Rich:
If this ring is associative, then it has a matrix representation.
Have you attempted to find the mapping?
Henry
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At 10:25 AM 4/30/03 -0700, Richard Schroeppel wrote:
>I've been wondering about doing a cubic analog of the quaternions.
>
>The idea is to have two generators Q and R with Q = cbrt(q) and
>R = cbrt(r), and the non-commutative multiplication rule R*Q = w Q*R
>(where w = cbrt(1) = (-1 + i sqrt3)/2 = e^(2 pi i/3)).
>For definiteness, I'm imagining q=2 and r=3.
>
>The result seems to be a nine-dimensional space, linear combinations
>of 1, Q, Q^2, R, R^2, Q R, Q^2 R, Q R^2, Q^2 R^2, with coefficients
>of the shape a + b w.
>
>The type of construction seems to guarantee associativity, needing
>only to check things like R Q^3 = Q^3 R -- required, since Q^3 is
>in the ground field which should commute with R, and true, since
>we pick up three factors of w when we move the R through the Qs
>while sorting the factors.
>
>There are some nice properties like (Q+R)^3 = Q^3 + R^3, which
>happens because the cross terms like QRR come in three orders and
>when reordered and collected have coefficient 1+w+w^2 = 0.
>
>I haven't proved "no zero divisors", which is an important theorem
>in the quaternion case. Or worked out the Norm formula, which is
>needed to compute reciprocals.
>
>One proof of the Four Square Theorem (every number is the sum of
>four squares) uses quternion arithmetic. Perhaps there's a proof
>of the Nine Cube Theorem lurking somewhere (with q=r=1 ?).
>
>Has anyone seen this stuff before?
>
>Rich rcs(a)cs.arizona.edu
