This question is related my question in An example of a Deligne–Lusztig variety for a general linear group, and the obtained answer.
Situation
I am working with the general linear group. Specifically, take $G_0=\text{GL}_{n,\mathbb{F}_q}$, and $G=G_0\times_{\mathbb{F}_q}\overline{\mathbb{F}_q}$ the base change. Let $X$ be the set of Borel subgroups of $G$. The Weyl group $W$ of $G$ is isomorphic to the symmetric group $S_3$, and we have a bijection between $W$ and the set of $G$-orbits on $X\times X$. Fix a Borel subset $B^+\in X$. In this case, take the upper triangular matrices. For $w\in W$, define $\mathcal{O}(w)$ to be the orbit of $(B^+,\dot wB^+\dot w^{-1})$ in $X\times X$. We say that $B_1,B_2\in X$ are in relative position $w$ if $(B_1,B_2)\in\mathcal{O}(w)$, and we write $B_1\xrightarrow{w}B_2$.
One can show that $X$ is isomorphic to the full flag variety $\mathcal{F}_n$, and we say that two vlags $V$ and $V'$ are in relative position $w$ if $$\dim V_i\cap U_j=\#(\{ 1,\dotsc,i \}\cap \{ w(1),\dots,w(j) \})$$ for any $i,j\in\{1,\dotsc,n\}$, and we write $V\xrightarrow{w}V'$.
My question
I am trying to connect the two notions of "relative position $w$". Take for instance $G=\text{GL}_{3}$ and $w=(1\ 2)\in S_3\cong W$.
Suppose that a flag $(V\colon V_1\subset V_2)\in \mathcal{F}_3$ is in relative position $w$ with the standard flag $V^+$, i.e. $V^+\xrightarrow{w}V$. By the defintion above, this means that $V_1\neq \langle e_1\rangle $ and $V_2=\langle e_1,e_2\rangle$. We know that $V^+$ corresponds to $B^+\in X$, and let $V$ correspond to some $B\in X$.
To show that $B^+\xrightarrow{w}B$, I need to show $B=\dot wB^+\dot w^{-1}$. Am I correct, or did I misunderstand something? I do not see a reason why $\dot w e_1=e_2$ should generate $V_1$, for example. A priori, we just know that $B=gB^+g^{-1}$ for some $g\in G$ and that $B$ fixes the vlag $V$.
Any help is greatly appreciated!
