# Little puzzle about intersections of Schubert cells.

I was reading chapter 6 in the book of Harris on Algebraic geometry and came to the following puzzle.

It seems to me that every Schubert cell in a Grassmanian is obtaining by cutting the Grassmanian by a certain plane in its Plucker embedding (hope this is correct, I got this idea from Harris' book)

Now, I was thinking always, that there is a whole theory of Littlwood Richardson coefficients that explains the intersection theory of Grassmanian and in particular explains what are intersection numbers of various cells.

My question is as follows. Why do we have this complicated theory of Littlwood Richardson coefficients? (is this because Schubert cells don't always have expected dimension ?) Naively, if all Schubert cells were obtained by intersecting Grassmanian transversally by some planes the only number that would pop up would be the degree of Grassmanian in its Plucker embedding.

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Incidentally, in any $G$-equivariant projective embedding of any $G/P$, the Schubert varieties are defined (even scheme-theoretically) by the intersection with a linear subspace; this is due to Ramanathan. – Allen Knutson Jan 4 '15 at 16:49

For example, think about $G(2,4)$. It is the quadric $p_{12} p_{34} - p_{13} p_{24} + p_{14} p_{23} =0$ in $\mathbb{P}^{\binom{4}{2}-1}$. There are two codimension $2$ Schubert cells: One of them is given by $p_{12}=p_{13} = p_{23}=0$ and the other is given by $p_{12}=p_{13}=p_{14}=0$. In each case, we are using $3$ linear equations to cut out something which is codimension $2$. If we just interesect $G(2,4)$ transversely with a codimension $2$-plane, we get something whose cohomology class is the sum of these two classes; if we intersect $G(2,4)$ with $p_{12}=p_{13}=0$ then we literally get the union of these two cycles.