I am looking for an explanation of why Triangle of Mahonian numbers T(n,k) form the rank of the vector space $H^k(GL\_n/B)$? 
With respect to the [property of Kendall-Mann numbers](https://mathoverflow.net/questions/46368/the-property-of-kendall-mann-numbers) where the statement appeared I wonder if there are any similar assymtotics in the area of Galois numbers?
Any explanations are highly welcomed.



Why Triangle of Mahonian numbers T(n,k) forms the rank of the vector space?
More generally, for any sequence $0 < k\_1 < k\_2 < \cdots < k\_r < n$ of positive integers, let $F(k\_1, k\_2, \ldots, k\_r; n)$ be the set of flags $V\_1 \subset V\_2 \subset \cdots V\_r \subset \mathbb{C}^n$ with $\dim V\_j = k\_j$. We have a map $F(k\_1, k\_2, \ldots, k\_{r-1}, k\_r; n) \to F(k\_1, k\_2, \ldots, k\_{r-1}; n)$ that forgets the subspace $V\_r$. Note that, if $k\_r = k\_{r-1}+1$, then this map is a $\mathbb{CP}^{n-k\_r}$ bundle, because specifying a $k$-plane containing a given $k-1$ plane $V \subset \mathbb{C}^n$ is the same as giving a line in $\mathbb{C}^n/V$. 


If $Y \to X$ is a $\mathbb{P}^j$ bundle, then the Poincare polynomials of $X$ and $Y$ are related by $\sum \dim H^k(Y) q^k = \left( \sum \dim H^k(X) q^k \right) (1+q^2+q^4+ \cdots + q^{2j})$. Consider the sequence of maps 
$$F(1,2,\ldots,n-2,n-1; n) \to F(1,2,\ldots,n-2; n) \to \cdots \to F(1,2; n) \to F(1;n) \to F(\emptyset;n)$$
and note that $F(\emptyset; n)$ is a single point.
We get that the Poincare polynomial of $Fl\_n = F(1,2,3,\cdots,n-1;n)$ is 
$$(1+q^2)(1+q^2+q^4) \cdots (1+q^2+\cdots +q^{2n-4}) (1+q^2+\cdots + q^{2n-4}+q^{2n-2}),$$
the well known generating function for inversions.


