Bruhat decomposition of a Lie group

Remarks

The coset $P \tilde w P$ does not depend on $\tilde w$ because two representatives differ by an element of $M \subseteq P$. A standard proof follows the algebraic case, with the minimal parabolic subgroup in place of a Borel subgroup. Disjointness uses that $W$ acts simply transitively on the positive systems of restricted roots.

Example: $G = \mathrm{SL}_2(\R)$, where $P$ is the group of upper triangular matrices and $W = \set{1, s}$ with $s$ represented by $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. A matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ lies in $P$ when $c = 0$ and in $PsP$ when $c \ne 0$. So $\mathrm{SL}_2(\R) = P \sqcup PsP$.

The Bruhat decomposition for reductive algebraic groups over an algebraically closed field is the analogue with a Borel subgroup in place of $P$. For a real group, $W$ is the Weyl group of the restricted root system, which can be smaller than the Weyl group of the complexified group.

Going deeper

For a student

Passing to the flag manifold $G/P$, the decomposition says that $P$ has finitely many orbits on $G/P$, one for each $w \in W$. The orbit of the longest element of $W$ is open and dense, and its complement has lower dimension. This open orbit is used to realize principal series representations on functions on $N^- = \theta(N)$.