Some Results on Zinbiel Algebras and Rota–Baxter Operators

Abstract

Rota–Baxter operators (RBOs) play a substantial role in many subfields of mathematics, especially in mathematical physics. In the article, RBOs on Zinbiel algebras (ZAs) and their sub-adjacent algebras are first investigated. Moreover, all the RBOs on two and three-dimensional ZAs are presented. Finally, ZAs are also realized in low dimensions of the RBOs of commutative associative algebras. It was found that not all ZAs can be attained in this way.

1. Introduction

Zinbiel algebras (ZAs) were defined in [1] and are dual to Leibniz algebras in the Koszul sense. A ZA is a commutative Dendriform algebra [2]. It is well known that any ZA in terms of the anti-commutator defined as $a\ast b=ab+ba$ is called a commutative associative algebra (CAA). Some interesting properties of ZAs were presented in [3,4,5,6,7]. In particular, the nilpotent property of an arbitrary finite-dimensional complex ZA was proved in [6]. Thus, the classification of complex ZAs up to the third dimension can be attained [6,8].

The Rota–Baxter operators (RBOs) were originally proposed to resolve an analytic problem [9]. Later, they were researched in several areas of mathematics [10,11,12,13,14,15]. In particular, some studies investigated RBOs on different algebras [16,17,18]. In the article, the RBOs on ZAs are focused on. First, the relationship between the RBO on ZA and the ones on its sub-adjacent CAA will be explored. Then, the RBOs on ZAs up to the third dimension based on the classification of ZAs will be determined [6]. Finally, the mutual realization of ZAs and the sub-adjacent CAAs with RBOs up to the third dimension will be investigated based on the derived result of the RBOs on the CAA [19]. Throughout the manuscript, all vector spaces and algebras are composed of finite dimensions over $\mathbb{C}$ unless stated otherwise.

2. Preliminary

Definition 1.

A ZA is a vector space $A$ with a bilinear map $(x,y)\to xy$ satisfying the associative property,

$$\left(xy\right)z=x\left(yz\right)+x\left(zy\right), \forall x,y\in A.$$

$A$ with $xy=0,\forall x,y\in A$ is a ZA, which is called a trivial ZA, otherwise, a nontrivial ZA.

For a ZA A, the anti-commutator is defined by

$$x\ast y=xy+yx,\forall x,y\in A,$$

and it satisfies the associative property,

$$\begin{array}{ll}(x\ast y)\ast z& =(xy+yx)\ast z=(xy+yx)z+z(xy+yx)\\ & =\left(xy\right)z+\left(yx\right)z+\left(zy\right)x\\ & =x\left(yz\right)+x\left(zy\right)+\left(yz\right)x+\left(zy\right)x\\ & =x\ast (yz+zy)\\ & =x\ast (y\ast z).\end{array}$$

So, the new product defines a CAA denoted by $\mathfrak{B}=\mathfrak{B}\left(A\right)$, which is called the sub-adjacent algebra (SAA) of A.

Definition 2.

Let $A$ be an algebra not necessarily associative. A linear operator $R:A\to$ $A$ is called an RBO on $A$ if $R$ satisfies Equation (1).

$$R\left(x\right)R\left(y\right)=R\left(R\right(x)y+xR(y\left)\right), \forall x,y\in A.$$

(1)

We denote the set of RBOs on an algebra $A$ by $\mathrm{R}\mathrm{B}\left(A\right)$. The relationship between the two sets, $RB\left(A\right)$ and $RB\left(\mathfrak{B}\right(A\left)\right)$, can be obtained as follows:

Proposition 1.

Let $R$ be an $RBO$ on a $ZA A$. Then, $R$ is an $RBO$ on its $SAA \mathfrak{B}\left(A\right)$. It is implied that $RB\left(A\right)\subseteq RB\left(\mathfrak{B}\right(A\left)\right)$.

Proof. 

For $x,y\in \mathfrak{B}\left(A\right),$ and $R\in RB\left(A\right)$, the subsequent equality is presented.

$$\begin{array}{ll}R\left(R\right(x\left)\right)\ast y+x\ast R\left(y\right))& =R\left(R\right(x)y+yR(x)+xR(y)+R(y\left)x\right)\\ & =R\left(R\right(x)y+yR(x\left)\right)+R\left(xR\right(y)+R(y\left)x\right)\\ & =R\left(x\right)R\left(y\right)+R\left(y\right)R\left(x\right)\\ & =R\left(x\right)\ast R\left(y\right).\end{array}$$

□

Next, new RBOs will be constructed from several aspects. Note that a derivation on a ZA $A$ is a linear operator $D:A\to A$ satisfying

$$D\left(xy\right)=D\left(x\right)y+xD\left(y\right),\forall x,y\in A.$$

Proposition 2.

Suppose that $A$ is a $ZA$ and $R:A\to A$ is an invertible operator. Then, $R$ is an $RBO$ on $A$ if and only if $R^{-1}$ is a derivation of $A$.

Alternatively, any derivation $D$ of a ZA $A$ is also a derivation of its $\mathrm{S}\mathrm{A}\mathrm{A} \mathfrak{B}\left(A\right)$, that is, $D$ satisfies

$$D(x\ast y)=D\left(x\right)\ast y+x\ast D\left(y\right),\forall x,y\in A.$$

Furthermore, invertible derivations or RBOs are simply constructed. Let $A$ be a ZA. $A$ is called graded if $A={\oplus }_{\lambda \in \mathsf{\Gamma }}{A}_{\lambda }$ as a direct summation of vector spaces, where $A_{\lambda }\ne 0$ and $A_{\alpha }{A}_{\beta }\subseteq A_{\alpha +\beta }$.

Proposition 3.

Let $A={\oplus }_{\lambda \in \Gamma }{A}_{\lambda }$ be a graded $ZA$. If $0\notin \Gamma$, then $A$ has an invertible derivation.

Proof. 

Suppose that $D:A\to A$ is a linear map defined by $D\left(x\right)=\lambda x$ for each $x\in A_{\lambda }$. Then, $D$ is a derivation of A. Furthermore, the invertibility of $D$ is realized since $0\notin \mathsf{\Gamma }$. □

Proposition 4.

Let $A=A_1\oplus A_2$ be a direct sum of two ideals of a $ZAA$. For any $RBO{R}_i$ on $A_i,(i=\mathrm{1,2})$, the linear map $R:A\to A$ is given by

$$R\left(x_1,x_2\right)=\left(R_1\left(x_1\right),R_2\left(x_2\right)\right)$$

(for any $x_1\in A_1,x_2\in A_2$) that defines an RBO on $A$.

Finally, how CAAs as well as ZAs generate new ZAs through their RBOs will be discussed. Firstly, a result is extracted from [20,21].

Lemma 1.

Suppose that $(B,\cdot )$ is a $CAA$ and $R$ is an $RBO$ [20,21]. Then, the following product

$$x\ast y=x\cdot R\left(y\right), \forall x,y\in B$$

(2)

defines $a ZA$.

Proposition 5.

Let $(B,\cdot )$ be a $CAA$. If $D$ is an invertible derivation on $B$, then there exist two isomorphic ZAs given by

$$x\ast y=x\cdot D^{-1}\left(y\right), \forall x,y\in B,$$

(3)

$$x\circ y=D^{-1}\left(D\right(x)\cdot y), \forall x,y\in B.$$

(4)

Proof. 

For each $x,y,z\in B$, by Propositions 2 and Lemma 1, $(B,\ast )$ is a ZA led by the subsequent equality.

$$\begin{array}{ll}x\ast (y\ast z+z\ast y)& =x{D}^{-1}\left(y{D}^{-1}\left(z\right)+z{D}^{-1}\left(y\right)\right)\\ & =x{D}^{-1}\left(y\right)D^{-1}\left(z\right)=\left(x{D}^{-1}y\right)D^{-1}\left(z\right)\\ & =(x\ast y)\ast D^{-1}\left(z\right)=(x\ast y)\ast z.\end{array}$$

$(B,\cdot )$ is a ZA led by the subsequent equality.

$$\begin{array}{ll}x\cdot (y\cdot z+z\cdot y)& =D^{-1}\left(D\right(x\left)\right(y\cdot z\left)\right)+D^{-1}\left(D\right(x\left)\right(z\cdot y\left)\right)\\ & =D^{-1}\left(D\left(x\right)D^{-1}\left(D\right(y\left)z\right)\right)+D^{-1}\left(D\left(x\right)D^{-1}\left(D\right(z\left)y\right)\right)\\ & =D^{-1}\left(D\right(x)\ast (D\left(y\right)z\left)\right)+D^{-1}\left(D\right(x)\ast (D\left(z\right)y\left)\right)\\ & =D^{-1}\left(D\right(x)\ast (D\left(y\right)\ast D\left(z\right)\left)\right)+D^{-1}\left(D\right(x)\ast (D\left(z\right)\ast D\left(y\right)\left)\right)\\ & =D^{-1}\left(\right(D\left(x\right)\ast D\left(y\right))\ast D(z\left)\right)=D^{-1}\left(D\right(x)\ast (D\left(y\right))\cdot z\\ & =D^{-1}\left(D\right(x\left)y\right)\cdot z=(x\cdot y)\cdot z.\end{array}$$

Alternatively, $(B,0)$ is a ZA whose product is induced by $(B,\ast )$ through the algebraic isomorphism $D$ defined by

$$x\circ y=D^{-1}\left(D\right(x)\ast D(y\left)\right)=D^{-1}\left(D\right(x)\cdot y),\forall x,y\in B.$$

Thus, the conclusion holds. □

Corollary 1.

Let $(B,\cdot )$ be a $CAA$ and $R$ be an $RBO$. Then, $R$ is an $RBO$ on the $ZA$ generated by Equation (2).

Proof. 

For each $x,y\in B$, the equality is satisfied as follows:

$$R\left(x\right)\ast R\left(y\right)=R\left(x\right)\cdot R^2\left(y\right)=R\left(R\left(x\right)\cdot R\left(y\right)+x\cdot R^2\left(y\right)\right)=R\left(R\right(x)\ast y+x\ast R(y\left)\right).$$

Thus, the $R$ becomes an $\mathrm{R}\mathrm{B}\mathrm{O}$ on $(B,\ast )$. □

Corollary 2.

Suppose that $(B,\cdot )$ is $ZA$ and $R$ is an $RBO$. Then, the product is given by

$$x\ast y=x\cdot R\left(y\right)+R\left(y\right)\cdot x, \forall x,y\in B,$$

which defines a new ZA. Furthermore, $R$ is still an RBO on $(B,\ast )$.

Proof. 

The first and second statements are followed by Propositions 1 and Lemma 1 and Corollary 1, respectively.

The ZA $(B,\ast )$ given above is the (1-st) double of $(B,\cdot )$ associated with the RB $R$. Additionally, for any $\mathrm{Z}\mathrm{A}(B,\cdot )$ with an $\mathrm{R}\mathrm{B}\mathrm{O} R$, a series of $\mathrm{Z}\mathrm{A}\mathrm{s}\left(B,{\ast }_i\right)$ can be defined as follows: $\left(B,{\ast }_0\right)=(B,\cdot )$ and a product on $\left(B,{\ast }_i\right)(i\ge 1)$ is given by

$$x{\ast }_{i}y=x{\ast }_{i-1}R\left(y\right)+R\left(y\right){\ast }_{i-1}x,\forall x,y\in B.$$

$(B,\cdot )$ is called the $i$-th double of $(B,\cdot )$. It is the $(1$-st$)$ double of $\left(B,{\ast }_{i-1}\right)$ associated with $R$. □

Proposition 6.

Suppose that $(B,\cdot )$ is a $ZA$ and $R$ is an $RBO$. Then, for any $i\ge 0$, Equation (5) is attained.

$$x{\ast }_{i+1}y=\sum _{k=0}^i C_i^k\left\{R^k\left(x\right),R^{i+1-k}\left(y\right)\right\},\forall x,y\in B,$$

(5)

where $a\cdot b+b\cdot a$ is denoted $by \{a,b\}$ for any $a,b\in B$.

Proof. 

The conclusion is proved by induction on $i$.

Equation (5) holds for $i=1$.

Now, suppose that it holds for a generic $i$, i.e.,

$$x{\ast }_{i+1}y=\sum _{k=0}^i C_i^k\left\{R^k\left(x\right),R^{i+1-k}\left(y\right)\right\}$$

(6)

Then, Equation (6) leads to

$$\begin{array}{ll}& x{\ast }_{i+2}y=x{\ast }_{i+1}R\left(y\right)+R\left(y\right){\ast }_{i+1}x\\ =& {\displaystyle \sum _{k=0}^i}{C}_i^k\left\{R^k\left(x\right),R^{i+2-k}\left(y\right)\right\}+{\displaystyle \sum _{k=0}^i}{C}_i^k\left\{R^{k+1}\left(y\right),R^{i+1-k}\left(x\right)\right\}\\ =& {\displaystyle \sum _{k=0}^i}{C}_i^k\left\{R^k\left(x\right),R^{i+2-k}\left(y\right)\right\}+{\displaystyle \sum _{k=0}^i}{C}_i^k\left\{R^{i+1-k}\left(x\right),R^{k+1}\left(y\right)\right\}\\ =& {\displaystyle \sum _{k=0}^i}{C}_i^k\left\{R^k\left(x\right),R^{i+2-k}\left(y\right)\right\}+{\displaystyle \sum _{t=0}^{i+1}}{C}_i^{i+1-t}\left\{R^t\left(x\right),R^{i+2-t}\left(y\right)\right\}\\ =& C_i^0\left\{x,R^{i+2}\left(y\right)\right\}+{\displaystyle \sum _{k=1}^i}{C}_i^k\left\{R^k\left(x\right),R^{i+2-k}\left(y\right)\right\}\\ & +C_i^{i+1-i-1}\left\{R^{i+1}\left(x\right),R\left(y\right)\right\}+{\displaystyle \sum _{t=1}^i}{C}_i^{i+1-t}\left\{R^t\left(x\right),R^{i+2-t}\left(y\right)\right\}\\ =& C_i^0\left\{x,R^{i+2}\left(y\right)\right\}+C_i^0\left\{R^{i+1}\left(x\right),R\left(y\right)\right\}+{\displaystyle \sum _{k=1}^i}\left(C_i^k+C_i^{i+1-k}\right)\left\{R^k\left(x\right),R^{i+2-k}\left(y\right)\right\}\\ =& \left\{x,R^{i+2}\left(y\right)\right\}+\left\{R^{i+1}\left(x\right)+R\left(y\right)\right\}+{\displaystyle \sum _{k=1}^i}{C}_{i+1}^k\left\{R^k\left(x\right),R^{i+2-k}\left(y\right)\right\}\\ & {\displaystyle \sum _{k=0}^{i+1}}{C}_{i+1}^k\left\{R^k\left(x\right),R^{i+2-k}\left(y\right)\right\}.\end{array}$$

So, Equation (5) holds for any $i$. □

Corollary 3.

Suppose that $(B,\cdot )$ is a $ZA$ and $R$ is an $RBO$. If $R$ is nilpotent, then there exists $a$ positive integer $N$ such that $\left(B,{\ast }_n\right)$ are trivial for $n>N$.

Proof. 

Set $R^m=0$. For any $n\ge 2m-1$ and $k\le n$, either $k\ge m$ or $n-k\le m$ holds. Hence, by Equation (5), $x{\ast }_{n}y=0$ for any $x,y\in B$. □

3. RBOs of Low-Dimensional ZAs

All the RBOs on two- and three-dimensional ZAs will be presented. Suppose that the set $\left\{e_1,e_2,\dots ,e_n\right\}$ is a basis of ZA A. $e_i{e}_j={\sum }_{k=1}^n{c}_{ij}^k{e}_k$ is set. Any RBO $R$ could then be characterized by a matrix $\left(r_{ij}\right).R\left(e_i\right)={\sum }_{j=1}^n r_{ji}{e}_j$ and $r_{ij}^{\prime }$ satisfies Equation (7).

$$\sum _{k,l,m}^n\left(c_{kl}^m{r}_{ik}{r}_{jl}-c_{kj}^l{r}_{ik}{r}_{lm}-c_{il}^k{r}_{jl}{r}_{km}\right)=0,i,j=\mathrm{1,2},\dots ,n.$$

(7)

Next, the classification results of ZAs up to the third dimension in the literature are presented.

Lemma 2.

Let $A$ be a $ZA$ with up to the third dimension; then, it must be isomorphic to one of the following cases (just list the nonzero product for nontrivial cases) [6,8].

$$dim A=1. e_1{e}_1=0;$$

$$dim A= 2. T_1:e_i{e}_j=0 \mathrm{and} T_2:e_1{e}_1=e_2;$$

$$dim A=3. A_1:e_i{e}_j=0; A_2:e_1{e}_1=e_3; A_3:e_1{e}_1=e_3, e_2{e}_2=e_3; A_4:e_1{e}_2=\frac{1}{2}{e}_3, e_2{e}_1=-\frac{1}{2}{e}_3;$$

$$A_5:e_2{e}_1=-e_3; A_6:e_1{e}_1=e_3, e_1{e}_2=e_3, e_2{e}_2=\lambda e_3, \lambda \ne 0; A_7:e_1{e}_1=e_2, e_1{e}_2=\frac{1}{2}{e}_3, e_2{e}_1=e_3.$$

In the first dimension, there is only the trivial ZA (the products being zero). In this case, any linear transformation is an RBO. In the second dimension, there are two ZAs: one is the trivial ZA $T_1$ whose RBOs are all linear transformations and another $T_2$ is given by the nonzero products.

$$e_1{e}_1=e_2.$$

By Equation (7), the subsequent equations are attained.

$$\left\{\begin{array}{l}{r}_{12}^2-2r_{11}{r}_{22}=0\\ 2r_{11}{r}_{12}=0\\ r_{12}\left(r_{11}-r_{22}\right)=0\\ r_{12}^2=0\end{array}\right.$$

So, the subsequent results are attained.

• $r_{12}=0,r_{11}\ne 0,r_{11}=2r_{22}$.
• $r_{12}=0,r_{11}=0,r_{22}\in \mathbb{C}$.

The new ZAs pertinent to the RBO R can be attained as follows:

For case $1,R=\left(\begin{array}{cc}2r_{22}& 0\\ r_{21}& r_{22}\end{array}\right)\left(r_{22}\ne 0\right)$. Then,

$$\begin{array}{ll}{e}_1\ast e_1& =e_{1}R\left(e_1\right)+R\left(e_1\right)e_1\\ & =e_1\left(2r_{22}{e}_1+r_{21}{e}_2\right)+\left(2r_{22}{e}_1+r_{21}{e}_2\right)e_1\\ & =4r_{22}{e}_2\end{array}$$

Similarly, $e_1\ast e_2=0,e_2\ast e_1=0,e_2\ast e_2=0$ are attaiened. When $4r_{22}{e}_2$ is taken as $e_2$, $e_1\ast e_1=e_2$ becomes a nonzero product. As discussed above, $e_1{\ast }_k{e}_1=e_2,k\in \mathbb{N}$ is attained.

For case $2,R=\left(\begin{array}{cc}0& 0\\ r_{21}& r_{22}\end{array}\right)$. $e_i\ast e_j=0,i,j=\mathrm{1,2}$ is attained. So $e_i\ast e_j=0$ and $e_i{\ast }_k{e}_j=0$, where $i,j=\mathrm{1,2},k\in \mathbb{N}$.

Based on the above arguments, Theorem 1 is stated.

Theorem 1.

Let $A$ be the nontrivial two-dimensional $ZA{T}_2$ with an $RBO R$. If $=$ $\left(\begin{array}{cc}2r_{22}& 0\\ r_{21}& r_{22}\end{array}\right)\left(r_{22}\ne 0\right)$, for each $i\ge 1$, the $i$-th double pertinent to $R$ has an isomorphism to A. If $R=\left(\begin{array}{cc}0& 0\\ r_{21}& r_{22}\end{array}\right)$, the $i$-th double is related to $R$ that becomes trivial for each $i\ge 1$.

By Equation (7), for three-dimensional ZA algebras, calculations are performed one at a time, and 103 of them are obtained by the subsequent steps.

• For $A_2$, the nonzero products are

  $$e_1{e}_1=e_3.$$
  A set of equations is attained as follows:

  $$\left\{\begin{array}{l}{r}_{11}{r}_{23}=0\\ r_{11}^2-2r_{11}{r}_{33}=0\\ r_{12}=r_{13}=0\end{array}\right.$$
  Then, the subsequent results are attained.

  (1)

  $r_{12}=r_{13}=r_{11}=0, r_{ij}\in \mathbb{C}, i=\mathrm{2,3}; j=\mathrm{1,2},3$.

  (2)

  $r_{12}=r_{13}=0, r_{11}\ne 0, r_{ij}\in \mathbb{C}, i=\mathrm{2,3}; j=\mathrm{1,2},3$.

• For $A_3$, the nonzero products are

  $$e_1{e}_1=e_3, e_2{e}_2=e_3.$$
  A set of equations is attained as follows:

  $$\left\{\begin{array}{l}{r}_{13}=r_{23}=0,\\ 2r_{11}{r}_{33}=r_{11}^2+r_{21}^2,\\ 2r_{11}{r}_{33}=r_{12}^2+r_{22}^2,\\ \left(r_{12}+r_{21}\right)r_{33}=r_{12}{r}_{11}+r_{22}{r}_{21}.\end{array}\right.$$
  Then, the subsequent results are attained.

  (1)

  $r_{13}=r_{23}=0, r_{11}=r_{22}, r_{12}=\mp \alpha , r_{21}=\pm \alpha , \alpha =\sqrt{-r_{22}\left(r_{22}-2r_{33}\right)}, r_{22}, r_{3j}\in \mathbb{C}, j=$ $\mathrm{1,2},3$.

  (2)

  $r_{13}=r_{23}=0, r_{11}=-r_{22}+2r_{33}, r_{12}=\pm \alpha , r_{21}=\pm \alpha , \alpha =\sqrt{-r_{22}\left(r_{22}-2r_{33}\right)}, r_{22}, r_{3j}\in$ $\mathbb{C}, j=\mathrm{1,2},3$.

• For $A_4$, the nonzero products are

  $$e_1{e}_2=\frac{1}{2}{e}_3, e_2{e}_1=-\frac{1}{2}{e}_3.$$
  A set of equations is attained as follows:

  $$\left\{\begin{array}{l}{r}_{13}=r_{23}=0,\\ \left(r_{11}+r_{22}{r}_{33}=r_{11}{r}_{22}-r_{21}{r}_{12}\right.\end{array}\right.$$
  Then, the subsequent outcomes are attained.

  (1)

  $r_{22}\ne r_{33}, r_{13}=r_{23}=0, r_{11}=\frac{r_{12}{r}_{21}+r_{22}{r}_{33}}{r_{22}-r_{33}}, r_{21}, r_{22}, r_{3j}\in \mathbb{C}, j=\mathrm{1,2},3$.

  (2)

  $r_{22}=r_{33}, r_{13}=r_{23}=r_{21}=0, r_{11}, r_{12}, r_{21}, r_{22}, r_{3j}\in \mathbb{C}, j=\mathrm{1,2}$.

  (3)

  $r_{22}=r_{33}, r_{13}=r_{23}=0, r_{12}=-\frac{r_{22}^2}{r_{21}}, r_{11}, r_{21}, r_{22}, r_{3j}\in \mathbb{C}, j=\mathrm{1,2}$.

• For $A_5$, the nonzero products are

  $$e_2{e}_1=-e_3.$$
  A set of equations is attained as follows:

  $$\left\{\begin{array}{l}{r}_{13}=r_{23}=0\\ r_{21}{r}_{12}=0\\ r_{12}{r}_{33}=r_{12}{r}_{22}\\ r_{21}{r}_{33}=r_{21}{r}_{11}\\ \left(r_{11}+r_{22})r_{33}=r_{11}{r}_{22}\right.\end{array}\right.$$
  Then the subsequent results are obtained.

  (1)

  $r_{11}=r_{12}=r_{13}=r_{23}=r_{33}=0, r_{21}, r_{22}, r_{3j}\in \mathbb{C}, j=\mathrm{1,2}$.

  (2)

  $r_{21}=r_{22}=r_{13}=r_{23}=r_{33}=0, r_{11}, r_{12}, r_{31}=0, r_{32}\in \mathbb{C}$.

  (3)

  $r_{22}\ne r_{33}, r_{11}=\frac{r_{22}{r}_{33}}{r_{22}-r_{33}}, r_{12}=r_{13}=r_{21}=r_{23}=0, r_{22}, r_{3j}\in \mathbb{C}, j=\mathrm{1,2},3$.

• For $A_6$, the nonzero products are

  $$e_1{e}_1=e_3, e_1{e}_2=e_3, e_2{e}_2=\lambda e_3, \lambda \ne 0.$$
  A set of equations is attained as follows:

  $$\left\{\begin{array}{l}{r}_{13}=r_{23}=0\\ \left(2r_{11}+r_{21}\right)r_{33}=r_{11}^2+r_{11}{r}_{21}+\lambda r_{21}^2\\ \left(r_{12}+r_{11}+r_{22}+\lambda r_{21}\right)r_{33}=r_{11}{r}_{12}+r_{11}{r}_{22}+\lambda r_{21}{r}_{22}\\ \left(r_{12}+\lambda r_{21}\right)r_{33}=r_{12}{r}_{11}+r_{12}{r}_{21}+\lambda r_{22}{r}_{21}\\ \left(r_{12}+2\lambda r_{22}\right)r_{33}=r_{12}^2+r_{12}{r}_{22}+\lambda r_{22}^2\end{array}\right.$$
  Then, the following solutions are reached:

  $$r_{13}=r_{23}=0, r_{11}=\frac{1r_{33}-r_{21}}{2}\pm \alpha , r_{12}\frac{r_{33}-r_{22}}{2}+\pm \beta , r_{21}, r_{22}, r_{3j}\in \mathbb{C}, j=\mathrm{1,2},3,$$

  for $\alpha =\frac{\sqrt{-4\lambda r_{21}^2+r_{21}^2+4r_{33}^2}}{2}, \beta =\frac{\sqrt{-4\lambda r_{22}^2+8\lambda r_{22}{r}_{33}+r_{22}^2-2r_{22}{r}_{33}+r_{33}^2}}{2}$
• For $A_7$, the nonzero products are

  $$e_1{e}_1=e_2, e_1{e}_2=\frac{1}{2}{e}_3, e_2{e}_1=e_3.$$
  A set of equations is attained as follows:

  $$\left\{\begin{array}{l}{r}_{12}=r_{13}=r_{23}=0,\\ 2r_{11}{r}_{22}=r_{11}^2,\\ 2r_{11}{r}_{32}+\frac{3}{2}{r}_{21}{r}_{33}=\frac{3}{2}{r}_{11}{r}_{21},\\ \left(r_{22}+r_{11}\right)r_{33}=r_{22}{r}_{11.}\end{array}\right.$$
  Then, the following outcomes are reached:

  (1)

  $r_{ij}=0, r_{3k}\in \mathbb{C}, i=\mathrm{1,2}; j,k=\mathrm{1,2},3$.

  (2)

  $r_{1j}=0, r_{23}=r_{33}=0, r_{11}, r_{21}, r_{22}, r_{31}, r_{32}\in \mathbb{C}$.

  (3)

  $r_{11}=3r_{33}, r_{21}=2r_{32}, r_{22}=\frac{3}{2}{r}_{33}, r_{3j}\in \mathbb{C}, j=\mathrm{1,2},3$.

  The arguments presented above lead to Theorem 2.

Theorem 2.

The subsequent mathematical expressions depict the RBOs of the ZAs with three dimensions in

Table 1. ZAA and RBOs with three dimensions.

ZAA	RBOs
$\left(A_1\right)e_i{e}_j=0$	$\left(\begin{array}{lll}{r}_{11}& r_{12}& r_{13}\\ r_{21}& r_{22}& r_{21}\\ r_{31}& r_{32}& r_{33}\end{array}\right)$
$\left(A_2\right)e_1{e}_1=e_3$	$\left(\begin{array}{ccc}0& 0& 0\\ r_{21}& r_{22}& r_{23}\\ r_{31}& r_{32}& r_{33}\end{array}\right),\left(\begin{array}{ccc}{r}_{11}& 0& 0\\ r_{21}& r_{22}& r_{23}\\ r_{31}& r_{32}& \frac{1}{2}{r}_{11}\end{array}\right)\left(r_{11}\ne 0\right)$
$\left(A_3\right)\left\{\begin{array}{l}{e}_1{e}_1=e_3\\ e_2{e}_2=e_3\end{array}\right.$	$\begin{array}{ll}& \left(\begin{array}{ccc}-r_{22}+r_{33}& \alpha & 0\\ \alpha & r_{22}& 0\\ r_{31}& r_{32}& r_{33}\end{array}\right),\left(\begin{array}{ccc}-r_{22}+r_{33}& -\alpha & 0\\ -\alpha & r_{22}& 0\\ r_{31}& r_{32}& r_{33}\end{array}\right)\\ & \left(\begin{array}{ccc}{r}_{22}& \alpha & 0\\ -\alpha & r_{22}& 0\\ r_{31}& r_{32}& r_{33}\end{array}\right),\left(\begin{array}{ccc}{r}_{22}& -\alpha & 0\\ \alpha & r_{22}& 0\\ r_{31}& r_{32}& r_{33}\end{array}\right)\left(\alpha =\sqrt{-r_{22}\left(r_{22}-r_{33}\right)}\right)\end{array}$
$\left(A_4\right)\left\{\begin{array}{c}{e}_1{e}_2=\frac{1}{2}{e}_3\\ e_2{e}_1=-\frac{1}{2}{e}_3\end{array}\right.$	$\begin{array}{ll}& \left(\begin{array}{ccc}{r}_{11}& r_{12}& 0\\ 0& r_{22}& 0\\ r_{31}& r_{32}& r_{33}\end{array}\right),\left(\begin{array}{ccc}{r}_{11}& \frac{-r_{22}^2}{r_{21}}& 0\\ r_{21}& r_{22}& 0\\ r_{31}& r_{32}& r_{22}\end{array}\right)\\ & \left(\begin{array}{ccc}\frac{r_{12}{r}_{21}+r_{22}{r}_{33}}{r_{22}-r_{33}}& r_{12}& 0\\ r_{21}& r_{22}& 0\\ r_{31}& r_{32}& r_{33}\end{array}\right)\left(r_{22}\ne r_{33}\right)\end{array}$
$\left(A_5\right)e_2{e}_1=-e_3$	$\left(\begin{array}{ccc}{r}_{11}& r_{12}& 0\\ 0& 0& 0\\ r_{31}& r_{32}& 0\end{array}\right),\left(\begin{array}{ccc}0& 0& 0\\ r_{21}& r_{22}& 0\\ r_{31}& r_{32}& 0\end{array}\right),\left(\begin{array}{ccc}\frac{r_{22}{r}_{33}}{r_{22}-r_{33}}& 0& 0\\ 0& r_{22}& 0\\ r_{31}& r_{32}& r_{33}\end{array}\right)\left(r_{22}\ne r_{33}\right)$
$\left(A_6\right)\left\{\begin{array}{c}{e}_1{e}_1=e_3\\ e_1{e}_2=e_3\\ e_2{e}_2=\lambda e_3,\lambda \ne 0\end{array}\right.$	$\begin{array}{ll}& \left(\begin{array}{ccc}\frac{2r_{33}-r_{21}}{2}\pm \alpha & \frac{r_{33}-r_{22}}{2}+\beta & 0\\ r_{21}& r_{22}& 0\\ r_{31}& r_{32}& r_{33}\end{array}\right),\left(\begin{array}{ccc}\frac{2r_{33}-r_{21}}{2}\pm \alpha & \frac{r_{33}-r_{22}}{2}-\beta & 0\\ r_{21}& r_{22}& 0\\ r_{31}& r_{32}& r_{33}\end{array}\right)\\ & \alpha =\frac{1}{2}\sqrt{-4\lambda r_{21}^2+r_{21}^2+4r_{33}^2}\\ & \beta =\frac{1}{2}\sqrt{-4\lambda r_{22}^2+8\lambda r_{22}{r}_{33}+r_{22}^2-2r_{22}{r}_{33}+r_{33}^2}\end{array}$
$\left(A_7\right)\left\{\begin{array}{c}{e}_1{e}_1=e_2\\ e_1{e}_2=\frac{1}{2}{e}_3\\ e_2{e}_1=e_3\end{array}\right.$	$\left(\begin{array}{ccc}0& 0& 0\\ r_{21}& r_{22}& 0\\ r_{31}& r_{32}& 0\end{array}\right),\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ r_{31}& r_{32}& r_{33}\end{array}\right),\left(\begin{array}{ccc}3r_{33}& 0& 0\\ 2r_{32}& \frac{3}{2}{r}_{33}& 0\\ r_{31}& r_{32}& r_{33}\end{array}\right)$

.

With the notations presented above, the straightforward conclusion is reached as follows.

Corollary 4.

Let $A$ be a three-dimensional ZA.

• If $A$ is of type $A_1$ or $A_4$, then its $i$-th double associated to any $RBOsR$ is trivial, for each $i\ge 1$;
• If $A$ is one of the types $A_2,A_3,A_5,A_6$, and $A_7$, then there are nonzero $RBO$ s $R$ such that the associated $i$-th doubles are trivial, for each $i\ge 1$:

  $$\begin{array}{ccc}{A}_2:\left(\begin{array}{ccc}0& 0& 0\\ r_{21}& r_{22}& r_{21}\\ r_{31}& r_{32}& r_{33}\end{array}\right);& A_3:\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ r_{31}& r_{32}& 0\end{array}\right);& A_5:\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ r_{31}& r_{32}& 0\end{array}\right);\end{array}$$

  $$\begin{array}{ccc}{A}_6:\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ r_{31}& r_{32}& r_{33}\end{array}\right);& A_7:\left(\begin{array}{ccc}0& & 0\\ 0& 0& 0\\ r_{31}& r_{32}& 0\end{array}\right).& \end{array}$$

Proof. 

Based on the above results of RBOs and Corollary 2, the new ZAs are calculated.

(1)

$A_1:e_2{e}_1=-e_3$.

If $R=\left(\begin{array}{lll}{r}_{11}& r_{12}& r_{13}\\ r_{21}& r_{22}& r_{23}\\ r_{31}& r_{32}& r_{33}\end{array}\right)$, the subsequent equality is easily attained.

$$e_i{\ast }_1{e}_j=e_{i}R\left(e_j\right)+R\left(e_j\right)e_i=0$$

Then, $e_i{\ast }_k{e}_j=0$, where $i,j=\mathrm{1,2},3,$ and $k\in \mathbb{N}$.

(2)

$A_2:e_1{e}_1=e_3$.

If $R=\left(\begin{array}{ccc}0& 0& 0\\ r_{21}& r_{22}& r_{23}\\ r_{31}& r_{32}& r_{33}\end{array}\right)$, the new product appears as zero, so $\left(A,{\ast }_1\right)$ and $\left(A,{\ast }_k\right),k\in \mathbb{N}$, are of type $A_1$.

If $R=\left(\begin{array}{ccc}{r}_{11}& 0& 0\\ r_{21}& r_{22}& r_{23}\\ r_{31}& r_{32}& \frac{1}{2}{r}_{11}\end{array}\right)\left(r_{11}\ne 0\right)$, the following nonzero product is attained:

$$\begin{array}{ll}{e}_1{\ast }_1{e}_1& =e_{1}R\left(e_1\right)+R\left(e_1\right)e_1\\ & =e_1\left(r_{11}{e}_1+r_{21}{e}_2+r_{31}{e}_3\right)+\left(r_{11}{e}_1+r_{21}{e}_2+r_{31}{e}_3\right)e_1\\ & =2r_{11}{e}_3.\end{array}$$

Then, the new $\left(A,{\ast }_1\right)$ is of type $A_2$.

(3)

$A_3:e_1{e}_1=e_3,e_2{e}_2=e_3$.

If $R=\left(\begin{array}{ccc}-r_{22}+r_{33}& \mp \alpha & 0\\ \pm \alpha & r_{22}& 0\\ r_{31}& r_{32}& r_{33}\end{array}\right)$ or $R=\left(\begin{array}{ccc}-r_{22}+r_{33}& \mp \alpha & 0\\ \pm \alpha & r_{22}& 0\\ r_{31}& r_{32}& r_{33}\end{array}\right)$, where $\alpha =$ $\sqrt{-r_{22}\left(r_{22}-r_{33}\right)}$, the subsequent equality is attained.

$$e_1{\ast }_1{e}_1=2r_{11}{e}_3, e_1{\ast }_1{e}_2=2r_{12}{e}_3, e_2{\ast }_1{e}_1=2r_{21}{e}_3, e_2{\ast }_1{e}_2=2r_{22}{e}_3.$$

If $r_{22}=0$, namely, $R=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ r_{31}& r_{32}& 0\end{array}\right)$, it is obtained that $\left(A,{\ast }_k\right),k\in \mathbb{N}$, are of type $A_1$ by the equations above.

(4)

$A_4:e_1{e}_2=\frac{1}{2}{e}_3, e_2{e}_1=-\frac{1}{2}{e}_3$.

It is known that $R$ has the following three cases:

$$\left(\begin{array}{ccc}{r}_{11}& r_{12}& 0\\ 0& r_{22}& 0\\ r_{31}& r_{32}& r_{33}\end{array}\right),\left(\begin{array}{ccc}{r}_{11}& \frac{-r_{22}^2}{r_{21}}& 0\\ r_{21}& r_{22}& 0\\ r_{31}& r_{32}& r_{22}\end{array}\right),\left(\begin{array}{ccc}\frac{r_{12}{r}_{21}+r_{22}{r}_{33}}{r_{22}-r_{33}}& r_{12}& 0\\ r_{21}& r_{22}& 0\\ r_{31}& r_{32}& r_{33}\end{array}\right)\left(r_{22}\ne r_{33}\right)$$

No matter which situation $R$ takes, $e_i{\ast }_1{e}_j=0,i,j=\mathrm{1,2},3$ is obtained. Then, by induction, $\left(A,{\ast }_k\right),k\in \mathbb{N}$, are of type $A_1$.

(5)

$A_5:e_2{e}_1=-e_3$.

If $R=\left(\begin{array}{ccc}{r}_{11}& r_{12}& 0\\ 0& 0& 0\\ r_{31}& r_{32}& 0\end{array}\right)$, then the nonzero product is attained as follows:

$$e_1{\ast }_1{e}_1=-r_{21}{e}_3, e_1{\ast }_1{e}_2=-r_{22}{e}_3.$$

If $R=\left(\begin{array}{ccc}0& 0& 0\\ r_{21}& r_{22}& 0\\ r_{31}& r_{32}& 0\end{array}\right)$, the following nonzero product is attained:

$$e_2{\ast }_1{e}_1=-r_{11}{e}_3, 2_1{\ast }_1{e}_2=-r_{12}{e}_3$$

If $R=\left(\begin{array}{ccc}\frac{r_{22}{r}_{33}}{r_{22}-r_{33}}& 0& 0\\ 0& r_{22}& 0\\ r_{31}& r_{32}& r_{33}\end{array}\right)\left(r_{22}\ne r_{33}\right)$, the following nonzero product is attained:

$$e_1{\ast }_1{e}_2=-r_{22}{e}_3, e_2{\ast }_1{e}_1=-r_{11}{e}_3$$

In summary, if $\left(A,{\ast }_k\right),k\in \mathbb{N}$ are of type $A_1$, $R=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ r_{31}& r_{32}& 0\end{array}\right)$ must be taken.

(6)

$A_6:e_1{e}_1=e_3, e_1{e}_2=e_3, e_2{e}_2=\lambda e_3,\lambda \ne 0$.

If $R=\left(\begin{array}{ccc}\frac{2r_{33}-r_{21}}{2}\pm \alpha & \frac{r_{33}-r_{22}}{2}+\beta & 0\\ r_{21}& r_{22}& 0\\ r_{31}& r_{32}& r_{33}\end{array}\right)$ or $R=\left(\begin{array}{ccc}\frac{2r_{33}-r_{21}}{2}\pm \alpha & \frac{r_{33}-r_{22}}{2}-\beta & 0\\ r_{21}& r_{22}& 0\\ r_{31}& r_{32}& r_{33}\end{array}\right)$, where $\alpha =\frac{1}{2}\sqrt{-4\lambda r_{21}^2+r_{21}^2+4r_{33}^2}$,

$$\beta =\frac{1}{2}\sqrt{-4\lambda r_{22}^2+8\lambda r_{22}{r}_{33}+r_{22}^2-2r_{22}{r}_{33}+r_{33}^2}.$$

The nonzero product is attained.

$$e_1{\ast }_1{e}_1=\left(2r_{11}+r_{21}\right)e_3, e_1{\ast }_1{e}_2=\left(2r_{12}+r_{22}\right)e_3,$$

$$e_1{\ast }_1{e}_1=\left(2\lambda r_{21}+r_{11}\right)e_3, e_2{\ast }_1{e}_2=\left(2\lambda r_{22}+r_{12}\right)e_3$$

Let $r_{11}$ and $r_{12}$ be the corresponding numbers in the matrix.

If $r_{11}=r_{12}=r_{21}=r_{22}=0,\left(A,{\ast }_k\right),k\in \mathbb{N}$ are of type $A_1$. Then, $R$ can only have the following form: $R=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ r_{31}& r_{32}& r_{33}\end{array}\right)$.

(7)

$A_7:e_1{e}_1=e_2, e_1{e}_2=\frac{1}{2}{e}_3, e_2{e}_1=e_3$.

If $R=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ r_{31}& r_{32}& r_{33}\end{array}\right)$, $e_i{\ast }_1{e}_j=0,i,j=\mathrm{1,2},3$ is attained. Then, $\left(A,{\ast }_k\right),k\in \mathbb{N}$, are of type $A_1$.

If $R=\left(\begin{array}{ccc}0& 0& 0\\ r_{21}& r_{22}& 0\\ r_{31}& r_{32}& 0\end{array}\right)$, the following nonzero product is attained:

$$e_1{\ast }_1{e}_1=\frac{3}{2}{r}_{21}{e}_3, e_1{\ast }_1{e}_2=\frac{3}{2}{r}_{22}{e}_3$$

If $R=\left(\begin{array}{ccc}3r_{33}& 0& 0\\ 2r_{32}& \frac{3}{2}{r}_{33}& 0\\ r_{31}& r_{32}& r_{33}\end{array}\right)$, the following nonzero product is attained:

$$e_1{\ast }_1{e}_1=6r_{33}{e}_2+3r_{32}{e}_3, e_1{\ast }_1{e}_2=\frac{1}{2}{r}_{22}{e}_2+r_{22}{e}_3, e_2{\ast }_1{e}_1=\frac{3}{2}{r}_{11}{e}_3.$$

When three cases are considered and if $R=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ r_{31}& r_{32}& 0\end{array}\right)$, $e_i{\ast }_1{e}_j=0,i,j=\mathrm{1,2},3$ is attained. Then, $\left(A,{\ast }_k\right),k\in \mathbb{N}$, are of type $A_1$. □

4. From CAAs to ZAs

The categorization of both the two- and three-dimensional complex CAAs and their RBOs is known [19]. Then, by Lemma 1 the corresponding ZAs are attained. In the following tables, the CAAs and their RBOs are listed in the first and second columns, respectively, and the corresponding ZAs in the sense of an isomorphism are presented in the third column. Moreover, these RBOs from [19] are sets denoted by $R\left(e_i\right)={\sum }_{j=1}^n r_{ij}{e}_j$.

Theorem 3.

The corresponding ZAs (in the sense of an isomorphism) are obtained from the RBOs on two-dimensional CAAs through Equation (2) in following

Table 2. CAA B, RBOs R, Type of ZA(B,∗).

$\mathit{C}\mathit{A}\mathit{A}\mathit{B}$	$\mathbf{RBOs} \mathit{R}$	$\mathbf{Type} \mathbf{of} \mathbf{Z}\mathbf{A}(\mathit{B},\mathbf{\ast })$
$\left(B_1\right)\left\{\begin{array}{l}{e}_1{e}_1=e_2\\ e_2{e}_2=e_2\end{array}\right.$	$\left(\begin{array}{ll}0& 0\\ 0& 0\end{array}\right)$	$T_1$
$\left(B_2\right)\left\{\begin{array}{c}{e}_1{e}_2=e_2{e}_1=e_1\\ e_2{e}_2=e_2\end{array}\right.$	$\left(\begin{array}{cc}0& 0\\ r_{21}& 0\end{array}\right)$	$r_{21}=0,T_1$ $r_{21}\ne 0,T_2$
$\left(B_3\right)e_1{e}_1=e_1$	$\left(\begin{array}{ll}0& r_{12}\\ 0& r_{21}\end{array}\right)$	$T_1$
$\left(B_4\right)e_i{e}_j=0$	$\left(\begin{array}{cc}{r}_{11}& r_{12}\\ r_{21}& r_{22}\end{array}\right)$	$T_1$
$\left(B_5\right)e_1{e}_1=e_2$	$\left(\begin{array}{cc}0& r_{12}\\ 0& r_{22}\end{array}\right)$	$T_1$
	$\left(\begin{array}{cc}2r_{22}& r_{12}\\ 0& r_{22}\end{array}\right)\left(r_{22}\ne \frac{1}{2}\right)$	$r_{22}=0,T_1$ $r_{22}\ne 0,T_2$

.

Proof. 

For $B_2$, by Equation (2), the subsequent equality is attained.

$$e_2\ast e_2=e_{2}R\left(e_2\right)=e_2\left(r_{21}{e}_1\right)=r_{21}{e}_1$$

and others are zero. Then $(B,\ast )$ is of type $T_1$ if $r_{21}=0$.

If $r_{21}\ne 0$, a new $e_1$ is taken as $r_{21}{e}_1$, and $(B,\ast )$ is of type $T_2$.

For the other cases, they all can be attained similarly. □

Remark 1.

All the two-dimensional ZAs can be obtained from two-dimensional CAAs and their RBOs through Equation (2).

The corresponding ZAs (in the sense of an isomorphism) obtained from the RBOs on three-dimensional CAAs through Equation (2) are summarized in

Table 3. CAA B, RBOs R, Type of ZA (B,∗).

$\mathbf{CAA} \mathit{B}$	$\mathbf{RBOs} \mathit{R}$	$\mathbf{Type} \mathbf{of} \mathbf{ZA} (\mathit{B},\mathbf{\ast })$
$\left(B_1\right)e_i{e}_j=0$	$\left(\begin{array}{lll}{r}_{11}& r_{12}& r_{13}\\ r_{21}& r_{22}& r_{23}\\ r_{31}& r_{32}& r_{33}\end{array}\right)$	$A_1$
$\left(B_2\right)e_3{e}_3=e_1$	$\left(\begin{array}{lll}{r}_{11}& r_{12}& 0\\ r_{21}& r_{22}& 0\\ r_{31}& r_{32}& 0\end{array}\right)$	$A_1$
	$\left(\begin{array}{ccc}{r}_{31}& 0& 0\\ r_{21}& r_{22}& 0\\ r_{31}& r_{32}& r_{11}\end{array}\right)\left(r_{11}\ne 0\right)$	$A_2$
$\left(B_3\right)\left\{\begin{array}{l}{e}_2{e}_2=e_1\\ e_3{e}_3=e_1\end{array}\right.$	$\left(\begin{array}{ccc}0& 0& 0\\ r_{21}& r_{22}& \pm \sqrt{-1}{r}_{22}\\ r_{31}& \mp \sqrt{-1}{r}_{33}& r_{33}\end{array}\right)$	$r_{22}=r_{33}=0,A_1$ $r_{22}{r}_{33}\ne 0,A_5$ $r_{22}=0,r_{33}\ne 0,A_5$ $r_{22}\ne 0,r_{33}=0,A_5$
	$\left(\begin{array}{ccc}{r}_{11}& 0& 0\\ r_{21}& r_{22}^{\pm }& r_{23}\\ r_{31}& r_{23}& r_{33}^{\pm }\end{array}\right)\left(r_{11}\ne 0\right)$ $r_{22}^{\pm }=r_{33}^{\pm }=r_{11}\pm {\left(r_{11}^2-r_{23}^2\right)}^{\frac{1}{2}}$	$r_{22}^{\pm }+{ }^{\pm }{r}_{33}=r_{23}=0,A_1$ $r_{22}^{\pm }+r_{33}^{\pm }=2r_{23},r_{22}^{\pm }+r_{33}^{\pm }\ne -2r_{23},A_2$ $r_{22}^{\pm }+r_{33}^{\pm }\ne 2r_{23},r_{22}^{\pm }+r_{33}^{\pm }=-2r_{23},A_2$ $r_{22}^{\pm }+r_{33}^{\pm }\ne 2r_{23},r_{22}^{\pm }+r_{33}^{\pm }\ne -2r_{23},A_3$
	$\left(\begin{array}{ccc}{r}_{11}& 0& 0\\ r_{21}& r_{22}^{\pm }& r_{23}\\ r_{31}& r_{23}& r_{33}^{\mp }\end{array}\right)\left(r_{11}\ne 0\right)$ $r_{22}^{\pm }=r_{33}^{\pm }=r_{11}\pm {\left(r_{11}^2-r_{23}^2\right)}^{\frac{1}{2}}$	$r_{22}^{\pm }+{ }^{\mp }{r}_{33}=r_{23}=0,A_1$ $r_{22}^{\pm }+r_{33}^{\mp }=2r_{23},r_{22}^{\pm }+r_{33}^{\mp }\ne -2r_{23},A_2$ $r_{22}^{\pm }+r_{33}^{\mp }\ne 2r_{23},r_{22}^{\pm }+r_{33}^{\mp }=-2r_{23},A_2$ $r_{22}^{\pm }+r_{33}^{\mp }\ne 2r_{23},r_{22}^{\pm }+r_{33}^{\mp }\ne -2r_{23},A_3$
$\left(B_4\right)\left\{\begin{array}{c}{e}_2{e}_3=e_3{e}_2=e_1\\ e_3{e}_3=e_2\end{array}\right.$	$\left(\begin{array}{ccc}0& 0& 0\\ r_{21}& r_{22}& 0\\ r_{31}& r_{32}& 0\end{array}\right)$	$r_{22}=r_{33}=0,A_1$ $r_{22}{r}_{33}\ne 0,A_2$ $r_{22}=0,r_{32}\ne 0,A_2$ $r_{22}\ne 0,r_{32}=0,A_5$
	$\left(\begin{array}{lll}{r}_{11}& 0& 0\\ r_{21}& 0& 0\\ r_{31}& 0& 0\end{array}\right)\left(r_{11}\ne 0\right)$	$A_1$
	$\left(\begin{array}{ccc}\frac{2}{3}{r}_{22}& 0& 0\\ \frac{2}{3}{r}_{32}& r_{22}& 0\\ r_{31}& r_{32}& 2r_{22}\end{array}\right)\left(r_{22}\ne 0\right)$	$A_7$
$\left(B_5\right)\left\{\begin{array}{l}{e}_1{e}_1=e_1\\ e_2{e}_2=e_2\\ e_3{e}_3=e_3\end{array}\right.$	$\left(\begin{array}{lll}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$	$A_1$
$\left(B_6\right)\left\{\begin{array}{l}{e}_2{e}_2=e_2\\ e_3{e}_3=e_3\end{array}\right.$	$\left(\begin{array}{lll}{r}_{11}& 0& 0\\ r_{21}& 0& 0\\ r_{31}& 0& 0\end{array}\right)$	$A_1$
$\left(B_7\right)\left\{\begin{array}{c}{e}_1{e}_3=e_3{e}_1=e_1\\ e_2{e}_2=e_2\\ e_3{e}_3=e_3\end{array}\right.$	$\left(\begin{array}{ccc}0& 0& 0\\ r_{21}& 0& 0\\ r_{31}& 0& 0\end{array}\right)$	$r_{21}=r_{31}=0,A_1$ $r_{21}{r}_{31}\ne 0,A_5$ $r_{21}=0,r_{31}\ne 0,A_2$ $r_{21}\ne 0,r_{31}=0,A_5$
$\left(B_8\right)e_3{e}_3=e_3$	$\left(\begin{array}{lll}{r}_{11}& r_{12}& 0\\ r_{21}& r_{22}& 0\\ r_{31}& r_{32}& 0\end{array}\right)$	$A_1$
$\left(B_9\right)\left\{\begin{array}{c}{e}_1{e}_3=e_3{e}_1=e_1\\ e_3{e}_3=e_3\end{array}\right.$	$\left(\begin{array}{ccc}0& 0& 0\\ r_{21}& r_{22}& 0\\ r_{31}& r_{32}& 0\end{array}\right)$	$\mathrm{the} \mathrm{same} \mathrm{as} B_7$
	$\left(\begin{array}{lll}0& r_{12}& 0\\ 0& r_{22}& 0\\ 0& r_{32}& 0\end{array}\right)\left(r_{12}\ne 0\right)$	$A_1$
$\left(B_{10}\right)\left\{\begin{array}{c}{e}_1{e}_3=e_3{e}_1=e_1\\ e_2{e}_3=e_3{e}_2=e_2\\ e_3{e}_3=e_3\end{array}\right.$	$\left(\begin{array}{ccc}0& 0& 0\\ r_{21}& 0& 0\\ r_{31}& 0& 0\end{array}\right)$	$\mathrm{the} \mathrm{same} \mathrm{as} B_7$
	$\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ r_{31}& r_{32}& 0\end{array}\right)\left(r_{32}\ne 0\right)$	$A_2$
	$\left(\begin{array}{ccc}{r}_{11}& r_{12}& 0\\ \frac{-r_{11}^2}{r_{12}}& -r_{11}& 0\\ \frac{r_{32}{r}_{11}}{r_{12}}& r_{32}& 0\end{array}\right)\left(r_{12}\ne 0\right)$	$A_5$
$\left(B_{11}\right)\left\{\begin{array}{l}{e}_1{e}_1=e_2\\ e_3{e}_3=e_3\end{array}\right.$	$\left(\begin{array}{ccc}{r}_{11}& r_{12}& 0\\ 0& 0& 0\\ 0& r_{32}& 0\end{array}\right)$	$\begin{array}{c}{r}_{11}=0,A_1\\ r_{11}\ne 0,A_2\end{array}$
	$\left(\begin{array}{ccc}{r}_{11}& r_{12}& 0\\ 0& 2r_{11}& 0\\ r_{31}& r_{32}& 0\end{array}\right)\left(r_{11}\ne 0\right)$	$\begin{array}{c}{r}_{31}=0,A_1\\ r_{31}\ne 0,A_5\end{array}$
$\left(B_{12}\right)\left\{\begin{array}{c}{e}_1{e}_1=e_2\\ e_1{e}_3=e_3{e}_1=e_1\\ e_2{e}_3=e_3{e}_2=e_2\\ e_3{e}_3=e_3\end{array}\right.$	$\left(\begin{array}{ccc}0& r_{12}& 0\\ 0& 0& 0\\ 0& r_{32}& 0\end{array}\right)$	$\begin{array}{c}{r}_{12}\ne 0,A_5\\ r_{12}=0,r_{32}\ne 0,A_2\\ r_{12}=0,r_{32}=0,A_1\end{array}$
	$\left(\begin{array}{ccc}0& r_{12}& 0\\ 0& 0& 0\\ 2r_{12}& r_{32}& 0\end{array}\right)\left(r_{12}\ne 0\right)$	$A_7$

.

Proof. 

$B_3,e_2{e}_2=e_1,e_3{e}_3=e_1$ and $R=\left(\begin{array}{ccc}0& 0& 0\\ r_{21}& r_{22}& \pm \sqrt{-1}{r}_{22}\\ r_{31}& \mp \sqrt{-1}{r}_{33}& r_{33}\end{array}\right)$ are proved as follows.

• If $r_{22}{r}_{33}\ne 0$, $\mathrm{Z}\mathrm{A}(B,\ast )$, the nonzero product, is attained.

  $$\begin{array}{ll}& e_2\ast e_2=e_{2}R\left(e_2\right)=e_2\left(r_{21}{e}_1+r_{22}{e}_2+r_{23}{e}_3\right)=r_{22}{e}_1,\\ & e_2\ast e_3=e_{2}R\left(e_3\right)=e_2\left(r_{31}{e}_1+r_{32}{e}_2+r_{33}{e}_3\right)=r_{32}{e}_1=\mp r_{33}{e}_1,\\ & e_3\ast e_2=e_{3}R\left(e_2\right)=e_3\left(r_{21}{e}_1+r_{22}{e}_2+r_{23}{e}_3\right)=r_{23}{e}_1=\pm r_{22}{e}_1,\\ & e_3\ast e_3=e_{3}R\left(e_3\right)=e_3\left(r_{31}{e}_1+r_{32}{e}_2+r_{33}{e}_3\right)=r_{33}{e}_1.\end{array}$$

Let $e_3^{\prime }=e_2\pm \sqrt{-1}{e}_3$. Then, for $e_1,e_2,e_3^{\prime }$, the subsequent results are attained.

$$\begin{array}{ll}{e}_3^{\prime }\ast e_3^{\prime }& =\left(e_2\pm \sqrt{-1}{e}_3\right)\ast \left(e_2\pm \sqrt{-1}{e}_3\right)\\ & =r_{22}{e}_1+r_{33}{e}_1-r_{22}{e}_1-r_{33}{e}_1\\ & =0\\ e_3^{\prime }\ast e_2& =\left(e_2\pm \sqrt{-1}{e}_3\right)\ast e_2\\ & =r_{22}{e}_1-r_{22}{e}_1\\ & =0.\end{array}$$

Then the nonzero product is as follows:

$$e_2\ast e_2=r_{22}{e}_1,e_2\ast e_3^{\prime }=2r_{22}{e}_1.$$

Let $e_2^{\prime }=\frac{1}{\sqrt{r_{22}}}{e}_2$ and $e_3^{″}=\frac{1}{2\sqrt{r_{22}}}{e}_3^{\prime }$. Then the nonzero product is as follows:

$$e_2^{\prime }\ast e_2^{\prime }=e_1,e_2^{\prime }\ast e_3^{″}=e_1.$$

When $e_2^{″}=e_2^{\prime }-e_3^{″}$ is taken, then the subsequent result is attained.

$$\begin{array}{ll}{e}_2^{″}\ast e_2^{″}& =\left(e_2^{\prime }-e_3^{″}\right)\ast \left(e_2^{\prime }-e_3^{″}\right)=e_1-e_1=0,\\ e_2^{″}\ast e_3^{‴}& =\left(e_2^{\prime }-e_3^{″}\right)\ast e_3^{‴}=e_1,\\ e_3^{″}\ast e_2^{″}& =e_3^{‴}\ast \left(e_2^{\prime }-e_3^{″}\right)=0.\end{array}$$

Namely, $(B,\ast )$ with the nonzero product is $e_2\ast e_3=e_1$. By changing $e_3$ to $-e_3$ and swapping $-e_3$ and $e_1$, $e_2\ast e_1=-e_3$ is attained, which means $(B,\ast )$ is of type $A_5$.

2.

$r_{22}=0,r_{33}\ne 0$. The following nonzero product is attained:

$$e_2\ast e_3=\mp \sqrt{-1}{e}_1, e_3\ast e_3=r_{33}{e}_1.$$

Let $e_1^{\prime }=r_{33}{e}_1$. Then,

$$e_2\ast e_3=\mp \sqrt{-1}{e}_1^{\prime }, e_3\ast e_3=r_{33}{e}_1^{\prime }.$$

where $e_3^{\prime }=e_3\mp \sqrt{-1}{e}_2$. It follows that

$$\begin{array}{ll}& e_2\ast e_3^{\prime }=\mp \sqrt{-1}{e}_1^{\prime }\\ & e_3^{\prime }\ast e_3^{\prime }=0,\\ & e_3^{\prime }\ast e_2=0.\end{array}$$

Then, $(B,\ast )$ is of type $A_5$ by simply reordering and tuning coefficients.

The cases $r_{22}\ne 0,r_{33}=0$ and $r_{22}=0,r_{33}\ne 0$ are symmetric. So, if $r_{22}\ne 0,r_{33}=0$, $(B,\ast )$ is of type $A_5$. If $r_{22}=r_{33}=0$, $(B,\ast )$ is of type $A_1$.

For $B_3,e_2{e}_2=e_1,e_3{e}_3=e_1$ and $=\left(\begin{array}{ccc}{r}_{11}& 0& 0\\ r_{21}& r_{22}^{\pm }& r_{23}\\ r_{31}& r_{23}& r_{33}^{\pm }\left(\mathrm{o}\mathrm{r}{r}_{33}^{\mp }\right)\end{array}\right)$ $\left(r_{11}\ne 0\right)$, where $r_{22}^{\pm }=r_{33}^{\pm }=r_{11}\pm {\left(r_{11}^2-r_{23}^2\right)}^{\frac{1}{2}}$.

For $r_{33}^{\pm }$, the subsequent equality is attained.

$$e_2\ast e_2=r_{22}^{\pm }{e}_1, e_2\ast e_3=r_{23}{e}_1, e_3\ast e_2=r_{23}{e}_1, e_3\ast e_3=r_{33}^{\pm }{e}_1$$

The others are zero. Let $e_2^{\prime }=e_2-e_3, e_3^{\prime }=e_2+e_3$. Then,

$$\begin{array}{ll}& e_2^{\prime }\ast e_3^{\prime }=e_3^{\prime }\ast e_2^{\prime }=0,\\ & e_2^{\prime }\ast e_2^{\prime }=\left(r_{22}^{\pm }+r_{33}^{\pm }-2r_{23}\right)e_1,\\ & e_3^{\prime }\ast e_3^{\prime }=\left(r_{22}^{\pm }+r_{33}^{\pm }+2r_{23}\right)e_1\end{array}$$

The others are zero.

The following cases are observable:

(1)

$r_{22}+r_{33}=r_{23}=0$. So, $(B,\ast )$ is obviously of type $A_1$.

(2)

$r_{22}^{\pm }+r_{33}^{\pm }=2r_{23},r_{22}^{\pm }+r_{33}^{\pm }\ne -2r_{23}$. So, $(B,\ast )$ is of type $A_2$.

(3)

${ r}_{22}^{\pm }+r_{33}^{\pm }\ne 2r_{23},r_{22}^{\pm }+r_{33}^{\pm }=-2r_{23}$. So, $(B,\ast )$ is of type $A_2$.

(4)

${ }^{\pm }{r}_{22}+{ }^{\pm }{r}_{33}\ne 2r_{23},r_{22}^{\pm }+r_{33}^{\pm }\ne -2r_{23}$. In this case, $(B,\ast )$ is of type $A_3$.

If the same argument is applied to $r_{33}^{\mp }$, similar outcomes are attained as above, so $r_{33}^{\pm }$ is replaced by $r_{33}^{\mp }$.

For the other CAAs, $B_i$, the results can be similarly attained. □

Remark 2.

The ZAs of the types $A_4$ and $A_6$ cannot be attained from three-dimensional CAAs and their RBOs through Equation (2).

5. Conclusions

In summary, we have obtained the RBO on a ZA that must be the one on its sub-adjacent algebra. We provide all the RBOs on two- and three-dimensional ZAs. Finally, ZAs are also realized in low dimensions of the RBOs of commutative associative algebras, and not all ZAs can be obtained in this way.