3 Finitely Generated Gyrovector Subspaces Let (G, ⊕, ⊗) Be a Gyrovector Space
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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 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain. Finitely generated gyrovector subspaces and orthogonal gyrodecomposition in the M¨obius gyrovector space 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 M¨obius gyrovector space coincides with the intersection of the vector subspace generated by the same generators and the M¨obius 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 M¨obius addition. Ungar extended the M¨obius addition in the complex disc to the ball of an arbitrary real inner product space, and observed that the ball endowed with the M¨obius 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, M¨obiusgyrovector spaces form the setting for the Poincar´eball 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 prod- uct 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 ele- ments 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, gyroad- dition 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.