Alternativity and reciprocity in the Cayley-Dickson algebra

0.21 MB·English

by  S. Kuwata

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