Re: cubic quarternion group

*To*: mathgroup at smc.vnet.net*Subject*: [mg53522] Re: cubic quarternion group*From*: Roger Bagula <tftn at earthlink.net>*Date*: Sat, 15 Jan 2005 21:07:51 -0500 (EST)*References*: <cs5an4$3q6$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Can anybody get mathematica to solve these for 2by2 matrices instead of complex numbers? Like Weinberg's U(1)*SU(2) weak force group. Creutz defines SU(2) as g={a0+I*a*sigma| a0+a^2=1,sigma=Pauli matrices} sigma1={{0,1},{1,0}} sigma2={{0,-I},{I,0}} sigma3={{1,0},{0,-1}} sigma^2=-1 This was before everyone realized that the quaternion definition was more fundamental/ number like. What we need is 2by2 matrices that have cube equal to -1. Roger L. Bagula wrote: > (* cubic quarternion group: *) > (* the fourth dimension without leaving the complex plane*) > (* a cyclotomic x^3+1=0 type of group *) > (* angles are {0,Pi,-2*Pi/3.Pi/6}*) > (* it is possible to map Hamilton quaternions to the complex plane using > this type of group*) > (* using a substitution conformal map: i->icube,j->jcube,k->kcube*) > (* this group owes it's origin to number theory quadratic fields and > Pell equations*) > (* when taken as polygon in the plane , it's "moment" as a center of > rotation is not zero *) > (* it is a boomerang type group for this reason*) > (* Roger L. Bagula 11 Jan 2005*) > i=-1 > j=1+(-1)^(1/3) > k=-(-1)^(1/3) > Simplify[ExpandAll[(1+i+j+k)^3]] > N[j] > N[k] > (* group average/ center*) > N[(1+i+j+k)/4] > (* subgroup centers*) > N[(1+i)/2] > N[(j+k)/2] > Clear[i,j,k] > Solve[{ (1+i+j+k)^3-1==0,i^3+1==0,j^3+1==0,k^3+1==0},{i,j,k}] > Simplify[Expand[(z-1)*(z+1)*(z+(-1)^(1/3))*(z-(1+(-1)^(1/3)))]] > Solve[(-1+z^2)*(-I*Sqrt[3]-z-z^2)==0,z] > Arg[1] > Arg[-1] > Arg[1+(-1)^(1/3)] > Arg[-(-1)^(1/3)] >