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Sextenions (6D hypercomplex numbers)

  1. Feb 20, 2016 #1
    So I have an interest in hypercomplex numbers and Clifford Algebras and was wondering a few months ago about other hypercomplex numbers besides the celebrated Quaternions and Octonions. I tried to construct a 5D complex number system using a Cayley table but noticed that entries in rows and columns were redundant. But what about a 6D complex number system? Well I managed to come up with a proposal using a Cayley table(and I do not claim to be the true originator) and here it is:


    Now the rule to generate this Cayley table is i2=j2=k2=l2=r2=-1

    And ijklr = (ij)(kl)r = -1

    Now this is a non-associative algebra because (for example)

    i((jk)(lr))= i(-rk) = i2 = -1
    and (i(jk)(lr) = jk = -r

    Has anyone else seen an Algebra like this before? I googled "sextenions" and found one article from 11 years ago where it mentioned a 6D complex number system but did not present a Cayley multiplication table for it.
  2. jcsd
  3. Feb 20, 2016 #2


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    How do we know we are supposed to read ijklr as (ij)(kl)r?

    (i(jk))(lr) = (i(-r))k = jk= -r

    It is easy to come up with multiplication tables, but do you get useful properties from it?
    They are not even alternative, a property octonions have.
  4. Feb 21, 2016 #3

    You're right that they're neither associative nor alternative.

    Using the associator [x,y,z] = (xy)z-x(yz), we get

    [i,j,k] = (ij)k-i(jk) = (k)k-i(-r) = k2+(ir=-j) = -(1+j)
    [k,j,i] = (kj)i-k(ji) = j - k(-k) = j+k2 = -(1-j)
    [i,k,j] = (ik)j-i(kj) = lj - (ir) = i+j
    [k,i,j] = (ki)j - k(ij) = 1-i
    [j,i,k] = (ji)k-j(ik) = 1+i
    [j,k,i] = (jk)i - j(ki) = -(i+j)

    And ∀(a,b,c,d,e,f,g,x,y,z,u,v,w)∈ℝ1(not all zero),
    let S = a +(σ = bi+cj+dk+el+fr), T = x +(τ = yi+zj+uk+vl+wr)

    S⊗T(the sextenion product) =
    + (bx)i -(by)+(bz)k+(bu)l -(bv)r -(bw)j
    + (cx)j -(cy)k -(cz) -(cu)r -(cv)i+(cw)l
    + (dx)k-(dy)l+(dz)r-(du)+(dv)j+(dw)i
    + (ex)l+(ey)r+(ez)i-(eu)j-(ev)+(ew)k
    + (fx)r+(fy)j-(fz)l-(fu)i-(fv)k-(fw)

    = ax + xσ + aτ -(σ⋅τ) + σ◊τ

    Where σ◊τ = (dw-cv+ez-fu)i +(dv-bw+fy-eu)j +(bz-cy+ew-fv)k +(bu-dy+cw-fz)l +(dz-bv+ey-cu)r .

    Now ∀S(where S is a Sextenion), we define S* = a-bi-cj-dk-el-fr, so

    S⊗S* = (a2)-(ab)i-(ac)j-(ad)k-(ae)l-(af)r
    + (ab)i +(b2)-(bc)k-(bd)l +(be)r +(bf)j
    + (ac)j -(cb)k -(c2) +(cd)r +(ce)i+(cf)l
    + (ad)k+(db)l-(cd)r+(d2)-(de)j-(df)i
    + (ae)l-(be)r-(ce)i+(ed)j+(e2)-(ef)k
    + (af)r-(bf)j-(cf)l+(df)i+(ef)k-(f2)

    = a2+b2+c2+d2+e2+f2

    Thus, if we define |S|2 = a2+b2+c2+d2+e2+f2,

    then ∀S = a+bi+cj+dk+el+fr, [tex]S^{-1} = \frac{S^*}{|S|^2}[/tex] since
    [tex]S ⊗ \frac{S^*}{|S|^2} = \frac{S^*}{|S|^2} ⊗ S = \frac{|S|^2}{|S|^2} = 1[/tex] . So the Sextenions are indeed a division algebra, which certainly is a useful property.
  5. Feb 21, 2016 #4
    Nope--for ##S^-1## to have the form as described above, you would need ##[S*, S, x]## to vanish for all ##x##, which doesn't even hold for the basis elements.

    Are you familiar with Hurwitz's theorem?
  6. Feb 21, 2016 #5
    Come to think of it, Hurwitz's theorem isn't appropriate here. In any case, the existence of a two-sided inverse isn't strong enough to imply the division algebra property. The sedenions are an example of this.
  7. Feb 21, 2016 #6
    The sedenions have zero-divisors. Haven't checked if the Sextenions do. In any case, according to the definition of a division algebra D:

    ∀a ∈ D & ∀(b≠0) ∈ D, ∃ Unique elements {x},{y} ∈ D such that a = bx & a = yb.

    Now for a = i and b = k, i = k*r and i = -r*k so x = r and y = -r. You can check to see this holds for all pairs of basis elements. So the Sextenions are neither associative nor alternative but unlike the Sedenions they do (appear)to meet the definition of a division algebra.
    And for all basis elements (i,j,k,l,r), the left and right product of each of these elements with its negative signed element = +1.
  8. Feb 22, 2016 #7


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    No division algebra has zero-divisors. This is a direct consequence of the definition for a=0 (as the elements x and y have to be unique, and x=0 and y=0 work).
  9. Feb 22, 2016 #8
    Just because the basis elements form a loop doesn't mean that the algebra's collection of units does, as well.
    All of the Cayley-Dickson loops are alternative, but the algebras associated with them have zero divisors beyond the octonions.
  10. Feb 23, 2016 #9
    It's an important theorem about division algebras, proved only in the 1950s, that all division algebras over the reals must be of dimension 1, 2, 4, or 8. (It had already been known for several decades that the dimension has to be a power of 2.)

    The usual examples are those created by the Cayley-Dickson doubling construction from the reals: the reals R, the complexes C, the quaternions H, and the octonions O.

    It's often erroneously claimed that these are the only division algebras over R. That's not the case. Even in dimension 2 there exist division algebras that are not isomorphic to the complexes.

    But if you toss in the extra condition that the division algebra must be normed — i.e., it must possess a norm |⋅| with the property that

    |xy| = |x| |y|

    for all x, y in the division algebra, then indeed R, C, H, and O are the only normed real division algebras. (Likewise if the division algebras are required to be alternative, as suggested above, these four are the only examples.)
  11. Feb 26, 2016 #10

    Where did the normed property come into play and have you tested to see whether or not the Algebra I proposed satisfies this property?
    I will be happy to do so when I have a bit more time(as it's gonna be very, very messy since each Sextenion has 6 terms! :-p)
  12. Feb 26, 2016 #11
    By the first theorem I quoted, any real algebra not of dimension 1, 2, 4, or 8 will necessarily have zero-divisors, i.e., two nonzero elements x, y such that

    xy = 0.​

    If this algebra were a normed algebra with |⋅|, we would have

    |xy| = |x| |y|.​

    But an element has norm 0 if and only if it is the zero element. Hence this would lead to a contradiction.
    Last edited: Feb 26, 2016
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