The zero divisors of the Cayley-Dickson algebras over the real numbers Guillermo Moreno Abstract. In this paper we describe algebraically the zero divisors of the n Cayley - Dickson algebras A n= R2 for n 4 over the real numbers. Introduction. The Cayley-Dickson algebra A nover R is an algebra struc- n ture on R 2 given inductively by the formulae: n 2n-1 2n-1 Let x = (x1; x2) and y = (y1; y2) in R 2 = R x R then xy = (x1y1 - __y2x2; y2x1 + x2__y1) where __x= (__x 1; -x2): Therefore A 0= R ; A1 = C complex numbers, A 2= H the Hamilton quater- nions, A 3= O the Cayley octonians, etc. This four algebras: R ; C; H and O are known as the classical Cayley-Dickson algebras and their main distinctive feature of these is: n Hurwitz Theorem: Let "k k" denote the euclidean norm in R 2. Then ______________________________ Classification 1991 AMS 17A99 Accepted for publication at Bol. Soc. Mat. Mex. 1 k xy k=k x kk y k 8x and y in A n if and only if n = 0; 1; 2; 3. (See [5] and [7]). That is, for n 4 this norm-preserving formula is not true in general and this opens the possibility of the existence of zero_divisors_in A nfor n 4 i.e. x and y non-zero elements in A nsuch that xy = 0 e.g. let x = e1 + e10 and y = e15- e4 in R 16= A4 where e0; e1; e2; . .;.e15 is the canonical basis in R * *16. In this paper we study the zero divisors in A nfor n 4. From the algebraic side these algebras are non-commutative for n 2 and non-associative for n 3. Moreover A 3is alternative_: x2y = x(xy) and xy2 = (xy)y for all x and y in A 3 and A nfor n 4 is flexible_: x(yx) = (xy)x for all x and y in A n. Clearly Associative ) Alternative ) Flexible and the backwards implications are not true. Introducing the "associator symbol" (a; b; c) = (ab)c - a(bc) we have that alternativity property in A 3is equivalent to the alternativity of the associator symbol in A3 via polarization; flexibility means that (x; y; x) = 0 for all x and y in A n. We will see that the absence of the norm-preserving property, the non- alternativity and the presence of zero divisors in A n for n 4 are very strongly related and in fact they determine one to each other. Usually the zero divisors are present in different algebraic contexts 2 1) the trivial one: in the direct sum of algebras i.e. if A and B are algebras over R: A B has a lot of zero divisors e.g. the ones of the form (a; 0) and (0; b). 2) In an associative (alternative) context as in Mn(R ) the n x n matrices over R where a god-given invariant called determinant says that 0 6= A 2 Mn(R ) is a zero divisor if and only if detA = 0. 3) In a non-associative (non-alternative) context which becomes rather difficult, this is our case. This paper has two chapters, in chapter one we give a general description of the zero divisors for A n (n 4) studying the linear transformations Ra; La : A n! A n right and left multiplication respectively for 0 6= a 2 A n fixed and we prove that each a 6= 0 (doubly pure) induce a direct sum decomposition of A n where one summand is a copy of H the quaternions, other summand is the elements of An which "alternate" with a and one third summand is the annihilator of A "KerLa = KerRa" (see Theorem I.15). This also implies that dimKerLa 0 mod 4 and dimKerLa 2n - 4. In Chapter II we construct some type of zero divisors (called special ones), in A 4are all of them, which are pairs of alternatives elements, we describe all of them completely for n 4 and relate these zero divisors with the multiplicative monomorphisms of A 3to A n n 3. n+1 In a second paper we will describe the topology of the subset of R 2 n 2n ZDn1= {(x; y) 2 R2 x R |xy = 0 and |x| = |y| = 1} 3 and relate this with suitable Stiefel manifolds. (Second paper is under review, the preprint is available under request). We work also on a third paper on the subject: "Applications" where we use the algebraic and topological parts to construct bilinear normed maps, construct generalized Hopf maps, etc. We thank to Fred Cohen, Sam Gitler and K.Y. Lam for many illumi- nating conversations and to Isidoro Gitler and Jose Martinez Bernal for the painstaking discussion on the linear algebra related with this subject. 4 I. Basic properties of the Cayley-Dickson Algebras and its zero divisors n A ndenotes R 2 with the Cayley-Dickson multiplication (n 1): n 2n-1 2n-1 x = (x1; x2); y = (y1; y2) in R 2 = R x R xy = (x1y1 - __y2x2; y2x1 + x2__y1) with __x = (__x 1; -x2) e0 = (1; 0; . .;.0) 2 An denotes the unit element. The Euclidian norm and inner product are given by k x k2= x__x= __xx =k x1 k2 + k x2 k2 and 1 __ __ = __(xy + yx ); 2 respectively. The_trace_is For x 2 An t(x) = x + __x; i.e. t(x) = 2(real part ofx) and t(x) = 2: Let x and y be in A n; x is orthogonal to y (x ? y) if and only if x__y= -y__xor t(x__y) = 0 because ___xy= __y_x. Therefore elements of zero trace (purely imaginaries) are orthogonal if and only if they anti-commute i.e. x ? y () xy = -yx 5 for x and y in A nwith t(x) = 0 i.e. __x= -x. Thus for all x 2 An we have the characteristic equation: x2 - t(x)x+ k x k2= 0 because x2 - (x + __x)x + __xx = 0: Definition: The associator_of a; b and c in A nis (a; b; c) = (ab)c - a(bc) which is linear in each variable. For n = 0; 1; 2 the associatior is indentically zero. For A 3= O the octonian numbers the associatior is alternative i.e. for all x and y in A 3 x2y = x(xy) and xy2 = (xy)y then 0 = (x + y; x + y; z)= (x; x; z) + (y; y; z) + (x; y; z) + (y; x; z) then 0 = (x; y; z) + (y; x; z). ..(x; y; z) = -(y; x; z): Similarly (x; y; z) = -(x; z; y) = (z; x; y) = -(z; y; x). A nfor n 4 is flexible i.e x(yx) = (xy)x for all x and y in A n. This means that (x; y; x) = 0 or equivalently (x; y; z) = -(z; y; x) for all x; y and z in A n. 6 Notice that (x; y; z) = 0 if one of the entries x; y or z is real, that is, * *its imaginary part is zero. Lemma 1.1. For all x; y and z in A n i)-(x; y; z) = (__x; y; z) = (x; __y; z) = (x; y; __z) ii)t((x; y; z)) = 0 iii)t(xy - yx) = 0 Proof. For n 2 the assertions are trivial. Let n 3 and x = xr+xI where xr = real part of x and xI = imaginary part of x so __x= xr - xI. (__x; y; z)= (xr - xI; y; z) = (xr; y; z) - (xI; y; z) = 0 - (xI; y; z) = -(xr; y; z) - (xI; y; z) = -(xr + xI; y; z) = -(x; y; z): Similarly -(x; y; z) = (x; __y; z) = (x; y; __z). So we prove i). To prove ii) we observe that ________ _______ ______ ______ ______ (x; y; z)= -x(yz) + (xy)z = -(z y)x + z(y x) = = -(__z; __y; __x) = (z; y; x) = -(x; y; z) by i) and flexibility. ________ Therefore (x; y; z) + (x; y; z)= 0 so ii) is proved. If x or y are real xy = yx and t(xy - yx) = 0. Suppose x and y are pure imaginary then ________ ___ ___ ____ ___ xy - yx = xy - yx = y x - xy = yx - xy . .. __________ (xy - yx) + (xy - yx) = xy - yx + yx - xy = 0: 7 On the other hand the symbol [x; y] = xy - yx is bilinear, therefore decom- posing x and y in the real and imaginary parts respectively, we are done with (iii). Q.E.D. Lemma 1.2. For all x; y; z and w in A n. x(y; z; w) + (x; y; z)w = (xy; z; w) - (x; yz; w) + (x; y; zw) Proof. (Adem's paper [1]). Lemma 1.3. For all x; y and z in A n = = <__yx; z> Proof. 2 = t(x(___yz)) = t(x(__z_y)) = t((x__z)__y) = 2 by Lemma 1.1 ii). Similarly 2<__yx; z>= t((__yx)__z) = t(__y(x__z)) = t((x__z)__y) = = 2 using Lemma 1.1. iii) and ii). Q.E.D. 8 Lemma 1.4. For all x and y in A n. k xy k=k __xy k Proof. k xy k2= and k __xy k2= <__xy; __xy> using Lemma 1.3. 1 __ __ = <__x(xy); y> = <(__x(xy))__y; e0> = __t((x(xy))y) 2 1 __ __ <__xy; __xy>= = <(x(__xy))__y; e0> = __t((x(xy))y)) 2 thus 1 __ __ __ k xy k2 - k __xy k2= __t[(x(xy) - x(xy))y] 2 1 __ __ __ __ __ = __t[(-(x; x; y) + (xx)y + (x; x; y) - (xx)y)y] 2 1 __ 2 2 __ = __t[((x; x; y) + (x; x; y)+ k x k y- k x k y)y] 2 1 __ = __t(0y) = 0: 2 . ..k xy k2=k __xy k2 and k xy k=k __xy k Q.E.D. Corollary 1.5. For all x and y in A n. k xy k=k __xy k=k x__yk=k yx k : Proof. Follows from the lemma recalling that ____ __ k z k=k __zk 8z 2 An so k xy k=k __xy k=k xy k=k yx k=k yx k Q.E.D. 9 Corollary 1.6. (Elementary facts on zero divisors). For x and y in A n n 4. 1. xy = 0 , yx = 0 , __xy = 0 , x__y= 0. 2. If x 6= 0 and xy = 0 then t(y) = 0. 3. If y 6= 0 and xy = 0 then t(x) = 0. 4. x2 = 0 if and only if x = 0. Proof. 1. By Corollary 1.5. k xy k=k yx k=k __xy k=k x__yk so xy = 0 , yx = 0 , __xy = 0 , x__y= 0. 2. xt(y) = x(y + __y) = xy + x__y= 0 if xy = 0 but t(y) is a real number so t(y) = 0. 3. t(x)y = (x + __x)y = xy + __xy = 0 if xy = 0 but t(x) is a real number so t(x) = 0. 4. If x2 = x . x = 0 then t(x) = 0 and x = -__x and k x k2= __xx = -x2 . .. x2 = 0 , x = 0: Q.E.D. Notation: OA n = {x 2 An |t(x) = 0}. Definition. 0 6= x 2 OA n is a zero divisor if there exists 0 6= y 2 OA n with xy = 0. 10 Notice that we don't have to distinguish between left and right zero di- visors because xy = 0 if and only if yx = 0. For x 2 An we define the following linear transformations. Lx; Rx : An ! A n by Lx(y) = xy and Rx(y) = yx 8 y 2 An clearly Lx and Rx are linear. Proposition 1.7. For x 2 OA n Lx and Rx are skew-symmetric. Proof. Let y and z in A n. Then = = = = : = = = = : Q.E.D. Remark: x 2 OA n is a zero divisor if and only if {0} 6= Ker Lx and {0} 6= Ker Rx also notice that Ker Lx = Ker Rx. Lemma 1.8. If x 2 An is zero divisor and x = (x1; x2) in A n-1x An-1 then x=..(x2; x1) is zero divisor in A n. Proof. Consider y 6= 0 in A nsuch that xy = 0 and y = (y1; y2) so xy = (x1; x2)(y1; y2) = (x1y1 - __y2x2; y2x1 + x2__y1) = (0; 0): 11 Consider the following product in A n-1x An-1 = An (-__y2; y1)(x2; x1)=(-__y2x2 - __x1y1; -x1__y2+ y1__x2) ____________ = (x1y1 - __y2x2; x2y1 + y2x1) Recall that t(x) = t(x1) = 0. Therefore (-__y2; y1)(x2; x1) = (0; 0) and x= (x2; x1) is a zero divisor. Q.E.D. Corollary 1.9. If x 2 An is a zero divisor and x = (x1; x2) then t(x) = t(x1) = t(x2) = 0. Proof. We already know that t(x) = t(x1) = 0. Now x is a zero divisor then t(x) = t(x2) = 0. Q.E.D. Definition. a 2 A n with a = (a1; a2) 2 A n-1 x A n-1 is doubly pure if t(a1) = 0 and t(a2) = 0. So any zero divisor is doubly pure. Let e0 be the neutral element in A n-1 so the neutral element in A n is e0 = (e0; 0) 2 An-1 x An-1. Denote by ee0the element ee0..=(0; e0) 2 A n-1x An-1 i.e. ee0= e2n-1 in n the canonical basis of R 2. Lemma 1.10 Let a 2 OA n i.e. t(a) = 0. Then a is doubly pure if and only 12 if a ? ee0. Proof. Calculating aee0= (a1; a2)(0; e0) = (-a2; a1) and ee0a = (0; e0)(a1; a2) = (-__a2; __a1) = (-__a2; -a1) because t(a) = 0 and __a 1= -a1. If a is doubly pure t(a2) = 0 and __a2= -a2 so ee0a = (a2; -a1) = -aee0. Conversely if a ? ee0then aee0= -ee0a and (-a2; a1) = (__a2; a1) then __a2= -a2 and t(a2) = 0. Q.E.D. Right multiplication by "ee0" defines an orthogonal transformation of de- terminant one whose square is minus the identity linear transformation in An. For x 2 An denotes by ex= Ree0(x) = (-x2; x1) if x = (x1; x2) so k exk=k x k and eex= -x. n And RTee0= -Ree0= R-1ee0and det(Ree0) = (-1)2 = 1. Proposition 1.11 For a = (a1; a2) and b = (b1; b2) doubly pure elements in An we have: 1) aee0= ea; ee0a = -ea. 2) aea= - k a k2 ee0; eaa =k a k2 ee0 so a ? ea. 3) eab = -aeb. 13 4) If a ? b then eab + eba = 0. 5) If ea? b then ab = ebea. Proof. 1) Follows from definition. 2) aea= (a1; a2)(-a2; a1) = (-a1a2 + a1a2; a21+ a22) = (0; - k a k2) = - k a k2 ee0 eaa = (-a2; a1)(a1; a2) = (-a2a1 + a2a1; -a22- a21) = (0; k a k2) =k a k2 ee0. Since t(a) = t(ea) = 0 then aea= -eaa and a ? ea. 3) eab = (-a2; a1)(b1; b2) = (-a2b1 + b2a1; -b2a2 - a1b1) so eeab= (a1b1 + b2a2; b2a1 - a2b1) = ab . .. eab = -aeb. 4) If ab = -ba then eab = -aeb= eba= -eba . .. eab + eba = 0. 5) If ea? b then 0 = aeb+ bea= -aeb+ eba= ab + ebea= ab - ebaand 0 = eeab + ebea= -ab + ebea. Q.E.D. Corollary 1.12 For 0 6= a 2 An (doubly pure) ax = 0 if and only if aex= 0 8x. Proof. ax = 0 , xa = 0 , fxa= 0 , exa = 0 , aex= 0. Q.E.D. Theorem 1.13 For a 2 An doubly pure of norm one and n 2. The vector space generated by {e0; ea; a; ee0} is multiplicatively closed and isomorphic as 14 algebra to A 2= H the quaternions. Proof. Construct the following multiplication table | |e0 ea a ee0 aee0 = ea _____|____________________ | e0 ||e0 ea a ee0 ee0a = -ea | ea ||ea -e0 ee0 -a since eaee0= eea= -a | a ||a -ee0 -e0 ea ee0ea= -eea= a | ee0||ee0 a -ea -e0 aea = -ee0 and eaa = ee0 Q.E.D. Recall that KerLra = KerLa for all r 6= 0 in R and that KerLa 6= {0} only if a is doubly pure non-zero element for n 4. From now on we assume that n 4; k a k= 1 and a is doubly pure element in A n. H adenotes the copy of the quaternions inside of An generated by {e0; ea; a;* * ee0}. Let Ta : An ! An be given by Ta(y) = (a; a; y). Because the associator is tri-linear, Ta is a linear map and because H ais associative Ta(H a) = {0}. On the other hand Ta(y) = a2y - a(ay) = -y - L2a(y) so Ta = -I - L2aand since La is anti-symmetric then Ta is symmetric n H?adenotes the orthogonal complement of H ain A n= R2 . Lemma 1.14 Ta(H ?a) H ?a. Proof. Let 0 6= y 2 H ?athen Ta(y) = (a; a; y). 15 By Lemma 1.2. a(a; a; y) + (a; a; a)y = (a2; a; y) - (a; a2; y) + (a; a; ay) . .. then a(a; a; y) = (a; a; ay) because a2 = -e0 and ________ = t(a(a; a; y)) = -t(a(a; a; y)) = -t((a; a; ay)) = 0 because the trace of any asociator is zero (Lemma 1.1). Similarly ea(a; a; y) + (ea; a; a)y = (Sum of Associators) since (ea; a; a) = 0 because H a is associative then ea? (a; a; y). Finally ee0(a; a; y) + (ee0; a; a)y = (Sum of Associators) since (ee0; a; a) = 0 because H a is associative then ee0? (a; a; y). Therefore (a; a; y) 2 H ?aif y 2 H ?a Q.E.D. eTadenotes the restriction of Ta to H ?aso eTa: H ?a! H ?a; eTa= -I - L2ais symmetric. Theorem 1.15 For each a 2 A n of norm one and doubly pure we have a direct sum decomposition. n M A n= R2 ~=H a KerTea KerLa V where V = {x 2 An |a(ax) = -2x}; > 0; 6= 1 16 Proof. Recall that La is skew-symmetric and consequently L2ais symmetric, negative semidefinite, that is for, all x 2 An = - = -|La(x)|2 0 Therefore a(ax) = 0 , ax = 0 for x 2 An On the other hand being L2asymmetric no positive implies that all its eigenvalues are negative real numbers. So there exist {0; 1; . .;.s} R with 0 = 0 and 1 = 1 and eigenspaces V = {x 2 H ?a|a(ax) = -2x}: Recall that L2a|H a = -I. And V0 = KerL2a= KerLa and V1 = KerTea. Q.E.D. Theorem 1.16. For the V 's as in the last theorem we have that for > 0 dimR V 0 mod 4: Proof. CLAIM 1. For x 2 V with > 0 y ..=-1(ax) belongs V , and y ? x. ay = -1a(ax) = -1(-2x) = -x and a(ay) = -ax = -2(-1ax) = -2y. Now = <-1ax; x> = -1 = 0. CLAIM 2. If x 2 V then ex2 V . 17 For this we observe that: Since x 2 H ?a; a ? x; a ? exand x is doubly pure then applying Proposition 1.11 (3) twice and noting that a(ax) = (xa)a by flexibility we have that a(aex) = (exa)a = -((gxa)a) = ag(ax)= -2ex. Therefore if we take 0 6= x 2 V for 6= 0 we may construct one orthog- onal set {x; y; ex; ey} inside of V . Thus dimR V 0 mod 4 > 0 Q.E.D. Corollary 1.17. 2n - 4 dimR KerLa 0 mod 4 Proof. Since dimH a = 4; 2n - 4 dimR KerLa and the last two theorems implies that dimR KerLa 0 mod 4: Q.E.D. 18 II. Zero divisors with alternative entries Throughout this chapter a and b are nonzero elements in A nfor n 3. Definition: a 2 An is alternative_if (a; a; x) = 0 for all x 2 An . The real elements (R e0) are alternative and since the associator symbol is linear in each variable we have that one element in A nis alternative if and only if it's imaginarie part is alternative. Therefore we restrict ourselves to the trace zero of norm one alternative elements in A n. By Schafer [8] we know that the canonical basis {e0; e1; . .;.e2n-1} consists of alternative elements. Lemma 2.1. If a 2 An is alternative with t(a) = 0 and |a| = 1 then i)KerLa = {0} ii)La 2 SO(2n) iii)ax = y has a unique solution for all y 2 An iv) L2a= -I Proof. 0 = (a; a; x) = -x - a(ax) then a(ax) = -x and if ax = 0 then a(ax) = -x = 0 and KerLa = {0} and La is one to one. Now La is skew- symmetric so 0 = <(a; a; x); x> = <-x - a(ax); x> = - - 19 = -|x|2 + = -|x|2 + |ax|2 then |La(x)| = |x| for all x 2 An0and La 2 O(n).1Therefore L-1a= LTa= -La 0 I and L2a= -I and La is similar to @ A for I and 0 in M2n-1(R ) which -I 0 has determinant one. Q.E.D. Also we have similar results for Ra, right multiplication. Definition: a 2 An is special_if a is alternative, t(a) = 0 and |a| = 1. If a 2 A nis special then {e0; a} generates a copy a A 1= C inside of A n. So the special elements can be regarded as a closed subset of the manifold of the 2n x 2 matrices of rank two with real coeficients. Definition. {a; b} A n is a special couple if both a and b are special elements in A nand a is orthogonal to b (a ? b) i.e. ab = -ba. Notation: V (a; b) denotes the vector space generated by the set {e0; a; b; ab}; obviously V (a; b) is fourth-dimensional: a ? ab and b ? ab. Proposition 2.2. V (a; b) is multiplicatively closed and isomorphic to A2 = H the quaternions. _ Proof: Since a ? b; t(ab) = -t(ab) = 0 and |ab|2 = = <-a(ab); b> = = |b|2 = 1. Therefore t(ab) = 0 and |ab| = 1. {e0; a; b; ab} is an 20 orthonormal set. Clearly a(ab) = a2b = -b and b(ab) = -b(ba) = -b2a = a. So we have the following multiplication table | |e0 a b ab ____|______________________ | e0 ||e0 a b ab | a ||e0 -e0 ab -b | b || b -ab -e0 a | ab ||ab b -a -e0 Q.E.D. Therefore each special couple induce a multiplicative monomorphism from A2 = H to A nand the set of special couples can be regarded as closed subset of the 2n x 4 matrices with real entries and rank four. Depending on {a; b} special couple we define S : An ! An by S(y) = (a; y; b) First of all S is a linear map and V (a; b) KerS because V (a; b) ~= H is associative. Also S = RbLa - LaRb = [Rb; La] is skew-symmetric because Rb and La are skew-symmetric. Therefore S2(y) = S(S(y)) = 0 if and only if S(y) = 0 and S : V (a; b)? ! V (a; b)? because S(y) 2 V (a; b) for y 2 V (a; b). We are interested in calcu- late KerS and ImS for {a; b} special couple and S restricted to V (a; b)? . Theorem 2.3. For {a; b} special couple and S and V (a; b) as above we have that Ker(S : V (a; b)? ! V (a; b)? ) = KerLa+b KerLa-b: 21 Proof. First of all we notice that KerLa+b \ KerLa-b = {0}. If 0 = (a + b)z = (a - b)z then az + bz = az - bz = 0 and 2bz = 0 and z = 0 because b is alternative. Now if (a + b)x = 0 then ax = -bx = xb for x 2 V (a; b)? then (a; x; b) = (ax)b - a(xb) = (xb)b - a(ax) = xb2 - a2x = 0. Similarly if (a - b)x = 0 then ax = bx = -xb for x 2 V (a; b)? . Thus (a; x; b) = (ax)b - a(xb) = -(xb)b + a(ax) = -xb2 + a2x = 0. Therefore KerLa+b KerLa-b KerS. Now if y 2 V (a; b)? with S(y) = 0 we have that (a + b)(y - a(yb))= ay + by - a2yb - b(a(yb)) = ay + by - by + ((ay)b)b = ay + ayb2 = ay - ay = 0 because a(yb) = (ay)b and b ? [(ay)b]. Similarly (a - b)(y + a(yb)) = 0 if S(y) = 0 and y 2 V (a; b)? . Since y = 1_2[(y - a(yb)) + (y + a(yb)] we have that: If S(y) = 0 and y 2 V (a; b)? then y 2 KerLa+b KerLa-b and we are done. Q.E.D. Corollary 2.4 dimR KerS = 2dimKerLa+b 0 mod 8. Proof. By the proof of the Theorem 2.3. we know that the elements in KerLa+b are of the form [y - (a(yb)] and the elements in KerLa-b are of the 22 form [y + (a(yb))] for (a; y; b) = 0. Therefore the assignment y + a(yb) 7! y - a(yb) defines a linear isomorphism between KerLa+b and KerLa-b. Therefore dimR KerS = 2dimKerLa+b that by Theorem 1.15 is congruent with 0 mod- ule 8. Lemma 2.5 For {a; b} special couple in A nand y 2 V (a; b)? we have that (a + b; a + b; y) = -(a; y; b) + 2(ay)b = (a; y; b) + 2a(yb) in short Ta+b = -S + 2RbLa = S + 2LaRb on V (a; b)? Proof. (a + b; a + b; y)= (a; a; y) + (b; b; y) + (a; b; y) + (b; a; y) = 0 + 0 + (ab + ba)y - a(by) - b(ay) = a(yb) + (ay)b = -(a; y; b) + 2(ay)b = (a; y; b) + 2a(yb) Q.E.D. Corollary 2.6 Let {a; b} be one special couple and y 2 V (a; b)? in A n. (a + b; a + b; y) = 0 if and only if 1 (a; b)(-__S(y); y) = (0; 0) in A n+1; 2 23 Proof. By Lemma 2.5 Ta+b(y) = 0 , a(yb) = -(ay)b then (a; b)(a(yb); y)= (a(a(yb)) + yb; ya - b(a(yb))) = (a2yb + yb; ya - ((ay)b)b) = (-yb + yb; ya - (ay)b2) = (0; 0) because y 2 V (a; b)? . Conversely if (a; b)(x; y) = (0; 0) for y 2 V (a; b)? then ax + yb = 0 and ya - bx = 0 so ax = -yb ) a(ax) = a2x = -x = -a(yb) . ..x = a(yb) bx = ya ) ay = xb ) (ay)b = (xb)b = xb2 = -x: Therefore a(yb) + (ay)b = x - x = 0. And by the Lemma 2.5 Ta+b(y) = 0 and x = 1_2S(y) Q.E.D. Theorem 2.7 Let {a; b} be one special couple in A nand L(a;b): An+1 ! An+1 then dimR KerL(a;b) 2n - 4 - 2dimR KerLa+b Proof. By the last corollary KerL(a;b)~=Ker[Ta+b : V (a; b)? ! V (a; b)? ] but KerTa+b ImS restricted to V (a; b)? and ImS ~=(KerS)? = (KerLa+b KerLa-b)? (by Theorem 2.3). Also we know that KerLa+b ~=KerLa-b and dimR KerL(a;b) dimR A n- dimR V (a; b) - 2dimR KerLa+b. 24 Q.E.D. Examples: For n = 3. All element in A 3are alternative so KerTa+b = A 3for all a and b. Then dimKerL(a;b)= 4 for all special couple with S(y) 6= 0 for some y 2 V (a; b)? . For n 4. The top dimension 2n - 4 is always realizable. For instance if a is one special element and doubly pure in A nthen eais also special and V (a; ea) = H athen KerTa+ea= {0} so dimKerL(a;ea)= 2n - 4. More generally if {a; b} is one special couple in An with (a+b) alternative then KerL(a;b)~=V (a; b)? i.e. dimKerL(a;b)= 2n - 4. Now we analyze a more general case: Given a and b alternative non-zero elements in A n. Under what conditions on a and b; (a; b) is a zero divisor in A n+1? Suppose that (a; b)(x; y) = (0; 0) for x 6= 0 and y 6= 0. Then 1) t(a) = t(b) = 0 (Corollary 1.9) 2) |a| = |b| ax = -yb ) |a||x| = |b||y| and ya = bx ) |a||y| = |x||b| so |a|2|y| = |a||x|b| = |b|2|y| and |a| = |b|: Notice that without loosing generality we may assume that |a| = |b| = 1 from now on. 25 3) a and b are linearly independent. Suppose that a = b for 6= 0 in R so ax = -yb ) bx = -yb and ya = bx ) yb = bx so (bx) = (yb) = 2yb = -yb ) 2 = -1 which is a contradiction. 4) (a; y; b) = -2x and (a; x; b) = 2y. Recall that (a; b)(x; y) = (0; 0) , (a; b)(-y; x) = (0; 0). Thus ax = -yb and xa = -by ya = bx ay = xb Then ax = -yb ) a2x = -a(yb) ) x = a(yb) ay = xb ) (ay)b = xb2 ) x = -(ay)b Also xa = -by ) b(xa) = -b2y = y ya = bx ) ya2 = (bx)a ) -y = (bx)a Therefore (a; y; b) = (ay)b-a(yb) = -x-x = -2x and (a; x; b) = -(b; x; a) = -[(bx)a - b(xa)] = 2y. Notice that we also show that x ? y because S ..=(a; -; b) is skew- symmetric and that x and y belongs to {e0; a; b; ab}? because S(e0) = S(a) = S(b) = 0 and S(ab) = -aS(b) = 0. 26 5) Ta+b(y) ..=(a + b; a + b; y) = -2y and L2a+b(y) ..=(a + b)[(a + b)y] = -2y (a + b)[(a + b)y] = (a + b)[ay + by] = a(ay) + b(by) + b(ay) +a(by) = -y - y + x - x = -2y because b(ay) = -(ay)b = x and a(by) = -a(yb) = -x. And Ta+b(y) = (a + b)2y - L2a+b(y) = (a2 + b2 + ab + ba)y + 2y = -2y - 2y + 2y = -2y We collect enough necessary conditions to stablish. Theorem 2.9 Let a and b alternatives elements in An of zero trace and norm one, then (a; b) 2 An+1 is a zero divisor if and only if = -2 is an eigenvalue of L2a+b Proof. )) Follows from the conditions enlisted above. () Let y 6= 0 in A nsuch that L2a+b(y) = -2y then -2y = (a + b)[(a + b)y] = a2y + by2 + a(by) + b(ay) = -2y + a(by) + b(ay) 27 therefore a(by) = -b(ay): On the other hand L2a-b(y) = (a - b)[(a - b)y] = (a2 + b2)y - (b(ay) + a(by)) = -2y: then y ? (a + b) and y ? (a - b) and y ? a and y ? b Consider the following product in A n+1 (a; b)(-a(by); y)= (-a(a(by)) + yb; ya + b(a(by)) = (-a2(by) + yb; ya - b(b(ay)) = (by + yb; ya - b2(ay) = (-yb + yb; ya - ya) = (0; 0) because a(by) = -b(ay); a ? y and b ? y. Q.E.D. Remark: Condition 5) above shows that Theorem 2.9 implies Corollary 2.6 it is the case when = 0. Corollary 2.10 For a and b alternative elements in A n of zero trace and norm one dimKerL(a;b) 2n - 4 - 2dimKerLa+b 28 Proof. Notice that a(by) = -b(ay) , S(y) = -2(ay)b for y 2 V (a; b)? such that (a; b)(x; y) = (0; 0) then KerL(a;b) (KerS)? ~= (KerLa+b KerLa-b)? then dimKerL(a;b) 2n - 4 - 2dimKerLa+b because KerLa+b ~=KerLa-b. Q.E.D. Remarks: Notice L2a+b(y) = -2y for a; b as in the Theorem 2.9 and y 6= 0 implies that a and b are linearly independent. If it would a = b or a = -b then L2a+b= -4I and L2a-b= 0. There are examples of couples of alternatives linearly independent ele- ments (of norm one and zero trace) in A n which no form a zero divisor in An+1. For instances put a = e1 and b = e1+e2_p_2in A3 then (a; b) 2 A4 is neith* *er a zero divisor nor an alternative element. (We thank to Paul Yiu for this piece of information). Definition. A triple {a; y; b} in A nis special if it is an orthonormal set and i) (a; b) is an special couple ii) (ay)b = -a(yb) Definition. A zero divisor (a; b) in A n+1 is special if (a; b)(x; y) = (0; 0) implies that {a; y; b} is one special triple. Proposition 2.11 Let a and b are alternatives of norm one in A n with 29 (a; b)(x; y) = (0; 0) in A n+1. If (a + b) is alternative then (a; b) is an spe* *cial zero divisor. Proof. By Theorem 2.9) Ta+b(y) = -2y. So 0 = Ta+b(y) ) a ? b and {a; b} is one special couple on the other hand (ay)b = -(ay)b by 4). Q.E.D. Corollary 2.12 Any zero divisor (up to norm) in A 4is special zero divisor. Consider the vector subspace in A n V (a; y; b) ..=<{e0; a; b; ab; (ay)b; yb; ay; y}> Theorem 2.13 V (a; y; b) is a copy of A 3= O the octonian numbers inside of A nif and only if {a; y; b} is an special triple. Proof. Let's make the following assignment. e0 a b ab (ay)b yb ay y | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | e0; e1; e2; e3; e4 e5 e6 e7 we may easily see that this define an homomorphism of A 3to A nif {a; y; b} is an special triple. Notice in fact that this a monomorphism from A 3to A n. Conversely is easy to see that {e1; e7; e2} form in special triple in A nfor n 3. Q.E.D. 30 Corollary 2.14 The set of zero divisor in A 4(with entries of norm one) are homemorphic to Aut(A 3) = G2 the exceptional Lie group of rank two. Remarks: Notice that if {a; y; b} is an special triple then {b; y; a} is an special triple and {a; b; y} is no necessarily an special triple. e.g. a = e1 b = e2 y = e15 in A 4. (e1; e15; e2) = 2e12 and (e1e15)e2 = -e1(e15e2) = e12 but (e1; e2; e15) = 0 and (e1e2)e15 = e1(e2e15). REFERENCES 1. Adem, Construction of Some Normed Maps. Bol. Soc. Mat. Mexicana (1975), 59-75. 2. F.R. Cohen. On Whitehead Squares, Cayley-Dickson algebras and rational functions. Bol. Soc. Mat. Mexicana 37(1992), 55-62. 3. P. Eakin, A. Sathaye. On Automorphisms and Derivations of Cayley- Dickson Algebras. Journal of Algebra (1990). 4. P. Eakin, A. Sathaye, Yiu's Conjecture. Unpublished, University of Kentucky (1992). 5. Kantor-Solodovnikov. Hypercomplex Numbers. Springer-Verlag (1989). 6. S. Khalil. The Cayley-Dickson Algebras. (Master's Thesis 1993), Florida Atlantic University. 7. Koecher-Remmert. Numbers. Part B. Graduate Texts in Mathematics 123, Springer-Verlag 1991. 8. R.D. Schafer. On algebras formed by the Cayley-Dickson process. American Journal of Mathematics, (1954), 435-446. 9. W. Whitehead. Elements of Homotopy Theory. (Appendix A). Grad- uate text in Mathematics 61, Springer-Verlag. Departamento de Matematicas Centro de Investigacion y de Estudios Avanzados del I.P.N. Apdo. Post. 14-740,07000, Mexico, D.F., Mexico e-mail: gmoreno@math.cinvestav.mx 31