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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.

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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.

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  • $\begingroup$ Thanks for your answer ! I just need to read a bit more for be sure I understood. I don't know yet Littlewood-Richardson rule, and the purpose of the exercise was probably to do the computations by hand. So I don't really understand the last line but I guess now I need to think a bit more about it. $\endgroup$
    – user378546
    Commented Mar 28, 2017 at 15:06
  • $\begingroup$ Schubert Calculus is a way to compute products in the cohomology ring of the Grassmanian. These products can be translated into informations of the intersection of certain Schubert varieties if you use flags in general position (such as opposite flags). By the way the most important part here is that it's like you tried to intersect the varieties in the affine part and you didn't find any point, but you can think as they actually intersect at infinity because we are working in a projective space. A Schubert variety is a union of disjoint affine cells of decreasing dimension. $\endgroup$
    – Maffred
    Commented Mar 28, 2017 at 15:12
  • $\begingroup$ You intersected the bigger cells but you forgot to check for the small ones which do belong to the Schubert varieties. $\endgroup$
    – Maffred
    Commented Mar 28, 2017 at 15:14
  • $\begingroup$ You can think as if you try to find the intersection of two spheres which meet at a point. You think the two spheres as union of the intersection point and a a cell of dimension 2. You check the intersection between the two cells and you don't find any point! But they actually intersect in the $0$-cells. $\endgroup$
    – Maffred
    Commented Mar 28, 2017 at 15:18
  • $\begingroup$ Ok I understand. But here could you explain to me why $e_1 \oplus e_4 \in \sigma_{1,1}' ?$ The closure should be remplacing some $1$ by $0$ but I will never got any positive coefficient in $e_1$ right ? Or am I mistaken again ? $\endgroup$
    – user378546
    Commented Mar 28, 2017 at 15:46

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