quaternions and octonions mutated to allow zero-divisors

Discussion in 'Math Research' started by rokirovka, Apr 3, 2007.

  1. rokirovka

    rokirovka Guest

    I recently read John Baez's paper on the octonions (http://
    math.ucr.edu/home/baez/octonions/) and in particular the section on
    the Cayley-Dickson Construction of the complex numbers, quaternions,
    octonions, sedenions, etc. (http://math.ucr.edu/home/baez/octonions/
    node5.html).

    If the octonion (a0+a1e1+a2e2+a3e3+a4e4+a5e5+a6e6+a7e7) is expressed
    as the ordered set of real numbers (a0, a1, a2, a3, a4, a5, a6, a7),
    then after expanding all the terms of octonion multiplication in the
    Cayley-Dickson Construction, one gets the following result:

    (a0,a1,a2,a3,a4,a5,a6,a7)(b0,b1,b2,b3,b4,b5,b6,b7) =

    (a0b0 -- a1b1 -- a2b2 -- a3b3 -- a4b4 -- a5b5 -- a6b6 -- a7b7,
    a0b1 + a1b0 + a3b2 -- a2b3 + a5b4 -- a4b5 + a6b7 -- a7b6,
    a0b2 + a2b0 + a1b3 -- a3b1 + a6b4 -- a4b6 + a7b5 -- b7a5,
    a0b3 + a3b0 + a2b1 -- a1b2 + a7b4 -- a4b7 + a5b6 -- a6b5,
    a0b4 + a4b0 + a1b5 -- a5b1 + a2b6 -- a6b2 + a3b7 -- a7b3,
    a0b5 + a5b0 + a4b1 -- a1b4 + a2b7 -- a7b2 + a6b3 -- a3b6,
    a0b6 + a6b0 + a4b2 -- a2b4 + a3b5 -- a5b3 + a7b1 -- a1b7,
    a0b7 + a7b0 + a4b3 -- a3b4 + a1b6 -- a6b1 + a5b2 -- a2b5)

    Since the sedenions and all higher-dimensional algebras created by the
    Cayley-Dickson Construction have zero-divisors, for example in the
    sedenions (e1 + e10)(e15 -- e4) = 0, I was looking for a way to
    demonstrate that no such zero-divisors of the form
    (ex + ey)(ez -- ew) = 0 exist in the octonions simply by looking at
    the above expansion. In doing so I discovered that if the signs are
    switched on a single pair of terms in the above expansion, for example
    if instead of + a2b1 -- a1b2 we had + a1b2 -- a2b1, then in this 8-
    dimensional algebra we would have zero-divisors, for example (e1 + e5)
    (e2 -- e6) = + e4 -- e4 -- e7 + e7 = 0. What's more, if the signs are
    switched on the terms a3b2, a2b3, a1b3, and a3b1 as well, we get an
    algebra with still more zero-divisors of this form, and in which the
    coefficients a1, b1, etc., still appear four times with positive sign
    and four times with negative sign in the expansion. Note that in this
    example, the octonions e1, e2, e3 whose coefficients' signs have been
    switched constitute together with 1 a subset of quaternions within the
    octonions.

    I call these octonions with one or more pairs of switched signs
    mutated octonions, because the transformation from octonions to them
    reminds me of a mutation of DNA in genetics. I have verified that
    these mutated octonions, like the sedenions, are a *-algebra with a
    conjugation
    (a0, a1, a2, a3, a4, a5, a6, a7)* = (a0, -a1, -a2, -a3, -a4, -a5, -a6,
    -a7) which satisfies a**=a and (ab)* = b*a*, and that they are nicely
    normed since a + a* is in the reals and aa* = a*a > 0 for all nonzero
    a in the mutated octonion algebras. (See the section of Baez's paper
    on the Cayley-Dickson Construction for these definitions.)

    While the octonions with three pairs of switched signs appear more
    symmetrical than the octonions with a single pair of switched signs,
    in a certain way the single-pair mutation makes more sense. When the
    signs of the coefficients of e1, e2, and e3 are all switched, products
    that equal +/-2e1, +/-2e2, and +/-2e3 in the octonions all become 0.
    For example (e1 + e4)(e3 -- e6) = 2e2 in the octonions, 0 in the three-
    mutation octonions; (e3 + e7)(e2 -- e6) = 2e1 in the octonions, 0 in
    the three-mutation octonions; and (e2 + e6)(e1 -- e5) = 2e3 in the
    octonions, 0 in the three-mutation octonions. But when only the signs
    of the products e1e2 and e2e1 are switched from -e3 and +e3
    respectively to +e3 and -e3 respectively, then products that equal
    +/-2e3 in the octonions become 0, but products that equal +/-2e1 and
    +/-2e2 in the octonions are unaffected. Thinking in terms of the
    genetic mutation concept, this is a mutation of e1 and e2 that
    produces a relative increase in the amount of them by reducing the
    amount of e3.

    Quaternions can also be mutated in this way. Mutating three pairs of
    terms in the expansion of the quaternion product would just create
    another set of normal quaternions, but mutating one pair of terms
    works the same way as in the case of octonions. The mutated quaternion
    multiplication table is as follows:

    (1)(1) = 1 ; (1)(e1) = e1 ; (1)(e2) = e2 ; (1)(e3) = e3
    (e1)(1) = e1 ; (e1)(e1) = -1 ; (e1)(e2) = +e3 ; (e1)(e3) = +e2
    (e2)(1) = e2 ; (e2)(e1) = -e3 ; (e2)(e2) = -1 ; (e2)(e3) = -e1
    (e3)(1) = e3 ; (e3)(e1) = -e2 ; (e3)(e2) = +e1 ; (e3)(e3) = -1

    The properties of mutated quaternions are rather remarkable. Like the
    sedenions and the mutated octonions, they are a *-algebra with a
    conjugation (a0, a1, a2, a3)* = (a0, -a1, -a2, -a3) which satisfies
    a**=a and (ab)* = b*a*, and they are nicely normed. The really
    interesting thing about them is the patterns formed by the products of
    two mutated quaternions with two nonzero +/-1 coefficients. In the
    normed division algebras, the reals, complex numbers, quaternions, and
    octonions, the product of two n-ions each with two nonzero +/-1
    coefficients always has either four nonzero coefficients or one
    nonzero coefficient. But in the mutated quaternions, as in the mutated
    octonions and in the sedenions, the product of two of them each with
    two nonzero +/-1 coefficients can have either zero, one, two, or four
    nonzero coefficients. What's remarkable about the mutated quaternions
    is not only their zero-divisors, the products with zero nonzero
    coefficients, but also the products with two nonzero coefficients, and
    the relationship between the two sets of products.

    The first noteworthy observation is that in the mutated quaternions,
    in each pair of zero-divisors one of them must have a nonzero real
    part. Here are the products of two mutated quaternions with two
    nonzero +/-1 coefficients that have zero nonzero coefficients, i.e.,
    the zero-divisors:

    (1 + e1)(e2 -- e3) = 0 (e2 + e3)(1 + e1) = 0
    (1 -- e1)(e2 + e3) = 0 (e2 -- e3)(1 -- e1) = 0

    (1 + e2)(e1 + e3) = 0 (e1 -- e3)(1 + e2) = 0
    (1 -- e2)(e1 -- e3) = 0 (e1 + e3)(1 -- e2) = 0

    Now the relationship between these products and the products with two
    nonzero coefficients: they are generated by switching the sign of e1
    or e2 in each of the (1 +/- e1) and (1 +/- e2) terms of the above zero-
    divisors. But that's not all: in each case, the resulting product
    equals exactly two times the (e2 +/- e3) or (e1 +/- e3) term!

    (1 -- e1)(e2 -- e3) = 2e2 -- 2e3 (e2 + e3)(1 -- e1) = 2e2 + 2e3
    (1 + e1)(e2 + e3) = 2e2 + 2e3 (e2 -- e3)(1 + e1) = 2e2 -- 2e3

    (1 -- e2)(e1 + e3) = 2e1 + 2e3 (e1 -- e3)(1 -- e2) = 2e1 -- 2e3
    (1 + e2)(e1 -- e3) = 2e1 -- 2e3 (e1 + e3)(1 + e2) = 2e1 + 2e3

    Thus, wherever (1 + e1) is a zero-divisor, (1 -- e1) in effect
    multiplies the other term by 2, and vice versa, and the same holds for
    (1 + e2) and (1 -- e2). Thus we can say that in the (1 +/- e1) and (1
    +/- e2) terms of the above equations, e1 and e2 are functioning
    exactly like real numbers +/-1 ! Where (1 + e1) in effect multiplies
    by 0 and (1 -- e1) multiplies by 2, e1 is functioning as the real
    number -1, and where (1 -- e1) multiplies by 0 and (1 + e1) multiplies
    by 2, e1 is functioning as the real number 1.

    We can define precisely where e1 and e2 function as 1 and -1 as
    follows:

    If (1 +/- e1) is the left-hand multiplier and e2 and e3 have the same
    sign, e1 functions as 1.
    If (1 +/- e1) is the left-hand multiplier and e2 and e3 have opposite
    signs, e1 functions as -1.
    If (1 +/- e1) is the right-hand multiplier and e2 and e3 have the same
    sign, e1 functions as -1.
    If (1 +/- e1) is the right-hand multiplier and e2 and e3 have opposite
    signs, e1 functions as 1.
    If (1 +/- e2) is the left-hand multiplier and e1 and e3 have the same
    sign, e2 functions as -1.
    If (1 +/- e2) is the left-hand multiplier and e1 and e3 have opposite
    signs, e2 functions as 1.
    If (1 +/- e2) is the right-hand multiplier and e1 and e3 have the same
    sign, e2 functions as 1.
    If (1 +/- e2) is the right-hand multiplier and e1 and e3 have opposite
    signs, e2 functions as -1.
     
    rokirovka, Apr 3, 2007
    #1
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