# Case2i3i

## 3136 days ago by jimlb

Here we give calculations for computing the volume of Case (2i,3i). Recall the coset representatives for $P\backslash GSp_4(\mathbb{F}_p)$.

Of course, we have the identity matrix as our first representative.  The next set were given by the following matrix as $z$ runs over $\mathbb{F}_p$.

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

We now introduce a general element in $P$:

var('a11,a12,a21,a22,b11,b12,b21,b22,d11,d12,d21,d22');
 (a11, a12, a21, a22, b11, b12, b21, b22, d11, d12, d21, d22) (a11, a12, a21, a22, b11, b12, b21, b22, d11, d12, d21, d22)
g=Matrix(SR,[[a11,a12,b11,b12],[a21,a22,b21,b22],[0,0,d11,d12],[0,0,d21,d22]]);g
 [a11 a12 b11 b12] [a21 a22 b21 b22] [ 0 0 d11 d12] [ 0 0 d21 d22] [a11 a12 b11 b12] [a21 a22 b21 b22] [ 0 0 d11 d12] [ 0 0 d21 d22]

Thus, a typical element in $PZ$ looks like:

g*Z;
 [ a11 -b12 b11 b12*z + a12] [ a21 -b22 b21 b22*z + a22] [ 0 -d12 d11 d12*z] [ 0 -d22 d21 d22*z] [ a11 -b12 b11 b12*z + a12] [ a21 -b22 b21 b22*z + a22] [ 0 -d12 d11 d12*z] [ 0 -d22 d21 d22*z]

Clearly, nothing from these cosets contribute to this case.  We move on to the next set of cosets, which run over $y,z \in \mathbb{F}_p$.

Y = Matrix(SR,[[0,1,0,0],[0,0,1,0],[0,0,y,1],[-1,y,z,0]]);Y
 [ 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]
g*Y;
 $\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} -b_{12} & b_{12} y + a_{11} & b_{11} y + b_{12} z + a_{12} & b_{11} \\ -b_{22} & b_{22} y + a_{21} & b_{21} y + b_{22} z + a_{22} & b_{21} \\ -d_{12} & d_{12} y & d_{11} y + d_{12} z & d_{11} \\ -d_{22} & d_{22} y & d_{21} y + d_{22} z & d_{21} \end{array}\right)$

Note that if we consider the bottom right matrix, this has determinant 0 still so these cosets are not in our case either.

The last set of cosets are the only ones that can contribute to our case.  For this, we let $x,y,z \in \mathbb{F}_p$.

X= Matrix(SR,[[0,0,0,1],[0,0,1,0],[0,-1,y,x],[-1,0,z,y]]);X
 $\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & y & x \\ -1 & 0 & z & y \end{array}\right)$
g*X
 $\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr} -b_{12} & -b_{11} & b_{11} y + b_{12} z + a_{12} & b_{11} x + b_{12} y + a_{11} \\ -b_{22} & -b_{21} & b_{21} y + b_{22} z + a_{22} & b_{21} x + b_{22} y + a_{21} \\ -d_{12} & -d_{11} & d_{11} y + d_{12} z & d_{11} x + d_{12} y \\ -d_{22} & -d_{21} & d_{21} y + d_{22} z & d_{21} x + d_{22} y \end{array}\right)$

Here we have that $D$ is invertible, so the determinant of $D(g*X)$ is nonzero.  However, since we are interested only in those elements where the $(2,1)$ entry is not 0, we want to count those $g$ so that $d_{22} \neq 0$.  However, this has already been calculated in the earlier volume write-up.  In particular, we recall there are $(p-1)(p^3-p)$ elements in $GL_2(\mathbb{F}_p)$.  If we assume $d_{22} = 0$, then there are $p$ choices for $d_{11}$.  We then have to choose $d_{12}$ and $d_{21}$ both not equal to 0, so there are $(p-1)$ choices for each of them.  Thus, there are $p(p-1)^2$ choices for the elements we don't want, i.e., there are $(p-1)(p^3 - p) - p(p-1)^2 = p^2(p-1)^2$ choices for our $D$.  Recall elements in the Siegel parabolic can be written as $\begin{pmatrix} A & 0 \\ 0 & \alpha\, ^{t}A^{-1} \end{pmatrix} \begin{pmatrix} 1 & S \\ 0 & 1 \end{pmatrix}$ with $S$ a symmetric 2 by 2 matrix.  The choice of $D$ fixes $A$ up to nonzero scalar and $S$ is independent. Thus, there are $p^2(p-1)^2$ choices for $D$ that work, which give $(p-1)$ choices for $A$ and $p^3$ choices for $S$.  This gives a total of $p^5 (p-1)^3$ choices for elements in $P$ that work. There are $p^3$ choices for $x,y,z$, so there are $p^8 (p-1)^3$ choices of elements in $G$ that satisfy $det C \neq 0$ and $c_{21} \neq 0$.