An Alternate Cayley-Dickson Product

Abstract

Although the Cayley-Dickson algebras are twisted group algebras, little attention has been paid to the nature of the Cayley-Dickson twist. One reason is that the twist appears to be highly chaotic and there are other interesting things about the algebras to focus attention upon. However, if one uses a doubling product for the algebras different from yet equivalent to the ones commonly used and if one uses a numbering of the basis vectors different from the standard basis a quite beautiful and highly periodic twist emerges. This leads easily to a simple closed form formula for the product of any two basis vectors of a Cayley-Dickson algebra.

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AN ALTERNATE CAYLEY-DICKSON PRODUCT
JOHN W. BALES
Abstract. Although the Cayley-Dickson algebras are twisted group
algebras, little attention has been paid to the nature of the Cayley-
Dickson twist. One reason is that the twist appears to be highly
chaotic and there are other interesting things about the algebras to
focus attention upon. However, if one uses a doubling product for
the algebras different from yet equivalent to the ones commonly
used and if one uses a numbering of the basis vectors different
from the standard basis a quite beautiful and highly periodic twist
emerges. This leads easily to a simple closed form equation for the
product of any two basis vectors of a Cayley-Dickson algebra.
1. Introduction
The purpose of this paper is to give a closed form formula for the
product of any two basis vectors of a Cayley-Dickson algebra.1
The complex numbers are constructed by a doubling product on the
set of real numbers:
(a, b)(c, d) = (ac −bd, ad +bc) (1)
To produce the quaternions by a doubling product on the complex
numbers requires that one take conjugation into consideration in such
a way that, for real numbers the product reduces to the one above.
There are eight (and only eight) distinct Cayley-Dickson doubling
products [4] which accomplish this. For each of the eight, the conjugate
of an ordered pair (a, b) is defined recursively by
(a, b)∗= (a∗,−b) (2)
2000 Mathematics Subject Classification. 16S99,16W99.
Key words and phrases. Cayley-Dickson,doubling product,twisted group prod-
uct,fractal,twist tree.
1After publication the author discovered that this is not a Cayley-Dickson dou-
bling product and retracted the paper.
1

2 JOHN W. BALES
The eight doubling products are:
P0: (a, b)(c, d) = (ca −b∗d, da∗+bc)
P1: (a, b)(c, d) = (ca −db∗, a∗d+cb)
P2: (a, b)(c, d) = (ac −b∗d, da∗+bc)
P3: (a, b)(c, d) = (ac −db∗, a∗d+cb)
P⊤
0: (a, b)(c, d) = (ca −bd∗, ad +c∗b)
P⊤
1: (a, b)(c, d) = (ca −d∗b, da +bc∗)
P⊤
2: (a, b)(c, d) = (ac −bd∗, ad +c∗b)
P⊤
3: (a, b)(c, d) = (ac −d∗b, da +bc∗)
Only two of these eight, P3and P⊤
3have been investigated. The eight
algebras resulting from these products are isomorphic [4] and all have
the same elements and the same unit basis vectors e0, e1, e2,··· , en,···.
The basis vectors will be defined below. The eight products may be
arranged in four transpose pairs P0,P⊤
0,P1,P⊤
1,P2,P⊤
2,P3,P⊤
3. They
are transposes in the sense that, given two basis vectors ep,eq, it is
the case that P(ep, eq) = P⊤(eq, ep). This holds for each of the four
product pairs. So the multiplication table of the basis vectors for a
product Pis the transpose of the multiplication table of its transpose
P⊤. Given a basis vector epand a basis vector eqthere is only one r
for which it is the case that either P(ep, eq) = eror P(ep, eq) = −er,
For any pand qthe value of rwill be the same for all eight of the
products and is denoted by p⊕q(which happens to also equal q⊕p),
but whether the product of epand eqis ep⊕qor −ep⊕qwill depend upon
which of the eight products is used.
Let Wdenote the set of non-negative integers. For each of the eight
products there is a corresponding twist function [7, 10, 11] ω:W×W→
{−1,1}such that for each p, q ∈W,P(ep, eq) = ω(p, q)ep⊕q
Historically, researchers have been focused on the properties of the
Cayley-Dickson algebras and not on the nature of the twist ω. One
reason for this is that there seemed little rhyme or reason to ω. The
fact that different researches numbered the basis vectors differently did
not help the situation. Furthermore, for P3and P⊤
3the function ω
is particularly inscrutable. However, in [3] a heuristic Cayley-Dickson
tree method was described for computing ω(p, q) for the product P3.
In [4] the products P0,P⊤
0,P1,P⊤
1,P2, and P⊤
2were derived. Further
investigation has shown that for the product P2(and its corresponding

AN ALTERNATE CAYLEY-DICKSON PRODUCT 3
transpose) there is a simple closed form formula for ω. That is the
subject of this paper.
2. Background
Each real number xis identified with the infinite sequence x, 0,0,···
and an ordered pair of two infinite sequences x=x0, x1, x2, x3,··· and
y=y0, y1, y2, y3,··· is equated with the shuffled sequence
(x, y) = x0, y0, x1, y1, x2, y2,···
Only real number sequences terminating in a string of zeros are considered—
that is, finite real sequences.
The basis for this space is chosen to be
e0= 1,0,0,0,···
e1= 0,1,0,0,0,···
e2= 0,0,1,0,0,0,···
.
.
.
This basis differs from bases commonly used by other researchers. To
distinguish this basis from others we call it the ‘shuffle basis.’ The
shuffle basis vectors satisfy
e0= 1
e2k= (ek,0)
e2k+1 = (0, ek)
The conjugate of a sequence xis x∗=x0,−x1,−x2,−x3,··· thus
(x, y)∗= (x∗,−y)
If p, q < 2Nare positive integers let p⊕qdenote the ‘bit-wise exclu-
sive or’ of the binary representations of pand q. This is equivalent to
the sum of pand qin ZN
2.
The non-negative integers are an abelian group with respect to the
operation ⊕with identity 0.

4 JOHN W. BALES
The twist functions for each of the eight doubling products satify the
following [4].
epeq=ω(p, q)ep⊕q(3)
ω(p, 0) = 1 (4)
ω(0, q) = 1 (5)
ω(p, p) = −1 for p > 0 (6)
ω(p, q) = −ω(q, p) provided 0 6=p6=q6= 0 (7)
3. Properties of ω2
The following properties of ωare peculiar to the product P2.
If 2N≤p < q < 2N+1 then ω2(p, q) = 1 (8)
If 2N≤p < 2N+1 ≤qthen ω2(p, q) = (−1)⌊q/2N⌋(9)
Figure 1 on page 4 shows ω2(p, q) for Z7
2×Z7
2. The rows and columns
of the matrix are numbered 0 through 127 with gray cells representing
ω(p, q) = 1 and white cells representing ω(p, q) = −1.
Figure 1. ωfor Z7
2×Z7
2
4. Traversing the Cayley-Dickson ω2Tree
In order to validate equations (8) and (9) we will traverse the ω-tree
[3] in Figure 2 on page 6 associated with the doubling product P2.

AN ALTERNATE CAYLEY-DICKSON PRODUCT 5
Others have used such ωtree maps to research properties of Cayley-
Dickson algebras [8].
In order to find ω2(p, q) for non-negative integers pand qit will be
necessary to shuffle their bits. In order not to confuse this process with
the shuffling (x, y) of sequences xand y, the shuffle of integers pand
qwill be denoted using square brackets. So, for example [111,101] =
11,10,11 and [11,11101] = [00011,11101] = 01,01,01,10,11. Notice
that the shuffled binary numbers have been rendered as a sequence
of binary doublets. Each doublet, beginning with the leftmost, is an
instruction for traversing the ω2-tree beginning with the top Cnode. A
0 is an instruction to move down a left branch and a 1 is an instruction
to move down a right branch of the tree.
To find ω2(p, q) by this method, traverse the tree using instruction
sequence [p, q]. Terminating at −1 or −Dmeans that ω(p, q) = −1.
Terminating at any other node means that ω2(p, q) = 1. An important
property of ω2is that once either 1 or −1 is reached it is unnecessary
to continue traversing the tree and it will always be the case that
ω2(p, q) = 1 or ω2(p, q) = −1, respectively.
As an example of how one traverses the ω2-tree, let us find the basis
vector product e3e14. First, 3 = 0011Band 14 = 1110B. So 3 ⊕14 =
1101B= 13. So e3e14 =ω2(3,14)e13. Now [3,14] = 01,01,11,10. Using
this sequence of doublets to traverse the ω-tree gives us T,T,−1,−1.
So e3e14 =−e13. One may stop, of course, with the first −1 encoun-
tered.
Next, it will be seen how to use the ω2-tree in Figure 2 on page 6 to
validate equations (8) and (9).
Beginning with equation (8). Suppose 2N≤p < q < 2N+1. Then
[p, q] = 11,···. The doublets following the first will be either 00,01,10,
or 11. The first doublet 11 moves to node −D. Subsequent doublets
of either 00 or 11 remain at node −D. Since p < q there must occur a
doublet 01 and it must occur prior to any potential doublet 10. But 01
moves from node −Dto node 1. Thus ω2(p, q) = 1 verifying equation
(8).
For equation (9) suppose 2N≤p < 2N+1 ≤q. Then it is either
the case that [p, q] = 01,··· ,10,··· or it is the case that [p, q] =
01,··· ,11,··· (where the 10 and 11 doublets are the bits of pand q
corresponding to 2N). In either case the first ellipsis consists of binary
doublets of the form 00 or 01 so we are at a Tnode until arriving
at either the doublet 10 in which case ω(p, q) = 1 or we arrive at the
doublet 11 in which case ω(p, q) = −1. In the first case, ⌊q/2N⌋is

6 JOHN W. BALES
even and in the second case ⌊q/2N⌋is odd. So in either case ω(p, q) =
(−1)⌊q/2N⌋verifying equation (9).
Let us illustrate the use of equations (8) and (9) with a couple of
examples.
Find e35e55 .
Since 35 = 100011Band 55 = 110111Bthen 35 ⊕55 = 10100B= 20.
So e35e55 =ω2(35,55)e20 . And since 25≤35 <55 <26it follows from
equation 9 that ω2(35,55) = 1. So e35 e55 =e20.
Find e87e340 .
Convert 87 = 001010111Band 340 = 101010100B, so 87 ⊕340 =
100000011B= 259. So e87e340 =ω2(87,340)e259 . Since 64 ≤87 <128,
and 128 ≤340 and 340
64 = 5 then ω2(87,340) = (−1)5=−1. So
e87e340 =−e259
Find e51e12 .
First, e51e12 =−e12 e51. 12 = 001100Band 51 = 110011Bso 12⊕51 =
111111B= 63. So e51 e12 =−e12e51 =−ω2(12,51)e63. Since 8 ≤12 <
16 ≤51 and 51
8= 6, ω2(12,51) = (−1)6= 1. So e51 e12 =−e63.
0←C→1
C T
TT1−1
L
L−1 L 1
−D
−D 1 −1−D
Figure 2. Twist tree for ω2
Lest the reader get eye strain from trying to verify the results for
Z7
2×Z7
2using Figure 1 on page 4, the ω2table for Z5
2×Z5
2is provided
in Figure 3 on page 7. Recall that gray cells denote ω2(p, q) = 1 and
white cells denote ω2(p, q) = −1.
To give a better indication of the fractal nature of ω2, Figure 4 on
page 8 is the 1024 ×1024 bit-mapped image of ω2for Z10
2×Z10
2. For
comparison we also provide the corresponding image of the ‘inscrutable’
ω3for Z10
2×Z10
2in Figure 5 on page 8 to give a visual indication of
why no one has searched for a simple formula for it.
5. Conclusion
The problem historically with finding a simple closed form equation
for the product of two Cayley-Dickson basis vectors has been caused by

AN ALTERNATE CAYLEY-DICKSON PRODUCT 7
Figure 3. ω2for Z5
2×Z5
2
various approaches to the algebras. One issue is that only two of the
eight Cayley-Dickson doubling products have been used [12, 6, 2, 3, 5]
each of which is the transpose of the other.
(a, b)(c, d) = (ac −db∗, a∗d+cb) (10)
(a, b)(c, d) = (ac −d∗b, da +bc∗) (11)
Unfortunately, the ωmatrix of these two is sufficiently chaotic to dis-
suade further investigation. Furthermore, a different way of numbering
the basis vectors has traditionally been used which further scrambles
the ωmatrix. These issues have conspired to inhibit investigation into
ω.
Now we see that if the doubling product
P2: (a, b)(c, d) = (ac −b∗d, da∗+bc)

8 JOHN W. BALES
is used and if the basis vectors are indexed over the group (W, ⊕) (the
‘shuffle’ basis) a natural inverse fractal pattern emerges leading to the
simple result in the theorem on page 9.
Figure 4. ω2for Z10
2×Z10
2
Figure 5. ω3for Z10
2×Z10
2

AN ALTERNATE CAYLEY-DICKSON PRODUCT 9
Theorem 5.1. If 2N≤p < q < 2N+1 then epeq=ep⊕q.
If 2N≤p < 2N+1 ≤qthen epeq= (−1)⌊q/2N⌋ep⊕q.
Combined with e0= 1, e2
p=−1 for p > 0 and epeq=−eqepfor
06=p6=q6= 0 we have a simple closed formulation for the product of
any two Cayley-Dickson basis vectors.
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Department of Mathematics, Tuskegee University, Tuskegee, AL
36088, USA
E-mail address:john.w.bales@gmail.com