Re: matrix substitution-> theta1 minimal Pisot quotient group
- To: mathgroup at smc.vnet.net
- Subject: [mg67912] Re: matrix substitution-> theta1 minimal Pisot quotient group
- From: Paul Abbott <paul at physics.uwa.edu.au>
- Date: Wed, 12 Jul 2006 05:06:29 -0400 (EDT)
- Organization: The University of Western Australia
- References: <e665nv$n43$1@smc.vnet.net> <e7585m$l4b$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
In article <e7585m$l4b$1 at smc.vnet.net>,
Roger Bagula <rlbagula at sbcglobal.net> wrote:
> This kind of group works is daunting for just about everybody!
What do you mean by "groups works" ?
I expect that it is not daunting for experts in Group Theory. I'm no
expert on SU(3). Also, I've found that my third year undergraduate
students can handle basic questions on such groups, given sufficient
information and context.
> I used the this Russian substitution method
What Russian substitution method? Give a reference!
> on a well known matrix group Gell-Mann su(3):
Surely you mean the matrix representation of the group SU(3), not the
algebra su(3)?
> a = {{1, 0}, {0, 1}};
> b = {{0, 1}, {-1, 0}};
> c = {{0, 0}, {0, 0}};
> s1 = {{c, a, c}, {a, c, c}, {c, c, c}};
> s2 = {{c, -b, c}, {b, c, c}, {c, c, c}};
> s3 = {{a, c, c}, {c, -a, c}, {c, c, c}};
> s4 = {{c, c, a}, {c, c, c}, {a, c, c}};
> s5 = {{c, c, -b}, {c, c, c}, {b, c, c}};
> s6 = {{c, c, c}, {c, c, a}, {c, a, c}};
> s7 = {{c, c, c}, {c, c, -b}, {c, b, c}};
> s8 = {{a, c, c}, {c, a, c}, {c, c, -2*a}}/Sqrt[3];
>
> MatrixForm[s1]
> MatrixForm[s2]
> MatrixForm[s3]
> MatrixForm[s4]
> MatrixForm[s5]
> MatrixForm[s6]
> MatrixForm[s7]
> MatrixForm[s8]
Instead of using MatrixForm, try setting your Default Output FormatType
to TraditionalForm -- hopefully, you will be pleasantly suprised to see
matrices appear on output, as if by magic ...
Now, did you use BlockMatrix
> I got
> s1={{0,0,1,0,0,0},
> {0,0,0,1,0,0},
> {1,0,0,0,0,0},
> {0,1,0,0,0,0},
> {0,0,0,0,0,0},
> {0,0,0,0,0,0}}
> s2={{0,0,0,-1,0,0},
> {0,0,1,0,0,0},
> {0,1,0,0,0,0},
> {-1,0,0,0,0,0},
> {0,0,0,0,0,0},
> {0,0,0,0,0,0}}
> s3={{1,0,0,0,0,0},
> {0,1,0,0,0,0},
> {0,0,-1,0,0,0},
> {0,0,0,-1,0,0},
> {0,0,0,0,0,0},
> {0,0,0,0,0,0}}
> s4={{0,0,1,0,0,0},
> {0,0,0,1,0,0},
> {1,0,0,0,0,0},
> {0,1,0,0,0,0},
> {0,0,0,0,0,0},
> {0,0,0,0,0,0}}
> s5={{0,0,0,0,1,0},
> {0,0,0,0,0,1},
> {0,0,0,0,0,0},
> {0,0,0,0,0,0},
> {1,0,0,0,0,0},
> {0,1,0,0,0,0}}
> s6={{0,0,0,0,0,-1},
> {0,0,0,0,1,0},
> {0,0,0,0,0,0},
> {0,0,0,0,0,0},
> {0,1,0,0,0,0},
> {-1,0,0,0,0,0}}
> s7={{0,0,0,0,0,0},
> {0,0,0,0,0,0},
> {0,0,0,0,1,0},
> {0,0,0,0,0,1},
> {0,0,1,0,0,0},
> {0,0,0,1,0,0}}
> s8={{1,0,0,0,0,0},
> {0,1,0,0,0,0},
> {0,0,1,0,0,0},
> {0,0,0,1,0,0},
> {0,0,0,0,-2,0},
> {0,0,0,0,0,-2}}/Sqrt[3]
>
> I welcome someone to check my calculations.
You have clearly explained what calculation you have just done.
To obtain real 6 x 6 matrix group by substitution, I would proceed as
follows. First, you are, effectively, using the following replacements:
subs = {1 -> {{1, 0}, {0, 1}},
I -> {{0, 1}, {-1, 0}},
0 -> {{0, 0}, {0, 0}}};
that is a 2 x 2 matrix representation of 1, I, and 0. Written this way,
everthing is much clearer. Then, after loading
<< LinearAlgebra`
I would compute the real 6 x 6 matrix via replacement rules. For example,
s1 = BlockMatrix @ ({{0, 1, 0}, {1, 0, 0}, {0, 0, 0}} /. subs]
and, indeed, I do get the same answer as you.
> These matrices might be useful in real number calculations for strong
> field interactions.
I don't see why? Any computation that can be done with the complex 3x3
matrices can be done with the real 6x6 ones. What advantages do you
foresee?
> It might be possible that double group for this representation can be found.
Perhaps it would be helpful to readers if you explained (or referenced)
what the double group is, and why it would be useful.
Cheers,
Paul
_______________________________________________________________________
Paul Abbott Phone: 61 8 6488 2734
School of Physics, M013 Fax: +61 8 6488 1014
The University of Western Australia (CRICOS Provider No 00126G)
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