Question about Bruhat decomposition for $SL_2$.

Question

I am reading the lecture notes. On page 11, let $G=SL_2$. It is said that $G = NwP \cup P = PwN \cup PwNw^{−1}$, where $P= \left\{ \left( \begin{matrix} a & x \\ 0 & 1/a \end{matrix} \right) \right\}$, $N= \left\{ \left( \begin{matrix} 1 & x \\ 0 & 1 \end{matrix} \right) \right\}$, $w= \left( \begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right) $. It is easy to see that $G = NwP \cup P$. I am trying to show that $NwP \cup P = PwN \cup PwNw^{−1}$. It seems that $PwN = NwP$ but $P \neq PwNw^{−1}$. How to show that $NwP \cup P = PwN \cup PwNw^{−1}$? Thank you very much.

Answer 1

Since $G$ is stable under inversion (it's a group), we have

$$\begin{eqnarray*} G &=& G^{-1} \\ &=& (NwP)^{-1} \cup P^{-1} \\ &=& PwN \cup P \end{eqnarray*} $$ Where $P = P^{-1}$ also because $P$ is a group, and similarly

$$(NwP)^{-1} = P w^{-1} N = P (-w) N = P w N$$ because $-1 \in P$. Now $P \subset P w N w^{-1}$ so $G = PwN \cup P \subset PwN \cup P w N w^{-1} \subset G$, which yields

$$G = PwN \cup P w N w^{-1}$$