# 3-IDEMPOTENT ALGEBRA

FARID MESSELMI

ABSTRACT. The purpose of this paper is to present a novel concept in the field of hypercomplex numbers, termed 3-idempotent numbers. We start by defining this concept and subsequently explore its algebraic properties. Furthermore, we will try to extend the exponential function to the 3-idempotent algebra.

## 1. INTRODUCTION

A hypercomplex number is defined as an element of a finite-dimensional unital algebra over the real number field. The study of hypercomplex numbers in the late 19th century established the foundation for modern group representation theory, see [6, 7].

In 2-dimensional case, it was proven that three are exactly three types of unital algebras over the real numbers up to isomorphism, namely the ordinary complex numbers, the dual numbers and the split (or hyperbolic numbers). Notably, every 2-dimensional unital algebra over the reals is both associative and commutative.

In the 16th century, the Italian mathematicians Jérôme Cardan, Raphaël Bombelli, and Tartaglia introduced complex numbers to solve cubic equations using Cardan's formulas, even utilizing numbers with negative squares, for more details about complexe numbers see the references [4, 1].

The theory of dual numbers was initially introduced by W. K. Clifford in 1873. Later, in 1891, E. Study recognized that the associative algebra of dual numbers was perfect for describing the group of motions in three-dimensional space, dual numbers have been explored in numerous papers, namely [2, 5, 8, 10].

Split (or hyperbolic) numbers were first introduced in 1848 when James Cockle presented his Tessarines. In 1882, W. K. Clifford used split numbers to represent sums of spins, naming the elements motors. During the twentieth century, split numbers became a widely-used tool for describing Lorentz transformations in special relativity, as velocity changes between reference frames were elegantly captured by hyperbolic rotations. Split numbers were considered in [3, 9].

The main object of this paper is to explore the study of a new concept in the field of hypercomplex that will be called 3-idempotent algebra and denoted by $I_2$. The paper is structured as follows. In the second section, we introduce the concept of 3-idempotent algebra. This involves considering a 3-idempotent unit in order to construct a 3-dimensional algebra over the field of real numbers. We will emphasize its principal algebraic properties, including the matrix representation,

<footnote n="2000">2000 Mathematics Subject Classification. 15A66, 30G35.</footnote>

<footnote>Key words and phrases. 3-idempotent algebra, 3-idempotent numbers, 3-idempotent exponential function.</footnote>
the null basis, and their ideals. Additionally, we establish some intriguing results. The third section will be consecrated to studying the 3-idempotent exponential function, defined as an extension of the real one.

## 2. 3-IDEMPOTENT ALGEBRA

We define a 3-idempotent number $z$ as an ordered 3-tuple of real numbers $(x_0, x_1, x_2)$ associated with the real unit 1 and the powers of the 3-idempotent unit $\lambda$, such that $\lambda$ satisfies $\lambda^3 = \lambda$ and $\lambda \neq 0, 1$. Therefore, a 3-idempotent number has the form

$$z = x_0 + x_1\lambda + x_2\lambda^2, \qquad (2.1)$$

and we admit that $\lambda^0 = 1$.

We denote by $\mathbb{I}_2$ the set of 3-idempotent numbers, that is,

$$\mathbb{I}_2 = \{z = x_0 + x_1\lambda + x_2\lambda^2 \mid (x_0, x_1, x_2) \in \mathbb{R}^3 \text{ and } \lambda^3 = \lambda\} \qquad (2.2)$$

There are various manner to choose the multisplit unit number $\lambda$. For instance, we can choose the real matrix

$$\lambda = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}. \qquad (2.3)$$

Doing so enables us to write down immediately the rules for addition and multiplication as follows

$$(x_0 + x_1\lambda + x_2\lambda^2) + (y_0 + y_1\lambda + y_2\lambda^2) = (x_0 + y_0) + (x_1 + y_1)\lambda + (x_2 + y_2)\lambda^2, \qquad (2.4)$$

$$(x_0 + x_1\lambda + x_2\lambda^2) \cdot (y_0 + y_1\lambda + y_2\lambda^2) = x_0y_0 + (x_0y_1 + x_1y_0 + x_2y_1)\lambda + (x_0y_2 + x_1y_1 + x_2y_0 + x_2y_2)\lambda^2. \qquad (2.5)$$

If $z = x_0 + x_1\lambda + x_2\lambda^2$ is a 3-idempotent number, we will denote by real $(z)$ the real part of $z$ given by

$$\text{real}(z) = x_0. \qquad (2.6)$$

Thus, the 3-idempotent numbers form a commutative ring with characteristic 0. Moreover the inherited multiplication gives the multisplit 3-idempotent numbers the structure of 3-dimensional associative, commutative and unitary generalized Clifford Algebra.

In abstract algebra terms, the algebra of 3-idempotent numbers can be described as the polynomial ring $\mathbb{R}[X]$ quotiented with the ideal generated by the polynomial $X^3 - X$, meaning that

$$\mathbb{I}_2 \approx \frac{\mathbb{R}[X]}{\langle X^3 - X \rangle}. \qquad (2.7)$$

It is also noteworthy to verify that every 3-idempotent number has matrix representation. For this aim, let us denote by $\mathcal{G}_3(\mathbb{R})$ the subset of $\mathcal{M}_3(\mathbb{R})$ given by

$$\mathcal{G}_3(\mathbb{R}) = \left\{ A \in \mathcal{M}_3(\mathbb{R}) \mid A = \begin{bmatrix} a_0 & 0 & 0 \\ a_1 & a_0 + a_2 & a_1 \\ a_2 & a_1 & a_0 + a_2 \end{bmatrix} \right\} \qquad (2.8)$$
Obviously, $\mathcal{G}_3(\mathbb{R})$ is a subring of $\mathcal{M}_3(\mathbb{R})$ that forms a 3-dimensional associative, commutative and unitary algebra, Introduce now the map given by

$$ \mathcal{R} : \mathbb{I}_2 \to \mathcal{G}_3(\mathbb{R}), \\ \mathcal{R}(x_0 + x_1\lambda + x_2\lambda^2) = \begin{bmatrix} x_0 & 0 & 0 \\ x_1 & x_0 + x_2 & x_1 \\ x_2 & x_1 & x_0 + x_2 \end{bmatrix} \qquad (2.9) $$

It is staiforward to show that $\mathcal{R}$ is an isomorphism of algebras. The matrice $\mathcal{R}$ is called the matrix representation of 3-idempotent numbers.

Denoting now by $\mathcal{G}_3^0(\mathbb{R})$ the subset of $\mathcal{G}_3(\mathbb{R})$ defined as

$$ \mathcal{G}_3^0(\mathbb{R}) = \{A \in \mathcal{G}_3(\mathbb{R}) \mid \det A = 0\}. \qquad (2.10) $$

It is evident that $\mathcal{G}_3(\mathbb{R}) - \mathcal{G}_3^0(\mathbb{R})$ is a subgroup of $GL(3)$. The null part of $\mathbb{I}_2$ denoted by $\mathcal{D}_2$ is the set

$$ \mathcal{D}_2 = \mathcal{R}^{-1}(\mathcal{G}_3^0(\mathbb{R})), \qquad (2.11) $$

consisting of the zero divisors of $\mathbb{I}_2$. As consequence, the subset of $\mathbb{I}_2$ given by $\mathbb{I}_2^* = \mathbb{I}_2 - \mathcal{D}_2$ is a multiplicative group corresponding to the unit group of the ring $\mathbb{I}_2$.

A simple calculation shows that

$$ \det \mathcal{R}(z) = x_0(x_0 - x_1 + x_2)(x_0 + x_1 + x_2). \qquad (2.12) $$

Hence, the null part $\mathcal{D}_2$ can be also written

$$ \mathcal{D}_2 = J_0 \cup J_1 \cup J_2, \qquad (2.13) $$

where

$$ \left\{ \begin{array}{l} J_0 = \{z = x_1\lambda + x_2\lambda^2\}, \\ J_1 = \{z = x_0 + x_1\lambda - (x_0 + x_1)\lambda^2\}, \\ J_2 = \{z = x_0 + x_1\lambda + (x_1 - x_0)\lambda^2\}. \end{array} \right. \qquad (2.14) $$

**Proposition 1.** The sets $J_0$, $J_1$ and $J_2$ are ideals of the ring $\mathbb{I}_2$. Moreover, the following identities holed

$$ \left\{ \begin{array}{l} J_0 = \langle \lambda \rangle, \\ J_1 = \langle 1 - \lambda \rangle, \\ J_2 = \langle 1 + \lambda \rangle. \end{array} \right. \qquad (2.15) $$

The proof follows immediately.

Let us denote by $e_0$, $e_1$ and $e_2$ the elements of $\mathbb{I}_2$ given by

$$ \left\{ \begin{array}{l} e_0 = 1 - \lambda^2, \\ e_1 = \frac{1}{2}(-\lambda + \lambda^2), \\ e_2 = \frac{1}{2}(\lambda + \lambda^2). \end{array} \right. \qquad (2.16) $$

Through simple algebraic manipulations, we can derive the following result, which illustrates some properties of $e_0$, $e_1$ and $e_2$.
**Proposition 2.** The elements $e_0, e_1$ and $e_2$ satisfy the following

$$ egin{cases} e_0, e_1, e_2 \in \mathcal{D}_2 \\ e_0^3 = e_0^2 = e_0, \\ e_1^3 = e_1^2 = e_1, \\ e_2^3 = e_2^2 = e_2, \\ e_0 e_1 = e_0 e_2 = e_1 e_2 = 0. \end{cases} \tag{2.17} $$

The set $\{e_0, e_1, e_2\}$ will be called the null basis of $\mathbb{I}_2$.

*Proof.* Let $z = x_0 + x_1\lambda + x_2\lambda^2$ be an element of $\mathbb{I}_2$. So, $z$ can be written in the basis $\{e_0, e_1, e_2\}$

$$ z = \varphi_0(z)e_0 + \varphi_1(z)e_1 + \varphi_2(z)e_2. \tag{2.18} $$

Hence, we can infer

$$ egin{aligned} z &= \varphi_0(z)(1 - \lambda^2) + \frac{1}{2}\varphi_1(z)(-\lambda + \lambda^2) + \frac{1}{2}\varphi_2(z)(\lambda + \lambda^2) \\ &= \varphi_0(z) + \frac{1}{2}(-\varphi_1(z) + \varphi_2(z))\lambda + \frac{1}{2}(-2\varphi_0(z) + \varphi_1(z) + \varphi_2(z)). \end{aligned} $$

Then, since $z = x_0 + x_1\lambda + x_2\lambda^2$, we obtain the below formulas

$$ egin{cases} \varphi_0(z) = x_0, \\ \varphi_1(z) = x_0 + x_1 + x_2, \\ \varphi_1(z) = x_0 - x_1 + x_2. \end{cases} \tag{2.19} $$

□

Some other properties are subject of the following Theorem.

**Theorem 3. 1.** The maps $\varphi_i : \mathbb{I}_2 \to \mathbb{R}$, $i = 0, 1, 2$ are homomorphism of ring.

2. The kernels of $\varphi_i$ are given by

$$ \ker(\varphi_i) = J_0, \quad i = 0, 1, 2. \tag{2.20} $$

3. The following identities hold

$$ egin{cases} J_1 \cap J_2 = \langle e_0 \rangle, \\ J_0 \cap J_2 = \langle e_1 \rangle, \\ J_0 \cap J_1 = \langle e_2 \rangle. \end{cases} \tag{2.21} $$

4. The below isomorphism holds true

$$ \mathbb{I}_2 \simeq \frac{\mathbb{I}_2}{\langle e_0 \rangle} \times \frac{\mathbb{I}_2}{\langle e_1 \rangle} \times \frac{\mathbb{I}_2}{\langle e_2 \rangle}, \tag{2.22} $$

via the mapping

$$ z \in \mathbb{I}_2 \mapsto (z + \langle e_0 \rangle, z + \langle e_1 \rangle, z + \langle e_2 \rangle). \tag{2.23} $$

The three first assertions can be easily obtained. For the fourth one, it is straightforward to remark the ideals $\langle e_0 \rangle$, $\langle e_1 \rangle$ and $\langle e_2 \rangle$ are pairwise coprime. Thus, in order to conclude the proof it is enough to applied the Chinese remainder Theorem.
Let us introduce the map given by

$$ \left\{ \begin{array}{l} \varphi : \mathbb{I}_2 \to \mathbb{R} \times \mathbb{H}, \\ z = x_0 + x_1 \lambda + x_2 \lambda^2 \mapsto \varphi(z) = (x_0, x_1 + x_2 + x_3 h), \end{array} \right. \qquad (2.24) $$

where $\mathbb{H}$ represents the Clifford algebra of split numbers and $h$ is the unit split element satisfying $h^2 = 1$, see for more details about split numbers the references. The below result follows immediately.

**Proposition 4.** The map $\varphi$ is an isomorphism of algebras.

Denote now by $\tilde{\mathbb{I}}_2$ the subset of $\mathbb{I}_2$ defined as follows.

$$ \tilde{\mathbb{I}}_2 = \{z = x_0 + x_2 \lambda^2 \in \mathbb{I}_2 \mid (x_0, x_2) \in \mathbb{R}^2\}. \qquad (2.25) $$

**Theorem 5.** $\tilde{\mathbb{I}}_2$ is a subalgebra of $\mathbb{I}_2$. In addition, the following isomorphism holds

$$ \tilde{\mathbb{I}}_2 \simeq \mathbb{H}. \qquad (2.26) $$

*Proof.* Let $z = x_0 + x_2 \lambda^2$, $z' = x'_0 + x'_2 \lambda^2 \in \tilde{\mathbb{I}}_2$. It is clear that

$$ \left\{ \begin{array}{l} z + z' = x_0 + x'_0 + (x_2 + x'_2) \lambda^2 \in \tilde{\mathbb{I}}_2, \\ zz' = x_0 x'_0 + (x_0 x'_2 + x'_0 x_2 + x_2 x'_2) \lambda^2 \in \tilde{\mathbb{I}}_2. \end{array} \right. $$

We deduce that $\tilde{\mathbb{I}}_2$ is a subalgebra of $\mathbb{I}_2$. Furthermore, let us consider the map given by

$$ \left\{ \begin{array}{l} \psi : \tilde{\mathbb{I}}_2 \to \mathbb{H}, \\ z = x_0 + x_2 \lambda^2 \mapsto \psi(z) = x_0 + \frac{x_2}{2} + \frac{x_2}{2} h. \end{array} \right. $$

We have

$$ \begin{aligned} \psi(z+z') &= \psi(x_0 + x'_0 + (x_2+x'_2)\lambda^2) \\ &= x_0 + x'_0 + \frac{x_2+x'_2}{2} + \frac{x_2+x'_2}{2}h \\ &= \psi(z) + \psi(z'). \end{aligned} $$

In addition

$$ \begin{aligned} \psi(zz') &= \psi(x_0x'_0 + (x_0x'_2 + x'_0x_2 + x_2x'_2)\lambda^2) \\ &= x_0x'_0 + \frac{x_0x'_2 + x'_0x_2 + x_2x'_2}{2} + \frac{x_0x'_2 + x'_0x_2 + x_2x'_2}{2}h. \end{aligned} $$

Moreover, we find

$$ \begin{aligned} \psi(z)\psi(z') &= \left(x_0 + \frac{x_2}{2} + \frac{x_2}{2}h\right) \left(x'_0 + \frac{x'_2}{2} + \frac{x'_2}{2}h\right) \\ &= x_0x'_0 + x_0x'_0 + \frac{x_0x'_2 + x'_0x_2 + x_2x'_2}{2} + \frac{x_0x'_2 + x'_0x_2 + x_2x'_2}{2}h. \end{aligned} $$

As consequence, one can conclude that

$$ \psi(zz') = \psi(z)\psi(z'). $$

□
Further, the concept of conjugation can be also generalized to 3-idempotent numbers. This is the subject of the following definition.

**Definition 1.** For all $z \in \mathbb{I}_2^*$, the conjugate of $z$ denoted by $\bar{z}$ is the 3-idempotent number that satisfies

$$z\bar{z} = \det(\mathcal{R}(z)). \qquad (2.27)$$

The following result suggests some useful properties of the conjugate.

**Proposition 6.** The following statements hold

1. $\det(\mathcal{R}(\bar{z})) = \det(\mathcal{R}(z))^2$.

2. $\bar{z}$ can be calculated employing the formula

$$\bar{z} = (x_0 + x_2)^2 - x_1^2 - x_0x_1\lambda + (x_1^2 - x_2(x_0 + x_2))\lambda^2. \qquad (2.28)$$

3. The conjugation mapping

$$\left\{ \begin{array}{l} \mathcal{C}: \mathbb{I}_2^* \to \mathbb{I}_2^*, \\ z \mapsto \mathcal{C}(z) = \bar{z}, \end{array} \right. \qquad (2.29)$$

is an endomorphism of group having as Kernel

$$\ker(\mathcal{C}) = \{-1, +1\}. \qquad (2.30)$$

*Proof.* Since $z\bar{z} = \det(\mathcal{R}(z))$, we find

$$\det \mathcal{R}(z) \det \mathcal{R}(\bar{z}) = \det (\mathcal{R}(z))^3.$$

So, the first assertion follows. The second one can be easily obtained. To prove third assertion, let us consider two elements $z_1, z_2 \in \mathbb{I}_2^*$. Here we can write

$$\begin{aligned} z_1 z_2 \mathcal{C}(z_1 z_2) &= \det(\mathcal{R}(z_1 z_2)) \\ &= \det(\mathcal{R}(z_1)) \det(\mathcal{R}(z_2)) \\ &= z_1 z_2 \mathcal{C}(z_1) \mathcal{C}(z_2). \end{aligned}$$

This eventually yields the desired result. One the other hand, if $z \in \ker(\mathcal{C})$ we find

$$z = \det(\mathcal{R}(z)) \in \mathbb{R}.$$

It implies that

$$z = z^3.$$

Thus, the result is complete.

□

The concept of conjugate enables us to construct a structure of modulus over the multisplit algebra $\mathbb{I}_2$, given by map

$$\left\{ \begin{array}{l} \mathcal{N}: \mathbb{I}_2 \to \mathbb{R}_+, \\ \mathcal{N}(z) = |\det(\mathcal{R}(z))|^{\frac{1}{3}} = |z\bar{z}|^{\frac{1}{3}}. \end{array} \right. \qquad (2.31)$$

There is no chance that the modulus $\mathcal{N}$ induces a norm over the algebra $\mathbb{I}_2$. It is obvious that $\mathcal{N}$ satisfies the following properties

$$\left\{ \begin{array}{l} \mathcal{N}(z_1 z_2) = \mathcal{N}(z_1) \mathcal{N}(z_2) \quad \forall z_1, z_2 \in \mathbb{I}_2, \\ \mathcal{N}(z) = 0 \quad \text{iff } z \in \mathcal{D}_2. \end{array} \right. \qquad (2.32)$$
Thus, we can define the 3-idempotent disk and sphere of center $t \in \mathbb{I}_2$ and radius $r > 0$, respectively, by

$$D(t, r) = \{z \in \mathbb{I}_2 \mid \mathcal{N}(z-t) < r\}, \qquad (2.33)$$

$$S(t, r) = \{z \in \mathbb{I}_2 \mid \mathcal{N}(z-t) = r\}. \qquad (2.34)$$

It is worth noting that the algebra $\mathbb{I}_2$ endowed with the topology generated by the modulus $\mathcal{N}$ is not s Hausdorff space.

By virtue of the definition (2.38), we can affirm the following.

**Proposition 7.** The map

$$\left\{ \begin{array}{l} \mathbb{I}_2^* \to \mathbb{R}_+^*, \\ z \mapsto \mathcal{N}(z), \end{array} \right. \qquad (2.35)$$

is a homomorphism of groups where his kernel is given by

$$\ker(\mathcal{N}) = S(0, 1). \qquad (2.36)$$

The proof of this proposition is analogous to that of the previous one.

We introduce now the concepts of positive and negatives parts of $\mathbb{I}_2^*$ as follows.

$$\left\{ \begin{array}{l} \mathbb{I}_2^+ = \{z \in \mathbb{I}_2^* \mid \det(\mathcal{R}(z)) > 0\}, \\ \mathbb{I}_2^- = \{z \in \mathbb{I}_2^* \mid \det(\mathcal{R}(z)) < 0\}. \end{array} \right. \qquad (2.37)$$

In the following some interesting properties are collected and the proof directly follows.

**Proposition 8.** 1. $\mathbb{R}_+^* \subset \mathbb{I}_2^+$ and $\mathbb{R}_-^* \subset \mathbb{I}_2^-$.

2. If $z \in \mathbb{I}_2^+$ then $-z \in \mathbb{I}_2^-$ and if $z \in \mathbb{I}_2^-$ then $-z \in \mathbb{I}_2^+$.

3. If $z_1, z_2 \in \mathbb{I}_2^+$ then $z_1 z_2 \in \mathbb{I}_2^+$.

4. If $z \in \mathbb{I}_2^+$ then $\frac{1}{z} \in \mathbb{I}_2^+$.

5. $\forall z \in \mathbb{I}_2^*$ we have $\bar{z} \in \mathbb{I}_2^+$.

In particular the set $\mathbb{I}_2^+$ is a subgroup of $\mathbb{I}_2^*$.

The notion of absolute value can be also generalized to 3-idempotent numbers, as described in the following definition.

**Definition 2.** Given $z \in \mathbb{I}_2^*$. The absolute value of a $z$ is defined by

$$|z| = \begin{cases} z & \text{if } z \in \mathbb{I}_2^+; \\ -z & \text{if } z \in \mathbb{I}_2^- \end{cases} \qquad (2.38)$$

Obviously, the map $|\cdot|$ defines a homomorphism of groups between $\mathbb{I}_2^*$ and $\mathbb{I}_2^+$, whose

$$\ker(|\cdot|) = \{-1, +1\}. \qquad (2.39)$$

We introduce in the following the notion of positive unit sphere given by

$$S^+(0, 1) = \{z \in S(0, 1) \mid \det(\mathcal{R}(z)) > 0\}. \qquad (2.40)$$

It is straightforward to verify that $S^+(0, 1)$ is a subgroup of $S(0, 1)$. Moreover, we have

$$S^+(0, 1) \simeq \frac{S(0, 1)}{\{-1, +1\}}. \qquad (2.41)$$

The below proposition can be easily derived.
**Proposition 9. 1.** The subset denoted by G and given by

$$G = \{z = \pm 1 + x_1\lambda + x_2\lambda^2 \in \mathbb{I}_2^*\}, \qquad (2.42)$$

is a subgroup of $\mathbb{I}_2^*$.

In addition, the subset

$$G_+ = \{z = 1 + x_1\lambda + x_2\lambda^2 \in \mathbb{I}_2^*\}, \qquad (2.43)$$

is subgroup of G.

2. The subset E given by

$$E = \{z = \pm 1 + x_2\lambda^2 \in \mathbb{I}_2^*\}, \qquad (2.44)$$

is a subgroup of G.

In addition, the subset

$$E_+ = \{z = 1 + x_2\lambda^2 \in \mathbb{I}_2^*\}, \qquad (2.45)$$

is subgroup of E.

3. EXPONENTIAL FUNCTION

The 3–idempotent exponential function can be obtained as an extension of the exponential real function to the algebra of multisplit numbers. Nevertheless, we prefer to use the exponential of matrices.

Let $A \in \mathcal{G}_3(\mathbb{R})$ and suppose that $\|A\| < +\infty$ for some norm. It is well known that the exponential of A can be defined by the series

$$\exp(A) = e^A = \sum_{n=0}^{+\infty} \frac{1}{n!}A^n. \qquad (3.1)$$

In addition, the series converges normally in each bounded domain of $\mathcal{G}_3(\mathbb{R})$.

Since $A \in \mathcal{G}_3(\mathbb{R})$, we can verify that for all $n \in \mathbb{N}$ we have $A^n \in \mathcal{G}_3(\mathbb{R})$. Thus, by passage to the limit it follows that

$$\exp(A) \in \mathcal{G}_3(\mathbb{R}). \qquad (3.2)$$

Introducing now the function E of 3–idempotent variable defined for each $z \in \mathbb{I}_2$ by

$$E(z) = (\mathcal{R}^{-1} \circ \exp \circ \mathcal{R}) (z). \qquad (3.3)$$

By virtue of this definition, one has

$$\mathcal{R}(E(z)) = \exp(\mathcal{R}(z)), \qquad (3.4)$$

**Definition 3.** The 3–idempotent exponential function is defined as

$$\begin{cases} E : \mathbb{I}_2 \to \mathbb{I}_2 \\ z \mapsto E(z) \quad \forall z \in \mathbb{I}_2. \end{cases} \qquad (3.5)$$

Some properties of the 3–idempotent exponential function are collected in the following.

**Proposition 10. 1.** $E(z_1 + z_2) = E(z_1)E(z_2) \quad \forall z_1, z_2 \in \mathbb{I}_2$.

2. $E(0) = 1$.

3. $E(-z) = \frac{1}{E(z)} \quad \forall z \in \mathbb{I}_2$.

4. $E(z) = e^z \quad \forall z \in \mathbb{R}$.

5. $E(z) \in \mathbb{I}_2^+ \quad \forall z \in \mathbb{I}_2$.
Proof. 1. For every $z_1, z_2 \in \mathbb{I}_2$ we get, keeping in mind formula (3.4)

$$ \mathcal{R}(E(z_1 + z_2)) = e^{\mathcal{R}(z_1 + z_2)}. $$

Since, $\mathcal{R}(z_1)\mathcal{R}(z_2) = \mathcal{R}(z_2)\mathcal{R}(z_1)$, we find

$$ \begin{aligned} \mathcal{R}(E(z_1 + z_2)) &= e^{\mathcal{R}(z_1)}e^{\mathcal{R}(z_2)} \\ &= \mathcal{R}(E(z_1))\mathcal{R}(E(z_2)) \\ &= \mathcal{R}(E(z_1)E(z_2)). \end{aligned} $$

Therefore, the proof of the first statement is completed.

2. We have

$$ \begin{aligned} \mathcal{R}(E(0)) &= e^{\mathcal{R}(0)} \\ &= I_3 = \mathcal{R}(1). \end{aligned} $$

3. For every $z \in \mathbb{I}_2$, using the previous property, we obtain

$$ 1 = E(z)E(-z). $$

4. If $z \in \mathbb{R}$, one can find

$$ \begin{aligned} \mathcal{R}(E(z)) &= e^{zI_3} \\ &= e^z I_3 \\ &= e^z \mathcal{R}(1) \\ &= \mathcal{R}(e^z). \end{aligned} $$

So, $E(z) = e^z$.

5. Since $\det \mathcal{R}(E(z)) = \det e^{\mathcal{R}(z)}$, we get

$$ \begin{aligned} \det \mathcal{R}(E(z)) &= e^{\mathrm{tr}(\mathcal{R}(z))} \\ &= e^{3x_0+2x_2} > 0. \end{aligned} $$

Thus, $E(z) \in \mathbb{I}_2^+$.

This allows us to achieve the proof.

<figure><img src="image_1.png" /></figure>

In particular, the two groupes ($\mathbb{I}_2$, $+$) and ($\mathbb{I}_2^+$, $\cdot$) are isomorphic by means of the exponential function.

We will write $e^z$ instead of $E(z)$.

**Proposition 11.** For every $z = x_0 + x_1\lambda + x_2\lambda^2 \in \mathbb{I}_2$ the exponential function can be written as follows

$$ e^z = e^{x_0} (1 + e^{x_2} \sinh(x_1) \lambda + (e^{x_2} \cosh(x_1) - 1) \lambda^2). \quad (3.6) $$

Proof. Proposition 10 allows us to get

$$ \begin{aligned} \mathcal{R}(e^z) &= e^{\mathcal{R}(z)} \\ &= \sum_{n=0}^{+\infty} \frac{1}{n!} \mathcal{R}(z)^n \\ &= \sum_{n=0}^{+\infty} \mathcal{R}\left(\frac{1}{n!}z^n\right) \\ &= \mathcal{R}\left(\sum_{n=0}^{+\infty} \frac{1}{n!}z^n\right). \end{aligned} $$
So, we can infer

$$e^z = \sum_{n=0}^{+\infty} \frac{1}{n!} z^n.$$

Hence, for $z = x_0 + x_1\lambda + x_2\lambda^2 \in \mathbb{I}_2$, we get

$$ \begin{aligned} e^z &= e^{x_0} e^{x_1 \lambda} e^{x_2 \lambda^2} \\ &= e^{x_0} \sum_{n=0}^{+\infty} \frac{x_1^n \lambda^n}{n!} \sum_{n=0}^{+\infty} \frac{x_2^n \lambda^{2n}}{n!} \\ &= e^{x_0} \left( 1 + \sum_{n=0}^{+\infty} \frac{x_1^{2n+1}}{(2n+1)!} \lambda + \sum_{n=1}^{+\infty} \frac{x_1^{2n}}{(2n)!} \lambda^2 \right) (1 + (e^{x_2} - 1) \lambda^2) \\ &= e^{x_0} (1 + \sinh(x_1) \lambda + (\cosh(x_1) - 1) \lambda^2) (1 + (e^{x_2} - 1) \lambda^2) \\ &= e^{x_0} (1 + e^{x_2} \sinh(x_1) \lambda + e^{x_2} (\cosh(x_1) - 1) \lambda^2). \end{aligned} $$

This enables us to complete the proof. □

**Proposition 12.** Let us denote by $(G_+ \cap \mathbb{I}_2^+)_+$ the subset of $G_+ \cap \mathbb{I}_2^+$ given by

$$(G_+ \cap \mathbb{I}_2^+)_+ = \{z = 1 + x_1\lambda + x_2\lambda^2 \in G_+ \cap \mathbb{I}_2^+ \mid x_2 > -1\}. \quad (3.7)$$

Then,

1. $(G_+ \cap \mathbb{I}_2^+)_+$ is a subgroup of $G_+ \cap \mathbb{I}_2^+$.

2. $((G_+ \cap \mathbb{I}_2^+)_+ , \cdot) \simeq (\mathbb{R}^2, +)$.

*Proof.** 1. Denote by $\varphi$ the map

$$ \left\{ \begin{array}{l} \varphi : \mathbb{R}^2 \to \mathbb{I}_2^+, \\ (x_1, x_2) \mapsto e^{x_1 \lambda + x_2 \lambda^2}. \end{array} \right. \qquad (3.8) $$

Keepin' in mind Proposition 11 we can infer

$$e^{x_1 \lambda + x_2 \lambda^2} = 1 + e^{x_2} \sinh(x_1) \lambda + (e^{x_2} \cosh(x_1) - 1) \lambda^2.$$

Which asserts that $\mathrm{Im} \varphi \subset (G_+ \cap \mathbb{I}_2^+)_+$. Conversely, let $z = 1 + y_1\lambda + y_2\lambda^2 \in (G_+ \cap \mathbb{I}_2^+)_+$.

If $\varphi(x_1, x_2) = 1 + y_1\lambda + y_2\lambda^2$, then

$$ \left\{ \begin{array}{l} (y_2+1)^2 - y_1^2 > 0, \\ e^{x_2} \sinh(x_1) = y_1, \\ e^{x_2} \cosh(x_1) - 1 = y_2. \end{array} \right. $$

This leads to

$$ \left\{ \begin{array}{l} y_1 = \sinh^{-1} \frac{x_1}{\sqrt{(x_2+1)^2 - x_1^2}}, \\ y_2 = \frac{1}{2} \log \left( (x_2+1)^2 - x_1^2 \right). \end{array} \right. $$

Consequently,

$$\mathrm{Im} \varphi = (G_+ \cap \mathbb{I}_2^+)_+.$$

Moreover, it is straightforward that the set $\mathrm{Im} \varphi$ is a subgroup of $G_+ \cap \mathbb{I}_2^+$. Thus, We conclude that $(G_+ \cap \mathbb{I}_2^+)_+$ is a subgroup of $(G_+ \cap \mathbb{I}_2^+)_+$ and $((G_+ \cap \mathbb{I}_2^+)_+ , \cdot) \simeq (\mathbb{R}^2, +)$. □
Furthermore, we can also deduce that

$$ (G_+ \cap S^+(0,1), \cdot) \simeq (\mathbb{R}, +) \qquad (3.9) $$

via the map

$$ \left\{ \begin{array}{l} \mathbb{R} \to G_+ \cap S^+(0,1), \\ x_1 \mapsto e^{x_1 \lambda}, \end{array} \right. \qquad (3.10) $$

and

$$ (E_+ \cap S^+(0,1), \cdot) \simeq (\mathbb{R}, +) \qquad (3.11) $$

via the map

$$ \left\{ \begin{array}{l} \mathbb{R} \to E_+ \cap S^+(0,1), \\ x_2 \mapsto e^{x_2 \lambda^2}. \end{array} \right. \qquad (3.12) $$

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FARID MESSELMI, DEPARTMENT OF MATHEMATICS AND LDMM LABORATORY, UNIVERSITY OF DJELFA, ALGERIA

E-mail address: foudimath@yahoo.fr