac{1+2b^2}{4b}}\nu \mu \otimes \mu -{\displaystyle \frac{i}{2}}\nu \otimes \nu -{\displaystyle \frac{bi}{2}}g\mu \otimes g+{\displaystyle \frac{b}{2}}g\nu \otimes g\hfill \\ & & -{\displaystyle \frac{ib}{2}}g\otimes g\mu +{\displaystyle \frac{i}{2}}g\mu \otimes g\mu +{\displaystyle \frac{b}{2}}g\otimes g\nu +{\displaystyle \frac{b}{2}}\mu \nu \otimes g-{\displaystyle \frac{ib}{2}}\mu \otimes g\mu \hfill \\ & & +{\displaystyle \frac{b}{2}}\mu \otimes g\nu +{\displaystyle \frac{b}{2}}\nu \mu \otimes g+{\displaystyle \frac{ib}{2}}{\nu }^2\otimes g+{\displaystyle \frac{b}{2}}\nu \otimes g\mu -{\displaystyle \frac{1}{2}}\nu \mu \otimes g\mu \hfill \\ & & +{\displaystyle \frac{ib}{2}}\nu \otimes g\nu -{\displaystyle \frac{1}{4}}1\otimes \mu \nu -{\displaystyle \frac{ib}{2}}g\mu \otimes \mu +{\displaystyle \frac{b}{2}}g\nu \otimes \mu +{\displaystyle \frac{b}{2}}g\otimes \mu \nu \hfill \\ & & +{\displaystyle \frac{b}{2}}\mu \nu \otimes \mu +{\displaystyle \frac{b}{2}}\mu \otimes \mu \nu +{\displaystyle \frac{ib}{2}}{\nu }^2\otimes \mu +{\displaystyle \frac{ib}{2}}\nu \otimes \mu \nu \hfill \\ & & -{\displaystyle \frac{1}{4}}1\otimes \nu \mu +{\displaystyle \frac{1}{4b}}\mu \otimes \nu \mu -{\displaystyle \frac{i}{4}}1\otimes {\nu }^2+{\displaystyle \frac{b}{2}}g\mu \otimes \nu +{\displaystyle \frac{ib}{2}}g\nu \otimes \nu \hfill \\ & & +{\displaystyle \frac{b}{2}}g\otimes \nu \mu -{\displaystyle \frac{1}{2}}g\mu \otimes \nu \mu +{\displaystyle \frac{ib}{2}}g\otimes {\nu }^2+{\displaystyle \frac{ib}{2}}\mu \nu \otimes \nu +{\displaystyle \frac{b}{2}}\mu \otimes \nu \mu +{\displaystyle \frac{ib}{2}}\mu \otimes {\nu }^2\hfill \\ & & +{\displaystyle \frac{bi}{2}}\nu \mu \otimes \nu -{\displaystyle \frac{b}{2}}{\nu }^2\otimes \nu +{\displaystyle \frac{ib}{2}}\nu \otimes \nu \mu -{\displaystyle \frac{i}{2}}\nu \mu \otimes \nu \mu -{\displaystyle \frac{b}{2}}\nu \otimes {\nu }^2\hfill \\ & =& -{\displaystyle \frac{1}{2}}\mu \otimes 1-{\displaystyle \frac{i}{2}}\nu \otimes 1+bg\otimes g+b\mu \otimes g+ib\nu \otimes g-{\displaystyle \frac{1}{2}}1\otimes \mu +bg\otimes \mu \hfill \\ & & +(b+{\displaystyle \frac{1}{2b}})\mu \otimes \mu +ib\nu \otimes \mu -{\displaystyle \frac{i}{2}}1\otimes \nu +ibg\otimes \nu +ib\mu \otimes \nu -b\nu \otimes \nu \hfill \\ & =& ∆(\mu +g)=∆\left(g\nu \right).\hfill \end{array}$$

 □

Lemma 3.

The counit ε is given in Lemma 1. Then, ε satisfies the conditions in (2).

Proof. 

Since

$$\begin{array}{ccc}\hfill \epsilon \left(\mu g_{\left(1\right)}\right)\epsilon \left(g_{\left(2\right)}\nu \right)& =& -{\displaystyle \frac{1}{2}}\epsilon \left(\mu \mu \right)\epsilon \left(\nu \right)-{\displaystyle \frac{i}{2}}\epsilon \left(\mu \nu \right)\epsilon \left(\nu \right)+b\epsilon \left(\mu g\right)\epsilon \left(g\nu \right)\hfill \\ & & +b\epsilon \left(\mu \mu \right)\epsilon \left(g\nu \right)+ib\epsilon \left(\mu \nu \right)\epsilon \left(g\nu \right)-{\displaystyle \frac{1}{2}}\epsilon \left(\mu \right)\epsilon \left(\mu \nu \right)\hfill \\ & & +b\epsilon \left(\mu g\right)\epsilon \left(\mu \nu \right)+b\epsilon \left(\mu \mu \right)\epsilon \left(\mu \nu \right)+ib\epsilon \left(\mu \nu \right)\epsilon \left(\mu \nu \right)\hfill \\ & & -{\displaystyle \frac{i}{2}}\epsilon \left(\mu \right)\epsilon \left(\nu \nu \right)+ib\epsilon \left(\mu g\right)\epsilon \left(\nu \nu \right)+ib\epsilon \left(\mu g\right)\epsilon \left(\nu \nu \right)\hfill \\ & & +ib\epsilon \left(\mu \mu \right)\epsilon \left(\nu \nu \right)-b\epsilon \left(\mu \nu \right)\epsilon \left(\nu \nu \right)\hfill \\ & =& {\displaystyle \frac{i}{2}}\epsilon \left(\mu \right)\epsilon \left(\nu \right)+b\epsilon (\nu +1)\epsilon (\mu +g)+ib\epsilon (-\mu )\epsilon (\mu +g)\hfill \\ & & -{\displaystyle \frac{1}{2}}\epsilon \left(\mu \right)\epsilon (-\mu )+b\epsilon (\nu +1)\epsilon (-\mu )+ib\epsilon (-\mu )\epsilon (-\mu )\hfill \\ & & -{\displaystyle \frac{i}{2}}\epsilon \left(\mu \right)\epsilon \left(1\right)+ib\epsilon (\nu +1)\epsilon \left(1\right)+ib\epsilon (\nu +1)\epsilon \left(1\right)\hfill \\ & & -b\epsilon (-\mu )\epsilon \left(1\right)\hfill \\ & =& -2b^2+b(2+2ib)(2b+{\displaystyle \frac{1}{b}})+ib(-2b)(2b+{\displaystyle \frac{1}{b}})\hfill \\ & & -{\displaystyle \frac{1}{2}}2b(-2b)+b(2ib+2)(-2b)+ib(-2b)(-2b)\hfill \\ & & -{\displaystyle \frac{i}{2}}4b+ib(2ib+2)2+ib(2ib+2)2+4b^2\hfill \\ & =& 2ib+2\hfill \\ & =& \epsilon \left(\mu g\nu \right)\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \epsilon \left(\mu g_{\left(2\right)}\right)\epsilon \left(g_{\left(1\right)}\nu \right)& =& -{\displaystyle \frac{1}{2}}\epsilon \left(\mu \right)\epsilon \left(\mu \nu \right)-{\displaystyle \frac{i}{2}}\epsilon \left(\mu \right)\epsilon \left(\nu \nu \right)+b\epsilon \left(\mu g\right)\epsilon \left(g\nu \right)\hfill \\ & & +b\epsilon \left(\mu g\right)\epsilon \left(\mu \nu \right)+ib\epsilon \left(\mu g\right)\epsilon \left(\nu \nu \right)-{\displaystyle \frac{1}{2}}\epsilon \left(\mu \mu \right)\epsilon \left(\nu \right)\hfill \\ & & +b\epsilon \left(\mu \mu \right)\epsilon \left(g\nu \right)+b\epsilon \left(\mu \mu \right)\epsilon \left(\mu \nu \right)+ib\epsilon \left(\mu \mu \right)\epsilon \left(\nu \nu \right)\hfill \\ & & -{\displaystyle \frac{i}{2}}\epsilon \left(\mu \nu \right)\epsilon \left(\nu \right)+ib\epsilon \left(\mu \nu \right)\epsilon \left(g\nu \right)+ib\epsilon \left(\mu \nu \right)\epsilon \left(g\nu \right)\hfill \\ & & +ib\epsilon \left(\mu \nu \right)\epsilon \left(\mu \nu \right)-b\epsilon \left(\mu \nu \right)\epsilon \left(\nu \nu \right)\hfill \\ & =& -{\displaystyle \frac{1}{2}}\epsilon \left(\mu \right)\epsilon \left(\mu \nu \right)-{\displaystyle \frac{i}{2}}\epsilon \left(\mu \right)\epsilon \left(\nu \nu \right)+b\epsilon \left(\mu g\right)\epsilon \left(g\nu \right)\hfill \\ & & +b\epsilon \left(\mu g\right)\epsilon \left(\mu \nu \right)+ib\epsilon \left(\mu g\right)\epsilon \left(\nu \nu \right)+ib\epsilon \left(\mu \mu \right)\epsilon \left(\nu \nu \right)\hfill \\ & & -{\displaystyle \frac{i}{2}}\epsilon \left(\mu \nu \right)\epsilon \left(\nu \right)+ib\epsilon \left(\mu \nu \right)\epsilon \left(g\nu \right)+ib\epsilon \left(\mu \nu \right)\epsilon \left(g\nu \right)\hfill \\ & & +ib\epsilon \left(\mu \nu \right)\epsilon \left(\mu \nu \right)-b\epsilon \left(\mu \nu \right)\epsilon \left(\nu \nu \right)\hfill \\ & =& -{\displaystyle \frac{1}{2}}\epsilon \left(\mu \right)\epsilon (-\mu )-{\displaystyle \frac{i}{2}}\epsilon \left(\mu \right)\epsilon \left(1\right)+b\epsilon (\nu +1)\epsilon (\mu +g)\hfill \\ & & +b\epsilon (\nu +1)\epsilon (-\mu )+ib\epsilon (\nu +1)\epsilon \left(1\right)\hfill \\ & & -{\displaystyle \frac{i}{2}}\epsilon (-\mu )\epsilon \left(\nu \right)+ib\epsilon (-\mu )\epsilon (\mu +g)+ib\epsilon (-\mu )\epsilon (\mu +g)\hfill \\ & & +ib\epsilon (-\mu )\epsilon (-\mu )-b\epsilon (-\mu )\epsilon \left(1\right)\hfill \\ & =& 2b^2-2ib+b(2+2ib)(2b+{\displaystyle \frac{1}{b}})+b(2+2ib)(-2b)+2ib(2+2ib)\hfill \\ & & {\displaystyle \frac{i}{2}}2b2ib-ib2b(2b+{\displaystyle \frac{1}{b}})-ib2b(2b+{\displaystyle \frac{1}{b}})+ib2b2b+4b^2\hfill \\ & =& 2ib+2\hfill \\ & =& \epsilon \left(\mu g\nu \right),\hfill \end{array}$$

it follows that

$$\epsilon \left(\mu g_{\left(2\right)}\right)\epsilon \left(g_{\left(1\right)}\nu \right)=\epsilon \left(\mu g\nu \right)=\epsilon \left(\mu g_{\left(1\right)}\right)\epsilon \left(g_{\left(2\right)}\nu \right).$$

Similarly, the condition in (2) holds for the others. □

From Lemmas 1–3, we present a broad class of bialgebra structures on $\mathbb{K}$. The following theorem reveals that $\mathbb{K}$ endowed with suitable antipode forms a weak Hopf algebra.

Theorem 1.

The coalgebra structure on $\mathbb{K}$ is given in Lemma 1. $\mathbb{K}$ is a weak Hopf algebra with the antipode given by

$$S\left(1\right)=1,S\left(g\right)=(-1+{\displaystyle \frac{1}{2b^2}})\mu -i\nu ,$$

$$S\left(\mu \right)=2b^{2}g+2b^2\mu +2i{b}^2\nu ,$$

$$S\left(\nu \right)=2i{b}^{2}g-(i-2i{b}^2)\mu +(1-2b^2)\nu .$$

Proof. 

In order to verify that $\mathbb{K}$ is a weak Hopf algebra, we need to check that the condition (4) holds. In fact, since

$$\begin{array}{ccc}\hfill {\nu }_{\left(1\right)}S\left({\nu }_{\left(2\right)}\right)& =& -{\displaystyle \frac{i}{2}}\mu S\left(1\right)+{\displaystyle \frac{1}{2}}\nu S\left(1\right)-{\displaystyle \frac{i}{2}}S\left(\mu \right)+{\displaystyle \frac{i}{2b}}\mu S\left(\mu \right)+{\displaystyle \frac{1}{2}}S\left(\nu \right)\hfill \\ & =& -{\displaystyle \frac{i}{2}}\mu +{\displaystyle \frac{1}{2}}\nu -{\displaystyle \frac{i}{2}}(2b^{2}g+2b^2\mu +2i{b}^2\nu )\hfill \\ & & +{\displaystyle \frac{i}{2b}}\mu (2b^{2}g+2b^2\mu +2i{b}^2\nu )+{\displaystyle \frac{1}{2}}(2i{b}^{2}g-(i-2i{b}^2)\mu +(1-2b^2)\nu )\hfill \\ & =& (b-i)\mu +(1+ib)\nu +ib\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \epsilon \left(1_{\left(1\right)}\nu \right)1_{\left(2\right)}& =& {\displaystyle \frac{1}{2}}\epsilon \left(\nu \right)1-{\displaystyle \frac{1}{2}}\epsilon \left(\mu \nu \right)\mu -{\displaystyle \frac{i}{2}}\epsilon \left(\nu \nu \right)\mu -{\displaystyle \frac{i}{2}}\epsilon \left(\mu \nu \right)\nu +{\displaystyle \frac{1}{2}}\epsilon \left(\nu \nu \right)\nu \hfill \\ & =& {\displaystyle \frac{1}{2}}\epsilon \left(\nu \right)1-{\displaystyle \frac{1}{2}}\epsilon (-\mu )\mu -{\displaystyle \frac{i}{2}}\epsilon \left(1\right)\mu -{\displaystyle \frac{i}{2}}\epsilon (-\mu )\nu +{\displaystyle \frac{1}{2}}\epsilon \left(1\right)\nu \hfill \\ & =& (b-i)\mu +(1+ib)\nu +ib,\hfill \end{array}$$

this leads to

$${\nu }_{\left(1\right)}S\left({\nu }_{\left(2\right)}\right)=\epsilon \left(1_{\left(1\right)}\nu \right)1_{\left(2\right)}.$$

Since

$$\begin{array}{ccc}\hfill S\left({\nu }_{\left(1\right)}\right){\nu }_{\left(2\right)}& =& -{\displaystyle \frac{i}{2}}S\left(\mu \right)+{\displaystyle \frac{1}{2}}S\left(\nu \right)-{\displaystyle \frac{i}{2}}\mu +{\displaystyle \frac{i}{2b}}S\left(\mu \right)\mu +{\displaystyle \frac{1}{2}}\nu \hfill \\ & =& -{\displaystyle \frac{i}{2}}(2b^{2}g+2b^2\mu +2i{b}^2\nu )+{\displaystyle \frac{1}{2}}(2i{b}^{2}g-(i-2i{b}^2)\mu +(1-2b^2)\nu )\hfill \\ & & -{\displaystyle \frac{i}{2}}\mu +{\displaystyle \frac{i}{2b}}(2b^{2}g+2b^2\mu +2i{b}^2\nu )\mu +{\displaystyle \frac{1}{2}}\nu \hfill \\ & =& bi+(1-bi)\nu -(b+i)\mu \hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill 1_{\left(1\right)}\epsilon \left(\nu 1_{\left(2\right)}\right)& =& {\displaystyle \frac{1}{2}}1\epsilon \left(\nu 1\right)-{\displaystyle \frac{1}{2}}\mu \epsilon \left(\nu \mu \right)-{\displaystyle \frac{i}{2}}\nu \epsilon \left(\nu \mu \right)-{\displaystyle \frac{i}{2}}\mu \epsilon \left(\nu \nu \right)+{\displaystyle \frac{1}{2}}\epsilon \left(\nu \nu \right)\nu \hfill \\ & =& {\displaystyle \frac{1}{2}}1\epsilon \left(\nu 1\right)-{\displaystyle \frac{1}{2}}\mu \epsilon \left(\mu \right)-{\displaystyle \frac{i}{2}}\nu \epsilon \left(\mu \right)-{\displaystyle \frac{i}{2}}\mu \epsilon \left(1\right)+{\displaystyle \frac{1}{2}}\epsilon \left(1\right)\nu \hfill \\ & =& bi+(1-bi)\nu -(b+i)\mu ,\hfill \end{array}$$

it follows that

$$S\left({\nu }_{\left(1\right)}\right){\nu }_{\left(2\right)}=1_{\left(1\right)}\epsilon \left(\nu 1_{\left(2\right)}\right).$$

The others can be checked similarly. □

From Theorem 1, $\mathbb{K}$ has been found to be a weak Hopf algebra. The source and target algebras of $\mathbb{K}$ will be discussed next. First, we discuss the target algebra ${\mathbb{K}}_t$.By using a straightforward computation, we have

$${\epsilon }_t\left(\nu \right)=(b-i)\mu +(1+ib)\nu +ib,{\epsilon }_t\left(\mu \right)=b1-bi\mu +b\nu ,$$

$${\epsilon }_t\left(g\right)={\displaystyle \frac{1}{2b}}1+({\displaystyle \frac{i}{2b}}-1)\mu -(i+{\displaystyle \frac{1}{2b}})\nu .$$

Set $Q=\mathbb{C}<1,(\nu -i\mu )>$. Since

$${\epsilon }_t\left(g\right)={\displaystyle \frac{1}{2b}}-(i+{\displaystyle \frac{1}{2b}})(\nu -i\mu ),{\epsilon }_t\left(\mu \right)=b1+b(\nu -i\mu ),$$

$${\epsilon }_t\left(\nu \right)=bi1+(1+bi)(\nu -i\mu ),$$

it follows that ${\epsilon }_t\left(\mathbb{K}\right)\subseteq Q$. Since

$$01+0{\epsilon }_t\left(g\right)-i{\epsilon }_t\left(\mu \right)+{\epsilon }_t\left(\nu \right)=\nu -i\mu ,$$

we have $Q\subseteq {\epsilon }_t\left(\mathbb{K}\right)$. Thus

$${\epsilon }_t\left(\mathbb{K}\right)=\mathbb{C}<1,(\nu -i\mu )>.$$

We discuss the source algebra ${\epsilon }_s\left(\mathbb{K}\right)$. First, we have

$${\epsilon }_s\left(g\right)={\displaystyle \frac{1}{2b}}-(1+{\displaystyle \frac{i}{2b}})\mu +({\displaystyle \frac{1}{2b}}-i)\nu ,{\epsilon }_s\left(\mu \right)=b+ib\mu -b\nu ,$$

$${\epsilon }_s\left(\nu \right)=bi-(b+i)\mu +(1-bi)\nu .$$

Since

$${\epsilon }_s\left(g\right)={\displaystyle \frac{1}{2b}}+({\displaystyle \frac{1}{2b}}-i)(\nu -i\mu ),{\epsilon }_s\left(\mu \right)=b1-b(\nu -i\mu ),$$

$${\epsilon }_s\left(\nu \right)=bi1+(1-bi)(\nu -i\mu ),$$

it follows that ${\epsilon }_s\left(\mathbb{K}\right)\subseteq Q$. Since

$$01+0{\epsilon }_s\left(g\right)-i{\epsilon }_s\left(\mu \right)+{\epsilon }_s\left(\nu \right)=\nu -i\mu ,$$

we have $Q\subseteq {\epsilon }_s\left(\mathbb{K}\right)$. Thus,

$${\epsilon }_s\left(\mathbb{K}\right)=\mathbb{C}<1,(\nu -i\mu )>.$$

Through the above discussion, we know that ${\epsilon }_s\left(\mathbb{K}\right)={\epsilon }_t\left(\mathbb{K}\right)$.

By using Proposition 2.11 in [1], we can obtain the following result.

Proposition 1.

The subalgebra $\mathbb{C}<1,(\nu -i\mu )>$ of $\mathbb{K}$ is separable algebra with the separable idempotent given by

$$\begin{array}{ccc}\hfill q& =& {\displaystyle \frac{1}{2}}1\otimes 1-{\displaystyle \frac{1}{2}}\mu \otimes \mu -{\displaystyle \frac{i}{2}}\nu \otimes \mu -{\displaystyle \frac{i}{2}}\mu \otimes \nu +{\displaystyle \frac{1}{2}}\nu \otimes \nu .\hfill \end{array}$$

Proof. 

The desired q is

$$\begin{array}{ccc}\hfill q& =& S\left(1_{\left(1\right)}\right)\otimes 1_{\left(2\right)}\hfill \\ & =& {\displaystyle \frac{1}{2}}1-{\displaystyle \frac{1}{2}}S\left(\mu \right)\mu -{\displaystyle \frac{i}{2}}S\left(\nu \right)\mu -{\displaystyle \frac{i}{2}}S\left(\mu \right)\nu +{\displaystyle \frac{1}{2}}S\left(\nu \right)\nu \hfill \\ & =& {\displaystyle \frac{1}{2}}1-{\displaystyle \frac{1}{2}}(2b^{2}g+2b^2\mu +2i{b}^2\nu )\mu \hfill \\ & & -{\displaystyle \frac{i}{2}}(2i{b}^{2}g-(i-2i{b}^2)\mu +(1-2b^2)\nu )\mu -{\displaystyle \frac{i}{2}}(2b^{2}g+2b^2\mu +2i{b}^2\nu )\nu \hfill \\ & & +{\displaystyle \frac{1}{2}}(2i{b}^{2}g-(i-2i{b}^2)\mu +(1-2b^2)\nu )\nu \hfill \\ & =& {\displaystyle \frac{1}{2}}1\otimes 1-{\displaystyle \frac{1}{2}}\mu \otimes \mu -{\displaystyle \frac{i}{2}}\nu \otimes \mu -{\displaystyle \frac{i}{2}}\mu \otimes \nu +{\displaystyle \frac{1}{2}}\nu \otimes \nu .\hfill \end{array}$$

 □

4. Integrals on $\mathbb{K}$

In this section, we discuss left and right integrals on $\mathbb{K}$.

4.1. Left Integrals in $\mathbb{K}$

Set

$$l=k_{1}1+k_{2}g+k_3\mu +k_4\nu .$$

If l is a left integral, we have

$$xl={\epsilon }_t\left(x\right)l,x\in \{g,\mu ,\nu \}.$$

(5)

From (5), we obtain the following system of equations:

$$2b{k}_3+2ib{k}_4-k_1-i{k}_2+k_4=0,$$

(6)

$$k_1+k_4=\left({\displaystyle \frac{1}{b}}+i\right)k_2,$$

(7)

$$(2b-i)k_1+(-1-2ib)k_2+i\left(2b{k}_3+k_4\right)=0,$$

(8)

$$(1+2ib)k_1+(2b-i)k_2-2b{k}_3-k_4=0,$$

(9)

$$k_2=b\left(k_1-i{k}_2+k_4\right),$$

(10)

$$(1+ib)k_1+b{k}_2+(-1-ib)k_4=2b{k}_3,$$

(11)

$$ib\left(k_2-2k_3\right)+(b-i)k_4=(b-i)k_1.$$

(12)

By solving the system of Equations (6)–(12), we obtain the following result.

Theorem 2.

The left integral on $\mathbb{K}$ is

$$I^L\left(\mathbb{K}\right)=\mathbb{C}<\left({\displaystyle \frac{1}{b}}+i\right)\mu -\nu +1,-{\displaystyle \frac{(1-2b(b-i\left)\right)\mu }{2b^2}}+{\displaystyle \frac{(1+ib)\nu }{b}}+g>.$$

Furthermore, $\mathbb{K}$ is semisimple.

Proof. 

The solutions to the system of Equations (6)–(12) are

$$k_3=-{\displaystyle \frac{k_2-2b(b-i)\left(k_2+i{k}_1\right)}{2b^2}},k_4=-k_1+{\displaystyle \frac{(1+ib)k_2}{b}}.$$

Hence, all left integrals are

$$\begin{array}{ccc}& & k_{1}1+k_{2}g+(-{\displaystyle \frac{k_2-2b(b-i)\left(k_2+i{k}_1\right)}{2b^2}})\mu +(-k_1+{\displaystyle \frac{(1+ib)k_2}{b}})\nu \hfill \\ & =& k_1(\left({\displaystyle \frac{1}{b}}+i\right)\mu -\nu +1)+k_2(-{\displaystyle \frac{(1-2b(b-i\left)\right)\mu }{2b^2}}+{\displaystyle \frac{(1+ib)\nu }{b}}+g),\hfill \end{array}$$

yielding the desired result.

Let $l={\displaystyle \frac{1}{2}}(\left({\displaystyle \frac{1}{b}}+i\right)\mu -\nu +1)$. Then, we have

$$\begin{array}{ccc}\hfill {\epsilon }_t\left(l\right)& =& {\epsilon }_t({\displaystyle \frac{1}{2}}(\left({\displaystyle \frac{1}{b}}+i\right)\mu -\nu +1))\hfill \\ & =& {\displaystyle \frac{1}{2}}({\displaystyle \frac{1}{b}}+i){\epsilon }_t\left(\mu \right)-{\displaystyle \frac{1}{2}}{\epsilon }_t\left(\nu \right)+{\displaystyle \frac{1}{2}}\hfill \\ & =& {\displaystyle \frac{1}{2}}({\displaystyle \frac{1}{b}}+i)(b1-bi\mu +b\nu )-{\displaystyle \frac{1}{2}}((b-i)\mu +(1+ib)\nu +ib)+{\displaystyle \frac{1}{2}}\hfill \\ & =& 1.\hfill \end{array}$$

Hence, l is a normalized left integral. By using Theorem 3.13 in [1], we find that $\mathbb{K}$ is semisimple. □

4.2. Right Integrals in $\mathbb{K}$

Set

$$r=k_{1}1+k_{2}g+k_3\mu +k_4\nu .$$

If r is a right integral, we have

$$rx=r{\epsilon }_s\left(x\right),x\in \{g,\mu ,\nu \}.$$

(13)

From (13), we obtain the following system of equations:

$$-2b\left(k_3+i{k}_4\right)+k_1-i{k}_2+k_4=0,$$

(14)

$$k_1+\left(-{\displaystyle \frac{1}{b}}+i\right)k_2=k_4,$$

(15)

$$(2b+i)k_1+i\left((2b+i)k_2-2b{k}_3+k_4\right)=0,$$

(16)

$$(1-2ib)k_1+(2b+i)k_2-2b{k}_3+k_4=0,$$

(17)

$$-ib{k}_1+b{k}_2-ib{k}_4+k_1+k_4=2b{k}_3,$$

(18)

$$b{k}_1+i(b+i)k_2=b{k}_4,$$

(19)

$$(b+i)k_1+ib\left(k_2-2k_3\right)+(b+i)k_4=0.$$

(20)

By solving Equations (14)–(20), we obtain the following result.

Theorem 3.

The right integral space of $\mathbb{K}$ is

$$I^R\left(\mathbb{K}\right)=\mathbb{C}<\left({\displaystyle \frac{1}{b}}-i\right)\mu +\nu +1,{\displaystyle \frac{(-1+2b(b+i\left)\right)}{2b^2}}\mu +{\displaystyle \frac{(ib-1)}{b}}\nu +g>.$$

Proof. 

The solutions to the system of Equations (14)–(20) are

$$k_3=-{\displaystyle \frac{k_2+2ib(b+i)\left(k_1+i{k}_2\right)}{2b^2}},k_4=k_1+{\displaystyle \frac{i(b+i)k_2}{b}}.$$

Thus, all the right integrals are

$$\begin{array}{ccc}& & k_{1}1+k_{2}g+(-{\displaystyle \frac{k_2+2ib(b+i)\left(k_1+i{k}_2\right)}{2b^2}})\mu +(k_1+{\displaystyle \frac{i(b+i)k_2}{b}})\nu \hfill \\ & =& k_1(\left({\displaystyle \frac{1}{b}}-i\right)\mu +\nu +1)+k_2({\displaystyle \frac{(-1+2b(b+i\left)\right)}{2b^2}}\mu +{\displaystyle \frac{(ib-1)}{b}}\nu +g),\hfill \end{array}$$

which gives the desired result. □

5. Conclusions

By introducing a specific coalgebra structure for hybrid numbers, we successfully transform hybrid numbers into a bialgebra, thereby constructing a weak Hopf algebra. We compute the source algebra and target algebra of such a weak Hopf algebra and find them to be the same separable algebra. Furthermore, we describe the left and right integral spaces of this weak Hopf algebra and prove that it is semisimple. By discussing the weak Hopf algebra structure of hybrid numbers, we effectively combine hybrid numbers with weak Hopf algebras, opening up new directions for studying hybrid numbers. In the future, we will explore other aspects of such weak Hopf algebra, such as quasitriangular structures and weak Hopf algebra actions.