Let $G:=GL_2(\Bbb C)$, $B$ and $T$ be the subgroup consisting of all upper triangular and diagonal matrices in $G$, respectively. Set $w:= \left( \begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)$ and $w'= \left( \begin{array}{cc} 1 & 1\\ 1 & 0 \end{array}\right)$.
I want to prove the following:
$(1)$ $G=B\cup BwB, \quad B\cap BwB= \emptyset ,\quad BwB=TwB\cup Tw'B,\quad TwB\cap Tw'B=\emptyset$.
I can prove all but except the third, $BwB=TwB\cup Tw'B$.
$(2)$ Suppose that $G$ acts on a set $M:=G/B\times G/B\times G/B$ as follows: $$ G\times M \ni (g,\gamma) \longmapsto g\cdot \gamma \in M,$$ where $g\cdot \gamma = ((gx)B,(gy)B,(gz)B),\gamma =(xB,yB,zB), (x,y,z\in G)$. Determine $G$-orbit decomposition w.r.t the above action.
I cannot see why this is related to $(1)$.
Any help would be much appreciated.
The action of $G$ by multiplication on the the left cosets of $B$ is equivalent to the projective action of $G$ on the $1$-dimensional subspaces of ${\mathbb C}^2$, since $B$ is precisely the stabilizer of $\langle (1\ 0)^{\rm T} \rangle$.
Now $w$ maps $\langle (1\ 0)^{\rm T} \rangle$ to $\langle (0\ 1)^{\rm T} \rangle$, which is stabilized by $T$, so $TwB=wB$.
And $w_1$ maps $\langle (1\ 0)^{\rm T} \rangle$ to $\langle (1\ 1)^{\rm T} \rangle$, which can be mapped by elements of $T$ to any other $1$-dimensional subspace $\langle (a\ b)^{\rm T} \rangle$ of ${\mathbb C}^2$.
So we have the third equality $BwB = TwB \cup Tw_1B$.