Coset Reps

3161 days ago by jimlb

var('p,x,z,y') 
       
(p, x, z, y)
(p, x, z, y)
b2=Matrix(SR,[[p,0,0,0],[-x,1,0,z],[0,0,1,x],[0,0,0,p]]);b2 
       
[ p  0  0  0]
[-x  1  0  z]
[ 0  0  1  x]
[ 0  0  0  p]
[ p  0  0  0]
[-x  1  0  z]
[ 0  0  1  x]
[ 0  0  0  p]
J=Matrix(SR,[[0,0,-1,0],[0,0,0,-1],[1,0,0,0],[0,1,0,0]]);J; 
       
[ 0  0 -1  0]
[ 0  0  0 -1]
[ 1  0  0  0]
[ 0  1  0  0]
[ 0  0 -1  0]
[ 0  0  0 -1]
[ 1  0  0  0]
[ 0  1  0  0]
b2.transpose()*J*b2 
       
[ 0  0 -p  0]
[ 0  0  0 -p]
[ p  0  0  0]
[ 0  p  0  0]
[ 0  0 -p  0]
[ 0  0  0 -p]
[ p  0  0  0]
[ 0  p  0  0]

This gives $\mu(b_2) = p$, as expected.  Thus, the new coset representative $b_2^{*} = \mu(b_2) b_2^{-1}$ is given by:

p * b2.inverse() 
       
[ 1  0  0  0]
[ x  p  0 -z]
[ 0  0  p -x]
[ 0  0  0  1]
[ 1  0  0  0]
[ x  p  0 -z]
[ 0  0  p -x]
[ 0  0  0  1]
b1=Matrix(SR,[[1,0,x,y],[0,1,y,z],[0,0,p,0],[0,0,0,p]]);b1 
       
[1 0 x y]
[0 1 y z]
[0 0 p 0]
[0 0 0 p]
[1 0 x y]
[0 1 y z]
[0 0 p 0]
[0 0 0 p]
b1.transpose()*J*b1 
       
[ 0  0 -p  0]
[ 0  0  0 -p]
[ p  0  0  0]
[ 0  p  0  0]
[ 0  0 -p  0]
[ 0  0  0 -p]
[ p  0  0  0]
[ 0  p  0  0]

This gives $\mu(b_1) = p$, as expected.  Thus, the new coset representative $b_1^{*} = \mu(b_1) b_1^{-1}$ is given by:

p*(b1.inverse()) 
       
[ p  0 -x -y]
[ 0  p -y -z]
[ 0  0  1  0]
[ 0  0  0  1]
[ p  0 -x -y]
[ 0  p -y -z]
[ 0  0  1  0]
[ 0  0  0  1]
b3=Matrix(SR,[[1,0,x,0],[0,p,0,0],[0,0,p,0],[0,0,0,1]]);b3 
       
[1 0 x 0]
[0 p 0 0]
[0 0 p 0]
[0 0 0 1]
[1 0 x 0]
[0 p 0 0]
[0 0 p 0]
[0 0 0 1]
b3.transpose()*J*b3 
       
[ 0  0 -p  0]
[ 0  0  0 -p]
[ p  0  0  0]
[ 0  p  0  0]
[ 0  0 -p  0]
[ 0  0  0 -p]
[ p  0  0  0]
[ 0  p  0  0]

This gives $\mu(b_3) = p$, as expected.  Thus, the new coset representative $b_3^{*} = \mu(b_3) b_3^{-1}$ is given by:

p*(b3.inverse()) 
       
[ p  0 -x  0]
[ 0  1  0  0]
[ 0  0  1  0]
[ 0  0  0  p]
[ p  0 -x  0]
[ 0  1  0  0]
[ 0  0  1  0]
[ 0  0  0  p]
b4 = Matrix(SR,[[p,0,0,0],[0,p,0,0],[0,0,1,0],[0,0,0,1]]);b4 
       
[p 0 0 0]
[0 p 0 0]
[0 0 1 0]
[0 0 0 1]
[p 0 0 0]
[0 p 0 0]
[0 0 1 0]
[0 0 0 1]
b4.transpose()*J*b4 
       
[ 0  0 -p  0]
[ 0  0  0 -p]
[ p  0  0  0]
[ 0  p  0  0]
[ 0  0 -p  0]
[ 0  0  0 -p]
[ p  0  0  0]
[ 0  p  0  0]

This gives $\mu(b_4) = p$, as expected.  Thus, the new coset representative $b_4^{*} = \mu(b_4) b_4^{-1}$ is given by:

p*(b4.inverse()) 
       
[1 0 0 0]
[0 1 0 0]
[0 0 p 0]
[0 0 0 p]
[1 0 0 0]
[0 1 0 0]
[0 0 p 0]
[0 0 0 p]