Why Triangle of Mahonian numbers T(n,k) forms the rank of the vector space?

Question

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 where the statement appeared I wonder if there are any similar assymtotics in the area of Galois numbers? Any explanations are highly welcomed.

Answer 1

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.