I want to compute intersection of the Schubert cell $\sigma_{(3,0)}$ with all the cells $\sigma_{a_1, a_2}$ in the grassmanian $G(2,5)$. I am not sure I am doing correctly but I can't see my mistake. The notations are from the book 3264 and all that by Eisenbud and Harris.
By definition $\sigma_{3,0}$ is the set of 2-planes spanned by the vectors $\begin{pmatrix} 1 & 0 & 0& 0& 0\\ 0 & * & * &* & 1 \end{pmatrix}$.
Let's intersect it with for example $\sigma_{1,1}$ which is the set of $2$-planes on the form $\begin{pmatrix} * & * & 1& 0& 0\\ * & * & 0 & 1 & 0 \end{pmatrix}$.
For obtaining a transverse intersection I take the opposite flag $F_i' = \mathbb C\{e_n, e_{n-1}, \dots, e_{n-i}\}$. In this expression $\sigma_{1,1}'$ is $\begin{pmatrix} 0 & 0 & 1 & * & *\\ 0 & 1 & 0 & * & * \end{pmatrix}$.
But a $2$-plane in $\sigma_{1,1}'$ will never contains $e_1$, i.e $\sigma_{3,0} \cap \sigma_{1,1}' = \emptyset$. This is surely not correct since $\text{codim} (\sigma_{3,0}) + \text{codim} (\sigma_{1,1}) = 5 < 6$. Can someone explain to me where is my mistake ? Thanks in advance.
Your mistake is when you say: "By definition $\sigma_{3,0}$ is the set of 2-planes spanned by the vectors $\begin{pmatrix} 1 & 0 & 0& 0& 0\\ 0 & * & * &* & 1 \end{pmatrix}$".
This is not true! The spaces spanned by those matrices represent only the main cell, in fact the Schubert variety is the union of the cells $\begin{pmatrix} 1 & 0 & 0& 0& 0\\ 0 & * & * &* & 1 \end{pmatrix}$ $\begin{pmatrix} 1 & 0 & 0& 0& 0\\ 0 & * & * &1 & 0 \end{pmatrix}$ $\begin{pmatrix} 1 & 0 & 0& 0& 0\\ 0 & * & 1 &0 & 0 \end{pmatrix}$ $\begin{pmatrix} 1 & 0 & 0& 0& 0\\ 0 & 1 & 0 &0 & 0 \end{pmatrix}$. The same applies for the other variety, thus as you can see $\langle e_1, e_4 \rangle$ is a point in both of them.