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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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    Okay hopefully what I meant is now clear and that the Schubert varieties David constructed below are a counterexample.2012-03-07

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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 $e_2$.

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    If you happen to know the answer, do you know if we can put some restriction on the sort of flags that are chosen to prevent this behavior? Here your flags are very similar--you're just reordering the same basis vectors. I'm not sure what a good way to measure how different flags are though.2012-03-07