3 Finitely Generated Gyrovector Subspaces Let (G, ⊕, ⊗) Be a Gyrovector Space

Accepted Manuscript

Finitely generated gyrovector subspaces and orthogonal gyrodecomposition in the Möbius gyrovector space

Toshikazu Abe, Keiichi Watanabe

PII: S0022-247X(16)30729-6 DOI: http://dx.doi.org/10.1016/j.jmaa.2016.11.039 Reference: YJMAA 20892

To appear in: Journal of Mathematical Analysis and Applications

Received date: 3 October 2016

Please cite this article in press as: T. Abe, K. Watanabe, Finitely generated gyrovector subspaces and orthogonal gyrodecomposition in the Möbius gyrovector space, J. Math. Anal. Appl. (2017), http://dx.doi.org/10.1016/j.jmaa.2016.11.039

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Toshikazu Abe and Keiichi Watanabe

Toshikazu Abe: College of Engineering, Ibaraki University, Hitachi, Ibaraki 316- 8511, Japan E-mail address: [email protected] Keiichi Watanabe: Department of Mathematics, Faculty of Science, Niigata Uni- versity, Niigata 950–2181, Japan E-mail address: [email protected]

Key words and phrases. M¨obius gyrogroup, M¨obius gyrovector space

Mathematical Subject Classiﬁcation (2010). Primary 51M10; Secondary 20N05, 46C99, 51P05.

Abstract. We will show that any finitely generated gyrovector subspace in the Möbius gyrovector space coincides with the intersection of the vector subspace generated by the same generators and the Möbius ball. As an application, we will present a notion of orthogonal gyrodecomposition and clarify the relation to the orthogonal decomposition.

1 Introduction A gyrogroup (resp. gyrocommutative gyrogroup) is one of the most natural generalization of a group (resp. commutative group). Ungar has intensively stud- ied gyrogroups since 1988. The first known gyrogroup was the ball of Euclidean space R3 endowed with Einstein’s velocity addition associated with the special theory of relativity. Another example of a gyrogroup is the complex unit disc D = {a ∈ C : |a| < 1} endowed with the Möbius addition. Ungar extended the Möbius addition in the complex disc to the ball of an arbitrary real inner product space, and observed that the ball endowed with the Möbius addition turns out to be a gyrocommutative gyrogroup. Some gyrocommutative gyrogroups admit scalar multiplication, turning them- selves into gyrovector spaces. The latter, in turn, are analogous to vector spaces just as gyrogroups are analogous to groups. Indeed, gyrovector spaces provide

1 the setting for hyperbolic geometry just as vector spaces provide setting for Eu- clidean geometry. In particular, Möbiusgyrovector spaces form the setting for the Poincaréball model of hyperbolic geometry while, similarly, Einstein gyrovector spaces form the setting for the Beltrami-Klein ball model. The readers may consult with [U1] and references therein for general informations of gyrovector spaces. In [AH], the first author and Hatori introduced a notion of generalized gyrovector spaces (GGVs in short) in a way analogous to the generalization of inner product spaces into normed spaces. Accordingly, the notion of a GGV is a common generalization of the notion of a real normed space and the notion of a gyrovector space. Then, they showed that the gyrocommutative gyrogroup of all positive in- vertible elements in a unital C∗-algebra is indeed a GGV without using the theory of K-loops, and presented a Mazur-Ulam theorem for GGVs as well. It still continues to speculate more satisfactory axioms for generalizations of gyrovector spaces. Since gyro-operations are not commutative, associative or dis- tributive in general, lots of elementary problems remain unsolved. Very recently, the first author raised the following question in a conference (cf. [A]).

Question 1.1. Let (G, ⊕, ⊗) be a gyrovector space, or any kind of generalization of gyrovector space and a1, a2 ∈ G. Can we have

{r1 ⊗a1 ⊕ r2 ⊗a2 ; r1,r2 ∈ R} = {λ2 ⊗a2 ⊕ λ1 ⊗a1 ; λ1,λ2 ∈ R} ?

r⊗(r1 ⊗a1 ⊕ r2 ⊗a2) ∈{λ1 ⊗a1 ⊕ λ2 ⊗a2 ; λ1,λ2 ∈ R} ?

Such problems seem to be quite fundamental and important for developing the theory of generalization of gyrovector spaces, since one of the virtue of gyrovector spaces is that they have the nature which are fully analogous to vector spaces. In this paper we will consider this problem and its natural extension in the M¨obius gyrovector space (Vs, ⊕ , ⊗ ). M M

2 Gyrovector subspaces generated by two elements Assume s = 1 without loss of generality. The M¨obius addition and the M¨obius scalar multiplication are given by the equations

(1 + 2a·b + ||b||2) a +(1−||a||2) b a ⊕ b = M 1+2a·b + ||a||2||b||2 a r ⊗ a =tanh r tanh−1 ||a|| (if a = 0),r⊗ 0 = 0 M ||a|| M

2 for any a, b ∈ V1,r∈ R. We simply denote ⊕ , ⊗ by ⊕, ⊗, respectively. If ⊕ and M M ⊗ appear in a formula simultaneously, we always give priority to the operation ⊗, that is,

r1 ⊗a1 ⊕ r2 ⊗a2 =(r1 ⊗a1) ⊕ (r2 ⊗a2), and the parentheses will be omitted in such cases. Note that, in general, gyroaddition does not distribute with (gyro) scalar multiplication ([U1, (6.5)]): r⊗(a ⊕ b) = r⊗a ⊕ r⊗b. In this section, we will give an aﬃrmative answer to Question 1.1 for the M¨obius gyrovector space via the equation

{r1 ⊗a1 ⊕ r2 ⊗a2 ; r1,r2 ∈ R} = {λ1a1 + λ2a2; λ1,λ2 ∈ R}∩V1 for a1, a2 ∈ V1. Indeed, one of the inclusion (⊂) in the above equation is obvious, since r1 ⊗a1 ⊕ r2 ⊗a2 is a linear combination of a1, a2 by the deﬁnitions of ⊕, ⊗, and it belongs to V1 which is closed under these operations. The following theorem asserts that the other inclusion (⊃) holds as well.

Theorem 2.1. Let (V1, ⊕, ⊗) be the Mo¨bius gyrovector space and 0 = a1, a2 ∈ V1. a1 a2 Put α = · . Suppose that 0 = t1,t2 ∈ R satisfy the condition ||a1|| ||a2|| a1 a2 t1 + t2 < 1. ||a1|| ||a2||

(I) If 2αt2 + t1 =0 , we put 2 2 2 2 2 2 t1 +2αt1t2 + t2 +1− (t1 +2αt1t2 + t2 +1) − 8αt1t2 − 4t1 c1 = 2(2αt2 + t1) 2 2 2 2 2 2 t1 +2αt1t2 + t2 − 1+ (t1 +2αt1t2 + t2 +1) − 8αt1t2 − 4t1 c2 = . 2t2

(II) If 2αt2 + t1 =0,weput t c 1 1 = 2 t2 +1

c2 = t1.

Then we have 0 < |c1|, |c2| < 1 and a1 a2 t1 + t2 = r1 ⊗a1 ⊕ r2 ⊗a2, ||a1|| ||a2|| where −1 −1 tanh c1 tanh c2 r1 = −1 and r2 = −1 . tanh ||a1|| tanh ||a2||

3 In order to prove this theorem, we should express the element r1 ⊗a1 ⊕ r2 ⊗a2 as a linear combination of a1, a2 by the deﬁnition of operations ⊕, ⊗:

−1 a1 −1 a2 r1 ⊗a1 ⊕ r2 ⊗a2 = tanh(r1 tanh ||a1||) ⊕ tanh(r2 tanh ||a2||) ||a1|| ||a2||

a1 a2 = c1 ⊕ c2 ||a1|| ||a2||

2 a1 2 a2 (1+2c1c2α + c2 ) c1 +(1− c1 ) c2 ||a1|| ||a2|| . = 2 2 1+2c1c2α + c1 c2

Thus the problem is that, for t1,t2 satisfying the assumption of Theorem 2.1, whether there exists a pair c1,c2 which satisﬁes the conditions |c1|, |c2| < 1and

2 a1 2 a2 a a (1 + 2c1c2α + c2 ) c1 +(1− c1 ) c2 t 1 t 2 ||a1|| ||a2|| . 1 + 2 = 2 2 ||a1|| ||a2|| 1+2c1c2α + c1 c2 If this is the case, then we would have 2 2 2 1+2c1c2α + c1 c2 t1 = 1+2c1c2α + c2 c1 (1) 2 2 2 1+2c1c2α + c1 c2 t2 =(1− c1 )c2. (2)

The equation (2) yields

2 2 1 2 1+2c1c2α + c1 c2 = (1 − c1 )c2, (3) t2 from which we have t1 2 2 (1 − c1 )c2 = 1+2c1c2α + c2 c1 (4) t2

t1 1 by substituting (3) to (1). Putting β = ,γ= , then the equations (3) and (4) t2 t2 can be rewritten to a system of equations:

2 2 2 c1 c2 +(γc1 +2αc1 − γ)c2 +1=0

2 2 c1c2 + ((2α + β)c1 − β)c2 + c1 =0.

Therefore, we should consider the system of equations (5) and (6) of Theorem 2.4 in the sequel. We need two inequalities which are satisﬁed by the coeﬃcients of the system of equations.

Lemma 2.2. Suppose that 0 = t1,t2 ∈ R satisfy the condition a1 a2 t1 + t2 < 1. ||a1|| ||a2||

4 If we put

a1 a2 t1 1 α = · ,β= ,γ= ||a1|| ||a2|| t2 t2 as before, then we have the inequality

1+β(2α + β) <γ2.

Proof. It is easy to see that t t β α β 1 α 1 1 t 2 αt t t 2 1+ (2 + )=1+ 2 + = 2 2 +2 1 2 + 1 t2 t2 t2 a a 2 1 · t 1 t 2 < 1 γ2. = 2 1 + 2 2 = t2 ||a1|| ||a2|| t2

Lemma 2.3. Suppose that α, β, γ ∈ R, −1 ≤ α ≤ 1 and 1+β(2α + β) <γ2. Then we have the inequality

4(α + β)2γ2 < (1 + β(2α + β)+γ2)2.

Proof. Note that 1 + β(2α + β)=(α + β)2 +1− α2 ≥ 0. It follows from Arithmetic-Geometric mean inequality and its equality condition that {1+β(2α + β)} + γ2 {1+β(2α + β)}·γ2 < . 2 Thus we have

4(α + β)2γ2 ≤ 4{1+β(2α + β)}γ2 < (1 + β(2α + β)+γ2)2 as desired.

Theorem 2.4. Consider the following system of equations for real numbers: ⎧ ⎨x2y2 +(γx2 +2αx − γ)y +1=0 (5) ⎩ xy2 + ((2α + β)x2 − β)y + x =0 (6)

Suppose that −1 ≤ α ≤ 1,β=0 and 1+β(2α + β) <γ2.

(I) If 2α + β =0 , then 1+β(2α + β)+γ2 − (1 + β(2α + β)+γ2)2 − 4(2α + β)βγ2 x = (7) 2(2α + β)γ 1+β(2α + β) − γ2 + (1 + β(2α + β)+γ2)2 − 4(2α + β)βγ2 y = (8) 2γ

5 is a unique pair of the solution of the system of the equations (5), (6) which satisﬁes 0 < |x|, |y| < 1. Moreover, 1+β(2α + β)+γ2 + (1 + β(2α + β)+γ2)2 − 4(2α + β)βγ2 x = (9) 2(2α + β)γ 1+β(2α + β) − γ2 − (1 + β(2α + β)+γ2)2 − 4(2α + β)βγ2 y = (10) 2γ is a unique pair of the solution of the system of the equations (5), (6) which satisﬁes |x|, |y| > 1.

(II) If 2α + β =0, then βγ x = 1+γ2

y 1 = γ is a unique pair of the solution of the system of the equations (5), (6) which satisﬁes 0 < |x|, |y| < 1.

Proof. Suppose that a pair x, y is a solution of the system of the equations (5), (6), and that |x|= 1. By the assumption β =0,itiseasytoseethat x =0.

(I) The case of 2α + β = 0. Obviously we have − α β x2 β y − x y2 ( (2 + ) + ) = x by (6). Substituting it to (5), a simple calculation shows that

{(2α + β)x(1 − x2) − γ(1 − x2)}y +(1− x2)=0, which yields that 1 y = (11) −(2α + β)x + γ together with the assumption |x|= 1. Substituting (11) to (6), since the terms of x3 cancel each other, we obtain the following quadratic equation:

(2α + β)γx2 − (1 + β(2α + β)+γ2)x + βγ =0, so that we have 1+β(2α + β)+γ2 ± (1 + β(2α + β)+γ2)2 − 4(2α + β)βγ2 x = . 2(2α + β)γ

6 Using (11) again, it is straightforward to deduce that 1+β(2α + β) − γ2 ∓ (1 + β(2α + β)+γ2)2 − 4(2α + β)βγ2 y = . 2γ All of these x and y are real numbers, and comparing their absolute values with 1 is one of the most crucial points in the article. The inside of the square root in the above formulae can be rewritten as follows:

2 (1 + β(2α + β)+γ2)2 − 4(2α + β)βγ2 = 1+β(2α + β) − γ2 +4γ2 (12)

For y in the formula (8), |y| < 1 is obviously equivalent to that

2 1+β(2α + β) − γ2 + (1 + β(2α + β)+γ2)2 − 4(2α + β)βγ2 < 4γ2.

A straightforward calculation shows that the above inequality is equivalent to the following inequality: 2 (1 + β(2α + β) − γ2) (1 + β(2α + β) − γ2)+ (1 + β(2α + β) − γ2) +4γ2 < 0, which actually holds, since the 1st factor is negative and the 2nd factor is positive. Thus y in the formula (8) satisﬁes that |y| < 1.

Next, we should consider x in the formula (7). |x| < 1 is obviously equivalent to that 2 1+β(2α + β)+γ2 − (1 + β(2α + β)+γ2)2 − 4(2α + β)βγ2 < 4(2α + β)2γ2.

It is immediate to see that the above inequality is equivalent to the next one:

(1 + β(2α + β)+γ2)2 − 2(2α + β)βγ2 − 2(2α + β)2γ2 < (1 + β(2α + β)+γ2) (1 + β(2α + β)+γ2)2 − 4(2α + β)βγ2. (13)

If the left-hand side of the above inequality is negative, then the inequality holds, since the right-hand side is positive. If the left-hand side of the above inequality is non-negative, then the inequality is equivalent to the next one:

2 (1 + β(2α + β)+γ2)2 − 2(2α + β)βγ2 − 2(2α + β)2γ2 < (1 + β(2α + β)+γ2)2 (1 + β(2α + β)+γ2)2 − 4(2α + β)βγ2 .

An elementary calculation shows that the above inequality is equivalent to the inequality

{β +(2α + β)}2γ2 − (1 + β(2α + β)+γ2)2 < 0,

7 which is valid by Lemma 2.3. Therefore, x in the formula (7) satisﬁes that |x| < 1.

Now, we should show that x, y in the formulae (9), (10) respectively satisfy that |x|, |y| > 1. By the assumption 1 + β(2α + β) <γ2 and the formula (12), we have 1+β(2α + β) − γ2 − (1 + β(2α + β)+γ2)2 − 4(2α + β)βγ2 < −2|γ|, which yields |y| > 1. For x in the formula (9), it is straightforward to deduce that 1 1+β(2α + β)+γ2 − (1 + β(2α + β)+γ2)2 − 4(2α + β)βγ2 = , x 2βγ and so |x| > 1 is obviously equivalent to that

2 1+β(2α + β)+γ2 − (1 + β(2α + β)+γ2)2 − 4(2α + β)βγ2 < 4β2γ2.

It is immediate to see that the above inequality is equivalent to the following:

2 1+β(2α + β)+γ2 − 4(α + β)βγ2 < 1+β(2α + β)+γ2 (1 + β(2α + β)+γ2)2 − 4(2α + β)βγ2 (14)

2 { 1+β(2α + β)+γ2 − 4(α + β)βγ2}2

2 < 1+β(2α + β)+γ2 {(1 + β(2α + β)+γ2)2 − 4(2α + β)βγ2}

An elementary calculation shows that the above inequality is equivalent to the inequality

2 4(α + β)2γ2 < 1+β(2α + β)+γ2 which is valid by Lemma 2.3. Therefore, x in the formula (9) satisﬁes that |x| > 1.

Conversely, it is not diﬃcult to verify that two pairs (7), (8) and (9), (10) satisfy the system of the equations (5), (6) via (11).

(II) The case of 2α + β = 0. Then the system of the equations (5), (6) becomes x2y2 +(γx2 +2αx − γ)y +1=0

xy2 − βy + x =0.

8 βy − x y2 The 2nd equation requires that = x . By substituting it to the 1st equa- 1 βγ tion, we have y = and x = . By the assumption, we have 1 <γ2, hence γ 1+γ2 |y| < 1. Moreover, it follows from β = −2α and γ = ±1that|βγ|≤2|γ| < 1+γ2, hence |x| < 1. βγ 1 Conversely, it is easy to verify that the pair , satisﬁes the system of the 1+γ2 γ equations (5), (6). This completes the proof.

Remark 2.5. Suppose that the pair of real numbers x, y satisﬁes the system of the equations (5), (6) and |x| = 1. Then the system reduces to one equation

y2 +2αy +1=0 or y2 − 2αy +1=0, and each of them has a solution y of real number only if α = ±1. If α = 1, then two pairs (1, −1) and (−1, 1) are all of the real solutions of the system of the equations (5), (6). If α = −1, then two pairs (1, 1) and (−1, −1) are all of the real solutions of the system of the equations (5), (6).

Remark 2.6. The case that each left-hand side of (13) or (14) is negative actually exists. The case that it is non-negative actually exists too.

Corollary 2.7. In the Mo¨bius gyrovector space, we have

{r1 ⊗a1 ⊕ r2 ⊗a2 ; r1,r2 ∈ R} = {λ2 ⊗a2 ⊕ λ1 ⊗a1 ; λ1,λ2 ∈ R}

r⊗(r1 ⊗a1 ⊕ r2 ⊗a2) ∈{λ1 ⊗a1 ⊕ λ2 ⊗a2 ; λ1,λ2 ∈ R}, for a1, a2 ∈ V1,r,r1,r2 ∈ R.

Proof. It immediately follows from Theorem 2.1 and its preceding argument.

3 Finitely generated gyrovector subspaces Let (G, ⊕, ⊗) be a gyrovector space. A nonempty subset M of G is a gyrovector subspace of G if M is closed under gyrovector space addition and scalar multiplication, that is, a, b ∈ M and r ∈ R imply a ⊕ b ∈ M and r⊗a ∈ M.

9 For any nonempty subset A of G, the intersection of all gyrovector subspaces G A A of which contain is said to be the gyrovector subspace generated by ,and denoted by g A,thatis, gA = {M; A ⊂ M, M is a gyrovector subspace of G} .

Let n be a natural number and a1, ··· , an ∈ G.Put

A0 = {a1, a2, ··· , an} A {a ⊕ b,r⊗a a, b ∈∪k A ,r ∈ R} k+1 = ; j=0 j for any nonnegative integer k. We may describe the gyrovector subspace generated by ﬁnitely many elements {a1, ··· , an} in terms of Ak.

Proposition 3.1. ∞ g {a1, ··· , an} = Ak k=1 ∞ Proof. Let a, b ∈ Ak. There exist natural numbers k1,k2 satisfying that k=1 a ∈ A , b ∈ A k ≤ k A k1 k2 . We may assume 1 2. It follows from the deﬁnition of k ∞ ∞ a ⊕ b,r⊗a ∈ A ⊂ A A that k2 +1 k.Thus k is a gyrovector subspace which k=1 k=1 contains a1, ··· , an,soweobtain ∞ g {a1, ··· , an}⊂ Ak. k=1

On the other hand, if M is a gyrovector subspace which contains a1, ··· , an,then it is obvious that Ak ⊂ M for every nonnegative integer k. Hence ∞ g {a1, ··· , an}⊃ Ak. k=1 This completes the proof.

When we calculate some terms of G combined by gyroaddition ⊕,wehaveto repeat gyroaddition ⊕ according to a prescribed order by parentheses. Starting at a speciﬁed pair of two elements next to each other, add the result to another next element, and so on. For example, let n =4and(i1,i2,i3,i4)=(1, 4, 2, 3).

10 If we write parentheses into the formula c1 ⊕ c4 ⊕ c2 ⊕ c3 to specify the order of gyroaddition, there are 5 possibilities as follows:

c1 ⊕{c4 ⊕ (c2 ⊕ c3)}

(c1 ⊕ c4) ⊕ (c2 ⊕ c3)

c1 ⊕{(c4 ⊕ c2) ⊕ c3}

{c1 ⊕ (c4 ⊕ c2)}⊕c3

{(c1 ⊕ c4) ⊕ c2}⊕c3

Now we should introduce a notation.

Deﬁnition 3.2. Let (i1, ··· ,in) be a permutation of (1, ··· ,n). Then

c ⊕···⊕c “ i1 in ”

c ⊕···⊕c is deﬁned to be the formula i1 in to which a speciﬁc order of gyroaddition is given by parentheses.

The next theorem says that any finitely generated gyrovector subspace in the Möbius gyrovector space coincides with the intersection of the vector subspace generated by the same generators and the Möbius ball.

Theorem 3.3. Let (V1, ⊕, ⊗) be the Mo¨bius gyrovector space, a1, ··· , an ∈ V1 and let (i1, ··· ,in) be a permutation of (1, ··· ,n). For an arbitrary given order of c ⊕···⊕c gyroaddition of i1 in , g{a , ··· , a } { r ⊗a ⊕···⊕r ⊗a r , ··· ,r ∈ R} 1 n = “ i1 i1 in in ”; i1 in a1 an = t1 + ···+ tn ; t1, ··· ,tn ∈ R ∩ V1 ||a1|| ||an||

Proof. By the deﬁnitions of ⊕, ⊗ and the fact that (V1, ⊕, ⊗) is a gyrovector space, it is obvious that

{ r ⊗a ⊕···⊕r ⊗a r , ··· ,r ∈ R} “ i1 i1 in in ”; i1 in g ⊂ {a1, ··· , an} a1 an ⊂ t1 + ···+ tn ; t1, ··· ,tn ∈ R ∩ V1 ||a1|| ||an||

11 For n = 2, the converse inclusion holds by Theorem 2.1. Assume that theorem is valid up to n,thenumberofai’s. Let a1, ··· , an+1 ∈ V1, (i1, ··· ,in+1)bea permutation of (1, ··· ,n+ 1) and let t1, ··· ,tn+1 ∈ R satisfy that

a1 an+1 t1 + ···+ tn+1 ∈ V1. ||a1|| ||an+1|| c ⊕···⊕c Suppose that an order of gyroaddition of i1 in+1 is given by parentheses. The last operation has a form of c ⊕···⊕c ⊕ c ⊕···⊕c , ( i1 im ) im+1 in+1

c ⊕ ··· ⊕ c c ⊕···⊕c and the order of operation of i1 im (resp. im+1 in+1 )isthe c ⊕···⊕c restriction of that of i1 in+1 . ai We can rearrange the usual vector addition of ti as follows: ||ai||

a1 an+1 t1 + ···+ tn+1 ||a1|| ||an+1|| a a a a = t i1 + ···+ t im + t im+1 + ···+ t in+1 , i1 ||a || im ||a || im+1 ||a || in+1 ||a || i1 im im+1 in+1 and we put a a a = t i1 + ···+ t im , i1 ||a || im ||a || i1 im a a b = t im+1 + ···+ t in+1 . im+1 ||a || in+1 ||a || im+1 in+1 a b Obviously, we may assume a, b = 0, and put a = , b = . Then, by 2||a|| 2||b|| the hypothesis of the induction, we can express a and b as

a r ⊗a ⊕···⊕r ⊗a = i1 i1 im im b r ⊗a ⊕···⊕r ⊗a , = im+1 im+1 in+1 in+1

a b c ⊕···⊕c and the order of gyroaddition of (resp. ) coincides with that of i1 im c ⊕···⊕c (resp. im+1 in+1 ). Obviously, we have a a a b t 1 ··· t n+1 a b ||a|| ||b|| . 1 + + n+1 = + = + ||a1|| ||an+1|| ||a || ||b || The assumption that the norm of this element is less than 1 and the hypothesis of

12 the induction up to n, we can see that there exist ra,rb ∈ R satisfying a b ||a|| + ||b|| = ra ⊗a ⊕ rb ⊗b ||a|| ||b|| r ⊗ r ⊗a ⊕···⊕r ⊗a ⊕ r ⊗ r ⊗a ⊕···⊕r ⊗a = a ( i1 i1 im im ) b im+1 im+1 in+1 in+1 r ⊗a ⊕···⊕r ⊗a ⊕ r ⊗a ⊕···⊕r ⊗a = i1 i1 im im im+1 im+1 in+1 in+1 r ⊗a ⊕···⊕r ⊗a , =“ i1 i1 in+1 in+1 ” r , ··· ,r where i1 in+1 are some suitable real numbers. This expression coincides with c ⊕···⊕c the given order of gyroaddition of i1 in+1 at the beginning. This completes the proof.

Corollary 3.4. Let (V1, ⊕E, ⊗E) be the Einstein gyrovector space, a1, ··· , an ∈

V1 and let (i1, ··· ,in) be a permutation of (1, ··· ,n). For an arbitrary given order c ⊕ ···⊕ c of gyroaddition of i1 in , E E g{a , ··· , a } { r ⊗ a ⊕ ···⊕ r ⊗ a r , ··· ,r ∈ R} 1 n = “ i1 i1 in in ”; i1 in E E E E a1 an = t1 + ···+ tn ; t1, ··· ,tn ∈ R ∩ V1 ||a1|| ||an||

Proof. It immediately follows from the fact that the Einstein and M¨obiusspaces are isomorphic each other as gyrovector spaces (cf. [U1, Section 6.21]).

4 Orthogonal gyrodecomposition In this section, we will give a notion of orthogonal gyrodecomposition with respect to relatively closed gyrovector subspaces, which is analogous to the notion of the orthogonal decomposition with respect to closed linear subspaces in a Hilbert space. It is an application of the results which are established in the preceding sections, and one can ﬁnd that the orthogonal gyrodecomposition is easily obtained by the (ordinary) orthogonal decomposition in the M¨obiusgyrovector space. Throughout this section, we assume that V is a real Hilbert space. For any nonempty subset A of V,wedenotebylinA (resp. clinA) the linear (resp. closed linear) subspace generated by A, that is the intersection of all linear (resp. closed linear) subspaces of V which contain A.WealsodenotebyA⊥ the orthogonal complement of A in V,thatis A⊥ = {x ∈ V; x·a =0 forall a ∈ A}.

Let (V1, ⊕, ⊗)betheM¨obius gyrovector space, and let M be a gyrovector subspace of V1 which is topologically relatively closed.

13 Lemma 4.1.

clinM ∩ V1 = M

Proof. The inclusion (⊃) is trivial. Since M is relatively closed, there exists a closed subset F of V such that M = V1 ∩ F . In order to show the reverse inclusion, take an arbitrary element x ∈ clinM ∩ V1. Then there exists a sequence

{xn}⊂linM satisfying that ||xn − x|| → 0. Since ||x|| < 1, we may assume

||xn|| < 1. Each xn can be expressed by the form

xn = λ1xn,1 + ···+ λkxn,k, where xn,j ∈ M and λj ∈ R. Then, by Theorem 3.3, we can rewrite it to the form of

“ r1 ⊗xn,1 ⊕···⊕rk ⊗xn,k”, which belongs to M, since M is a gyrovector subspace. Thus we have xn ∈ M = V1 ∩ F . Now we can conclude that x ∈ V1 ∩ F = M. This completes the proof.

Theorem 4.2. (orthogonal gyrodecomposition) Let (V1, ⊕, ⊗) be the Mo¨bius gyrovector space, and let M be a gyrovector subspace of V1 which is topologically relatively closed. Suppose that

⊥ x = x1 + x2, x1 ∈ clinM, x2 ∈ M is the (ordinary) orthogonal decomposition of an arbitrary element x ∈ V1 with respect to the closed linear subspace clinM. Then there exists a unique pair (y, z) satisfying that

⊥ x = y ⊕ z, y ∈ M, z ∈ M ∩ V1.

Moreover, if x1, x2 = 0, then these elements y, z are determined by

y = λ1x1, z = λ2x2, where 2 2 2 2 2 2 ||x1|| + ||x2|| +1− (||x1|| + ||x2|| +1) − 4||x1|| λ 1 = 2 2||x1|| 2 2 2 2 2 2 ||x1|| + ||x2|| − 1+ (||x1|| + ||x2|| +1) − 4||x1|| λ . 2 = 2 (15) 2||x2||

Additionally, the inequalities 0 <λ1 < 1 and λ2 > 1 hold.

14 2 2 2 Proof. It follows from the Pythagorian identity ||x1|| + ||x2|| = ||x|| < 1that x1, x2 ∈ V1. Applying Theorem 2.1, there exists r1,r2 ∈ R satisfying that

x1 x2 x = ||x1|| + ||x2|| = r1 ⊗x1 ⊕ r2 ⊗x2. ||x1|| ||x2||

Put

y = r1 ⊗x1, z = r2 ⊗x2.

By the deﬁnition of ⊗, it is obvious that y, z ∈ V1 and rj ⊗xj = λjxj for some suitable λj ∈ R (j =1, 2). Furthermore, we have y = λ1x1 ∈ clinM ∩ V1 = M by x1 x2 Lemma 4.1. In order to check the formulae (15), note that α = · =0 ||x1|| ||x2|| and 2 2 2 2 2 2 ||x1|| + ||x2|| +1− (||x1|| + ||x2|| +1) − 4||x1|| c1 = 2||x1|| 2 2 2 2 2 2 ||x1|| + ||x2|| − 1+ (||x1|| + ||x2|| +1) − 4||x1|| c2 = . 2||x2|| It follows that −1 x1 c1 y = r1 ⊗x1 =tanh r1 tanh ||x1|| = x1, ||x1|| ||x1|| hence we have 2 2 2 2 2 2 c1 ||x1|| + ||x2|| +1− (||x1|| + ||x2|| +1) − 4||x1|| λ . 1 = = 2 ||x1|| 2||x1||

The second identity of the formulae (15) is similarly obtained. Clearly 0 <λ1. The inequality λ1 < 1 is equivalent to the next one: 2 2 2 2 2 2 −||x1|| + ||x2|| +1< (||x1|| + ||x2|| +1) − 4||x1|| , and it is easy to see that the above inequality actually holds by considering the square of both sides. The inside of the square root can be rewritten as follows (cf. (12)):

2 2 2 2 2 2 2 2 (||x1|| + ||x2|| +1) − 4||x1|| =(||x1|| + ||x2|| − 1) +4||x2|| .

Then the inequality λ2 > 1 is equivalent to the next one: 2 2 2 2 2 2 (||x1|| + ||x2|| − 1) +4||x2|| > −||x1|| + ||x2|| +1, and the above inequality actually holds.

15 Finally, we will show the uniqueness of the pair (y, z). Suppose there exists another pair (y, z) satisfying that

⊥ x = y ⊕ z , y ∈ M, z ∈ M ∩ V1.

Then, it follows from the deﬁnition of ⊕ and the relations y·z = y ·z =0that (1 + ||z||2) y +(1−||y||2) z (1 + ||z||2) y +(1−||y||2) z = , 1+||y||2||z||2 1+||y||2||z||2 which yields the system of equations: 1+||z||2 1+||z||2 y = y (16) 1+||y||2||z||2 1+||y||2||z||2 1 −||y||2 1 −||y||2 z = z (17) 1+||y||2||z||2 1+||y||2||z||2

Put a = ||y||,b= ||z||,c1 = ||y ||,c2 = ||z ||. By taking the norm, we obtain the following system of equations of real numbers: b2 c 2 1+ a 1+ 2 c 2 2 = 2 2 1 (18) 1+a b 1+c1 c2 − a2 − c 2 1 b 1 1 c 2 2 = 2 2 2 (19) 1+a b 1+c1 c2 If we put a(1 + b2) (1 − a2)b t1 = and t2 = , 1+a2b2 1+a2b2 then the above system of equations turns to be a special case of (1), (2). Further, put t a b2 a2b2 β 1 (1 + ) γ 1 1+ . = = 2 = = 2 t2 (1 − a )b t2 (1 − a )b Then it is easy to see that (1 − a2)(1 − b2)(1 + a2b2) γ2 − (1 + β2)= > 0, (1 − a2)2b2 so we can apply Theorem 2.4, hence 1+β2 + γ2 − (1 + β2 + γ2)2 − 4β2γ2 c1 = 2βγ 1+β2 − γ2 + (1 + β2 + γ2)2 − 4β2γ2 c2 = 2γ

16 is a unique pair of the solution of the system of the equations (18), (19) which satisﬁes 0 < |c1|, |c2| < 1. It is an elementary calculation to check that

(1 + b2)(1 + a2b2) (1 + β2 + γ2)2 − 4β2γ2 = , (1 − a2)b2 and now it is straightforward to see that c1 = a, c2 = b. This means that ||y|| = ||y||, ||z|| = ||z|| and, the system of equations (16), (17) implies that y = y, z = z. This completes the proof.

References [A] Toshikazu Abe, On gyrolinear spaces (Oral presentation), the 90th Yonezawa Mathematics Seminar, July 2nd, 2016.

[AH] Toshikazu Abe and Osamu Hatori, Generalized gyrovector spaces and a Mazur-Ulam theorem, Publ. Math. Debrecen 87 (2015), no. 3-4, 393–413.

[U1] Abraham Albert Ungar, Analytic Hyperbolic Geometry and Albert Einstein’s Special Theory of Relativity, World Scientiﬁc Publishing Co. Pte. Ltd., Sin- gapore, 2008.

[U2] Abraham Albert Ungar, From M¨obius to gyrogroups, Amer. Math. Monthly 115 (2008), no. 2, 138–144.