---

Division Algebras, Galois Fields, Quadratic Residues

arXiv:hep-th/9302113v1 23 Feb 1993BRX TH-344hep-th/930211323 February 1993Division Algebras, Galois Fields, Quadratic ResiduesGeoffrey DixonDepartment of PhysicsBrandeis UniversityWaltham, MA 02254AbstractIntended for mathematical physicists interested in applications of the divi-sion algebras to physics, this article highlights some of their more elegantproperties with connections to the theories of Galois fields and quadraticresidues.e-mail: Dixon@binah.cc.brandeis.edu

The reals, R, complexes, C, quaternions, Q, and octonions, O, are thenormed division algebras, proven by Hurwitz [1] to be the only ones oftheir kind. My interest in these algebras arises from a faith I share withmany mathematical physicists that they are intimitely linked to the designof our physical reality [2,3] (and if they are not, well they ought to be,and it is a shame they are not).In searching for the key to that link Ihave encountered many of the most beautiful properties of these algebras,including connections to Galois theory and to the theory of quadratic residuecodes.

The former connections highlight the elegant cyclic multiplicationrules of Q and O, and in combination with the latter connections theyprovide another explanation for the uniqueness of the collection.The octonion algebra, O, is often developed as an extension of the quater-nion algebra, Q. Let qi, i=1,2,3, be a conventional basis for the hypercomplexquaternions.

These elements associate, anticommute, and satisfy q2i = −1.The multiplication table for Q is then determined byqiqi+1 = qi+2,(1)i=1,2,3, all indices modulo 3, from 1 to 3.Relabel these quaternion units ei, i=1,2,3, and introduce a new unit, e7,anticommuting with each of the ei, which satisfies e27 = −1. Define threemore units:e4 = e1e7, e5 = e2e7, e6 = e3e7.

(2)Let O be the real algebra generated from the ea a=1,...,7, such that {q1 →ea,q2 →eb, q3 →ec} defines an injection of Q into O for (a,b,c)=(1,2,3),(1,7,4), (2,7,5), (3,7,6), (1,6,5), (2,4,6), (3,5,4).Therefore, for example,e1(e7e5) = e1e2 = e3 = −(−e3) = −e4e5 = −(e1e7)e5.So unlike thecomplexes and quaternions, the octonions are nonassociative.Like C and Q, however, O is a division algebra, and it is normed. Inparticular, if x = x0 + xaea, (sum a=1,...,7), and x† = x0 −xaea (an antiau-tomorphism), then∥x∥2 = x†x =7Xa=0xaxa(3)defines the square of the norm of x (so x−1 = x†/∥x∥2).This octonion multiplication is not, however, the most natural, and itwill not be employed in here.

Again let ea, a = 1, ..., 7, represent the hyper-complex units, but now adopt the cyclic multiplication rule:eaea+1 = ea+5,(4)1

a=1,...,7, all indices modulo 7, from 1 to 7 (the right-hand side could bechanged to ea+3, which generates an alternative multiplication table for O,dual to the first in a sense outlined below). In particular,{q1 →ea, q2 →ea+1, q3 →ea+5}(5)define injections of Q into O for a=1,...,7.

I am accustomed to using thesymbol e0 to represent unity, and I bother to remember that although 7 = 0mod 7, e7 ̸= e0, and in the multiplication rule (4) the indices range from 1to 7, and the index 0 is not subject to the rule. (In [3] ∞is used as the indexfor unity, and this has advantages, which I find intermittently persuasive.

)This octonion multiplication has some very nice properties. For example,if eaeb = ec, then e(2a)e(2b) = e(2c).

(6)(6) in combination with (4) immediately implieseaea+2 = ea+3,eaea+4 = ea+6(7)(so eaea+2n = ea−2n+1, or eaea+b = [b3 mod 7]ea−2b4, b = 1, ..., 6, where b3out front provides the sign of the product (modulo 7, 13 = 23 = 43 = 1, and33 = 53 = 63 = −1 )).These modulo 7 periodicity properties are reflected in the full multipli-cation table:1e1e2e3e4e5e6e7e1−1e6e4−e3e7−e2−e5e2−e6−1e7e5−e4e1−e3e3−e4−e7−1e1e6−e5e2e4e3−e5−e1−1e2e7−e6e5−e7e4−e6−e2−1e3e1e6e2−e1e5−e7−e3−1e4e7e5e3−e2e6−e1−e4−1. (8)2

The naturalness of this table is reflected in the matrix of its signs :O =111111111−111−11−1−11−1−111−11−11−1−1−111−1111−1−1−111−11−11−1−1−11111−11−1−1−11111−11−1−1−1. (9)(This is what is called a normalized Hadamard matrix of order 8 [4].

It isnormalized because the first row and column are all 1’s, and it is a Hadamardmatrix in containing only 1’s and -1’s, and in satisfying OO† = 8I, whereO† is the transpose of O, and I is the 8x8 identity matrix.) Note that ifa ̸= 0, b ̸= 0, then the components Oa,b = Oa+1,b+1, indices from 1 to 7,modulo 7 (first row and column of O are given the index 0).Let Oa be the ath row of O, a=0,1,...,7, and define the productOa • Ob = Oa,bOc,(10)where the components of Oc are Oc,d = Oa,dOb,d, for each d=0,1,...,7, andOa,b gives a sign to the product.

For example,O1 • O2 = +[1 · 1, (−1) · (−1), 1 · (−1), 1 · 1, (−1) · 1, 1 · (−1), (−1) · 1, (−1) · (−1)]= +[1, 1, −1, 1, −1, −1, −1, 1] = O6,where the plus sign out front arises from the component O1,2 = +1. Theresulting multiplication table of the Oa is exactly the same as (8), givingrise to the obvious isomorphism ea →Oa, a = 0, 1, ..., 7.The quaternion algebra arises in exactly the same way from the signmatrixQ =11111−11−11−1−1111−1−1.

(11)Likewise the complexes arise fromC ="111−1#. (12)3

(These are normalized Hadamard matrices of order 4 and 2. )The arrays used above are connected with Galois fields.

The real num-bers are the paradigm for mathematical field theory. There is addition (andsubtraction), an additive identity ,0, and every element x has an additiveinverse ,−x.

There is multiplication (and division), a multiplicative identity,1, and every element x ̸= 0 has a multiplicative inverse ,x−1. Multiplica-tion by zero gives zero, and for all x ̸= 0 and y ̸= 0, we also have xy ̸= 0(no divisors of zero).

Finally, xy = yx (commutative), and x(yz) = (xy)z(associative).R is an infinite field, but there also exist finite fields. For any prime pthere exist (unique up to isomorphism) fields of order pk for all k = 1, 2, 3, ...,denoted GF(pk) (G for Galois, their ill-fated founder, F for field).

For noother positive integers are there fields of that order.The pk elements of GF(pk) are easily written: {0, 1, h, h2, ..., hpk−2}.That is, the multiplication of GF(pk) is cyclic and for all x ̸= 0 in GF(pk),xpk−1 −1 = 0(13)(ie., hpk−1 = 1).All that remains then is to construct an addition table for GF(pk) con-sistent with its being a field. This problem can be reduced to finding what iscalled a Galois sequence for GF(pk), which consists of pk −1 elements of Zp(the integers modulo p).

Its further properties can be best illustrated by anexample. (Mathematicians have a more elaborate development in terms ofpolynomials and quotient modules; the elements of a Galois sequence appearin that context as coefficients of a polynomial.

)[ 01120221 ] is a Galois sequence for GF(32 = 9). Weidentify it with h0 = 1, the multiplicative identity of GF(9), and we’llidentify its kth cyclic permutation with hk.

That is,4

h1 = [ 10112022 ],h2 = [ 21011202 ],h3 = [ 22101120 ],h4 = [ 02210112 ],h5 = [ 20221011 ],h6 = [ 12022101 ],h7 = [ 11202210 ],h8 = [ 01120221 ],(14)where h8 = h0 = 1 gets us back where we started (any cyclic permutation ofthe initial sequence would have been a valid starting point). Notice that thefirst k = 2 elements of each sequence are unique, and can be used as labels forthe elements (we are using instead the exponents).

And notice that by ad-joining to this collection the zero sequence, 0 = [ 00000000 ],we have a set of pk = 32 = 9 vectors (sequences), each pk −1 = 32 −1 = 8-dimensional over Zp = Z3, and that the set is closed with respect to Z3vector addition. For example, using +p to represent modulo p addition,h2 +3 h4 = [ 21011202 ] +3 [ 02210112 ]= [ 20221011 ] = h5.A full addition table for GF(9) resulting from this sequence is listed below:0h1h2h3h4h5h6h7h8h1h5h8h4h60h3h2h7h2h8h6h1h5h70h4h3h3h4h1h7h2h6h80h5h4h6h5h2h8h3h7h10h50h7h6h3h1h4h8h2h6h30h8h7h4h2h5h1h7h2h40h1h8h3h3h6h8h7h3h50h2h1h4h4(15)5

(recall that h8 = 1). Note that hk +3 hk = hk+4 and hk +3 hk +3 hk =hk +3 hk+4 = 0.

Also, hk +3 hk+1 = hk+7. Because for any x and y in anyGF(3m),(x +3 y)3 = x3 +3 y3,(16)cubing the last equation above results in hk +3 hk+3 = hk+5 (exponents aretaken modulo 8 from 1 to 8, and although strictly speaking the exponentsk cube to 3k, because 3 and 8 are relatively prime we are allowed to replace3k by k in constructing new addition rules), and cubing this leads backto hk +3 hk+1 = hk+7.There is also, hk +3 hk+2 = hk+3, which cubedyields, hk +3 hk+6 = hk+1, and also hk +3 hk+5 = hk+2, which cubed yields,hk +3 hk+7 = hk+6.Of more interest to us here are the fields GF(2n), n = 1, 2, 3.

In partic-ular, a Galois sequence for GF(21) is [ 1 ], for GF(22) is [ 011 ], andfor GF(23) is [ 0010111 ]. In this last case we definee1 = [ 1001011 ],e2 = [ 1100101 ],e3 = [ 1110010 ],e4 = [ 0111001 ],e5 = [ 1011100 ],e6 = [ 0101110 ],e7 = [ 0010111 ].

(17)Addition in this case can also be completely described by cyclic equationsin the ea. To begin with,ea +2 ea = 0(18)(every element is its own additive inverse).

Also,ea +2 ea+1 = ea+5. (19)Since in the p = 2 case(x +2 y)2 = x2 +2 y2,(20)squaring the above addition rule leads to a new rule,ea +2 ea+2 = ea+3,(21)6

and squaring this leads toea +2 ea+4 = ea+6(22)(exponents are taken modulo 7 from 1 to 7).The link of GF(8) to the octonions should now be obvious. The matrixof signs in (9), used to construct an octonion multiplication, could have beenreplaced by the following matrix of elements of Z2 (ie., 0’s and 1’s):O′ =0000000001001011011001010111001000111001010111000010111000010111.

(23)Note that (−1)O′ab = Oab (see (9)), so if we defineO′a ∗O′b = (−1)O′ab[O′a +2 O′b],(24)then we have once again created an octonion product, where this time therows of O′ are identified with the basis of the octonions. Note!

We haveused GF(8) addition to create an octonion multiplication. The first row ofO′ is the multiplicative identity of O, and we must create a new 0 to playthe role of the additive identity of O.

With respect to O addition, the rowsof O′ are now treated as linearly independent, a basis for a real algebra.Relabel the rows of O′ as ea, a = 0, 1, ..., 7. So the exponents of GF(8) in(17) are mapped into the subscripts of the octonions.

Because the octonionproduct (now denoted just eaeb) is derived directly from the GF(8) addi-tion, the exponent rules (19,21,22) are valid for the octonion product, therules now applied to subscripts (see (4,7)). In addition, the index doublingautomorphism for the octonions (6) is now seen to follow from (20).

[Note: The sum rules (19,21,22) for GF(8) correspond to (4,7), butin general we can only make such correspondences up to a sign. For ex-ample, while it is true in GF(8) that ea +2 ea+5 = ea+1, in O we haveeaea+5 = −ea+1.

Index doubling is also tricky, and in Q it works out slightlydifferently. ][Also note: In GF(8), e7 = e0 = 1.

The reason that it was listed as e7 in(17) is to make the correspondence e7 →e7 of GF(8) to O. Therefore, since7

e0 = e7, we have e0 →e7, too! That is, e0 = 1 has no correspondence toany power of e1 ∈GF(8).

At this point the the notation e∞= 1 becomesincreasingly attractive. ][Finally note: the transpose of O′ also results in a valid GF(8) additionand O multiplication.

In this case, however, eaea+1 = −ea+3 in O. Exceptfor the sign change, this is the dual multiplication mentioned above. If wereplace (24) byO′a ∗O′b = (−1)O′ba[O′a +2 O′b],we generate the O multiplication rule, eaea+1 = −ea+5; and if we use thetranspose of O′, the rule eaea+1 = ea+3.

]Having made the correspondence between GF(8) addition and O mul-tiplication, one is naturally led to consider the role of GF(8) multiplicationin O. Since in GF(8), eaeb = ea+b, this operation on the indices of O is justa cyclic shift (of the index a for a = 1, ..., 7; e0 is left unaltered).

Let S bethe O automorphism that shifts the indices of eb, b = 1, ..., 7 by 1. So Sashifts the O indices by a, and S7 = S0 is the identity map.

Let φ be thezero map, mapping all x ∈O to 0. Obviously this collection of eight mapscan be made into the field GF(8) if given the appropriate addition.

Thismay or may not be of interest, but this is as far down that road as I amwilling to go at present.In the quaternion case one makes a correspondence with GF(4). Every-thing works out much the same, save that (20) doesn’t give rise to as simplea relation in Q as it did in O.

By inspection we see in this case thatif qiqj = qk, then q(2i)q(2j) = −q(2k). (25)Index quadrupalling gets us back to qiqj = qk, since 4=1 mod 3.

Hence inO, eaeb = ec could not imply e(2a)e(2b) = −e(2c), since 23 = 8 = 1 mod 7,and three (an odd number of) applications of index doubling must get usback to eaeb = ec.The binary matrix generating both Q multiplication and GF(4) additionisQ′ =0000010101100011. (26)In both O′ and Q′, the first row of each after the zeroth must be eitherthe one shown, or the first row of the respective transposes, for algebras8

isomorphic to O and Q to result from the process outlined. In particular,considerB =0000001101010110.

(27)[B11 B12 B13] = [0 1 1] is also a Galois sequence for GF(4), but in this casethe algebra multiplicationBi ∗Bj = (−1)Bij[Bi +2 Bj](28)does not result in Q, but rather an algebra isomorphic to that generatedby the adjoint elements, qL1qR3, qL2qR2, qL3qR1. Here the subscripts L andR denote multiplication from the left and right on Q.

Since qLiqRj[x] =qixqj = qRjqLi[x], it is apparent left adjoint multiplication commutes withright. (This is not the case for O, which is complicated by nonassociativity[2,5].

)Addition on GF(2n) can be turned into an algebra multiplication in theway outlined for n > 3 as well. For example, letg1 = [ 100010011010111 ],15-dimensional over Z2.

This is a Galois sequence for GF(16), and it canbe used to construct a new 16-dimensional algebra, extending the sequence,R, C, Q, O (this is distinct from the Cayley-Dickson prescription, which isfounded on the inclusion property, and in fact O is not a subalgebra of thisnew 16-dimensional algebra, which is noncommutative, nonassociative, andnonalternative; in [6] binary sequences are used to construct the Cayley-Dickson multiplication rules, as well as those of Clifford algebras).One final path down which I have no intention of travelling far: weshould be able to construct algebras in like manner from any GF(pn), forany prime p. For example, take the hk, k = 1, ..., 8, in GF(9) listed in (14),and map them to hk, k = 1, ..., 8, part of a basis for a new algebra. Mapthe zero sequence to 1, completing the basis.

Form the stacked sequences in(14) into a matrix, H (8 × 8). If hi +3 hj = hk in GF(9), then definehihj = exp[2πiHij/3]hk.

(29)If j −i = 4 mod 8, then replace hk by 1. Here we have yet another algebra,but at this point I’m just spewing out ideas without a clear notion of their9

interest or viability, so I’ll shift directions a bit in hopes of bringing orderout of chaos.It would seem in light of the material presented to this point that thedivision algebras are four out of an infinite collection of possible algebrasconstructable in like manner. And it is a collection, not a sequence.

High-lighting this is the fact that the first rows of Q′ and O′ (ignoring the intitial0’s) had to be [ 101 ] and [ 1001011 ] for Q and O withthe multiplication rules we are adopting to result (see (27,28)). Completelydifferent algebras result from most of the other cyclic permutations of thesesequences.We could also have begun with the dual sequences (beginning with thesame element, but in reverse order), [ 110 ] and [ 1110100 ].These sequences also give rise to Q and O, and they are Galois sequences forGF(4) and GF(8).

They are in addition quadratic residue codes of lengths3 and 7 over GF(2) [4]. For example, the quadratic residues modulo 7 are02 = 72 = 0, 12 = 62 = 1, 22 = 52 = 4, 32 = 42 = 2, so confusinglyrenumbering the positions of the sequence above 0 to 6, we see that the1’s appear in the 0, 1, 2, and 4 positions, which are determined by thequadratic residues.

Likewise, modulo 3, 02 = 32 = 0, 12 = 22 = 1, and the1’s of [ 110 ] appear in the 0 and 1 positions. The quadratic residuecode of length 1 over GF(2) is [ 1 ], also the Galois sequence of GF2, andassociated with C.There are no other examples of quadratic residue codes of any primelength p over GF(2) that correspond to Galois sequences.

To even have achance we must have a code of length 2k −1, and 2k −1 must be prime. So15 is out.

The quadratic residue code of length 31 is[1110110111100010101110000100100],and a Galois sequence, equal to (29) in the first 7 places, is[1110110001111100110100100001010].Let U be the 31x31 matrix formed of the first of these sequences and all itscyclic permutations, and let V be the 31x31 matrix formed from the second.The first has the nice property shared by all quadratic residue codes overGF(2) that(−1)Uab = −(−1)Uba, a ̸= b. (30)In the 22−1 = 3 and 23−1 = 7 cases this gives rise to the noncommutativity10

among the imaginary basis elements (̸= 1) of Q and O, which together with(−1)Uaa = −1(31)ensures that Q and O are division algebras (replace U by the appropriate3x3 and 7x7 matrices).But unfortunately the rows of U are not closedunder Z2 addition. Those of V are are closed under Z2 addition.

For alla, b ∈{1, ..., 31}, a ̸= b,Va +2 Vb = Vc,(32)for some c. So V gives rise to an algebra, but because(−1Vab ̸= −(−1)Vba, a ̸= b,(33)in general, there will be divisors of zero, and the algebra is not a divisionalgebra. Requiring of our generating sequences that they be both Galoisand quadratic residue is a heavy restriction, and the division algebras arethe only algebras that result.The quaternion and octonion codes/Galois sequences arise in other con-texts.

For example, they are useful in constructing the special lattices D4and E8 [4] associated with the integral quaternions and octonions, and theyarise in connection with projective geometry [7].Finally, it is my belief that the laws of Nature will be found to accreteabout the most special, select, and generative of mathematical objects andideas (a kind of hypervariational principle) that spawned my interest in thedivision algebras. At the very least it can not be doubted they are special,select, and generative.AcknowledgementI would like to thank Paul Monsky for information on Galois sequences.References:1.

A. Hurwitz, ¨Uber die Composition der quadratischen Formen von beliebigvielen Variablen, Nachrichten von der Gesellschaft der Wissenschaften zuG¨ottingen, 309 (1898).2. G.M.

Dixon, Derivation of the Standard Model, Il Nuovo Cimento 105B,349(1990).3. M. G¨unaydin and F. G¨ursey, Phys.

Rev. D 10, 674(1983).P.

Goddard, W. Nahm, D.I. Olive, H. Ruegg and A. Schwimmer, Comm.11

Math. Phys.

112, 385(1987).F. Smith, Hermitian Jordan Triple Systems, the Standard Model plus Grav-ity, and αE = 1/137.03608, hep-th 9302030.4.

J.H. Conway, N.J.A.

Sloane, ”Sphere Packings, Lattices and Groups”,Springer-Verlag, 2nd edition, 1991.5.C.A. Manogue, J. Schray, Finite Lorentz Transformations, Automor-phisms, and Division Algebras, hep-th 9302044.6.

P-E. Hagmark and P. Lounesto, Walsh Functions, Clifford Algebras, andCayley-Dickson Process, in ”Clifford Algebras and Their Applications inMathematical Physics”, D. Reidel Publishing Company, 531(1986).7. R. Shaw, Symmetry, ”Mathematical Perspectives: Four Recent InauguralLectures”, Hull University Press, 77(1991).12

---

출처: arXiv:9302.113 • 원문 보기