I am reading the lecture notes. On page 3, formulas (1.7), (1.8), (1.9), let $P$ be the standard Borel subgroup, we have $P=MN$. Why $M, N$ must be (1.8), (1.9)?
I know that $P$ should be a upper triangular matrix and we decompose $P=MN$ such that $M$ is a diagonal matrix, $N$ is a unipotent upper triangular matrix. But why $M, N$ must be (1.8), (1.9)? Thank you very much.
Edit: $$ M=\left\{ \left( \begin{matrix} t & 0 & 0 \\ 0 & s & 0 \\ 0 & 0 & \bar{t}^{-1} \end{matrix} \right) \mid t\in E^{\times}, s \in E^{1} \right\}, \\ N=\left\{ \left( \begin{matrix} 1 & p & z \\ 0 & 1 & -\bar{p} \\ 0 & 0 & 1 \end{matrix} \right) \mid z + \bar{z} = -p \bar{p} \right\}, $$ where $E^1=\ker N_{E/F}$, $E/F$ is a quadratic extension.
Try multiplying out matrices and you'll see that it works. If the question is actually "ok, $P=MN$, but why are these the Levi factor and unipotent radical", then heres a way of seeing it, admittedly by cheating slightly. Because unitary groups are classical, they come equipped with a standard embedding into $GL(n)$ (in fact, that's how the notes define them, $n=3$ here). Then your Borel $P$ is the intersection of the Borel in $GL(n)$ consisting of upper triangular matrices with $U(2,1)$. Similarly, its Levi factor and unipotent radical are obtained by intersecting with $U(2,1)$. It shouldn't be hard to see that this give the desired groups.