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Article

A Gyrogeometric Mean in the Einstein Gyrogroup

1
Niigata Pref. Takada Senior High School, Yasuduka Branch School, Yasuduka 942-0411, Japan
2
Institute of Science and Technology, Niigata University, Niigata 950-2181, Japan
*
Author to whom correspondence should be addressed.
Symmetry 2020, 12(8), 1333; https://doi.org/10.3390/sym12081333
Submission received: 20 July 2020 / Revised: 3 August 2020 / Accepted: 4 August 2020 / Published: 10 August 2020
(This article belongs to the Special Issue Symmetry and Geometry in Physics)

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 V be a real inner product space. For a positive real number s we denote V s the s-ball of V , i.e.,
V s = { v V : v < s } .
The Einstein addition E on V s is a binary operation on V s given by the equation
u E v = 1 1 + u · v s 2 u + 1 γ u v + 1 s 2 γ u 1 + γ u ( u · v ) u ,
where γ u is the gamma factor
γ u = 1 1 u 2 s 2
in V s , where · and · are the inner product and the norm of V respectively. By the definition of E , u E v V s for every pair u , v 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 ( V 1 , 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 E on ( V 1 , E ) is given by the equation
r E a = ( 1 + a ) r ( 1 a ) r ( 1 + a ) r + ( 1 a ) r a a = tanh ( r tanh 1 a ) a a ,
where r R , a V 1 { 0 } ; and r E 0 = 0 . By Theorem 6.84 in [1], ( V 1 , E , E ) is a gyrovector space, which is called an Einstein gyrovector space.
Ungar [1] (pp. 172–173) defined the gyromidpoint P a b m of two elements in a gyrovector space. For a , b V 1 we have
P a b m = γ a a + γ b b γ a + γ b .
Ungar also defined the gyrocentroid C a b c m of three elements a , b , c V 1 as
C a b c m = γ a a + γ b b + γ c c γ a + γ b + γ c .
The gyromidpoint corresponds to the average of two velocities in the special theory of relativity. On the other hand the gyrocentroid C a b c m does not satisfy a certain desirable property one would expected for means; by a simple calculation we have C a b c m = γ c γ c + 2 c 1 3 E c for a = b = 0 and c 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 ( V 1 , E , E )

We define the set V 1 = { ± v : v V 1 } R which coincides with the open interval ( 1 , 1 ) . V 1 admits the addition and the scalar multiplication given by the following:
a b = a + b 1 + a b r a = tanh ( r tanh 1 a )
where a , b V 1 and r R . Please note that the triplet ( V 1 , , ) is a real one-dimensional space.
The gyrometric is defined by
d ( a , b ) = a E b V 1 ,
where a , b V 1 and a E b = a E ( b ) . The gyrometric needs not be a metric. It satisfies the following [1] (p. 158).
Proposition 1.
(1) 
For every pair a , b V 1 , d ( a , b ) 0 The equality d ( a , b ) = 0 holds if and only if a = b .
(2) 
d ( a , b ) = d ( b , a ) for any a , b V 1 .
(3) 
The gyrotriangle inequality:
d ( a , b ) d ( a , c ) d ( c , b )
holds for any a , b , c in V 1 .
We define the metric δ on V 1 induced by the gyrometric d. Put the map f : V 1 R by f ( x ) = tanh 1 ( x ) . For any a , b V 1 and r R , the map f satisfies the following.
(F1)
f ( a b ) = f ( a ) + f ( b )
(F2)
f ( r a ) = r f ( a )
Let the map δ on V 1 be given by
δ ( a , b ) = f ( d ( a , b ) )
for a , b V 1 .
Lemma 1.
The inequality
1 γ a ( 1 a · b ) b a a E b 1 1 a · b b a
holds for every pair a , b V 1 .
Proof. 
Recall the equations (3.177) and (3.178) [1] (pp. 88–89):
γ a E b = γ a γ b ( 1 + a · b ) , γ a E b = γ a γ b ( 1 a · b ) .
By γ a E b = 1 1 a E b we have
a E b = 1 1 γ a E b 2 = 1 1 γ a γ b ( 1 a · b ) = 1 ( 1 a 2 ) ( 1 b 2 ) ( 1 a · b ) 2 = ( 1 a · b ) 2 ( 1 a 2 ) ( 1 b 2 ) 1 a · b = b a 2 a 2 b 2 + ( a · b ) 2 1 a · b .
Since a · b a b then we have
a E b 1 1 a · b b a .
Next we calculate a E b 2 1 γ a ( 1 a · b ) b a 2 .
b a 2 a 2 b 2 + ( a · b ) 2 ( 1 a · b ) 2 1 a 2 ( 1 a · b ) 2 b a 2 = 1 ( 1 a · b ) 2 b a 2 a 2 b 2 + ( a · b ) 2 ( 1 a 2 ) b a 2 = 1 ( 1 a · b ) 2 ( a · b ) 2 + a 4 2 a 2 ( a · b ) = 1 ( 1 a · b ) 2 ( a 2 a · b ) 2 0 .
Thus, we have the desired inequalities and conclude the proof. □
Proposition 2.
( V 1 , δ ) is a complete metric space.
Proof. 
We first prove that ( V 1 , δ ) is a metric space. By (1) and (2) of Proposition 1, it is trivial that δ ( a , b ) = 0 a = b and δ ( a , b ) = δ ( b , a ) for every a , b V 1 . By (3) of Proposition 1 and the monotonicity of f, the inequality f ( d ( a , b ) ) f ( d ( a , c ) d ( c , b ) ) for every a , b , c V 1 . By (F1) we have
δ ( a , b ) δ ( a , c ) + δ ( c , b ) .
As V is complete, we have by Lemma 1 and the definition of δ ( · , · ) that ( V 1 , δ ) is complete. □
We recall the gyroline and the gyrosegment [1] (Definition 6.19). Let a , b be elements of V 1 . The gyroline through a and b is defined by
L ( a , b ) = { a E t E ( E a E b ) : t R } .
A gyrosegment with endpoints a and b is denoted by
S ( a , b ) = { a E t E ( E a E b ) : 0 t 1 } .
The point a # t b = a E t E ( E a E b ) is called the gyro t-point on a gyroline or gyrosegment. We abbreviate a # 1 2 b by a # b . Please note that a # b = b # a for every pair a , b V 1 .
Theorem 1.
For any a , b , c V 1 we have
d ( a # b , a # c ) 1 2 d ( b , c ) .
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):
γ a E b = ( γ a + γ b ) 2 γ a b + 1 1 , γ 1 2 E a = 1 + γ a 2 .
Hence
γ a # b = γ a + γ b 2 ( γ a b + 1 )
holds. By a simple calculation, we have
1 ( a # b ) · ( a # c ) = 1 + γ a b + γ b c + γ c a ( γ a + γ b ) ( γ a + γ c ) .
Hence we have
γ ( a # b ) E ( a # c ) = γ a # b γ a # c ( 1 ( a # b ) · ( a # c ) ) = 1 + γ a b + γ b c + γ c a 2 ( γ a b + 1 ) ( γ a c + 1 )
and
d 2 ( a # b , a # c ) = 1 1 γ ( a # b ) E ( a # c ) 2 = 1 4 ( γ a b + 1 ) ( γ a c + 1 ) ( 1 + γ a b + γ b c + γ c a ) 2 .
We also have
1 2 d ( b , c ) = tanh 1 2 tanh 1 d ( b , c ) = γ b E c 1 + γ b E c d ( b , c ) = γ b E c 1 1 + γ b E c ,
where γ d ( b , c ) = γ b E c . Hence we have
1 2 d ( b , c ) 2 d 2 ( a # b , a # c )
= γ b E c 1 1 + γ b E c 1 4 ( γ a b + 1 ) ( γ a c + 1 ) ( 1 + γ a b + γ b c + γ c a ) 2 = 4 ( γ a b + 1 ) ( γ a c + 1 ) ( 1 + γ a b + γ b c + γ c a ) 2 2 1 + γ b E c = 2 2 ( γ a b + 1 ) ( γ b E c + 1 ) ( γ c a + 1 ) ( 1 + γ a b + γ b c + γ c a ) 2 ( 1 + γ b E c ) ( 1 + γ a b + γ b c + γ c a ) 2
= 2 2 γ a b γ b c γ c a ( γ a b 2 + γ b c 2 + γ c a 2 ) + 1 ( 1 + γ b E c ) ( 1 + γ a b + γ b c + γ c a ) 2 .
Let A = E c E a , B = E c E b . It is well defined by ( E c E a ) E ( E c E b ) = gyr [ E c , a ] ( a E b ) = a E b . We calculate the numerator of (8);
2 γ a b γ b c γ c a ( γ a b 2 + γ b c 2 + γ c a 2 ) + 1 = 2 γ A E B γ A γ B ( γ A E B 2 + γ A 2 + γ B 2 ) + 1 = 2 γ A 2 γ B 2 ( 1 A · B ) ( γ A 2 γ B 2 ( 1 A · B ) 2 + γ A 2 + γ B 2 ) + 1 = γ A 2 γ B 2 ( 1 ( A · B ) 2 ) ( γ A 2 + γ B 2 ) + 1 = 1 ( A · B ) 2 ( 1 A 2 ) ( 1 B 2 ) 1 ( 1 A 2 ) + 1 ( 1 B 2 ) + 1 = A 2 B 2 ( A · B ) 2 ( 1 A 2 ) ( 1 B 2 ) 0 .
We conclude a proof of Theorem 1. □
By Theorem 1 and the monotonicity of f ( x ) = tanh 1 ( x ) , we have
δ ( a # b , a # c ) 1 2 δ ( b , c ) .
By the triangle inequality, we have
δ ( a # b , c # d ) δ ( a # b , a # d ) + δ ( a # d , c # d ) 1 2 δ ( b , d ) + 1 2 δ ( a , c ) .
Moreover, since the map g ( t ) = δ ( a # t b , c # t d ) is continuous, we infer that g is convex, i.e.,
δ ( a # t b , c # t d ) ( 1 t ) δ ( a , c ) + t δ ( b , d ) .
Letting a = c we have
δ ( a # t b , a # t c ) t δ ( b , c ) .

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 V 1 . We say that A is gyroconvex if S ( a , b ) A for any a , b A . Let X be a non-empty subset of V 1 .
conv ( X ) = { C V 1 : X C a n d C i s gyr o c o n v e x s e t } .
We call conv ( X ) the gyroconvex hull of X.
Please note that the gyroconvex hull of a non-empty set X V is gyroconvex.
Lemma 2.
Let a , b V 1 . Then the gyrosegment S ( a , b ) is gyroconvex. The gyroconvex hull conv ( { a , b } ) coincides with S ( a , b ) .
Proof. 
Let P j be an arbitrary point in S ( a , b ) for j = 1 , 2 . There exists 0 t j 1 such that
P j = a E t j E ( E a E b )
for j = 1 , 2 . We may assume that t 1 t 2 . where 0 t 1 t 2 1 . Then we have t [ 0 , 1 ] , P 1 # t P 2 S ( a , b ) . In fact, by the Equation (6.63) in [1] (p. 167) we have
P 1 E t E ( E P 1 E P 2 ) = a E ( t 1 + ( t 1 + t 2 ) t ) E ( E a E b ) .
Since t 1 t 1 + ( t 1 + t 2 ) t t 2 , we have P 1 # t P 2 S ( a , b ) . Thus, S ( P 1 , P 2 ) S ( a , b ) for every pair P 1 and P 2 in S ( a , b ) . Thus, S ( a , b ) is gyroconvex. □
Let C 0 be a non-empty subset of V 1 . We define a sequence { C n } of a non-empty subset of V 1 by induction. Suppose that C n 1 is defined. Put
C n = a , b C n 1 S ( a , b ) .
Proposition 3.
Let C 0 be a non-empty subset of V 1 . Then
conv ( C 0 ) = n = 0 C n .
Proof. 
We prove that k = 0 C n is gyroconvex. Let a , b k = 0 C n . Since C k C k + 1 for every k N { 0 } , there exists a positive integer n 0 with a , b C n 0 . Then by the definition of C n 0 + 1 we have S ( a , b ) C n 0 + 1 k = 0 C n . Thus, n = 0 C n is a gyroconvex set. As C 0 k = 0 C n , we have conv ( C 0 ) n = 0 C n .
We prove n = 0 C n conv ( C 0 ) . For any a , b C 0 , S ( a , b ) conv ( C 0 ) . Hence C 1 conv ( X ) . Similarly, assuming that C n conv ( C 0 ) for any n N { 0 } we have C n + 1 conv ( C 0 ) . So for arbitrary nonnegative integer n, C n conv ( C 0 ) ; n = 0 C n conv ( C 0 ) . □

4. The Gyrogeometric Mean

Let ϕ X V 1 . We define diam ( X ) = sup { δ ( x , y ) : x , y X } . First we prove the following.
Proposition 4.
Suppose that x 0 , y 0 , x 1 , y 1 V 1 . If the inequality diam ( { x 0 , y 0 , x 1 , y 1 } ) M holds for a positive real number M, then the inequality δ ( x , y ) M holds for arbitrary points x S ( x 0 , x 1 ) and y S ( y 0 , y 1 ) .
Proof. 
Put x = x 0 # t x 1 and y = y 0 # s y 1 , where 0 s t 1 . We have
d ( y 0 # t x 1 , y 0 # s x 1 ) = { y 0 E t E ( E y 0 E x 1 ) } E { y 0 E s E ( E y 0 E x 1 ) } = gyr [ E y 0 , t E ( E y 0 E x 1 ) ] { t E ( E y 0 E x 1 ) E s E ( E y 0 E x 1 ) } = ( t s ) E y 0 E x 1 .
Applying this equality, we have by (11) and (12) that
δ ( x , y ) δ ( x , y 0 # t x 1 ) + δ ( y 0 # t x 1 , y 0 # s x 1 ) + δ ( y 0 # s x 1 , y ) = δ ( x 0 # t x 1 , y 0 # t x 1 ) + ( t s ) δ ( y 0 , x 1 ) + δ ( y 0 # s x 1 , y 0 # s y 1 ) ( 1 t ) δ ( x 0 , y 0 ) + ( t s ) δ ( y 0 , x 1 ) + s δ ( x 1 , y 1 ) ( 1 t ) M + ( t s ) M + s M = M .
Lemma 3.
Let X V be a non-empty set. Then
diam ( X ) = diam ( conv ( X ) ) = diam ( conv ( X ) ¯ ) .
Proof. 
First we prove diam ( X ) = diam ( conv ( X ) ) . By Proposition 3, conv ( X ) = n = 0 C n where C 0 = X . For any positive integer k let x , y be arbitrary points in C k . Then there exist x 0 , x 1 , y 0 , y 1 C k 1 such that x S ( x 0 , x 1 ) , y S ( y 0 , y 1 ) . Put M = diam ( C k 1 ) then by Proposition 4.1 we have
δ ( x , y ) M = diam ( C k 1 ) ,
whence
diam ( C k ) diam ( C k 1 ) .
Thus, for arbitrary n N , diam ( C n ) diam ( C 0 ) = diam ( X ) . It follows that for any x , y conv ( X ) = n = 0 C n , we have
δ ( x , y ) diam ( X ) .
Therefore
diam ( conv ( X ) ) diam ( X ) .
The converse inequality is trivial, hence we have diam ( conv ( X ) ) = diam ( X ) .
Next we prove diam ( conv ( X ) ) = diam ( conv ( X ) ¯ ) . For any pair x , y conv ( X ) ¯ , there exist sequences { x n } and { y n } in conv ( X ) such that x n , y n converge to x , y respectively. Letting n for δ ( x n , y n ) diam ( conv ( X ) ) , we have δ ( x , y ) diam ( X ) . Thus, diam ( conv ( X ) ¯ ) diam ( conv ( X ) ) . The converse inequality is trivial, hence diam ( conv ( X ) ) = diam ( conv ( X ) ¯ ) holds. □
Lemma 4.
Suppose that K is a gyroconvex subset of V . Then the closure K ¯ of K is gyroconvex.
Proof. 
For any x , y K ¯ , there exist { x n } , { y n } K such that x n , y n converge to x , y respectively. We show x n # t y n converges to x # t y for arbitrary 0 t 1 . By (11) we have
δ ( x n # t y n , x # t y ) ( 1 t ) δ ( x n , x ) + t δ ( y n , y ) .
By n , then δ ( x n # t y n , x # y ) 0 . Thus, x # t y K ¯ . Hence K ¯ is gyroconvex. □
Let n be a positive integer. Let X n be the set of all subsets of V 1 whose number of elements is exactly n. We define, by induction, the sequence { G n } n = 2 of the maps G n : X n V 1 which satisfy the following two conditions ( p n ) and ( q n );
( p n )
G n ( Δ ) conv ( Δ ) ¯ for every Δ X n ,
( q n )
δ ( G n ( Δ ) , G n ( Δ ) ) 1 n i = 1 n δ ( a i , a i ) for every pair Δ = { a 1 , , a n } and Δ = { a 1 , , a n } in X n .
First, put G 2 ( { a , b } ) = a # b for { a , b } X 2 . As conv ( { a , b } ) = S ( a , b ) by Lemma 2 we obtain that G 2 ( { a , b } ) conv ( { a , b } ) ; ( p 2 ) holds. Let Δ = { a , b } , Δ = { c , d } X 2 . Then by (10) we have
δ ( G 2 ( Δ ) , G 2 ( Δ ) ) = δ ( a # b , c # d ) 1 2 { δ ( a , c ) + δ ( b , d ) } ,
which is ( q 2 ).
Suppose now that the map G k : X k V 1 which satisfies ( p k ) and ( q k ) is defined. We will define G k + 1 : X k + 1 V 1 which satisfies ( p k + 1 ) and ( q k + 1 ). Let Δ 0 = { a 1 0 , , a k + 1 0 } X k + 1 . For a positive integer m we define Δ m X k + 1 which satisfies that Δ m conv ( Δ m 1 ) ¯ by induction on m. For every 1 i k + 1 , put
a i 1 = G k ( { a 1 0 , , a i 1 0 , a i + 1 0 , , a k + 1 0 } ) .
Please note that a i 1 is well defined since { a 1 0 , , a i 1 0 , a i + 1 0 , , a k + 1 0 } X k and we have assumed that the map G k is defined. By the condition ( p k ) we have that
a i 1 conv ( { a 1 0 , , a i 1 0 , a i + 1 0 , , a k + 1 0 } ) ¯
for every 1 i k + 1 . Put Δ 1 = { a 1 1 , , a k + 1 1 } . Then Δ 1 X k + 1 and Δ 1 conv ( Δ 0 ) ¯ since { a 1 0 , , a i 1 0 , a i + 1 0 , , a k + 1 0 } Δ 0 for every 1 i k + 1 . Suppose that Δ l = { a 1 l , , a k + 1 l } X k + 1 such that Δ l conv ( Δ l 1 ) ¯ is defined. For every 1 i k + 1 , put
a i l + 1 = G k ( { a 1 l , , a i 1 l , a i + 1 l , , a k + 1 l } ) .
As in the same way as the above, a i l + 1 is well defined for 1 i k + 1 , and Δ l + 1 = { a 1 l + 1 , , a k + 1 l + 1 } X k + 1 satisfies that Δ l + 1 conv ( Δ l ) ¯ . Hence, by induction, we have defined a sequence { Δ m } m = 1 X k + 1 such that Δ m conv ( Δ m 1 ) ¯ . Applying ( q k ) for Δ = { a 1 m , , a i 1 m , a i + 1 m , , a k + 1 m } and Δ = { a 1 m , , a j 1 m , a j + 1 m , , a k + 1 m } , we infer that
δ ( a i m + 1 , a j m + 1 ) = δ ( G k ( Δ ) , G k ( Δ ) ) = δ ( G k ( { a 1 m , , a i 1 m , a i + 1 m , , a k + 1 m } ) , G k ( { a 1 m , , a j 1 m , a j + 1 m , , a k + 1 m } ) ) = δ ( G k ( { a 1 m , , a i 1 m , a i + 1 m , , a j 1 m , a j + 1 m , , a k + 1 m , a j m } ) , G k ( { a 1 m , , a i 1 m , a i + 1 m , , a j 1 m , a j + 1 m , , a k + 1 m , a i m } ) ) 1 k δ ( a j m , a i m ) = 1 k δ ( a i m , a j m )
Then by Lemma 3 we obtain
diam ( conv ( Δ m + 1 ) ¯ ) 1 k diam ( conv ( Δ m ) ¯ )
for every positive integer m. By Cantor’s intersection theorem there exists a unique M V 1 with
{ M } = m conv ( Δ m ) ¯ .
As a i m conv ( Δ m 1 ¯ ) for every 1 i k + 1 , we infer that lim m a i m = M for every 1 i k + 1 . Put G k + 1 ( Δ 0 ) = M . Then the map G k + 1 : X k + 1 V 1 is well defined, and G k + 1 ( Δ 0 ) = M conv ( Δ 0 ) ¯ ; G k + 1 satisfies the condition ( p n ).
Next we prove that the map G k + 1 satisfies the condition ( q k + 1 );
δ ( G k + 1 ( Δ ) , G k + 1 ( Δ ) ) 1 k + 1 j = 1 k + 1 δ ( a j , a j ) ,
where Δ = { a 1 0 , , a k + 1 0 } , Δ = { a 1 0 , , a k + 1 0 } X k + 1 . Let m be a positive integer. We define a i m and a i m for every 1 i k + 1 as in the same way as before. As ( q k ) holds for G k , we have
δ ( a i m , a i m ) = δ ( G k ( { a 1 m 1 , a 2 m 1 , , a i 1 m 1 , a i + 1 m 1 , , a k + 1 m 1 } ) , G k ( { a 1 m 1 , a 2 m 1 , , a i 1 m 1 , a i + 1 m 1 , , a k + 1 m 1 } ) ) 1 k j = 1 , j i k + 1 δ ( a j m 1 , a j m 1 ) .
By summing up the above inequalities with respect to 1 i k + 1 we have
i = 1 k + 1 δ ( a i m , a i m ) i = 1 k + 1 1 k j = 1 , j i k + 1 δ ( a j m 1 , a j m 1 ) = j = 1 k + 1 δ ( a j m 1 , a j m 1 ) .
for every positive integer m. Thus, we have
i = 1 k + 1 δ ( a i m , a i m ) j = 1 k + 1 δ ( a j , a j ) .
Letting m , since lim m a i m = G k + 1 ( Δ ) , lim m a i m = G k + 1 ( Δ ) , we have
i = 1 k + 1 δ ( G k + 1 ( Δ ) , G k + 1 ( Δ ) ) j = 1 k + 1 δ ( a j , a j ) ,
hence
δ ( G k + 1 ( Δ ) ) , G k + 1 ( Δ ) ) 1 k + 1 ( j = 1 k + 1 δ ( a j , a j ) ) .
So, the condition ( q k + 1 ) holds for the map G k + 1 . We conclude that the map G n : X n 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 V 1 .
Definition 2.
Let Δ = { a 1 , , a n } V 1 . We call that G n ( Δ ) the gyrogeometric mean of Δ.
Due to the definition, the gyrogeometric mean of { a , b } V 1 is a # b . The gyrocentroid C a b c m is defined by applying the internal division points on the usual lines which makes the inconvenience such as C a b c m = γ c γ c + 2 c 1 3 E c for a = 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 V 1 for a positive integer n.
Proposition 5.
Let x V 1 and D n = { a 1 , a 2 , , a n } X n . Put x E D n = { x E a 1 , x E a 2 , , x E a n } . Then the following holds:
G n ( x E D n ) = x E G n ( D n ) .
Proof. 
We prove the equality (13) by induction on n. For n = 2 , G 2 ( { a 1 , a 2 } ) = P a 1 a 2 m . By Theorem 6.37 in [1] (p. 175) we have
G 2 ( { x E a 1 , x E a 2 } ) = x E G 2 ( { a 1 , a 2 } ) .
Assume that (13) holds for n = k . Let D k + 1 m = { a 1 m , a 2 m , , a k + 1 m } where a i m = G k ( { a 1 m 1 , a 2 m 1 , , a i 1 m 1 , a i + 1 m 1 , , a k + 1 } ) for 1 i k + 1 . By the assumption we have
G k ( { x E a 1 m 1 , x E a 2 m 1 , , x E a i 1 m 1 , x E a i + 1 m 1 , , x E a k + 1 m 1 } ) = x E G k ( { a 1 m 1 , a 2 m 1 , , a i 1 m 1 , a i + 1 m 1 , , a k + 1 m 1 } ) = x E a i m
for every 1 i k + 1 . Then for x E D k + 1 = { x E a 1 , x E a 2 , , x E a k + 1 } we have
x E D k + 1 m = { G k ( { x E a 1 m 1 , , x E a i 1 m 1 , x E a i + 1 m 1 , , x E a k + 1 m 1 } ) : 1 i k + 1 } = { x E a i m : i = 1 , 2 , , k + 1 } .
We prove that x E D k + 1 m { x E G k + 1 ( D k + 1 ) } as m . By a simple calculation we have
d ( x E a i m , x E G k + 1 ( D k + 1 ) ) = gyr [ x , a i m ] ( a i m E G ( D k + 1 ) ) = a i m E G ( D k + 1 ) 0
as m . Hence we have δ ( x E a i m , x E G k + 1 ( D k + 1 ) ) 0 as m . Thus, x E a i m x E G ( D k + 1 ) as m for 1 i k + 1 . We conclude that G n ( x E D n ) = x E G n ( D n ) . □
For a = b = 0 and c in V 1 , G 3 ( { a , b , c } ) = 1 3 E c . More generally, the gyrogeometric mean satisfies the following.
Proposition 6.
Let Δ = { t 1 E a , t 2 E a , , t n E a } V 1 .
G n ( Δ ) = t 1 + t 2 + + t n n E a
In the case of n = 2 , it is proved by the following calculation.
G 2 ( t 1 E a , t 2 E a ) = t 1 E a # t 2 E a = 1 2 E { ( t 1 E a ) E ( t 2 E a ) } = 1 2 E { ( t 1 E a ) E gyr [ t 1 E a , E t 2 E a ] ( t 2 E a ) } = 1 2 E { ( t 1 E a ) E ( t 2 E a ) } = 1 2 E { ( t 1 + t 2 ) a } = t 1 + t 2 2 E a .
Proposition 6 is proved by induction on n.
In Section 6 V s = { v V : v < s } with appropriate operation is a gyrocommutative gyrogroup, which is also called the Einstein gyrogroup. The gyrogeometric mean is defined for V s similarly. If s or v V s such that v is small enough, γ v 1 . So, in the case,
G n ( { a 1 , a 2 , , a n } ) a 1 + a 2 + + a n n
is hold. It is simply proved by induction.

6. Proof that ( V s , E ) Is a Gyrocommutative Gyrogroup

A magma ( G , ) is a non-empty set G with a binary operation ⊕. A magma ( G , ) is a gyrogroup if its binary operation ⊕ satisfies the following axioms (G1) through (G5):
(G1)
There exists a left identity 0 in G such that
0 a = a
for all a G .
(G2)
For each a G there exists a left inverse a G such that
a a = 0 .
(G3)
For any a , b , c G there exists a unique element gyr [ a , b ] c G such that the binary operation obeys the left gyroassociative law
a ( b c ) = ( a b ) gyr [ a , b ] c .
(G4)
The map gyr [ a , b ] : G G given by c gyr [ a , b ] c is an automorphism of the magma ( G , ) . It is called a gyroautomorphism. gyr [ a , b ] generated by a , b G is called a gyration.
(G5)
The gyroautomorphism gyr [ a , b ] generated by any a , b G satisfies the left loop property:
gyr [ a , b ] = gyr [ a b , b ] .
The gyrogroup ( G , ) is called gyrocommutative if the following (G6) holds for every pair a , b G
(G6)
a b = gyr [ a , b ] ( b a ) .
We prove that the Einstein gyrogroup ( V s , 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 u E ( v E w ) = ( u E v ) E gyr [ u , v ] w holds for all u , v , w V s . First, we prove the left cancellation law which is given by the equation
E a E ( a E b ) = b ,
for all a , b V s . Put D u v = 1 + u · v s 2 for any u , v V s and put
x = a E b = 1 D a b a + 1 γ a b + 1 s 2 γ a 1 + γ a ( a · b ) a .
Put 1 + 1 s 2 ( ( a ) · x ) = D ( a ) x . We have
E a E x = 1 D ( a ) x ( a ) + 1 γ a x + 1 s 2 γ a 1 + γ a ( a · x ) a = 1 D ( a ) x ( a ) + 1 γ a D a b ( 1 + γ a D a b 1 + γ a a + 1 γ a b ) + 1 s 2 γ a 1 + γ a ( a · x ) a .
We compute,
D ( a ) x = 1 + 1 s 2 ( ( a ) · x ) = 1 1 D a b s 2 a 2 + 1 γ a ( a · b ) + 1 s 2 γ a 1 + γ a ( a · b ) a 2 = 1 1 D a b γ a 2 1 γ a 2 + 1 γ a ( a · b ) s 2 + γ a 1 + γ a ( a · b ) s 2 γ a 2 1 γ a 2 = 1 1 D a b 1 1 γ a 2 + 1 γ a ( D a b 1 ) + γ a 1 γ a ( D a b 1 ) = 1 γ a 2 D a b ,
and
a · x s 2 = 1 D ( a ) x = γ a 2 D a b 1 γ a 2 D a b .
Hence we have
E a E ( a E b ) = E a E x = γ a 2 D a b { ( a ) + 1 γ a D a b ( 1 + γ a D a b 1 + γ a a + 1 γ a b ) + γ a 1 + γ a γ a 2 D a b 1 γ a 2 D a b a } = γ a 2 D a b a + γ a ( 1 + γ a D a b 1 + γ a a + 1 γ a b ) + γ a 1 + γ a ( γ a 2 D a b 1 ) a = b + γ a 2 D a b + γ a + γ a 2 D a b 1 + γ a + γ a 3 D a b γ a 1 + γ a a = b .
Next, we prove the following equation
gyr [ u , v ] w = E ( u E v ) E ( u E ( v E w ) ) .
It is known in [5] ((2.84), (2.85)) that the Equation (14) can be rewritten as
gyr [ u , v ] w = w + A u + B v D
by applying computer algebra, where
A = 1 s 2 γ u 2 γ u + 1 ( γ v 1 ) ( u · w ) + 1 s 2 γ u γ v ( v · w ) + 2 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) ( v · w ) B = 1 s 2 γ v γ v + 1 γ u ( γ v + 1 ) ( u · w ) + ( γ u 1 ) γ v ( v · w ) D = γ u γ v ( 1 + u · v s 2 ) + 1 .
We prove (15) without applying computer algebra. Put
z = E ( u E v ) = 1 1 + u · v s 2 u + 1 γ u v + 1 s 2 γ u 1 + γ u ( u · v ) u = 1 + γ u D uv D uv ( 1 + γ u ) u 1 γ u D uv v , x = ( v E w ) = 1 + γ v D vw D vw ( 1 + γ v ) v + 1 γ v D vw w .
Put also
y = u E ( v E w ) = u E x = 1 + γ u D ux D ux ( 1 + γ u ) u + 1 γ u D ux x = 1 + γ u D ux D ux ( 1 + γ u ) u + 1 + γ v D vw D ux D vw γ u ( 1 + γ v ) v + 1 D ux D vw γ u γ v w .
Then gyr [ u , v ] w is given by the following:
gyr [ u , v ] w = z E y = 1 + γ z D zy D zy ( 1 + γ z ) z + 1 γ z D zy y = ( 1 + γ z D zy ) ( 1 + γ u D uv ) D zy D uv ( 1 + γ z ) ( 1 + γ u ) u 1 + γ z D zy D zy D uv ( 1 + γ z ) γ u v + 1 + γ u D ux D zy D ux γ z ( 1 + γ u ) u + 1 + γ v D vw D zy D ux γ z γ u ( 1 + γ v ) v + 1 D zy D ux D vw γ z γ u γ v w .
We will calculate each coefficient of u , v , w of the equation above.
We prove that the coefficient of w is 1, i.e., D zy D ux D vw γ u γ v γ z is 1. The equation a · b s 2 = D a b 1 holds for all a , b V s . Applying this equation, we have
D ux = 1 + u · x s 2 = 1 + 1 s 2 1 + γ v D vw D vw ( 1 + γ v ) ( u · v ) + 1 γ v D vw ( u · w ) = D vw γ v ( 1 + γ v ) + γ v ( 1 + γ v D vw ) ( D uv 1 ) + ( 1 + γ v ) ( D uw 1 ) D vw γ v ( 1 + γ v ) = γ v 2 D uv D vw + γ v D uv + γ v D vw + γ v D uw + D uw 2 γ v 1 D vw γ v ( 1 + γ v ) , D zy = 1 + z · y s 2 = 1 1 s 2 { ( 1 + γ u D uv ) ( 1 + γ u D ux ) D uv D ux ( 1 + γ u ) 2 u 2 + ( 1 + γ u D uv ) ( 1 + γ v D vw ) D uv D ux D vw γ u ( 1 + γ u ) ( 1 + γ v ) ( u · v ) + 1 + γ u D uv D uv D ux D vw γ u γ v ( 1 + γ u ) ( u · w ) + 1 + γ u D ux D uv D ux γ u ( 1 + γ u ) ( u · v ) + 1 + γ v D vw D uv D ux D vw γ u 2 ( 1 + γ v ) v 2 + v · w D uv D ux D vw γ u 2 γ v } .
Multiplying D ux D vw from the right-hand side of the last equation, and applying the gamma factor, we have
D zy ( D ux D vw ) = D ux D vw ( 1 + γ u D uv ) ( 1 + γ u D ux ) ( γ u 1 ) D vw D uv ( 1 + γ u ) γ u 2 ( 1 + γ u D uv ) ( 1 + γ v D vw ) ( D uv 1 ) D uv γ u ( 1 + γ u ) ( 1 + γ v ) ( 1 + γ u D uv ) ( D uw 1 ) D uv γ u γ v ( 1 + γ u ) ( 1 + γ u D ux ) ( D uv 1 ) D vw D uv γ u ( 1 + γ u ) ( 1 + γ v D vw ) ( γ v 1 ) D uv γ u 2 γ v 2 D vw 1 D uv γ u 2 γ v .
Dividing the common denominator D uv γ u 2 γ v 2 ( 1 + γ u ) ( 1 + γ v ) and multiplying γ u γ v γ z to D zy D ux D vw , where γ z = γ E ( u E v ) = γ u E v = D uv γ u γ v , we have
D zy D ux D vw γ u γ v γ z = 1 ( 1 + γ u ) ( 1 + γ v ) { D uv D ux D vw γ u 2 γ v 2 ( 1 + γ u ) ( 1 + γ v ) ( 1 + γ u D uv ) ( 1 + γ u D ux ) ( γ u 1 ) D vw γ v 2 ( 1 + γ v ) ( 1 + γ u D uv ) ( 1 + γ v D vw ) ( D uv 1 ) γ u γ v 2 ( 1 + γ u D uv ) ( D uw 1 ) ( 1 + γ v ) γ u γ v ( 1 + γ u D ux ) ( D uv 1 ) D vw γ u γ v 2 ( 1 + γ v D vw ) ( γ v 1 ) ( 1 + γ u ) ( 1 + γ v ) ( D vw 1 ) γ v ( 1 + γ u ) ( 1 + γ v ) } .
We compute { · } of the Equation (16).
{ · } = D uv D ux D vw γ u 2 γ v 2 + D uv D ux D vw γ u 3 γ v 2 + D uv D ux D vw γ u 2 γ v 3 + D uv D ux D vw γ u 3 γ v 3 D vw γ u γ v 2 + D vw γ v 3 D vw γ u γ v 3 + D vw γ v 2 D vw D ux γ u 2 γ v 2 + D vw D ux γ u γ v 3 D vw D ux γ u 2 γ v 3 + D vw D ux γ u γ v 2 D uv D vw γ u 2 γ v 2 + D uv D vw γ u γ v 3 D uv D vw γ u 2 γ v 3 + D uv D vw γ u γ v 2 D uv D vw D ux γ u 3 γ v 2 + D uv D vw D ux γ u 2 γ v 3 ̲ D uv D vw D ux γ u 3 γ v 3 + D uv D vw D ux γ u 2 γ v 2 ̲ + γ u γ v 2 + D uv γ u 2 γ v 2 + D vw γ u γ v 3 + D uv D vw γ u 2 γ v 3 D uv γ u γ v 2 D uv 2 γ u 2 γ v 2 D uv D vw γ u γ v 3 D uv 2 D vw γ u 2 γ v 3 + γ u γ v + γ u γ v 2 + D uv γ u 2 γ v + D uv γ u 2 γ v 2 D uw γ u γ v D uw γ u γ v 2 D uw D uv γ u 2 γ v D uw D uv γ u 2 γ v 2 + D vw γ u γ v 2 + D vw γ u γ v 3 + D vw D ux γ u 2 γ v 2 + D vw D ux γ u 2 γ v 3 D uv D vw γ u γ v 2 D uv D vw γ u γ v 3 D uv D vw D ux γ u 2 γ v 2 D uv D vw D ux γ u 2 γ v 3 γ v 2 + 1 γ u γ v 2 + γ u D vw γ v 3 + D vw γ v D vw γ u γ v 3 + D vw γ u γ v + γ v + γ u γ v + γ v 2 + γ u γ v 2 D vw γ v D vw γ u γ v D vw γ v 2 D vw γ u γ v 2 .
Applying
D ux D vw γ v ( 1 + γ v ) = γ v 2 D uv D vw + γ v D uv + γ v D vw + γ v D uw + D uw 2 γ v 1
for underline items, we infer that
D uv D vw D ux γ u 2 γ v 3 ̲ + D uv D vw D ux γ u 2 γ v 2 ̲ = D uv γ u 2 γ v D ux D vw γ v ( 1 + γ v ) = D uv 2 D vw γ u 2 γ v 3 + D uv 2 γ u 2 γ v 2 + D uv D vw γ u 2 γ v 2 + D uw D uv γ u 2 γ v 2 + D uw D uv γ u 2 γ v 2 D uv γ u 2 γ v 2 + D uv γ u 2 .
So, we have
{ · } = D vw γ u γ v 2 D vw γ u γ v 3 D ux D vw γ v ( 1 + γ v ) ( γ u 2 γ v γ u γ v ) + 2 γ u γ v 2 + D vw γ u γ v 3 D uv γ u γ v 2 D uv D vw γ u γ v 3 + γ u γ v D uw γ u γ v D uw γ u γ v 2 + D ux D vw γ v ( 1 + γ v ) γ u 2 γ v + 1 + γ u + γ v + γ u γ v = ( 1 + γ u ) ( 1 + γ v ) ,
hence we have D zy D ux D vw γ u γ v γ z = 1 .
Next, we prove that coefficient of u is A D .
We have 1 + γ z = 1 + D uv γ u γ v = D . Then we compute the coefficient of u applying the equation D zy D ux D vw γ u γ v γ z = 1 .
( 1 + γ z D zy ) ( 1 + γ u D uv ) D zy D uv ( 1 + γ z ) ( 1 + γ u ) + 1 + γ u D ux D zy D ux γ z ( 1 + γ u ) = ( 1 + γ z D zy ) ( 1 + γ u D uv ) γ z D ux + ( 1 + γ u D ux ) ( 1 + γ z ) D uv D zy D ux D uv D γ z ( 1 + γ u ) = D vw γ u γ v D D uv ( 1 + γ u ) { ( D ux γ z + D ux D uv γ u γ z + D zy D ux γ z 2 + D zy D ux D uv γ z 2 γ u ) + D uv + D uv γ z + D ux D uv γ u + D ux D uv γ u γ z } = D vw γ u γ v D D uv ( 1 + γ u ) D uv + D uv 2 γ u γ v + D ux D uv γ u ( 1 γ v ) D uv D vw D uv 2 γ u D vw = γ u γ v D ( 1 + γ u ) { D vw + D uv D vw γ u γ v 1 γ u D uv + γ u ( 1 γ v ) γ v 2 D uv D vw + γ v D uv + γ v D vw + γ v D uw + D uw 2 γ v 1 γ v ( 1 + γ v ) } = γ u γ v D ( 1 + γ u ) { γ u γ v ( γ v 1 ) D uw + ( 1 γ u ( γ v 1 ) 1 + γ v ) D vw ( 1 + γ v 1 1 + γ v ) γ u D uv + ( 1 γ v 1 1 + γ v ) γ u γ v D uv D vw 1 + γ u ( γ v 1 ) ( 2 γ v + 1 ) γ v ( 1 + γ v ) }
By D uv = 1 + u · v s 2 ,
= 1 D { 1 s 2 γ u 2 γ v + 1 ( γ u 1 ) ( u · v ) γ u 2 γ v + 1 ( γ v 1 ) + 1 s 2 γ u γ v ( 1 + γ u + γ v γ u γ v ) ( γ u + 1 ) ( γ v + 1 ) ( v · w ) + γ u γ v ( 1 + γ u + γ v γ u γ v ) ( γ u + 1 ) ( γ v + 1 ) 1 s 2 2 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) 1 s 2 2 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) + 2 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( 1 + ( u · v ) s 2 + ( v · w ) s 2 + 1 s 4 ( u · v ) ( v · w ) ) γ u γ v 1 + γ u + γ u 2 ( 2 γ v 2 γ v 1 ) ( γ u + 1 ) ( γ v + 1 ) } = 1 D 1 s 2 γ u 2 γ v + 1 ( γ v 1 ) ( u · v ) + 1 s 2 γ u γ v ( v · w ) + 2 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) ( v · w ) = A D .
Finally, we prove that the coefficient of v is B D .
1 + γ z D zy D zy D uv ( 1 + γ z ) γ u + 1 + γ v D vw D zy D ux γ z γ u ( 1 + γ v ) = ( 1 + γ z D zy ) D ux D vw γ z ( 1 + γ v ) + ( 1 + γ v D vw ) D uv ( 1 + γ z ) D zy D ux D uv D vw D γ z γ u ( 1 + γ v ) = γ v D D uv ( 1 + γ v ) { D ux D vw γ z ( 1 + γ v ) D zy D ux D vw γ z 2 ( 1 + γ v ) + D uv ( 1 + γ z ) + D uv D vw γ v ( 1 + γ z ) } .
Using (16) and D zy D ux D vw γ z = 1 γ u γ v , then we have
= γ v D D uv ( 1 + γ v ) { ( γ z γ v D uv D vw + γ z D uv + γ z D vw + γ z D uw + γ z γ v D uw γ z γ v ( 2 γ v + 1 ) ) γ z ( 1 + γ v ) γ u γ v + D uv ( 1 + γ z ) + D uv D vw γ v ( 1 + γ z ) } = 1 D D uv ( 1 + γ v ) { γ v 2 ( 1 + ( u · v ) s 2 + ( v · w ) s 2 + 1 s 4 ( u · v ) ( v · w ) ) γ u γ v 2 ( 1 + ( u · v ) s 2 + ( v · w ) s 2 + 1 s 4 ( u · v ) ( v · w ) ) γ u γ v 2 ( 1 + ( u · v ) s 2 + ( u · w ) s 2 + 1 s 4 ( u · v ) ( u · w ) ) γ u γ v ( 1 + ( u · v ) s 2 + ( u · w ) s 2 + 1 s 4 ( u · v ) ( u · w ) ) + γ u γ v ( 2 γ v + 1 ) ( 1 + ( u · v ) s 2 ) γ v 2 ( 1 + ( u · v ) s 2 ) } = γ v s 2 D ( 1 + γ v ) γ u ( γ v + 1 ) ( u · w ) + ( γ v 1 ) γ u ( v · w ) = B D .
Hence gyr [ u , v ] w = w + A u + B v D holds. By applying the left cancellation law for the Equation (14), we obtain (G3).
We prove (G4). We prove that gyr [ u , v ] is automorphism for every pair u , v V . To prove (G4), we first show the gyration preserves, the inner product of V and the norm. So, we compute
gyr [ u , v ] a · gyr [ u , v ] b = a · b
for all a , b , u , v V s . By applying the Equation (15), we have
gyr [ u , v ] a = a + A a u + B a v D
and
gyr [ u , v ] b = b + A b u + B b v D
respectively, where
A a = 1 s 2 γ u 2 γ u + 1 ( γ v 1 ) ( u · a ) + 1 s 2 γ u γ v ( v · a ) + 2 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) ( v · a ) , B a = 1 s 2 γ v γ v + 1 γ u ( γ v + 1 ) ( u · a ) + ( γ u 1 ) γ v ( v · a ) .
The terms A b and B b are defined in the similar way an A a and B b respectively. Then we have
gyr [ u , v ] a · gyr [ u , v ] b = a · b + A a ( u · b ) + B a ( v · b ) D + A b ( u · a ) + B b ( v · a ) D + 1 D 2 { A a A b u 2 + A a B b ( u · v ) + A b B a ( u · v ) + B a B b v 2 } ,
We show that terms other than a · b of the right-hand side of the Equation (18) equal to zero.
A a ( u · b ) + B a ( v · b ) = 1 s 2 γ u 2 γ u + 1 ( γ v 1 ) ( u · a ) ( u · b ) + 1 s 2 γ u γ v ( v · a ) ( u · b ) + 2 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) ( v · a ) ( u · b ) 1 s 2 γ u γ v ( u · a ) ( v · b ) 1 s 2 γ v 2 γ v + 1 ( γ u 1 ) ( v · a ) ( v · b ) . A b ( u · a ) + B b ( v · a ) = 1 s 2 γ u 2 γ u + 1 ( γ v 1 ) ( u · a ) ( u · b ) + 1 s 2 γ u γ v ( u · a ) ( v · b ) + 2 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) ( u · a ) ( v · b ) 1 s 2 γ u γ v ( v · a ) ( u · b ) 1 s 2 γ v 2 γ v + 1 ( γ u 1 ) ( v · a ) ( v · b ) . A a ( u · b ) + B a ( v · b ) + A b ( u · a ) + B b ( v · a ) = 2 s 2 γ u 2 γ u + 1 ( γ v 1 ) ( u · a ) ( u · b ) + 2 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) ( u · a ) ( v · b ) 2 s 2 γ u 2 γ u + 1 ( γ v 1 ) ( v · a ) ( u · b ) + 2 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) ( v · a ) ( v · b ) .
Then we compute each terms of the sum
A a A b u 2 + A a B b ( u · v ) + A b B a ( u · v ) + B a B b v 2 .
A a A b u 2 = u 2 { 1 s 4 γ u 4 ( γ u + 1 ) 2 ( γ v 1 ) 2 ( u · a ) ( u · b ) 1 s 4 γ u 2 γ u + 1 ( γ v 1 ) γ u γ v ( u · a ) ( v · b ) 2 s 6 γ u 2 γ u + 1 ( γ v 1 ) γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) ( u · a ) ( u · b ) 1 s 4 γ u γ v γ u 2 γ u + 1 ( γ v 1 ) ( v · a ) ( u · b ) + 1 s 4 γ u 2 γ v 2 ( v · a ) ( v · b ) + 2 s 6 γ u γ v γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) ( v · a ) ( v · b ) 2 s 6 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) γ u 2 γ u + 1 ( γ v 1 ) ( u · v ) ( v · a ) ( u · b ) + 2 s 6 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) γ u γ v ( u · v ) ( v · a ) ( v · b ) + 4 s 8 γ u 4 γ v 4 ( γ u + 1 ) 2 ( γ v + 1 ) 2 ( u · v ) 2 ( v · a ) ( v · b ) } .
By u 2 s 2 = γ u 2 1 γ u 2 this equation is rewritten in the following.
A a A b u 2 = 1 s 2 γ u 2 ( γ u + 1 ) ( γ v 1 ) 2 ( γ u 1 ) ( u · a ) ( u · b ) 1 s 2 γ u γ v ( γ u 1 ) ( γ v 1 ) ( u · a ) ( v · b ) 2 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( γ u 1 ) ( γ v 1 ) ( u · v ) ( u · a ) ( u · b ) 1 s 2 γ u γ v ( γ u 1 ) ( γ v 1 ) ( v · a ) ( u · b ) + 1 s 2 γ v 2 ( γ u 2 1 ) ( v · a ) ( v · b ) + 4 s 4 γ u γ v 3 ( γ v + 1 ) ( γ u 1 ) ( u · v ) ( v · a ) ( v · b ) 2 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( γ u 1 ) ( γ v 1 ) ( u · v ) ( v · a ) ( u · b ) ) + 4 s 6 γ u 2 γ v 4 ( γ u + 1 ) ( γ v + 1 ) 2 ( γ u 1 ) ( u · v ) 2 ( v · a ) ( v · b ) .
Then we obtain
A a B b ( u · v ) = 1 s 4 γ u 3 γ v ( γ u + 1 ) 2 ( γ v 1 ) ( u · v ) ( u · a ) ( u · b ) + 1 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v 1 ) ( γ u 1 ) ( γ v 1 ) ( u · v ) ( u · a ) ( v · b ) 1 s 4 γ u 2 γ v 2 ( u · v ) ( v · a ) ( u · b ) 1 s 4 γ u γ v 3 ( γ v + 1 ) ( γ u 1 ) ( u · v ) ( v · a ) ( v · b ) 2 s 6 γ u 3 γ v 3 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) 2 ( v · a ) ( u · b ) 2 s 6 γ u 2 γ v 4 ( γ u + 1 ) ( γ v + 1 ) 2 ( γ u 1 ) ( u · v ) 2 ( v · a ) ( v · b ) . A b B a ( u · v ) = 1 s 4 γ u 3 γ v ( γ u + 1 ) 2 ( γ v 1 ) ( u · v ) ( u · a ) ( u · b ) + 1 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v 1 ) ( γ u 1 ) ( γ v 1 ) ( u · v ) ( u · b ) ( v · a ) 1 s 4 γ u 2 γ v 2 ( u · v ) ( v · b ) ( u · a ) 1 s 4 γ u γ v 3 ( γ v + 1 ) ( γ u 1 ) ( u · v ) ( v · a ) ( v · b ) 2 s 6 γ u 3 γ v 3 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) 2 ( v · b ) ( u · a ) 2 s 6 γ u 2 γ v 4 ( γ u + 1 ) ( γ v + 1 ) 2 ( γ u 1 ) ( u · v ) 2 ( v · a ) ( v · b ) .
Calculating B a B b v 2 in a way similar to the calculation of A a A b u 2 , we have
B a B b v 2 = 1 s 2 γ u 2 ( γ v 2 1 ) ( u · a ) ( u · b ) + 1 s 2 γ u γ v ( γ u 1 ) ( γ v 1 ) ( u · a ) ( v · b ) + 1 s 2 γ u γ v ( γ u 1 ) ( γ v 1 ) ( v · a ) ( u · b ) + 1 s 2 γ v 2 γ v + 1 ( γ u 1 ) 2 ( γ v 1 ) ( v · a ) ( v · b ) .
Hence, comparing the Equations (19) with (22) we have
A a A b u 2 + A a B b ( u · v ) + A b B a ( u · v ) + B a B b v 2 = 1 s 2 γ u 2 γ u + 1 ( γ v 1 ) ( u · a ) ( u · b ) { ( γ u + 1 ) ( γ v + 1 ) + ( γ u 1 ) ( γ v 1 ) + 2 γ u γ v u · v s 2 } 1 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) ( u · a ) ( v · b ) { ( γ u + 1 ) ( γ v + 1 ) + ( γ u 1 ) ( γ v 1 ) + 2 γ u γ v u · v s 2 } + 1 s 2 γ u 2 γ u + 1 ( γ v 1 ) ( v · a ) ( u · b ) { ( γ u + 1 ) ( γ v + 1 ) + ( γ u 1 ) ( γ v 1 ) + 2 γ u γ v u · v s 2 } 1 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) ( v · a ) ( v · b { ( γ u + 1 ) ( γ v + 1 ) + ( γ u 1 ) ( γ v 1 ) + 2 γ u γ v u · v s 2 } = D ( A a ( u · b ) + B a ( v · b ) + A b ( u · a ) + B b ( v · a ) ) .
We conclude that gyr [ u , v ] a · gyr [ u , v ] b = a · b .
To prove that gyr [ u , v ] is a homomorphism for all u , v V s , we show
gyr [ u , v ] ( x E y ) = gyr [ u , v ] x E gyr [ u , v ] y
for all x , y V s . Applying (15) we have
gyr [ u , v ] ( x E y ) = x E y + 1 D ( A x E y u + B x E y v ) .
Put
x E y = 1 + γ x D xy D xy ( 1 + γ x ) x + 1 γ x D xy y = E xy x + F xy y .
By a simple calculation we infer that
A x E y = E xy A x + F xy A y B x E y = E xy B x + F xy B y .
We have
gyr [ u , v ] ( x E y ) = E xy x + F xy y + 1 D ( E xy A x + F xy A y ) u + ( E xy B x + F xy B y ) v ) = E xy x + 1 D ( A x u + B x v ) + F xy y + 1 D ( A y u + B y v ) = E xy gyr [ u , v ] x + F xy gyr [ u , v ] y .
Then the right-hand side of the Equation (24) is rewritten as the following equation.
gyr [ u , v ] x E gyr [ u , v ] y = E gyr [ u , v ] x gyr [ u , v ] y gyr [ u , v ] x + F gyr [ u , v ] x gyr [ u , v ] y gyr [ u , v ] y .
Since gyr [ u , v ] preserves the inner product and the norm of V , we have
γ gyr [ u , v ] x = γ x D gyr [ u , v ] x gyr [ u , v ] y = D xy ,
so that
E gyr [ u , v ] x gyr [ u , v ] y = E xy
and
F gyr [ u , v ] x gyr [ u , v ] y = F xy .
Hence gyr [ u , v ] ( x E y ) = gyr [ u , v ] x E gyr [ u , v ] y . We conclude that gyr [ u , v ] is a homomorphism.
We observe that gyr [ u , v ] is bijective for every pair of u , v V s . To prove this, we compute gyr [ u , v ] ( gyr [ v , u ] w ) = w for every w V s . We denote
A ab ( c ) = 1 s 2 γ a 2 γ a + 1 ( γ b 1 ) ( a · c ) + 1 s 2 γ a γ b ( b · c ) + 2 s 4 γ a 2 γ b 2 ( γ a + 1 ) ( γ b + 1 ) ( a · b ) ( b · c ) B ab ( c ) = 1 s 2 γ b γ b + 1 γ a ( γ b + 1 ) ( a · c ) + ( γ a 1 ) γ b ( b · c ) ,
where a , b , c V s . Then applying (15), we have
gyr [ u , v ] ( gyr [ v , u ] w ) = gyr [ u , v ] ( w + A vu ( w ) v + B vu ( w ) u D ) = w + A vu ( w ) v + B vu ( w ) u D + 1 D A uv ( w + A vu ( w ) v + B vu ( w ) u D ) u + B uv ( w + A vu ( w ) v + B vu ( w ) u D ) v .
We compute
A uv ( w + A vu ( w ) v + B vu ( w ) u D ) = 1 s 2 γ u 2 γ u + 1 ( γ v 1 ) ( u · ( w + A vu ( w ) v + B vu ( w ) u D ) ) + 1 s 2 γ u γ v ( v · ( w + A vu ( w ) v + B vu ( w ) u D ) ) + 2 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) ( v · ( w + A vu ( w ) v + B vu ( w ) u D ) ) = A uv ( w ) + 1 D A vu ( w ) A uv ( v ) + 1 D B vu ( w ) A uv ( u )
and
B uv ( w + A vu ( w ) v + B vu ( w ) u D ) = 1 s 2 γ v γ v + 1 { γ u ( γ v + 1 ) ( u · ( w + A vu ( w ) v + B vu ( w ) u D ) ) + ( γ u 1 ) γ v ( v · ( w + A vu ( w ) v + B vu ( w ) u D ) ) } = B uv ( w ) + 1 D A vu ( w ) B uv ( v ) + 1 D B vu ( w ) B uv ( u ) .
So, we have
gyr [ u , v ] ( gyr [ v , u ] w ) = w + B vu ( w ) D + A uv ( w ) D + 1 D 2 ( A vu ( w ) A uv ( v ) + B vu ( w ) A uv ( u ) ) u + A vu ( w ) D + B uv ( w ) D + 1 D 2 ( A vu ( w ) B uv ( v ) + B vu ( w ) B uv ( u ) ) v .
We show that the coefficients of u and v vanish. We compute the coefficient of u .
B vu ( w ) + A uv ( w ) = 2 s 2 γ u 2 γ u + 1 ( γ v 1 ) ( u · w ) + 2 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) ( v · w ) .
We also have
A vu ( w ) A uv ( v ) + B vu ( w ) A uv ( u ) = 1 s 2 γ v 2 γ v + 1 ( γ u 1 ) ( v · w ) 1 s 2 γ u 2 γ u + 1 ( γ v 1 ) ( u · v ) 1 s 2 γ v 2 γ v + 1 ( γ u 1 ) ( v · w ) 1 s 2 γ u γ v v 2 1 s 2 γ v 2 γ v + 1 ( γ u 1 ) ( v · w ) 2 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) v 2 1 s 2 γ u γ v ( u · w ) γ u 2 γ u + 1 ( γ v 1 ) ( u · v ) + 1 s 2 γ u γ v ( u · w ) 1 s 2 γ u γ v v 2 + 1 s 2 γ u γ v ( u · w ) 2 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) v 2 2 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) ( v · w ) 1 s 2 γ u 2 γ u + 1 ( γ v 1 ) ( u · v ) + 2 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) ( v · w ) 1 s 2 γ u γ v v 2 + 2 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) ( v · w ) 2 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) v 2 + 1 s 2 γ u γ v ( v · w ) 1 s 2 γ u 2 γ u + 1 ( γ v 1 ) u 2 1 s 2 γ u γ v ( v · w ) 1 s 2 γ u γ v ( u · v ) 1 s 2 γ u γ v ( v · w ) 2 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) 2 + 1 s 2 γ u 2 γ u + 1 ( γ v 1 ) ( u · w ) γ u 2 γ u + 1 ( γ v 1 ) u 2 1 s 2 γ u 2 γ u + 1 ( γ v 1 ) ( u · w ) 1 s 2 γ u γ v ( u · v ) 1 s 2 γ u 2 γ u + 1 ( γ v 1 ) ( u · w ) 2 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) 2 .
Applying the gamma identity, we have the following.
A vu ( w ) A uv ( v ) + B vu ( w ) A uv ( u ) = 1 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) ( v · w ) γ u γ v + γ u + γ v 1 2 γ u γ v ( u · v ) s 2 1 s 2 γ u γ v ( v · w ) γ u γ v ( u · v ) s 2 + 1 s 2 γ u 2 ( γ v 2 1 ) + γ u 2 ( γ u 1 ) ( γ v 1 ) 2 γ u + 1 + 2 s 4 γ u 2 γ u + 1 ( γ v 1 ) γ u γ v ( u · v ) ( u · w ) = 1 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) ( v · w ) ( γ u 1 ) ( γ v 1 ) ( γ u + 1 ) ( γ v + 1 ) 2 γ u γ v ( u · v ) s 2 + 2 s 4 γ u 2 γ u + 1 ( γ v 1 ) ( u · w ) ( γ u γ v + 1 ) + γ u γ v ( u · v ) s 2 = D ( B vu ( w ) + A uv ( w ) ) .
So, the coefficient of u vanishes.
We compute the coefficient of v .
A vu ( w ) + B uv ( w ) = 2 s 2 γ v 2 γ v + 1 ( γ u 1 ) ( v · w ) + 2 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) ( u · w ) .
We also have
A vu ( w ) B uv ( v ) + B vu ( w ) B uv ( u ) = 1 s 2 γ v 2 γ v + 1 ( γ u 1 ) ( v · w ) 1 s 2 γ u γ v ( u · v ) + 1 s 2 γ v 2 γ v + 1 ( γ u 1 ) ( v · w ) 1 s 2 γ v 2 γ v + 1 ( γ u 1 ) v 2 1 s 2 γ u γ v ( u · w ) 1 s 2 γ u γ v ( u · v ) 1 s 2 γ u γ v ( u · w ) 1 s 2 γ v 2 γ v + 1 ( γ u 1 ) v 2 2 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) ( u · w ) 1 s 2 γ u γ v ( u · v ) 2 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) ( u · w ) 1 s 2 γ v 2 γ v + 1 ( γ u 1 ) v 2 + 1 s 2 γ u γ v ( v · w ) 1 s 2 γ u γ v u 2 + 1 s 2 γ u γ v ( v · w ) 1 s 2 γ v 2 γ v + 1 ( γ u 1 ) ( u · v ) + 1 s 2 γ u 2 γ u + 1 ( γ v 1 ) ( u · w ) 1 s 2 γ u γ v u 2 + 1 s 2 γ u 2 γ u + 1 ( γ v 1 ) ( u · w ) 1 s 2 γ v 2 γ v + 1 ( γ u 1 ) ( u · v ) = 2 s 2 γ v 2 γ v + 1 ( γ u 1 ) ( v · w ) γ u γ v ( u · v ) s 2 + ( γ u 1 ) ( γ v 1 ) + ( γ u + 1 ) ( γ v + 1 ) 2 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) ( u · w ) { ( γ u + 1 ) ( γ v + 1 ) + ( γ u 1 ) ( γ v 1 ) + γ u γ v ( u · v ) s 2 } = D ( A vu ( w ) + B uv ( w ) ) .
So, the coefficient of v vanishes. Thus, gyr [ u , v ] ( gyr [ v , u ] w ) = w holds for every w V s . Changing u and v , gyr [ v , u ] ( gyr [ u , v ] w ) = w also holds for every w V s . We conclude that gyr [ u , v ] is bijective. Thus, gyr [ u , v ] is an automorphism; a proof of (G4) is complete.
To prove (G5) we first observe gyr [ u E v , v ] = gyr [ u , v ] for every pair u , v V s . Let u , v , w V s . Applying (15) we have that
gyr [ u E v , v ] w = w + A u E v + B v D = w + A D 1 + γ u D uv D uv ( 1 + γ u ) u + 1 D A D uv γ u + B v ,
where
A = 1 s 2 γ u E v 2 ( γ u E v + 1 ) ( γ v 1 ) ( u E v · w ) + 1 s 2 γ u E v γ v ( v · w ) + 2 s 4 γ u E v 2 γ v 2 ( γ u E v + 1 ) ( γ v + 1 ) ( u E v · v ) ( v · w ) B = 1 s 2 γ v γ v + 1 γ u E v ( γ v + 1 ) ( u E v · w ) + ( γ u E v 1 ) γ v ( v · w ) D = γ ( u E v ) E v + 1 = γ u E v γ v ( 1 + ( u E v ) · v s 2 ) + 1 .
We prove that gyr [ u E v , v ] w = gyr [ u , v ] w for every u , v , w V s . We have
( u E v ) · w = 1 + γ u D uv D uv ( 1 + γ u ) ( u · w ) + 1 D uv γ u ( v · w ) , ( u E v ) · v = 1 + γ u D uv D uv ( 1 + γ u ) ( u · v ) + 1 D uv γ u v 2 , γ u E v + 1 = D .
Then A is computed as in the following.
A = 1 γ u E v + 1 { γ u E v 2 ( γ v 1 ) 1 + γ u D uv D uv ( 1 + γ u ) ( u · w ) s 2 γ u E v 2 ( γ v 1 ) 1 D uv γ u ( v · w ) s 2 + 1 s 2 γ u E v γ v ( γ u E v + 1 ) ( v · w ) + 2 s 4 γ u E v 2 γ v 2 ( γ v + 1 ) 1 + γ u D uv D uv ( 1 + γ u ) ( u · v ) ( v · w ) + 2 s 4 γ u E v 2 γ v 2 ( γ v + 1 ) 1 D uv γ u ( v · w ) v 2 } = 1 D { γ u 2 γ v 2 D uv 2 ( γ v 1 ) 1 + γ u D uv D uv ( 1 + γ u ) ( u · w ) s 2 γ u 2 γ v 2 D uv 2 ( γ v 1 ) 1 D uv γ u ( v · w ) s 2 + 1 s 2 γ u γ v 2 D uv ( γ u γ v D uv + 1 ) ( v · w ) + 2 s 4 γ u 2 γ v 4 D uv 2 ( γ v + 1 ) 1 + γ u D uv D uv ( 1 + γ u ) ( u · v ) ( v · w ) + 2 s 2 γ u 2 γ v 2 D uv 2 ( γ v + 1 ) 1 D uv γ u γ v 2 1 γ v 2 ( v · w ) } = 1 D { D uv γ v 2 ( 1 + γ u D uv ) 1 s 2 γ u 2 γ u + 1 ( γ v 1 ) ( u · w ) + D uv γ v 2 ( 1 + γ u D uv ) 1 s 2 γ u γ v ( v · w ) + D uv γ v 2 ( 1 + γ u D uv ) 2 s 4 γ u 2 γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) ( v · w ) } = D uv γ v 2 ( 1 + γ u D uv ) A D .
D can be computed as in the following.
D = γ u γ v 2 D uv 1 + 1 + γ u D uv D uv ( 1 + γ u ) ( D uv 1 ) + 1 D uv γ u γ v 2 1 γ v 2 + 1 = γ u γ v 2 ( 2 D uv 1 + γ u D uv 2 ) 1 + γ u + γ v 2 = γ v 2 ( 1 + γ u D uv ) 2 1 + γ u
Hence A D 1 + γ u D uv D uv ( 1 + γ u ) = A D .
Next, we compute the coefficient of v . B can be computed as in the following.
B = 1 s 2 γ v γ v + 1 { γ u γ v D uv ( γ v + 1 ) 1 + γ u D uv D uv ( 1 + γ u ) ( u · w ) + γ u γ v D uv ( γ v + 1 ) 1 D uv γ u ( v · w ) + ( γ u γ v D uv 1 ) γ v ( v · w ) } = 1 s 2 γ v 2 ( 1 + γ u D uv ) γ u 1 + γ u ( u · w ) + γ v 1 + γ v ( v · w ) .
The Equation (25) is rewritten by A 1 D uv γ u = A D γ v 2 ( 1 + γ u D uv ) γ u . Then
A 1 D uv γ u + B = γ v 2 ( 1 + γ u D uv ) D { 1 s 2 γ u γ u + 1 ( γ v 1 ) ( u · w ) + 1 s 2 γ v ( v · w ) + 2 s 4 γ u γ v 2 ( γ u + 1 ) ( γ v + 1 ) ( u · v ) ( v · w ) D 1 s 2 γ u 1 + γ u ( u · w ) + 1 s 2 γ v 1 + γ v ( v · w ) } = γ v 2 ( 1 + γ u D uv ) D ( 1 + γ u ) { 1 s 2 ( γ u ( γ v 1 ) + ( γ u γ v D uv + 1 ) γ u ) ( u · w ) + 1 s 2 γ v ( 1 + γ u ) ( γ u γ v D uv + 1 ) γ v 1 + γ v ( 1 + γ u ) ( v · w ) + 2 s 2 γ u γ v 2 ( γ v + 1 ) ( D uv 1 ) ( v · w ) } = γ v 2 ( 1 + γ u D uv ) D ( 1 + γ u ) { 1 s 2 γ u γ v ( 1 + γ u D uv ) ( u · w ) + 1 s 2 γ v 2 1 + γ v ( 1 + γ u D uv ) ( 1 γ u ) ( v · w ) } = D D B . <