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 $f(p^n)$)



May Schubert cell intersection with opposite big cell polynomial count?
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**](http://dx.doi.org/10.1007/BF01388520), Invent. Math. 79, 499-511 (1985). [ZBL0563.14023](https://zbmath.org/?q=an:0563.14023).


For very explicit formulas, you might also like


*Marsh, R. J.; Rietsch, K.*, [**Parametrizations of flag varieties**](http://dx.doi.org/10.1090/S1088-4165-04-00230-4), Represent. Theory 8, 212-242 (2004). [ZBL1053.14057](https://zbmath.org/?q=an:1053.14057).


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.


