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