Re: cubic quarternion group

*To*: mathgroup at smc.vnet.net*Subject*: [mg53583] Re: cubic quarternion group*From*: Roger Bagula <tftn at earthlink.net>*Date*: Wed, 19 Jan 2005 01:59:37 -0500 (EST)*References*: <cs5an4$3q6$1@smc.vnet.net> <cscphp$dvi$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

"I" times the Pauli sigma matrices gives -1 determinant matrices. If you take the square root of them , you get i^4=-1, j^4=-1 and k^4=-1 matrices. I also figured out cubic 2by2 versions of that quaternion group last night too. i4={{1,i},{i,1}/Sqrt[2] j4={{1,1},{-1,1}}/Sqrt[2] k4={{(-1)^(1/4),0},{0,-(1)^(3/4)}} and the i3={{1/2,-Sqrt[3]/2},{Sqrt[3]/2,1/2}} j3={{-1,0},0,-1}} k3={{1/2,Sqrt[3]/2},{-Sqrt[3]/2,1/2}} The (1+i+j+k)3 works, but it doesn't work for n=4. g(n)=w+i(n)*x+y*j(n)+z*k(n) i(n)^n=-1 j(n)^n=-1 k(n)^n=-1 n Generalization of quaternions that seems to work after a fashion Roger Bagula wrote: > 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)] >> > > -- Respectfully, Roger L. Bagula tftn at earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : alternative email: rlbtftn at netscape.net URL : http://home.earthlink.net/~tftn