-----Original Message----- From: Richard Schroeppel [mailto:rcs@CS.Arizona.EDU] Sent: Wednesday, April 30, 2003 1:25 PM To: math-fun@mailman.xmission.com Subject: [math-fun] cubic quaternions

I've been wondering about doing a cubic analog of the quaternions. 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.

I think that it's guaranteed that you will have zero divisors. I think there's a theorem that the only finite-dimensional division algebras over the reals are the complex numbers, the quaternions, and the octonions. But I might be confused. Or there might be some property that you need to assume before you get limited to these three, that you don't have and don't need for your "cubic quaterions". How do you check for zero divisors? You can express elements of your structure as 9-vectors, and multiplication as a 9x9 matrix. A nonzero determinant is necessary, but not sufficient, for no zero divisors. If the multiplication was commutative, then the matrix would be symmetric, and you would have no zero divisors just if the matrix is positive definite. But what is the corresponding criterion for assymmetric matrices? Andy Latto andy.latto@pobox.com