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)]
>