# Orthogonal Gyroexpansion in Möbius Gyrovector Spaces

Authors: Keiichi Watanabe
Journal: Journal of Function Spaces
 (2017)

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## Abstract

We investigate the Möbius gyrovector spaces which are open balls centered at the origin in a real Hilbert space with the Möbius addition, the Möbius scalar multiplication, and the Poincaré metric introduced by Ungar. In particular, for an arbitrary point, we can easily obtain the unique closest point in any closed gyrovector subspace, by using the ordinary orthogonal decomposition. Further, we show that each element has the orthogonal gyroexpansion with respect to any orthogonal basis in a Möbius gyrovector space, which is similar to each element in a Hilbert space having the orthogonal expansion with respect to any orthonormal basis. Moreover, we present a concrete procedure to calculate the gyrocoefficients of the orthogonal gyroexpansion.


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## Body

## 1. Introduction

A. Ungar initiated study on gyrogroups and gyrovector spaces (cf. [1]). Gyrovector spaces are generalized vector spaces, with which they share important analogies, just as gyrogroups are analogous to groups. 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 open unit disc in the complex plain endowed with the Möbius addition. Ungar extended these gyroadditions to the ball of an arbitrary real inner product space, introduced a common gyroscalar multiplication, and observed that the ball endowed with gyrooperations are gyrovector spaces (cf. [2, 3]). He describes that gyrovector spaces provide the setting for hyperbolic geometry just as vector spaces provide the setting for Euclidean geometry. In particular, Möbius gyrovector spaces form the setting for the Poincaré ball model of hyperbolic geometry, and similarly, Einstein gyrovector spaces form the setting for the Beltrami-Klein ball model. Readers may consult [4, 5] and the references therein for general information about gyrogroups and gyrovector spaces.Gyrooperations are generally not commutative, associative, or distributive. Thus the theory of gyrovector spaces falls within the general area of nonlinear functional analysis. They are enjoying algebraic rules such as left and right gyroassociative, gyrocommutative, scalar distributive, and scalar associative laws, so there exist rich symmetrical structures which we should clarify precisely. Many elementary problems are still unsolved. We refer to [6–10] as examples of recent papers for gyrovector spaces, their generalizations, and related matters.In [8], Abe and the author of the present article showed that any finitely generated gyrovector subspace in the Möbius gyrovector space coincides with the intersection of the linear subspace generated by the same generators and the Möbius ball. As an application, they presented a notion of orthogonal gyrodecomposition and clarified the relation to the ordinary orthogonal decomposition.The importance of the orthogonal expansion of each vector with respect to an orthonormal basis in a Hilbert space cannot be overemphasized in both theory and application of functional analysis. In this paper we will introduce a concept of orthogonal gyroexpansion of each element with respect to an orthogonal basis in a Möbius gyrovector space and reveal analogies that it shares with its classical counterpart. Such problems seem to be quite fundamental and important for developing pure and applied mathematics, since one of the virtues of gyrovector spaces is that they have properties which are fully analogous to vector space properties. Moreover, the gyrocoefficients of the orthogonal gyroexpansion can be concretely calculated by a procedure that is given here.The paper is organized as follows. Section2 is the preliminaries. In Section 3, we introduce a notion of gyrolinear independency for finite sets in a gyrovector space and show that it coincides with the notion of the linear independency. In Section 4, we give a notion of orthogonal gyroexpansions with respect to a complete orthogonal sequence in the Möbius gyrovector space, and we present an explicit procedure to obtain the orthogonal gyroexpansions.


## 2. Preliminaries

Let us briefly recall the definitions of two models of gyrovector spaces, that is, the Möbius and Einstein gyrovector spaces. For precise definitions and basic results of gyrocommutative gyrogroups and gyrovector spaces, see [4].LetV=(V,+,·) be a real inner product space with a binary operation + and a positive definite inner product · and let Vs be the ball (1)Vs=a∈V:a<sfor any fixed s>0.Definition 1 (see [4, Definitions  3.40 and  6.83]).
The Möbius addition⊕M and the Möbius scalar multiplication ⊗M are given by the equations (2)a⊕Mb=1+2/s2a·b+1/s2b2a+1-1/s2a2b1+2/s2a·b+1/s4a2b2r⊗Ma=stanh⁡rtanh-1⁡asaaifa≠0,r⊗M0=0for any a,b∈Vs, r∈R. The addition ⊕M and scalar multiplication ⊗M for the set Vs=±a;a∈Vs in the axiom (VV) of gyrovector space are defined by the equations (3)a⊕Mb=a+b1+1/s2abr⊗Ma=stanh⁡rtanh-1⁡asfor any a,b∈Vs,r∈R.We simply denote⊕M,⊗M by ⊕,⊗, respectively. If several kinds of operations appear in a formula simultaneously, we always give priority by the following order: (i) ordinary scalar multiplication; (ii) gyroscalar multiplication ⊗; (iii) gyroaddition ⊕; that is, (4)r1⊗w1a1⊕r2⊗w2a2=r1⊗w1a1⊕r2⊗w2a2,and the parentheses are omitted in such cases. In general, we note that gyroaddition does not distribute with (both ordinary and gyro) scalar multiplications:(5)ta⊕b≠ta⊕tb,r⊗a⊕b≠r⊗a⊕r⊗b.In the limit of larges, s→∞, the ball Vs expands to the whole space V. The next proposition suggests that each result in linear analysis can be restored from the counterpart in gyrolinear analysis.Proposition 2 (see [4, p. 78]).
The Möbius addition (resp., Möbius scalar multiplication) reduces to the vector addition (resp., scalar multiplication) ass→∞; that is, (6)a⊕b⟶a+bs⟶∞r⊗a⟶ras⟶∞.Definition 3 (see [4, Definitions  3.45 and  6.86]).
The Einstein addition⊕E and the Einstein scalar multiplication ⊗E are given by the equations (7)a⊕Eb=11+a·b/s2a+1γab+1s2γa1+γaa·bar⊗Ea=stanh⁡rtanh-1⁡asaaifa≠0,r⊗E0=0for any a,b∈Vs,r∈R, where γa=1/1-a2/s2.Note that each of the Einstein scalar multiplication and the operations on the setVs is identical to the corresponding operation for the Möbius gyrovector spaces.Definition 4 (see [4, Definition  6.88]).
An isomorphism from a gyrovector space(G1,⊕1,⊗1) to a gyrovector space (G2,⊕2,⊗2) is a bijective map ϕ:G1→G2 that preserves gyrooperations and keeps the inner product of normalized elements invariant; that is, (8)ϕa⊕1b=ϕa⊕2ϕbϕr⊗1a=r⊗2ϕaϕaϕa·ϕbϕb=aa·bbifa,b≠0for any a,b∈G1,r∈R.Theorem 5 (see [4, Table  6.1]).
LetϕEM:Vs→Vs be the map defined by the equation (9)ϕEMa=2⊗afor any a∈Vs. Then ϕEM is an isomorphism from the Möbius gyrovector space to the Einstein gyrovector space.Thus, most of results established for the Möbius gyrovector spaces in the sequel can be transformed to corresponding results for the Einstein gyrovector spaces by the isomorphism stated above.


## 3. Gyrolinear Independency

We begin with consideration of a counterpart in a gyrovector space to the notion of linearly independent sets in a linear space.Definition 6.
A finite subseta1,…,an⊂Vs is gyrolinearly independent if, for any permutation (i1,…,in) of 1,…,n and for any order of gyroaddition, the following implication holds: (10)ri1⊗ai1⊕⋯⊕rin⊗ain=0⟹r1=⋯=rn=0.Example 7.
LetV=R2 with the Euclidean inner product and s=1. If we identify V1 with the open unit disc D=z∈C;z<1 in the complex plain C by (x,y)=x+iy, then it is well-known that the Möbius addition reduces to(11)a⊕b=a+b1+a¯bfor any a,b∈D (cf. [4, (3.127)]). If we take (12)a=i2,b=-25-25i,c=12,then (13)a⊕b⊕c=0,a⊕b⊕c=4+16i53-8i.This means that a,b,c is not gyrolinearly independent, and if we put (14)r1=tanh-1-33+689/20tanh-11/2,r2=tanh-117-689/20tanh-11/2,then it is readily checked that (15)r1⊗c⊕r2⊗a=-33+68920⊕17-68920i=-33+689/20+17-689/20i1+-33+689/20·17-689/20i=b.It is immediate to see the following lemma by the fact that1⊗a=a and 0⊗a=0 and Definition 6. We omit the proof.Lemma 8.
Leta1,…,an⊂Vs be gyrolinearly independent. Then (i)
each element is nonzero;(ii)
any subset is also gyrolinearly independent.Lemma 9.
Suppose thata1,a2 is linearly independent in Vs and (16)r1⊗a1⊕r2⊗a2=λ1⊗a1⊕λ2⊗a2Then one has r1=λ1 and r2=λ2.Proof.
Without loss of generality, we may assume thats=1, rj≠0, and λj≠0. If we put (17)α=a1a1·a2a2,cj=tanh⁡rjtanh-1aj,dj=tanh⁡λjtanh-1ajfor j=1,2, then, from the definitions of ⊕,⊗, it follows that -1≤α≤1, 0<|cj|, |dj|<1, and (18)r1⊗a1⊕r2⊗a2=λ1⊗a1⊕λ2⊗a2=t1a1a1+t2a2a2,where we put (19)t1=1+2c1c2α+c22c11+2c1c2α+c12c22=1+2d1d2α+d22d11+2d1d2α+d12d22t2=1-c12c21+2c1c2α+c12c22=1-d12d21+2d1d2α+d12d22.This means that (c1,c2) and (d1,d2) are solutions to the system of equations (20)x2y2+γx2+2αx-γy+1=0xy2+2α+βx2-βy+x=0,where we put β=t1/t2 and γ=1/t2. Then, we have β≠0 and 1+β(2α+β)<γ2 by [8, Lemma  2.2]. So we can apply [8, Theorem  2.4] to obtain that cj=dj, which yields that rj=λj for j=1,2. This completes the proof.Theorem 10.
Leta1,…,an be a linearly independent set in Vs. Suppose that two gyrolinear combinations r1⊗a1⊕⋯⊕rn⊗an, λ1⊗a1⊕⋯⊕λn⊗an are given the same order of gyroaddition and (21)r1⊗a1⊕⋯⊕rn⊗an=λ1⊗a1⊕⋯⊕λn⊗anThen one has rj=λj(j=1,…,n).Proof.
Without loss of generality, we may assume thats=1. Assume that the theorem is valid up to n. Let a1,…,an+1 be a linearly independent set in V1 and let the following formula (22)r1⊗a1⊕⋯⊕rm⊗am⊕rm+1⊗am+1⊕⋯⊕rn+1⊗an+1=λ1⊗a1⊕⋯⊕λm⊗am⊕λm+1⊗am+1⊕⋯⊕λn+1⊗an+1show the latest gyroadditions. Put (23)a=r1⊗a1⊕⋯⊕rm⊗amb=rm+1⊗am+1⊕⋯⊕rn+1⊗an+1a′=λ1⊗a1⊕⋯⊕λm⊗amb′=λm+1⊗am+1⊕⋯⊕λn+1⊗an+1.Then a,a′ (resp., b,b′) belong to the linear span of a1,…,am (resp., am+1,…,an+1). If a=0, then we have b=a′⊕b′. By [8, Theorem  3.3], we can express b,a′,b′ of the form (24)b=tm+1am+1am+1+⋯+tn+1an+1an+1a′=t1′a1a1+⋯+tm′amamb′=tm+1′am+1am+1+⋯+tn+1′an+1an+1.By the definition of ⊕, it follows that (25)tm+1am+1am+1+⋯+tn+1an+1an+1=c1a1a1+⋯+cn+1an+1an+1,where we put (26)cj=1+2a′·b′+b′2tj′1+2a′·b′+a′2b′2j=1,…,m1-a′2tj′1+2a′·b′+a′2b′2j=m+1,…,n+1.Since a1,…,an+1 is linearly independent, we have c1=⋯=cm=0, which implies that t1′=⋯=tm′=0; that is, a′=0. By the assumption of our induction, it follows that rj=λj for all j.
Similarly, we may assume thata,a′,b,b′≠0, so a,b is linearly independent. By the definition of ⊕, we can rewrite the equation(27)a⊕b=a′⊕b′as (28)t1aa+t2bb=t1′a′a′+t2′b′b′,so we obtain that (29)t1aa=t1′a′a′,t2bb=t2′b′b′.Therefore, (27) can be changed to the following equation: (30)1⊗a⊕1⊗b=r1⊗a⊕r2⊗b,where (31)r1=tanh-1t1a′/t1′tanh-1a,r2=tanh-1t2b′/t2′tanh-1b.By the previous lemma, we can conclude that 1=r1=r2, which implies that a=a′, b=b′. Then, the assumption of our induction shows that rj=λj(j=1,…,n+1). This completes the proof.Theorem 11.
For any finite subset inVs, two notions of linearly independent and gyrolinearly independent coincide.Proof.
(
⇒
) It immediately follows from the previous theorem.
(
⇐
) We may assume that s=1. Assume that the theorem is valid up to n, the number of elements of the finite set. Suppose that a1,…,an+1⊂V1 is gyrolinearly independent and (32)t1a1a1+⋯+tn+1an+1an+1=0.By Lemma 8(ii) and the assumption of our induction, it suffices to show that tn+1=0. On the contrary, assume that tn+1≠0. Then, it is obvious that (33)t1a1a1+⋯+tnanan≠0.Take a positive number M satisfying that (34)t1Ma1a1+⋯+tnManan∈V1,tn+1Man+1an+1∈V1.
Thus we have(35)aaa+tn+1Man+1an+1=0,where we put (36)a=t1Ma1a1+⋯+tnManan.From [8, Theorem  2.1], we can rewrite (35) in the form of(37)r⊗a⊕rn+1⊗an+1=0.We can also rewrite r⊗a in the form of (38)r⊗a=r1⊗a1⊕⋯⊕rn⊗anby using [8, Theorem  3.3], so we obtain the following equation: (39)r1⊗a1⊕⋯⊕rn⊗an⊕rn+1⊗an+1=0.Since a1,…,an+1 is assumed to be gyrolinearly independent, we can conclude that r1=⋯=rn+1=0, which implies that tn+1=0. This is a contradiction and completes the proof.Although the contents in the rest of this section are actually known and used repeatedly in [4], we give their proofs for the convenience of readers.Lemma 12 (see also [10, Proposition  2.3]).
(40)
a
⊕
b
2
=
a
2
+
2
a
·
b
+
b
2
1
+
2
/
s
2
a
·
b
+
1
/
s
4
a
2
b
2for any a,b∈Vs.Proof.
By using the definition of⊕, one can easily calculate the inner product of a⊕b with itself to obtain (41)a⊕b2=11+2/s2a·b+1/s4a2b22·1+2s2a·b+1s2b22a2+21+2s2a·b+1s2b21-1s2a2a·b+1-1s2a22b2.If we put u=a/s, v=b/s, then it is easy to factorize the second factor as (42)1+2u·v+v22su2+21+2u·v+v21-u2s2u·v+1-u22sv2=s2u2+2u·v+v21+2u·v+u2v2=a2+2a·b+b21+2s2a·b+1s4a2b2;hence we can conclude identity (40).Definition 13 (see [4, Definition  2.7, (2.1)]).
Recall that the inverse element ofa is denoted by ⊖a in a gyrogroup, and one uses the notation (43)a⊖b=a⊕⊖bas in group theory.Lemma 14.
The following formulae hold:(i)
a/s⊕b/s=a⊕b/s.(ii)
0<1-2/s2a·b+1/s4a2b2<22.(iii)
a-b≤2a⊖b,
for anya,b∈Vs, where ⊕ in the left-hand side of (i) is in the space V1.Proof.
(i) It immediately follows from the definition of⊕.
(ii) By the Cauchy-Schwarz inequality, we have(44)0<1-1s2ab2≤1-2s2a·b+1s4a2b2≤1+2s2ab+1s4a2b2=1+1s2ab2<22.(iii) From (ii) just established, identity (40) in Lemma 12, and the fact that ⊖a=-a in Vs, we have (45)a-b222≤a2-2a·b+b21-2/s2a·b+1/s4a2b2=a⊖b2.This completes the proof.Lemma 15 (see [4, Theorem  8.33]).
Let{cj}j=1n be an orthogonal set in Vs. Then, for any permutation (i1,…,in) of the numbers 1,…,n and any order of gyroaddition for ci1⊕⋯⊕cin, the following equality holds: (46)ci1⊕⋯⊕cin2s=ci12s⊕⋯⊕cin2s.Here, ⊕ in the right-hand side are in the space Vs.Proof.
The previous lemma (i) shows that(47)as⊕bs2=a⊕bs2=a⊕b2s2.for any a,b in Vs. On the other hand, if a,b is orthogonal, then it follows from identity (40) in Lemma 12 that (48)as⊕bs2=a/s2+b/s21+a/s2b/s2=1s·a2/s+b2/s1+1/s2a2/sb2/s=1sa2s⊕b2s.Thus the theorem holds for n=2.
Assume that the theorem is valid up ton. Let {cj}j=1n+1 be an orthogonal set in Vs and let the following equation(49)ci1⊕⋯⊕cin+1=ci1⊕⋯⊕cim⊕cim+1⊕⋯⊕cin+1show the latest gyroaddition ⊕. If we put (50)a=ci1⊕⋯⊕cim,b=cim+1⊕⋯⊕cin+1,then a,b is orthogonal. From the case of n=2, it follows that (51)ci1⊕⋯⊕cin+12s=a⊕b2s=a2s⊕b2s=ci1⊕⋯⊕cim2s⊕cim+1⊕⋯⊕cin+12s.Due to the assumption of our induction, we can conclude that (52)=ci12s⊕⋯⊕cim2s⊕cim+12s⊕⋯⊕cin+12s.This completes the proof.


## 4. The Poincaré Metric and Orthogonal Gyroexpansion in the Möbius Gyrovector Space

In this section, we give a notion of orthogonal gyroexpansions with respect to a complete orthogonal sequence in the Möbius gyrovector space, which is fully analogous to the notion of the orthogonal expansions with respect to a complete orthonormal sequence in a Hilbert space. It is an application of the orthogonal gyrodecomposition which was established in [8, Theorem  4.2], and we present an explicit procedure to obtain the orthogonal gyroexpansions in the Möbius gyrovector space.Definition 16 (see [4, Definition  6.8, 6.17 (6.286) and (6.293)]).
The Möbius gyrodistance functiond on a Möbius gyrovector space (Vs,⊕,⊗) is defined by the equation (53)da,b=b⊖a.Moreover, the Poincaré distance function h on the ball Vs is introduced by the equation (54)ha,b=tanh-1da,bsfor any a,b∈Vs. Then h satisfies the triangle inequality [4, (6.294)], so that (Vs,h) is a metric space.Remark 17.
As pointed out in [4, 6.17, p.217], the distance function h(a,b) is the obvious generalization into the ball Vs of the well-known Poincaré distance function on the disc D. There are a number of literatures dealing with relationship between hyperbolic geometry and Hilbert spaces. In particular, Goebel and Reich [11] introduced the hyperbolic metric ρ on the open unit ball of a complex Hilbert space, from a viewpoint of holomorphic function theory. They developed the study of the Hilbert ball, which leads to research on ρ-convexity, nonexpansive mappings, fixed point theorems, and so forth, and [11] is cited in many bibliography such as [12]. The definition of ρ is equivalent to (55)ρx,y=tanh-11-σx,y1/2,where (56)σx,y=1-x21-y21-x,y2for any elements x,y in the Hilbert ball. If we identify R2 with C, then it is easy to see that h and ρ coincide with the Poincaré metric on D. In general, however, h and ρ do not coincide for higher dimensional spaces. We clarify the relationship between h and ρ below.Lemma 18.
LetV be a real inner product space. Then the norm of the Einstein addition of two elements is given by the equation(57)a⊕Eb2=11+a·b/s22a2+b2+2a·b-1s2a2b2+1s2a·b2for any a,b∈Vs.Proof.
At first, consider the cases=1. From the definition of ⊕E, it is easy to calculate the inner product of a⊕Eb with itself as follows: (58)1+a·b2a⊕Eb2=a+1γab+γa1+γaa·ba,a+1γab+γa1+γaa·ba=a2+1γa2b2+γa1+γa2a·b2a2+2γaa·b+2γa1+γaa·ba2+21+γaa·b2=a2+1-a2b2+1/1-a21+1/1-a22a·b2a2+21-a2a·b+2·1/1-a21+1/1-a2a·ba2+21+1/1-a2a·b2=a2+1-a2b2+11-a2+12a·b2a2+21-a2a·b+21-a2+1a·ba2+21-a21-a2+1a·b2=a2+1-a2b2+1-1-a2a22a·b2a2+21-a2a·b+21-1-a2a2a·ba2+21-a21-1-a2a2a·b2=a2+b2+2a·b-a2b2+a·b2.Thus the lemma holds for s=1. For general s>0, let a,b∈Vs. If we put u=a/s and v=b/s, then it is immediate to see that(59)u⊕Ev=a⊕Ebs,and we can easily deduce identity (57) by applying the case s=1 to u,v. This completes the proof.Theorem 19.
Let one use the notationsρ and σ in Vs by (60)ρa,b=tanh-11-σa,b1/2,where (61)σa,b=s2-a2s2-b2s2-a·b2for a,b∈Vs. Then the following identities hold: (i)
a⊖Eb/s=1-σ(a,b)1/2.(ii)
hE(a,b)=tanh-1a⊖Eb/s=ρ(a,b).(iii)
2h(a,b)=2tanh-1⁡a⊖b/s=ρ(2⊗a,2⊗b).Proof.
(i) and (ii) immediately follow from the previous lemma. (iii) It is not difficult to see that we may assumes=1. By (ii) just established, it suffices to show that(62)2⊗a⊖b=2⊗a⊖E2⊗b.Note that (63)2⊗a=1+a2-1-a21+a2+1-a2=2a1+a2for real number 0≤a<1. For any a,b∈V1, if we put a=a, b=b, and α=a/a·b/b, then, by the definition or the axioms of gyrovector spaces, we have (64)2⊗a=2⊗a=2a1+a2,2⊗a·2⊗b=2a1+a2·2b1+b2·α.By identity (40) in Lemma 12, (65)2⊗a⊖b=2⊗a2-2a·b+b21-2a·b+a2b2=2a2-2abα+b2/1-2abα+a2b21+a2-2abα+b2/1-2abα+a2b2=2a2-2abα+b21-2abα+a2b21-2abα+a2b2+a2-2abα+b2.On the other hand, identity (57) in the previous lemma shows that (66)2⊗a⊖E2⊗b=2⊗a2+2⊗b2-22⊗a·2⊗b-2⊗a22⊗b2+2⊗a·2⊗b21-2⊗a·2⊗b2=2a/1+a22+2b/1+b22-2·2a/1+a2·2b/1+b2·α-2a/1+a222b/1+b22+2a/1+a2·2b/1+b2·α21-2a/1+a2·2b/1+b2·α=2a2-2abα+b21-2abα+a2b21+a21+b2-4abα.This completes the proof.In the rest of the paper, we should concentrate to investigate the Möbius ball endowed with the Poincaré metrich introduced by Ungar. We can perform gyrolinear algebraic operations which behave quite well for orthogonal sequences in the Möbius gyrovector spaces, like as linear algebraic ones in Hilbert spaces.Lemma 20.
For any sequence{an}n=1∞ and any element a in Vs, (i)
h(an,a)→0(n→∞)⇔d(an,a)→0(n→∞),(ii)
h(an,am)→0(n,m→∞)⇔d(an,am)→0(n,m→∞).Proof.
It is obvious, because bothtanh and tanh-1 are uniformly continuous on a neighborhood of 0.Lemma 21.
For any fixeda∈Vs, the map Vs∋x↦x·a∈(-s2,s2) is continuous, where one considers the metric h on both sets.Proof.
Suppose thath(xn,x)→0. Then, from the previous lemma and Lemma 14(iii), it follows that xn-x≤2xn⊖x→0. Therefore, we have (67)xn·a⊖x·a=xn·a-x·a1-1/s4xn·ax·a⟶0n⟶∞.This implies that h(xn·a,x·a)→0.We should make sure of two definitions here. One of them is quite usual; another is very natural.For any nonempty subsetA of Vs, we denote A⊥ as the orthogonal complement of A in V; that is,(68)A⊥=x∈V;x·a=0∀a∈A.A nonempty subsetM of Vs is a gyrovector subspace if M is closed under gyroaddition and gyroscalar multiplication; that is, a,b∈M and r∈R imply that a⊕b∈M and r⊗a∈M.Lemma 22.
A
⊥
∩
V
s is an h-closed gyrovector subspace.Proof.
From the definitions of⊕ and ⊗, it is immediate to see that A⊥∩Vs forms a gyrovector subspace. Moreover, A⊥∩Vs is obviously h-closed by the previous lemma.Lemma 23.
Ifan,a∈Vs and an-a→0, then h(an,a)→0.Proof.
It suffices to show thatd(an,a)→0. By the assumption a<s, we can obtain (69)dan,a2=an⊖a2=an2-2an·a+a21-2/s2an·a+1/s4an2a2⟶0n⟶∞,where we used identity (40) in Lemma 12.Proposition 24.
LetM be a gyrovector subspace of Vs. Then the closure M¯h with respect to the metric h is also a gyrovector subspace.Proof.
Suppose thata,b∈M¯h. There exist sequences {an},{bn}⊂M such that h(an,a),h(bn,b)→0. By Lemmas 20 and 14(iii), we have an-a,bn-b→0. From the definitions of ⊕,⊗, it is easy to see that (an⊕bn)-(a⊕b)→0 and r⊗an-r⊗a→0. By Lemma 23, it follows that h(an⊕bn,a⊕b)→0 and h(r⊗an,r⊗a)→0. Since an⊕bn,r⊗an∈M, we can conclude that a⊕b,r⊗a∈M¯h. This completes the proof.Lemma 25.
Any finitely generated gyrovector subspace ish-closed.Proof.
LetM be a gyrovector subspace generated by nonzero elements a1,…,an in Vs. For an arbitrary element x∈M¯h, there exists a sequence xkk=1∞⊂M such that h(xk,x)→0. Then, from Lemmas 20(i) and 14(iii), it follows that xk-x→0. By [8, Theorem  3.3], we have (70)M=t1a1a1+⋯+tnanan;t1,…,tn∈R∩Vs.Since t1a1/a1+⋯+tnan/an;t1,…,tn∈R is a finite dimensional linear subspace, it is closed with respect to the norm topology. Therefore x∈M. This completes the proof.From now on, we assume that the carrierV of the Möbius gyrovector space Vs is complete as a metric space with respect to the norm induced by the inner product. Thus, V is a real Hilbert space.Theorem 26.
LetV be a real Hilbert space. Then (Vs,h) is a complete metric space.Although this fact is well-known and it can be deduced by existing results and Theorem19, we give a direct proof here in order to show how gyrovector space approach is fully analogous to vector space approach.Proof.
Without loss of generality, we may assume thats=1. Suppose that {an}n=1∞ is a Cauchy sequence in (V1,h). From Lemmas 14(iii) and 20(ii), it follows that (71)an-am≤2an⊖am=2dan,am⟶0n,m⟶∞,which implies that {an} is a Cauchy sequence with respect to the norm of V. Hence there exists a unique element a∈V such that a≤1,an-a→0. In order to show that a<1, on the contrary, we assume that a=1. By the assumption that {an} is a Cauchy sequence in (V1,h), there exists a natural number m0 such that (72)dam+p,am2<12for any m≥m0 and any p. On the other hand, from identity (40) in Lemma 12, we have (73)dam+p,am2=am+p2-2am+p·am+am21-2am+p·am+am+p2am2.Now we fix m≥m0 and let p→∞. Then, from the fact that am+p→a and the assumption a=1, we can obtain (74)12≥a2-2a·am+am21-2a·am+a2am2=1,which is a contradiction. This implies that a<1. By Lemma 23, the proof is complete.Theorem 27.
LetV be a real Hilbert space. If A is a closed subset in (Vs,h), then A is relatively closed in (Vs,·). Therefore, the orthogonal gyrodecomposition is applicable to h-closed gyrovector subspaces in the sense of [8, Theorem  4.2].Proof.
Denote byA¯ the closure of A with respect to the norm topology. It suffices to show that A=A¯∩Vs. One of the inclusions (⊂) is trivial. If x∈A¯∩Vs, then there exists a sequence {xn}⊂A such that xn-x→0. So we can apply Lemma 23 to obtain that h(xn,x)→0. Hence x∈A and this completes the proof.Theorem 28.
LetM be a gyrovector subspace of Vs and x∈Vs. Then one has (75)x∈M⊥⟺x⊖m≥xm∈M.Proof.
(
⇒
) Suppose that x∈M⊥∩Vs,m∈M. By identity (40) in Lemma 12, we obtain (76)x⊖m2=x2+m21+1/s4x2m2≥x2.(⇐) Suppose that m∈M. For an arbitrary positive real number t, take (77)r=tx·mx·m.Since r⊗m∈M, it follows from identity (40) in Lemma 12 that (78)x2≤x⊖r⊗m2=x2-2x·r⊗m+r⊗m21-2/s2x·r⊗m+1/s4x2r⊗m2.By the axiom (V7), we have r⊗m=r⊗m=t⊗m and the inequality (79)x21-2s2x·r⊗m+1s4x2t⊗m2≤x2-2x·r⊗m+t⊗m2,which yields the following inequality: (80)2x·r⊗m≤1+1s2x2t⊗m2;namely, (81)2stanh⁡rtanh-1⁡msx·mm≤1+1s2x2stanh⁡ttanh-1⁡ms2.Note that x·m and r have the same signature, so we have (82)tanhrtanh-1⁡msx·m=tanhttanh-1⁡msx·m.Therefore, we can obtain the inequality (83)2x·mm≤1+1s2x2stanh⁡ttanh-1⁡ms.Since t>0 is arbitrary, we can let t→+0 and conclude that x·m=0. This completes the proof.Lemma 29.
In a gyrocommutative gyrogroup, one has(84)a⊖b⊕c=gyra,⊖b⊖b⊕a⊖c.This lemma can be obtained if we puta,b and c as ⊖b,a and ⊖c, respectively, in [4, Theorem  3.9]. However, we give a proof for the convenience of readers by using gyroautomorphic inverse property ⊖(a⊕b)=⊖a⊖b, left gyroassociative law (G3), gyrocommutativity (G6), and gyroautomorphism (G4).Proof.
(85)
a
⊖
b
⊕
c
=
a
⊕
⊖
b
⊖
c
=
a
⊕
⊖
b
⊕
g
y
r
a
,
⊖
b
⊖
c
=
g
y
r
a
,
⊖
b
⊖
b
⊕
a
⊕
g
y
r
a
,
⊖
b
⊖
c
=
g
y
r
a
,
⊖
b
⊖
b
⊕
a
⊕
⊖
c
.Theorem 30.
LetM be an h-closed gyrovector subspace of Vs and x∈Vs. (i)
Let(86)x=y⊕z,y∈M,z∈M⊥∩Vsbe the orthogonal gyrodecomposition ofx∈Vs by Theorem 27 and [8, Theorem  4.2]. Then y is the closest point to x in M. Thus y satisfies the identity(87)hx,y=infm∈Mhx,m.(ii)
Conversely, lety be the closest point to x in M; namely, y is an element in M satisfying identity (87). Then(88)x=y⊕⊖y⊕xis the orthogonal gyrodecomposition. Thus⊖y⊕x∈M⊥∩Vs.Proof.
Note thath(a1,a2)≤h(b1,b2) if and only if a1⊖a2≤b1⊖b2.
(i) Suppose thatx=y⊕z,y∈M,z∈M⊥∩Vs. For any m∈M, we have (89)x⊖m=x⊖y⊕⊖y⊕m=gyrx,⊖y⊖y⊕x⊖⊖y⊕m=gyrx,⊖yz⊖⊖y⊕mby the previous lemma. From ⊖y⊕x=z∈M⊥∩Vs,⊖y⊕m∈M and Theorem 28, it follows that (90)x⊖y=z≤z⊖⊖y⊕m=gyrx,⊖yz⊖⊖y⊕m=x⊖m,because each gyroautomorphism preserves the norm. Since m∈M is arbitrary, we can conclude x⊖y≤infm∈Mx⊖m, and the opposite inequality trivially holds. Thus y satisfies identity (87).
(ii) Putz=⊖y⊕x. For any m∈M, we have (91)x⊖y⊕m=gyrx,⊖y⊖y⊕x⊖m=gyrx,⊖yz⊖mby the previous lemma. From y⊕m∈M and identity (87), it follows that (92)z=⊖y⊕x=x⊖y≤x⊖y⊕m=gyrx,⊖yz⊖m=z⊖m,because each gyroautomorphism preserves the norm. Thus we can apply Theorem 28 and obtain that z∈M⊥. This completes the proof.The following lemma plays a key role in our orthogonal gyroexpansion.Lemma 31.
Ifu,v,w is an orthogonal set in Vs, then the associative law holds; that is, (93)u⊕v⊕w=u⊕v⊕w.Proof.
By [4, (3.147), (3.148)], the gyration in the Möbius gyrovector spaces Vs can be expressed by the equation (94)gyru,vw=w+2Au+BvD,where (95)A=-1s4u·wv2+1s2v·w+2s4u·vv·wB=-1s4v·wu2-1s2u·wD=1+2s2u·v+1s4u2v2for all u,v,w∈Vs. See also [10, Proposition  2.14] for a proof by hand calculation. If u,v,w is orthogonal, then we have A=B=0, so that gyr[u,v]w=w. This completes the proof.Definition 32.
(i) Let{an}n be a sequence in Vs. One says that a series (96)a1⊕a2⊕a3⊕⋯⊕an⊕⋯converges if there exists an element x∈Vs such that h(x,xn)→0(n→∞), where the sequence {xn}n is defined recursively by x1=a1 and xn=xn-1⊕an. In this case, we say the series converges to x and denote(97)x=a1⊕a2⊕a3⊕⋯⊕an⊕⋯.(ii) Let {an}n be a sequence in R with |an|<s for all n. We say that a series (98)∑n=1∞an⊕=a1⊕a2⊕⋯⊕an⊕⋯converges if there exists x∈R with |x|<s such that xn→x, where the sequence {xn}n is defined recursively by x1=a1 and xn=xn-1⊕an. In this case, we say the series converges to x and denote(99)x=∑n=1∞an⊕.Theorem 33.
Let{en}n=1∞ be an orthonormal sequence in a real Hilbert space V. Let {wn}n=1∞ be a sequence in R such that 0<wn<s for all n. For any sequence {rn}n=1∞ in R, the following are equivalent: (i)
The series(100)r1⊗w1e1⊕r2⊗w2e2⊕⋯⊕rn⊗wnen⊕⋯converges to an elementx∈Vs.(ii)
The series∑n=1∞‍⊕rn⊗wn2/s converges to x∈R with |x|<s.Note that parentheses are not necessary in the formula in (i) above by Lemma31.Proof.
(
i
)
⇒
(
i
i
). Put (101)xn=r1⊗w1e1⊕⋯⊕rn⊗wnenx=r1⊗w1e1⊕r2⊗w2e2⊕⋯⊕rn⊗wnen⊕⋯.From Lemma 15, it follows that (102)xn2s=r1⊗w1e12s⊕⋯⊕rn⊗wnen2s=r1⊗w12s⊕⋯⊕rn⊗wn2s.By the assumption, we have h(x,xn)→0. It follows from Lemma 20(i) and Lemma 14(iii) that xn→x. Thus we have (103)limn→∞⁡∑j=1n⊕⁡rj⊗wj2s=x2s<s.(ii)⇒(i). Suppose x=∑n=1∞‍⊕rn⊗wn2/s<s. We put (104)xn=r1⊗w1e1⊕⋯⊕rn⊗wnen,cn=∑j=1n⊕⁡rj⊗wj2s.By the assumption, for any ε>0, there exists a number n0 such that (105)n≥n0⟹0≤x⊖cn<ε.The last inequality implies that x-cn/1-1/s2xcn<ε. For n0≤m<n, by Lemma 15, (106)dxn,xm2=⊖xm⊕xn2=rm+1⊗wm+1em+1⊕⋯⊕rn⊗wnen2=cn⊖cm=cn-cm1-1/s2cncm≤x-cm1-1/s2xcm<ε.Note that the strict inequality x<s is crucial in the argument above. This implies that {xn}n is a Cauchy sequence with respect to the metric h by Lemma 20(ii). Since (Vs,h) is complete by Theorem 26, there exists a unique element x∈Vs such that h(x,xn)→0. This completes the proof.Example 34.
Consider the sequence{an}n=1∞ in R defined by an=1/2n. For s=1, it is easy to see that(107)xn=a1⊕⋯⊕an=1-1n+1n=1,2,….Put rn=tanh-11/2n/tanh-11/2. Then, we have rn⊗1/2=tanhrntanh-11/2=1/2n. It follows that (108)∑j=1n⊕⁡rj⊗122=∑j=1n⊕⁡12j=1-1n+1,which does not converge to an element x∈R with |x|<1. This example can be considered as a counterpart in the Möbius gyrovector space to the series ∑n=1∞‍1/2n.Theorem 35.
Let{en}n=1∞ be a complete orthonormal sequence in a real Hilbert space V. Let {wn}n=1∞ be a sequence in R such that 0<wn<s for all n. Then, for any x∈Vs, we have the orthogonal gyroexpansion (109)x=r1⊗w1e1⊕r2⊗w2e2⊕⋯⊕rn⊗wnen⊕⋯,where the sequence of gyrocoefficients {rn}n=1∞ is determined by the following equations:(110)xn=x·en,xn1=∑j=1nxjej,xn2=∑j=n+1∞xjejuj=μj-12⋯μ12xjejj=2,3,…u1=x1e1=x11vj=μj-12⋯μ12xj2j=2,3,…v1=x12μj1=uj2+vj2+s2-uj2+vj2+s22-4s2uj22uj2μj2=uj2+vj2-s2+uj2+vj2+s22-4s2uj22vj2rj=tanh-1μj1μj-12⋯μ12xj/stanh-1wj/sfor all j,n=1,2,…. If xj=0, then we do not define μj(1) but define as rj=0 and continue the procedure. If vn=0, then we do not define μn(2) but define as rj=0 for all j≥n+1 and finish the procedure.Proof.
It is not difficult to see that we may assumes=1. It is obvious that the series ∑n=1∞‍xnen converges to x in the norm topology and that (111)x=xn1+xn2is the orthogonal decomposition with respect to the closed linear subspace generated by e1,…,en. Let (112)x=yn⊕zn,yn∈Mn,zn∈Mn⊥∩V1be the orthogonal gyrodecomposition with respect to Mn by Theorem 27 and [8, Theorem  4.2], where Mn is an h-closed gyrovector subspace generated by {w1e1,…,wnen}. Then, yn,zn are given by the equations (113)yn=λn1xn1,zn=λn2xn2,where (114)λn1=x2+1-x2+12-4xn122xn12λn2=x2-1+x2+12-4xn122xn22.Since xn(1)-x→0, (115)λn1⟶x2+1-x2+12-4x22x2=1,so that we have (116)yn=λn1xn1⟶xin the norm topology, which implies that h(yn,x)→0.
Next, we expressyn in the form of a gyrolinear combination (117)yn=r1⊗w1e1⊕r2⊗w2e2⊕⋯⊕rn⊗wnenand present a concrete procedure to seek the gyrocoefficients rn.
Forn=1, by using the above decomposition, we take r1=tanh-1λ1(1)x1/tanh-1w1. It follows that (118)r1⊗w1e1=tanh⁡r1tanh-1w1e1w1e1w1e1=λ11x1e1=λ11x11=y1,x=r1⊗w1e1⊕z1.
Suppose that we proceed up to then-th step and obtain the quantities, identities (110) for j=1,…,n, and (119)x=r1⊗w1e1⊕⋯⊕rn⊗wnen⊕μn2vn.Now, (120)μn2vn=μn2⋯μ12xn+1en+1+μn2⋯μ12∑j=n+2∞xjej=un+1+vn+1is the orthogonal decomposition with respect to the finite dimensional linear subspace generated by {e1,…,en+1}. Let (121)μn2vn=yn+1′⊕zn+1′,yn+1′∈Mn+1,zn+1′∈Mn+1⊥∩V1be the orthogonal gyrodecomposition with respect to Mn+1. Then, yn+1′,zn+1′ are given by the equations (122)yn+1′=μn+11un+1,zn+1′=μn+12vn+1,where (123)μn+11=un+12+vn+12+1-un+12+vn+12+12-4un+122un+12μn+12=un+12+vn+12-1+un+12+vn+12+12-4un+122vn+12.By taking (124)rn+1=tanh-1μn+11μn2⋯μ12xn+1tanh-1wn+1,we have (125)x=r1⊗w1e1⊕⋯⊕rn⊗wnen⊕yn+1′⊕zn+1′=r1⊗w1e1⊕⋯⊕rn⊗wnen⊕μn+11un+1⊕μn+12vn+1=r1⊗w1e1⊕⋯⊕rn⊗wnen⊕rn+1⊗wn+1en+1⊕μn+12vn+1.Thus, we can inductively take a sequence {rn}n=1∞ by the procedure above.
Finally, from the uniqueness of the orthogonal gyrodecomposition with respect to theh-closed gyrovector subspace Mn, it follows that (126)yn=r1⊗w1e1⊕⋯⊕rn⊗wnenand the series converges as follows: (127)x=r1⊗w1e1⊕⋯⊕rn⊗wnen⊕⋯.This completes the proof.Theorem 36.
Let{en}n=1∞ be an orthonormal sequence in a real Hilbert space V. Let {wn}n=1∞ be a sequence in R such that 0<wn<s for all n. Then the following are equivalent: (i)
{en}n=1∞ is complete.(ii)
Theh-closed gyrovector subspace generated by wnenn=1∞ coincides with Vs.(iii)
x2=∑n=1∞‍⊕rn⊗wn2/s for all x∈Vs,
where{rn}n=1∞ is the sequence determined by identities (110).Proof.
It is easy to deduce implications(i)⇒(ii) and (i)⇒(iii) from the previous theorem.
(
i
i
i
)
⇒
(
i
) Suppose that there exists an element x≠0 such that x·en=0 for all n. By multiplying nonzero scalar, we may assume that x∈Vs. Then we have xn=0, hence rn=0 for all n. Thus (iii) is violated.
(
i
i
)
⇒
(
i
) Suppose that there exists an element x such that x·en=0 for all n. We may assume that x∈Vs. Then {x}⊥∩Vs is an h-closed gyrovector subspace by Lemma 22. Since it contains wnen for all n, it coincides with Vs by the assumption. Therefore, we have x∈{x}⊥∩Vs, which implies that x=0. This completes the proof.



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*Source: 1518254-2017-12-13.xml*