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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 $n$?

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  • $\begingroup$ I think you should be taking the first Chern classes of the duals of quotients of adjacent bundles in your flag. $\endgroup$ Commented Jul 21, 2014 at 15:11
  • $\begingroup$ No Peter, I am interested in classes $\sigma_i$. If one replaces $\sigma_i$ by $\sigma_i-\sigma_{i-1}$ as you suggest, the question becomes completely different (and easy to answer) $\endgroup$
    – aglearner
    Commented Jul 21, 2014 at 15:15
  • $\begingroup$ As hinted in the previous comments, this is a purely combinatorial problem. We have $H^*(F_n,\mathbb{Q})=\mathbb{Q}[t_1,\ldots ,t_n]/(s_1,\ldots ,s_n)$, where the $t_i$'s are the first Chern classes of the successive tautological quotient line bundles and $s_i$ is the $i$-th symmetric function. The question is to find the order of nilpotence of $t_1(t_1+t_2)\ldots (t_1+\ldots +t_{n-1})$ in that ring (and same question with the last factor deleted). $\endgroup$
    – abx
    Commented Jul 21, 2014 at 15:34
  • $\begingroup$ abx, yes thank you, I understand (I added one more tag - combinatorics). I ask this question here because for the moment I can not answer this combinatorial question. $\endgroup$
    – aglearner
    Commented Jul 21, 2014 at 15:46

1 Answer 1

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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 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.

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  • $\begingroup$ Thank you very much David! I will need a bit of time to digest your proof :) $\endgroup$
    – aglearner
    Commented Jul 21, 2014 at 22:04
  • $\begingroup$ I finally got it :) $\endgroup$
    – aglearner
    Commented Jul 22, 2014 at 10:04

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