Re: [math-fun] cubic quaternions
But if Q^3 is a scalar q, you have QRQQ = w RQQQ = w q R QRQQ = ww QQQR = w^2 q R so w must equal w^2 -> w can only be 0,1 I was looking at the same thing you described while trying to get an extension of geometric algebra (the even grades of 3-d geometric algebra are the quaternions) where instead of the product rule e_i^2 = (e_i . e_i) + (e_i /\ e_i) = 1+0 = 1 we have e_i^3 = 1 but I spotted the problem above and couldn't think of a solution for it (the commutative case being boring...) It reminded me of anyons, which only work in 2-D, so I think it might be possible to get an arbitrary phase on the exchange operator if you restrict yourself to powers of e_1 and e_2. Rich 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)). -- Mike Stay staym@clear.net.nz http://www.xaim.com/staym
http://members.rogers.com/jveng/etc/cuban-coins1.jpg Couldn't resist. --Ed Pegg Jr, www.mathpuzzle.com
Mensaje citado por: Mike Stay <staym@clear.net.nz>:
But if Q^3 is a scalar q, you have
QRQQ = w RQQQ = w q R QRQQ = ww QQQR = w^2 q R so w must equal w^2 -> w can only be 0,1
wrong! QRQQ = wwQQRQ = wwwwQQQR so w must equal w^4 -> w is a cube root of unity; choose whichever you want to get the algebra you want. - hvm ------------------------------------------------- Obtén tu correo en www.correo.unam.mx UNAMonos Comunicándonos
Mensaje citado por: mcintosh@servidor.unam.mx:
wrong!
meesa wrong too. if QR = w RQ then RQ = w^{-1} QR and so w better not be zero. if w^3 = 1 holds, the rest of my comment applies as posted, but then it is silly to use it to conclude that w = (cube root of 1). In fact w could be anything but zero, but context dictates that cube root. Mensaje citado por: Michael Reid <reid@math.arizone.edu>:
... which is the continuous cohomology group ...
Ah, yes ... those cohomologies. Another application of exotic quaternions is to the symmetry group of a magnetic lattice; this and related things are the subject of some articles by J. Zak in Journal of Mathematical Physics and elsewhere.
divisors of zero
how are you going to avoid them if you have nontrivial matrices and orthogonal idempotents? - hvm ------------------------------------------------- Obtén tu correo en www.correo.unam.mx UNAMonos Comunicándonos
participants (3)
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Ed Pegg Jr -
mcintosh@servidor.unam.mx -
Mike Stay