Alternative elements in the Cayley-Dickson
algebras
Guillermo Moreno
Abstract
n
We describe the alternative elements in A n = R 2 the Cayley-
Dickson algebras for n 4. Also we "measure" the failure of A n
with n 4 of being a normed algebra in terms of the alternative
elements.
Introduction.
n
An = R2 denotes the Cayley-Dickson algebra over R the real numbers.
For n = 0; 1; 2; 3 A n is the real, complex, quaternion and octonion num-
bers, respectively (denoted by R ; C; H and O , respectively). These algebras
are normed i.e.
||xy|| = ||x||||y||:
n
For all x and y in R 2 n = 0; 1; 2; 3. Also A n is alternative for n =
0; 1; 2; 3 , i.e.
x2y = x(xy) and xy2 = (xy)y
for all x and y in A n.
As is well known these algebras are defined inductively by the Cayley-
Dickson process:
x = (x1; x2) y = (y1; y2) in A n x An = An+1
xy = (x1y1 - __y2x2; y2x1 + x2__y1) and
__x= (__x
1; -x2):
For n 4; An is a neither normed nor alternative algebra, but An is flexible,
(see [3]) i.e.
(xy)x = x(yx) for all x and y in An
1
In this paper we characterize the subset of A n consisting of alternative el-
ements, i.e. {a 2 A n|a(ax) = a2x for all x in A n} and in terms of this
characterization we "measure" the failure of A n (n 4) of being normed .
Introducing the associator notation
(a; b; c) := (ab)c - a(bc):
We have that a in A nis alternative if and only if (a; a; x) = 0 for all x in A*
* n
and A n is flexible means: (a; x; b) = 0 for all x in A n whenever a and b are
linearly dependent.
In x2 we prove that (a; x; b) = 0 for all x in A n n 4 if and only if a
and b are linearly dependent.(For a and b pure elements)
This is known by the experts as Yiu's conjecture [1].
Then we use the positive solution to Yiu's conjecture to characterize the
alternative elements.
Here is an outline of the main ideas.
First of all we notice that if La : An ! An denotes the left multiplication
by a in A n then La is a linear transformation. We know that for a pure
element i.e. a ? e0 := (1; 0; : :;:) La is skew-symmetric so L2ais symmetric
non-positive definite [2].
But a is alternative if and only if its imaginary part is alternative and for
a pure alternative element L2a= a2I = -||a||2I i.e. L2ahas all its e igenvalues
equal to -||a||2.
By direct calculation we see that for a and b pure elements in A n
L2(a;b): An+1 ! An+1
is given by
L2(a;b)(x; y) = (A(x) - S(y); A(y) + S(x))
where
A = L2a+ R2b: An ! An
and
S = (a; -; b) = RbLa - LaRb : An ! An
where Rb is the right multiplication by b (see Lemma 3.2) So if (a; b) is
alternative in A n+1then (A(x) - S(y); A(y) + S(x)) = (a; b)2(x; y) for all x
and y then S(x) = 0 for all x in A nthen a and b are linearly dependent and
A(x) = (a2 + b2)x for all x so a and b are alternatives (see Theorem 3.3)
2
Thus alternative elements in A n+1 are "made from" the alternative ele-
ments in A n. In spite of the fact that in A 3 all element is alternative, in
A4 they form a "very small" subset, this set remains "constant" during the
doubling process.
In x4 we define the property of being strongly alternative: a in A n is
strongly alternative if (a; a; x) = (a; x; x) = 0 for all x in A n n 4. They
form even a smaller subset, namely {re0 + see0|r and s in R } andnee0= e2n-1
in A nwhere {e0; e1; : :;:e2n-1} is the canonical basis of R 2.
In x5 we study the properties of alternativity and strongly alternative
locally, i.e. between two elements in A n.
This give us a criterion to know which elements generate associative and
alternative normed subalgebras inside of A n n 4.
In this paper, sequel of [2], we use that the canonical basis {e0; e1n: :;:e*
*2n-1}
consists of alternative elements and that the Euclidean structure in R 2 and
the C - D algebra structure are related by
2 = x__y+ y__x
||x||2 = x__xfor all x and y in A n. (See [3]).
x1. Pure and doubly pure elements.
For a = (a1; a2) 2 An-1 x An-1 = An the trace is
tn : An ! A0 = R
tn(a) = a + __a:
Definition. We say that a 2 An is pure if
tn(a) = 0 i:e: __a= -a:
Notice that a is pure in A nif and only if a1 is pure in Ann-1.
Let {e0; e1; : :;:e2n-1} be the canonical basis in R 2 so e0 = (1; 0; : :;:0)
is the unit in the algebra A n.
Notation. 0A n:= {a 2 An |a is pure } = {e0}? .
For a = (a1; a2) 2 An-1 x An-1 = An we denote
ea= (-a2; a1) 2 An-1 x An-1 = An :
3
Note that ||a||2 = ||ea||2 = ||a1||2 + ||a2||2.
In terms of the decomposition A n= An-1 x An-1
e0 = (e0; 0); e1 = (e1; 0); : :;:e2n-1-1= (e2n-1-1; 0);
e2n-1= (0; e0); e2n-1+1= (0; e1); : :;:e2n-1 = (0; e2n-1-1);
we have that eei= e2n-1+ifor 0 i 2n-1, and
eei= -ei-2n-1for 2n-1 < i 2n - 1; so ee0= e2n-1 and ea= aee0.
Definition. An element a = (a1; a2) 2 A n-1x An-1 = A n is doubly pure
if both coordinates a1 and a2 are pure elements in A n-1 i.e. tn-1(a1) =
tn-1(a2) = 0. Notice that a in A n is doubly pure if and only if a and eaare
pure elements in A n.
Notice that for any a pure element in A n, a = rc + see0for r and s real
numbers and c doubly pure element in A n.
Notation. Aen := {a 2 An |a is doubly pure }.
Notice that eAn= {e0; ee0}? =oA n-1xoAn-1n An .
Now the Euclidean product in A n= R2 is given by
2 = tn(x__y) = x__y+ y__x
and the Euclidean norm by ||x||2 = x__x. For a = (a1; a2) 2 An-1 x An-1 = An
doubly pure we have that a 2 {e0; ee0}? so ea= aee0= -ee0a and eaa + aea=
(aee0)a - a(ee0a) = (a; ee0; a) = 0 (by flexibility) then ea? a.
Lemma 1.1. Let a = (a1; a2) and x = (x1; x2) be elements in An-1 xA n-1 =
An. If x is doubly pure then eax = -afx.
Proof. Define c = a1x1 + x2a2 and d = x2a1 - a2x1 in A n-1so ax = (a1x1 +
x2a2; x2a1-a2x1) = (c; d) and eax = (-a2x1+x2a1; -x2a2-a1x1) = (+d; -c)
and fax= (-d; c) so eax = -afx. Q.E.D.
Corollary 1.2. For a and x doubly pure elements in A nwe have that
1) eax + exa = 0 if and only if a ? x.
2) ax - exea= 0 if and only if ea? x.
3) eax = 0 if and only if ax = 0.
4
Proof.
1) a ? x if and only if ax = -xa , fax= -xfa, eax = -exa.
2) ea? x , eax = -xea, feax= -xfea, ax = exea.
3) 0 = ax , 0 = fax, 0 = eax. Q.E.D.
Corollary 1.3. For 0 6= a in A ndoubly pure element and n 3 the vector
subspace of A ngenerated by {e0; ea; a; ee0} is a copy of the quaternions A 2.
Proof. Since a 2 {e0; ee0}? then ea2 {e0; ee0}? and a ? easo {e0; ea; a; ee0} is
an orthogonal set of fourt vectors in A n for n 3. (we denote it by H a).
Now we suppose that ||a|| = 1 otherwise we take _a_||a||. From Lemma 1.1 and
Corollary 1.2 we have the following quaternion multiplication table.
|e ea a ee
____|_0__________________0_
e0 ||e0 ea a ee0
ea ||ea -e0 -ee0 -a
a ||a ee0 -e0 ea
ee0||ee0 a -ea -e0
here e0 $ 1;bi$ ea; ^j$ a and bk$ ee0. Q.E.D.
Proposition 1.4. For a and b doubly pure elements in A n for n 3 we
have:
1) eab = aebif and only if a ? b and ea? b.
2) If (a; ee0; b) = 0 then ab = re0 + see0for r and s in R .
Proof. Let a = (a1; a2) and b = (b1; b2) be in oAn-1xoA n-1. If r = a1b1+b2a2
and s = b2a1 - a2b1 in A n-1 then ab = (r; s) 2 A n-1 x A n-1 = A n. So
fab= (-s; r) and eab = (s; -r). On the other hand aeb= (a1; a2)(-b2; b1) =
(-a1b2 + b1a2; b1a1 + a2b2) so aeb= (-__s; __r). Therefore eab = aeb, s = -__sa*
*nd
__r= -r , t
n-1(s) = 0 and tn-1(r) = 0 , tn(ab) = 0 and tn(eab) = 0 , a ? b
and ea? b so we prove 1).
To prove 2) we see that (a; ee0; b) = eab + aebso if 0 = (a; ee0; b) in A nt*
*hen
(0; 0) = (s; -r) + (-__s; __r) = (s - __s; -r + __r) in A n-1 x An-1
5
so __s= s and __r= r and s and r are real numbers therefore ab = re0 + see0.
Q.E.D.
Corollary 1.5. For a and b doubly pure elements in A nfor n 3, we have
that (
0 if b 2 H a
-(ee0; a; b) = (a; ee0; b) = ?
2eab if b 2 H a
Proof. Since H a is associative -(ee0; a; b) = (a; ee0; b) = 0 for b 2 H a. If
b 2 H ?athen a ? b and ea? b and by Proposition 1.4 (2)
(a; ee0; b) = eab + aeb= 2eab = -(ee0a)b + ee0(ab) = -(ee0; a; b):
Q.E.D.
Remark. Notice that also we prove that for a and b doubly pures, if b 2 H a
then ab = re0 + see0for r and s in R .
Lemma 1.6. Let a be a doubly pure element in A nfor n 3 we have that
(
0 if x 2 H a
(ea; x; a) = ?
-2a(aex) if x 2 H a
Proof. Since H a= <{e0; ea; a; ee0}> is associative
(ea; x; a) = 0 for x 2 H a:
Suppose that 0 6= x 2 H ?athen x is doubly pure and a ? x and ea? x. By
Proposition 1.4 (1) and flexibility (eax)a = (aex)a = a(exa) = -a(aex) (recall
that a ? ex). Since a ? x then ax is pure and ea? x implies that ax is doubly
pure then applying Proposition 1.4 (1) to a and xa we have that
ea(xa) = a(xfa) = -a(exa) = a(aex)
because a ? xa and ea? xa (recall that right multiplication by any pure
element is a skew symmetric linear transformation). Therefore (ea; x; a) =
(eax)a - ea(xa) = -2a(aex). Q.E.D.
x2. Yiu's conjecture.
In this section we give an affirmative answer to following question: Is the
converse of the following (trivial) statement true?
6
If a and b are linearly dependent in A nwith n 3 then (a; x; b) = 0 for
all x in A n.
We first show the case for a and b doubly pure elements and then we
proceed with the general case:
Lemma 2.1. Let a 2 A nbe a doubly pure element with n 4. If ax = 0
for all x 2 H ?a then a = 0.
Proof. Let `"' denote the basic element e2n-2in An i.e. " = e4 in A4 " = e8
in A 5; : :;:etc. Therefore " is an alternative element so " can't be a zero
divisor. Suppose that ax = 0 for all x 2 H ?a An and n 4. If " 2 H ?athen
a" = 0 and a = 0. If " 2 H athen H a= H "and ax = 0 if and only if "x = 0
i.e. x = 0 and H ?a= {0} and A n = H a but n 4 and 2n > 4 = dimH a
(contradiction). Let's suppose that " = ("0+ "00) 2 H a H ?awith "006= 0 so
a"00= 0.
Therefore a" = a"0+ a"00= a"0+ 0 = a"02 H a.
Since a is doubly pure, a" ? a and a" ? easo we have a" 2 H a implies
that a" = re0 + see0for some r and s in R , and a(a") = ra + sea.
On the other hand a(a") = a(a"0) = a2"0 because H a is associative so
-||a||2"0 = a2" = ra + seaand -||a||2"0"00= (ra + sea)"00= ra"00+ sea"00=
0 + 0 = 0 (Lemma 1.1) and "0"00= 0 unless a = 0.If a 6= 0 then "0" =
"0("0+ "00) = "02+ 0 = -||"0||2 2 R so " and "0 are linearly dependent and
" 2 H a. But this is impossible,so a = 0. Q.E.D.
Notice that in this proof we use only the fact that " is an alternative
element of norm one.
Theorem 2.2. For a and b non-zero doubly pure elements in A nand n 4
we have: If (a; y; b) = 0 for all y 2 An then a and b are linearly dependent.
Proof. We proceed by contradiction. Suppose that a and b are linearly
independent. Without loss of generality we can suppose that a is orthogonal
to b because, by flexibility, (a; y; b) = (a; y; b - ra) for all y and r = **_. On
the other hand if (a; ee0; b) = 0 then by (Proposition 1.4 (2)) ab = pe0 + qee0
for p and q in R , but a ? b so p = 0 and ab = qee0then ea and b are
linearly dependent. Now (a; x; ea) = -(ea; x; a) = 0 for all x 2 An implies that
a(aex) = 0 for all x 2 H ?a(Lemma 1.6) so ax = 0 for all x 2 H ?a(recall that
7
KerLa = KerL2a). By Lemma 2.1 a = 0 which is a contradiction. Q.E.D.
Now we proceed with the general case: Since any associator with one
entrie equal to e0 automatically vanishes we have to prove that if (ff; x; fi) *
*= 0
for all x 2 An then ff and fi are linearly dependent for ff and fi pure.
But ff = a + pee0and fi = b + qee0where a and b are doubly pure elements
and p and q are real numbers.
From now on we suppose that: (ff; x; fi) = 0 8x 2 An (n 4)
Suppose that b = 0.
Thus (a + pee0; x; ee0) = (a; x; ee0) + p(ee0; x; ee0) = (a; x; ee0) but (a; x;*
* ee0) =
fax- aex= -eax - aex= -2eax for all x 2 H ?a.
So (ff; x; fi) = 0 if and only if ax = 0 for all x 2 H ?aand by Lemma 2.1
a = 0 and ff and fi are linearly dependent. The argument is similar for a = 0.
Suppose that a 6= 0 and b 6= 0
(ff; x; fi) = (a + pee0; x; b + qee0) = (a; x; b) + (qa - pb; x; ee0): *
* (*)
Put x = ee0
(ff; ee0; fi) = (a; ee0; b) + (qa - pb; ee0; ee0) = (a; ee0; b) + 0:
So, (ff; ee0; fi) = (a; ee0; b) = 0 then by Proposition 1.4 (2)
ab = re0 + see0
for r and s real numbers. Put x = a
(ff; a; fi) = (a; a; b) + (qa - pb; a; ee0) = (a; a; b) + q(a; a; ee0) + p(b*
*; a; ee0)
but (a; a; ee0) = 0 because H ais associative and
(b; a; ee0) = fba- bea= -eba - bea= -(b; ee0; a) = (a; ee0; b) = 0
Therefore (ff; a; fi) = (a; a; b) = 0 and
a2b = a(ab) = a(re0 + see0) = ra + sea
Since a 6= 0 b = ua + veawhere
r s
u = ___; v = ___
a2 a2
8
and substituting in (*) above we have
0 = (ff; x; fi) = (a; x; b) + (qa - pb; x; ee0)
= (a; x; ua + vea) + (qa - p(ua + vea); x; ee0)
= u(a; x; a) + v(a; x; ea) + (q - pu)(a; x; ee0) - pv(ea; x; ee0)
= 0 + v(a; x; ea) + (q - pu)(a; x; ee0) + pv(ea; x; ee0):
Now we have
(a; x; ea)= (a; x; ee0) = (ea; x; ee0) = 0 for x 2 H aand
(a; x; ea)= -(ea; x; a) = 2a(aex) for x 2 H ?a(Lemma 1.6)
(a; x; ee0)= fax- aex= -eax - aex= -2eax for x 2 H ?a
(ea; x; ee0)= (efax) - eaex= ax - xa = 2ax for x 2 H ?a:
Now multiplication by ee0and `a' are skew-symmetric linear transformations
so {-eax = fax; ax; a(aex)} is an orthogonal subset.
By Lemma 2.1 ax = 0; eax = 0 and a(aex) = 0 for all x 2 H ?aonly if
a = 0.Therefore 0 = (ff; x; fi) for all x in A nimplies that:
v = 0 q - pu = 0 and pv = 0
then s = 0 q = pu.
To finish we argue as follows:
Since b = ua then b + qee0= ua + puee0= u(a + pee0), i.e. fi = uff and ff
and fi are linearly dependent, so we proved Yiu's conjecture:
Theorem 2.3. If a and b are non-zero pure elements in A nfor n 4 such
that (a; x; b) = 0 for all x in A nthen a and b are linearly dependent.
x3. Alternative elements in A n n 4.
Definition. a 2 An is an alternative element if (a; a; x) = 0 for all x in A n.
It is known [3] that the elements in the canonical basis are alternative.
Clearly a scalar multiple of an alternative element is alternative but the
sum of two alternative elements is not necessarily alternative.
Because the associator symbol vanish if one of the entries is real, i.e. be-
longs to Re0, then an element is alternative if and only if its pure (imaginary)
9
part is alternative, Therefore we need to characterize the pure (non-zero) al-
ternative elements.
Proposition 3.1. Let a 2 An (n 4) be a pure element with
a = rc + see0
r 6= 0 and s in R and c doubly pure. Then a is alternative if and only if c is
alternative. (We define `c' as the doubly pure part of a).
Proof. For all x in A n
(a; a; x)= (rc + see0; rc + see0; x)
= r2(c; c; x) + s2(ee0; ee0; x) + rs(c; ee0; x) + rs(ee0; c; x):
But ee0is alternative so (ee0; ee0; x) = 0 and by Corollary 1.5 (c; ee0; x) +
(ee0; c; x) = 0 so (a; a; x) = r2(c; c; x).
If r = 0 then a = see0and a is alternative.
If r 6= 0 then (a; a; x) = 0 if and only if (c; c; x) = 0 for all x 2 An . Q*
*.E.D.
Because of Proposition 3.1 we will focus on the doubly pure alternative
elements in A nfor n 4.
For a and b pure elements in A nand n 3 we denote La; Rb : A n! A n
the left and right multiplication by a and b and L(a;b): An+1 ! An+1 the left
multiplication by (a; b) in A n+1. Note that by flexibility (a; a; x) = -(x; a;*
* a)
and a(ax) = (xa)a i.e. L2a= R2a.
Notation. For a and b fixed pure elements
A := L2a+ R2b
S := (a; -; b) = RbLa - LaRb = [Rb; La]:
Lemma 3.2. For a and b pure elements in A n n 3.
L2(a;b)(x; y) = (A(x) - S(y); A(y) + S(x))
for (x; y) 2 An x An = An+1 .
Proof. (Direct calculation). We have (a; b)(x; y) = (ax - __yb; ya + b__x) and
10
(a; b)[(a; b)(x; y)] = (a;_b)(ax_- __yb; ya + b__x) ________
= (a(ax - __yb) - (ya + b__x)b; (ya + b__x)a + b(ax - __yb))
= (a(ax) - a(__yb) - (___ya)b + (xb)b; (ya)a + (b__x)a + b(___ax) + b(*
*by))
= (a(ax) + (xb)b + (a__y)b - a(__yb); a(ay) + (yb)b + (b__x)a - b(__xa*
*))
= (L2a(x) + R2b(x) + S(__y); L2a(y) + R2b(y) - S(__x))
= (A(x) - S(y); A(y) + S(x)):
_
because (a; __x; b) = -(a; x; b) for all x, __a= -a,b = -b and b(by) = (yb)b.
Q.E.D.
Remark. Notice if we interchange the role of a and b we have that
L2(b;a)(x; y) = (A(x) + S(y); A(y) - S(x))
by flexibility.
Theorem 3.3. For a and b pure elements in A n, for n 3, consider the
following statements:
(i)(a; b) is an alternative element in A n+1.
(ii)a and b are alternative elements in A n.
(iii)a and b are linearly dependent in A n.
Then (i) if and only if (ii) and (iii).
Proof. Suppose that (a; b) is alternative in A n+1. Then Lemma 3.2 implies
L2(a;b)(x; y) = (A(x) - S(y); A(y) + S(x)) = (a; b)2(x; y); and
A(x) - S(y) = (a2 + b2)x
A(y) + S(x) = (a2 + b2)y
for all x and y in A n.
Let's put x = y and substract,so
S(x) = 0 for all x 2 An :
11
By the positive solution to Yiu's conjecture Theorem 2.3 we have that a and
b are linearly dependent. Now A(x) := (L2a+ R2b)(x) = (a2 + b2)x and a and
b linearly dependent implies that a(ax) = a2x and (xb)b = b2x, i.e. a and
b are alternative in A n. Conversely if a and b are linearly dependent, then
S(x) := (a; x; b) = 0 for all x 2 A n and L2aand R2bare multiples of each
other. So L2(a;b)(x; y) = (A(x); A(y)) = ((L2a+ R2b)(x); (L2a+ R2b)(y)) but a
and b are also alternative so L2a(x) = a2x and R2b(x) = b2x for all x in A n.
So L2(a;b)(x; y) = (a2 + b2)(x; y) for all (x; y) in A n+1and (a; b) is alterna*
*tive
in A n+1. Q.E.D.
Notation.
Altn = {a 2 An |a is alternative}
Alt0n = {a 2oA n|a is alternative}
Agltn = {a 2 eAn|a is alternative}
i.e. while Altn denote the subset of An consisting of alternative elements, Alt*
*0n
and gAltndenote the pure and doubly pure alternative elements respectively
in A n.
So by the decompositions
An =oA n Re0 = (Aenx Ree0) x Re0
we have Altn ~=Alt0nx R and gAltnx R = Alt0n. Now Alt3 = A 3; Alt03=o A3
and gAlt3= eA3so gAlt4= {(ra; sa) 2 A 3x A3|r and s in R } and gAltn+1=
{(ra; sa) 2 Alt0nx Alt0n|r and s in R } i.e. gAltn+1is a 'cone' on Alt0n
with an extra point (r; s) = (0; 0).
x4. Strong alternativity
Definition. An element a in A nwith n 3 is strongly alternative if it is an
alternative element (i.e. (a; a; x) = 0 for all x in A n) and also (a; x; x) =*
* 0
for all x in A n.
Example. The element a = e1 is alternative but a = e1 is not strongly
alternative element in A 4. Take x = e4 + e15 in A 4so
12
(x; x; a)= (e4 + e15; e4 + e15; e1)
= (e4; e4; e1) + (e15; e15; e1) + (e4; e15; e1) + (e15; e4; e1)
= 0 + 0 + (e4e15+ e15e4)e1 - (e4(e15e1) + e15(e4e1))
= 0 + 0 + 0 - e4e14+ e15e5
= e10+ e10
= 2e10:
Therefore among the alternative elements there are "few" strongly alternative
elements.
Example. An element ee0in A nis strongly alternative for n 3.
If x is doubly pure element in A nthen (x; x; ee0) = 0 because H x is asso-
ciative.
For x + ree0with x doubly pure we have
(x + ree0; x + ree0; ee0)=(x; x; ee0) + r(x; ee0; ee0) + r(ee0; x; ee0) + r2(ee*
*0; ee0; ee0)
= 0
because ee0is alternative.Therefore ee0is strongly alternative.
We will show that the only strongly alternative elements are of the form
re0 + see0for r and s in R .
Lemma 4.1. Let a be a non-zero element in A nfor n 3.
If (x; a; y) = 0 for all x and y in A nthen a = re0 for r 2 R.
Proof. First of all we notice that a has to be an alternative element because
if x = a then (a; a; y) = 0 for all y in A n.
Now (x; a; y) = 0 for all x and y if and only if the pure part of a also
has this property, i.e. write a = b + re0 for b pure element and r 2 R then
(x; a; y) = (x; b; y) + (x; re0; y) but (x; re0; y) = 0 for all x and y in A n *
*and
r in R . Since b is also alternative its doubly pure part is alternative so if
b = c + see0with c doubly pure and s in R then c is alternative. Setting
x = ee0we have that
0 = (ee0; b; y) = (ee0; c + see0; y) = (ee0; c; y) + s(ee0; ee0; y)
= -2ecy + 0
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for all y in A n. But c is alternative so c = 0. Therefore b = see0with s in R .
By hypothesis (x; b; y) = s(x; ee0; y) = 0 for all x and y, so s = 0 and b = 0 *
*in
An. So if (x; a; y) = 0 for all x and y in A nthen a has imaginary part equal
to zero. Q.E.D.
Theorem 4.2. If ff 2 A n+1 n 3 is a pure strongly alternative element
then ff = ee0for some 2 R.
Proof. First suppose that ff is doubly pure, and strongly alternative in An+1.
We want to show that ff = 0.
Using theorem 3.3, since ff is alternative then ff = (ra; ta) 2 A nx An
with a an alternative pure element in A n and r and t fixed real numbers,
not both equal to zero (otherwise we are done with ff = 0). Now for all
fl = (x; y) 2oA nxoAn we have that
L2fl(ff) = (A(ra) - S(ta); A(ta) + S(ra))
where A = L2x+R2yand S = (x; -; y) (see x3). Since ff is strongly alternative
then
L2fl(ff) = fl2ff forall fl 2 eAn+1
thus we have the following system of equations
rA(a) - tS(a) = (x2 + y2)ra
tA(a) + rS(a) = (x2 + y2)ta:
If r = 0 or t = 0 then S(a) = 0. If r 6= 0 and t 6= 0 then also S(a) = 0 because
multiplying the first equation by `t', the second by `r' and substracting we
have that
(r2 + t2)S(a) = 0:
Therefore S(a) = 0. But a is pure element so by Lemma 4.1 a = 0 in A nand
ff = 0 in A n+1.
Suppose now that ff = fi + ee0with fi double pure in A n+1 and in R .
Since (ff; x; x) = 0 for all x in A n+1 then 0 = (fi + ee0; x; x) = (fi; x; x) +
(ee0; x; x) but (ee0; x; x) = 0 for all x (see example above). Therefore 0 =
(fi; x; x) for all x in A n+1and fi = 0 so ff = ee0for in R . Q.E.D.
Corollary 4.3. If ' 2 Aut(A n) is an algebra authomorphism of An for n 4
then '(ee0) = ee0and '(ea) = 'g(a)for all a 2 An .
Proof. If ' 2 Aut(A n) then '(e0) = e0 and ' preserves real parts.
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Now ' preserves the properties of being alternative and strongly alterna-
tive so '(ee0) = ee0for some in R and ' 2 Aut(A n) but
1 = ||'(ee0)||2 = ||ee0||2 = 2||ee0|| = 2 so = 1: Q.E.D.
Remark. From the last corollary we can deduce the group of authomorphism
of A n+1in terms of the group of authomorphism of A nfor n 3.
For ' 2 Aut(A n+1); ' is completly determined by the action on "the even
part of A n+1" i.e. {(a; 0) 2 An x An} which is isomorphic to A nand the sign
'(ee0) = ee0.
Recall that (a; b) = (a; 0) + (b; 0)(0; e0) in A nx An.
Eakin-Sathaye already calculate this group [1]
Aut(A n+1) = Aut(A n) x 3 n 3
where 3 is the symmetric group with 6 elements given by ; o : A n+1! A n+1is o(x; y) = (x; -y) then o(ee0) = -ee0and
(x; 0) = (x; 0) and (0; x) = (x; 0)ff where ff = -1_2e0 - 1_2ee0.
x5 Local alternativity.
Definition. a and b are in A nwith n 4. We say that a is normed with b if
||ab|| = ||a||||b||:
We say that a alternate with b if
(a; a; b) = 0:
Way say that a alternate strongly with b if (a; a; b) = 0 and (a; b; b) = 0.
Now a is a normed element if ||ax|| = ||a||||x|| for all x in A n.
Notice that since `e0' is normed, alternative and strongly alternative el-
ement we have that a is normed with b, a alternate with b and a alternate
strongly with b if the pure part of a is normed, alternate and strongly alter-
nate with b.
Now if a is a pure element in A n with n 4 then a alternative with b
implies that a is normed with b for b in A n.
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To see this we argue as follows:
||ab||2= ** = -** = -**** = -a2****
= ||a||2||b||2 for b in A n:
The converse is not true.
Example. a = e1 + e10 and b = e15 in A 4.
(a; a; b)= (e1; e1; e15) + (e10; e10; e15) + (e1; e10; e15) + (e10; e1; e1*
*5)
= 0 + 0 + [(e1e10) + (e10e1)]e15- (e1(e10e15) + e10(e1e15))
= 0 + 0 + 0 - (e1e5 - e10e14)
= -(-e4 - e4)
= 2e4:
On the other hand
||ab||2 = ||(e1 + e10)e15||2 = |e1e15+ e10e15||2 = || - e14+ e5||2
= ||e5||2 + ||e14||2 - 2
= 2
||a||2||b||2= ||e1 + e10||2||e15||2 = ||e1||2 + ||e10||2 + 2
= 2
so ||ab|| = ||a||||b||.
Notice that if ||ab|| = ||a||||b|| with a pure then
-**** = -a2**** = ||a||2||b||2 = ||ab||2 = ** = -**:
Therefore **** = 0 and (a; a; b) ? b which is equivalent to (a; b; b)*
* ? a.
Also if a alternate strongly with b then a alternate with b (trivially) but the
converse is also not true. (See example in x4). Therefore
"a alternate strongly with b"
+
"a alternate with b"
+
"a is normed with b"
and the converse of both implications are not true. Notice that since A 3is
an alternative normed algebra the three concepts are equivalent in A 3
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Since ee0in A n; n 4, is a strongly alternative element, we have that a
alternate strongly with b if the doubly pure part of a strongly alternate with
b and viceversa. So we are interested in these relations between couples of
doubly pure elements in A nfor n 4.
Now for a (non-zero) doubly pure element we have that if b 2 H athen a
alternate strongly with b because H ais associative.
Theorem 5.1.
Let a and b be (non-zero) doubly pure elements in A nwith n 4.
If b 2 H ?aand a alternate strongly with b of A n.
Then i) The vector subspace of A ngenerated by
{e0; a; b; ab}:
is multiplicatively closed and isomorphic to A 2= H .
ii) The vector subspace of A ngenerated by
{e0; a; b; ab; eab; -eb; ea; ee0}
is multiplicatively closed and isomorphic to A 3= O .
Proof. Without loss of generality we assume that ||a|| = ||b|| = 1 (otherwise
we take _a_||a||and _b_||b||).
Now construct a multiplication table
|e a b ab
____|_0____________________
e0 ||e0 a b ab
a || a -e0 ab -b
b || b -ab -e0 a
ab ||ab b -a -e0
Since b ? a; ab = -ba and (a; a; b) = (a; b; b) = 0 we have that a(ab) = a2b =
-b; b(ab) = -b(ba) = -b2a = a; (ab)a = -(ba)a = -ba2 = b; (ab)b = ab2 =
-a and (ab)2 = -||ab||2 = -** = ** = **** = a2||b||2 = -e0.
This multiplication table is the one of A 2 by the identification e0 $
e0; e1 $ a; e2 $ b; e3 $ ab. So we are done with i).
ii) is a routine calculation using i) and Lemma 1.1 and Corollary 1.2.
Q.E.D.
Remarks: Given a non-zero element a in A n n 4 there exists b such
that: a ? b; ||a|| = ||b|| and a alternate strongly with b.
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To see this take a of norm one (otherwise we take _a_||a||) and write a =
rc + see0where c is the doubly pure part of a and r and s are in R with
r2 + s2 = 1.
Define b = sc - ree0.By flexibility and because ee0is strongly alternative
is easy to see that (a; a; b) = (a; b; b) = 0 and ||a|| = ||b|| and a ? b by
construction.
Definition. Let B be a subset of A n n 4 B is a normed subset of A n
if ||xy|| = ||x||||y|| for all x and y in B. Thus a is normed with b means that
{a; b} is a normed subset of A n.
Theorem 5.2 Let a and b be non-zero pure elements in A n n 4.
1) a alternate with b if and only if {a; b} and {a; ab} are normed subsets.
2) a alternate strongly with b if and only if {a; b; ab} is a normed subset.
Proof. 1) We know that if (a; a; b) = 0 then ||ab|| = ||a||||b|| for any b in A*
* n.
But a(a; a; b) = a(a2b-a(ab)) = a2ab-a(a(ab)) = (a; a; ab) so if (a; a; b) =
0 then (a; a; ab) = 0 therefore {a; b} and {a; ab} are normed subsets when a
alternate with b.
Conversely suppose that ||ab|| = ||a||||b|| and that ||a(ab)|| = ||a||||ab||*
* =
||a||2||b||. Then
||(a; a; b)||2=<(a; a; b); (a; a; b)>
=
= a4**** + - 2
= ||a||4||b||2 + ||a(ab)||2 + 2||a||2****
= ||a||4||b||2 + ||a||4||b||2 - 2||a||2**
= 2||a||4||b||2 - 2||a||2||ab||2
= 0
Therefore (a; a; b) = 0 and we are done with 1).
To prove 2) we apply 1) to {a; b}; {a; ab} and to {b; a}; {b; ba} but ||ba||*
* =
||ab|| so (a; a; b) = (a; b; b) = 0 if and only if {a; b; ab} is a normed set. *
*Q.E.D.
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References
[1]P. Eakin - A. Sathaye. On the automorphisms of the Cayley-Dickson
algebras and its derivations. Journal of algebra 129, 263-278 (1990).
[2]G. Moreno. The zero divisors of the Cayley-Dickson algebras over the
real numbers. Boletin de la Sociedad Matematica Mexicana (3), Vol.
4, 13-27, 1998.
[3]R.D. Schafer. On the algebras formed by the Cayley-Dickson process.
American Jorunal of Math. 76, 1954, 435-446.
Guillermo Moreno Rodriguez
Centro de Investigacion
(CINVESTAV del IPN)
MEXICO
gmoreno@math.cinvestav.mx
and
Department of Mathematics
University of Oregon
Eugene, Oregon, U.S.A.
moreno@math.uoregon.edu
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