Paramodular_Lemma_5.1

1754 days ago by jimlb

Q2=matrix([[1,0,0,0,0,0,0,0],[0,1,0,0,0,0,0,0],[0,0,0,0,0,0,1,0],[0,0,0,0,0,0,0,1],[0,0,0,0,1,0,1,0],[0,0,0,0,0,1,0,1],[1,0,-1,0,0,0,0,0],[0,1,0,-1,0,0,0,0]]);Q2 
       
N=4;s=[]; for (i,j) in cartesian_product([range(N),range(N)]): s.append('d%d%d' %(i,j)) a = var(s); D = matrix(SR,N,N,a); D 
       
N=4;s=[]; for (i,j) in cartesian_product([range(N),range(N)]): s.append('x%d%d' %(i,j)) a = var(s); X = matrix(SR,N,N,a); X 
       

As our matrices will be in $\textrm{Sp}_4(F)$, we can take the parabolic element to be of the form $\begin{pmatrix} ^{t}D^{-1} & X \\ 0 & D \end{pmatrix}$. 

A=(D.det()*D.transpose().inverse()).simplify_full();A 
       
O4=matrix(4);O4;O2=matrix(2);O2;I2=matrix.identity(2);I2 
       


We now create our element $P \in P_8(F) \subset \textrm{Sp}_8(F)$. 

P=block_matrix(2,2,[A,X,O4,D]);P; 
       
N=4;s=[]; for (i,j) in cartesian_product([range(N),range(N)]): s.append('h%d%d' %(i,j)) var(s); 
       
h1=matrix([[h00,p*h01], [h10,h11]]);h2=matrix([[h02,h03],[h12,h13/p]]);h3=matrix([[h20,p*h21],[p*h30,p*h31]]);h4=matrix([[h22,h23],[p*h32,h33]]); 
       
N=4;s=[]; for (i,j) in cartesian_product([range(N),range(N)]): s.append('k%d%d' %(i,j)) var(s); 
       
k1=matrix([[k00,p*k01], [k10,k11]]);k2=matrix([[k02,k03],[k12,k13/p]]);k3=matrix([[k20,p*k21],[p*k30,p*k31]]);k4=matrix([[k22,k23],[p*k32,k33]]); 
       
HK=block_matrix(4,4,[h1,O2,-h2,O2,O2,k1,O2,k2,-h3,O2,h4,O2,O2,k3,O2,k4]);HK 
       
N=4;s=[]; for (i,j) in cartesian_product([range(N),range(N)]): s.append('H%d%d' %(i,j)) var(s); var('p'); 
       
H1=matrix([[H00,p*H01], [H10,H11]]);H2=matrix([[H02,H03],[H12,H13/p]]);H3=matrix([[H20,p*H21],[p*H30,p*H31]]);H4=matrix([[H22,H23],[p*H32,H33]]); 
       
HI=block_matrix(4,4,[H1,O2,-H2,O2,O2,I2,O2,O2,-H3,O2,H4,O2,O2,O2,O2,I2]);HI 
       

We set $Z = (h_1,k_1) = Q_2^{-1}P Q_2 (H,1)$ and use that this is in $\Gamma_2[\varpi] \times \Gamma_2[\varpi]$ to try and deduce $H \in \Gamma_2[\varpi]$. 

Z=Q2.inverse()*P*Q2*HI;Z 
       

We start by simplifying the third and fourth columns. 

(HK[0,2],Z[0,2]);(HK[0,3], Z[0,3]);(HK[1,2],Z[1,2]);(HK[1,3],Z[1,3]); 
       



Z=Z.substitute(x02==0).substitute(x03==0).substitute(x12==0).substitute(x13==0);Z; 
       
(HK[4,2],Z[4,2]);(HK[4,3], Z[4,3]);(HK[5,2],Z[5,2]);(HK[5,3],Z[5,3]); 
       



Z=Z.substitute(x22==d02).substitute(x23==d03).substitute(x32==d12).substitute(x33==d13);Z; 
       
Z=Z.substitute(d22==k00).substitute(d23==p*k01).substitute(d32==k10).substitute(d33==k11).substitute(d02==-k20).substitute(d03==-p*k21).substitute(d12==-p*k30).substitute(d02==-p*k31).substitute(d13==-p*k31);Z; 
       

Now the 8th column. 

Z=(Z.substitute(x31==(Z[5,7]==0).solve(x31)[0].right()).substitute(x11==(Z[1,7]==0).solve(x11)[0].right()).substitute(x01==(Z[0,7]==0).solve(x01)[0].right()).substitute(x21==(Z[4,7]==0).solve(x21)[0].right())).simplify_full();Z 
       
Z=Z.substitute(d21==-k03).substitute(d31==-k13/p).substitute(d01==k23).substitute(d11==k33);Z; 
       

On to column 7. 

Z=(Z.substitute(x00==(Z[0,6]==0).solve(x00)[0].right()).substitute(x10==(Z[1,6]==0).solve(x10)[0].right()).substitute(x20==(Z[4,6]==0).solve(x20)[0].right()).substitute(x30==(Z[5,6]==0).solve(x30)[0].right())).simplify_full();Z 
       
Z=(Z.substitute(d20==-k02).substitute(d30==-k12).substitute(d00==k22).substitute(d10==p*k32)).simplify_full();Z; 
       

Recall we have that $Z[2,0]=Z[3,0]=Z[6,0]=Z[7,0]=0$.

Z[0,0]=Z[0,0]-Z[2,0]; 
       
Z[2,0]=0;Z[1,0]=Z[1,0]-Z[3,0];Z.simplify_full() 
       
Z[3,0]=0;Z[4,0]=Z[4,0]+Z[6,0];Z[5,0]=Z[5,0]+Z[7,0];Z.simplify_full() 
       
Z[6,0]=0;Z[7,0]=0; 
       
Z.simplify_full(); 
       

Now we make use of the fact that $Z[2,1]=Z[3,1]=Z[6,1]=Z[7,1]=0$.

Z[0,1]=Z[0,1]-Z[2,1];Z[1,1]=Z[1,1]-Z[3,1];Z[4,1]=Z[4,1]+Z[6,1];Z[5,1]=Z[5,1]+Z[7,1];Z.simplify_full() 
       
Z[2,1]=0; Z[3,1]=0;Z[6,1]=0;Z[7,1]=0;Z.simplify_full() 
       

We now use that $Z[2,4]=Z[3,4]=Z[6,4]=Z[7,4]=0$. 

Z[0,4]=Z[0,4]-Z[2,4];Z[1,4]=Z[1,4]-Z[3,4];Z[4,4]=Z[4,4]+Z[6,4];Z[5,4]=Z[5,4]+Z[7,4];Z.simplify_full(); 
       
Z[2,4]=0;Z[3,4]=0;Z[6,4]=0;Z[7,4]=0;Z.simplify_full() 
       

We now use that $Z[2,5]=Z[3,5]=Z[6,5]=Z[7,5]=0$. 

Z[0,5]=Z[0,5]-Z[2,5];Z[1,5]=Z[1,5]-Z[3,5];Z[4,5]=Z[4,5]+Z[6,5];Z[5,5]=Z[5,5]+Z[7,5];Z.simplify_full(); 
       
Z[2,5]=0;Z[3,5]=0;Z[6,5]=0;Z[7,5]=0;Z.simplify_full() 
       

Recalling this matrix is equal to our $HK$ matrix above, this gives 16 equations in the 16 unknowns $H_{00}, \dots, H_{33}$. In terms of how we defined the $H_{ij}$ earlier, we just need to show they are integers to have the result.

We can actually set this up as  four separate matrix equations. The first one we want to solve is $\begin{pmatrix} k_{00} & p k_{01} & k_{02} & p k_{03} \\ k_{10}& k_{11} & k_{12} & k_{13} \\ k_{20} & p k_{21} & k_{22} & p k_{23} \\k_{30} & k _{31} & k_{32} & k_{33} \end{pmatrix}\begin{pmatrix} H_{00} \\ H_{10} \\ H_{20} \\ H_{30} \end{pmatrix} = \begin{pmatrix} h_{00} \\h_{10} \\h_{20} \\ h_{30} \end{pmatrix}$.  Note the $h_{ij}$ and $k_{ij}$ are integral by how they were defined, so if we can show that the matrix with the $k_{ij}$ has determinant  $\pm 1$, we have the result for $H_{00}, H_{10},H_{20}, H_{30}$.

A=matrix([[k00,p*k01,k02,p*k03],[k10,k11,k12,k13],[k20,p*k21,k22,p*k23],[k30,k31,k32,k33]]);A 
       
K=block_matrix(2,2,[k1,k2,k3,4]);K 
       
(K.determinant()- A.determinant()).simplify_full() 
       

As the matrix $K \in \Gamma_{2}[\varpi]$, we know $\det(K) = \pm 1$.  This gives the desired result for $H_{00}, H_{10}, H_{20} H_{30}$.  Now we must check the rest by the same method.

A=matrix([[k00,k01,k02,k03],[p*k10,k11,p*k12,k13],[k20,k21,k22,k23],[p*k30,k31,p*k32,k33]]);A 
       
(K.determinant()- A.determinant()).simplify_full() 
       

This gives that $H_{01}, H_{11}, H_{21},$ and $H_{31}$ are integral as desired. 

One obtains the same matrix of coefficients for $H_{02}, H_{12}, H_{22}, H_{32}$ as $H_{00}, H_{10}, H_{20}, H_{30}$ so this case is already done. The matrix of coefficients for $H_{03}, H_{13}, H_{23}, H_{33}$ is the same as that of $H_{01}, H_{11}, H_{21}, H_{31}$ so this case is done as well.  Combining all of this, we get the version of Lemma 5.1 for our paramodular section.