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 $\mathbb{F}_p$.
(x, y, z) (x, y, 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 $P$:
(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 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:
[ 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 $y,z \in \mathbb{F}_p$.
[ 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] 

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 $x,y,z \in \mathbb{F}_p$.


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 writeup. In particular, we recall there are $(p1)(p^3p)$ 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 $(p1)$ choices for each of them. Thus, there are $p(p1)^2$ choices for the elements we don't want, i.e., there are $(p1)(p^3  p)  p(p1)^2 = p^2(p1)^2$ 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 $p^2(p1)^2$ choices for $D$ that work, which give $(p1)$ choices for $A$ and $p^3$ choices for $S$. This gives a total of $p^5 (p1)^3$ choices for elements in $P$ that work. There are $p^3$ choices for $x,y,z$, so there are $p^8 (p1)^3$ choices of elements in $G$ that satisfy $det C \neq 0$ and $c_{21} \neq 0$.
