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MathGroup Archive 2006

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Re: matrix substitution--> Gell-Mann su(3) ->repartitioned

  • To: mathgroup at smc.vnet.net
  • Subject: [mg67357] Re: matrix substitution--> Gell-Mann su(3) ->repartitioned
  • From: Roger Bagula <rlbagula at sbcglobal.net>
  • Date: Tue, 20 Jun 2006 02:14:22 -0400 (EDT)
  • References: <e665nv$n43$1@smc.vnet.net> <e7583n$l3s$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

By trial and error I figured out a clunky method of repartitioning the 
matrices for this group:

s1a = Flatten[Table[{Flatten[Table[s1[[
      n, m]][[1, i]], {n,
           1, 3}, {i, 1, 2}]], Flatten[Table[s1[[n, m]][[2,
            i]], {n, 1, 3}, {i, 1, 2}]]}, {m, 1, 3}], 1]
s2a = Flatten[Table[{
          Flatten[Table[s2[[n, m]][[1, i]], {n, 1, 3}, {i, 1,
            2}]], Flatten[Table[s2[[n, m]][[2, i]], {n, 1,
      3}, {i, 1, 2}]]}, {m, 1, 3}], 1]
s3a = Flatten[Table[{Flatten[Table[s3[[
      n, m]][[1, i]], {n,
           1, 3}, {i, 1, 2}]], Flatten[Table[s3[[n, m]][[2,
            i]], {n, 1, 3}, {i, 1, 2}]]}, {m, 1, 3}], 1]
s4a = Flatten[Table[{
          Flatten[Table[s4[[n, m]][[1, i]], {n, 1, 3}, {i, 1,
            2}]], Flatten[Table[s4[[n, m]][[2, i]], {n, 1,
      3}, {i, 1, 2}]]}, {m, 1, 3}], 1]
s5a = Flatten[Table[{Flatten[Table[s5[[
      n, m]][[1, i]], {n,
           1, 3}, {i, 1, 2}]], Flatten[Table[s5[[n, m]][[2,
            i]], {n, 1, 3}, {i, 1, 2}]]}, {m, 1, 3}], 1]
s6a = Flatten[Table[{
          Flatten[Table[s6[[n, m]][[1, i]], {n, 1, 3}, {i, 1,
            2}]], Flatten[Table[s6[[n, m]][[2, i]], {n, 1,
      3}, {i, 1, 2}]]}, {m, 1, 3}], 1]
s7a = Flatten[Table[{Flatten[Table[s7[[
      n, m]][[1, i]], {n,
           1, 3}, {i, 1, 2}]], Flatten[Table[s7[[n, m]][[2, i]], {n, 1,
            3}, {i, 1, 2}]]}, {m, 1, 3}], 1]
s8a = Flatten[Table[{Flatten[Table[s8[[n, m]][[1, i]], {n,
  1, 3}, {i, 1, 2}]], Flatten[
      Table[s8[[n, m]][[2, i]], {n, 1, 3}, {i, 1, 2}]]}, {m, 1, 3}], 1]

These appear to work...
I had to reread the matrices in the right order and Flatten at the right 
places!
Roger Bagula wrote:

>This kind of group works is daunting for just about everybody!
>It's one of the reasons I bless Mathematica everyday!
>
>
>I used the this Russian substitution method on a well known matrix group 
>Gell-Mann 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]
>
>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.
>These matrices might  be useful in real number calculations for strong 
>field interactions.
>It might be possible that double group for this representation can be found.
>
>  
>


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