Analysis of Complex Entities in Algebra B

1. Products in Quaternion and ${\mathbb{C}}^2$ Algebras

The product of quaternions is defined as [2]:

$$(a_1,a_2,a_3,a_4)(b_1,b_2,b_3,b_4)\equiv$$

$$(a_1{b}_1-a_2{b}_2-a_3{b}_3-a_4{b}_4,a_1{b}_2+a_2{b}_1+a_3{b}_4-a_4{b}_3,a_1{b}_3-a_2{b}_4+a_3{b}_1+a_4{b}_2,a_1{b}_4+a_2{b}_3-a_3{b}_2+a_4{b}_1).$$

And the product in ${\mathbb{C}}^2$ is defined as [3,6]:

$$(a_1,a_2)(b_1,b_2)\equiv (a_1{b}_1-a_2{b}_2,a_1{b}_2+a_2{b}_1).$$

2. Product in Algebra B

The product ⊙ in algebra B is defined as [1]:

$$(a_1,a_2,a_3)\odot (b_1,b_2,b_3)\equiv (a_1{b}_1-a_2{b}_2-a_3{b}_3,a_1{b}_2+a_2{b}_1+a_3{b}_2,a_1{b}_3+a_3{b}_2+a_3{b}_1).$$

Both the quaternionic product and the product in ${\mathbb{C}}^2$ are different from the product in B. It is not sufficient to nullify one or another component to derive algebra B from quaternions. Similarly, it does not make sense to impose an additional dimension to ${\mathbb{C}}^2$ to reach the product of algebra B. Therefore, from algebra B, the complex elements that arise must be different from the complex elements of the set ${\mathbb{C}}^2$ and the quaternions.

3. Representations in Quaternion and ${\mathbb{C}}^2$ Algebras

Quaternions can be represented as:

$$(a_1,a_2,a_3,a_4)=a_1+a_{2}i+a_{3}j+a_{4}k,i^2=j^2=k^2=-1.$$

And complex numbers in ${\mathbb{C}}^2$ as:

$$(a_1,a_2)=a_1+a_{2}i,i^2=-1.$$

In algebra B, Bermejo proposes the presence of the representation [1]:

$$(a_1,a_2,a_3)=a_1+a_{2}i+a_{3}j,$$

though he does not define it strictly, i.e., from an algebra isomorphism.

4. Defining Treons

Motivated by investigating this equivalent representation, we seek to demonstrate that this representation is mathematically correct. This implies that there must exist an isomorphism between the structure $(a_1,a_2,a_3)$ and $a_1+a_{2}i+a_{3}j$; both referenced by Bermejo in his definition and analysis of algebra B[1]. We assume that the field over which the algebra is defined is the real field $\mathbb{R}$. Accordingly, we will seek an isomorphism between the real field ${\mathbb{R}}^3$ and the field of complex entities with structure $a_1+a_{2}i+a_{3}j$ which we will call "treons" or "treonic numbers" to differentiate from the term "trions" used in various disciplines and from hypercomplex numbers called "ternions".

We define treons as:

$$a_1+a_{2}i+a_{3}j,$$

such that $a_1,a_2,a_3\in \mathbb{R}$ and $i^2=j^2=-1$.

We assumed that the treonic elements $a_1+a_{2}i+a_{3}j$ are elements of an arbitrary algebra A.

5. Addition and Product in Treons

The addition + in A is:

$$(a_1+a_{2}i+a_{3}j)+(b_1+b_{2}i+b_{3}j)=(a_1+b_1)+(a_2+b_2)i+(a_3+b_3)j,$$

and the product ⊗ in A is:

$$(a_1+a_{2}i+a_{3}j)\otimes (b_1+b_{2}i+b_{3}j)=(a_1{b}_1-a_2{b}_2-a_3{b}_3)+(a_1{b}_2+a_2{b}_1)i+(a_1{b}_3+a_3{b}_1)j+a_{2}i{b}_{3}j+a_{3}j{b}_{2}i.$$

Note that we have not imposed a definition, but have simply grouped the terms in the addition + by factoring out i and j without altering their action on the right on the elements of the field. We have also considered the distributive property of the product ⊗ over +.

6. Isomorphism with ${\mathbb{R}}^3$

Now, we analyze ${\mathbb{R}}^3$: The addition + in algebra B is defined as:

$$(a_1,a_2,a_3)+(b_1,b_2,b_3)\equiv (a_1+b_1,a_2+b_2,a_3+b_3).$$

While the product ⊙ in B was previously defined as:

$$(a_1,a_2,a_3)\odot (b_1,b_2,b_3)\equiv (a_1{b}_1-a_2{b}_2-a_3{b}_3,a_1{b}_2+a_2{b}_1+a_3{b}_2,a_1{b}_3+a_3{b}_2+a_3{b}_1).$$

Under this product operation, an expression of the type $a_{3}j{b}_{2}i$ can be expressed as [1]:

$$\begin{array}{cc}\hfill a_{3}j\odot b_{2}i& =a_3{b}_{2}j\odot i\hfill \\ \hfill & =\left(a_3{b}_2\right)((0,0,1)\odot (0,1,0))\hfill \\ \hfill & =a_3{b}_2(0,1,1)\hfill \\ \hfill & =(0,a_3{b}_2,a_3{b}_2)\hfill \\ \hfill & =a_3{b}_2(0,1,0)+a_3{b}_2(0,0,1)\hfill \\ \hfill & =a_3{b}_{2}i+a_3{b}_{2}j.\hfill \end{array}$$

Since in algebra B[1] we have the property of orthomulearity, $ai\odot bj=0$, expressions of the type $a_{2}i{b}_{3}j$ would remain:

$$a_{2}i{b}_{3}j=a_2{b}_{3}i\odot j=0.$$

Taking into account that under the product in B, $a_{3}j{b}_{2}i=a_3{b}_{2}i+a_3{b}_{2}j$ and $a_{2}i{b}_{3}j=0$, then,

$$(a_1+a_{2}i+a_{3}j)\otimes (b_1+b_{2}i+b_{3}j)=(a_1{b}_1-a_2{b}_2-a_3{b}_3)+(a_1{b}_2+a_2{b}_1+a_3{b}_2)i+(a_1{b}_3+a_3{b}_1+a_3{b}_2)j.$$

This has a preserved structure with respect to the product in B:

$$(a_1,a_2,a_3)\odot (b_1,b_2,b_3)=(a_1{b}_1-a_2{b}_2-a_3{b}_3,a_1{b}_2+a_2{b}_1+a_3{b}_2,a_1{b}_3+a_3{b}_2+a_3{b}_1).$$

Thus, for $(c_1,c_2,c_3)=c_1+c_{2}i+c_{3}j$, such that $(c_1,c_2,c_3)\in {\mathbb{R}}^3$ and $(c_1+c_{2}i+c_{3}j)\in A$, we can consider:

$$c_1=\mathrm{Re}\left(c\right),c_2={\mathrm{Im}}_1\left(c\right),c_3={\mathrm{Im}}_2\left(c\right),$$

where $\mathrm{Re}\left(c\right)$ is the real part of c, ${\mathrm{Im}}_1\left(c\right)$ is the first imaginary part of c, and ${\mathrm{Im}}_2\left(c\right)$ is the second imaginary part of c.

With this, we have sufficient data to define an isomorphism $\Phi$ such that:

$$\Phi :A\to {\mathbb{R}}^3,$$

$$\Phi (a_1+a_{2}i+a_{3}j)=(a_1,a_2,a_3).$$

7. Verification of Isomorphism Properties

7.1. Preservation of Addition

$$\Phi ((a_1+a_{2}i+a_{3}j)+(b_1+b_{2}i+b_{3}j))=$$

$$\Phi (a_1+b_1+(a_2+b_2)i+(a_3+b_3)j)=(a_1+b_1,a_2+b_2,a_3+b_3).$$

7.2. Preservation of Product

$$\Phi ((a_1+a_{2}i+a_{3}j)\otimes (b_1+b_{2}i+b_{3}j))=$$

$$\Phi ((a_1{b}_1-a_2{b}_2-a_3{b}_3)+(a_1{b}_2+a_2{b}_1+a_3{b}_2)i+(a_1{b}_3+a_3{b}_2+a_3{b}_1)j)=$$

$$(a_1{b}_1-a_2{b}_2-a_3{b}_3,a_1{b}_2+a_2{b}_1+a_3{b}_2,a_1{b}_3+a_3{b}_2+a_3{b}_1).$$

7.3. Isomorphism Verification

7.3.1. Morphism Verification

$\Phi$ is an algebra morphism (homomorphism) if [7,8,9]:

$$1)\Phi :A\to B,$$

$$\Phi ((a_1+a_{2}i+a_{3}j)\otimes (b_1+b_{2}i+b_{3}j))=\Phi ((a_1,a_2,a_3)\odot \Phi (b_1,b_2,b_3)),$$

and moreover:

$$2)\Phi {\mathrm{id}}_A={\mathrm{id}}_B,$$

where ${\mathrm{id}}_A$ is the identity under the product ⊗ in A, and ${\mathrm{id}}_B$ is the identity under the product ⊙ in B.

The first condition holds:

$$\Phi ((a_1+a_{2}i+a_{3}j)\otimes (b_1+b_{2}i+b_{3}j))=(a_1{b}_1-a_2{b}_2-a_3{b}_3,a_1{b}_2+a_2{b}_1+a_3{b}_2,a_1{b}_3+a_3{b}_1+a_3{b}_2),$$

$$\Phi (a_1,a_2,a_3)\odot \Phi (b_1,b_2,b_3)=(a_1{b}_1-a_2{b}_2-a_3{b}_3,a_1{b}_2+a_2{b}_1+a_3{b}_2,a_1{b}_3+a_3{b}_2+a_3{b}_1),$$

where we used $\Phi \left(a_k\right)\Phi \left(b_l\right)=(a_k+0i+0j)(b_l+0i+0j)=a_k{b}_l$.

The second condition also holds:

$$\Phi {\mathrm{id}}_A={\mathrm{id}}_B,$$

The identity in A is defined as ${\mathrm{id}}_A\equiv (1+0i+0j)$:

$$(a_1+a_{2}i+a_{3}j)\otimes (1+0i+0j)=(1+0i+0j)\otimes (a_1+a_{2}i+a_{3}j)=(a_1+a_{2}i+a_{3}j),$$

Therefore:

$$\Phi (1+0i+0j)=(1,0,0).$$

Since in algebra B we have ${\mathrm{id}}_B=(1,0,0)$, it is verified that $\Phi$ is an algebra morphism between A and B.

For $\Phi$ to be an isomorphism, $\Phi$ must be both a monomorphism and an epimorphism [7,8,10].

7.3.2. Monomorphism Verification (Injectivity)

Assuming $\Phi \left(a\right)=\Phi \left(b\right)$. Then:

$$\Phi (a_1+a_{2}i+a_{3}j)=\Phi (b_1+b_{2}i+b_{3}j)\Rightarrow (a_1,a_2,a_3)=(b_1,b_2,b_3).$$

This implies that:

$$a_1=b_1,a_2=b_2,a_3=b_3.$$

Therefore:

$$a_1+a_{2}i+a_{3}j=b_1+b_{2}i+b_{3}j.$$

Thus, $\Phi$ is an monomorphism.

Through the kernel, we can equally verify this. We define the kernel of $\Phi$ as the set of elements in A that map to the identity element of addition in B [7,8,10]. In our case:

$$\mathrm{Ker}(\Phi )=\{a\in A\mid \Phi (a)=0\}.$$

For $a=a_1+a_{2}i+a_{3}j$, we have $\Phi \left(a\right)=(a_1,a_2,a_3)$. Therefore:

$$\Phi \left(a\right)=0\Rightarrow (a_1,a_2,a_3)=(0,0,0).$$

This implies that:

$$a_1=0,a_2=0,a_3=0.$$

Therefore:

$$a=0+0i+0j=0.$$

Thus, the only element in the kernel of $\Phi$ is the identity element of addition in A.

The fact that $ker(\Phi )$ is trivial (i.e., $ker(\Phi )=\left\{0\right\}$) implies that the homomorphism is injective [7,8,10].

7.3.3. Epimorphism Verification (Surjectivity)

To verify that $\Phi$ is an epimorphism, we considered any $(a_1,a_2,a_3)\in {\mathbb{R}}^3$. There exists a treon $a_1+a_{2}i+a_{3}j\in A$ such that:

$$\Phi (a_1+a_{2}i+a_{3}j)=(a_1,a_2,a_3).$$

Thus, $\Phi$ is surjective, as by definition every element in ${\mathbb{R}}^3$ has a preimage in A. Thus, by definition, it is an epimorphism.

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With all this, we have the necessary tools to tackle the search for Euler’s identity in the context of algebra B. The isomorphism of algebras allows us to conduct a well-defined analysis of algebra B in the form of treons. With these, we proceed to deduce the form of Euler’s identity in our algebra.

8. Analysis of Treons and Their Complex Conjugates

8.1. Definition of a Complex Entity

Considering Bermejo’s work on algebra B [1], a complex entity can be described as:

$$b\equiv b_1·(1,0,0)+b_2·(0,1,0)+b_3·(0,0,1)=(b_1,b_2,b_3),$$

where $b_1=\Re \left(b\right)$, $b_2={\Im }_1\left(b\right)$, and $b_3={\Im }_2\left(b\right)$. $\Re \left(b\right)$ is the real part of b, ${\Im }_1\left(b\right)$ is the first imaginary part of b, and ${\Im }_2\left(b\right)$ is the second imaginary part of b. It should be noted that the definition is equivalent to the one we articulated in Section 6: Isomorphism with ${\mathbb{R}}^3$.

8.2. Definition of the Complex Conjugate

The complex conjugate, $b^{*}$, was obtained by changing $i\equiv (0,1,0)$ or $j\equiv (0,0,1)$ [1] to their respective additive inverses in algebra B. Thus:

$$\begin{array}{cc}\hfill & b^{\left({*}_i\right)}=b_1·(1,0,0)-b_2·(0,1,0)+b_3·(0,0,1),\hfill \\ \hfill & b^{\left({*}_j\right)}=b_1·(1,0,0)+b_2·(0,1,0)-b_3·(0,0,1),\hfill \\ \hfill & b^{\left({*}_{i,j}\right)}=b_1(1,0,0)-b_2(0,1,0)-b_3(0,0,1),\hfill \end{array}$$

where $b^{\left({*}_i\right)}$ denotes conjugation in i, and $b^{\left({*}_j\right)}$ denotes conjugation in j.

8.3. Powers of a Complex Entity

We define the product $b^2\equiv b\odot b$ as:

$$b^2=(b_1,b_2,b_3)\odot (b_1,b_2,b_3)=(b_1·b_1-b_2·b_2-b_3·b_3,b_1·b_2+b_2·b_1+b_3·b_2,b_1·b_3+b_3·b_2+b_3·b_1).$$

Hence, simplifying the notation:

$$b_1^2=b_1·b_1\equiv b_1{b}_1.$$

Thus, $b^2$ becomes:

$$b^2=(b_1^2-b_2^2-b_3^2,2b_1{b}_2+b_2{b}_3,2b_1{b}_3+b_3{b}_2),$$

where we use the notation $2b$ to denote $b+b$.

We define higher powers as follows:

$$\begin{array}{cc}\hfill b^3& \equiv (b\odot b)\odot b=(b_1^2-b_2^2-b_3^2,2b_1{b}_2+b_2{b}_3,2b_1{b}_3+b_3{b}_2)\odot b,\hfill \\ \hfill b^4& \equiv \left(\right(b\odot b)\odot b)\odot b,\hfill \\ \hfill b^5& \equiv \left(\right((b\odot b)\odot b)\odot b)\odot b,\hfill \end{array}$$

and so on.

We also define the quantity $\langle b^2\rangle \equiv b\odot b^{\left({*}_{i,j}\right)}$. Therefore:

$$\langle b^2\rangle =(b_1,b_2,b_3)\odot {(b_1,b_2,b_3)}^{\left({*}_{i,j}\right)}={\left(\right|b|}^2,2b_1{b}_2+b_2{b}_3,2b_1{b}_3+b_3{b}_2),$$

where ${\left|b\right|}^2\equiv b_1^2+b_2^2+b_3^2$, which we call the "squared norm" of the vector. Note that ${\left|b\right|}^2$ results from the definition of the product of the vector b with its complex conjugate $\left({*}_{i,j}\right)$, as in the field of complex numbers [11]. However, in algebra B, this quantity appears in the first component (the real component) of the vector.

9. Derivation of Euler’s Identity

9.1. Step 1: Power Series

We perform the following power series expansion [12]:

$$\sum _0^{\infty }\frac{b^n}{n!}\equiv \frac{b^0}{0!}+\frac{b^1}{1!}+\frac{b^2}{2!}+\frac{b^3}{3!}+\cdots +\frac{b^n}{n!}.$$

where ! denotes some kind of "factorial" in the field F of algebra B [1]; therefore, the multiplication of the factorial corresponds to the multiplication · defined for the elements of the field. On the other hand, we have: 0 is the identity element of addition in F, 1 is the identity element of multiplication in F, 2 is the successor of 1, 3 is the successor of 2, and so on. We also define: $0!\equiv 1$, $1!\equiv 1$, $b^0\equiv (1,0,0)$, and $b^1\equiv b$. Assuming that the field F is the real field $\mathbb{R}$, then ${\sum }_{n=0}^{\infty }\frac{b^n}{n!}$ represents the Taylor series expansion of $e^b$ [12,13].

9.2. Step 2: Taylor Series for $e^{i{b}_2}$

We utilize the powers of the components of b multiplied by their corresponding imaginary units to perform a Taylor series expansion [13], considering algebra B over the real field $\mathbb{R}$. For the case of the first imaginary component, we have:

$$e^{i{b}_2}=\sum _0^{\infty }\frac{{\left(i{b}_2\right)}^n}{n!}=\frac{{\left(i{b}_2\right)}^0}{0!}+\frac{{\left(i{b}_2\right)}^1}{1!}+\frac{{\left(i{b}_2\right)}^2}{2!}+\frac{{\left(i{b}_2\right)}^3}{3!}+\cdots +\frac{{\left(i{b}_2\right)}^n}{n!},$$

$$=(1,0,0)+b_2(0,1,0)+\frac{{\left(b_2(0,1,0)\right)}^2}{2!}+\frac{{\left(b_2(0,1,0)\right)}^3}{3!}+\cdots +\frac{{\left(b_2(0,1,0)\right)}^n}{n!}.$$

Note that ${\left(i{b}_2\right)}^0={\left(b_2(0,1,0)\right)}^0\equiv (1,0,0)$ and ${\left(i{b}_2\right)}^1=b_2(0,1,0)\equiv \left(i{b}_2\right)$.

9.2.1. Analyzing the Powers of $(0,1,0)$

$$\begin{array}{cc}\hfill i^0& \equiv (1,0,0)\equiv i{d}_{\odot }(i{d}_{\odot }\mathrm{denotes}\mathrm{the}\mathrm{identity}\mathrm{element}\mathrm{of}\mathrm{the}\mathrm{product}\odot \mathrm{in}\mathrm{algebra}\mathrm{B}),\hfill \\ \hfill i^1& \equiv (0,1,0),\hfill \\ \hfill i^2& \equiv (0,1,0)\odot (0,1,0)=(-1,0,0)=-i{d}_{\odot },\hfill \\ \hfill i^3& \equiv ((0,1,0)\odot (0,1,0))\odot (0,1,0)=(-1,0,0)\odot (0,1,0)=(0,-1,0)=-i=i^{*},\hfill \\ \hfill i^4& =(0,-1,0)\odot (0,1,0)=(-1,0,0)=-i{d}_{\odot },\hfill \\ \hfill i^5& =(-1,0,0)\odot (0,1,0)=(0,-1,0)=i^{*},\hfill \\ \hfill i^6& =(0,-1,0)\odot (0,1,0)=i^4=(-1,0,0)=-i{d}_{\odot },\hfill \\ \hfill i^7& =(-1,0,0)\odot (0,1,0)=i^5=i^{*}.\hfill \end{array}$$

The sequence is cyclic and has the form, from the second power: $-i{d}_{\odot },-i,-i{d}_{\odot },-i,-i{d}_{\odot },-i$. Therefore, the power series expansion becomes:

$$e^{i{b}_2}=\sum _0^{\infty }\frac{{\left(i{b}_2\right)}^n}{n!}=i{d}_{\odot }+b_{2}i-i{d}_{\odot }\left(\frac{b_2^2}{2!}+\frac{b_2^4}{4!}+\frac{b_2^6}{6!}+\cdots \right)+i^{*}\left(\frac{b_2^3}{3!}+\frac{b_2^5}{5!}+\cdots \right),$$

$$=i{d}_{\odot }+b_{2}i-i{d}_{\odot }cosh\left(b_2\right)-isinh\left(b_2\right),$$

where $cosh\left(\right)$ and $sinh\left(\right)$ denote hyperbolic cosines and sines [14], respectively.

9.2.2. Analyzing the Powers of $(0,0,1)$

$$\begin{array}{cc}\hfill j^0& \equiv (1,0,0)\equiv i{d}_{\odot },\hfill \\ \hfill j^1& \equiv (0,0,1),\hfill \\ \hfill j^2& \equiv (0,0,1)\odot (0,0,1)=(-1,0,0)=-i{d}_{\odot },\hfill \\ \hfill j^3& \equiv ((0,0,1)\odot (0,0,1))\odot (0,0,1)=(-1,0,0)\odot (0,0,1)=(0,0,-1)=-j=j^{*},\hfill \\ \hfill j^4& =(0,0,-1)\odot (0,0,1)=(-1,0,0)=-i{d}_{\odot },\hfill \\ \hfill j^5& =(-1,0,0)\odot (0,0,1)=(0,0,-1)=j^{*},\hfill \\ \hfill j^6& =(0,-1,0)\odot (0,0,1)=i^4=(-1,0,0)=-i{d}_{\odot },\hfill \\ \hfill j^7& =(-1,0,0)\odot (0,0,1)=j^5=j^{*}.\hfill \end{array}$$

Thus:

$$e^{j{b}_3}=i{d}_{\odot }+b_{3}j-i{d}_{\odot }cosh\left(b_3\right)-jsinh\left(b_3\right).$$

For the full vector, with its two imaginary components, we have:

$$e^{(b_1+b_{2}i+b_{3}j)}=e^{b_1}(i{d}_{\odot }+b_{2}i-i{d}_{\odot }cosh\left(b_2\right)-isinh\left(b_2\right))(i{d}_{\odot }+b_{3}j-i{d}_{\odot }cosh\left(b_3\right)-jsinh\left(b_3\right)).$$

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This demonstrates that Euler’s identity in algebra B, or equivalently in the algebra of treons, possesses a structure that preserves the connection between the exponential function and trigonometric functions. However, instead of the standard sine and cosine functions, hyperbolic sine and hyperbolic cosine functions appear.