CAYLEY-DICKSON AND CLIFFORD ALGEBRAS AS TWISTED GROUP ALGEBRAS (2003)

8/6/2019 CAYLEY-DICKSON AND CLIFFORD ALGEBRAS AS TWISTED GROUP ALGEBRAS (2003)

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CAYLEY-DICKSON AND CLIFFORD ALGEBRAS AS TWISTEDGROUP ALGEBRAS (2003)

JOHN W. BALES

Abstract. Given a nite group G, a set of basis vectors B = {i p | p G }and a sign function or twist : G G { 1, 1}, there is a twistedgroup algebra dened on the set V of all linear combinations of elements of B over a eld F such that if p, q G, then i p i q = ( p, q ) i pq . This product isextended to V by distribution. Examples of such twisted group algebras arethe Cayley-Dickson algebras and Clifford algebras. It is conjectured that theHilbert Space 2 of square summable sequences is a Cayley-Dickson algebra.

1. Introduction

In 1972 I took a complex analysis course taught by William R. R. Transue usinga text by J. S. McNerney An Introduction to Analytic Functions with TheoreticalImplications[3]. An exercise in this book asked the student to decide whether thescheme for multiplying two ordered pairs of real numbers could be extended tohigher dimensional spaces. At the time, I was also taking a course taught by CokeReed on the Hilbert space 2 of square summable sequences of real numbers. Idecided to investigate whether a product akin to the product of complex numberscould be extended in a meaningful way to 2 . My idea was to do this by equating anordered pair of sequences with the shuffling of the two sequences. Being unawareat the time of the Cayley-Dickson construction I naively constructed a sequence of spaces utilizing the product

(1.1) (a, b ) (c, d ) = ( ac bd, ad + bc)

This construction led to a sequence i0 = 1 , i 1 = i, i 2 , i 3 , of unit basis vectorsfor 2 satisfying the twisted product

(1.2) i p iq = ( p,q)i pqwhere pq was the bit-wise exclusive or of the binary representations of p and qand

(1.3) ( p,q) = ( 1)

where pq is the bit-wise and function of p and q and < r > represents the sumof the bits in the binary representation of r . Since the matrix associated with thisfunction is a Hadamaard matrix, I called this the Hadamaard sign function. (Iwas unaware of the term twist and of twisted group algebras at the time since mytraining was in general topology, not algebra.) Seeing that the four-dimensionalHadamaard space created by this construction was not the quaternions, but a

2000 Mathematics Subject Classication. 16S99,16W99.Key words and phrases. twisted group algebra, sign function, Cayley-Dickson algebra, Clifford

algebra, quaternions, octonions, sedenions, geometric algebra, Hilbert space.1


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2 JOHN W. BALES

space with zero divisors, I abandoned the project in 1972. In 1983, while cullingsome of my papers, I came across the notes I had made on this problem and began

to look at it again. At that time I found a nice way to represent a twisted productusing inner products and conjugates (Theorem 4.10). I gave a short talk on thoseresults at a meeting of the Alabama Academy of Sciences in 1992 at TuskegeeUniversity.

Last year I read John Baez article The Octonions [1] where I learned for the rsttime of twisted group algebras and of Cayley-Dickson and Clifford algebras. Thismotivated me to look at Cayley-Dickson and Clifford algebras. The resulting workis presented in this paper beginning with the section titled The Cayley-DicksonConstruction. The sections prior to that section are the most general results of mywork prior to 1985 not specically tied to the Hadamaard twist.

2. Twisted Group Algebras

Let V denote an n -dimensional vector space over the eld F . Let G denotea group of order n. Let B = {i p | p G} denote a set of unit basis vectors forV. Then, for each x V, there exist elements {x p | x p F , p G} such thatx = pG x p i p .Dene a product on the elements of B and their negatives in the following manner.

Let : G G { 1, 1} denote a sign function on G. Then for p, q G denethe product of i p and iq as follows.

Denition 2.1.i p iq = ( p,q)i pq

Extend this product to V in the natural way. That is,

Denition 2.2.

xy = pG

x p i pqG

yq iq

= pG qG

x pyq i p iq

= pG qG

x pyq ( p,q)i pq

In dening the product this way, one gets the closure and distributive propertiesfor free, as well as (cx )y = x(cy) = c(xy )

This product transforms the vector space V into a twisted group algebra . Theproperties of the algebra depend upon the properties of the twist and the propertiesof G.

Notation 2.3. Given a group G , twist on G and eld F , let [G,, F ] denotethe corresponding twisted group algebra. If F = R , abbreviate this notation [ G, ] .

3. Twists and Field Properties

Let G denote a nite group, F a eld and a twist on G . Let V = [G,, F ].

Denition 3.1. If ( p,e) = (e, p ) = 1 where e is the identity element of G,then is said to an identive twist on G.


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CAYLEY-DICKSON AND CLIFFORD ALGEBRAS AS TWISTED GROUP ALGEBRAS (2003) 3

Theorem 3.2. If is identive, then i e is the identity element, 1, of V.

Proof. If x V, thenxi e = pG x p i p i e = pG x p i p i e = pG x p ( p,e)i pe = pG x p i p = x.The proof for i e x = x is similar. Thus, ie is the identity element, 1, of V

Denition 3.3. If ( p,q) ( pq, r ) = ( p,qr ) (q, r ) for p,q,r G, then is saidto be an associative twist on G.

Theorem 3.4. If p ,q,r G and is associative, then i p (iq i r ) = ( i p iq ) i r .

Proof.

i p (iq i r ) = i p ( (q, r )iqr )= (q, r )i p iqr = ( p, qr ) (q, r )i p(qr )= ( p,q) ( pq,r )i ( pq ) r= ( p,q)i pq i r= ( i p iq ) i r .

Theorem 3.5. If x ,y,z V and if is associative, then x (yz ) = ( xy ) z.

Proof. yz = qG r G (yq zr ) iq i r , so

x(yz ) = pG

x p i pqG rG

(yq zr ) iq i r

= pG qG rG

x p (yq zr ) i p (iq i r )

= pG qG rG

(x p yq ) zr (i p iq ) i r

= pG qG

(x p yq) i p iqrG

zr i r

= ( xy ) z

.

Denition 3.6. If (e, e ) = 1 , then is said to be a positive twist. Otherwise, is a negative twist.

Theorem 3.7. If is associative, then (e, p ) = ( p,e) = (e, e ).Proof. (e, e ) (e, p ) = (e, e ) (ee,p )= (e,ep ) (e, p ) = (e, p ) (e, p ) = 1 .Thus (e, p ) = (e, e ). ( p,e) (e, e ) = ( p, ee) (e, e )= ( p,e) ( pe,e ) = ( p,e) ( p,e) = 1 .Thus ( p,e) = (e, e )

Corollary 3.8. Every positive associative twist is identive.


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4 JOHN W. BALES

Theorem 3.9. The twist is associative if and only if, for p ,q,r G ( p,q) (q, r ) = ( p,qr ) ( pq,r )

Proof. Multiplying each side of the equation ( p,q) ( pq, r ) = ( p, qr ) (q, r ) by (q, r ) ( pq, r ) yields ( p,q) (q, r ) = ( p,qr ) ( pq,r ). So the two conditions are equivalent.

Theorem 3.10. If is positive and associative, then < V, + , > is a ring with unity.

Proof. Follows immediately from Theorems 3.2 and 3.5 and Corollary 3.8.

Theorem 3.11. For each p G, p,p 1 i p 1 and p 1 , p i p 1 are right and left inverses, respectively, of i p .

Proof.

i p p,p 1

i p 1

= p,p 1

i p i p 1

= p,p 1 p,p 1 i pp 1

= i e = 1

p 1 , p i p 1 i p = p 1 , p i p 1 i p= p 1 , p p 1 , p i pp 1

= i e = 1

Denition 3.12. If p,p 1 = p 1 , p for p G, then is an invertivetwist on G.

Theorem 3.13. If p G and if is invertive, then i p has an inversei 1 p = p,p 1 i p 1 = p 1 , p i p 1 .

Proof. Follows immediately from Theorem 3.11 and Denition 3.12.

Denition 3.14. If is invertive, and x V, then let x= pG x p i 1 p denote

the conjugate of x.

Theorem 3.15. If is an invertive twist on G, and if x, y V, then (i) x= pG p

1 , p x p 1 i p(ii) x= x

(iii) (x + y)= x+ y(iv) (cx )= cx for all c F .

Proof.(i) Let q 1 = p. Then

x=qG

xq i 1q

=qG

xq q, q 1 iq 1

= pG

p 1 , p x p 1 i p


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(ii) Let z = x. Then

z = pG

p 1 , p x p 1 i p

= pG

z p i p

where z p = p 1 , p x p 1 . Then z p 1 = p,p 1 x p , and

z p 1 = p,p 1 x p . So

x= z

= pG

p 1 , p z p 1 i p

= pG

p 1 , p p,p 1 x p i p

= pG

x p i p = x

(iii)

(x + y)= pG

(x p + y p)i 1 p

= pG

x p + y p i 1 p

= pG

x p i 1 p +

pG

y p i 1 p

= x+ y.

(iv)

(cx )= pG

(cx p )i 1 p

= pG

cx p i 1 p

= c pG

x p i 1 p

= cx

4. Proper Sign FunctionsDenition 4.1. The statement that the twist on G is proper means that if p,q G, then

(1) ( p,q) q, q 1 = pq,q 1

(2) p 1 , p ( p,q) = p 1 , pq .

Theorem 4.2. Every positive associative twist is proper.

Proof.


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6 JOHN W. BALES

(1) ( p,q) q, q 1 = pq,q 1 p,qq 1

= pq, q 1 ( p,e) = pq, q 1 (e, e ) = pq,q 1

(2) p 1 , p ( p,q) = p 1 p,q p 1 , pq= (e, q) p 1 , pq = (e, e ) p 1 , pq = sgn p 1 , pq

Theorem 4.3. Every proper twist is positive, identive and invertive.

Proof. Suppose is proper.Then ( p,e) e, e 1 = pe, e 1 . That is, ( p,e) (e, e ) = ( p,e). So, (e, e ) =1. Thus, is positive.Furthermore, (e, q) q, q 1 = eq,q 1 = q, q 1 . So, (e, q ) = 1 .Also, p 1 , p ( p,e) = p 1 , pe = p 1 , p . So ( p,e) = 1 . Thus, isidentive.And since p,p 1 p 1 , p = pp 1 , p = (e, p ) = 1 , it follows that p,p 1 = p 1 , p . So is invertive.

Theorem 4.4. If is a proper twist on G, then (i p iq )= iqi p for all p, q G.

Proof. Since i p iq = ( p,q)i pq , (i p iq )= ( ( p,q)i pq )= ( p,q)( i pq )

= ( p,q)i 1 pq = ( p,q) ( pq) 1 , pq i ( pq ) 1 .

On the other hand,

iqi p= i 1q i 1 p = q

1 , q iq 1 p 1 , p i p 1

= q 1 , q p 1 , p q 1 , p 1 iq 1 p 1

= q 1 , q p 1 , p q 1 , p 1 i ( pq ) 1

Therefore, in order to show that ( i p iq )= iqi p, it is sufficient to show that ( p,q) ( pq) 1 , pq = q 1 , q p 1 , p q 1 , p 1 .Beginning with the expression on the left,

( p,q) ( pq) 1 , pq = ( p,q) q, q 1 q, q 1 ( pq) 1 , pq

= pq,q 1 q, q 1 ( pq) 1 , pq

= ( pq) 1 , pq pq, q 1 q, q 1

= ( pq) 1 , p q, q 1

= ( pq) 1

, p p,p 1

p,p 1

q, q 1

= q 1 p 1 , p p,p 1 p,p 1 q, q 1

= q 1 , p 1 p,p 1 q, q 1

= q 1 , p 1 p 1 , p q 1 , q

Theorem 4.5. If is a proper twist on G and if x, y V, then (xy )= yx.


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Proof.

(xy )= pG

x p i pqG

yq iq

= pG qG

x p yq i p iq

= pG qG

(x p yq i p iq )

=qG pG

yqx pi 1q i 1 p

=qG

yqi 1q pG

x pi 1 p

= yx

Denition 4.6. The inner product of elements x and y in V is x, y = pG x p y p .

Theorem 4.7. [1 ( p,p)] x, i p x = 0 provided is proper and p = p 1 .

Proof. i px = q xq i p iq = q ( p,q)x q i pq = r ( p, pr )x pr i r where q = pr.

2 x, i px = 2r

( p,pr )x r x pr

= 2 ( p,p)r

( p,r )x r x pr

= ( p,p)r

( p,r )x r x pr + ( p,p)q

( p,q)x q x pq

= ( p,p)r

( p,r )x r x pr + ( p,p)r

( p, pr )x pr x r

= ( p,p)r

( p,r )x r x pr + ( p,p) ( p,p)r

( p,r )x r x pr

= [ ( p,p) + 1]r

( p,r )x r x pr

= [ ( p,p) + 1] ( p,p)r

( p,pr )x r x pr

= [1 + ( p,p)] x, i p x

Thus [1 ( p,p)] x, i px = 0 .

Corollary 4.8. If p = p 1 , is proper and ( p,p) = 1, and if x V , thenx, i px = 0 .

Theorem 4.9. [1 ( p,p)] x ,x i p = 0 provided is proper and p = p 1

Proof. The proof is similar to that of Theorem 4.7.


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8 JOHN W. BALES

Theorem 4.10. If is a proper twist on G, and if x, y V, then

xy =rG

x, i r y i r =rG

y, x i r i r

Proof.(i) y= sG s, s

1 ys 1 i s . Let p = rs. Then

i r y=sG

s, s 1 ys 1 i r i s

=sG

(r, s ) s, s 1 ys 1 i rs

=sG

rs,s 1 ys 1 i rs

= pG

p,p 1r y p 1 r i p

Thus,

x, i r y = pG

p,p 1r x p y

p 1 r

= pG

p,p 1r x p y p 1 r

(ii) x= sG s 1 , s xs 1 i s . Let q = sr. Then

xi r =sG

s 1 , s xs 1 i s i r

=sG

s 1 , s (s, r )xs 1 i sr

=sG

s 1 , sr xs 1 i sr

=qG

rq 1 , q xrq 1 iq

So y, x i r = qG rq 1 , q x rq 1 yq .

(iii)

xy = pG qG

x p yq i p iq

= pG qG

( p,q)x p yq i pq

=qG pG

( p,q)x p yq i pq

Let pq = r. Then p = rq 1 and q = p 1r.Thus,xy = rG qG rq

1 , q x rq 1 yq i r = rG y, xi r i r .

Andxy = rG pG p,p

1r x p y p 1 r i r = rG x, i r y i r .

Corollary 4.11. If is proper, then xy,i r = x, i r y = xi r , y .


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CAYLEY-DICKSON AND CLIFFORD ALGEBRAS AS TWISTED GROUP ALGEBRAS (2003)11

Suppose the principle is true for n = k. Let n = k + 1 . Let 0 p < 2n and0 q < 2n . Then there are numbers r and s such that 0 r < 2k and 0 s < 2k

and such that one of the following is true: p = 2 r, q = 2 s p = 2 r, q = 2 s + 1 p = 2 r + 1 , q = 2 s p = 2 r + 1 , q = 2 s + 1

(1) Assume p = 2 r, q = 2 s. Then

i p iq = i2r i2s = ( i r i s , 0)= ( (r, s )i rs , 0) = (r, s )( i rs , 0)= (r, s )i2rs = (2r, 2s)i (2 r )(2 s )

=

( p,q

)i pq

provided (2r, 2s) = (r, s ).(2) Assume p = 2 r, q = 2 s + 1 . Then i p iq = i2r i2s +1 = (0 , ir i s ).

If r = 0 , then

i p iq = (0, i r i s ) = (0, (r, s )i rs )= (r, s )i2rs +1 = (2r, 2s + 1) i (2 r )(2 s +1)= ( p,q)i pq

provided (2r, 2s + 1) = (r, s ) when r = 0 .If r = 0 , then

i p iq = i0 i2s +1 = (0 , i 0 i s )= (0 , (0, s )i s ) = (0, s )i2s +1= (0, 2s + 1) i pq = ( p,q)i pq

provided (0, 2s + 1) = (0, s ).(3) Assume p = 2 r + 1 , q = 2 s. Then

i p iq = i2r +1 i2s = (0 , i s i r )= (s, r )(0 , i sr ) = (s, r )i2sr +1= (2r + 1 , 2s)i (2 r +1)(2 s ) = ( p,q)i pq

provided (2r + 1 , 2s ) = (s, r ).(4) Assume p = 2 r + 1 , q = 2 s + 1 . Then i p iq = i2r +1 i2s +1 = (i s ir , 0). If

r = 0 , then

i p iq = ( i s i r , 0) = (s, r )( i sr , 0)= (s, r )i2sr = (2r + 1 , 2s + 1) i (2 r +1)(2 s +1)= ( p,q)i pq


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12 JOHN W. BALES

provided (2r + 1 , 2s + 1) = (s, r ) when r = 0 .If r = 0 , then

i p iq = i1 i2s +1 = (i s i0 , 0)= (i s i0 , 0) = (s, 0)( i s , 0)= (s, 0)i2s = (1, 2s + 1) i1(2 s +1)= ( p,q)i pq

provided (1, 2s + 1) = (s, 0).Thus, the principle is true for n = k + 1 provided the twist is dened as requiredin these four cases.

Corollary 8.2. The requisite properties of the sign function are(1) (0, 0) = 1(2) (2r, 2s) = (r, s )(3) If r = 0 then (2r, 2s + 1) = (r, s )(4) (0, 2s + 1) = (0, s ) = 1(5) (2r + 1 , 2s ) = (s, r )(6) If r = 0 then (2r + 1 , 2s + 1) = (s, r )(7) (1, 2s + 1) = (s, 0) = 1

Let us apply Theorem 8.1 and Corollary 8.2 to the example of nding the Cayley-Dickson product of the basis vectors i9 and i11 . The process is easier if 9 and 11 arewritten in their binary representations 1001 and 1011. Their product under the bit-wise exclusive or group operation is 0010, or 2. Thus, i9 i7 = (1001, 1011) i0010 .The twist can be worked out using Corollary 8.2 as follows:

(1001, 1011) = (101, 100) (by Corollary8.2.6)= (10, 10) (by Corollary8.2.5)= (1, 1) (by Corollary 8.2.1)= 1 (by Corollary 8.2.7)

Thus, i9 i11 = i2 .The following establishes the quaternion properties of the twist

Theorem 8.3. If 0 = p = q = 0 then (1) ( p,p) = 1(2) ( p,q) = (q, p)(3) ( p,q) = (q,pq) = ( pq,p)

Proof. These follow by induction from Corollary 8.2.

Theorem 8.4. If p, q G, then (1) ( p,q) (q, q) = ( pq, q)(2) ( p,p) ( p,q) = ( p,pq)

Thus is a proper twist.

Proof. If either p or q is 0, or if p = q, the results are immediate from Corollary8.2. If 0 = p = q = 0 , the results follow from Theorem 8.3.


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14 JOHN W. BALES

Denition 9.9. If x, y 2 , dene the dyadic convolution of x and y as

x y =r p

x p y pr i r .

Remark 9.10. The convolution is simply the product which results from the trivial twist ( p,q) = 1 for all p, q G.

Conjecture 9.11. If x, y 2, then x y 2 .

Denition 9.12. If x, y 2 , dene the commutator of x and y as

[x, y ] = xy yx.

Theorem 9.13. If x, y 2 , then

[x, y ] =r p

[ ( p, pr ) ( pr,p )]x p y pr i r

=r> 1 0


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x, y i r = p

x p i p ,q

(q, q)yq iq i r

= p

x p i p ,q

(q, r ) (q, q)yq iqr

= p

x p i p ,q

(q,qr )yq i rq

= p

x p i p , p

( pr,p )y pr i p

=

p

( pr,p )x p y pr

Thus,

[x, y ] =r p

[ ( p, pr ) ( pr,p )]x p y pr i r

Then

2[x, y ] =r p

( ( p, pr ) ( pr,p )) x p y pr i r +q

( (q,qr ) (qr,q )) x q yqr i r

=r p

( ( p, pr ) ( pr,p )) x p y pr i r + p

( ( pr,p ) ( p, pr )) x pr y p i r

=r p

[ ( pr,p ) ( p, pr )] (x pr y p x p y pr ) i r

=r p

( p,p) [ (r, p ) ( p,r )] (x pr y p x p y pr ) i r

If r = 0 or p = 0 or r = p , then (r, p ) ( p,r ) = 0 . If p = 0 then ( p,p) = 1.And if 0 = p = r = 0, then ( p,r ) = (r, p ).

So,

2[x, y ] =r> 0 0

0 0


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16 JOHN W. BALES

10. Clifford Algebra

In Clifford algebra, the same basis vectors B = {i p | pG} will be used, as wellas the same group G of non-negative integers with group operation the bit-wiseexclusive or operation. Only the twists will differ.

In Clifford algebra, the basis vectors are called blades. Each blade has a nu-merical grade.

i0 = 1 is the unit scalar, and is a 0-blade.i1 , i 2 , i 4 , , i 2n are 1-blades, or vectors in Clifford algebra parlance.i3 , i 5 , i 6 , are 2-blades or bi-vectors. The common characteristic of the sub-

scripts is the fact that the sum of the bits of the binary representations of thesubscripts is two.

i7 , i 11 , i 13 , i 14 , are 3-blades or tri-vectors, etc.The grade of a blade equals the sum of the bits of its subscript.As was the case with Cayley-Dickson algebras, this is not the standard notation.

However, it has the advantage that the product of basis vectors satises i p iq =( p,q)i pq for a suitably dened Clifford twist .

In the standard notation, 1-blades or vectorsare denoted e1 , e2 , e3 , , whereas2-blades or bivectors are denoted e12 , e13 , e23 , etc.

Translating from the e -notation to the i -notation is straightforward. For ex-ample, the 3-blade e134 translates as i13 since the binary representation of 13 is1101 with bits 1, 3 and 4 set. The 2-blade e23 = i6 since the binary representationof 6 is 110, with bits 2 and 3 set.

Stated more formally, the i notation is related to the e notation in thefollowing way: If pk {0, 1} for 0 k < n, and if p = k pk 2

k then i p =k e(k +1) p k .There are four fundamental multiplication properties of 1-blades.

(1) The square of 1-blades is 1.(2) The product of 1-blades is anticommutative.(3) The product of 1-blades is associative.(4) Every n -blade can be factored into the product of n distinct 1-blades.

The convention is that, if j < k, then ej ek = ejk , thus ek e j = ejk .Any two n -blades may be multiplied by rst factoring them into 1-blades. For

example, the product of e134 and e23 , is computed as follows:

e134 e23 = e1e3e4e2e3= e1e4e3e2e3= e1e4e2e3e3= e1e4e2

= e1e2e4= e124

Since e134 = i13 and e23 = i6 , and the bit-wise exclusive or of 13 and 6 is 11,the same product using the i notation is

i13 i6 = (13, 6)i11so evidently, (13, 6) = 1.

As in the case of the Cayley-Dickson product, the function may be denedrecursively.


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Since the grade of a blade i p equals the sum of the bits of p, it will be convenientto have a notation for the sum of the bits of a binary number.

Denition 10.1. If p is a binary number, let p denote the sum of the bits of p.

The sum of the bits function can be dened recursively as follows:

(1) 0 = 0(2) 2 p = p(3) 2 p + 1 = p + 1

Lemma 10.2.e1 i2 p = i2 p+1

Lemma 10.3.e1 i2 p+1 = i2 p

Lemma 10.4.i2 p e1 = ( 1) p i2 p+1

Lemma 10.5.i2 p+1 e1 = ( 1) p i2 p

Theorem 10.6. There is a twist ( p,q) mapping G G into { 1, 1} such that if p, q G, then i p iq = ( p,q)i pq .

Proof. Let G n = { p | 0 p < 2n } with group operation bit-wise exclusive or asin the Cayley-Dickson algebras.

We begin by noticing that i0 i0 = (0, 0)i0 = 1 provided (0, 0) = 1 .This denes the twist for G0 .If p and q are in G n +1 , but not G n , then there are elements u and v in G n

such that one of the following is true:

(1) p = 2 u and q = 2 v(2) p = 2 u and q = 2 v + 1(3) p = 2 u + 1 and q = 2 v(4) p = 2 u + 1 and q = 2 v + 1

Assume is dened for u, v G n , then consider these four cases in order.

(1) p = 2 u and q = 2 v

i p iq = i2u i2v= (u, v )i2uv= (2u, 2v)i (2 u )(2 v )= ( p,q)i pq

provided (2u, 2v) = (u, v ).


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18 JOHN W. BALES

(2) p = 2 u and q = 2 v + 1i p iq = i2u i2v +1

= i2u e1 i2v

= ( 1) u e1 i2u i2v

= ( 1) u e1 (2u, 2v)i2uv

= ( 1) u (u, v )i2uv +1= (2u, 2v + 1) i2uv +1= ( p,q)i pq

provided (2u, 2v + 1) = ( 1) u (u, v ).(3) p = 2 u + 1 and q = 2 v

i p iq = i2u +1 i2v

= e1 i2u i2v= e1 (u, v )i2uv= (u, v )i2uv +1= (2u + 1 , v)i2uv +1= ( p,q)i pq

provided (2u + 1 , 2v) = (u, v ).(4) p = 2 u + 1 and q = 2 v + 1

i p iq = i2u +1 i2v +1= e1 i2u e1 i2v

= ( 1) u e1e1 i2u i2v

= ( 1) u (u, v )i2uv= (2u + 1 , 2v + 1) i2uv= ( p,q)i pq

provided (2u + 1 , 2v + 1) = ( 1) u (u, v ).

Corollary 10.7. Assume p, q G n . The Clifford algebra twist can be denedrecursively as follows:

(1) (0, 0) = 1(2) (2 p, 2q) = (2 p + 1 , 2q) = ( p,q)(3) (2 p, 2q + 1) = (2 p + 1 , 2q + 1) = ( 1) p ( p,q)

Lemma 10.8. ( 1) u ( 1) v = ( 1) uv

Proof. This follows from the fact that u + v = uv + 2 u v where u vrepresents the bitwise and of u and v.

Theorem 10.9. The Clifford twist is associative.

Proof. By Denition 3.3, the twist : G G { 1, 1} is associative provided( p,q) ( pq, r ) = ( p, qr ) (q, r ) for p,q,r G. This is true for G0 since (0, 0) =1.


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Suppose (u, v ) (uv,w ) = (u,vw ) (v, w ) for u,v,w G n . Let p,q,r be inG n +1 but not in G n . Then there are u,v,w G n such that one of the following

eight cases is true.(1) p = 2 u, q = 2 v, r = 2 w

Then ( p,q) ( pq, r ) = (2u, 2v) (2uv, 2w)= (u, v ) (uv,w )= (u,vw ) (v, w )= (2u, 2vw) (2v, 2w)= ( p, qr ) (q, r )

(2) p = 2 u, q = 2 v, r = 2 w + 1

Then ( p,q) ( pq, r ) = (2u, 2v) (2uv, 2w + 1)

= (u, v )( 1) uv (uv,w )

= ( 1) uv (u,vw ) (v, w )

= ( 1) u (u,vw )( 1) v (v, w )= (2u, 2vw + 1) (2v, 2w + 1)= ( p, qr ) (q, r )

(3) p = 2 u, q = 2 v + 1 , r = 2 w

Then ( p,q) ( pq,r ) = (2u, 2v + 1) (2uv + 1 , 2w)

= ( 1) u (u, v )( 1) uv (uv,w )

= ( 1) u ( 1) u (u,vw )( 1) v (v, w )= (2u, 2vw + 1) (2v + 1 , 2w)= ( p,qr ) (q, r )

(4) p = 2 u, q = 2 v + 1 , r = 2 w + 1

Then ( p,q) ( pq,r ) = (2u, 2v + 1) (2uv + 1 , 2w + 1)

= ( 1) u (u, v )( 1) uv (uv,w )

= ( 1) u ( 1) u (u,vw )( 1) v (v, w )= (2u, 2vw ) (2v + 1 , 2w + 1)= ( p,qr ) (q, r )

(5) p = 2 u + 1 , q = 2 v, r = 2 w

Then ( p,q) ( pq,r ) = (2u + 1 , 2v) (2uv + 1 , 2w)= (u, v ) (uv,w )= (u,vw ) (v, w )= (2u + 1 , 2vw ) (2v, 2w)= ( p,qr ) (q, r )


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20 JOHN W. BALES

(6) p = 2 u + 1 , q = 2 v, r = 2 w + 1

Then ( p,q) ( pq,r ) = (2u + 1 , 2v) (2uv + 1 , 2w + 1)

= (u, v )( 1) vw (uv,w )

= ( 1) v (u,vw )( 1) w (v, w )= (2u + 1 , 2vw + 1) (2v, 2w + 1)= ( p,qr ) (q, r )

(7) p = 2 u + 1 , q = 2 v + 1 , r = 2 w

Then ( p,q) ( pq,r ) = (2u + 1 , 2v + 1) (2uv, 2w)

= ( 1) u (u, v ) (uv,w )

= ( 1) u (u,vw ) (v, w )

= (2u + 1 , 2vw + 1) (2v + 1 , 2w)= ( p,qr ) (q, r )

(8) p = 2 u + 1 , q = 2 v + 1 , r = 2 w + 1

Then ( p,q) ( pq,r ) = (2u + 1 , 2v + 1) (2uv, 2w + 1)

= ( 1) u (u, v )( 1) uv (uv,w )

= ( 1) u ( 1) u ( 1) v (u,vw ) (v, w )

= (u,vw )( 1) v (v, w )= (2u + 1 , 2vw ) (2v + 1 , 2w + 1)= ( p, qr ) (q, r )

Thus is associative for G n +1 if it is associative for G n establishing the associa-tivity of for all G k .

By Theorem 4.2, is proper, thus, if x, y Cn , where Cn is the n -dimensionalClifford algebra, then by Theorem 4.10 the product xy may be computed from thesum

xy =rG n

x, i r y i r

=rG n

xi r , y i r

where the conjugate of a multivector x is x= pG n ( p,p)x p i p by Theorem3.15 part i.

Theorem 10.10. If p G n , then ( p,p) = ( 1)s , where s is the triangular number T p =

p [ p 1]2 .

Proof. The proof is by induction. It is true for G0 , since if p = 0 then p = 0 sos = 0 , thus ( p,p) = 1 = ( 1)s .

Assume the relation is true for G n . Let p G n +1 . Then there is a u G n suchthat either p = 2 u or p = 2 u + 1 .


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Suppose p = 2 u. Then u G n , so (u, u ) = ( 1)s , where s = u [ u 1]2 . Butu = 2u and (u, u ) = (2u, 2u) = ( p,p). So ( p,p) = p [ p 1]2 .

Suppose p = 2 u + 1 . Then( p,p) = (2u + 1 , 2u + 1)

= ( 1) u (u, u )

= ( 1) u ( 1)t where t =u [ u 1]

2

= ( 1)s where s = u +u [ u 1]

2Thus,

s =[ u + 1] u

2

= p

[ p

1]2Thus for p G n +1 , ( p,p) = ( 1)s , where s = p [ p 1]2 .

Corollary 10.11. If x Cn , p G n and if T p is an odd triangular number,then x, i p x = 0 .

Theorem 10.12. (q, p) = ( p,q)( 1) p q pq

Proof. The multiblades i p and iq contain exactly p q 1-blade factors in com-mon. Assume p q = 0 . Factor i p and iq into the product of 1-blades. To getfrom the arrangement i p iq to the arrangement iq i p , swap the left-most 1-bladefactor of iq with each of the 1-blade factors of i p resulting in p changes of sign.Repeat this for each of the q 1-blade factors of iq from left to right until each1-blade of iq has been swapped with each of the p 1-blades of i p . This results in p q changes of sign to reverse i p and iq . If p q = 0 , then the sign will fail tochange for a total of pq times. Thus, in general, there will be p q pqchanges of sign in reversing the product of i p and iq .

It is easily veried that ( 1) p q and ( 1) pq are associative twists on G.The latter of these two is the Hadamaard twist. Interestingly, this is the twist whichresults if, instead of the Cayley-Dickson construction, one denes the product of anordered pair as

(a, b )(c, d ) = ( ac bd, ad + bc)Another interesting twist which is not associative, but proper is ( 1) pq .

11. ConclusionMy primary purpose in writing this paper was to document some of my notes.There are many questions and avenues of further research in the area of proper

twists on groups and their resulting twisted algebras.The question whether the Hilbert space 2 is a Cayley-Dickson algebra is open,

so far as I know, and would be resolved by a proof of Conjecture 9.11. In fact, aresolution of this conjecture in the affirmative would render 2 closed under anyproper twist dened on the group G of non-negative integers under the exclusiveor product.


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22 JOHN W. BALES

The set S P (G ) of proper twists on an arbitrary group G is an abelian group.The set S A (G ) of associative twists is a subgroup of S P (G ) . Do these groups

have any interesting properties?References

References

[1] John C. Baez, The Octonions , Bulletin of the American Mathematical Society, Vol. 39 No.2, (2002) 145-205

[2] Leo Dorst, Geometric Algebra (based on Clifford Algebra) (1999)http://staff.science.uva.nl/ leo/clifford/talknew.pdf

[3] J. S. McNerney, An Introduction to Analytic Functions with Theoretical Implications , (1972)Houston: University of Houston.

Department of Mathematics, Tuskegee University, Tuskegee, AL 36088, USAE-mail address : [email protected]

CAYLEY-DICKSON AND CLIFFORD ALGEBRAS AS TWISTED GROUP ALGEBRAS (2003)

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