This question was posted at [MSE](https://math.stackexchange.com/questions/3764904/containment-of-bruhat-cells-on-flag-variety), but it did not receive any answer there.


Let $G$ be a connected semisimple algebraic group over $\mathbb C$, $X$ the flag variety of $G$, $B\_0$ a Borel subgroup, $\mathbb O$ a $B\_0$-orbit on $X$.


**Question**: Can we always find a Borel subgroup $B\_x$ so that the open $B\_x$-orbit $\mathbb O'$ contains $\mathbb O$?


The obvious attempt would be to choose a point $\bar x \in \mathbb O$, and then take $B\_x$ to be a Borel in opposite relative position to $B\_{\bar x}$. But not all choices work. For example when $\mathbb O$ is the open orbit, choices are pretty restricted, and I'm failing to find a construction that works in general.


Any help would be appreciated.



Containment of Bruhat cells on flag variety
I'd say that the relevant fact here is as follows. For two Borels $B\_1$ and $B\_2$ with a common maximal torus $T$ let $x\_1$ be the unique $T$-fixed point in the open $B\_1$-orbit. Then the $B\_2$-orbit $B\_2x\_1$ lies in the open $B\_1$-orbit. For instance, you can see this by proving that
(a) for every $x\in B\_2x\_1$ the orbit closure $\overline{Tx}$ contains $x\_1$ and
(b) the open $B\_1$-orbit consists precisely of those points $x$ for which $x\_1\in\overline{Tx}$.
(These are fairly basic properties of Schubert decompositions.)


In your setting this means that you can choose any maximal torus $T\subset B\_0$, let $x'$ be the unique $T$-fixed point in $\mathbb O$ and consider the unique Borel $B\supset T$ for which the orbit $Bx'$ is open. This $B$ will be your $B\_x$. In fact, this $B$ is the opposite (with respect to $T$) Borel of the stabilizer of $x'$. So in terms of your last paragraph you can choose any $\bar x$ but you have to choose the opposite Borel correctly: it must intersect $B\_0$ in a maximal torus that fixes $\bar x$.


