# Intersection of the Irreducible Components of Intersections of Schubert Varieties

Let $K$ be an algebraically closed field and $G$ be the Grassmannian of $k$ planes in some $l$ dimensional vector space $V$ over $K$. Let $V_1\subsetneq ... \subsetneq V_l$ be a flag for $V$. A Schubert variety in $G$ for our flag is $S_{a_1,...,a_k}:=\{\Lambda\in G:dim(\Lambda\cap V_{l-k+i-a_i})\geq i,\ \forall i\}$. A Schubert variety has codimension $\sum a_i$ in $G$. Call a Schubert variety special if $a_i = 0$ for $i>1$.

Let $S_1,...,S_n$ be special Schubert varieties of $Gr$ and let $V_1,V_2$ be distinct irreducible components of $\cap_i S_i$. My question is, must $V_1\cap V_2 = \emptyset$? If so are there any conditions we can impose for this intersection must be empty?

Thanks

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What is a special Schubert varieties? I thought every Schubert variety was special in its own way... –  Stephen Mar 6 '12 at 23:47
But of course they are all special. First I'm not sure it's relevant but it's the situation I'm in so I thought I might as well include it. Writing a general Schubert variety as $V_{a_1,...,a_N}$ then a special Schubert variety is of the form $V_{a_1,0,...,0}$. Hopefully that notation makes sense to everyone it doesn't appear consistent in the literature to me. –  user16544 Mar 7 '12 at 0:08
Ryan, especially since David also is not sure what you mean by special, would you mind editing the question to make the definition of $V_{a_1,\dots,a_n}$ more precise? The kids these days seem to be using several different indexing conventions for Schubert varieties. –  Stephen Mar 7 '12 at 1:57
Okay hopefully what I meant is now clear and that the Schubert varieties David constructed below are a counterexample. –  user16544 Mar 7 '12 at 18:44

I don't think I fully understand the question, but see if this answers it. Look at the Grassmannian $G(2,4)$. For any $2$-plane $F$, the set of all $2$-planes $L$ with $F \cap L \neq 0$ is a Schubert variety, corresponding to the partition $(1)$. (Is that what you mean by special?) Let's call it $X(F)$.
Let $e_1$, $e_2$, $e_3$, $e_4$ be a basis for four-space. Then $X(\mathrm{Span}(e_1, e_2)) \cap X(\mathrm{Span}(e_2, e_3))$ has two components. The first component corresponds to $L$ contained in $\mathrm{Span}(e_1, e_2, e_3)$, the second corresponds to $L$ containing $e_2$. These two components meet along a curve, parametrizing those $L$ which are both contained in $\mathrm{Span}(e_1, e_2, e_3)$ and contain $e_2$.