Nonempty intersection in Grassmannian Where can I find a proof of the following fact:
If $X_1$ and $X_2$ are subvarieties of $\mathbb{G}(k,n)$ of codimension $c_1$ and $c_2$ satisfying $c_1+c_2<n+1-2k$, then the intersection $X_1\cap X_2$ is nonempty.
(This is Lemma 3.13 in the paper Hypersurfaces of Low Degree, and it was said that it is a standard fact and stated without proof or reference.)
 A: By recursion on $k$.
If $k=1$, this is Bezout's theorem in projective space. Now let $k>1$. We have the Pl\"ucker embedding $\mathbb{G}(k,n) \subset \mathbb{P}(\bigwedge^k \mathbb{C}^n)$. Let $p$ be a general point in $\mathbb{C}^n$. Let $Y_i = \mathbb{P}(\bigwedge^{k-1} \mathbb{C}^n/\langle p\rangle) \cap X_i$.
The variety $Y_i$ is the variety of subspaces parametrized by $X_i$ which contains the point $p$. The codimension $Y_i$ in $\mathbb{G}(k-1,n-1) = \mathbb{P}(\bigwedge^{k-1} \mathbb{C}^n/\langle p\rangle) \cap \mathbb{G}(k,n)$ is $c_i$ (because $p$ is generic).
By hypothesis, we have $c_1 + c_2 < n+1 - 2k$, so that $c_1 + c_2 < (n-1)+1 - 2(k-1)$. Hence by hyptohesis, we have $Y_1 \cap Y_2 \neq \emptyset$. 
Let $L \subset \mathbb{C}^{n}/\langle p \rangle$ be a linear space of dimension $k-1$ which is  in $Y_1$ and $Y_2$. Then, by construction, the inverse image of $L$ in $\mathbb{C}^n$ is in $X_1$ and $X_2$. 
This concludes the recursion.
EDIT : correction of a misleading typo about codimensions
A: Use Borel's theorem (about the existence of fixed points for actions of solvable groups) on the relevant Hilbert scheme, to degenerate $X_1$ to being $B$-invariant and $X_2$ to being $B_-$-invariant. Now they are schemy unions of Schubert varieties resp. opposite Schubert varieties.
This reduces the problem to the case of a Schubert variety intersecting an opposite Schubert variety.
Those varieties are indexed by partition resp. partition rotated $180^\circ$ inside $k\times (n-k)$, with #boxes = codimension, and the varieties intersect iff the partitions don't. The cheapest way (in terms of #boxes) to get two to intersect is for one partition to be a single whole row, the other a single whole column. I'm apparently confused though, because that would give a bound of $n$, not $n-2k$.
