I am reading notes of Michel Brion on spherical varieties.


Consider a reductive group $G$, a Borel $B$ in $G$, a finite dimensional $G$-module $M$ and a closed orbit $Y$ of $G$ in $\mathbb{P}(M)$. The stabilisator of every point of $Y$ is a parabolic subgroup of $G$, so we can fix a $y \in Y$ such that $By$ is open in $Y$ (by Bruhat-decomposition). Also choose $m \in M$ such that $y=[m]$.


My Question is first of all: why is there an $\eta \in (M^\*)^{(B)}$ such that $\eta(m)=1$. And why are the parabolic groups $G\_{[\eta]}$ and $G\_y$ opposed?



Why are these parabolic subgroups opposed?
Let's start with $x \in Y$ a $B$-fixed vector. Then it's a high weight line in $M$, so the linear span $N \leq M$ of $Y$ is an irreducible subrepresentation of $M$. 


Do you mind if I assume we're in characteristic $0$, as Brion does? Then we can split $M = N \oplus N'$, and $M^\* = N^\* \oplus N'^\*$.` 
(Which is to say, you can reduce to the case $M$ irreducible, if that helps understand the situation.) So it's sufficient (and in the general case, necessary!) to look for your $\eta$ inside $N^\*$.


Your first question is then why the $\eta \in (N^\*)^B$, unique up to scale, has $\eta(m) \neq 0$. If it were zero,
$$ 0 = \eta(m) = (b\cdot \eta)(m) = \eta(b^{-1}\cdot m) \qquad \forall b\in B $$
from which we learn that $\eta$ annihilates a spanning set of $N$, so all of $N$. But $\eta \in N^\*$, ` so it's zero, contradiction.


I'm not used to describing parabolic subgroups as "opposed" but I assume it means that $G\_{[\eta]} G\_y$ is dense in $G$. By definition of $\eta$, $G\_{[\eta]} \geq B$, so it's enough to show $B G\_y$ is dense in $G$, or that $B G\_y / G\_y$ is dense in $G/G\_y$. This was exactly the condition used to pick $y$, that $B y$ was dense in $Y \cong G/G\_y$.


