Relative position of flags and the Robinson-Schensted correspondence

Question

This question is related my question in An example of a Deligne–Lusztig variety for a general linear group, and the obtained answer.

I am currently reading Steinberg, Robert, An occurrence of the Robinson-Schensted correspondence, J. Algebra 113, No. 2, 523-528 (1988). ZBL0653.20039., but there are some parts that I do not understand.

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 $B^+\in X$. 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 $B_1\xrightarrow{w}B_2$.

My Question

I assume that two Borel subgroups $B_1$ and $B_2$ give two tableaux $T_1$ and $T_2$ respectively. The mentioned article associates to the pair $(T_1,T_2)$ a permutation $w(T_1,T_2)\in W$. What I want to say is that $B_1\xrightarrow{w}B_2$ if and only if $w=w(T_1,T_2)$. Is this true? My confusion is probably coming from the unipotent transformation $u$ mentioned in the article. How can a generic element of an irreducible component of $\mathcal F_u$ (the variety of flags fixed by $u$) be identified by a Borel subgroup $B\in X$?

Any help is appreciated.

Answer 1

Your initial description is missing the choice of an unipotent$~u$, which is essential to having any correspondence between Borel subgroups and tableaux. Fixing$~u$ and denoting $\def\Fu{\mathcal{F}_u}\Fu$ the set of fixed points of the action of $u$ on $X$, one can associate a Young tableau (whose shape is the Jordan type of$~u$) to any $B\in\Fu$. This can be used to bijectively map the set of irreducible components of $\Fu$ to the the set of standard Young tableaux of that shape, but the set of Borels that map to a given tableau en general form only dense subset of the irreducible component corresponding to that tableau. (Given that irreducible components are not connected components, this is unavoidable; in fact $\Fu$ is connected.) So the initial map only gives the tableau for an irreducible component for a Borel chosen sufficiently generically in the component (being in the complement of other components suffices).

Similarly, the relative position of pairs of Borels in a given pair of irreducible components is only equal to the Weyl group element associated to that pair of components for sufficiently generic pairs. Here being in the complement of other irreducible components does not suffice: clearly for the generic relative position of a component to itself, one should avoid choosing the second Borel equal to the first (if one can), no matter how generic the first one is.