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Alternativity and reciprocity in the Cayley-Dickson algebra PDF

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Alternativity and reciprocity in the Cayley-Dickson algebra 6 S. Kuwata, H. Fujii, and A. Nakashima 0 0 Faculty of Information Sciences, Hiroshima City University, 2 Asaminami-ku, Hiroshima 731-3194, Japan n a J Abstract 6 Wecalculate theeigenvalueρof themultiplication mappingR on the 1 Cayley-Dickson algebra An. If the element in An is composed of a pair v of alternative elements in An−1, half the eigenvectors of R in An are still 3 eigenvectors in the subspace which is isomorphic to An−1. The invariant 3 0 under the reciprocal transformation An×An ∋ (x,y) 7→ (−y,x) plays a 1 fundamentalroleinsimplifyingthefunctionalform ofρ. Ifsomephysical 0 field can be identified with the eigenspace of R, with an injective map 6 from the field to a scalar quantity (such as a mass) m, then there is a 0 one-to-one map π :m 7→ρ. As an example, the electro-weak gauge field / canberegardedastheeigenspaceofR,whereπimpliesthattheW-boson h mass is less than theZ-boson mass, as in thestandard model. t - p e 1 Introduction h : v TheCayley-DicksonalgebraA overafieldF isthealgebrastructureF2n given i n X inductively by a pair of A through the Cayley-Dickson process [1]. If the n−1 r base field F is taken as the real number R, A1,A2, and A3 correspond to the a complex number C, the quaternion H, and the octonion O, respectively. For n < 4, A satisfies the alternative law: x2y = x(xy) and xy2 = (xy)y, so that n A is a composition algebra, xy = x y for all x,y in A (n < 4), that n n is, a normed division algebra.kThuks, thke kmkulktiplication map R : A A by x n n → R :y yx (with x =1) belongs to SO(2n) for n<4. x The7→reciprocaltkraknsformationisgivenbyA A (x,y) ( y,x). Orig- n n × ∋ 7→ − inally, the reciprocal transformation was given by the rotation of the phase space as (q,p) ( p,q), where q,p represent position and momentum, re- 7→ − spectively. As emphasized by Born [2], the law of nature is invariant un- der the reciprocal transformation. For example, the bosonic hamiltonian for a harmonic oscillator 1(q2 +p2), the fermionic one i(qp pq), and the cor- 2 2 − responding (anti)commutation relations ([q,p] = i,[q,q] = [p,p] = 0) and ( q,p = 0, q,q = p,p = 1) are all invariant under (q,p) ( p,q). Thus, { } { } { } 7→ − 1 the invariant under the reciprocal transformation plays a fundamental role in restricting the theory. For a physicalapplication, A has been utilized by many authors [3, 4, 5,6, n 7, 8, 9, 10, 11]. However, most of the studies are based on the division algebra (except in Ref. [12], where it is discussed that the gravitational force may be embedded in the enlarged space of A ). One of the reasons of adopting A for 4 n n<4isthatitiscloselyrelatedtothe(simple)Liegroup. ThesimpleLiegroups arecategorizedintotwotypes: classicalandexceptionalones. Theclassicalones, denoted by SO(n), SU(n), and Sp(n), represent the isometric transformation in a n-dimensional vector space over R, C, and H, respectively. Note that the isometry over O is not a Lie group, due to the lack of associativity; the multiplication over O does not form a group. However,O is still a composition algebra due to alternativity, so that O is related to the simple Lie group of an exceptional type as [13] G =Aut(O), F =Isom(OP2), 2 ∼ 4 ∼ E =Isom((C O)P2), E =Isom((H O)P2), E =Isom((O O)P2), 6 ∼ ⊗ 7 ∼ ⊗ 8 ∼ ⊗ where Aut, Isom and KP2 represent the automorphism, isometry and the pro- jective plane over K, respectively. For n > 3, A is not directly related to the n simpleLiealgebra. ThisconverselyimpliesthatA (forn>3)isanaturalcan- n didate of an algebra for describing the breaking of the isometry of the simple Lie group. Theaimofthispaperisfirsttoobtaintheisometry(orrotationalsymmetry) breaking of the multiplication map R , and then to apply the result to some x rotational symmetry breaking system. If the field (or state) of the isometry breaking system can be identified with the eigenspace of R , with an injective x map from the field and some scalar quantity q (called an “order parameter”), thenitisfoundthatthereisaone-to-onemapπ fromq tothe eigenvalueofR . x As an example, we will show that the electro-weak gauge field can be regarded astheeigenspaceofR . Inthiscase,theone-to-onemapπ betweenq (here,the x gaugebosonmass)andtheeigenvalueofR impliesthat(undersomereasonable x condition) the W-boson mass is less than the Z-boson mass, as in the standard model. InSec.2,wefirstreviewthe basicpropertiesofthe Cayley-Dicksonalgebra, and then calculate the eigenvalue of R , to find that half the eigenvectorsof (x,y) R inA (forn<4)arestilleigenvectorsinthesubspacewhichisisomor- (x,y) n+1 phic to A . This is due to the alternativityofx andy in A . To guaranteethis n n propertyforalln,werequirethattheelementinA shouldbe composedofa n+1 pair of alternative elements in A . In Sec. 3, we show how the invariant under n the reciprocal transformation simplifies the functional form of the eigenvalue. In Sec. 4, we find that the eigenvalue of R can be interpreted as the order x parameterin some isometrybreaking field, as long as the fieldcan be identified with the eigenspace of R . In Sec. 5, we give summary. x 2 2 Cayley-Dickson algebra 2.1 Basic property We review the basic properties of the Cayley-Dickson algebra A over the real n numbers R. For x = (x ,x ) and y = (y ,y ) in R2n = R2n−1 R2n−1, the 1 2 1 2 × multiplication is inductively given by [1, 14] xy =(x y y¯ x , y x +x y¯ ), with x¯=(x¯ , x ), (1) 1 1 2 2 2 1 2 1 1 2 − − so that xy = y¯x¯ for all x,y in A . The Euclidean norm and the inner product n in R2n are given by x 2 =xx¯=x¯x, k k x,y = 1(xy¯+yx¯)= (xy¯), h i 2 ℜ respectively, where (x) representsthe realpart of x, that is, (x)= 1(x+x¯). For n=0,1,2, andℜ3, A represents R,C,H, and O, respectiveℜly. 2 n Notation 1 Denote the commutator by [x, y]=xy yx, and the associator by − [x, y, z]=(xy)z x(yz). − The basicpropertiesofA aresummarizedinTable1. Duetothe flexibility n of A , we obtain for all x,y,z in A the following identity [14] n n x,yz = xz¯,y = y¯x,z . (2) h i h i h i Moreover,the flexibility leads to [15] x¯(xy)=(yx¯)x, (3) which is less popular, compared to the flexible law. However, Eq. (3) is useful in obtaining Eq. (6). Notation 2 For x in A , the right and left multiplications are denoted by R n x and L , respectively, that is, R ,L :A A by x x x n n → yR =yx, yL =xy. x x Table 1: Basic properties of A . n n Property Identity 0 Self-conjugate x=x¯ 0,1 Commutative [x, y]=0 0,1,2 Associative [x, y, z]=0 0,1,2,3 Alternative [x, x, y]=0 All Flexible [x, y, x]=0 3 SomemayfeelsomewhatuneasyaboutthenotationlikeyR ,ratherthanausual x notation like R (y). If we regard y A = R2n as a row (not column) vector x n in R2n, and R as the corresponding∈2n 2n matrix, yR =yx corresponds to x x × y R = yx in the Dirac notation. From the recursion formula of Eq. (1), R x x h | h | and L are decomposed into x R L L ηL R = x1 x2 , L = x1 x2 , (4) x L R x ηR R (cid:18) − x¯2 x¯1 (cid:19) (cid:18) − x2 x1 (cid:19) where η :A A by xη =x¯, so that the correspondingrepresentationmatrix n n → is given by diag(1, 1, 1,..., 1). From Eq. (4), it is found that − − − tR =R , tL =L , (5) x x¯ x x¯ where the superscript t represents the transposition. Now we examine whether R (with x = 1) belongs to SO(2n) or not. x k k In this case, it is sufficient to analyze the algebra structure of R R instead x x¯ of R itself. This is because we have y,y yR ,yR = y,(yR )R = x x x x x¯ h i → h i h i y,y(R R ) from Eq. (2). Since R R (= R tR ) is positive semidefinite in x x¯ x x¯ x x Rh 2n, its eigeinvalue turns out to be (non-negative)real,so that the vector space R2n can be decomposed into the eigenspace of R R . If y is the eigenvector x x¯ of R R (with its eigenvalue ν), then we obtain y 2 ν y 2, so that the x x¯ k k → k k violation of R SO(2n) can be checked through the relation of ν = 1. In x ∈ 6 what follows, we will obtain the eigenvalue (together with the corresponding eigenvector) of R R . x x¯ Notation 3 For x in A , N =R R and N¯ =L L . n x x x¯ x x x¯ We have the following identity for all x in A : n N =N =N¯ =N¯ , (6) x x¯ x x¯ whereN =N¯ isderivedfromEq.(3). Substituting Eq.(4)intoN andusing x¯ x x Eq. (6), we obtain [16] A B Nx = Bx1,x2 −A x1,x2 , (7) (cid:18) x1,x2 x1,x2 (cid:19) where A = N +N and B = [R , L ]. Using the determinant x1,x2 x1 x2 x1,x2 x1 x2 A B identity − = A+iB A iB = A+iB tA itB , andnoticing that B A | || − | | || − | (cid:12) (cid:12) Ax1,x2 is(cid:12)(cid:12)symmetric(cid:12)(cid:12)and that Bx1,x2 is antisymmetric, we find that (cid:12) (cid:12) N νI = C νI 2, (8) | x− | | x1,x2 − | whereC =A +iB , sothatthe eigenvalueofN is equaltothatof x1,x2 x1,x2 x1,x2 x C [whilethecorrespondingeigenvectorrofN ,whichisdoublydegenerate, x1,x2 x is given by r = (d ,d ) and (d , d ), where d and d represent the real and 1 2 2 1 1 2 − imaginary parts of the eigenvector of C , respectively]. x1,x2 Thus, in the following, we concentrate on the calculation of the eigenvalue of C . x1,x2 4 Notation 4 For x,y in A , we set C′ =C ( x 2+ y 2)I, where x 2+ n x,y x,y− k k k k k k y 2 representsthediagonalelementsofC ,whichisobtainedinductively. We x,y k k also denote by S the set of eigenvalues of C′ . For later convenience, if the n x,y element si ∈ Sn has the same functional form as s′i ∈ Sn′ (n′ 6= n), we regard them as the same element. Forx,yinA withn<4,C′ turnsouttobeiB . Forn<3,B isvanishing n x,y x,y x,y due to the associativity of A , so that all the elements of S is vanishing. For n n n =3, S is calculated in a somewhat brute-force way [16]. To summarize, we n have S = 0 , 0 { } S = 0,0 , 1 { } S = 0,0,0,0 , 2 { } S = 0,0,0,0, 2x y, 2x y , 3 { ± | × | ± | × |} where x y2 = x 2 y 2 x,y 2, with the bold face letter x representing the ima|gi×nary| parkt okf xk, ktha−t his, xi= x (x). For x A (with n < 4), it n −ℜ ∈ is found from Eq. (8) that all the eigenvalues of N are given by x 2, due to x k k the vanishing of all the elements of S ,S and S . Thus, we obtain xy 2 = 0 1 2 x,xN = x,x y 2 = x 2 y 2 forallx,y inA (withn<4). Thisikndickates y n hthat A i(forh n<ik4)kis thke nkorkmked division algebra, as is well known. n ItshouldbenotedthatthesetsS (forn<4)satisfythefollowinginclusion n relation: S S S S . 0 1 2 3 ⊂ ⊂ ⊂ However, S S does not, in general hold for n 4 in general; there is n−1 n ⊂ ≥ a counterexample. The relation of S S implies that half of the eigen- n−1 n vectors of N in A are still eigenvec⊂tors in the subspace V′ = (x,y) = (x,y) n+1 (x ,x ,y ,y ) A2 = A4 x = y = 0 , which is isomorphic to{A . To be1mo2re1str2ict,∈letn(p,q) n=−1(|p2,p ;q2,q ) i}n A2 = A4 be the eigennvector 1 2 1 2 n n−1 of N with its eigenvalue ν , that is, (p,q)N = ν (p,q). In the sub- (x,y) x,y (x,y) x,y spaceV′,theeigenequationturnsouttobe(p ,q )N =ν′ (p ,q ),where 1 1 (x1,y1) x,y 1 1 ν′ = lim ν . Suppose that [ν ( x 2+ y 2)] S belongs also x,y x2,y2→0 x,y x,y − k k k k ∈ n to S . Then, we find that lim ν is the same functional form of ν n−1 x2,y2→0 x,y x,y where x and y are replaced by x and y , respectively, because the element in 1 1 S having the same functional form as the element in S is regarded as the n n−1 same. Thus, we obtain ν′ = ν , which indicates that the eigenvector of x,y x1,y1 N(x,y) in An+1 is still the eigenvector of N(x1,y1) in the subspace V′ (∼=An), as long as S S . n−1 n ⊂ In the following subsection, we deal with the case where S S is n−1 n ⊂ satisfied for all n. This guarantees that half of the eigenvectors of N in (x,y) A remain the eigenvectors in the subspace which is isomorphic to A . n+1 n 5 2.2 Alternative entry Recall that for n < 4, where S S is satisfied, A is alternative; recall n−1 n n also that for n 4, where S ⊂S does not hold in general, A is not n−1 n n alternative. This≥implies that if the⊂elements x,y are “alternative” in A , the n inclusionrelationS S is satisfied. Fortunately,this is indeedtrue, which n−1 n ⊂ will be proven in the following. Definition 5 The element x in A is alternative, if [a, a, x] = 0 for all x in n A . n Lemma 6 For x,y in A (for n 3), n ≥ xqy B =0, x,y ⇔ where x=x (x), y=y (y), and xqy represents x y =0. −ℜ −ℜ | × | Proof. ) Since B is independent of (x) and (y), there is no loss of x,y generality⇒that we can set y = kx (or x =ℜ ky), whℜere k R. Due to the flexibility of A , we find that B = 0 for all x in A . ∈Thus, we obtain n x,x n B =kB (or kB )=0. x,y x,x y,y ) Notice that B = 0 z A , [y, z, x] = 0. Then, we show the con- x,y n ⇐traposition such that if x⇔/ y∀, t∈hen z A , [y, z, x] = 0. From x / y, it n ∃ ∈ 6 follows that both x and y are nonzero, so that there is no loss of generality that we can choose x = e and y = c e +c e , where c ,c R with c = 0 1 1 1 2 2 1 2 2 so as to guarantee x / y. Since A (for n 3) is not an ass∈ociative algeb6 ra, n there always exists an element z A such≥that [e , z, e ] = 0. Thus, we ob- n 2 1 tain z A , [y, z, x] = 0, where∈use has been made of [e 6, z, e ] = 0 by the n 1 1 flexib∃ilit∈y of A . 6 (cid:3) n From Lemma 6, together with Eq. (1), we find that for x = (x ,x ) in 1 2 A =A A , n n−1 n−1 × x is alternative x ,x are alternative with B =0. (9) ⇔ 1 2 x1,x2 The repeated application of Eqs. (7) and (9) leads to N = x 2 for x A x n k k ∈ alternative, so that we obtain x,y are alternative in A C′ =iB . (10) n ⇒ x,y x,y Definition 7 For x,y in A , V(x,y) denotes the vector space generated by the n set e ,x,y,xy , where e represents the real element of A . V⊥(x,y) denotes 0 0 n the{orthogonal }complement in A =R2n. n Lemma 8 For x,y are alternative in A , we have n 1. For x/y, dimRV(x,y)=4; 2. For all v V(x,y), vB =0; x,y ∈ 3. For all w V⊥(x,y), wB = wB˜ , where B˜ :=[R , R ]. x,y x,y x,y x y ∈ − 6 Proof. 1. Thelinearindependenceof e ,x,y,xy isequivalenttothatof e ,x,y,xy , 0 0 { } { } so that we can take x and y to be pure imaginary. Suppose that c e + 0 0 c x+c y+c xy = 0, where c R. Then from Eq. (2), we obtain 0 = 1 2 3 i ∈ c e ,e =c x 2+c x,y =c x,y +c y 2 =c x,y +c x 2 y 2. 0 0 0 1 2 1 2 0 3 Thhus fori x / ky,kwe geth c i= 0 (hi =i0,1,2k,3k), so thhat tihe elekmkenktskin i e ,x,y,xy are linearly independent over R. 0 { } 2. It is sufficient to show that vB =0 for v =e ,x,y,xy. For v =e , this x,y 0 0 is trivial. For v = x,y, vB = 0 by definition of x,y being alternative. x,y (xy)B = [y, xy, x] = [x, xy, y] = (x(xy))y x((xy)y) = (x2y)y x,y − − − x(xy2) = x2y2 x2y2 = 0, where in the second equality, we have used the identity [x,−y, z] = [z, y, x] for all x,y, and z in A [this identity n − can be derived from the linearization of the flexibility [x, y, x]=0 under x x+λz (λ R)]. → ∈ 3. Since B and B˜ are independent of (x), (y), we can set x,y to x,y x,y ℜ ℜ be pure imaginary (notice that w is pure imaginary due to w e ). 0 ⊥ From xw + wx = (xw¯ + wx¯) = 2 x,w = 0, yw + wy = 0, and − − h i x(wy)+(wy)x =x(yw)+(wy)x=x¯(wy)+(wy)x=2 x¯,wy =2 x¯y¯,w = 2 xy,w =0,itfollowsthatw(B +B˜ )=[x, w, yh]+((wxi)y h(wy)x)i= x,y x,y h i − (xw+wx)y (x(wy)+(wy)x)=0. − (cid:3) Notation 9 For x and y in A e , ∆ := x,y I + 1(R R +R R ). n\{ 0} x,y h i 2 x y y x In a similar way, as in the proof of Lemma 8, we find that Lemma 10 For x and y are alternative in A e , n 0 \{ } 1. ∆ =∆ =0; x,x y,y 2. v V(x,y), v∆ =0. x,y ∀ ∈ Notation 11 Forx=(x ,x )in A =A A , x (for j =1,2,...) is given inductively by x 1 2=(x n ,xn−1×)nin−1A i1i2=...iAj A . i1...ij i1...ij1 i1...ij2 n−j n−j−1× n−1−j We denote x by x . 11...1 (k) ktimes Forx,y ar|e{za}lternativeinA (withx/y), itisconvenienttodecomposethe n vector space A =R2n by using V(x,y) and V⊥(x,y) as n n−1 A = V , (11) n k k=1 M 7 where V = r A r V(x,y) , 1 n { ∈ | ∈ } V = r A r V⊥(x,y);r ,r V(x ,y ) , 2 n 1 2 1 1 { ∈ | ∈ ∈ } V = r A r V⊥(x,y);r ,r V⊥(x ,y );r ,r ,r ,r V(x ,y , 3 n 1 2 1 1 11 12 21 22 11 11 { ∈ | ∈ ∈ ∈ } . . . n−3 V = r A r V⊥(x ,y ) r A r V(x ,y ) . n−1 { ∈ n| i1...ik ∈ (k) (k) }∩{ ∈ n| i1...in−2 ∈ (n−2) (n−2) } k=0 \ NowweevaluateS forxandy alternativeinA . Inthiscase,C′ =iB n n x,y x,y by Eq. (10). Recall that B is anti-symmetric, so that the eigenpolyno- x,y mial for iB is an even function, and that the eigenvalue of iB is a real x,y x,y number. Thus, S can be decomposed into two sets: S+ and S−, where n n n S+ (S−) represents the non-negative (non-positive) elements of S such that n n n S =S+ S− =S+ ( S+). n n ∪ n n ∪ − n Notation 12 For x,y in A , the set of the eigenvalues of (iB )2 is denoted n x,y by S˜ , so that S+ S+ is given by square-rooting the elements of S˜ . n n ∪ n n From Lemma 8, it is found that v V(x,y) is the eigenvector of (iB )2 with x,y ∈ its eigenvalue 0. Thus, the rest we have to do is to obtain the eigenvalue of (iB )2 whose eigenvector belongs to V⊥(x,y). In this case, (iB )2 can be x,y x,y replaced by (iB˜ )2, which can be written using ∆ as x,y x,y (iB˜ )2 =(iB˜ )2 x,y x,y =2(R R R R +R R R R ) (R R +R R )2 x y y x y x x y x y y x − =4 x2 y2 ( x,y ∆ )2 , (12) x,y | | | | − h i− (cid:2) (cid:3) where we have used the (x) and (y)-independence of B and Lemma 10i). x,y ℜ ℜ The next step is to obtain the recursion formula for ∆ . The repeated x,y application of Eq. (7) yields for x A n ∈ n−3 N = x 2I + σ σ ...σ ( iσ ) B , x k k 0⊗ 0⊗ 0⊗ − 2 ⊗ xi1...ik−11,xi1...ik−12 kX=1 k−1times i1,.X..,ik−1 1 0| {z } 0 1 where σ = and iσ = . The linearization of N (= 0 0 1 2 1 0 x,x Rx,x¯ = Rx,x(cid:16)) by x (cid:17) x+λy An(cid:16) −e0 lea(cid:17)ds to − → ∈ \{ } n−3 2∆ = σ σ ...σ ( iσ ) (B +(x y)). x,y 0⊗ 0⊗ 0⊗ − 2 ⊗ xi1...ik−11,yi1...ik−12 ↔ Xk=1 k−1times i1,.X..,ik−1 (13) | {z } 8 Recall that we are dealing with the case of x,y being alternative in A . Then n fromLemma6andEq.(9),wefindthatx qx ,andtheanalogous i1...ik−11 i1...ik−12 relation for y, so that we can set x =κ(k)x , y =κ(k)y , (14) 11...12 1 (k) 11...12 2 (k) k−1 k−1 whereκ(k),κ(k) R(|fo{rz}k =1,2,...,n 3)|a{rze}certainparameters. Substituting 1 2 ∈ − Eq. (14) into Eq. (13), we obtain n−3 k−1 2∆ = a b σ ...σ (iσ ) B , (15) x,y (k) (i) 0⊗ 0⊗ 2 ⊗ x(k),y(k) kX=1 iY=1 k−1times where 0 b stands for 1, and a |,b {azre gi}ven by i=1 (i) (k) (i) Q a =κ(k) κ(k), b =1+κ(i)κ(i). (k) 2 − 1 (i) 1 2 From Eq. (15), the recursion formula for ∆ can be written as x,y 2∆ =a iσ B +2b σ ∆ (k =1,2,...,n 3), x(k−1),y(k−1) (k) 2⊗ x(k),y(k) (k) 0⊗ x(k),y(k) − (16) wherex denotesxinA e . NotethatEq.(16)isanalogoustoEq.(7). Re- (0) n 0 \{ } callthedeterminantidentity σ A+iσ B = A+iB A iB . Then,wefind 0 2 | ⊗ ⊗ | | || − | that all the eigenvalues of ∆ remain invariant, even if ∆ x(k−1),y(k−1) x(k−1),y(k−1) is replaced by ∆ ∆˜ := 1a iB +b ∆ . x(k−1),y(k−1) → x(k),y(k) ±2 (k) x(k),y(k) (k) x(k),y(k) From Lemmas 8 and 10, it follows that 0 for r V(x ,y ) , r∆˜ = ∈ (k) (k) x(k),y(k) ( ±21a(k) (iB˜x(k),y(k))2+b(k)∆x(k),y(k) (cid:2)for r ∈V⊥(x(k),y(k)(cid:3)) , q (17) (cid:2) (cid:3) where(iB˜ )2 canbe rewrittenusing∆ asinEq.(12). Eventually, x(k),y(k) x(k),y(k) all the eigenvalues of ∆ are given inductively by those of ∆ . x(k−1),y(k−1) x(k),y(k) Proposition 13 Forx,y alternativeinA ,thesetsS (forn=0,1,...)satisfy n n the inclusion relation S S S ... 0 1 2 ⊂ ⊂ ⊂ 9 Proof. It is sufficient to show that S˜ S˜ for n = 1,2,..., because S is n−1 n n given by square-rooting the element of S˜ .⊂In the case of x q y for x,y in A , n n whereB =0byLemma6,alltheelementsofS˜ (hencethoseofS )turnout x,y n n to be vanishing, so that the Proposition is trivial in this case. In what follows, we assume that x/y. We decompose the vector space A as in Eq. (11). For all r V =V(x,y), n 1 ∈ wheredimV =4,weobtainrB =0byLemma8ii),sothatV representsthe 1 x,y 1 eigenspace of (iB )2 whose eigenvalue is zero. It should be recalled here that x,y B =0forallx,y inA duetotheassociativityofA ,sothatS˜ = 0,0,0,0 . x,y 2 2 2 { } Hence, the eigenvalue of (iB )2 whose eigenvector belongs to V is given by x,y 1 S˜ . 2 If the eigenvector r = (r ,r ) in A = A A of (iB )2 belongs to 1 2 n n−1 n−1 x,y × V⊥(x,y),itseigenvalueturnsouttobetheeigenvalueof(iB˜ )2byLemma8iii), x,y so that we have only to calculate the eigenvalue of ∆ , hence that of ∆˜ . x,y x1,y1 If r further belongs to V (where dimV = 4) so that r ,r V(x ,y ), the 2 2 1 2 1 1 eigenvalue of ∆˜ turns out to be vanishing by Eq. (17) wit∈h k =1. On the x1,y1 other hand, the relation of ∆ = 0, in itself, holds for all x,y in A (where x,y 3 dimA = 8) by Eq. (15) or (16). Thus, the four eigenvalues of (iB )2 whose 3 x,y corresponding eigenvectors belong to V are given by the four elements in S˜ . 2 3 Recallthatamongthe elementsofS˜ ,the eigenvaluewhoseeigenvectorbelongs 3 toV isgivenbyS˜ ,sothattheeigenvalueof(iB )2 whoseeigenvaluebelongs 1 2 x,y to V is given by S˜ S˜ . 2 3 2 \ Inasimilarway,werepeatedlyapplyEqs.(16)and(17)andusetheidentity ∆ = 0, to find that the eigenvalue of (iB )2 whose eigenvector x(n−3),y(n−3) x,y belongs to V (k = 2,3,...,n 1) is given by S˜ S˜ , as is summarized in k k+1 k Table2. Thus,weobtainthein−clusionrelationS˜ S˜\ fork=2,3,...,n 1. k k+1 Since n can be chosen as any n 3, we get S˜ ⊂S˜ for k 2 (for k = 0−,1, k k+1 ≥ ⊂ ≥ the above inclusion relation is found to be satisfied from the direct calculation of S ). (cid:3) k Table 2: Eigenvalues of (iB )2 for x,y alternative in A = n−1V , with x,y n k=1 k x/y. L Eigenspace Eigenvalues dimV i V S˜ 4 1 2 V S˜ S˜ 4 2 3 2 \ V S˜ S˜ 8 3 4 3 \ . . . . . . . . . V S˜ S˜ 2n−1 n−1 n n−1 \ 10

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