A Gyrogeometric Mean in the Einstein Gyrogroup

Abstract

In this paper, we define a gyrogeometric mean on the Einstein gyrovector space. It satisfies several properties one would expect for means. For example, it is permutation-invariant and left-translation invariant. It is already known that the Einstein gyrogroup is a gyrocommutative gyrogroup. We give an alternative proof which depends only on an elementary calculation.

1. Introduction

Einstein addition is a binary operation that stems from his velocity composition law of relativistic admissible velocities. Einstein addition is neither commutative nor associative. Ungar initiated the study of gyrogroups and gyrovector spaces [1] associated with the Einstein addition in the theory of special relativity. A gyrocommutative gyrogroup is a gyrogroup which has weak associativity and commutativity. It is a generalization of a commutative group.

Let $\mathbb{V}$ be a real inner product space. For a positive real number s we denote ${\mathbb{V}}_s$ the s-ball of $\mathbb{V}$, i.e.,

$${\mathbb{V}}_s=\{\mathbf{v}\in \mathbb{V}:\parallel \mathbf{v}\parallel <s\}.$$

The Einstein addition ${\oplus }_E$ on ${\mathbb{V}}_s$ is a binary operation on ${\mathbb{V}}_s$ given by the equation

$$\mathbf{u}{\oplus }_E\mathbf{v}=\frac{1}{1+\frac{\mathbf{u}·\mathbf{v}}{s^2}}\left\{\mathbf{u}+\frac{1}{{\gamma }_{\mathbf{u}}}\mathbf{v}+\frac{1}{s^2}\frac{{\gamma }_{\mathbf{u}}}{1+{\gamma }_{\mathbf{u}}}(\mathbf{u}·\mathbf{v})\mathbf{u}\right\},$$

where ${\gamma }_{\mathbf{u}}$ is the gamma factor

$${\gamma }_{\mathbf{u}}=\frac{1}{\sqrt{1-\frac{{\parallel \mathbf{u}\parallel }^2}{s^2}}}$$

in ${\mathbb{V}}_s$, where · and $\parallel ·\parallel$ are the inner product and the norm of $\mathbb{V}$ respectively. By the definition of ${\oplus }_E$, $\mathbf{u}{\oplus }_E\mathbf{v}\in {\mathbb{V}}_s$ for every pair $\mathbf{u},\mathbf{v}\in {\mathbb{V}}_s$ by Theorem 3.46 and the identity (3.189) in [1].

In [1] (p. 88) Ungar described that “one can show by computer algebra that Einstein addition in the ball is a gyrocommutative gyrogroup operation, giving rise to the Einstein ball gyrogroup $({\mathbb{V}}_1,{\oplus }_E)$.” On the other hand, Suksumran and Wiboonton [2] gave a proof applying the theory of Clifford algebras, without using computer algebras. We give an elementary and direct proof in Section 6, which is lengthy but just by a simple calculation without applying any substantial theory of mathematics.

In the following up to Section 5 we assume $s=1$ just for simplicity. The Einstein scalar multiplication ${\otimes }_E$ on $({\mathbb{V}}_1,{\oplus }_E)$ is given by the equation

$$\begin{array}{ccc}\hfill r{\otimes }_E\mathbf{a}& \hfill =& \hfill \frac{{(1+\parallel \mathbf{a}\parallel )}^r-{(1-\parallel \mathbf{a}\parallel )}^r}{{(1+\parallel \mathbf{a}\parallel )}^r+{(1-\parallel \mathbf{a}\parallel )}^r}\frac{\mathbf{a}}{\parallel \mathbf{a}\parallel }\\ \hfill & \hfill =& tanh(r{tanh}^{-1}\parallel \mathbf{a}\parallel )\frac{\mathbf{a}}{\parallel \mathbf{a}\parallel },\hfill \end{array}$$

where $r\in \mathbb{R},\mathbf{a}\in {\mathbb{V}}_1\setminus \left\{\mathbf{0}\right\}$; and $r{\otimes }_E\mathbf{0}=\mathbf{0}$. By Theorem 6.84 in [1], $({\mathbb{V}}_1,{\oplus }_E,{\otimes }_E)$ is a gyrovector space, which is called an Einstein gyrovector space.

Ungar [1] (pp. 172–173) defined the gyromidpoint ${\mathbf{P}}_{\mathbf{a}\mathbf{b}}^m$ of two elements in a gyrovector space. For $\mathbf{a},\mathbf{b}\in {\mathbb{V}}_1$ we have

$${\mathbf{P}}_{\mathbf{a}\mathbf{b}}^m=\frac{{\gamma }_{\mathbf{a}}\mathbf{a}+{\gamma }_{\mathbf{b}}\mathbf{b}}{{\gamma }_{\mathbf{a}}+{\gamma }_{\mathbf{b}}}.$$

(1)

Ungar also defined the gyrocentroid ${\mathbb{C}}_{\mathbf{a}\mathbf{b}\mathbf{c}}^m$ of three elements $\mathbf{a},\mathbf{b},\mathbf{c}\in {\mathbb{V}}_1$ as

$${\mathbb{C}}_{\mathbf{a}\mathbf{b}\mathbf{c}}^m=\frac{{\gamma }_{\mathbf{a}}\mathbf{a}+{\gamma }_{\mathbf{b}}\mathbf{b}+{\gamma }_{\mathbf{c}}\mathbf{c}}{{\gamma }_{\mathbf{a}}+{\gamma }_{\mathbf{b}}+{\gamma }_{\mathbf{c}}}.$$

The gyromidpoint corresponds to the average of two velocities in the special theory of relativity. On the other hand the gyrocentroid ${\mathbb{C}}_{\mathbf{a}\mathbf{b}\mathbf{c}}^m$ does not satisfy a certain desirable property one would expected for means; by a simple calculation we have ${\mathbb{C}}_{\mathbf{a}\mathbf{b}\mathbf{c}}^m=\frac{{\gamma }_{\mathbf{c}}}{{\gamma }_{\mathbf{c}}+2}\mathbf{c}\ne \frac{1}{3}{\otimes }_E\mathbf{c}$ for $\mathbf{a}=\mathbf{b}=\mathbf{0}$ and $\mathbf{c}\ne 0$. In this paper, we propose an alternative definition of the mean of three or more elements, the gyrogeometric mean, and show that it has several properties one would expect for means. The gyrogeometric mean corresponding to the average of the velocities in the special relativity. It is symmetric in the sense that permutation-invariant by the definition of the gyrogeometric mean. It is translation invariant (Proposition 5). The main idea of the definitions come from the geometric mean for positive definite matrices by Bhatia and Holbrook [3] and Ando, Li and Mathias [4].

2. The Metric Space $({\mathbb{V}}_{\mathbf{1}},{\oplus }_{\mathbf{E}},{\otimes }_{\mathbf{E}})$

We define the set $\parallel {\mathbb{V}}_1\parallel =\{\pm \parallel \mathbf{v}\parallel :\mathbf{v}\in {\mathbb{V}}_1\}\subset \mathbb{R}$ which coincides with the open interval $(-1,1)$. $\parallel {\mathbb{V}}_1\parallel$ admits the addition ${\oplus }^{\prime }$ and the scalar multiplication ${\otimes }^{\prime }$ given by the following:

$$\begin{array}{ccc}\hfill a{\oplus }^{\prime }b& =& \frac{a+b}{1+ab}\hfill \\ \hfill r{\otimes }^{\prime }a& =& tanh\left(r{tanh}^{-1}a\right)\hfill \end{array}$$

where $a,b\in \parallel {\mathbb{V}}_1\parallel$ and $r\in \mathbb{R}$. Please note that the triplet $(\parallel {\mathbb{V}}_1\parallel ,{\oplus }^{\prime },{\otimes }^{\prime })$ is a real one-dimensional space.

The gyrometric is defined by

$$d(\mathbf{a},\mathbf{b})=\parallel \mathbf{a}{\ominus }_E\mathbf{b}\parallel \in \parallel {\mathbb{V}}_1\parallel ,$$

where $\mathbf{a},\mathbf{b}\in {\mathbb{V}}_1$ and $\mathbf{a}{\ominus }_E\mathbf{b}=\mathbf{a}{\oplus }_E(-\mathbf{b})$. The gyrometric needs not be a metric. It satisfies the following [1] (p. 158).

Proposition 1.

(1) 

For every pair $\mathbf{a},\mathbf{b}\in {\mathbb{V}}_1$, $d(\mathbf{a},\mathbf{b})\ge 0$ The equality $d(\mathbf{a},\mathbf{b})=0$ holds if and only if $\mathbf{a}=\mathbf{b}$.

(2) 

$d(\mathbf{a},\mathbf{b})=d(\mathbf{b},\mathbf{a})$ for any $\mathbf{a},\mathbf{b}\in {\mathbb{V}}_1$.

(3) 

The gyrotriangle inequality:

$$d(\mathbf{a},\mathbf{b})\le d(\mathbf{a},\mathbf{c}){\oplus }^{\prime }d(\mathbf{c},\mathbf{b})$$

holds for any $\mathbf{a},\mathbf{b},\mathbf{c}$ in ${\mathbb{V}}_1$.

We define the metric $\delta$ on ${\mathbb{V}}_1$ induced by the gyrometric d. Put the map $f:\parallel {\mathbb{V}}_1\parallel \to \mathbb{R}$ by $f\left(x\right)={tanh}^{-1}\left(x\right)$. For any $a,b\in \parallel {\mathbb{V}}_1\parallel$ and $r\in \mathbb{R}$, the map f satisfies the following.

(F1)

$f\left(a{\oplus }^{\prime }b\right)=f\left(a\right)+f\left(b\right)$

(F2)

$f\left(r{\otimes }^{\prime }a\right)=rf\left(a\right)$

Let the map $\delta$ on ${\mathbb{V}}_1$ be given by

$$\delta (\mathbf{a},\mathbf{b})=f\left(d\right(\mathbf{a},\mathbf{b}\left)\right)$$

for $\mathbf{a},\mathbf{b}\in {\mathbb{V}}_1$.

Lemma 1.

The inequality

$$\frac{1}{{\gamma }_{\mathbf{a}}(1-\mathbf{a}·\mathbf{b})}\parallel \mathbf{b}-\mathbf{a}\parallel \le \parallel \mathbf{a}{\ominus }_E\mathbf{b}\parallel \le \frac{1}{1-\mathbf{a}·\mathbf{b}}\parallel \mathbf{b}-\mathbf{a}\parallel$$

holds for every pair $\mathbf{a},\mathbf{b}\in {\mathbb{V}}_1$.

Proof. 

Recall the equations (3.177) and (3.178) [1] (pp. 88–89):

$$\begin{array}{ccc}\hfill {\gamma }_{\mathbf{a}{\oplus }_{\mathbf{E}}\mathbf{b}}& =& {\gamma }_{\mathbf{a}}{\gamma }_{\mathbf{b}}(1+\mathbf{a}·\mathbf{b}),\hfill \\ \hfill {\gamma }_{\mathbf{a}{\ominus }_{\mathbf{E}}\mathbf{b}}& =& {\gamma }_{\mathbf{a}}{\gamma }_{\mathbf{b}}(1-\mathbf{a}·\mathbf{b}).\hfill \end{array}$$

By ${\gamma }_{\mathbf{a}{\oplus }_{\mathbf{E}}\mathbf{b}}=\frac{1}{\sqrt{1-\parallel \mathbf{a}{\ominus }_E\mathbf{b}\parallel }}$ we have

$$\begin{array}{ccc}\hfill \parallel \mathbf{a}{\ominus }_E\mathbf{b}\parallel & \hfill =& \sqrt{1-\frac{1}{{{\gamma }_{\mathbf{a}{\oplus }_{\mathbf{E}}\mathbf{b}}}^2}}\hfill \\ \hfill & \hfill =& \sqrt{1-\frac{1}{{\gamma }_{\mathbf{a}}{\gamma }_{\mathbf{b}}(1-\mathbf{a}·\mathbf{b})}}\hfill \\ \hfill & \hfill =& \sqrt{1-\frac{{(1-\parallel \mathbf{a}\parallel }^2\left)\right(1-{\parallel \mathbf{b}\parallel }^2)}{{(1-\mathbf{a}·\mathbf{b})}^2}}\hfill \\ \hfill & \hfill =& \frac{\sqrt{{(1-\mathbf{a}·\mathbf{b})}^2{-(1-\parallel \mathbf{a}\parallel }^2\left)\right(1-{\parallel \mathbf{b}\parallel }^2)}}{1-\mathbf{a}·\mathbf{b}}\hfill \\ \hfill & \hfill =& \frac{\sqrt{{\parallel \mathbf{b}-\mathbf{a}\parallel }^2-{\parallel \mathbf{a}\parallel }^2{\parallel \mathbf{b}\parallel }^2+{(\mathbf{a}·\mathbf{b})}^2}}{1-\mathbf{a}·\mathbf{b}}.\hfill \end{array}$$

Since $\mathbf{a}·\mathbf{b}\le \parallel \mathbf{a}\parallel \parallel \mathbf{b}\parallel$ then we have

$$\parallel \mathbf{a}{\ominus }_E\mathbf{b}\parallel \le \frac{1}{1-\mathbf{a}·\mathbf{b}}\parallel \mathbf{b}-\mathbf{a}\parallel .$$

Next we calculate $\parallel \mathbf{a}{\ominus }_E{\mathbf{b}\parallel }^2-{\left\{\frac{1}{{\gamma }_{\mathbf{a}}(1-\mathbf{a}·\mathbf{b})}\parallel \mathbf{b}-\mathbf{a}\parallel \right\}}^2$.

$$\begin{array}{cc}& \frac{{\parallel \mathbf{b}-\mathbf{a}\parallel }^2-{\parallel \mathbf{a}\parallel }^2{\parallel \mathbf{b}\parallel }^2+{(\mathbf{a}·\mathbf{b})}^2}{{(1-\mathbf{a}·\mathbf{b})}^2}-\frac{1-{\parallel \mathbf{a}\parallel }^2}{{(1-\mathbf{a}·\mathbf{b})}^2}{\parallel \mathbf{b}-\mathbf{a}\parallel }^2\hfill \\ =& \frac{1}{{(1-\mathbf{a}·\mathbf{b})}^2}\left\{{\parallel \mathbf{b}-\mathbf{a}\parallel }^2-{\parallel \mathbf{a}\parallel }^2{\parallel \mathbf{b}\parallel }^2+{(\mathbf{a}·\mathbf{b})}^2{-(1-\parallel \mathbf{a}\parallel }^2{)\parallel \mathbf{b}-\mathbf{a}\parallel }^2\right\}\hfill \\ =& \frac{1}{{(1-\mathbf{a}·\mathbf{b})}^2}\left\{{(\mathbf{a}·\mathbf{b})}^2+{\parallel \mathbf{a}\parallel }^4-2{\parallel \mathbf{a}\parallel }^2(\mathbf{a}·\mathbf{b})\right\}\hfill \\ =& \frac{1}{{(1-\mathbf{a}·\mathbf{b})}^2}{(\parallel \mathbf{a}\parallel }^2{-\mathbf{a}·\mathbf{b})}^2\hfill \\ \ge & 0.\hfill \end{array}$$

Thus, we have the desired inequalities and conclude the proof. □

Proposition 2.

$({\mathbb{V}}_1,\delta )$ is a complete metric space.

Proof. 

We first prove that $({\mathbb{V}}_1,\delta )$ is a metric space. By (1) and (2) of Proposition 1, it is trivial that $\delta (\mathbf{a},\mathbf{b})=0\iff \mathbf{a}=\mathbf{b}$ and $\delta (\mathbf{a},\mathbf{b})=\delta (\mathbf{b},\mathbf{a})$ for every $\mathbf{a},\mathbf{b}\in {\mathbb{V}}_1$. By (3) of Proposition 1 and the monotonicity of f, the inequality $f\left(d\right(\mathbf{a},\mathbf{b}\left)\right)\le f\left(d\right(\mathbf{a},\mathbf{c})\oplus d(\mathbf{c},\mathbf{b}\left)\right)$ for every $\mathbf{a},\mathbf{b},\mathbf{c}\in {\mathbb{V}}_1$. By (F1) we have

$$\delta (\mathbf{a},\mathbf{b})\le \delta (\mathbf{a},\mathbf{c})+\delta (\mathbf{c},\mathbf{b}).$$

As $\mathbb{V}$ is complete, we have by Lemma 1 and the definition of $\delta (·,·)$ that $({\mathbb{V}}_1,\delta )$ is complete. □

We recall the gyroline and the gyrosegment [1] (Definition 6.19). Let $\mathbf{a},\mathbf{b}$ be elements of ${\mathbb{V}}_1$. The gyroline through $\mathbf{a}$ and $\mathbf{b}$ is defined by

$$L(\mathbf{a},\mathbf{b})=\{\mathbf{a}{\oplus }_{E}t{\otimes }_E\left({\ominus }_E\mathbf{a}{\oplus }_E\mathbf{b}\right):t\in \mathbb{R}\}.$$

A gyrosegment with endpoints $\mathbf{a}$ and $\mathbf{b}$ is denoted by

$$S(\mathbf{a},\mathbf{b})=\{\mathbf{a}{\oplus }_{E}t{\otimes }_E\left({\ominus }_E\mathbf{a}{\oplus }_E\mathbf{b}\right):0\le t\le 1\}.$$

The point $\mathbf{a}{\#}_t\mathbf{b}=\mathbf{a}{\oplus }_{E}t{\otimes }_E\left({\ominus }_E\mathbf{a}{\oplus }_E\mathbf{b}\right)$ is called the gyro t-point on a gyroline or gyrosegment. We abbreviate $\mathbf{a}{\#}_{\frac{1}{2}}\mathbf{b}$ by $\mathbf{a}\#\mathbf{b}$. Please note that $\mathbf{a}\#\mathbf{b}=\mathbf{b}\#\mathbf{a}$ for every pair $\mathbf{a},\mathbf{b}\in {\mathbb{V}}_1$.

Theorem 1.

For any $\mathbf{a},\mathbf{b},\mathbf{c}\in {\mathbb{V}}_1$ we have

$$d(\mathbf{a}\#\mathbf{b},\mathbf{a}\#\mathbf{c})\le \frac{1}{2}\otimes d(\mathbf{b},\mathbf{c}).$$

(2)

Proof. 

To begin with the proof of the inequality (2), we show an equation related to the gyrometric and gamma factor. Recall the equations (3.197) and (6.266) [1] (pp. 93, 209):

$$\begin{array}{ccc}\hfill {\gamma }_{\mathbf{a}{⊞}_{\mathbf{E}}\mathbf{b}}& =& \frac{{({\gamma }_{\mathbf{a}}+{\gamma }_{\mathbf{b}})}^2}{{\gamma }_{\mathbf{a}\ominus \mathbf{b}}+1}-1,\hfill \\ \hfill {\gamma }_{\frac{\mathbf{1}}{\mathbf{2}}{\otimes }_{\mathbf{E}}\mathbf{a}}& =& \sqrt{\frac{1+{\gamma }_{\mathbf{a}}}{2}}.\hfill \end{array}$$

Hence

$${\gamma }_{\mathbf{a}\#\mathbf{b}}=\frac{{\gamma }_{\mathbf{a}}+{\gamma }_{\mathbf{b}}}{\sqrt{2({\gamma }_{\mathbf{a}\ominus \mathbf{b}}+1)}}$$

(3)

holds. By a simple calculation, we have

$$1-(\mathbf{a}\#\mathbf{b})·(\mathbf{a}\#\mathbf{c})=\frac{1+{\gamma }_{\mathbf{a}\ominus \mathbf{b}}+{\gamma }_{\mathbf{b}\ominus \mathbf{c}}+{\gamma }_{\mathbf{c}\ominus \mathbf{a}}}{({\gamma }_{\mathbf{a}}+{\gamma }_{\mathbf{b}})({\gamma }_{\mathbf{a}}+{\gamma }_{\mathbf{c}})}.$$

(4)

Hence we have

$$\begin{array}{ccc}\hfill {\gamma }_{(\mathbf{a}\#\mathbf{b}){\ominus }_{\mathbf{E}}(\mathbf{a}\#\mathbf{c})}& \hfill =& \hfill {\gamma }_{\mathbf{a}\#\mathbf{b}}{\gamma }_{\mathbf{a}\#\mathbf{c}}(1-(\mathbf{a}\#\mathbf{b})·(\mathbf{a}\#\mathbf{c}))\\ \hfill & \hfill =& \frac{1+{\gamma }_{\mathbf{a}\ominus \mathbf{b}}+{\gamma }_{\mathbf{b}\ominus \mathbf{c}}+{\gamma }_{\mathbf{c}\ominus \mathbf{a}}}{2\sqrt{({\gamma }_{\mathbf{a}\ominus \mathbf{b}}+1)({\gamma }_{\mathbf{a}\ominus \mathbf{c}}+1)}}\hfill \end{array}$$

(5)

and

$$\begin{array}{ccc}\hfill d^2(\mathbf{a}\#\mathbf{b},\mathbf{a}\#\mathbf{c})& \hfill =& 1-\frac{1}{{{\gamma }_{(\mathbf{a}\#\mathbf{b}){\ominus }_{\mathbf{E}}(\mathbf{a}\#\mathbf{c})}}^2}\hfill \\ \hfill & \hfill =& \hfill 1-\frac{4({\gamma }_{\mathbf{a}\ominus \mathbf{b}}+1)({\gamma }_{\mathbf{a}\ominus \mathbf{c}}+1)}{{(1+{\gamma }_{\mathbf{a}\ominus \mathbf{b}}+{\gamma }_{\mathbf{b}\ominus \mathbf{c}}+{\gamma }_{\mathbf{c}\ominus \mathbf{a}})}^2}.\end{array}$$

(6)

We also have

$$\begin{array}{ccc}\hfill \frac{1}{2}\otimes d(\mathbf{b},\mathbf{c})& \hfill =& tanh\left(\frac{1}{2}{tanh}^{-1}d(\mathbf{b},\mathbf{c})\right)\hfill \\ \hfill & \hfill =& \frac{{\gamma }_{\mathbf{b}{\ominus }_{\mathbf{E}}\mathbf{c}}}{1+{\gamma }_{\mathbf{b}{\ominus }_{\mathbf{E}}\mathbf{c}}}d(\mathbf{b},\mathbf{c})\hfill \\ \hfill & \hfill =& \sqrt{\frac{{\gamma }_{\mathbf{b}{\ominus }_{\mathbf{E}}\mathbf{c}}-1}{1+{\gamma }_{\mathbf{b}{\ominus }_{\mathbf{E}}\mathbf{c}}}},\hfill \end{array}$$

where ${\gamma }_{\mathbf{d}(\mathbf{b},\mathbf{c})}={\gamma }_{\mathbf{b}{\ominus }_{\mathbf{E}}\mathbf{c}}$. Hence we have

$$\begin{array}{cc}& {\left(\frac{1}{2}\otimes d(\mathbf{b},\mathbf{c})\right)}^2-d^2(\mathbf{a}\#\mathbf{b},\mathbf{a}\#\mathbf{c})\hfill \end{array}$$

(7)

$$\begin{array}{cc}=& \frac{{\gamma }_{\mathbf{b}{\ominus }_{\mathbf{E}}\mathbf{c}}-1}{1+{\gamma }_{\mathbf{b}{\ominus }_{\mathbf{E}}\mathbf{c}}}-\left(1-\frac{4({\gamma }_{\mathbf{a}\ominus \mathbf{b}}+1)({\gamma }_{\mathbf{a}\ominus \mathbf{c}}+1)}{{(1+{\gamma }_{\mathbf{a}\ominus \mathbf{b}}+{\gamma }_{\mathbf{b}\ominus \mathbf{c}}+{\gamma }_{\mathbf{c}\ominus \mathbf{a}})}^2}\right)\hfill \\ =& \frac{4({\gamma }_{\mathbf{a}\ominus \mathbf{b}}+1)({\gamma }_{\mathbf{a}\ominus \mathbf{c}}+1)}{{(1+{\gamma }_{\mathbf{a}\ominus \mathbf{b}}+{\gamma }_{\mathbf{b}\ominus \mathbf{c}}+{\gamma }_{\mathbf{c}\ominus \mathbf{a}})}^2}-\frac{2}{1+{\gamma }_{\mathbf{b}{\ominus }_{\mathbf{E}}\mathbf{c}}}\hfill \\ =& \frac{2\left\{2({\gamma }_{\mathbf{a}\ominus \mathbf{b}}+1)({\gamma }_{\mathbf{b}{\ominus }_{\mathbf{E}}\mathbf{c}}+1)({\gamma }_{\mathbf{c}\ominus \mathbf{a}}+1)-{(1+{\gamma }_{\mathbf{a}\ominus \mathbf{b}}+{\gamma }_{\mathbf{b}\ominus \mathbf{c}}+{\gamma }_{\mathbf{c}\ominus \mathbf{a}})}^2\right\}}{(1+{\gamma }_{\mathbf{b}{\ominus }_{\mathbf{E}}\mathbf{c}}){(1+{\gamma }_{\mathbf{a}\ominus \mathbf{b}}+{\gamma }_{\mathbf{b}\ominus \mathbf{c}}+{\gamma }_{\mathbf{c}\ominus \mathbf{a}})}^2}\hfill \end{array}$$

$$\begin{array}{ccc}\hfill & =& \frac{2\left\{2{\gamma }_{\mathbf{a}\ominus \mathbf{b}}{\gamma }_{\mathbf{b}\ominus \mathbf{c}}{\gamma }_{\mathbf{c}\ominus \mathbf{a}}-({{\gamma }_{\mathbf{a}\ominus \mathbf{b}}}^2+{{\gamma }_{\mathbf{b}\ominus \mathbf{c}}}^2+{{\gamma }_{\mathbf{c}\ominus \mathbf{a}}}^2)+1\right\}}{(1+{\gamma }_{\mathbf{b}{\ominus }_{\mathbf{E}}\mathbf{c}}){(1+{\gamma }_{\mathbf{a}\ominus \mathbf{b}}+{\gamma }_{\mathbf{b}\ominus \mathbf{c}}+{\gamma }_{\mathbf{c}\ominus \mathbf{a}})}^2}.\hfill \end{array}$$

(8)

Let $\mathbf{A}={\ominus }_E\mathbf{c}{\oplus }_E\mathbf{a},\mathbf{B}={\ominus }_E\mathbf{c}{\oplus }_E\mathbf{b}$. It is well defined by $\parallel \left({\ominus }_E\mathbf{c}{\oplus }_E\mathbf{a}\right){\ominus }_E\left({\ominus }_E\mathbf{c}{\oplus }_E\mathbf{b}\right)\parallel =\parallel \mathrm{gyr}[{\ominus }_{\mathbf{E}}\mathbf{c},\mathbf{a}]\left(\mathbf{a}{\ominus }_E\mathbf{b}\right)\parallel =\parallel \mathbf{a}{\ominus }_E\mathbf{b}\parallel$. We calculate the numerator of (8);

$$\begin{array}{cc}& 2{\gamma }_{\mathbf{a}\ominus \mathbf{b}}{\gamma }_{\mathbf{b}\ominus \mathbf{c}}{\gamma }_{\mathbf{c}\ominus \mathbf{a}}-({{\gamma }_{\mathbf{a}\ominus \mathbf{b}}}^2+{{\gamma }_{\mathbf{b}\ominus \mathbf{c}}}^2+{{\gamma }_{\mathbf{c}\ominus \mathbf{a}}}^2)+1\hfill \\ =& 2{\gamma }_{\mathbf{A}{\ominus }_{\mathbf{E}}\mathbf{B}}{\gamma }_{\mathbf{A}}{\gamma }_{\mathbf{B}}-({{\gamma }_{\mathbf{A}{\ominus }_{\mathbf{E}}\mathbf{B}}}^2+{{\gamma }_{\mathbf{A}}}^2+{{\gamma }_{\mathbf{B}}}^2)+1\hfill \\ =& 2{{\gamma }_{\mathbf{A}}}^2{{\gamma }_{\mathbf{B}}}^2(1-\mathbf{A}·\mathbf{B})-({{\gamma }_{\mathbf{A}}}^2{{\gamma }_{\mathbf{B}}}^2{(1-\mathbf{A}·\mathbf{B})}^2+{{\gamma }_{\mathbf{A}}}^2+{{\gamma }_{\mathbf{B}}}^2)+1\hfill \\ =& {{\gamma }_{\mathbf{A}}}^2{{\gamma }_{\mathbf{B}}}^2(1-{(\mathbf{A}·\mathbf{B})}^2)-({{\gamma }_{\mathbf{A}}}^2+{{\gamma }_{\mathbf{B}}}^2)+1\hfill \\ =& \frac{1-{(\mathbf{A}·\mathbf{B})}^2}{{(1-\parallel \mathbf{A}\parallel }^2\left)\right(1-{\parallel \mathbf{B}\parallel }^2)}-\left(\frac{1}{(1-\parallel \mathbf{A}{\parallel }^2)}+\frac{1}{(1-\parallel \mathbf{B}{\parallel }^2)}\right)+1\hfill \\ =& \frac{{\parallel \mathbf{A}\parallel }^2{\parallel \mathbf{B}\parallel }^2-{(\mathbf{A}·\mathbf{B})}^2}{{(1-\parallel \mathbf{A}\parallel }^2\left)\right(1-{\parallel \mathbf{B}\parallel }^2)}\hfill \\ \ge & 0.\hfill \end{array}$$

We conclude a proof of Theorem 1. □

By Theorem 1 and the monotonicity of $f\left(x\right)={tanh}^{-1}\left(x\right)$, we have

$$\delta (\mathbf{a}\#\mathbf{b},\mathbf{a}\#\mathbf{c})\le \frac{1}{2}\delta (\mathbf{b},\mathbf{c}).$$

(9)

By the triangle inequality, we have

$$\begin{array}{ccc}\hfill \delta (\mathbf{a}\#\mathbf{b},\mathbf{c}\#\mathbf{d})& \hfill \le & \hfill \delta (\mathbf{a}\#\mathbf{b},\mathbf{a}\#\mathbf{d})+\delta (\mathbf{a}\#\mathbf{d},\mathbf{c}\#\mathbf{d})\\ \hfill & \hfill \le & \frac{1}{2}\delta (\mathbf{b},\mathbf{d})+\frac{1}{2}\delta (\mathbf{a},\mathbf{c}).\hfill \end{array}$$

(10)

Moreover, since the map $g\left(t\right)=\delta (\mathbf{a}{\#}_t\mathbf{b},\mathbf{c}{\#}_t\mathbf{d})$ is continuous, we infer that g is convex, i.e.,

$$\delta (\mathbf{a}{\#}_t\mathbf{b},\mathbf{c}{\#}_t\mathbf{d})\le (1-t)\delta (\mathbf{a},\mathbf{c})+t\delta (\mathbf{b},\mathbf{d}).$$

(11)

Letting $\mathbf{a}=\mathbf{c}$ we have

$$\delta (\mathbf{a}{\#}_t\mathbf{b},\mathbf{a}{\#}_t\mathbf{c})\le t\delta (\mathbf{b},\mathbf{c}).$$

(12)

3. The Gyroconvex Set and the Gyroconvex Hull in a Gyrovector Space

We define a gyroconvex set and a gyroconvex hull.

Definition 1.

Let A be a subset of ${\mathbb{V}}_1$. We say that A is gyroconvex if $S(\mathbf{a},\mathbf{b})\subset A$ for any $\mathbf{a},\mathbf{b}\in A$. Let X be a non-empty subset of ${\mathbb{V}}_1$.

$$conv\left(X\right)=\cap \{C\subset {\mathbb{V}}_1:X\subset CandCis\mathrm{gyr}oconvexset\}.$$

We call $conv\left(X\right)$ the gyroconvex hull of X.

Please note that the gyroconvex hull of a non-empty set $X\subset \mathbb{V}$ is gyroconvex.

Lemma 2.

Let $a,b\in {\mathbb{V}}_1$. Then the gyrosegment $S(\mathbf{a},\mathbf{b})$ is gyroconvex. The gyroconvex hull $conv\left(\right\{\mathbf{a},\mathbf{b}\left\}\right)$ coincides with $S(\mathbf{a},\mathbf{b})$.

Proof. 

Let ${\mathbf{P}}_j$ be an arbitrary point in $S(\mathbf{a},\mathbf{b})$ for $j=1,2$. There exists $0\le t_j\le 1$ such that

$${\mathbf{P}}_j=\mathbf{a}{\oplus }_E{t}_j{\otimes }_E\left({\ominus }_E\mathbf{a}{\oplus }_E\mathbf{b}\right)$$

for $j=1,2$. We may assume that $t_1\le t_2$. where $0\le t_1\le t_2\le 1$. Then we have $t\in [0,1]$, ${\mathbf{P}}_1{\#}_t{\mathbf{P}}_2\in S(\mathbf{a},\mathbf{b})$. In fact, by the Equation (6.63) in [1] (p. 167) we have

$${\mathbf{P}}_1{\oplus }_{E}t{\otimes }_E\left({\ominus }_E{\mathbf{P}}_1{\oplus }_E{\mathbf{P}}_2\right)=\mathbf{a}{\oplus }_E(t_1+(-t_1+t_2)t){\otimes }_E\left({\ominus }_E\mathbf{a}{\oplus }_E\mathbf{b}\right).$$

Since $t_1\le t_1+(-t_1+t_2)t\le t_2$, we have ${\mathbf{P}}_1{\#}_t{\mathbf{P}}_2\in S(\mathbf{a},\mathbf{b})$. Thus, $S({\mathbf{P}}_1,{\mathbf{P}}_2)\subset S(\mathbf{a},\mathbf{b})$ for every pair ${\mathbf{P}}_1$ and ${\mathbf{P}}_2$ in $S(\mathbf{a},\mathbf{b})$. Thus, $S(\mathbf{a},\mathbf{b})$ is gyroconvex. □

Let $C_0$ be a non-empty subset of ${\mathbb{V}}_1$. We define a sequence $\left\{C_n\right\}$ of a non-empty subset of ${\mathbb{V}}_1$ by induction. Suppose that $C_{n-1}$ is defined. Put

$$C_n=\bigcup _{\mathbf{a},\mathbf{b}\in C_{n-1}}S(\mathbf{a},\mathbf{b}).$$

Proposition 3.

Let $C_0$ be a non-empty subset of ${\mathbb{V}}_1$. Then

$$conv\left(C_0\right)=\bigcup _{n=0}^{\infty }{C}_n.$$

Proof. 

We prove that ${\cup }_{k=0}^{\infty }{C}_n$ is gyroconvex. Let $\mathbf{a},\mathbf{b}\in {\cup }_{k=0}^{\infty }{C}_n$. Since $C_k\subset C_{k+1}$ for every $k\in \mathbb{N}\cup \left\{0\right\}$, there exists a positive integer $n_0$ with $\mathbf{a},\mathbf{b}\in C_{n_0}$. Then by the definition of $C_{n_0+1}$ we have $S(\mathbf{a},\mathbf{b})\subset C_{n_0+1}\subset {\cup }_{k=0}^{\infty }{C}_n$. Thus, ${\cup }_{n=0}^{\infty }{C}_n$ is a gyroconvex set. As $C_0\subset {\cup }_{k=0}^{\infty }{C}_n$, we have $conv\left(C_0\right)\subset {\cup }_{n=0}^{\infty }{C}_n$.

We prove ${\cup }_{n=0}^{\infty }{C}_n\subset conv\left(C_0\right)$. For any $\mathbf{a},\mathbf{b}\in C_0$, $S(\mathbf{a},\mathbf{b})\subset conv\left(C_0\right)$. Hence $C_1\subset conv\left(X\right)$. Similarly, assuming that $C_n\subset conv\left(C_0\right)$ for any $n\in \mathbb{N}\cup \left\{0\right\}$ we have $C_{n+1}\subset conv\left(C_0\right)$. So for arbitrary nonnegative integer n, $C_n\subset conv\left(C_0\right)$; ${\cup }_{n=0}^{\infty }{C}_n\subset conv\left(C_0\right)$. □

4. The Gyrogeometric Mean

Let $\varphi \ne X\subset {\mathbb{V}}_1$. We define $diam\left(X\right)=sup\left\{\delta \right(\mathbf{x},\mathbf{y}):\mathbf{x},\mathbf{y}\in X\}$. First we prove the following.

Proposition 4.

Suppose that ${\mathbf{x}}_0,{\mathbf{y}}_0,{\mathbf{x}}_1,{\mathbf{y}}_1\in {\mathbb{V}}_1$. If the inequality $diam\left(\{{\mathbf{x}}_0,{\mathbf{y}}_0,{\mathbf{x}}_1,{\mathbf{y}}_1\}\right)\le M$ holds for a positive real number M, then the inequality $\delta (\mathbf{x},\mathbf{y})\le M$ holds for arbitrary points $\mathbf{x}\in S({\mathbf{x}}_0,{\mathbf{x}}_1)$ and $\mathbf{y}\in S({\mathbf{y}}_0,{\mathbf{y}}_1)$.

Proof. 

Put $\mathbf{x}={\mathbf{x}}_0{\#}_t{\mathbf{x}}_1$ and $\mathbf{y}={\mathbf{y}}_0{\#}_s{\mathbf{y}}_1$, where $0\le s\le t\le 1$. We have

$$\begin{array}{cc}& d({\mathbf{y}}_0{\#}_t{\mathbf{x}}_1,{\mathbf{y}}_0{\#}_s{\mathbf{x}}_1)\hfill \\ =& \parallel \left\{{\mathbf{y}}_0{\oplus }_{E}t{\otimes }_E\left({\ominus }_E{\mathbf{y}}_0{\oplus }_E{\mathbf{x}}_1\right)\right\}{\ominus }_E\left\{{\mathbf{y}}_0{\oplus }_{E}s{\otimes }_E\left({\ominus }_E{\mathbf{y}}_0{\oplus }_E{\mathbf{x}}_1\right)\right\}\parallel \hfill \\ =& \parallel \mathrm{gyr}[{\ominus }_{\mathit{E}}{\mathbf{y}}_{\mathbf{0}},\mathit{t}{\otimes }_{\mathit{E}}\left({\ominus }_{\mathit{E}}{\mathbf{y}}_{\mathbf{0}}{\oplus }_{\mathit{E}}{\mathbf{x}}_{\mathbf{1}}\right)]\left\{t{\otimes }_E\left({\ominus }_E{\mathbf{y}}_0{\oplus }_E{\mathbf{x}}_1\right){\ominus }_{E}s{\otimes }_E\left({\ominus }_E{\mathbf{y}}_0{\oplus }_E{\mathbf{x}}_1\right)\right\}\parallel \hfill \\ =& (t-s)\otimes \parallel {\ominus }_E{\mathbf{y}}_0{\oplus }_E{\mathbf{x}}_1\parallel .\hfill \end{array}$$

Applying this equality, we have by (11) and (12) that

$$\begin{array}{ccc}\hfill \delta (\mathbf{x},\mathbf{y})& \hfill \le & \delta (\mathbf{x},{\mathbf{y}}_0{\#}_t{\mathbf{x}}_1)+\delta ({\mathbf{y}}_0{\#}_t{\mathbf{x}}_1,{\mathbf{y}}_0{\#}_s{\mathbf{x}}_1)+\delta ({\mathbf{y}}_0{\#}_s{\mathbf{x}}_1,\mathbf{y})\hfill \\ \hfill & \hfill =& \delta ({\mathbf{x}}_0{\#}_t{\mathbf{x}}_1,{\mathbf{y}}_0{\#}_t{\mathbf{x}}_1)+(t-s)\delta ({\mathbf{y}}_0,{\mathbf{x}}_1)+\delta ({\mathbf{y}}_0{\#}_s{\mathbf{x}}_1,{\mathbf{y}}_0{\#}_s{\mathbf{y}}_1)\hfill \\ \hfill & \hfill \le & (1-t)\delta ({\mathbf{x}}_0,{\mathbf{y}}_0)+(t-s)\delta ({\mathbf{y}}_0,{\mathbf{x}}_1)+s\delta ({\mathbf{x}}_1,{\mathbf{y}}_1)\hfill \\ \hfill & \hfill \le & (1-t)M+(t-s)M+sM=M.\hfill \end{array}$$

□

Lemma 3.

Let $X\subset \mathbb{V}$ be a non-empty set. Then

$$diam\left(X\right)=diam(conv\left(X\right))=diam\left(\overline{conv\left(X\right)}\right).$$

Proof. 

First we prove $diam\left(X\right)=diam(conv(X\left)\right)$. By Proposition 3, $conv\left(X\right)={\bigcup }_{n=0}^{\infty }{C}_n$ where $C_0=X$. For any positive integer k let $\mathbf{x},\mathbf{y}$ be arbitrary points in $C_k$. Then there exist ${\mathbf{x}}_0,{\mathbf{x}}_1,{\mathbf{y}}_0,{\mathbf{y}}_1\in C_{k-1}$ such that $\mathbf{x}\in S({\mathbf{x}}_0,{\mathbf{x}}_1),\mathbf{y}\in S({\mathbf{y}}_0,{\mathbf{y}}_1)$. Put $M=diam\left(C_{k-1}\right)$ then by Proposition 4.1 we have

$$\delta (\mathbf{x},\mathbf{y})\le M=diam\left(C_{k-1}\right),$$

whence

$$diam\left(C_k\right)\le diam\left(C_{k-1}\right).$$

Thus, for arbitrary $n\in \mathbb{N}$, $diam\left(C_n\right)\le diam\left(C_0\right)=diam\left(X\right)$. It follows that for any $\mathbf{x},\mathbf{y}\in conv\left(X\right)={\bigcup }_{n=0}^{\infty }{C}_n$, we have

$$\delta (\mathbf{x},\mathbf{y})\le diam\left(X\right).$$

Therefore

$$diam(conv(X\left)\right)\le diam\left(X\right).$$

The converse inequality is trivial, hence we have $diam(conv(X\left)\right)=diam\left(X\right)$.

Next we prove $diam(conv\left(X\right))=diam\left(\overline{conv\left(X\right)}\right)$. For any pair $\mathbf{x},\mathbf{y}\in \overline{conv\left(X\right)}$, there exist sequences $\left\{{\mathbf{x}}_n\right\}$ and $\left\{{\mathbf{y}}_n\right\}$ in $conv\left(X\right)$ such that ${\mathbf{x}}_n,{\mathbf{y}}_n$ converge to $\mathbf{x},\mathbf{y}$ respectively. Letting $n\to \infty$ for $\delta ({\mathbf{x}}_n,{\mathbf{y}}_n)\le diam(conv\left(X\right))$, we have $\delta (\mathbf{x},\mathbf{y})\le diam\left(X\right)$. Thus, $diam\left(\overline{conv\left(X\right)}\right)\le diam(conv\left(X\right))$. The converse inequality is trivial, hence $diam(conv\left(X\right))=diam\left(\overline{conv\left(X\right)}\right)$ holds. □

Lemma 4.

Suppose that K is a gyroconvex subset of $\mathbb{V}$. Then the closure $\overline{K}$ of K is gyroconvex.

Proof. 

For any $\mathbf{x},\mathbf{y}\in \overline{K}$, there exist $\left\{{\mathbf{x}}_n\right\},\left\{{\mathbf{y}}_n\right\}\subset K$ such that ${\mathbf{x}}_n,{\mathbf{y}}_n$ converge to $\mathbf{x},\mathbf{y}$ respectively. We show ${\mathbf{x}}_n{\#}_t{\mathbf{y}}_n$ converges to $\mathbf{x}{\#}_t\mathbf{y}$ for arbitrary $0\le t\le 1$. By (11) we have

$$\delta ({\mathbf{x}}_n{\#}_t{\mathbf{y}}_n,\mathbf{x}{\#}_t\mathbf{y})\le (1-t)\delta ({\mathbf{x}}_n,\mathbf{x})+t\delta ({\mathbf{y}}_n,\mathbf{y}).$$

By $n\to \infty$, then $\delta ({\mathbf{x}}_n{\#}_t{\mathbf{y}}_n,\mathbf{x}\#\mathbf{y})\to 0$. Thus, $\mathbf{x}{\#}_t\mathbf{y}\in \overline{K}$. Hence $\overline{K}$ is gyroconvex. □

Let n be a positive integer. Let $X_n$ be the set of all subsets of ${\mathbb{V}}_1$ whose number of elements is exactly n. We define, by induction, the sequence ${\left\{G_n\right\}}_{n=2}^{\infty }$ of the maps $G_n:X_n\to {\mathbb{V}}_1$ which satisfy the following two conditions ($p_n$) and ($q_n$);

($p_n$)

$G_n(\Delta )\in \overline{conv(\Delta )}$ for every $\Delta \in X_n$,

($q_n$)

$\delta (G_n(\Delta ),G_n\left({\Delta }^{\prime }\right))\le \frac{1}{n}{\sum }_{i=1}^n\delta ({\mathbf{a}}_i,{\mathbf{a}}_i^{\prime })$ for every pair $\Delta =\{{\mathbf{a}}_1,\dots ,{\mathbf{a}}_n\}$ and ${\Delta }^{\prime }=\{{\mathbf{a}}_1^{\prime },\dots ,{\mathbf{a}}_n^{\prime }\}$ in $X_n$.

First, put $G_2\left(\{\mathbf{a},\mathbf{b}\}\right)=\mathbf{a}\#\mathbf{b}$ for $\{\mathbf{a},\mathbf{b}\}\in X_2$. As $conv\left(\right\{\mathbf{a},\mathbf{b}\left\}\right)=S(\mathbf{a},\mathbf{b})$ by Lemma 2 we obtain that $G_2\left(\{\mathbf{a},\mathbf{b}\}\right)\in conv\left(\{\mathbf{a},\mathbf{b}\}\right)$; ($p_2$) holds. Let $\Delta =\{\mathbf{a},\mathbf{b}\},{\Delta }^{\prime }=\{\mathbf{c},\mathbf{d}\}\in X_2$. Then by (10) we have

$$\delta (G_2(\Delta ),G_2\left({\Delta }^{\prime }\right))=\delta (\mathbf{a}\#\mathbf{b},\mathbf{c}\#\mathbf{d})\le \frac{1}{2}\{\delta (\mathbf{a},\mathbf{c})+\delta (\mathbf{b},\mathbf{d})\},$$

which is ($q_2$).

Suppose now that the map $G_k:X_k\to {\mathbb{V}}_1$ which satisfies ($p_k$) and ($q_k$) is defined. We will define $G_{k+1}:X_{k+1}\to {\mathbb{V}}_1$ which satisfies ($p_{k+1}$) and ($q_{k+1}$). Let ${\Delta }^0=\{{\mathbf{a}}_1^0,\dots ,{\mathbf{a}}_{k+1}^0\}\in X_{k+1}$. For a positive integer m we define ${\Delta }^m\in X_{k+1}$ which satisfies that ${\Delta }^m\subset \overline{conv\left({\Delta }^{m-1}\right)}$ by induction on m. For every $1\le i\le k+1$, put

$${\mathbf{a}}_i^1=G_k\left(\{{\mathbf{a}}_1^0,\dots ,{\mathbf{a}}_{i-1}^0,{\mathbf{a}}_{i+1}^0,\dots ,{\mathbf{a}}_{k+1}^0\}\right).$$

Please note that ${\mathbf{a}}_i^1$ is well defined since $\{{\mathbf{a}}_1^0,\dots ,{\mathbf{a}}_{i-1}^0,{\mathbf{a}}_{i+1}^0,\dots ,{\mathbf{a}}_{k+1}^0\}\in X_k$ and we have assumed that the map $G_k$ is defined. By the condition ($p_k$) we have that

$${\mathbf{a}}_i^1\in \overline{conv\left(\{{\mathbf{a}}_1^0,\dots ,{\mathbf{a}}_{i-1}^0,{\mathbf{a}}_{i+1}^0,\dots ,{\mathbf{a}}_{k+1}^0\}\right)}$$

for every $1\le i\le k+1$. Put ${\Delta }^1=\{{\mathbf{a}}_1^1,\dots ,{\mathbf{a}}_{k+1}^1\}$. Then ${\Delta }^1\in X_{k+1}$ and ${\Delta }^1\subset \overline{conv\left({\Delta }^0\right)}$ since $\{{\mathbf{a}}_1^0,\dots ,{\mathbf{a}}_{i-1}^0,{\mathbf{a}}_{i+1}^0,\dots ,{\mathbf{a}}_{k+1}^0\}\subset {\Delta }^0$ for every $1\le i\le k+1$. Suppose that ${\Delta }^l=\{{\mathbf{a}}_1^l,\dots ,{\mathbf{a}}_{k+1}^l\}\in X_{k+1}$ such that ${\Delta }^l\subset \overline{conv\left({\Delta }^{l-1}\right)}$ is defined. For every $1\le i\le k+1$, put

$${\mathbf{a}}_i^{l+1}=G_k\left(\{{\mathbf{a}}_1^l,\dots ,{\mathbf{a}}_{i-1}^l,{\mathbf{a}}_{i+1}^l,\dots ,{\mathbf{a}}_{k+1}^l\}\right).$$

As in the same way as the above, ${\mathbf{a}}_i^{l+1}$ is well defined for $1\le i\le k+1$, and ${\Delta }^{l+1}=\{{\mathbf{a}}_1^{l+1},\dots ,{\mathbf{a}}_{k+1}^{l+1}\}\in X_{k+1}$ satisfies that ${\Delta }^{l+1}\subset \overline{conv\left({\Delta }^l\right)}$. Hence, by induction, we have defined a sequence ${\left\{{\Delta }^m\right\}}_{m=1}^{\infty }\subset X_{k+1}$ such that ${\Delta }^m\subset \overline{conv\left({\Delta }^{m-1}\right)}$. Applying ($q_k$) for $\Delta =\{{\mathbf{a}}_1^m,\dots ,{\mathbf{a}}_{i-1}^m,{\mathbf{a}}_{i+1}^m,\dots ,{\mathbf{a}}_{k+1}^m\}$ and ${\Delta }^{\prime }=\{{\mathbf{a}}_1^m,\dots ,{\mathbf{a}}_{j-1}^m,{\mathbf{a}}_{j+1}^m,\dots ,{\mathbf{a}}_{k+1}^m\}$, we infer that

$$\begin{array}{c}\delta ({\mathbf{a}}_i^{m+1},{\mathbf{a}}_j^{m+1})=\delta (G_k(\Delta ),G_k\left({\Delta }^{\prime }\right))\hfill \\ =\delta (G_k\left(\{{\mathbf{a}}_1^m,\dots ,{\mathbf{a}}_{i-1}^m,{\mathbf{a}}_{i+1}^m,\dots ,{\mathbf{a}}_{k+1}^m\}\right),G_k\left(\{{\mathbf{a}}_1^m,\dots ,{\mathbf{a}}_{j-1}^m,{\mathbf{a}}_{j+1}^m,\dots ,{\mathbf{a}}_{k+1}^m\}\right))\\ =\delta (G_k\left(\{{\mathbf{a}}_1^m,\dots ,{\mathbf{a}}_{i-1}^m,{\mathbf{a}}_{i+1}^m,\dots ,{\mathbf{a}}_{j-1}^m,{\mathbf{a}}_{j+1}^m,\dots ,{\mathbf{a}}_{k+1}^m,{\mathbf{a}}_j^m\}\right),\\ G_k\left(\{{\mathbf{a}}_1^m,\dots ,{\mathbf{a}}_{i-1}^m,{\mathbf{a}}_{i+1}^m,\dots ,{\mathbf{a}}_{j-1}^m,{\mathbf{a}}_{j+1}^m,\dots ,{\mathbf{a}}_{k+1}^m,{\mathbf{a}}_i^m\}\right))\\ \hfill \le \frac{1}{k}\delta ({\mathbf{a}}_j^m,{\mathbf{a}}_i^m)=\frac{1}{k}\delta ({\mathbf{a}}_i^m,{\mathbf{a}}_j^m)\end{array}$$

Then by Lemma 3 we obtain

$$diam\left(\overline{conv\left({\Delta }^{m+1}\right)}\right)\le \frac{1}{k}diam\left(\overline{conv\left({\Delta }^m\right)}\right)$$

for every positive integer m. By Cantor’s intersection theorem there exists a unique $M_{\infty }\in {\mathbb{V}}_1$ with

$$\left\{M_{\infty }\right\}={\cap }_m\overline{conv\left({\Delta }^m\right)}.$$

As $a_i^m\in \overline{conv({\Delta }^{m-1}})$ for every $1\le i\le k+1$, we infer that ${lim}_{m\to \infty }{\mathbf{a}}_i^m=M_{\infty }$ for every $1\le i\le k+1$. Put $G_{k+1}\left({\Delta }^0\right)=M_{\infty }$. Then the map $G_{k+1}:X_{k+1}\to {\mathbb{V}}_1$ is well defined, and $G_{k+1}\left({\Delta }^0\right)=M_{\infty }\in \overline{conv\left({\Delta }^0\right)}$; $G_{k+1}$ satisfies the condition ($p_n$).

Next we prove that the map $G_{k+1}$ satisfies the condition ($q_{k+1}$);

$$\delta (G_{k+1}(\Delta ),G_{k+1}\left({\Delta }^{\prime }\right))\le \frac{1}{k+1}\left(\sum _{j=1}^{k+1}\delta ({\mathbf{a}}_j,{{\mathbf{a}}^{\prime }}_j)\right),$$

where $\Delta =\{{\mathbf{a}}_1^0,\dots ,{\mathbf{a}}_{k+1}^0\},{\Delta }^{\prime }=\{{{\mathbf{a}}^{\prime }}_1^0,\dots ,{{\mathbf{a}}^{\prime }}_{k+1}^0\}\in X_{k+1}$. Let m be a positive integer. We define ${\mathbf{a}}_i^m$ and ${{\mathbf{a}}^{\prime }}_i^m$ for every $1\le i\le k+1$ as in the same way as before. As ($q_k$) holds for $G_k$, we have

$$\begin{array}{c}\delta ({\mathbf{a}}_i^m,{{\mathbf{a}}^{\prime }}_i^m)\hfill \\ =\delta (G_k\left(\{{\mathbf{a}}_1^{m-1},{\mathbf{a}}_2^{m-1},\cdots ,{\mathbf{a}}_{i-1}^{m-1},{\mathbf{a}}_{i+1}^{m-1},\cdots ,{\mathbf{a}}_{k+1}^{m-1}\}\right),\\ G_k\left(\{{{\mathbf{a}}^{\prime }}_1^{m-1},{{\mathbf{a}}^{\prime }}_2^{m-1},\cdots ,{{\mathbf{a}}^{\prime }}_{i-1}^{m-1},{{\mathbf{a}}^{\prime }}_{i+1}^{m-1},\cdots ,{{\mathbf{a}}^{\prime }}_{k+1}^{m-1}\}\right))\\ \hfill \le \frac{1}{k}\left(\sum _{j=1,j\ne i}^{k+1}\delta ({\mathbf{a}}_j^{m-1},{{\mathbf{a}}^{\prime }}_j^{m-1})\right).\end{array}$$

By summing up the above inequalities with respect to $1\le i\le k+1$ we have

$$\sum _{i=1}^{k+1}\delta ({\mathbf{a}}_i^m,{{\mathbf{a}}^{\prime }}_i^m)\le \sum _{i=1}^{k+1}\left(\frac{1}{k}\left(\sum _{j=1,j\ne i}^{k+1}\delta ({\mathbf{a}}_j^{m-1},{{\mathbf{a}}^{\prime }}_j^{m-1})\right)\right)=\sum _{j=1}^{k+1}\delta ({\mathbf{a}}_j^{m-1},{{\mathbf{a}}^{\prime }}_j^{m-1}).$$

for every positive integer m. Thus, we have

$$\sum _{i=1}^{k+1}\delta ({\mathbf{a}}_i^m,{{\mathbf{a}}^{\prime }}_i^m)\le \sum _{j=1}^{k+1}\delta ({\mathbf{a}}_j,{{\mathbf{a}}^{\prime }}_j).$$

Letting $m\to \infty$, since ${lim}_{m\to \infty }{\mathbf{a}}_i^m=G_{k+1}(\Delta ),{lim}_{m\to \infty }{{\mathbf{a}}^{\prime }}_i^m=G_{k+1}\left({\Delta }^{\prime }\right)$, we have

$$\sum _{i=1}^{k+1}\delta (G_{k+1}(\Delta ),G_{k+1}\left({\Delta }^{\prime }\right))\le \sum _{j=1}^{k+1}\delta ({\mathbf{a}}_j,{{\mathbf{a}}^{\prime }}_j),$$

hence

$$\delta \left(G_{k+1}(\Delta )\right),G_{k+1}\left({\Delta }^{\prime }\right))\le \frac{1}{k+1}\left(\sum _{j=1}^{k+1}\delta ({\mathbf{a}}_j,{{\mathbf{a}}^{\prime }}_j)\right).$$

So, the condition ($q_{k+1}$) holds for the map $G_{k+1}$. We conclude that the map $G_n:X_n\to {\mathbb{V}}_1$ which satisfies the conditions ($p_n$) and ($q_n$) are defined by induction.

By applying the maps $G_n$ we define the gyrogeometric mean of n elements in ${\mathbb{V}}_1$.

Definition 2.

Let $\Delta =\{{\mathbf{a}}_1,\dots ,{\mathbf{a}}_n\}\subset {\mathbb{V}}_1$. We call that $G_n(\Delta )$ the gyrogeometric mean of Δ.

Due to the definition, the gyrogeometric mean of $\{\mathbf{a},\mathbf{b}\}\subset {\mathbb{V}}_1$ is $\mathbf{a}\#\mathbf{b}$. The gyrocentroid ${\mathbb{C}}_{\mathbf{a}\mathbf{b}\mathbf{c}}^m$ is defined by applying the internal division points on the usual lines which makes the inconvenience such as ${\mathbb{C}}_{\mathbf{a}\mathbf{b}\mathbf{c}}^m=\frac{{\gamma }_{\mathbf{c}}}{{\gamma }_{\mathbf{c}}+2}\mathbf{c}\ne \frac{1}{3}{\otimes }_E\mathbf{c}$ for $\mathbf{a}=\mathbf{b}=0$. The gyrogeometric mean is defined by applying the gyrolines and it resolve the inconvenience.

5. Properties of the Gyrogeometric Mean

The gyrogeometric mean satisfies certain desirable properties one would expect for means in general. For example, the permutation invariance and the left-translation invariance would be expected properties. It is trivial that the gyrogeometric mean is permutation-invariant. We prove that the gyrogeometric mean is left-translation invariant.

Recall that $X_n$ is the set of all n-points subset of ${\mathbb{V}}_1$ for a positive integer n.

Proposition 5.

Let $\mathbf{x}\in {\mathbb{V}}_1$ and $D_n=\{{\mathbf{a}}_1,{\mathbf{a}}_2,\cdots ,{\mathbf{a}}_n\}\in X_n$. Put $\mathbf{x}{\oplus }_E{D}_n=\{\mathbf{x}{\oplus }_E{\mathbf{a}}_1,\mathbf{x}{\oplus }_E{\mathbf{a}}_2,\cdots ,\mathbf{x}{\oplus }_E{\mathbf{a}}_n\}$. Then the following holds:

$$G_n\left(\mathbf{x}{\oplus }_E{D}_n\right)=\mathbf{x}{\oplus }_E{G}_n\left(D_n\right).$$

(13)

Proof. 

We prove the equality (13) by induction on n. For $n=2$, $G_2\left(\{{\mathbf{a}}_1,{\mathbf{a}}_2\}\right)={\mathbf{P}}_{{\mathbf{a}}_1{\mathbf{a}}_2}^m$. By Theorem 6.37 in [1] (p. 175) we have

$$G_2\left(\{\mathbf{x}{\oplus }_E{\mathbf{a}}_1,\mathbf{x}{\oplus }_E{\mathbf{a}}_2\}\right)=\mathbf{x}{\oplus }_E{G}_2\left(\{{\mathbf{a}}_1,{\mathbf{a}}_2\}\right).$$

Assume that (13) holds for $n=k$. Let $D_{k+1}^m=\{{\mathbf{a}}_1^m,{\mathbf{a}}_2^m,\cdots ,{\mathbf{a}}_{k+1}^m\}$ where ${\mathbf{a}}_i^m=G_k\left(\{{\mathbf{a}}_1^{m-1},{\mathbf{a}}_2^{m-1},\cdots ,{\mathbf{a}}_{i-1}^{m-1},{\mathbf{a}}_{i+1}^{m-1},\cdots ,{\mathbf{a}}_{k+1}\}\right)$ for $1\le i\le k+1$. By the assumption we have

$$\begin{array}{c}\hfill G_k\left(\{\mathbf{x}{\oplus }_E{\mathbf{a}}_1^{m-1},\mathbf{x}{\oplus }_E{\mathbf{a}}_2^{m-1},\cdots ,\mathbf{x}{\oplus }_E{\mathbf{a}}_{i-1}^{m-1},\mathbf{x}{\oplus }_E{\mathbf{a}}_{i+1}^{m-1},\cdots ,\mathbf{x}{\oplus }_E{\mathbf{a}}_{k+1}^{m-1}\}\right)\\ \hfill =\mathbf{x}{\oplus }_E{G}_k\left(\{{\mathbf{a}}_1^{m-1},{\mathbf{a}}_2^{m-1},\cdots ,{\mathbf{a}}_{i-1}^{m-1},{\mathbf{a}}_{i+1}^{m-1},\cdots ,{\mathbf{a}}_{k+1}^{m-1}\}\right)\\ \hfill =\mathbf{x}{\oplus }_E{\mathbf{a}}_i^m\end{array}$$

for every $1\le i\le k+1$. Then for $\mathbf{x}{\oplus }_E{D}_{k+1}=\{\mathbf{x}{\oplus }_E{\mathbf{a}}_1,\mathbf{x}{\oplus }_E{\mathbf{a}}_2,\cdots ,\mathbf{x}{\oplus }_E{\mathbf{a}}_{k+1}\}$ we have

$$\begin{array}{c}\mathbf{x}{\oplus }_E{D}_{k+1}^m\hfill \\ =\{G_k\left(\{\mathbf{x}{\oplus }_E{\mathbf{a}}_1^{m-1},\cdots ,\mathbf{x}{\oplus }_E{\mathbf{a}}_{i-1}^{m-1},\mathbf{x}{\oplus }_E{\mathbf{a}}_{i+1}^{m-1},\cdots ,\mathbf{x}{\oplus }_E{\mathbf{a}}_{k+1}^{m-1}\}\right):1\le i\le k+1\}\\ \hfill =\{\mathbf{x}{\oplus }_E{\mathbf{a}}_i^m:i=1,2,\cdots ,k+1\}.\end{array}$$

We prove that $\mathbf{x}{\oplus }_E{D}_{k+1}^m\to \left\{\mathbf{x}{\oplus }_E{G}_{k+1}\left(D_{k+1}\right)\right\}$ as $m\to \infty$. By a simple calculation we have

$$\begin{array}{c}d(\mathbf{x}{\oplus }_E{\mathbf{a}}_i^m,\mathbf{x}{\oplus }_E{G}_{k+1}\left(D_{k+1}\right))\hfill \\ \hfill =\parallel \mathrm{gyr}[\mathbf{x},{\mathbf{a}}_{\mathbf{i}}^{\mathbf{m}}]\left({\mathbf{a}}_i^m{\ominus }_{E}G\left(D_{k+1}\right)\right)\parallel =\parallel {\mathbf{a}}_i^m{\ominus }_{E}G\left(D_{k+1}\right)\parallel \to 0\end{array}$$

as $m\to \infty$. Hence we have $\delta (\mathbf{x}{\oplus }_E{\mathbf{a}}_i^m,\mathbf{x}{\oplus }_E{G}_{k+1}\left(D_{k+1}\right))\to 0$ as $m\to \infty$. Thus, $\mathbf{x}{\oplus }_E{\mathbf{a}}_i^m\to \mathbf{x}{\oplus }_{E}G\left(D_{k+1}\right)$ as $m\to \infty$ for $1\le i\le k+1$. We conclude that $G_n\left(\mathbf{x}{\oplus }_E{D}_n\right)=\mathbf{x}{\oplus }_E{G}_n\left(D_n\right)$. □

For $\mathbf{a}=\mathbf{b}=\mathbf{0}$ and $\mathbf{c}$ in ${\mathbb{V}}_1$, $G_3\left(\{\mathbf{a},\mathbf{b},\mathbf{c}\}\right)=\frac{1}{3}{\otimes }_E\mathbf{c}$. More generally, the gyrogeometric mean satisfies the following.

Proposition 6.

Let $\Delta =\{t_1{\otimes }_E\mathbf{a},t_2{\otimes }_E\mathbf{a},\cdots ,t_n{\otimes }_E\mathbf{a}\}\subset {\mathbb{V}}_1$.

$$G_n(\Delta )=\frac{t_1+t_2+\cdots +t_n}{n}{\otimes }_E\mathbf{a}$$

In the case of $n=2$, it is proved by the following calculation.

$$\begin{array}{cc}& G_2(t_1{\otimes }_E\mathbf{a},t_2{\otimes }_E\mathbf{a})\hfill \\ =& t_1{\otimes }_E\mathbf{a}\#t_2{\otimes }_E\mathbf{a}\hfill \\ =& \frac{1}{2}{\otimes }_E\left\{\left(t_1{\otimes }_E\mathbf{a}\right){⊞}_E\left(t_2{\otimes }_E\mathbf{a}\right)\right\}\hfill \\ =& \frac{1}{2}{\otimes }_E\left\{\left(t_1{\otimes }_E\mathbf{a}\right){\oplus }_E\mathrm{gyr}[{\mathbf{t}}_{\mathbf{1}}{\otimes }_{\mathbf{E}}\mathbf{a},{\ominus }_{\mathbf{E}}{\mathbf{t}}_{\mathbf{2}}{\otimes }_{\mathbf{E}}\mathbf{a}]\left(t_2{\otimes }_E\mathbf{a}\right)\right\}\hfill \\ =& \frac{1}{2}{\otimes }_E\left\{\left(t_1{\otimes }_E\mathbf{a}\right){\oplus }_E\left(t_2{\otimes }_E\mathbf{a}\right)\right\}\hfill \\ =& \frac{1}{2}{\otimes }_E\{(t_1+t_2)\otimes \mathbf{a}\}\hfill \\ =& \frac{t_1+t_2}{2}{\otimes }_E\mathbf{a}.\hfill \end{array}$$

Proposition 6 is proved by induction on n.

In Section 6 ${\mathbb{V}}_s=\{\mathbf{v}\in \mathbb{V}:\parallel \mathbf{v}\parallel <s\}$ with appropriate operation is a gyrocommutative gyrogroup, which is also called the Einstein gyrogroup. The gyrogeometric mean is defined for ${\mathbb{V}}_s$ similarly. If $s\to \infty$ or $\mathbf{v}\in {\mathbb{V}}_s$ such that $\parallel \mathbf{v}\parallel$ is small enough, ${\gamma }_{\mathbf{v}}\to 1$. So, in the case,

$$G_n\left(\{{\mathbf{a}}_1,{\mathbf{a}}_2,\dots ,{\mathbf{a}}_n\}\right)\to \frac{{\mathbf{a}}_1+{\mathbf{a}}_2+\dots +{\mathbf{a}}_n}{n}$$

is hold. It is simply proved by induction.

6. Proof that $({\mathbb{V}}_{\mathbf{s}},{\oplus }_{\mathbf{E}})$ Is a Gyrocommutative Gyrogroup

A magma $(G,\oplus )$ is a non-empty set G with a binary operation ⊕. A magma $(G,\oplus )$ is a gyrogroup if its binary operation ⊕ satisfies the following axioms (G1) through (G5):

(G1)

There exists a left identity $\mathbf{0}$ in G such that

$$\mathbf{0}\oplus \mathbf{a}=\mathbf{a}$$

for all $\mathbf{a}\in G$.

(G2)

For each $\mathbf{a}\in G$ there exists a left inverse $\ominus \mathbf{a}\in G$ such that

$$\ominus \mathbf{a}\oplus \mathbf{a}=\mathbf{0}.$$

(G3)

For any $\mathbf{a},\mathbf{b},\mathbf{c}\in G$ there exists a unique element $\mathrm{gyr}[\mathbf{a},\mathbf{b}]\mathbf{c}\in G$ such that the binary operation obeys the left gyroassociative law

$$\mathbf{a}\oplus (\mathbf{b}\oplus \mathbf{c})=(\mathbf{a}\oplus \mathbf{b})\oplus \mathrm{gyr}[\mathbf{a},\mathbf{b}]\mathbf{c}.$$

(G4)

The map $\mathrm{gyr}[\mathbf{a},\mathbf{b}]:G\mapsto G$ given by $\mathbf{c}\mapsto \mathrm{gyr}[\mathbf{a},\mathbf{b}]\mathbf{c}$ is an automorphism of the magma $(G,\oplus )$. It is called a gyroautomorphism. $\mathrm{gyr}[\mathbf{a},\mathbf{b}]$ generated by $\mathbf{a},\mathbf{b}\in G$ is called a gyration.

(G5)

The gyroautomorphism $\mathrm{gyr}[\mathbf{a},\mathbf{b}]$ generated by any $\mathbf{a},\mathbf{b}\in G$ satisfies the left loop property:

$$\mathrm{gyr}[\mathbf{a},\mathbf{b}]=\mathrm{gyr}[\mathbf{a}\oplus \mathbf{b},\mathbf{b}].$$

The gyrogroup $(G,\oplus )$ is called gyrocommutative if the following (G6) holds for every pair $\mathbf{a},\mathbf{b}\in G$

(G6)

$\mathbf{a}\oplus \mathbf{b}=\mathrm{gyr}[\mathbf{a},\mathbf{b}](\mathbf{b}\oplus \mathbf{a}).$

We prove that the Einstein gyrogroup $({\mathbb{V}}_s,{\oplus }_E)$ is in fact a gyrocommutative gyrogroup only by simple calculations. Proof of (G1) and (G2) are simple and omitted.

We prove (G3). We prove that $\mathbf{u}{\oplus }_E\left(\mathbf{v}{\oplus }_E\mathbf{w}\right)=\left(\mathbf{u}{\oplus }_E\mathbf{v}\right){\oplus }_E\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{w}$ holds for all $\mathbf{u},\mathbf{v},\mathbf{w}\in {\mathbb{V}}_s$. First, we prove the left cancellation law which is given by the equation

$${\ominus }_E\mathbf{a}{\oplus }_E\left(\mathbf{a}{\oplus }_E\mathbf{b}\right)=\mathbf{b},$$

for all $\mathbf{a},\mathbf{b}\in {\mathbb{V}}_s$. Put $D_{\mathbf{u}\mathbf{v}}=1+\frac{\mathbf{u}·\mathbf{v}}{s^2}$ for any $\mathbf{u},\mathbf{v}\in {\mathbb{V}}_s$ and put

$$\begin{array}{ccc}\hfill \mathbf{x}& =& \mathbf{a}{\oplus }_E\mathbf{b}\hfill \\ & =& \frac{1}{D_{\mathbf{a}\mathbf{b}}}\left\{\mathbf{a}+\frac{1}{{\gamma }_{\mathbf{a}}}\mathbf{b}+\frac{1}{s^2}\frac{{\gamma }_{\mathbf{a}}}{1+{\gamma }_{\mathbf{a}}}(\mathbf{a}·\mathbf{b})\mathbf{a}\right\}.\hfill \end{array}$$

Put $1+\frac{1}{s^2}((-\mathbf{a})·\mathbf{x})=D_{(-\mathbf{a})\mathbf{x}}$. We have

$$\begin{array}{ccc}\hfill {\ominus }_E\mathbf{a}{\oplus }_E\mathbf{x}& =& \frac{1}{D_{(-\mathbf{a})\mathbf{x}}}\left\{(-\mathbf{a})+\frac{1}{{\gamma }_{\mathbf{a}}}\mathbf{x}+\frac{1}{s^2}\frac{{\gamma }_{\mathbf{a}}}{1+{\gamma }_{\mathbf{a}}}(\mathbf{a}·\mathbf{x})\mathbf{a}\right\}\hfill \\ & =& \frac{1}{D_{(-\mathbf{a})\mathbf{x}}}\left\{(-\mathbf{a})+\frac{1}{{\gamma }_{\mathbf{a}}{D}_{\mathbf{a}\mathbf{b}}}(\frac{1+{\gamma }_{\mathbf{a}}{D}_{\mathbf{a}\mathbf{b}}}{1+{\gamma }_{\mathbf{a}}}\mathbf{a}+\frac{1}{{\gamma }_{\mathbf{a}}}\mathbf{b})+\frac{1}{s^2}\frac{{\gamma }_{\mathbf{a}}}{1+{\gamma }_{\mathbf{a}}}(\mathbf{a}·\mathbf{x})\mathbf{a}\right\}.\hfill \end{array}$$

We compute,

$$\begin{array}{ccc}\hfill D_{(-\mathbf{a})\mathbf{x}}& \hfill =& 1+\frac{1}{s^2}((-\mathbf{a})·\mathbf{x})\hfill \\ \hfill & \hfill =& 1-\frac{1}{D_{\mathbf{a}\mathbf{b}}{s}^2}\left\{{\parallel \mathbf{a}\parallel }^2+\frac{1}{{\gamma }_{\mathbf{a}}}(\mathbf{a}·\mathbf{b})+\frac{1}{s^2}\frac{{\gamma }_{\mathbf{a}}}{1+{\gamma }_{\mathbf{a}}}(\mathbf{a}·\mathbf{b}){\parallel \mathbf{a}\parallel }^2\right\}\hfill \\ \hfill & \hfill =& 1-\frac{1}{D_{\mathbf{a}\mathbf{b}}}\left\{\frac{{{\gamma }_{\mathbf{a}}}^2-1}{{{\gamma }_{\mathbf{a}}}^2}+\frac{1}{{\gamma }_{\mathbf{a}}}\frac{(\mathbf{a}·\mathbf{b})}{s^2}+\frac{{\gamma }_{\mathbf{a}}}{1+{\gamma }_{\mathbf{a}}}\frac{(\mathbf{a}·\mathbf{b})}{s^2}\frac{{{\gamma }_{\mathbf{a}}}^2-1}{{{\gamma }_{\mathbf{a}}}^2}\right\}\hfill \\ \hfill & \hfill =& 1-\frac{1}{D_{\mathbf{a}\mathbf{b}}}\left\{1-\frac{1}{{{\gamma }_{\mathbf{a}}}^2}+\frac{1}{{\gamma }_{\mathbf{a}}}(D_{\mathbf{a}\mathbf{b}}-1)+\frac{{\gamma }_{\mathbf{a}}-1}{{\gamma }_{\mathbf{a}}}(D_{\mathbf{a}\mathbf{b}}-1)\right\}\hfill \\ \hfill & \hfill =& \frac{1}{{{\gamma }_{\mathbf{a}}}^2{D}_{\mathbf{a}\mathbf{b}}},\hfill \end{array}$$

and

$$\frac{\mathbf{a}·\mathbf{x}}{s^2}=1-D_{(-\mathbf{a})\mathbf{x}}=\frac{{{\gamma }_{\mathbf{a}}}^2{D}_{\mathbf{a}\mathbf{b}}-1}{{{\gamma }_{\mathbf{a}}}^2{D}_{\mathbf{a}\mathbf{b}}}.$$

Hence we have

$$\begin{array}{cc}\hfill {\ominus }_E\mathbf{a}{\oplus }_E\left(\mathbf{a}{\oplus }_E\mathbf{b}\right)& ={\ominus }_E\mathbf{a}{\oplus }_E\mathbf{x}\hfill \\ \hfill & ={{\gamma }_{\mathbf{a}}}^2{D}_{\mathbf{a}\mathbf{b}}\{(-\mathbf{a})+\frac{1}{{\gamma }_{\mathbf{a}}{D}_{\mathbf{a}\mathbf{b}}}(\frac{1+{\gamma }_{\mathbf{a}}{D}_{\mathbf{a}\mathbf{b}}}{1+{\gamma }_{\mathbf{a}}}\mathbf{a}+\frac{1}{{\gamma }_{\mathbf{a}}}\mathbf{b})\hfill \\ \hfill & +\frac{{\gamma }_{\mathbf{a}}}{1+{\gamma }_{\mathbf{a}}}\frac{{{\gamma }_{\mathbf{a}}}^2{D}_{\mathbf{a}\mathbf{b}}-1}{{{\gamma }_{\mathbf{a}}}^2{D}_{\mathbf{a}\mathbf{b}}}\mathbf{a}\}\hfill \\ \hfill & =-{{\gamma }_{\mathbf{a}}}^2{D}_{\mathbf{a}\mathbf{b}}\mathbf{a}+{\gamma }_{\mathbf{a}}(\frac{1+{\gamma }_{\mathbf{a}}{D}_{\mathbf{a}\mathbf{b}}}{1+{\gamma }_{\mathbf{a}}}\mathbf{a}+\frac{1}{{\gamma }_{\mathbf{a}}}\mathbf{b})+\frac{{\gamma }_{\mathbf{a}}}{1+{\gamma }_{\mathbf{a}}}({{\gamma }_{\mathbf{a}}}^2{D}_{\mathbf{a}\mathbf{b}}-1)\mathbf{a}\hfill \\ \hfill & =\mathbf{b}+\left(-{{\gamma }_{\mathbf{a}}}^2{D}_{\mathbf{a}\mathbf{b}}+\frac{{\gamma }_{\mathbf{a}}+{{\gamma }_{\mathbf{a}}}^2{D}_{\mathbf{a}\mathbf{b}}}{1+{\gamma }_{\mathbf{a}}}+\frac{{{\gamma }_{\mathbf{a}}}^3{D}_{\mathbf{a}\mathbf{b}}-{\gamma }_{\mathbf{a}}}{1+{\gamma }_{\mathbf{a}}}\right)\mathbf{a}\hfill \\ \hfill & =\mathbf{b}.\hfill \end{array}$$

Next, we prove the following equation

$$\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{w}={\ominus }_E\left(\mathbf{u}{\oplus }_E\mathbf{v}\right){\oplus }_E\left(\mathbf{u}{\oplus }_E\left(\mathbf{v}{\oplus }_E\mathbf{w}\right)\right).$$

(14)

It is known in [5] ((2.84), (2.85)) that the Equation (14) can be rewritten as

$$\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{w}=\mathbf{w}+\frac{\mathrm{A}\mathbf{u}+\mathrm{B}\mathbf{v}}{\mathrm{D}}$$

(15)

by applying computer algebra, where

$$\begin{array}{ccc}\hfill A& =& -\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{w})+\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{v}·\mathbf{w})+\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{w})\hfill \\ \hfill B& =& -\frac{1}{s^2}\frac{{\gamma }_{\mathbf{v}}}{{\gamma }_{\mathbf{v}}+1}\left\{{\gamma }_{\mathbf{u}}({\gamma }_{\mathbf{v}}+1)(\mathbf{u}·\mathbf{w})+({\gamma }_{\mathbf{u}}-1){\gamma }_{\mathbf{v}}(\mathbf{v}·\mathbf{w})\right\}\hfill \\ \hfill D& =& {\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(1+\frac{\mathbf{u}·\mathbf{v}}{s^2})+1.\hfill \end{array}$$

We prove (15) without applying computer algebra. Put

$$\begin{array}{ccc}\hfill \mathbf{z}& \hfill =& {\ominus }_E\left(\mathbf{u}{\oplus }_E\mathbf{v}\right)\hfill \\ \hfill & \hfill =& -\frac{1}{1+\frac{\mathbf{u}·\mathbf{v}}{s^2}}\left\{\mathbf{u}+\frac{1}{{\gamma }_{\mathbf{u}}}\mathbf{v}+\frac{1}{s^2}\frac{{\gamma }_{\mathbf{u}}}{1+{\gamma }_{\mathbf{u}}}(\mathbf{u}·\mathbf{v})\mathbf{u}\right\}\hfill \\ & \hfill =& -\frac{1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}}}{D_{\mathbf{uv}}(1+{\gamma }_{\mathbf{u}})}\mathbf{u}-\frac{1}{{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}}}\mathbf{v},\hfill \\ \hfill \mathbf{x}& \hfill =& \left(\mathbf{v}{\oplus }_E\mathbf{w}\right)\hfill \\ \hfill & \hfill =& \frac{1+{\gamma }_{\mathbf{v}}{D}_{\mathbf{vw}}}{D_{\mathbf{vw}}(1+{\gamma }_{\mathbf{v}})}\mathbf{v}+\frac{1}{{\gamma }_{\mathbf{v}}{D}_{\mathbf{vw}}}\mathbf{w}.\hfill \end{array}$$

Put also

$$\begin{array}{ccc}\hfill \mathbf{y}& \hfill =& \mathbf{u}{\oplus }_E\left(\mathbf{v}{\oplus }_E\mathbf{w}\right)\hfill \\ \hfill & \hfill =& \mathbf{u}{\oplus }_E\mathbf{x}\hfill \\ \hfill & \hfill =& \frac{1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{ux}}}{D_{\mathbf{ux}}(1+{\gamma }_{\mathbf{u}})}\mathbf{u}+\frac{1}{{\gamma }_{\mathbf{u}}{D}_{\mathbf{ux}}}\mathbf{x}\hfill \\ \hfill & \hfill =& \frac{1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{ux}}}{D_{\mathbf{ux}}(1+{\gamma }_{\mathbf{u}})}\mathbf{u}+\frac{1+{\gamma }_{\mathbf{v}}{D}_{\mathbf{vw}}}{D_{\mathbf{ux}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{u}}(1+{\gamma }_{\mathbf{v}})}\mathbf{v}+\frac{1}{D_{\mathbf{ux}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}}\mathbf{w}.\hfill \end{array}$$

Then $\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{w}$ is given by the following:

$$\begin{array}{cc}\hfill \mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{w}& =\mathbf{z}{\oplus }_E\mathbf{y}\hfill \\ \hfill & =\frac{1+{\gamma }_{\mathbf{z}}{D}_{\mathbf{zy}}}{D_{\mathbf{zy}}(1+{\gamma }_{\mathbf{z}})}\mathbf{z}+\frac{1}{{\gamma }_{\mathbf{z}}{D}_{\mathbf{zy}}}\mathbf{y}\hfill \\ \hfill & =-\frac{(1+{\gamma }_{\mathbf{z}}{D}_{\mathbf{zy}})(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}})}{D_{\mathbf{zy}}{D}_{\mathbf{uv}}(1+{\gamma }_{\mathbf{z}})(1+{\gamma }_{\mathbf{u}})}\mathbf{u}-\frac{1+{\gamma }_{\mathbf{z}}{D}_{\mathbf{zy}}}{D_{\mathbf{zy}}{D}_{\mathbf{uv}}(1+{\gamma }_{\mathbf{z}}){\gamma }_{\mathbf{u}}}\mathbf{v}\hfill \\ \hfill & +\frac{1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{ux}}}{D_{\mathbf{zy}}{D}_{\mathbf{ux}}{\gamma }_{\mathbf{z}}(1+{\gamma }_{\mathbf{u}})}\mathbf{u}+\frac{1+{\gamma }_{\mathbf{v}}{D}_{\mathbf{vw}}}{D_{\mathbf{zy}}{D}_{\mathbf{ux}}{\gamma }_{\mathbf{z}}{\gamma }_{\mathbf{u}}(1+{\gamma }_{\mathbf{v}})}\mathbf{v}\hfill \\ \hfill & +\frac{1}{D_{\mathbf{zy}}{D}_{\mathbf{ux}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{z}}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}}\mathbf{w}.\hfill \end{array}$$

We will calculate each coefficient of $\mathbf{u},\mathbf{v},\mathbf{w}$ of the equation above.

We prove that the coefficient of $\mathbf{w}$ is 1, i.e., $D_{\mathbf{zy}}{D}_{\mathbf{ux}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}{\gamma }_{\mathbf{z}}$ is 1. The equation $\frac{\mathbf{a}·\mathbf{b}}{s^2}=D_{\mathbf{a}\mathbf{b}}-1$ holds for all $\mathbf{a},\mathbf{b}\in {\mathbb{V}}_s$. Applying this equation, we have

$$\begin{array}{cc}\hfill D_{\mathbf{ux}}& =1+\frac{\mathbf{u}·\mathbf{x}}{s^2}\hfill \\ \hfill & =1+\frac{1}{s^2}\left\{\frac{1+{\gamma }_{\mathbf{v}}{D}_{\mathbf{vw}}}{D_{\mathbf{vw}}(1+{\gamma }_{\mathbf{v}})}(\mathbf{u}·\mathbf{v})+\frac{1}{{\gamma }_{\mathbf{v}}{D}_{\mathbf{vw}}}(\mathbf{u}·\mathbf{w})\right\}\hfill \\ \hfill & =\frac{D_{\mathbf{vw}}{\gamma }_{\mathbf{v}}(1+{\gamma }_{\mathbf{v}})+{\gamma }_{\mathbf{v}}(1+{\gamma }_{\mathbf{v}}{D}_{\mathbf{vw}})(D_{\mathbf{uv}}-1)+(1+{\gamma }_{\mathbf{v}})(D_{\mathbf{uw}}-1)}{D_{\mathbf{vw}}{\gamma }_{\mathbf{v}}(1+{\gamma }_{\mathbf{v}})}\hfill \\ \hfill & =\frac{{{\gamma }_{\mathbf{v}}}^2{D}_{\mathbf{uv}}{D}_{\mathbf{vw}}+{\gamma }_{\mathbf{v}}{D}_{\mathbf{uv}}+{\gamma }_{\mathbf{v}}{D}_{\mathbf{vw}}+{\gamma }_{\mathbf{v}}{D}_{\mathbf{uw}}+D_{\mathbf{uw}}-2{\gamma }_{\mathbf{v}}-1}{D_{\mathbf{vw}}{\gamma }_{\mathbf{v}}(1+{\gamma }_{\mathbf{v}})},\hfill \\ \hfill D_{\mathbf{zy}}& =1+\frac{\mathbf{z}·\mathbf{y}}{s^2}\hfill \\ \hfill & =1-\frac{1}{s^2}\{\frac{(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}})(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{ux}})}{D_{\mathbf{uv}}{D}_{\mathbf{ux}}{(1+{\gamma }_{\mathbf{u}})}^2}{\parallel \mathbf{u}\parallel }^2+\frac{(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}})(1+{\gamma }_{\mathbf{v}}{D}_{\mathbf{vw}})}{D_{\mathbf{uv}}{D}_{\mathbf{ux}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{u}}(1+{\gamma }_{\mathbf{u}})(1+{\gamma }_{\mathbf{v}})}(\mathbf{u}·\mathbf{v})\hfill \\ \hfill & +\frac{1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}}}{D_{\mathbf{uv}}{D}_{\mathbf{ux}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(1+{\gamma }_{\mathbf{u}})}(\mathbf{u}·\mathbf{w})+\frac{1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{ux}}}{D_{\mathbf{uv}}{D}_{\mathbf{ux}}{\gamma }_{\mathbf{u}}(1+{\gamma }_{\mathbf{u}})}(\mathbf{u}·\mathbf{v})\hfill \\ \hfill & +\frac{1+{\gamma }_{\mathbf{v}}{D}_{\mathbf{vw}}}{D_{\mathbf{uv}}{D}_{\mathbf{ux}}{D}_{\mathbf{vw}}{{\gamma }_{\mathbf{u}}}^2(1+{\gamma }_{\mathbf{v}})}{\parallel \mathbf{v}\parallel }^2+\frac{\mathbf{v}·\mathbf{w}}{D_{\mathbf{uv}}{D}_{\mathbf{ux}}{D}_{\mathbf{vw}}{{\gamma }_{\mathbf{u}}}^2{\gamma }_{\mathbf{v}}}\}.\hfill \end{array}$$

Multiplying $D_{\mathbf{ux}}{D}_{\mathbf{vw}}$ from the right-hand side of the last equation, and applying the gamma factor, we have

$$\begin{array}{cc}\hfill & D_{\mathbf{zy}}\left(D_{\mathbf{ux}}{D}_{\mathbf{vw}}\right)\hfill \\ \hfill & =D_{\mathbf{ux}}{D}_{\mathbf{vw}}-\frac{(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}})(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{ux}})({\gamma }_{\mathbf{u}}-1)D_{\mathbf{vw}}}{D_{\mathbf{uv}}(1+{\gamma }_{\mathbf{u}}){{\gamma }_{\mathbf{u}}}^2}\hfill \\ \hfill & -\frac{(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}})(1+{\gamma }_{\mathbf{v}}{D}_{\mathbf{vw}})(D_{\mathbf{uv}}-1)}{D_{\mathbf{uv}}{\gamma }_{\mathbf{u}}(1+{\gamma }_{\mathbf{u}})(1+{\gamma }_{\mathbf{v}})}-\frac{(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}})(D_{\mathbf{uw}}-1)}{D_{\mathbf{uv}}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(1+{\gamma }_{\mathbf{u}})}\hfill \\ \hfill & -\frac{(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{ux}})(D_{\mathbf{uv}}-1)D_{\mathbf{vw}}}{D_{\mathbf{uv}}{\gamma }_{\mathbf{u}}(1+{\gamma }_{\mathbf{u}})}-\frac{(1+{\gamma }_{\mathbf{v}}{D}_{\mathbf{vw}})({\gamma }_{\mathbf{v}}-1)}{D_{\mathbf{uv}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}-\frac{D_{\mathbf{vw}}-1}{D_{\mathbf{uv}}{{\gamma }_{\mathbf{u}}}^2{\gamma }_{\mathbf{v}}}.\hfill \end{array}$$

Dividing the common denominator $D_{\mathbf{uv}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2(1+{\gamma }_{\mathbf{u}})(1+{\gamma }_{\mathbf{v}})$ and multiplying ${\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}{\gamma }_{\mathbf{z}}$ to $D_{\mathbf{zy}}{D}_{\mathbf{ux}}{D}_{\mathbf{vw}}$, where ${\gamma }_{\mathbf{z}}={\gamma }_{{\ominus }_{\mathbf{E}}\left(\mathbf{u}{\oplus }_{\mathbf{E}}\mathbf{v}\right)}={\gamma }_{\mathbf{u}{\oplus }_{\mathbf{E}}\mathbf{v}}=D_{\mathbf{uv}}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}$, we have

$$\begin{array}{cc}\hfill & D_{\mathbf{zy}}{D}_{\mathbf{ux}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}{\gamma }_{\mathbf{z}}\hfill \\ \hfill & =\frac{1}{(1+{\gamma }_{\mathbf{u}})(1+{\gamma }_{\mathbf{v}})}\{D_{\mathbf{uv}}{D}_{\mathbf{ux}}{D}_{\mathbf{vw}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2(1+{\gamma }_{\mathbf{u}})(1+{\gamma }_{\mathbf{v}})\hfill \\ \hfill & -(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}})(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{ux}})({\gamma }_{\mathbf{u}}-1)D_{\mathbf{vw}}{{\gamma }_{\mathbf{v}}}^2(1+{\gamma }_{\mathbf{v}})\hfill \\ \hfill & -(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}})(1+{\gamma }_{\mathbf{v}}{D}_{\mathbf{vw}})(D_{\mathbf{uv}}-1){\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^2-(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}})(D_{\mathbf{uw}}-1)(1+{\gamma }_{\mathbf{v}}){\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}\hfill \\ \hfill & -(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{ux}})(D_{\mathbf{uv}}-1)D_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^2-(1+{\gamma }_{\mathbf{v}}{D}_{\mathbf{vw}})({\gamma }_{\mathbf{v}}-1)(1+{\gamma }_{\mathbf{u}})(1+{\gamma }_{\mathbf{v}})\hfill \\ \hfill & -(D_{\mathbf{vw}}-1){\gamma }_{\mathbf{v}}(1+{\gamma }_{\mathbf{u}})(1+{\gamma }_{\mathbf{v}})\}.\hfill \end{array}$$

(16)

We compute $\{·\}$ of the Equation (16).

$$\begin{array}{cc}\hfill \{·\}& =D_{\mathbf{uv}}{D}_{\mathbf{ux}}{D}_{\mathbf{vw}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2+D_{\mathbf{uv}}{D}_{\mathbf{ux}}{D}_{\mathbf{vw}}{{\gamma }_{\mathbf{u}}}^3{{\gamma }_{\mathbf{v}}}^2+D_{\mathbf{uv}}{D}_{\mathbf{ux}}{D}_{\mathbf{vw}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^3\hfill \\ \hfill & +D_{\mathbf{uv}}{D}_{\mathbf{ux}}{D}_{\mathbf{vw}}{{\gamma }_{\mathbf{u}}}^3{{\gamma }_{\mathbf{v}}}^3-D_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^2+D_{\mathbf{vw}}{{\gamma }_{\mathbf{v}}}^3-D_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^3+D_{\mathbf{vw}}{{\gamma }_{\mathbf{v}}}^2\hfill \\ \hfill & -D_{\mathbf{vw}}{D}_{\mathbf{ux}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2+D_{\mathbf{vw}}{D}_{\mathbf{ux}}{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^3-D_{\mathbf{vw}}{D}_{\mathbf{ux}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^3+D_{\mathbf{vw}}{D}_{\mathbf{ux}}{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^2\hfill \\ \hfill & -D_{\mathbf{uv}}{D}_{\mathbf{vw}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2+D_{\mathbf{uv}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^3-D_{\mathbf{uv}}{D}_{\mathbf{vw}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^3+D_{\mathbf{uv}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^2\hfill \\ \hfill & -D_{\mathbf{uv}}{D}_{\mathbf{vw}}{D}_{\mathbf{ux}}{{\gamma }_{\mathbf{u}}}^3{{\gamma }_{\mathbf{v}}}^2+\underline{D_{\mathbf{uv}}{D}_{\mathbf{vw}}{D}_{\mathbf{ux}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^3}-D_{\mathbf{uv}}{D}_{\mathbf{vw}}{D}_{\mathbf{ux}}{{\gamma }_{\mathbf{u}}}^3{{\gamma }_{\mathbf{v}}}^3\hfill \\ \hfill & +\underline{D_{\mathbf{uv}}{D}_{\mathbf{vw}}{D}_{\mathbf{ux}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}+{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^2+D_{\mathbf{uv}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2+D_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^3+D_{\mathbf{uv}}{D}_{\mathbf{vw}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^3\hfill \\ \hfill & -D_{\mathbf{uv}}{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^2-D_{\mathbf{uv}}^2{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2-D_{\mathbf{uv}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^3-D_{\mathbf{uv}}^2{D}_{\mathbf{vw}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^3\hfill \\ \hfill & +{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}+{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^2+D_{\mathbf{uv}}{{\gamma }_{\mathbf{u}}}^2{\gamma }_{\mathbf{v}}+D_{\mathbf{uv}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2\hfill \\ \hfill & -D_{\mathbf{uw}}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}-D_{\mathbf{uw}}{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^2-D_{\mathbf{uw}}{D}_{\mathbf{uv}}{{\gamma }_{\mathbf{u}}}^2{\gamma }_{\mathbf{v}}-D_{\mathbf{uw}}{D}_{\mathbf{uv}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2\hfill \\ \hfill & +D_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^2+D_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^3+D_{\mathbf{vw}}{D}_{\mathbf{ux}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2+D_{\mathbf{vw}}{D}_{\mathbf{ux}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^3\hfill \\ \hfill & -D_{\mathbf{uv}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^2-D_{\mathbf{uv}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^3-D_{\mathbf{uv}}{D}_{\mathbf{vw}}{D}_{\mathbf{ux}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2\hfill \\ \hfill & -D_{\mathbf{uv}}{D}_{\mathbf{vw}}{D}_{\mathbf{ux}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^3-{{\gamma }_{\mathbf{v}}}^2+1-{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^2+{\gamma }_{\mathbf{u}}-D_{\mathbf{vw}}{{\gamma }_{\mathbf{v}}}^3+D_{\mathbf{vw}}{\gamma }_{\mathbf{v}}\hfill \\ \hfill & -D_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^3+D_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}+{\gamma }_{\mathbf{v}}+{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}+{{\gamma }_{\mathbf{v}}}^2+{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^2-D_{\mathbf{vw}}{\gamma }_{\mathbf{v}}-D_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}\hfill \\ \hfill & -D_{\mathbf{vw}}{{\gamma }_{\mathbf{v}}}^2-D_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^2.\hfill \end{array}$$

Applying

$$\begin{array}{cc}\hfill D_{\mathbf{ux}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{v}}(1+{\gamma }_{\mathbf{v}})& ={{\gamma }_{\mathbf{v}}}^2{D}_{\mathbf{uv}}{D}_{\mathbf{vw}}+{\gamma }_{\mathbf{v}}{D}_{\mathbf{uv}}+{\gamma }_{\mathbf{v}}{D}_{\mathbf{vw}}\hfill \\ \hfill & +{\gamma }_{\mathbf{v}}{D}_{\mathbf{uw}}+D_{\mathbf{uw}}-2{\gamma }_{\mathbf{v}}-1\hfill \end{array}$$

(17)

for underline items, we infer that

$$\begin{array}{cc}\hfill & \underline{D_{\mathbf{uv}}{D}_{\mathbf{vw}}{D}_{\mathbf{ux}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^3}+\underline{D_{\mathbf{uv}}{D}_{\mathbf{vw}}{D}_{\mathbf{ux}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}\hfill \\ \hfill & =D_{\mathbf{uv}}{{\gamma }_{\mathbf{u}}}^2{\gamma }_{\mathbf{v}}{D}_{\mathbf{ux}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{v}}(1+{\gamma }_{\mathbf{v}})\hfill \\ \hfill & =D_{\mathbf{uv}}^2{D}_{\mathbf{vw}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^3+D_{\mathbf{uv}}^2{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2+D_{\mathbf{uv}}{D}_{\mathbf{vw}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2+D_{\mathbf{uw}}{D}_{\mathbf{uv}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2\hfill \\ \hfill & +D_{\mathbf{uw}}{D}_{\mathbf{uv}}{{\gamma }_{\mathbf{u}}}^2{\gamma }_{\mathbf{v}}-2D_{\mathbf{uv}}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2+D_{\mathbf{uv}}{{\gamma }_{\mathbf{u}}}^2.\hfill \end{array}$$

So, we have

$$\begin{array}{cc}\hfill \{·\}& =-D_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^2-D_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^3-D_{\mathbf{ux}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{v}}(1+{\gamma }_{\mathbf{v}})({{\gamma }_{\mathbf{u}}}^2{\gamma }_{\mathbf{v}}-{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}})\hfill \\ \hfill & +2{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^2+D_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^3-D_{\mathbf{uv}}{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^2-D_{\mathbf{uv}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^3\hfill \\ \hfill & +{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}-D_{\mathbf{uw}}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}-D_{\mathbf{uw}}{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^2+D_{\mathbf{ux}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{v}}(1+{\gamma }_{\mathbf{v}}){{\gamma }_{\mathbf{u}}}^2{\gamma }_{\mathbf{v}}\hfill \\ \hfill & +1+{\gamma }_{\mathbf{u}}+{\gamma }_{\mathbf{v}}+{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}\hfill \\ \hfill & =(1+{\gamma }_{\mathbf{u}})(1+{\gamma }_{\mathbf{v}}),\hfill \end{array}$$

hence we have $D_{\mathbf{zy}}{D}_{\mathbf{ux}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}{\gamma }_{\mathbf{z}}=1$.

Next, we prove that coefficient of $\mathbf{u}$ is $\frac{A}{D}$.

We have $1+{\gamma }_{\mathbf{z}}=1+D_{\mathbf{uv}}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}=D$. Then we compute the coefficient of $\mathbf{u}$ applying the equation $D_{\mathbf{zy}}{D}_{\mathbf{ux}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}{\gamma }_{\mathbf{z}}=1$.

$$\begin{array}{c}\begin{array}{c}-\frac{(1+{\gamma }_{\mathbf{z}}{D}_{\mathbf{zy}})(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}})}{D_{\mathbf{zy}}{D}_{\mathbf{uv}}(1+{\gamma }_{\mathbf{z}})(1+{\gamma }_{\mathbf{u}})}+\frac{1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{ux}}}{D_{\mathbf{zy}}{D}_{\mathbf{ux}}{\gamma }_{\mathbf{z}}(1+{\gamma }_{\mathbf{u}})}\hfill \\ =\frac{-(1+{\gamma }_{\mathbf{z}}{D}_{\mathbf{zy}})(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}}){\gamma }_{\mathbf{z}}{D}_{\mathbf{ux}}+(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{ux}})(1+{\gamma }_{\mathbf{z}})D_{\mathbf{uv}}}{D_{\mathbf{zy}}{D}_{\mathbf{ux}}{D}_{\mathbf{uv}}D{\gamma }_{\mathbf{z}}(1+{\gamma }_{\mathbf{u}})}\hfill \\ =\frac{D_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}}{D{D}_{\mathbf{uv}}(1+{\gamma }_{\mathbf{u}})}\{-(D_{\mathbf{ux}}{\gamma }_{\mathbf{z}}+D_{\mathbf{ux}}{D}_{\mathbf{uv}}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{z}}+D_{\mathbf{zy}}{D}_{\mathbf{ux}}{{\gamma }_{\mathbf{z}}}^2+D_{\mathbf{zy}}{D}_{\mathbf{ux}}{D}_{\mathbf{uv}}{{\gamma }_{\mathbf{z}}}^2{\gamma }_{\mathbf{u}})\hfill \\ +D_{\mathbf{uv}}+D_{\mathbf{uv}}{\gamma }_{\mathbf{z}}+D_{\mathbf{ux}}{D}_{\mathbf{uv}}{\gamma }_{\mathbf{u}}+D_{\mathbf{ux}}{D}_{\mathbf{uv}}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{z}}\}\hfill \\ =\frac{D_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}}{D{D}_{\mathbf{uv}}(1+{\gamma }_{\mathbf{u}})}\left\{D_{\mathbf{uv}}+D_{\mathbf{uv}}^2{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}+D_{\mathbf{ux}}{D}_{\mathbf{uv}}{\gamma }_{\mathbf{u}}(1-{\gamma }_{\mathbf{v}})-\frac{D_{\mathbf{uv}}}{D_{\mathbf{vw}}}-\frac{D_{\mathbf{uv}}^2{\gamma }_{\mathbf{u}}}{D_{\mathbf{vw}}}\right\}\hfill \\ =\frac{{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}}{D(1+{\gamma }_{\mathbf{u}})}\{D_{\mathbf{vw}}+D_{\mathbf{uv}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}-1-{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}}\hfill \\ +{\gamma }_{\mathbf{u}}(1-{\gamma }_{\mathbf{v}})\frac{{{\gamma }_{\mathbf{v}}}^2{D}_{\mathbf{uv}}{D}_{\mathbf{vw}}+{\gamma }_{\mathbf{v}}{D}_{\mathbf{uv}}+{\gamma }_{\mathbf{v}}{D}_{\mathbf{vw}}+{\gamma }_{\mathbf{v}}{D}_{\mathbf{uw}}+D_{\mathbf{uw}}-2{\gamma }_{\mathbf{v}}-1}{{\gamma }_{\mathbf{v}}(1+{\gamma }_{\mathbf{v}})}\}\hfill \\ =\frac{{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}}{D(1+{\gamma }_{\mathbf{u}})}\{-\frac{{\gamma }_{\mathbf{u}}}{{\gamma }_{\mathbf{v}}}({\gamma }_{\mathbf{v}}-1)D_{\mathbf{uw}}+(1-\frac{{\gamma }_{\mathbf{u}}({\gamma }_{\mathbf{v}}-1)}{1+{\gamma }_{\mathbf{v}}})D_{\mathbf{vw}}\hfill \\ -(1+\frac{{\gamma }_{\mathbf{v}}-1}{1+{\gamma }_{\mathbf{v}}}){\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}}+(1-\frac{{\gamma }_{\mathbf{v}}-1}{1+{\gamma }_{\mathbf{v}}}){\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}{D}_{\mathbf{uv}}{D}_{\mathbf{vw}}-1+\frac{{\gamma }_{\mathbf{u}}({\gamma }_{\mathbf{v}}-1)(2{\gamma }_{\mathbf{v}}+1)}{{\gamma }_{\mathbf{v}}(1+{\gamma }_{\mathbf{v}})}\}\hfill \end{array}\end{array}$$

By $D_{\mathbf{uv}}=1+\frac{\mathbf{u}·\mathbf{v}}{s^2}$,

$$\begin{array}{c}\begin{array}{c}=\frac{1}{D}\{-\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{v}}+1}({\gamma }_{\mathbf{u}}-1)(\mathbf{u}·\mathbf{v})-\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{v}}+1}({\gamma }_{\mathbf{v}}-1)\hfill \\ +\frac{1}{s^2}\frac{{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(1+{\gamma }_{\mathbf{u}}+{\gamma }_{\mathbf{v}}-{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}})}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{v}·\mathbf{w})+\frac{{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(1+{\gamma }_{\mathbf{u}}+{\gamma }_{\mathbf{v}}-{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}})}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}\hfill \\ -\frac{1}{s^2}\frac{2{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})-\frac{1}{s^2}\frac{2{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}\hfill \\ +\frac{2{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(1+\frac{(\mathbf{u}·\mathbf{v})}{s^2}+\frac{(\mathbf{v}·\mathbf{w})}{s^2}+\frac{1}{s^4}(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{w}))\hfill \\ -\frac{{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}}{1+{\gamma }_{\mathbf{u}}}+\frac{{{\gamma }_{\mathbf{u}}}^2(2{{\gamma }_{\mathbf{v}}}^2-{\gamma }_{\mathbf{v}}-1)}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}\}\hfill \\ =\frac{1}{D}\left\{-\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{v}}+1}({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{v})+\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{v}·\mathbf{w})+\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{w})\right\}\hfill \\ =\frac{A}{D}.\hfill \end{array}\end{array}$$

Finally, we prove that the coefficient of $\mathbf{v}$ is $\frac{B}{D}$.

$$\begin{array}{c}\begin{array}{c}-\frac{1+{\gamma }_{\mathbf{z}}{D}_{\mathbf{zy}}}{D_{\mathbf{zy}}{D}_{\mathbf{uv}}(1+{\gamma }_{\mathbf{z}}){\gamma }_{\mathbf{u}}}+\frac{1+{\gamma }_{\mathbf{v}}{D}_{\mathbf{vw}}}{D_{\mathbf{zy}}{D}_{\mathbf{ux}}{\gamma }_{\mathbf{z}}{\gamma }_{\mathbf{u}}(1+{\gamma }_{\mathbf{v}})}\hfill \\ =\frac{-(1+{\gamma }_{\mathbf{z}}{D}_{\mathbf{zy}})D_{\mathbf{ux}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{z}}(1+{\gamma }_{\mathbf{v}})+(1+{\gamma }_{\mathbf{v}}{D}_{\mathbf{vw}})D_{\mathbf{uv}}(1+{\gamma }_{\mathbf{z}})}{D_{\mathbf{zy}}{D}_{\mathbf{ux}}{D}_{\mathbf{uv}}{D}_{\mathbf{vw}}D{\gamma }_{\mathbf{z}}{\gamma }_{\mathbf{u}}(1+{\gamma }_{\mathbf{v}})}\hfill \\ =\frac{{\gamma }_{\mathbf{v}}}{D{D}_{\mathbf{uv}}(1+{\gamma }_{\mathbf{v}})}\{-D_{\mathbf{ux}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{z}}(1+{\gamma }_{\mathbf{v}})-D_{\mathbf{zy}}{D}_{\mathbf{ux}}{D}_{\mathbf{vw}}{{\gamma }_{\mathbf{z}}}^2(1+{\gamma }_{\mathbf{v}})\hfill \\ +D_{\mathbf{uv}}(1+{\gamma }_{\mathbf{z}})+D_{\mathbf{uv}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{v}}(1+{\gamma }_{\mathbf{z}})\}.\hfill \end{array}\end{array}$$

Using (16) and $D_{\mathbf{zy}}{D}_{\mathbf{ux}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{z}}=\frac{1}{{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}}$, then we have

$$\begin{array}{c}\begin{array}{c}=\frac{{\gamma }_{\mathbf{v}}}{D{D}_{\mathbf{uv}}(1+{\gamma }_{\mathbf{v}})}\{-({\gamma }_{\mathbf{z}}{\gamma }_{\mathbf{v}}{D}_{\mathbf{uv}}{D}_{\mathbf{vw}}+{\gamma }_{\mathbf{z}}{D}_{\mathbf{uv}}+{\gamma }_{\mathbf{z}}{D}_{\mathbf{vw}}+{\gamma }_{\mathbf{z}}{D}_{\mathbf{uw}}+\frac{{\gamma }_{\mathbf{z}}}{{\gamma }_{\mathbf{v}}}{D}_{\mathbf{uw}}\hfill \\ -\frac{{\gamma }_{\mathbf{z}}}{{\gamma }_{\mathbf{v}}}(2{\gamma }_{\mathbf{v}}+1))-\frac{{\gamma }_{\mathbf{z}}(1+{\gamma }_{\mathbf{v}})}{{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}}+D_{\mathbf{uv}}(1+{\gamma }_{\mathbf{z}})+D_{\mathbf{uv}}{D}_{\mathbf{vw}}{\gamma }_{\mathbf{v}}(1+{\gamma }_{\mathbf{z}})\}\hfill \\ =\frac{1}{D{D}_{\mathbf{uv}}(1+{\gamma }_{\mathbf{v}})}\{{{\gamma }_{\mathbf{v}}}^2(1+\frac{(\mathbf{u}·\mathbf{v})}{s^2}+\frac{(\mathbf{v}·\mathbf{w})}{s^2}+\frac{1}{s^4}(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{w}))\hfill \\ -{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^2(1+\frac{(\mathbf{u}·\mathbf{v})}{s^2}+\frac{(\mathbf{v}·\mathbf{w})}{s^2}+\frac{1}{s^4}(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{w}))\hfill \\ -{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^2(1+\frac{(\mathbf{u}·\mathbf{v})}{s^2}+\frac{(\mathbf{u}·\mathbf{w})}{s^2}+\frac{1}{s^4}(\mathbf{u}·\mathbf{v})(\mathbf{u}·\mathbf{w}))\hfill \\ -{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(1+\frac{(\mathbf{u}·\mathbf{v})}{s^2}+\frac{(\mathbf{u}·\mathbf{w})}{s^2}+\frac{1}{s^4}(\mathbf{u}·\mathbf{v})(\mathbf{u}·\mathbf{w}))\hfill \\ +{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(2{\gamma }_{\mathbf{v}}+1)(1+\frac{(\mathbf{u}·\mathbf{v})}{s^2})-{{\gamma }_{\mathbf{v}}}^2(1+\frac{(\mathbf{u}·\mathbf{v})}{s^2})\}\hfill \\ =-\frac{{\gamma }_{\mathbf{v}}}{s^{2}D(1+{\gamma }_{\mathbf{v}})}\left\{{\gamma }_{\mathbf{u}}({\gamma }_{\mathbf{v}}+1)(\mathbf{u}·\mathbf{w})+({\gamma }_{\mathbf{v}}-1){\gamma }_{\mathbf{u}}(\mathbf{v}·\mathbf{w})\right\}\hfill \\ =\frac{B}{D}.\hfill \end{array}\end{array}$$

Hence $\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{w}=\mathbf{w}+\frac{\mathrm{A}\mathbf{u}+\mathrm{B}\mathbf{v}}{\mathrm{D}}$ holds. By applying the left cancellation law for the Equation (14), we obtain (G3).

We prove (G4). We prove that $\mathrm{gyr}[\mathbf{u},\mathbf{v}]$ is automorphism for every pair $\mathbf{u},\mathbf{v}\in \mathbb{V}$. To prove (G4), we first show the gyration preserves, the inner product of $\mathbb{V}$ and the norm. So, we compute

$$\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{a}·\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{b}=\mathbf{a}·\mathbf{b}$$

for all $\mathbf{a},\mathbf{b},\mathbf{u},\mathbf{v}\in {\mathbb{V}}_s$. By applying the Equation (15), we have

$$\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{a}=\mathbf{a}+\frac{A_{\mathbf{a}}\mathbf{u}+B_{\mathbf{a}}\mathbf{v}}{D}$$

and

$$\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{b}=\mathbf{b}+\frac{A_{\mathbf{b}}\mathbf{u}+B_{\mathbf{b}}\mathbf{v}}{D}$$

respectively, where

$$\begin{array}{cc}\hfill A_{\mathbf{a}}& =-\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{a})+\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{v}·\mathbf{a})+\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{a}),\hfill \\ \hfill B_{\mathbf{a}}& =-\frac{1}{s^2}\frac{{\gamma }_{\mathbf{v}}}{{\gamma }_{\mathbf{v}}+1}\left\{{\gamma }_{\mathbf{u}}({\gamma }_{\mathbf{v}}+1)(\mathbf{u}·\mathbf{a})+({\gamma }_{\mathbf{u}}-1){\gamma }_{\mathbf{v}}(\mathbf{v}·\mathbf{a})\right\}.\hfill \end{array}$$

The terms $A_{\mathbf{b}}$ and $B_{\mathbf{b}}$ are defined in the similar way an $A_{\mathbf{a}}$ and $B_{\mathbf{b}}$ respectively. Then we have

$$\begin{array}{cc}\hfill & \mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{a}·\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{b}=\mathbf{a}·\mathbf{b}+\frac{A_{\mathbf{a}}(\mathbf{u}·\mathbf{b})+B_{\mathbf{a}}(\mathbf{v}·\mathbf{b})}{D}+\frac{A_{\mathbf{b}}(\mathbf{u}·\mathbf{a})+B_{\mathbf{b}}(\mathbf{v}·\mathbf{a})}{D}\hfill \\ \hfill & +\frac{1}{D^2}\{A_{\mathbf{a}}{A}_{\mathbf{b}}{\parallel \mathbf{u}\parallel }^2+A_{\mathbf{a}}{B}_{\mathbf{b}}(\mathbf{u}·\mathbf{v})+A_{\mathbf{b}}{B}_{\mathbf{a}}(\mathbf{u}·\mathbf{v})+B_{\mathbf{a}}{B}_{\mathbf{b}}{\parallel \mathbf{v}\parallel }^2\},\hfill \end{array}$$

(18)

We show that terms other than $\mathbf{a}·\mathbf{b}$ of the right-hand side of the Equation (18) equal to zero.

$$\begin{array}{cc}\hfill & A_{\mathbf{a}}(\mathbf{u}·\mathbf{b})+B_{\mathbf{a}}(\mathbf{v}·\mathbf{b})\hfill \\ \hfill =& -\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{a})(\mathbf{u}·\mathbf{b})+\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{v}·\mathbf{a})(\mathbf{u}·\mathbf{b})\hfill \\ \hfill & +\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{a})(\mathbf{u}·\mathbf{b})\hfill \\ \hfill & -\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{u}·\mathbf{a})(\mathbf{v}·\mathbf{b})-\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{v}}}^2}{{\gamma }_{\mathbf{v}}+1}({\gamma }_{\mathbf{u}}-1)(\mathbf{v}·\mathbf{a})(\mathbf{v}·\mathbf{b}).\hfill \\ \hfill & A_b(\mathbf{u}·\mathbf{a})+B_b(\mathbf{v}·\mathbf{a})\hfill \\ \hfill =& -\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{a})(\mathbf{u}·\mathbf{b})+\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{u}·\mathbf{a})(\mathbf{v}·\mathbf{b})\hfill \\ \hfill & +\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})(\mathbf{u}·\mathbf{a})(\mathbf{v}·\mathbf{b})\hfill \\ \hfill & -\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{v}·\mathbf{a})(\mathbf{u}·\mathbf{b})-\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{v}}}^2}{{\gamma }_{\mathbf{v}}+1}({\gamma }_{\mathbf{u}}-1)(\mathbf{v}·\mathbf{a})(\mathbf{v}·\mathbf{b}).\hfill \\ \hfill & A_{\mathbf{a}}(\mathbf{u}·\mathbf{b})+B_{\mathbf{a}}(\mathbf{v}·\mathbf{b})+A_{\mathbf{b}}(\mathbf{u}·\mathbf{a})+B_{\mathbf{b}}(\mathbf{v}·\mathbf{a})\hfill \\ \hfill =& -\frac{2}{s^2}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{a})(\mathbf{u}·\mathbf{b})+\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})(\mathbf{u}·\mathbf{a})(\mathbf{v}·\mathbf{b})\hfill \\ \hfill & -\frac{2}{s^2}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1)(\mathbf{v}·\mathbf{a})(\mathbf{u}·\mathbf{b})+\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{a})(\mathbf{v}·\mathbf{b}).\hfill \end{array}$$

Then we compute each terms of the sum

$A_{\mathbf{a}}{A}_{\mathbf{b}}{\parallel \mathbf{u}\parallel }^2+A_{\mathbf{a}}{B}_{\mathbf{b}}(\mathbf{u}·\mathbf{v})+A_{\mathbf{b}}{B}_{\mathbf{a}}(\mathbf{u}·\mathbf{v})+B_{\mathbf{a}}{B}_{\mathbf{b}}{\parallel \mathbf{v}\parallel }^2$.

$$\begin{array}{cc}\hfill & A_{\mathbf{a}}{A}_{\mathbf{b}}{\parallel \mathbf{u}\parallel }^2\hfill \\ \hfill & ={\parallel \mathbf{u}\parallel }^2\{\frac{1}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^4}{{({\gamma }_{\mathbf{u}}+1)}^2}{({\gamma }_{\mathbf{v}}-1)}^2(\mathbf{u}·\mathbf{a})(\mathbf{u}·\mathbf{b})-\frac{1}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1){\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{u}·\mathbf{a})(\mathbf{v}·\mathbf{b})\hfill \\ \hfill & -\frac{2}{s^6}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1)\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})(\mathbf{u}·\mathbf{a})(\mathbf{u}·\mathbf{b})\hfill \\ \hfill & -\frac{1}{s^4}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1)(\mathbf{v}·\mathbf{a})(\mathbf{u}·\mathbf{b})+\frac{1}{s^4}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2(\mathbf{v}·\mathbf{a})(\mathbf{v}·\mathbf{b})\hfill \\ \hfill & +\frac{2}{s^6}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{a})(\mathbf{v}·\mathbf{b})\hfill \\ \hfill & -\frac{2}{s^6}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{a})(\mathbf{u}·\mathbf{b})\hfill \\ \hfill & +\frac{2}{s^6}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{a})(\mathbf{v}·\mathbf{b})\hfill \\ \hfill & +\frac{4}{s^8}\frac{{{\gamma }_{\mathbf{u}}}^4{{\gamma }_{\mathbf{v}}}^4}{{({\gamma }_{\mathbf{u}}+1)}^2{({\gamma }_{\mathbf{v}}+1)}^2}{(\mathbf{u}·\mathbf{v})}^2(\mathbf{v}·\mathbf{a})(\mathbf{v}·\mathbf{b})\}.\hfill \end{array}$$

(19)

By $\frac{{\parallel \mathbf{u}\parallel }^2}{s^2}=\frac{{{\gamma }_{\mathbf{u}}}^2-1}{{{\gamma }_{\mathbf{u}}}^2}$ this equation is rewritten in the following.

$$\begin{array}{cc}\hfill & A_{\mathbf{a}}{A}_{\mathbf{b}}{\parallel \mathbf{u}\parallel }^2\hfill \\ \hfill =& \frac{1}{s^2}\frac{{{\gamma }_{\mathbf{u}}}^2}{({\gamma }_{\mathbf{u}}+1)}{({\gamma }_{\mathbf{v}}-1)}^2({\gamma }_{\mathbf{u}}-1)(\mathbf{u}·\mathbf{a})(\mathbf{u}·\mathbf{b})-\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}({\gamma }_{\mathbf{u}}-1)({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{a})(\mathbf{v}·\mathbf{b})\hfill \\ \hfill & -\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}({\gamma }_{\mathbf{u}}-1)({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{v})(\mathbf{u}·\mathbf{a})(\mathbf{u}·\mathbf{b})\hfill \\ \hfill & -\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}({\gamma }_{\mathbf{u}}-1)({\gamma }_{\mathbf{v}}-1)(\mathbf{v}·\mathbf{a})(\mathbf{u}·\mathbf{b})+\frac{1}{s^2}{{\gamma }_{\mathbf{v}}}^2({{\gamma }_{\mathbf{u}}}^2-1)(\mathbf{v}·\mathbf{a})(\mathbf{v}·\mathbf{b})\hfill \\ \hfill & +\frac{4}{s^4}\frac{{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^3}{({\gamma }_{\mathbf{v}}+1)}({\gamma }_{\mathbf{u}}-1)(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{a})(\mathbf{v}·\mathbf{b})\hfill \\ \hfill & -\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}({\gamma }_{\mathbf{u}}-1)({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{a})(\mathbf{u}·\mathbf{b}))\hfill \\ \hfill & +\frac{4}{s^6}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^4}{({\gamma }_{\mathbf{u}}+1){({\gamma }_{\mathbf{v}}+1)}^2}({\gamma }_{\mathbf{u}}-1){(\mathbf{u}·\mathbf{v})}^2(\mathbf{v}·\mathbf{a})(\mathbf{v}·\mathbf{b}).\hfill \end{array}$$

(20)

Then we obtain

$$\begin{array}{cc}\hfill & A_{\mathbf{a}}{B}_{\mathbf{b}}(\mathbf{u}·\mathbf{v})\hfill \\ \hfill =& \frac{1}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^3{\gamma }_{\mathbf{v}}}{{({\gamma }_{\mathbf{u}}+1)}^2}({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{v})(\mathbf{u}·\mathbf{a})(\mathbf{u}·\mathbf{b})\hfill \\ \hfill & +\frac{1}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}-1)}({\gamma }_{\mathbf{u}}-1)({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{v})(\mathbf{u}·\mathbf{a})(\mathbf{v}·\mathbf{b})\hfill \\ \hfill & -\frac{1}{s^4}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{a})(\mathbf{u}·\mathbf{b})-\frac{1}{s^4}\frac{{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^3}{({\gamma }_{\mathbf{v}}+1)}({\gamma }_{\mathbf{u}}-1)(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{a})(\mathbf{v}·\mathbf{b})\hfill \\ \hfill & -\frac{2}{s^6}\frac{{{\gamma }_{\mathbf{u}}}^3{{\gamma }_{\mathbf{v}}}^3}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}{(\mathbf{u}·\mathbf{v})}^2(\mathbf{v}·\mathbf{a})(\mathbf{u}·\mathbf{b})\hfill \\ \hfill & -\frac{2}{s^6}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^4}{({\gamma }_{\mathbf{u}}+1){({\gamma }_{\mathbf{v}}+1)}^2}({\gamma }_{\mathbf{u}}-1){(\mathbf{u}·\mathbf{v})}^2(\mathbf{v}·\mathbf{a})(\mathbf{v}·\mathbf{b}).\hfill \\ \hfill & A_b{B}_a(\mathbf{u}·\mathbf{v})\hfill \\ \hfill =& \frac{1}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^3{\gamma }_{\mathbf{v}}}{{({\gamma }_{\mathbf{u}}+1)}^2}({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{v})(\mathbf{u}·\mathbf{a})(\mathbf{u}·\mathbf{b})\hfill \\ \hfill & +\frac{1}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}-1)}({\gamma }_{\mathbf{u}}-1)({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{v})(\mathbf{u}·\mathbf{b})(\mathbf{v}·\mathbf{a})\hfill \\ \hfill & -\frac{1}{s^4}{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{b})(\mathbf{u}·\mathbf{a})-\frac{1}{s^4}\frac{{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^3}{({\gamma }_{\mathbf{v}}+1)}({\gamma }_{\mathbf{u}}-1)(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{a})(\mathbf{v}·\mathbf{b})\hfill \\ \hfill & -\frac{2}{s^6}\frac{{{\gamma }_{\mathbf{u}}}^3{{\gamma }_{\mathbf{v}}}^3}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}{(\mathbf{u}·\mathbf{v})}^2(\mathbf{v}·\mathbf{b})(\mathbf{u}·\mathbf{a})\hfill \\ \hfill & -\frac{2}{s^6}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^4}{({\gamma }_{\mathbf{u}}+1){({\gamma }_{\mathbf{v}}+1)}^2}({\gamma }_{\mathbf{u}}-1){(\mathbf{u}·\mathbf{v})}^2(\mathbf{v}·\mathbf{a})(\mathbf{v}·\mathbf{b}).\hfill \end{array}$$

(21)

Calculating $B_{\mathbf{a}}{B}_{\mathbf{b}}{\parallel \mathbf{v}\parallel }^2$ in a way similar to the calculation of $A_{\mathbf{a}}{A}_{\mathbf{b}}{\parallel \mathbf{u}\parallel }^2$, we have

$$\begin{array}{cc}\hfill & B_{\mathbf{a}}{B}_{\mathbf{b}}{\parallel \mathbf{v}\parallel }^2\hfill \\ \hfill =& \frac{1}{s^2}{{\gamma }_{\mathbf{u}}}^2({{\gamma }_{\mathbf{v}}}^2-1)(\mathbf{u}·\mathbf{a})(\mathbf{u}·\mathbf{b})+\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}({\gamma }_{\mathbf{u}}-1)({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{a})(\mathbf{v}·\mathbf{b})\hfill \\ \hfill & +\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}({\gamma }_{\mathbf{u}}-1)({\gamma }_{\mathbf{v}}-1)(\mathbf{v}·\mathbf{a})(\mathbf{u}·\mathbf{b})\hfill \\ \hfill & +\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{v}}}^2}{{\gamma }_{\mathbf{v}}+1}{({\gamma }_{\mathbf{u}}-1)}^2({\gamma }_{\mathbf{v}}-1)(\mathbf{v}·\mathbf{a})(\mathbf{v}·\mathbf{b}).\hfill \end{array}$$

(22)

Hence, comparing the Equations (19) with (22) we have

$$\begin{array}{cc}\hfill & A_{\mathbf{a}}{A}_{\mathbf{b}}{\parallel \mathbf{u}\parallel }^2+A_{\mathbf{a}}{B}_{\mathbf{b}}(\mathbf{u}·\mathbf{v})+A_{\mathbf{b}}{B}_{\mathbf{a}}(\mathbf{u}·\mathbf{v})+B_{\mathbf{a}}{B}_{\mathbf{b}}{\parallel \mathbf{v}\parallel }^2\hfill \\ \hfill =& \frac{1}{s^2}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{a})(\mathbf{u}·\mathbf{b})\{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)\hfill \\ \hfill & +({\gamma }_{\mathbf{u}}-1)({\gamma }_{\mathbf{v}}-1)+2{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}\frac{\mathbf{u}·\mathbf{v}}{s^2}\}\hfill \\ \hfill & -\frac{1}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})(\mathbf{u}·\mathbf{a})(\mathbf{v}·\mathbf{b})\{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)\hfill \\ \hfill & +({\gamma }_{\mathbf{u}}-1)({\gamma }_{\mathbf{v}}-1)+2{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}\frac{\mathbf{u}·\mathbf{v}}{s^2}\}\hfill \\ \hfill & +\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1)(\mathbf{v}·\mathbf{a})(\mathbf{u}·\mathbf{b})\{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)\hfill \\ \hfill & +({\gamma }_{\mathbf{u}}-1)({\gamma }_{\mathbf{v}}-1)+2{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}\frac{\mathbf{u}·\mathbf{v}}{s^2}\}\hfill \\ \hfill & -\frac{1}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{a})(\mathbf{v}·\mathbf{b}\{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)\hfill \\ \hfill & +({\gamma }_{\mathbf{u}}-1)({\gamma }_{\mathbf{v}}-1)+2{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}\frac{\mathbf{u}·\mathbf{v}}{s^2}\}\hfill \\ \hfill =& -D(A_{\mathbf{a}}(\mathbf{u}·\mathbf{b})+B_{\mathbf{a}}(\mathbf{v}·\mathbf{b})+A_{\mathbf{b}}(\mathbf{u}·\mathbf{a})+B_{\mathbf{b}}(\mathbf{v}·\mathbf{a})).\hfill \end{array}$$

We conclude that $\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{a}·\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{b}=\mathbf{a}·\mathbf{b}$.

To prove that $\mathrm{gyr}[\mathbf{u},\mathbf{v}]$ is a homomorphism for all $\mathbf{u},\mathbf{v}\in {\mathbb{V}}_s$, we show

$$\mathrm{gyr}[\mathbf{u},\mathbf{v}]\left(\mathbf{x}{\oplus }_E\mathbf{y}\right)=\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{x}{\oplus }_E\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{y}$$

(23)

for all $\mathbf{x},\mathbf{y}\in {\mathbb{V}}_s$. Applying (15) we have

$$\mathrm{gyr}[\mathbf{u},\mathbf{v}]\left(\mathbf{x}{\oplus }_E\mathbf{y}\right)=\mathbf{x}{\oplus }_E\mathbf{y}+\frac{1}{D}(A_{\mathbf{x}{\oplus }_E\mathbf{y}}\mathbf{u}+B_{\mathbf{x}{\oplus }_E\mathbf{y}}\mathbf{v}).$$

Put

$$\mathbf{x}{\oplus }_E\mathbf{y}=\frac{1+{\gamma }_{\mathbf{x}}{D}_{\mathbf{xy}}}{D_{\mathbf{xy}}(1+{\gamma }_{\mathbf{x}})}\mathbf{x}+\frac{1}{{\gamma }_{\mathbf{x}}{D}_{\mathbf{xy}}}\mathbf{y}=E_{\mathbf{xy}}\mathbf{x}+F_{\mathbf{xy}}\mathbf{y}.$$

By a simple calculation we infer that

$$\begin{array}{cc}\hfill A_{\mathbf{x}{\oplus }_E\mathbf{y}}& =E_{\mathbf{xy}}{A}_{\mathbf{x}}+F_{\mathbf{xy}}{A}_{\mathbf{y}}\hfill \\ \hfill B_{\mathbf{x}{\oplus }_E\mathbf{y}}& =E_{\mathbf{xy}}{B}_{\mathbf{x}}+F_{\mathbf{xy}}{B}_{\mathbf{y}}.\hfill \end{array}$$

We have

$$\begin{array}{cc}\hfill & \mathrm{gyr}[\mathbf{u},\mathbf{v}]\left(\mathbf{x}{\oplus }_E\mathbf{y}\right)\hfill \\ \hfill & =E_{\mathbf{xy}}\mathbf{x}+F_{\mathbf{xy}}\mathbf{y}+\frac{1}{D}\left\{(E_{\mathbf{xy}}{A}_{\mathbf{x}}+F_{\mathbf{xy}}{A}_{\mathbf{y}})\mathbf{u}+(E_{\mathbf{xy}}{B}_{\mathbf{x}}+F_{\mathbf{xy}}{B}_{\mathbf{y}})\mathbf{v})\right\}\hfill \\ \hfill & =E_{\mathbf{xy}}\left\{\mathbf{x}+\frac{1}{D}(A_{\mathbf{x}}\mathbf{u}+B_{\mathbf{x}}\mathbf{v})\right\}+F_{\mathbf{xy}}\left\{\mathbf{y}+\frac{1}{D}(A_{\mathbf{y}}\mathbf{u}+B_{\mathbf{y}}\mathbf{v})\right\}\hfill \\ \hfill & =E_{\mathbf{xy}}\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{x}+F_{\mathbf{xy}}\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{y}.\hfill \end{array}$$

(24)

Then the right-hand side of the Equation (24) is rewritten as the following equation.

$$\begin{array}{cc}\hfill & \mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{x}{\oplus }_E\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{y}\hfill \\ \hfill & =E_{\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{x}\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{y}}\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{x}+F_{\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{x}\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{y}}\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{y}.\hfill \end{array}$$

Since $\mathrm{gyr}[\mathbf{u},\mathbf{v}]$ preserves the inner product and the norm of $\mathbb{V}$, we have

$$\begin{array}{cc}\hfill & {\gamma }_{\mathbf{gyr}[\mathbf{u},\mathbf{v}]\mathbf{x}}={\gamma }_{\mathbf{x}}\hfill \\ \hfill & D_{\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{x}\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{y}}=D_{\mathbf{xy}},\hfill \end{array}$$

so that

$$E_{\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{x}\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{y}}=E_{\mathbf{xy}}$$

and

$$F_{\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{x}\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{y}}=F_{\mathbf{xy}}.$$

Hence $\mathrm{gyr}[\mathbf{u},\mathbf{v}]\left(\mathbf{x}{\oplus }_E\mathbf{y}\right)=\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{x}{\oplus }_E\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{y}$. We conclude that $\mathrm{gyr}[\mathbf{u},\mathbf{v}]$ is a homomorphism.

We observe that $\mathrm{gyr}[\mathbf{u},\mathbf{v}]$ is bijective for every pair of $\mathbf{u},\mathbf{v}\in {\mathbb{V}}_s$. To prove this, we compute $\mathrm{gyr}[\mathbf{u},\mathbf{v}]\left(\mathrm{gyr}[\mathbf{v},\mathbf{u}]\mathbf{w}\right)=\mathbf{w}$ for every $\mathbf{w}\in {\mathbb{V}}_s$. We denote

$$\begin{array}{cc}\hfill A_{\mathbf{ab}}\left(\mathbf{c}\right)& =-\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{a}}}^2}{{\gamma }_{\mathbf{a}}+1}({\gamma }_{\mathbf{b}}-1)(\mathbf{a}·\mathbf{c})+\frac{1}{s^2}{\gamma }_{\mathbf{a}}{\gamma }_{\mathbf{b}}(\mathbf{b}·\mathbf{c})+\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{a}}}^2{{\gamma }_{\mathbf{b}}}^2}{({\gamma }_{\mathbf{a}}+1)({\gamma }_{\mathbf{b}}+1)}(\mathbf{a}·\mathbf{b})(\mathbf{b}·\mathbf{c})\hfill \\ \hfill B_{\mathbf{ab}}\left(\mathbf{c}\right)& =-\frac{1}{s^2}\frac{{\gamma }_{\mathbf{b}}}{{\gamma }_{\mathbf{b}}+1}\left\{{\gamma }_{\mathbf{a}}({\gamma }_{\mathbf{b}}+1)(\mathbf{a}·\mathbf{c})+({\gamma }_{\mathbf{a}}-1){\gamma }_{\mathbf{b}}(\mathbf{b}·\mathbf{c})\right\},\hfill \end{array}$$

where $\mathbf{a},\mathbf{b},\mathbf{c}\in {\mathbb{V}}_s$. Then applying (15), we have

$$\begin{array}{cc}\hfill & \mathrm{gyr}[\mathbf{u},\mathbf{v}]\left(\mathrm{gyr}[\mathbf{v},\mathbf{u}]\mathbf{w}\right)\hfill \\ \hfill =& \mathrm{gyr}[\mathbf{u},\mathbf{v}](\mathbf{w}+\frac{A_{\mathbf{vu}}\left(\mathbf{w}\right)\mathbf{v}+B_{\mathbf{vu}}\left(\mathbf{w}\right)\mathbf{u}}{D})\hfill \\ \hfill =& \mathbf{w}+\frac{A_{\mathbf{vu}}\left(\mathbf{w}\right)\mathbf{v}+B_{\mathbf{vu}}\left(\mathbf{w}\right)\mathbf{u}}{D}\hfill \\ \hfill & +\frac{1}{D}\left\{A_{\mathbf{uv}}(\mathbf{w}+\frac{A_{\mathbf{vu}}\left(\mathbf{w}\right)\mathbf{v}+B_{\mathbf{vu}}\left(\mathbf{w}\right)\mathbf{u}}{D})\mathbf{u}+B_{\mathbf{uv}}(\mathbf{w}+\frac{A_{\mathbf{vu}}\left(\mathbf{w}\right)\mathbf{v}+B_{\mathbf{vu}}\left(\mathbf{w}\right)\mathbf{u}}{D})\mathbf{v}\right\}.\hfill \end{array}$$

We compute

$$\begin{array}{cc}\hfill & A_{\mathbf{uv}}(\mathbf{w}+\frac{A_{\mathbf{vu}}\left(\mathbf{w}\right)\mathbf{v}+B_{\mathbf{vu}}\left(\mathbf{w}\right)\mathbf{u}}{D})\hfill \\ \hfill =& -\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·(\mathbf{w}+\frac{A_{\mathbf{vu}}\left(\mathbf{w}\right)\mathbf{v}+B_{\mathbf{vu}}\left(\mathbf{w}\right)\mathbf{u}}{D}))\hfill \\ \hfill & +\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{v}·(\mathbf{w}+\frac{A_{\mathbf{vu}}\left(\mathbf{w}\right)\mathbf{v}+B_{\mathbf{vu}}\left(\mathbf{w}\right)\mathbf{u}}{D}))\hfill \\ \hfill & +\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})(\mathbf{v}·(\mathbf{w}+\frac{A_{\mathbf{vu}}\left(\mathbf{w}\right)\mathbf{v}+B_{\mathbf{vu}}\left(\mathbf{w}\right)\mathbf{u}}{D}))\hfill \\ \hfill =& A_{\mathbf{uv}}\left(\mathbf{w}\right)+\frac{1}{D}{A}_{\mathbf{vu}}\left(\mathbf{w}\right)A_{\mathbf{uv}}\left(\mathbf{v}\right)+\frac{1}{D}{B}_{\mathbf{vu}}\left(\mathbf{w}\right)A_{\mathbf{uv}}\left(\mathbf{u}\right)\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & B_{\mathbf{uv}}(\mathbf{w}+\frac{A_{\mathbf{vu}}\left(\mathbf{w}\right)\mathbf{v}+B_{\mathbf{vu}}\left(\mathbf{w}\right)\mathbf{u}}{D})\hfill \\ \hfill =& -\frac{1}{s^2}\frac{{\gamma }_{\mathbf{v}}}{{\gamma }_{\mathbf{v}}+1}\{{\gamma }_{\mathbf{u}}({\gamma }_{\mathbf{v}}+1)(\mathbf{u}·(\mathbf{w}+\frac{A_{\mathbf{vu}}\left(\mathbf{w}\right)\mathbf{v}+B_{\mathbf{vu}}\left(\mathbf{w}\right)\mathbf{u}}{D}))\hfill \\ \hfill & +({\gamma }_{\mathbf{u}}-1){\gamma }_{\mathbf{v}}(\mathbf{v}·(\mathbf{w}+\frac{A_{\mathbf{vu}}\left(\mathbf{w}\right)\mathbf{v}+B_{\mathbf{vu}}\left(\mathbf{w}\right)\mathbf{u}}{D}))\}\hfill \\ \hfill & =B_{\mathbf{uv}}\left(\mathbf{w}\right)+\frac{1}{D}{A}_{\mathbf{vu}}\left(\mathbf{w}\right)B_{\mathbf{uv}}\left(\mathbf{v}\right)+\frac{1}{D}{B}_{\mathbf{vu}}\left(\mathbf{w}\right)B_{\mathbf{uv}}\left(\mathbf{u}\right).\hfill \end{array}$$

So, we have

$$\begin{array}{cc}\hfill & \mathrm{gyr}[\mathbf{u},\mathbf{v}]\left(\mathrm{gyr}[\mathbf{v},\mathbf{u}]\mathbf{w}\right)\hfill \\ \hfill =& \mathbf{w}+\left\{\frac{B_{\mathbf{vu}}\left(\mathbf{w}\right)}{D}+\frac{A_{\mathbf{uv}}\left(\mathbf{w}\right)}{D}+\frac{1}{D^2}(A_{\mathbf{vu}}\left(\mathbf{w}\right)A_{\mathbf{uv}}\left(\mathbf{v}\right)+B_{\mathbf{vu}}\left(\mathbf{w}\right)A_{\mathbf{uv}}\left(\mathbf{u}\right))\right\}\mathbf{u}\hfill \\ \hfill & +\left\{\frac{A_{\mathbf{vu}}\left(\mathbf{w}\right)}{D}+\frac{B_{\mathbf{uv}}\left(\mathbf{w}\right)}{D}+\frac{1}{D^2}(A_{\mathbf{vu}}\left(\mathbf{w}\right)B_{\mathbf{uv}}\left(\mathbf{v}\right)+B_{\mathbf{vu}}\left(\mathbf{w}\right)B_{\mathbf{uv}}\left(\mathbf{u}\right))\right\}\mathbf{v}.\hfill \end{array}$$

We show that the coefficients of $\mathbf{u}$ and $\mathbf{v}$ vanish. We compute the coefficient of $\mathbf{u}$.

$$\begin{array}{c}\hfill B_{\mathbf{vu}}\left(\mathbf{w}\right)+A_{\mathbf{uv}}\left(\mathbf{w}\right)=-\frac{2}{s^2}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{w})+\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{w}).\end{array}$$

We also have

$$\begin{array}{cc}\hfill & A_{\mathbf{vu}}\left(\mathbf{w}\right)A_{\mathbf{uv}}\left(\mathbf{v}\right)+B_{\mathbf{vu}}\left(\mathbf{w}\right)A_{\mathbf{uv}}\left(\mathbf{u}\right)\hfill \\ \hfill =& \frac{1}{s^2}\frac{{{\gamma }_{\mathbf{v}}}^2}{{\gamma }_{\mathbf{v}}+1}({\gamma }_{\mathbf{u}}-1)(\mathbf{v}·\mathbf{w})\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{v})-\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{v}}}^2}{{\gamma }_{\mathbf{v}}+1}({\gamma }_{\mathbf{u}}-1)(\mathbf{v}·\mathbf{w})\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}{\parallel \mathbf{v}\parallel }^2\hfill \\ \hfill & -\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{v}}}^2}{{\gamma }_{\mathbf{v}}+1}({\gamma }_{\mathbf{u}}-1)(\mathbf{v}·\mathbf{w})\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v}){\parallel \mathbf{v}\parallel }^2\hfill \\ \hfill & -\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{u}·\mathbf{w})\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{v})+\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{u}·\mathbf{w})\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}{\parallel \mathbf{v}\parallel }^2\hfill \\ \hfill & +\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{u}·\mathbf{w})\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v}){\parallel \mathbf{v}\parallel }^2\hfill \\ \hfill & -\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{w})\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{v})\hfill \\ \hfill & +\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{w})\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}{\parallel \mathbf{v}\parallel }^2\hfill \\ \hfill & +\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{w})\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v}){\parallel \mathbf{v}\parallel }^2\hfill \\ \hfill & +\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{v}·\mathbf{w})\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1){\parallel \mathbf{u}\parallel }^2-\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{v}·\mathbf{w})\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{u}·\mathbf{v})\hfill \\ \hfill & -\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{v}·\mathbf{w})\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}{(\mathbf{u}·\mathbf{v})}^2\hfill \\ \hfill & +\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{w})\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1){\parallel \mathbf{u}\parallel }^2\hfill \\ \hfill & -\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{w})\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{u}·\mathbf{v})\hfill \\ \hfill & -\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{w})\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}{(\mathbf{u}·\mathbf{v})}^2.\hfill \end{array}$$

Applying the gamma identity, we have the following.

$$\begin{array}{cc}\hfill & A_{\mathbf{vu}}\left(\mathbf{w}\right)A_{\mathbf{uv}}\left(\mathbf{v}\right)+B_{\mathbf{vu}}\left(\mathbf{w}\right)A_{\mathbf{uv}}\left(\mathbf{u}\right)\hfill \\ \hfill =& \frac{1}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{w})\left(-{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}+{\gamma }_{\mathbf{u}}+{\gamma }_{\mathbf{v}}-1-2{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}\frac{(\mathbf{u}·\mathbf{v})}{s^2}\right)\hfill \\ \hfill & -\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{v}·\mathbf{w}){\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}\frac{(\mathbf{u}·\mathbf{v})}{s^2}+\frac{1}{s^2}\left({{\gamma }_{\mathbf{u}}}^2({{\gamma }_{\mathbf{v}}}^2-1)+\frac{{{\gamma }_{\mathbf{u}}}^2({\gamma }_{\mathbf{u}}-1){({\gamma }_{\mathbf{v}}-1)}^2}{{\gamma }_{\mathbf{u}}+1}\right)\hfill \\ \hfill & +\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1){\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{u}·\mathbf{v})(\mathbf{u}·\mathbf{w})\hfill \\ \hfill =& \frac{1}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{w})\left(-({\gamma }_{\mathbf{u}}-1)({\gamma }_{\mathbf{v}}-1)-({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)-2{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}\frac{(\mathbf{u}·\mathbf{v})}{s^2}\right)\hfill \\ \hfill & +\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{w})\left(({\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}+1)+{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}\frac{(\mathbf{u}·\mathbf{v})}{s^2}\right)\hfill \\ \hfill & =-D(B_{\mathbf{vu}}\left(\mathbf{w}\right)+A_{\mathbf{uv}}\left(\mathbf{w}\right)).\hfill \end{array}$$

So, the coefficient of $\mathbf{u}$ vanishes.

We compute the coefficient of $\mathbf{v}$.

$$\begin{array}{c}\hfill A_{\mathbf{vu}}\left(\mathbf{w}\right)+B_{\mathbf{uv}}\left(\mathbf{w}\right)=-\frac{2}{s^2}\frac{{{\gamma }_{\mathbf{v}}}^2}{{\gamma }_{\mathbf{v}}+1}({\gamma }_{\mathbf{u}}-1)(\mathbf{v}·\mathbf{w})+\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})(\mathbf{u}·\mathbf{w}).\end{array}$$

We also have

$$\begin{array}{cc}\hfill & A_{\mathbf{vu}}\left(\mathbf{w}\right)B_{\mathbf{uv}}\left(\mathbf{v}\right)+B_{\mathbf{vu}}\left(\mathbf{w}\right)B_{\mathbf{uv}}\left(\mathbf{u}\right)\hfill \\ \hfill =& \frac{1}{s^2}\frac{{{\gamma }_{\mathbf{v}}}^2}{{\gamma }_{\mathbf{v}}+1}({\gamma }_{\mathbf{u}}-1)(\mathbf{v}·\mathbf{w})\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{u}·\mathbf{v})\hfill \\ \hfill & +\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{v}}}^2}{{\gamma }_{\mathbf{v}}+1}({\gamma }_{\mathbf{u}}-1)(\mathbf{v}·\mathbf{w})\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{v}}}^2}{{\gamma }_{\mathbf{v}}+1}({\gamma }_{\mathbf{u}}-1){\parallel \mathbf{v}\parallel }^2\hfill \\ \hfill & -\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{u}·\mathbf{w})\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{u}·\mathbf{v})-\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{u}·\mathbf{w})\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{v}}}^2}{{\gamma }_{\mathbf{v}}+1}({\gamma }_{\mathbf{u}}-1){\parallel \mathbf{v}\parallel }^2\hfill \\ \hfill & -\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})(\mathbf{u}·\mathbf{w})\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{u}·\mathbf{v})\hfill \\ \hfill & -\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})(\mathbf{u}·\mathbf{w})\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{v}}}^2}{{\gamma }_{\mathbf{v}}+1}({\gamma }_{\mathbf{u}}-1){\parallel \mathbf{v}\parallel }^2\hfill \\ \hfill & +\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{v}·\mathbf{w})\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}{\parallel \mathbf{u}\parallel }^2+\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{v}·\mathbf{w})\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{v}}}^2}{{\gamma }_{\mathbf{v}}+1}({\gamma }_{\mathbf{u}}-1)(\mathbf{u}·\mathbf{v})\hfill \\ \hfill & +\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{w})\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}{\parallel \mathbf{u}\parallel }^2\hfill \\ \hfill & +\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{w})\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{v}}}^2}{{\gamma }_{\mathbf{v}}+1}({\gamma }_{\mathbf{u}}-1)(\mathbf{u}·\mathbf{v})\hfill \\ \hfill & =\frac{2}{s^2}\frac{{{\gamma }_{\mathbf{v}}}^2}{{\gamma }_{\mathbf{v}}+1}({\gamma }_{\mathbf{u}}-1)(\mathbf{v}·\mathbf{w})\left\{{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}\frac{(\mathbf{u}·\mathbf{v})}{s^2}+({\gamma }_{\mathbf{u}}-1)({\gamma }_{\mathbf{v}}-1)+({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)\right\}\hfill \\ \hfill & -\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})(\mathbf{u}·\mathbf{w})\{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)\hfill \\ \hfill & +({\gamma }_{\mathbf{u}}-1)({\gamma }_{\mathbf{v}}-1)+{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}\frac{(\mathbf{u}·\mathbf{v})}{s^2}\}\hfill \\ \hfill & =-D(A_{\mathbf{vu}}\left(\mathbf{w}\right)+B_{\mathbf{uv}}\left(\mathbf{w}\right)).\hfill \end{array}$$

So, the coefficient of $\mathbf{v}$ vanishes. Thus, $\mathrm{gyr}[\mathbf{u},\mathbf{v}]\left(\mathrm{gyr}[\mathbf{v},\mathbf{u}]\mathbf{w}\right)=\mathbf{w}$ holds for every $\mathbf{w}\in {\mathbb{V}}_s$. Changing $\mathbf{u}$ and $\mathbf{v}$, $\mathrm{gyr}[\mathbf{v},\mathbf{u}]\left(\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{w}\right)=\mathbf{w}$ also holds for every $\mathbf{w}\in {\mathbb{V}}_s$. We conclude that $\mathrm{gyr}[\mathbf{u},\mathbf{v}]$ is bijective. Thus, $\mathrm{gyr}[\mathbf{u},\mathbf{v}]$ is an automorphism; a proof of (G4) is complete.

To prove (G5) we first observe $\mathrm{gyr}[\mathbf{u}{\oplus }_{\mathbf{E}}\mathbf{v},\mathbf{v}]=\mathrm{gyr}[\mathbf{u},\mathbf{v}]$ for every pair $\mathbf{u},\mathbf{v}\in {\mathbb{V}}_s$. Let $\mathbf{u},\mathbf{v},\mathbf{w}\in {\mathbb{V}}_s$. Applying (15) we have that

$$\begin{array}{cc}\hfill \mathrm{gyr}[\mathbf{u}{\oplus }_{\mathbf{E}}\mathbf{v},\mathbf{v}]\mathbf{w}=& \mathbf{w}+\frac{A^{\prime }\mathbf{u}{\oplus }_E\mathbf{v}+B^{\prime }\mathbf{v}}{D^{\prime }}\hfill \\ \hfill =& \mathbf{w}+\frac{A^{\prime }}{D^{\prime }}\frac{1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}}}{D_{\mathbf{uv}}(1+{\gamma }_{\mathbf{u}})}\mathbf{u}+\frac{1}{D^{\prime }}\left(\frac{A^{\prime }}{D_{\mathbf{uv}}{\gamma }_{\mathbf{u}}}+B^{\prime }\right)\mathbf{v},\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill A^{\prime }=& -\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{u}{\oplus }_{\mathbf{E}}\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}{\oplus }_{\mathbf{E}}\mathbf{v}}+1)}({\gamma }_{\mathbf{v}}-1)(\mathbf{u}{\oplus }_E\mathbf{v}·\mathbf{w})+\frac{1}{s^2}{\gamma }_{\mathbf{u}{\oplus }_{\mathbf{E}}\mathbf{v}}{\gamma }_{\mathbf{v}}(\mathbf{v}·\mathbf{w})\hfill \\ \hfill & +\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}{\oplus }_{\mathbf{E}}\mathbf{v}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}{\oplus }_{\mathbf{E}}\mathbf{v}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}{\oplus }_E\mathbf{v}·\mathbf{v})(\mathbf{v}·\mathbf{w})\hfill \\ \hfill B^{\prime }=& -\frac{1}{s^2}\frac{{\gamma }_{\mathbf{v}}}{{\gamma }_{\mathbf{v}}+1}\left\{{\gamma }_{\mathbf{u}{\oplus }_{\mathbf{E}}\mathbf{v}}({\gamma }_{\mathbf{v}}+1)(\mathbf{u}{\oplus }_E\mathbf{v}·\mathbf{w})+({\gamma }_{\mathbf{u}{\oplus }_{\mathbf{E}}\mathbf{v}}-1){\gamma }_{\mathbf{v}}(\mathbf{v}·\mathbf{w})\right\}\hfill \\ \hfill D^{\prime }=& {\gamma }_{\left(\mathbf{u}{\oplus }_{\mathbf{E}}\mathbf{v}\right){\oplus }_{\mathbf{E}}\mathbf{v}}+1\hfill \\ \hfill =& {\gamma }_{\mathbf{u}{\oplus }_{\mathbf{E}}\mathbf{v}}{\gamma }_{\mathbf{v}}(1+\frac{\left(\mathbf{u}{\oplus }_E\mathbf{v}\right)·\mathbf{v}}{s^2})+1.\hfill \end{array}$$

We prove that $\mathrm{gyr}[\mathbf{u}{\oplus }_{\mathbf{E}}\mathbf{v},\mathbf{v}]\mathbf{w}=\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{w}$ for every $\mathbf{u},\mathbf{v},\mathbf{w}\in {\mathbb{V}}_s$. We have

$$\begin{array}{cc}\hfill & \left(\mathbf{u}{\oplus }_E\mathbf{v}\right)·\mathbf{w}=\frac{1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}}}{D_{\mathbf{uv}}(1+{\gamma }_{\mathbf{u}})}(\mathbf{u}·\mathbf{w})+\frac{1}{D_{\mathbf{uv}}{\gamma }_{\mathbf{u}}}(\mathbf{v}·\mathbf{w}),\hfill \\ \hfill & \left(\mathbf{u}{\oplus }_E\mathbf{v}\right)·\mathbf{v}=\frac{1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}}}{D_{\mathbf{uv}}(1+{\gamma }_{\mathbf{u}})}(\mathbf{u}·\mathbf{v})+\frac{1}{D_{\mathbf{uv}}{\gamma }_{\mathbf{u}}}{\parallel \mathbf{v}\parallel }^2,\hfill \\ \hfill & {\gamma }_{\mathbf{u}{\oplus }_{\mathbf{E}}\mathbf{v}}+1=D.\hfill \end{array}$$

Then $A^{\prime }$ is computed as in the following.

$$\begin{array}{cc}\hfill A^{\prime }=& \frac{1}{{\gamma }_{\mathbf{u}{\oplus }_{\mathbf{E}}\mathbf{v}}+1}\{-{{\gamma }_{\mathbf{u}{\oplus }_{\mathbf{E}}\mathbf{v}}}^2({\gamma }_{\mathbf{v}}-1)\frac{1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}}}{D_{\mathbf{uv}}(1+{\gamma }_{\mathbf{u}})}\frac{(\mathbf{u}·\mathbf{w})}{s^2}-{{\gamma }_{\mathbf{u}{\oplus }_{\mathbf{E}}\mathbf{v}}}^2({\gamma }_{\mathbf{v}}-1)\frac{1}{D_{\mathbf{uv}}{\gamma }_{\mathbf{u}}}\frac{(\mathbf{v}·\mathbf{w})}{s^2}\hfill \\ \hfill & +\frac{1}{s^2}{\gamma }_{\mathbf{u}{\oplus }_{\mathbf{E}}\mathbf{v}}{\gamma }_{\mathbf{v}}({\gamma }_{\mathbf{u}{\oplus }_{\mathbf{E}}\mathbf{v}}+1)(\mathbf{v}·\mathbf{w})+\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}{\oplus }_{\mathbf{E}}\mathbf{v}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{v}}+1)}\frac{1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}}}{D_{\mathbf{uv}}(1+{\gamma }_{\mathbf{u}})}(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{w})\hfill \\ \hfill & +\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}{\oplus }_{\mathbf{E}}\mathbf{v}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{v}}+1)}\frac{1}{D_{\mathbf{uv}}{\gamma }_{\mathbf{u}}}(\mathbf{v}·\mathbf{w}){\parallel \mathbf{v}\parallel }^2\}\hfill \\ \hfill =& \frac{1}{D}\{-{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2{D}_{\mathbf{uv}}^2({\gamma }_{\mathbf{v}}-1)\frac{1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}}}{D_{\mathbf{uv}}(1+{\gamma }_{\mathbf{u}})}\frac{(\mathbf{u}·\mathbf{w})}{s^2}-{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2{D}_{\mathbf{uv}}^2({\gamma }_{\mathbf{v}}-1)\frac{1}{D_{\mathbf{uv}}{\gamma }_{\mathbf{u}}}\frac{(\mathbf{v}·\mathbf{w})}{s^2}\hfill \\ \hfill & +\frac{1}{s^2}{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^2{D}_{\mathbf{uv}}({\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}{D}_{\mathbf{uv}}+1)(\mathbf{v}·\mathbf{w})+\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^4{D}_{\mathbf{uv}}^2}{({\gamma }_{\mathbf{v}}+1)}\frac{1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}}}{D_{\mathbf{uv}}(1+{\gamma }_{\mathbf{u}})}(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{w})\hfill \\ \hfill & +\frac{2}{s^2}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2{D}_{\mathbf{uv}}^2}{({\gamma }_{\mathbf{v}}+1)}\frac{1}{D_{\mathbf{uv}}{\gamma }_{\mathbf{u}}}\frac{{{\gamma }_{\mathbf{v}}}^2-1}{{{\gamma }_{\mathbf{v}}}^2}(\mathbf{v}·\mathbf{w})\}\hfill \\ \hfill =& \frac{1}{D}\{-D_{\mathbf{uv}}{{\gamma }_{\mathbf{v}}}^2(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}})\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{u}}}^2}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{w})\hfill \\ \hfill & +D_{\mathbf{uv}}{{\gamma }_{\mathbf{v}}}^2(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}})\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(\mathbf{v}·\mathbf{w})\hfill \\ \hfill & +D_{\mathbf{uv}}{{\gamma }_{\mathbf{v}}}^2(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}})\frac{2}{s^4}\frac{{{\gamma }_{\mathbf{u}}}^2{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{w})\}\hfill \\ \hfill =& D_{\mathbf{uv}}{{\gamma }_{\mathbf{v}}}^2(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}})\frac{A}{D}.\hfill \end{array}$$

(25)

$D^{\prime }$ can be computed as in the following.

$$\begin{array}{cc}\hfill D^{\prime }=& {\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^2{D}_{\mathbf{uv}}\left\{1+\frac{1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}}}{D_{\mathbf{uv}}(1+{\gamma }_{\mathbf{u}})}(D_{\mathbf{uv}}-1)+\frac{1}{D_{\mathbf{uv}}{\gamma }_{\mathbf{u}}}\frac{{{\gamma }_{\mathbf{v}}}^2-1}{{{\gamma }_{\mathbf{v}}}^2}\right\}+1\hfill \\ \hfill =& \frac{{\gamma }_{\mathbf{u}}{\gamma }_{{\mathbf{v}}^{\mathbf{2}}}(2D_{\mathbf{uv}}-1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}}^2)}{1+{\gamma }_{\mathbf{u}}}+{{\gamma }_{\mathbf{v}}}^2\hfill \\ \hfill =& \frac{{{\gamma }_{\mathbf{v}}}^2{(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}})}^2}{1+{\gamma }_{\mathbf{u}}}\hfill \end{array}$$

Hence $\frac{A^{\prime }}{D^{\prime }}\frac{1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}}}{D_{\mathbf{uv}}(1+{\gamma }_{\mathbf{u}})}=\frac{A}{D}$.

Next, we compute the coefficient of $\mathbf{v}$. $B^{\prime }$ can be computed as in the following.

$$\begin{array}{cc}\hfill B^{\prime }=& -\frac{1}{s^2}\frac{{\gamma }_{\mathbf{v}}}{{\gamma }_{\mathbf{v}}+1}\{{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}{D}_{\mathbf{uv}}({\gamma }_{\mathbf{v}}+1)\frac{1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}}}{D_{\mathbf{uv}}(1+{\gamma }_{\mathbf{u}})}(\mathbf{u}·\mathbf{w})\hfill \\ \hfill & +{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}{D}_{\mathbf{uv}}({\gamma }_{\mathbf{v}}+1)\frac{1}{D_{\mathbf{uv}}{\gamma }_{\mathbf{u}}}(\mathbf{v}·\mathbf{w})+({\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}{D}_{\mathbf{uv}}-1){\gamma }_{\mathbf{v}}(\mathbf{v}·\mathbf{w})\}\hfill \\ \hfill =& -\frac{1}{s^2}{{\gamma }_{\mathbf{v}}}^2(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}})\left\{\frac{{\gamma }_{\mathbf{u}}}{1+{\gamma }_{\mathbf{u}}}(\mathbf{u}·\mathbf{w})+\frac{{\gamma }_{\mathbf{v}}}{1+{\gamma }_{\mathbf{v}}}(\mathbf{v}·\mathbf{w})\right\}.\hfill \end{array}$$

The Equation (25) is rewritten by $A^{\prime }\frac{1}{D_{\mathbf{uv}}{\gamma }_{\mathbf{u}}}=\frac{A}{D}\frac{{{\gamma }_{\mathbf{v}}}^2(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}})}{{\gamma }_{\mathbf{u}}}$. Then

$$\begin{array}{cc}\hfill & A^{\prime }\frac{1}{D_{\mathbf{uv}}{\gamma }_{\mathbf{u}}}+B^{\prime }\hfill \\ \hfill =& \frac{{{\gamma }_{\mathbf{v}}}^2(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}})}{D}\{-\frac{1}{s^2}\frac{{\gamma }_{\mathbf{u}}}{{\gamma }_{\mathbf{u}}+1}({\gamma }_{\mathbf{v}}-1)(\mathbf{u}·\mathbf{w})+\frac{1}{s^2}{\gamma }_{\mathbf{v}}(\mathbf{v}·\mathbf{w})\hfill \\ \hfill & +\frac{2}{s^4}\frac{{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{u}}+1)({\gamma }_{\mathbf{v}}+1)}(\mathbf{u}·\mathbf{v})(\mathbf{v}·\mathbf{w})-D\left(\frac{1}{s^2}\frac{{\gamma }_{\mathbf{u}}}{1+{\gamma }_{\mathbf{u}}}(\mathbf{u}·\mathbf{w})+\frac{1}{s^2}\frac{{\gamma }_{\mathbf{v}}}{1+{\gamma }_{\mathbf{v}}}(\mathbf{v}·\mathbf{w})\right)\}\hfill \\ \hfill =& \frac{{{\gamma }_{\mathbf{v}}}^2(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}})}{D(1+{\gamma }_{\mathbf{u}})}\{-\frac{1}{s^2}({\gamma }_{\mathbf{u}}({\gamma }_{\mathbf{v}}-1)+({\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}{D}_{\mathbf{uv}}+1){\gamma }_{\mathbf{u}})(\mathbf{u}·\mathbf{w})\hfill \\ \hfill & +\frac{1}{s^2}\left({\gamma }_{\mathbf{v}}(1+{\gamma }_{\mathbf{u}})-({\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}{D}_{\mathbf{uv}}+1)\frac{{\gamma }_{\mathbf{v}}}{1+{\gamma }_{\mathbf{v}}}(1+{\gamma }_{\mathbf{u}})\right)(\mathbf{v}·\mathbf{w})\hfill \\ \hfill & +\frac{2}{s^2}\frac{{\gamma }_{\mathbf{u}}{{\gamma }_{\mathbf{v}}}^2}{({\gamma }_{\mathbf{v}}+1)}(D_{\mathbf{uv}}-1)(\mathbf{v}·\mathbf{w})\}\hfill \\ \hfill =& \frac{{{\gamma }_{\mathbf{v}}}^2(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}})}{D(1+{\gamma }_{\mathbf{u}})}\{-\frac{1}{s^2}{\gamma }_{\mathbf{u}}{\gamma }_{\mathbf{v}}(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}})(\mathbf{u}·\mathbf{w})\hfill \\ \hfill & +\frac{1}{s^2}\frac{{{\gamma }_{\mathbf{v}}}^2}{1+{\gamma }_{\mathbf{v}}}(1+{\gamma }_{\mathbf{u}}{D}_{\mathbf{uv}})(1-{\gamma }_{\mathbf{u}})(\mathbf{v}·\mathbf{w})\}\hfill \\ \hfill =& \frac{D^{\prime }}{D}B.\hfill \end{array}$$

Hence $\frac{1}{D^{\prime }}(A^{\prime }\frac{1}{D_{\mathbf{uv}}{\gamma }_{\mathbf{u}}}+B^{\prime })=\frac{B}{D}$. We conclude that $\mathrm{gyr}[\mathbf{u}{\oplus }_{\mathbf{E}}\mathbf{v},\mathbf{v}]\mathbf{w}=\mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{w}$ for all $\mathbf{u},\mathbf{v},\mathbf{w}\in {\mathbb{V}}_s$, so (G5) holds.

We conclude that $({\mathbb{V}}_s,{\oplus }_E)$ is a gyrogroup. Finally, we prove that it is gyrocommutative.

We prove (G6). We prove that $\mathbf{a}\oplus \mathbf{b}=\mathrm{gyr}[\mathbf{a},\mathbf{b}](\mathbf{b}\oplus \mathbf{a})$ for all $\mathbf{a},\mathbf{b}\in {\mathbb{V}}_s$. Gyroautomorphic inverse property defined by Ungar in [1] (Definition 3.1, p. 51) is given by the equation

$${\ominus }_E\left(\mathbf{a}{\oplus }_E\mathbf{b}\right)={\ominus }_E\mathbf{a}{\ominus }_E\mathbf{b},$$

where $\mathbf{a},\mathbf{b}$ are arbitrary elements in ${\mathbb{V}}_s$. According to Theorem 3.2 in [1] (p. 51), $({\mathbb{V}}_s,{\oplus }_E)$ is gyrocommutative if and only if it has the gyroautomorphic inverse property. So, we observe the gyroautomorphic inverse property:

$$\begin{array}{ccc}\hfill {\ominus }_E\left(\mathbf{a}{\oplus }_E\mathbf{b}\right)& \hfill =& -\left(\mathbf{a}{\oplus }_E\mathbf{b}\right)\hfill \\ \hfill & \hfill =& -\frac{1}{1+\frac{\mathbf{a}·\mathbf{b}}{s^2}}\left\{\mathbf{a}+\frac{1}{{\gamma }_{\mathbf{a}}}\mathbf{b}+\frac{1}{s^2}\frac{{\gamma }_{\mathbf{a}}}{1+{\gamma }_{\mathbf{a}}}(\mathbf{a}·\mathbf{b})\mathbf{a}\right\},\hfill \\ \hfill {\ominus }_E\mathbf{a}{\ominus }_E\mathbf{b}& \hfill =& (-\mathbf{a}){\oplus }_E(-\mathbf{b})\hfill \\ \hfill & \hfill =& \frac{1}{1+\frac{\mathbf{a}·\mathbf{b}}{s^2}}\left\{-\mathbf{a}-\frac{1}{{\gamma }_{\mathbf{a}}}\mathbf{b}+\frac{1}{s^2}\frac{{\gamma }_{\mathbf{a}}}{1+{\gamma }_{\mathbf{a}}}(\mathbf{a}·\mathbf{b})(-\mathbf{a})\right\}\hfill \\ \hfill & \hfill =& {\ominus }_E\left(\mathbf{a}{\oplus }_E\mathbf{b}\right).\hfill \end{array}$$

Hence $({\mathbb{V}}_s,{\oplus }_E)$ is gyrocommutative.