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
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 $e_2$.