May Schubert cell intersection with opposite big cell polynomial count? Let $SL(n)$ be algebraic group defined over finite field $\mathbb{F}_{p^n}$, $B$ be Borel subgroup consist of upper triangular matrices and $T$ be maximal torus consist of diagonal matrices. Let $W$ be Weyl group, defined by $N_{G}T/T$. Let $G/B=\cup_{w\in W} B\cdot e_{w}$ be the Bruhat decomposition, where $e_{w}=wB$.  Denote $X:=B\cdot e_{w}$ be Schubert cell and $O^{-}$ be opposite big cell i.e.$B^{-}\cdot e_{id}$, where $B^{-}$ is the opposite Borel.
My question is how to calculate the number of $\mathbb{F}_{p^n}$-points of $Y:=X\cap O^{-}$. Is this one polynomial count?(i.e. there exists a polynomial $f$ with $\mathbb{Z}$-coefficients not depend on $n$, such that the number of $\mathbb{F}_{p^n}$-points is $f(p^n)$)
 A: Yes, this is polynomial point count. An intersection of a Bruhat cell and an opposite Bruhat cell is called a "Richardson variety", and Richardson varieties come with decompositions known as Deodhar decompositions. Each piece of the Deodhar decomposition is of the form $\mathbb{G}_m^{N-2k} \times \mathbb{A}^k$ where $N$ is the dimension of the Richardson, so it has $(q-1)^{N-2k} q^k$ points over $\mathbb{F}_q$, and adding them all up gives a polynomial in $q$.
To give the first nontrivial example, in $\text{GL}_3$, the big Schubert cell is flags which can be written as
$$\begin{bmatrix} 1&0&0 \\ x&1&0 \\ z&y&1 \\ \end{bmatrix} B.$$
The intersection with the opposite cell is the open locus $z(xy-z) \neq 0$.
We can decompose this as the union of the pieces where $x\neq 0$ and where $x = 0$, so we get
$$\left\{ \begin{bmatrix} 1&0&0 \\ x&1&0 \\ z&\tfrac{w+z}{x}&1 \\ \end{bmatrix} B : wxz \neq 0 \right\} \sqcup \left\{ \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ z&y&1 \\ \end{bmatrix} B : z \neq 0 \right\},$$
so we have
$$(q-1)^3 + (q-1)q$$
points.
The Deodhar stratification of $(B_+ w B_+ \cap B_- u B_+)/B_+$ depends on a choice of reduced word for $w$. Deodhar's original paper is
Deodhar, Vinay V., On some geometric aspects of Bruhat orderings. I: A finer decomposition of Bruhat cells, Invent. Math. 79, 499-511 (1985). ZBL0563.14023.
For very explicit formulas, you might also like
Marsh, R. J.; Rietsch, K., Parametrizations of flag varieties, Represent. Theory 8, 212-242 (2004). ZBL1053.14057.
In your particular case, you want $w = w_0$ and $u = e$. If you take the reduced word $(s_1 s_2 \cdots s_{n-1}) (s_1 s_2 \cdots s_{n-2}) \cdots (s_1 s_2) (s_1)$, Allen Knutson pointed out to me that the Deodhar stratification has a very explicit form -- it corresponds to taking a lower triangular matrix as above and specifying the ranks of the left-justified submatrices in consecutive rows. (In the example above, the two strata depend on whether the submatrix in row $2$ and column $1$ has rank $1$ or $0$.) I don't know whether Allen has written up this fact.
