Timeline for Properties of a general element of the intersection of two Schubert cycles

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Dec 6, 2019 at 10:10	comment	added	klerk		But I know for sure that $\Lambda$ belongs to $\Sigma_a \cap \Sigma_b$. Why does this must lie in $\Sigma_{\tilde{a}}\cap \Sigma_b$? If i take an element $[v] \in \mathbb{P}(V_{n-k+i-a_{i-1}})\cap \mathbb{P}(V_{n-i-b_{k-i}})$ it could be that an element of the fibre $\Lambda$ may lie in $\Sigma_a \setminus \Sigma_{\tilde{a}}$. Why does this cannot happen? Thank you for your help Libli.
Dec 6, 2019 at 8:30	comment	added	Libli		This is by definition of transverse flags and the definition of Schubert cycles!
Dec 4, 2019 at 12:02	comment	added	klerk		Why does the preimage of $$\mathbb{P}(V_{n-k+i-a_i-1}) \cap \mathbb{P}(W_{n-i-b_{k-i}}) \cup \mathbb{P}(V_{n-k+i-a_i}) \cap \mathbb{P}(W_{n-i-b_{k-i}-1})$$ in $\phi$ maps to $\Sigma_{\widetilde{a}} \cap \Sigma_b \cup \Sigma_{a} \cap \Sigma_{\widetilde{b}}$? To be more precise: let $[v] \in \mathbb{P}(V_{n-k+i-a_{i-1}}) \cap \mathbb{P}(V_{n-i-b_{k-i}})$, why does the preimage of this element consist of couples $(\Lambda, [v])$ with $\Lambda \in \Sigma_{\tilde{a}} \cap \Sigma_b$?
Dec 3, 2019 at 21:41	history	answered	Libli	CC BY-SA 4.0