# Computing intersection of cycles on the product of Grassmannians/Deligne-Lusztig varieties

My collaborators and I are preparing an interesting manuscript where the computation leads to something related to what we believe to be in the area of Schubert calculus; but none of us knows much about it. (It is a long story to explain how we ended up here.) We have a guess of what the answer is and we have tried but failed. Writing the paper with this formula as a conjecture is acceptable but certainly not ideal. It would be great if someone could help us with the problem, or point us to the right direction.

The problem is an intersection in the product of two Grassmannians (or a Deligne-Lusztig variety if it helps at all). Let $j < n/2$ be positive integers. The intersection happens in $\mathrm{Gr}(n,j) \times \mathrm{Gr}(n,j+1)$. Let $H_1$ and $H_2$ denote the universal subbundles of rank $j$ and $j+1$, respectively; we view them as the subbundle of a same vector bundle $V$ of rank $n$. We look at the intersection of four cycles $Z_1\cdot Z_2 \cdot c_{j}(?) \cdot c_{n-j-1}(??)$, where $c$'s are Chern classes of some bundles we specify below. This will eventually give us a number (of points / in the top cohomological degree). Let $p$ be a prime number and we view all Grassmannians over $\mathbb{F}_p$; and let $F$ denote the Frobenius pull-back. (This can be done over $\mathbb C$ if one prefers; then one just reinterprets all cycles algebraically with multiplicity $p$; then $p$ need not to be a prime.)

$Z_1$ is the cycle defined by the zero of the natural map $H_1 \to V/ H_2$. (One can compute this cycle using Thom-Porteous formula, but we are not sure if this is a good idea. Alternatively, one can consider a two-step Grassmannian; we are not sure if this is a good idea either.)

$Z_2$ is the cycle defined by the zero of the natural map $H_1 \to V / F(H_2)$. (So, one can say that we are looking at the locus where $H_1$ sits inside the intersection of $H_2$ and $F(H_2)$, so a Deligne-Lusztig variety; I am not sure if this helps at all.)

The first Chern class is the top Chern class $c_j$ of $H_1 \otimes (H_2/H_1)^\vee$. (This makes sense because $H_1 \subseteq H_2$; but we can also get it in an abstract way.)

The second Chern class is the top Chern class $c_{n-j-1}$ of the bundle $(F(H_2) / H_1) \otimes (V/F(H_2))^\vee$.

Now we multiply all these four cycles together. One can check that the dimension forces the intersections to be points.

Our conjectural answer to the question is $\pm$ times $$n p^{d(n,0)} + (n-2) p^{d(n-1,1)} \binom n1_p + (n-4) p^{d(n-2,2)} \binom n2_p + \cdots + (n-2i) p^{d(n-i,i)} \binom ni_p.$$ where $d(n,k) = \frac 12((2k-1)n-2k(k-1)-1)$ is the half dimension of the Grassmannian $\mathrm{Gr}(n,k) \times \mathrm{Gr}(n,k-1)$, and $\binom n\delta_p$ is the usual $p$-binomial expansion.

We have computed this for a few small numbers: e.g. $n=3, i=1$, the intersection number is $\pm(4p^2 + p+1)$; and $n=4, i=1$, the number is $\pm(6p^3+2p^2+2p +2)$. We have other reason to believe our conjectural formula. Also I know how to prove the formula when $p=1$ (as interpreted over $\mathbb C$), but I used a combinatorics formula to kill all $\binom n\delta_p$ terms, which doesn't hold if $p \neq 1$.