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

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.

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.