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Let $c,d<N$ be integers and consider the product of two Grassmannians $M=Gr(c,N)\times Gr(d,N)$. Define $S\subset M$ to be the set of the pairs $([A_{c\times N}],[B_{d\times N}])$ such that $AB^t=0$. Note that this equation is independent of the choices of $A$ and $B$ in classes $[A]$ and $[B]$. This is a subvariety of codimension $c\times d$ and its homology class can be written as $$ \sum_{(a,b)\in T} \sigma_a \times \sigma_b,$$ over some set $T$ of pairs of Schubert cycles of $Gr(c,N)$ and $Gr(d,N)$.

Question: What is the set $T$ in terms of $c,d$ and $N$? (Note that $a$ and $b$ are non-increasing sequences of $c$ and $d$ non-negative numbers, respectively.)

For example if $c=d=1$, this just a hypersurface of bidegree $(1,1)$ an is homologous to$\sigma_1 \times \sigma_0+ \sigma_0\times \sigma_1$, where $\sigma_0=\mathbb{P}^{N-1}$ and $\sigma_1=\mathbb{P}^{N-2}\subset \mathbb{P}^{N-1}$.

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Let $V$ be a vector space of dimension $N$. To be precise your factors are the Grassmannian of $c$-dimensional subspaces in $V$ and the Grassmannian of $d$-dimensional subspaces in $V^*$. let $U_c\subset V\otimes O$ and $U_d \subset V^*\otimes O$ be the tautological subbundles of rank $c$ and $d$ on these Grassmannians. Then $S$ is the zero locus of the composition $$ U_c \boxtimes O \to V \otimes (O \boxtimes O) \to O \boxtimes U_d^* $$ on $M$, i.e. in the zero locus of the canonical global section of the vector bundle $U_c^*\boxtimes U_d^*$ on $M$. The class of the zero locus is the top Chern class $$ \mathrm{c}_{cd}(U_c^*\boxtimes U_d^*). $$ You can express it as the polynomial in Chern classes of $U_c$ and $U_d$ and then in terms of Schubert classes. I am sure you can find an appropriate formula in Fulton's "Intersection theory".

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  • $\begingroup$ i will try to find it. $\endgroup$ Commented Dec 6, 2013 at 15:48
  • $\begingroup$ Just two questions: Is $\boxtimes$ just equal to the product of two bundles after pulling back via the projection maps into components? I don't remember its definition. Also, does $\boxtimes$ commute with direct sum, i.e. does $\oplus_i \oplus_j E_i \boxtimes E_j = \oplus_i E_i \boxtimes \oplus_j E_j$ hold? $\endgroup$ Commented Dec 6, 2013 at 19:26
  • $\begingroup$ Yes (the second follows from the first). $\endgroup$
    – Sasha
    Commented Dec 6, 2013 at 19:40

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