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Re: cubic quarternion group

"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 
group last night too.
and the
The (1+i+j+k)3 works, but it doesn't work for n=4.
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*)
>>(* group average/ center*)
>>(* subgroup centers*)
>>Solve[{ (1+i+j+k)^3-1==0,i^3+1==0,j^3+1==0,k^3+1==0},{i,j,k}]

Respectfully, Roger L. Bagula
tftn at, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 
619-5610814 :
alternative email: rlbtftn at

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