Coset Reps

3130 days ago by jimlb

Here I'll give the coset representatives that are given in Lemma 5.1.1 of Roberts-Schmidt.  Note that they use the symplectic form $\begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{pmatrix}$ as well as the transpose on the first factor. Therefore, we have to change bases to get the correct elements for our group.  Recall this amounts to conjugating by $P = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix}$.

var('x,y,z'); 
       
(x, y, z)
(x, y, z)
P=Matrix(SR,[[1,0,0,0],[0,1,0,0],[0,0,0,1],[0,0,1,0]]);P 
       
[1 0 0 0]
[0 1 0 0]
[0 0 0 1]
[0 0 1 0]
[1 0 0 0]
[0 1 0 0]
[0 0 0 1]
[0 0 1 0]

They give elements $s_1$ and $s_2$ that arise from the Bruhat decomposition.  We change those to $S_1$ and $S_2$ for us. In fact, $S_1 = s_1$, but $S_2$ is not the same as $s_2$.

s1=Matrix(SR,[[0,1,0,0],[1,0,0,0],[0,0,0,1],[0,0,1,0]]);s1 
       
[0 1 0 0]
[1 0 0 0]
[0 0 0 1]
[0 0 1 0]
[0 1 0 0]
[1 0 0 0]
[0 0 0 1]
[0 0 1 0]
s2=Matrix(SR,[[1,0,0,0],[0,0,1,0],[0,-1,0,0],[0,0,0,1]]);s2 
       
[ 1  0  0  0]
[ 0  0  1  0]
[ 0 -1  0  0]
[ 0  0  0  1]
[ 1  0  0  0]
[ 0  0  1  0]
[ 0 -1  0  0]
[ 0  0  0  1]
S1 = P*s1*P;S1 
       
[0 1 0 0]
[1 0 0 0]
[0 0 0 1]
[0 0 1 0]
[0 1 0 0]
[1 0 0 0]
[0 0 0 1]
[0 0 1 0]
S2=P*s2*P;S2 
       
[ 1  0  0  0]
[ 0  0  0  1]
[ 0  0  1  0]
[ 0 -1  0  0]
[ 1  0  0  0]
[ 0  0  0  1]
[ 0  0  1  0]
[ 0 -1  0  0]

They give representatives for $K_0(p^{n})\backslash Sp_4(F_p)$, so their first set of representatives reduces to just the identity, which is the same in either basis.  Define the matrices of the following form where $z \in F_p$:

g1 = Matrix(SR,[[1,0,0,0],[0,1,0,0],[0,z,1,0],[0,0,0,1]]);g1 
       
[1 0 0 0]
[0 1 0 0]
[0 z 1 0]
[0 0 0 1]
[1 0 0 0]
[0 1 0 0]
[0 z 1 0]
[0 0 0 1]

The second set of coset representatives are given by these matrices multiplied by $s_2$ on the right:

g1*s2; 
       
[ 1  0  0  0]
[ 0  0  1  0]
[ 0 -1  z  0]
[ 0  0  0  1]
[ 1  0  0  0]
[ 0  0  1  0]
[ 0 -1  z  0]
[ 0  0  0  1]

We now change this into our basis:

G1=P*g1*P;G1 
       
[1 0 0 0]
[0 1 0 0]
[0 0 1 0]
[0 z 0 1]
[1 0 0 0]
[0 1 0 0]
[0 0 1 0]
[0 z 0 1]
G1*S2 
       
[ 1  0  0  0]
[ 0  0  0  1]
[ 0  0  1  0]
[ 0 -1  0  z]
[ 1  0  0  0]
[ 0  0  0  1]
[ 0  0  1  0]
[ 0 -1  0  z]

Now consider matrices of the following form for $y,z$ in $F_p$:

g2=Matrix(SR,[[1,0,0,0],[0,1,0,0],[y,z,1,0],[0,y,0,1]]);g2 
       
[1 0 0 0]
[0 1 0 0]
[y z 1 0]
[0 y 0 1]
[1 0 0 0]
[0 1 0 0]
[y z 1 0]
[0 y 0 1]

The representatives here are given by these matrices multiplied on the right by $s_2 s_1$.  We change bases:

G2=P*g2*P;G2 
       
[1 0 0 0]
[0 1 0 0]
[0 y 1 0]
[y z 0 1]
[1 0 0 0]
[0 1 0 0]
[0 y 1 0]
[y z 0 1]
G2*S2*S1 
       
[ 0  1  0  0]
[ 0  0  1  0]
[ 0  0  y  1]
[-1  y  z  0]
[ 0  1  0  0]
[ 0  0  1  0]
[ 0  0  y  1]
[-1  y  z  0]

Finally, consider matrices of the following form for $x,y,z \in F_p$:

g3=Matrix(SR,[[1,0,0,0],[0,1,0,0],[y,z,1,0],[x,y,0,1]]);g3 
       
[1 0 0 0]
[0 1 0 0]
[y z 1 0]
[x y 0 1]
[1 0 0 0]
[0 1 0 0]
[y z 1 0]
[x y 0 1]

The representatives are given by these matrices multiplied on the right by $s_2 s_1 s_2$. We change bases:

G3= P*g3*P;G3 
       
[1 0 0 0]
[0 1 0 0]
[x y 1 0]
[y z 0 1]
[1 0 0 0]
[0 1 0 0]
[x y 1 0]
[y z 0 1]
G3*S2*S1*S2 
       
[ 0  0  0  1]
[ 0  0  1  0]
[ 0 -1  y  x]
[-1  0  z  y]
[ 0  0  0  1]
[ 0  0  1  0]
[ 0 -1  y  x]
[-1  0  z  y]