Equivalent form for the Bruhat decomposition Let $G$ be a reductive group and $B$ a Borel subgroup. The Bruhat decomposition allows us to write (where $W$ is the Weyl group):
$$ G/B = \coprod_{w\in W} BwB$$
Why is this form the same as looking at the $G$-orbit decomposition ($G$ acting diagonally):
$$ (G/B)^2 = \coprod_{w\in W} G\cdot (eB,wB)$$
I believe this generalizes. If $P_1,...,P_n$ are parabolic subgroups, then looking at the orbits of the diagonal $G$ action on $G/P_1\times ... \times G/P_n$ is the same as the orbit structure of the $P_1$ action on $G/P_2 \times ... \times G/P_n$.
 A: This is just an instance of the following general fact. Suppose a group $G$ acts transitively on two sets $X,Y$. Choose any $x_0\in X$ and $y_0\in Y$, and put $\def\Stab{\operatorname{Stab}_G}H_x=\Stab(x_0)$ and $H_y=\Stab(y_0)$. Then there is a bijection between on one hand the orbits of $G$ acting componentwise on $X\times Y$, and on the other hand the set of double cosets $H_x\backslash G/H_y$: the orbit of $(x,y)$ corresponds to the double coset $H_xg_1^{-1}g_2H_y$ where $g_1,g_2$ are any group elements with $g_1\cdot x_0=x$ and $g_2\cdot y_0=y$. Since the elements $g_1,g_2$ are determined up to right multiplication by $H_x$ respectively by $H_y$, the double coset associated to $(x,y)$ is well defined, and it does not change if $(x,y)$ is replaced by another point $(g\cdot x,g\cdot y)$ in its orbit. Moreover
any double coset $H_xgH_y$ corresponds to some orbit, namely that of the point $(x_0,g\cdot y_0)$, so the correspondence is surjective, and its injectivity is also easily checked.
Now in the case of the question one can take $X=Y$ the set $\def\B{\mathcal B}\B$ of Borel subgroups; since these Borel subgroups are their own normalisers, $\B$ is in bijection with $G/B$ once the particular Borel subgroup $B$ is chosen (via $gBg^{-1}\mapsto gB$), and one chooses $x_0=y_0=B$ as base point, for which $H_x=H_y=B$. Now the Bruhat decomposition says that $W$ provides a set of double coset representatives for $B\backslash G/B$, and by the general correspondence I indicated this also means that $\{\,(x_0,w\cdot y_0)\mid w\in W\,\}$ is a set of representatives for the diagonal $B$-orbits on $\B\times\B$. The element $(x_0,w\cdot y_0)$ is actually $(B,wBw^{-1})$ in $\B\times\B$, or under the map $\B\to G/B$ it is $(eB,wB)$ in $(G/B)^2$.
A: So, suppose 
$$G/B = \coprod_{w \in W} B w B.$$ 
Then, pick $(gB, hB) \in (G/B)^{2}.$ We can find $w$ such that $g^{-1}h \in BwB$. Say $g^{-1}hB = bwB$. Then, $b^{-1}g^{-1}h \in wB$ and hence,
$$(gB, hB) = gb(eB, b^{-1}g^{-1}hB) = gb(eB, wB).$$
Hence, $(G/B)^{2} \subseteq \coprod_{w \in W} G\cdot (eB, wB)$, and the reverse inclusion is obvious.
Now, suppose 
$$(G/B)^{2} \subseteq \coprod_{w \in W} G \cdot (eB, wB).$$
Pick $gB \in G/B$. Then, for some $w \in W$, and $g \in G$,
$$(eB, gB) = (g'B, g'wB)$$
and hence $g' \in B$ and hence $gB \in BwB.$ Hence, $G/B \subseteq \coprod_{w \in W} BwB$ and the reverse inclusion is trivial.
