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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?


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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
up vote 0 down vote accepted

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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Yes this is what I was looking for. That was much simplier than I thought. Thank you. I'll update the question with actual definitions. – user16544 Mar 7 '12 at 18:33
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. – user16544 Mar 7 '12 at 18:50

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