Consider the full flag variety $F\_n$ consisting of full flags in $\mathbb C^n$. There is a collection of tautological bundles on $F\_n$: 


$0=U\_0\subset U\_1\subset ...\subset U\_{n-1}\subset U\_n=\mathbb C^n\otimes O\_{F\_n}$.


Recall that for $i=1,...,n-1$ the classes $\sigma\_i=c\_1(U\_i^\*)\in H^2(F\_n)$ form a base in $H^2(F\_n)$ and moreover generate multiplicatively the cohomology ring of $F\_n$.


**Question.** Is it true that the cohomology classes $(\sigma\_1\cdot\sigma\_2\cdot...\cdot \sigma\_{n-1})^k$ and $(\sigma\_1\cdot\sigma\_2\cdot...\cdot \sigma\_{n-2})^k$ are non-zero for $k$ small enough with respect to $n$ (say for $k<\log(n)/100$)? Or at least, say for $k\le 10$ for large $n$?



Non-vanishing of elements in cohomology of full Flag varieties
Yes. I claim that $(\sigma\_1 \sigma\_2 \cdots \sigma\_{n-1})^{\lfloor n/2 \rfloor} \neq 0$, which is better than anything you ask for. (Here $\lfloor x \rfloor$ is $x$ rounded down to the nearest integer.) 


Recall that a basis for $H^{\ast}(FL\_n)$ is the Schubert classes $[X\_w]$, indexed by permutations $w$ in $S\_n$. We use $\ell$ for the length function on $S\_n$ and write $(i j)$ for the transposition that interchanges $i$ and $j$.


[Monk's formula](http://en.wikipedia.org/wiki/Monk's_formula) states that
$$\sigma\_r [X\_w] = \sum\_{\begin{matrix} \ell( w(ij)) = \ell(w)+1 \\ i \leq r < j \end{matrix}} X\_{w (ij)}.$$
The identity of $H^{\ast}(FL\_n)$ is the identity permutation. Thus, we see that the coefficient of $[X\_w]$ in $\sigma\_{r\_1} \sigma\_{r\_2} \cdots \sigma\_{r\_N}$ is the number of chains 
$$e = w\_0,\ w\_1,\ w\_2,\ \ldots,\ w\_N=w$$
so that $\ell(w\_k)=k$ and $w\_k$ is of the form $w\_{k-1} (i j)$, with $i \leq r\_k < j$.


In particular, if the word $(r\_1 \ r\_1+1) (r\_2\ r\_2+1) \cdots (r\_N \ r\_N+1)$ is reduced, meaning that the product of the first $k$ transpositions has length $k$, then $\sigma\_{r\_1} \sigma\_{r\_2} \cdots \sigma\_{r\_N} \neq 0$.


Now notice that
$${\Big (}(1 \ 2)(3\ 4)(5\ 6) \cdots (2\ 3)(4\ 5)(6\ 7) \cdots {\Big )}^{\lfloor n/2 \rfloor}$$
is reduced.


