# Coset Reps

## 3341 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]