I am reading the lecture notes and trying to understand a Levi decomposition of a parabolic subgroup.
I am trying to understand how to compute $N_{\beta}$ on page 12.
Let $G = Sp_4(K)$, where $K$ is a algebraically closed field. Take \begin{align} T_{\beta} = \{ diag(a,1,1,a^{-1}) : a \in K^{\times} \}. \end{align} By direct computation, we have \begin{align} M_{\beta} := Z_G(T_{\beta}) = \left\{ \left( \begin{matrix} d & 0 & 0 & 0 \\ 0 & c_{11} & c_{12} & 0 \\ 0 & c_{21} & c_{22} & 0 \\ 0 & 0 & 0 & d^{-1} \end{matrix} \right) : d \in K^{\times}, c_{11} c_{22} - c_{12} c_{21} = 1 \right\}. \end{align} We have $P_{\beta} = M_{\beta} \ltimes N_{\beta}$. How to show that \begin{align} N_{\beta} = \left\{ \left( \begin{matrix} 1 & x & y & z \\ 0 & 1 & 0 & y \\ 0 & 0 & 1 & -x \\ 0 & 0 & 0 & 1 \end{matrix} \right) : x,y,z \in K \right\}? \end{align}
Let \begin{align} u = \left(\begin{array}{cccc} 1 & z_{1,2} & z_{1,3} & z_{1,4}\\ 0 & 1 & z_{2,3} & z_{2,4}\\ 0 & 0 & 1 & z_{3,4}\\ 0 & 0 & 0 & 1 \end{array}\right) \in N_{\beta}. \end{align} Let $J = \left(\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{array}\right)$. By $u^T J u = J$, we have \begin{align} \left(\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & z_{1,2} + z_{3,4}\\ 0 & 0 & 0 & z_{2,3}\, z_{3,4} + z_{1,3} - z_{2,4}\\ 0 & - z_{1,2} - z_{3,4} & z_{2,4} - z_{1,3} - z_{2,3}\, z_{3,4} & 0 \end{array}\right) = 0. \end{align} Therefore $z_{34}=-z_{12}$. If we know that $z_{23}=0$, then $z_{24}=z_{13}$ and hence $u$ has the form $\left( \begin{matrix} 1 & x & y & z \\ 0 & 1 & 0 & y \\ 0 & 0 & 1 & -x \\ 0 & 0 & 0 & 1 \end{matrix} \right)$. But how to show that $z_{23}=0$?
Thank you very much.
Let $b_1 = \left(\begin{array}{cccc} x_{1,1} & x_{1,2} & x_{1,3} & x_{1,4}\\ 0 & x_{2,2} & x_{2,3} & x_{2,4}\\ 0 & 0 & x_{3,3} & x_{3,4}\\ 0 & 0 & 0 & x_{4,4} \end{array}\right) \in Sp_4$ and $b_2 = \left(\begin{array}{cccc} y_{1,1} & y_{1,2} & y_{1,3} & y_{1,4}\\ 0 & y_{2,2} & y_{2,3} & y_{2,4}\\ 0 & 0 & y_{3,3} & y_{3,4}\\ 0 & 0 & 0 & y_{4,4} \end{array}\right) \in Sp_4$. By $b_1^TJb_1=J$ and $b_2^TJb_2=J$, we have \begin{align} x_{11}x_{44}=1, x_{22}x_{33}=1, y_{11}y_{44}=1, y_{22}y_{33}=1. \end{align}
We have \begin{align} w_{\beta} = \left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 0 & 1 \end{array}\right), \end{align} and \begin{align} b_1 w_{\beta} b_2 = \left(\begin{array}{cccc} x_{1,1}\, y_{1,1} & x_{1,1}\, y_{1,2} - x_{1,3}\, y_{2,2} & x_{1,1}\, y_{1,3} + x_{1,2}\, y_{3,3} - x_{1,3}\, y_{2,3} & x_{1,1}\, y_{1,4} + x_{1,2}\, y_{3,4} - x_{1,3}\, y_{2,4} + x_{1,4}\, y_{4,4}\\ 0 & - x_{2,3}\, y_{2,2} & x_{2,2}\, y_{3,3} - x_{2,3}\, y_{2,3} & x_{2,2}\, y_{3,4} - x_{2,3}\, y_{2,4} + x_{2,4}\, y_{4,4}\\ 0 & - x_{3,3}\, y_{2,2} & - x_{3,3}\, y_{2,3} & x_{3,4}\, y_{4,4} - x_{3,3}\, y_{2,4}\\ 0 & 0 & 0 & x_{4,4}\, y_{4,4} \end{array}\right), \end{align} Let \begin{align} m = \left(\begin{array}{cccc} d & 0 & 0 & 0\\ 0 & c_{1,1} & c_{1,2} & 0\\ 0 & c_{2,1} & \frac{c_{1,2}\, c_{2,1} + 1}{c_{1,1}} & 0\\ 0 & 0 & 0 & \frac{1}{d} \end{array}\right) \in M_{\beta}. \end{align} Then \begin{align} m u = \left(\begin{array}{cccc} d & d\, z_{1,2} & d\, z_{1,3} & d\, z_{1,4}\\ 0 & c_{1,1} & c_{1,1}\, z_{2,3} + c_{1,2} & c_{1,1}\, z_{2,4} + c_{1,2}\, z_{3,4}\\ 0 & c_{2,1} & \frac{\left(c_{1,2}\, c_{2,1} + 1\right)}{c_{1,1}} + c_{2,1}\, z_{2,3} & c_{2,1}\, z_{2,4} + \frac{\left(c_{1,2}\, c_{2,1} + 1\right)\, z_{3,4}}{c_{1,1}}\\ 0 & 0 & 0 & \frac{1}{d} \end{array}\right). \end{align}
Since $P_\beta = Bw_{\beta}B = M_{\beta} N_{\beta}$ and $\det( submatrix(b_1 w_{\beta} b_2, [2,3], [2,3] )) = 1$, we must have $\det( submatrix(mu, [2,3], [2,3] )) = 1$. Hence $z_{23}=0$.