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A TREE FOR COMPUTING THE CAYLEY-DICKSON TWIST
JOHN W. BALES
Abstract. A universal twist γfor all finite-dimensional Cayley-Dickson al-
gebras is defined recursively and a tree diagram ‘computer’ is presented for
determining the value of γ(p, q) for any two non-negative integers pand q.
1. Introduction
The Cayley-Dickson algebras are a nested sequence {Ak}of algebras with Ak⊂
Ak+1 .A0=Rand for any k≥0 , Ak+1 consists of all ordered pairs of elements
of Akwith a conjugate defined by [8]
(1.1) (x, y)∗= (x∗,−y)
and multiplication by
(1.2) (a, b)(c, d) = (ac −db∗, a∗d+cb)
For k≥0 , the real number ais identified with the ordered pair (a, 0) in
Ak+1.Accordingly, scalar multiplication of an ordered pair (c, d) in Ak+1 by a
real number ais (a, 0)(c, d) = (ac, ad).The first few algebras in this sequence are
A0=Rthe reals, A1=Cthe complex numbers, A2=Hthe quaternions, A3=O
the octonions and A4=Sthe sedenions [2].
In the following development of the sequence of Cayley-Dickson algebras, every
element of each algebra will be identified with an infinite sequence of real numbers
terminating in a sequence of zeros. A real number ris identified with the sequence
r, 0,0,0,· · · .If each of x=x0, x1, x2,· · · and y=y0, y1, y2,· · · is an element in
one of the Cayley-Dickson algebras, then the ordered pair (x, y) is the sequence
constructed by “shuffling” the two sequences.
(1.3) (x, y) = x0, y0, x1, y1, x2, y2,···
For example, a complex number (a, b) is formed by shuffling the sequences
a, 0,0,0,· · · and b, 0,0,0,· · · to obtain (a, b) = a, b, 0,0,0,··· .The quaternion
((a, b),(c, d)) is formed by shuffling the sequences a, b, 0,0,0,· · · and c, d, 0,0,0,· · ·
to obtain the sequence a, c, b, d, 0,0,0· · · .This process may be repeated indefinitely
to represent all Cayley-Dickson elements as finite sequences terminated by an infi-
nite sequence of zeros. This construction of the elements of the algebras leads to a
universal Cayley-Dickson algebra Awhich is simply the union of all the algebras
with each Aka proper subspace of A.Furthermore, if a, b ∈A,then the ordered
pair (a, b)∈A.Since A⊆`2,the Hilbert space of square summable sequences,
and since `2is a completion of Ait is natural to ask whether the product defined
2000 Mathematics Subject Classification. 16S99,16W99.
Key words and phrases. twisted group algebra, Cayley-Dickson algebra, quaternions, octonions,
sedenions.
1
2 JOHN W. BALES
on Acan be extended to `2and whether `2is closed under that product. The
anwer to the first question is ‘yes’ and the answer to the second is unknown.
The sequences
i0= (1,0) = 1,0,0,0,· · ·
i1= (0,1) = 0,1,0,0,· · ·
i2= (i1,0) = 0,0,1,0,· · ·
i3= (0, i1) = 0,0,0,1,· · ·
.
.
.
i2n−1
form the canonical basis for Anand satisfy the identities
i2k= (ik,0)(1.4)
i2k+1 = (0, ik)(1.5)
i∗
p=(ipif p= 0
−ipif p > 0
(1.6)
One may establish immediately that i0= (1,0) is both the left and the right
identity for An.Furthermore, applying the Cayley-Dickson product (1.2) to all the
unit basis vectors yields the following identities:
i2pi2q= (ip,0) (iq,0) = (ipiq,0)(1.7)
i2pi2q+1 = (ip,0) (0, iq) = ¡0, i∗
piq¢
(1.8)
i2p+1i2q= (0, ip) (iq,0) = (0, iqip)(1.9)
i2p+1i2q+1 = (0, ip) (0, iq) = −¡iqi∗
p,0¢
(1.10)
For p=q= 0 this generates the multiplication table for complex numbers
(Table 1).
i0i1
i0i0i1
i1i1−i0
Table 1. Multiplication Table for Complex Number Basis Vectors
For 0 ≤p≤1 and 0 ≤q≤1 one obtains, using binary notation, the multipli-
cation table for quaternions (Table 2).
i00 i01 i10 i11
i00 i00 i01 i10 i11
i01 i01 −i00 i11 −i10
i10 i10 −i11 −i00 i01
i11 i11 i10 −i01 −i00
Table 2. Multiplication Table for Quaternion Basis Vectors
A TREE FOR COMPUTING THE CAYLEY-DICKSON TWIST 3
For 0 ≤p≤2 and 0 ≤q≤2 one obtains the multiplication table for octonions
(Table 3).
i000 i001 i010 i011 i100 i101 i110 i111
i000 i000 i001 i010 i011 i100 i101 i110 i111
i001 i001 −i000 i011 −i010 i101 −i100 i111 −i110
i010 i010 −i011 −i000 i001 i110 −i111 −i100 i101
i011 i011 i010 −i001 −i000 −i111 −i110 i101 i100
i100 i100 −i101 −i110 i111 −i000 i001 i010 −i011
i101 i101 i100 i111 i110 −i001 −i000 −i011 −i010
i110 i110 −i111 i100 −i101 −i010 i011 −i000 i001
i111 i111 i110 −i101 −i100 i011 i010 −i001 −i000
Table 3. Multiplication Table for Octonion Basis Vectors
Representing the basis vectors with binary subscripts reveals that the product
of ipand iqis a multiple of the basis vector subscripted by the sum of pand qin
Zn
2.This is equivalent to the bit-wise ‘exclusive or’ of the binary numbers pand q.
The multiple is either +1 or −1.The basis vectors are indexed by elements of the
group Zn
2with a ‘twist’ γ[6]. That is, there is a function γ:Zn
2×Zn
2→ {−1,1}
such that for all p, q ∈Zn
2,
(1.11) ipiq=γ(p, q)ipq
where pq represents the sum of pand qin the group Zn
2.The elements of the group
Zn
2may be regarded as integers ranging from 0 to 2n−1 with group operation the
‘exclusive or’ of the binary representions. This operation is equivalent to addition
in Zn
2.
If Gis a group and Kis a ring, then one may construct a ‘group algebra’ A(K)
consisting of all elements Pr∈Gkrirwhere kr∈Kand iris a basis vector in
A(K).The product of basis vectors ipand iqis ipiq=ipq .This scheme may be
modified by adding a ‘twist.’ A twist is a function αfrom G×Gto {−1,1}.For
the twisted group algebra, the product of the basis vectors is ipiq=α(p, q )ipq.
2. The recursive definition of the universal Cayley-Dickson twist γ
Theorem 2.1. There is a twist γ(p, q)mapping ∪Zk
2× ∪Zk
2onto {−1,1}such
that if p, q ∈ ∪Zk
2,then ipiq=γ(p, q)ipq .
Proof. Assume 0 ≤p < 2nand 0 ≤q < 2nand proceed by induction on nusing
(1.4)–(1.10).
If n= 0,then p=q= 0 and ipiq=i0i0=i0=γ(p, q)ipq provided γ(0,0) = 1.
Suppose the principle is true for n=k. Let 0 ≤p < 2k+1 and 0 ≤q < 2k+1
Then there are numbers rand ssuch that 0 ≤r < 2kand 0 ≤s < 2kand such
that one of the following is true:
•p= 2r, q = 2s
•p= 2r, q = 2s+ 1
•p= 2r+ 1, q = 2s
•p= 2r+ 1, q = 2s+ 1
4 JOHN W. BALES
(1) Assume p= 2r, q = 2s. Then
ipiq=i2ri2s= (iris,0)
= (γ(r, s)irs,0) = γ(r, s)(irs ,0)
=γ(r, s)i2rs =γ(2r, 2s)i(2r)(2s)
=γ(p, q)ipq
provided γ(2r, 2s) = γ(r, s).
(2) Assume p= 2r, q = 2s+ 1.Then ipiq=i2ri2s+1 = (0, i∗
ris).
If r6= 0,then
ipiq=−(0, iris) = −(0, γ(r, s)irs )
=−γ(r, s)i2rs+1 =γ(2r, 2s+ 1)i(2r)(2s+1)
=γ(p, q)ipq
provided γ(2r, 2s+ 1) = −γ(r, s) when r6= 0.
If r= 0,then
ipiq=i0i2s+1 = (0, i0is)
= (0, γ(0, s)is) = γ(0, s)i2s+1
=γ(0,2s+ 1)ipq =γ(p, q)ipq
provided γ(0,2s+ 1) = γ(0, s).
(3) Assume p= 2r+ 1, q = 2s. Then
ipiq=i2r+1i2s= (0, isir)
=γ(s, r)(0, isr ) = γ(s, r)i2sr+1
=γ(2r+ 1,2s)i(2r+1)(2s)=γ(p, q)ipq
provided γ(2r+ 1,2s) = γ(s, r).
(4) Assume p= 2r+ 1, q = 2s+ 1.Then ipiq=i2r+1 i2s+1 =−(isi∗
r,0).If
r6= 0,then
ipiq= (isir,0) = γ(s, r)(isr ,0)
=γ(s, r)i2sr =γ(2r+ 1,2s+ 1)i(2r+1)(2s+1)
=γ(p, q)ipq
provided γ(2r+ 1,2s+ 1) = γ(s, r) when r6= 0.
If r= 0,then
ipiq=i1i2s+1 =−(isi∗
0,0)
=−(isi0,0) = −γ(s, 0)(is,0)
=−γ(s, 0)i2s=γ(1,2s+ 1)i1(2s+1)
=γ(p, q)ipq
provided γ(1,2s+ 1) = −γ(s, 0).
Thus, the principle is true for n=k+ 1 provided the twist is defined as required
in these four cases. ¤
The properties of the twist γmay be summarized as follows:
A TREE FOR COMPUTING THE CAYLEY-DICKSON TWIST 5
γ(0,0) = γ(p, 0) = γ(0, q) = 1(2.1)
γ(2p, 2q) = γ(p, q)(2.2)
γ(2p+ 1,2q) = γ(q, p)(2.3)
γ(2p, 2q+ 1) = (−γ(p, q) if p6= 0
1 otherwise
(2.4)
γ(2p+ 1,2q+ 1) = (γ(q, p) if p6= 0
−1 otherwise
(2.5)
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1
1 -1 -1 1 1 -1 -1 1 1 -1 -1 1 1 -1 -1 1
1 1 -1 -1 -1 -1 1 1 -1 -1 1 1 -1 -1 1 1
1 -1 -1 1 -1 1 1 -1 1 -1 -1 1 -1 1 1 -1
1 1 1 1 -1 -1 -1 -1 -1 -1 1 1 1 1 -1 -1
1 -1 1 -1 -1 1 -1 1 -1 1 -1 1 1 -1 1 -1
1 1 -1 -1 1 1 -1 -1 1 1 1 1 -1 -1 -1 -1
1 -1 -1 1 -1 1 1 -1 -1 1 1 -1 1 -1 -1 1
1 1 1 1 1 1 -1 -1 -1 -1 -1 -1 -1 -1 1 1
1 -1 1 -1 1 -1 1 -1 -1 1 -1 1 -1 1 -1 1
1 1 -1 -1 -1 -1 -1 -1 1 1 -1 -1 1 1 1 1
1 -1 -1 1 1 -1 -1 1 -1 1 1 -1 -1 1 1 -1
1 1 1 1 -1 -1 1 1 1 1 -1 -1 -1 -1 -1 -1
1 -1 1 -1 -1 1 -1 1 1 -1 1 -1 -1 1 -1 1
1 1 -1 -1 1 1 1 1 -1 -1 -1 -1 1 1 -1 -1
Table 4. Sedenion Twist Table
Table 4 is the twist table for A4,the sedenions–the fourth algebra in the sequence
of algebras formed by the Cayley-Dickson process. The sedenions consist of all
ordered pairs of octonions. The rows and columns of table 4 are numbered 0
through 15. The entry in row pcolumn qis γ(p, q).The sedenion twist table
subsumes the twist tables for the octonions, quaternions, complex numbers and
reals in the following sense: The submatrix formed by rows 0–7 and columns 0–7
is the twist table for the octonions, rows 0–3 columns 0–3 the twist table for the
quaternions, etc. The table reveals a uniform structure which is common to the
twist tables of all higher order Cayley-Dickson algebras.
In general, the twist table for Zn
2may be partitioned into 2 ×2 matrices where
each 2 ×2 matrix is one of three matrices or the negatives of two of those three
matrices.
To see why this is the case, re-form the recursive definition 2.1–2.5 of the Cayley-
Dickson twist on ∪Zk
2as follows: For p, q ∈ ∪Zk
2and r, s ∈ {0,1},
γ(0,0) = 1(2.6)
γ(2p+r, 2q+s) = γ(p, q)Epq (r, s)(2.7)
6 JOHN W. BALES
where
Epq =µ1 1
1−1¶if p= 0(2.8)
=µ1−1
1 1 ¶if 0 6=p=qor p6=q= 0(2.9)
=µ1−1
−1−1¶if 0 6=p6=q6= 0.(2.10)
Define γ0= (1). Then, for each non-negative integer n,γn+1 is a partitioned
matrix defined by
(2.11) γn+1 = (γn(p, q)Epq )
This arrangement is interesting enough, but the structure of the table can also
be analysed in a different way using the same three matrices.
Theorem 2.2. For n > 0,the Cayley-Dickson twist table γncan be partitioned
into 2×2blocks of matrices A, B, C, −B, or −C., defined as follows:
A0=A=µ1 1
1−1¶,B=µ1−1
1 1 ¶,C=µ1−1
−1−1¶
Proof. To construct γn+1 from γn,replace each entry γ(p, q) in γnby the 2 ×2
block
(2.12) Epq =µγ(2p, 2q)γ(2p, 2q+ 1)
γ(2p+ 1,2q)γ(2p+ 1,2q+ 1) ¶.
When p= 0,
(2.13) E0q=µγ(0,2q)γ(0,2q+ 1)
γ(1,2q)γ(1,2q+ 1) ¶=µ1 1
1−1¶.
When p6= 0 we have
(2.14) Epq =µγ(p, q)−γ(p, q)
γ(q, p)γ(q, p)¶.
There are only five possible values for the 2 ×2 matrix Epq in (2.14). These
may be found by considering the following cases:
(1) 0 = p=q
(2) 0 = p6=q
(3) p6=q= 0
(4) p=q6= 0
(5) 0 6=p6=q6= 0
For the first two cases, Epq =E0q=µ1 1
1−1¶=A.
For the third case, Epq =µ1−1
1 1 ¶=B.
For the fourth case, Epq =µ−1 1
−1−1¶=−B.
For the fifth case, Epq =µ1−1
−1−1¶=Cwhen γ(p, q) = 1 and
A TREE FOR COMPUTING THE CAYLEY-DICKSON TWIST 7
Epq =µ−1 1
1 1 ¶=−Cwhen γ(p, q) = −1.
¤
The first few tables are displayed in Tables 5 through 8
γ1=µ1 1
1−1¶=A0
Table 5. Complex Twist Table
γ2=
1 1 1 1
.1−1 1 −1
1−1−1 1
1 1 −1−1
=µA0A
B−B¶
Table 6. Quaternion Twist Table
γ3=
1 1 1 1 1 1 1 1
1−1 1 −1 1 −1 1 −1
1−1−1 1 1 −1−1 1
1 1 −1−1−1−1 1 1
1−1−1 1 −1 1 1 −1
1 1 1 1 −1−1−1−1
1−1 1 −1−1 1 −1 1
1 1 −1−1 1 1 −1−1
=
A0AAA
B−B C −C
B−C−B C
B C −C−B
Table 7. Octonion Twist Table
For n > 0 , γncan be partitioned into 2 ×2 matrices, or blocks, consisting
of only A, B, −B, C, or −C. The first row of the partitioned γnwill consist
entirely of Ablocks. The first such Ablock in row 1 will be denoted A0. The first
column of the partitioned γn, with the exception of A0will consist of Bblocks.
All blocks occuring along the diagonal of the partitioned γn, with the exception
γ4=
A0A A A A A A A
B−B C −C C −C C −C
B−C−B C C −C−C C
B C −C−B−C−C C C
B−C−C C −B C C −C
B C C C −C−B−C−C
B−C C −C−C C −B C
B C −C−C C C −C−B
Table 8. Sedenion Twist Table
8 JOHN W. BALES
of A0will be −B. All other blocks of the partitioned γnwill consist of either C
or −C. Notice that the signs of the entries in γnare the same as the signs of the
corresponding blocks in γn+1 .
A function Tis defined mapping each of the six 2 ×2 blocks A0, A, B, −B,
Cand −Cinto a 4 ×4 block according to the following rules:
(2.15) T(A0) = µA0A
B−B¶
(2.16) T(A) = µA A
C−C¶
(2.17) T(B) = µB−C
B C. ¶
(2.18) T(−B) = µ−B C
−C−B¶
(2.19) T(C) = µC−C
−C−C¶
(2.20) T(−C) = µ−C C
C C ¶.
For a given twist table γn, let T(γn) denote the matrix which results by re-
placing each occurance of A0, A, B, −B , C, and −Cin γnwith the 2 ×2
blocks T(A0), T (A), T (B), T (−B), T (C),and T(−C),respectively. Then the
Cayley-Dickson twist tables can be defined recursively as follows:
(1) γ1=A0
(2) γn+1 =T(γn)
This process can be summarized by the tree in Figure 9 which for clarity is
broken into its components. The root of the tree is A0.
A0
A0AB –B
A
AAC –C
B
B –C B C
–B
–B C –C –B
C
C –C –C –C
–C
–C C C C
Table 9. Cayley-Dickson Twist Tree
A TREE FOR COMPUTING THE CAYLEY-DICKSON TWIST 9
The Cayley-Dickson twist for any pair of non-negative integers may be found
using this tree.
Example. Use the Cayley-Dickson tree to find γ(9,11) .
(1) First, rewrite 9 and 11 in binary form 1001 and 1011.
(2) Next, “shuffle” the two numbers to obtain 11000111. It will be easier to
use this bit string if it is separated by commas into pairs 11,00,01,11. Each
pair, beginning at the left of the string is a navigation instruction for the
Cayley-Dickson tree. A 0 is an instruction to move down the left branch
and a 1 is an instruction to move down the right branch.
(3) Beginning at the root A0and applying the first instruction 11 on the left
yields A0(11) = −B.
(4) Beginning at −Band applying the next instruction 00 yields −B(00) =
−B. (Notice that 00 always leaves the state unchanged.)
(5) Beginning at −Band applying the next instruction 01 yields −B(01) = C.
(6) Finally, beginning at Cand applying the final instruction 11 yields C(11) =
−C. Since the ‘coefficient’ of Cis −1 , γ(9,11) = −1.
(7) This particular traversal of the tree may be summarized as follows:
γ(9,11) = γ(1001,1011) →11,00,01,11 → −B, −B , C, −C→(−1)
3. Conclusion
The Cayley-Dickson twist for any pair of non-negative integers may be found
using the recursive definition (2.1) or by traversing the Cayley-Dickson tree.
The fact that a universal twist γon ∪Zk
2exists establishes that the set A=∪Ak
of all finite sequences of real numbers is a Cayley-Dickson algebra. The twist γon
the group ∪Zk
2is a proper [3] twist, meaning that it satisfies the properties
γ(p, q)γ¡q, q −1¢=γ¡pq, q−1¢
(3.1)
γ¡p−1, p¢γ(p, q) = γ¡p−1, pq¢
(3.2)
All properly twisted group algebras, including the Cayley-Dickson and Clifford
algebras, satisfy the adjoint properties [10, 3]. The adjoint properties state that for
elements x, y and zof the algebra,
hxy, zi=hy, x∗zi(3.3)
hx, yzi=hxz∗, yi(3.4)
from which it follows that the Cayley-Dickson product of finite sequences xand y
has [3] the fourier expansion
(3.5) xy =X
r
hxy, iriir=X
r
hx, iry∗iir=X
r
hy, x∗iriir
If the product in (3.5) is applied to sequences xand yin `2the resulting product
xy is a well-defined number sequence, but it is not obviously square summable. So
it remains to be seen whether `2is closed under the Cayley-Dickson product. If it
were, it would be a universal topologically complete Cayley-Dickson algebra.
10 JOHN W. BALES
References
[1] W. Ambrose Structure theorems for a special class of Banach algebras Trans. Amer. Math.
Soc. Vol. 57, (1945) 364-386
[2] J. Baez The Octonions Bul. Am. Math. Soc. Vol. 39, No. 2, (2001) 145-205
[3] J. Bales Properly twisted groups and their algebras Unpublished monograph (2006)
[4] M. Bremner, I. Hentzel Identities for algebras obtained from the Cayley-Dickson process
Comm. Algebra Vol. 29 (2001) 3523-3534
[5] R. Brown, On generalized Cayley-Dickson algebras Pacific J. Math. Vol. 20 (1967)
[6] R. Busby, H. Smith Representations of twisted group algebras Trans. Am. Math. Soc. Vol.
149, No. 2, (1970) 503-537
[7] R. Erdmann ¨
Uber verallgemeinerte Cayley-Dickson Algebren J. Reine Angew. Math. Vol. 250
(1971) 153-181
[8] N. Jacobson, Basic Algebra I, W. H. Freeman, San Francisco, 1974, 417-427
[9] K. McCrimmon Derivations and Cayley derivations of generalized Cayley-Dickson algebras
Pacific J. Math. Vol. 117 (1985) 163-182
[10] G. Moreno The zero divisors of the Cayley-Dickson algebras over the reals Bol. Soc. Mat.
Mex. 4, (1998) 13-27
[11] R. Schafer On the algebras formed by the Cayley-Dickson process Amer. J. Math. Vol. 76
(1954) 435-446
Department of Mathematics, Tuskegee University, Tuskegee, AL 36088, USA
E-mail address:jbales@tuskegee.edu