Let $\{v\_1,v\_2, \cdots , v\_n, w\_1,w\_2, \cdots ,w\_n\}$ be a basis of $\mathbb C^{2n}$. For a $n$-dimensional subspace $V \in Gr(n,\mathbb C^{2n})$ define another $n$ dimensional subspace $\bar{V} \in Gr(n, \mathbb C^{2n})$ as follows: 


If $V$ is generated by $n$ linearly independent vectors $x\_i\in \mathbb C^{2n}$ of the form $\sum\_{j=1}^ka\_{i\_j}v\_{i\_j}+\sum\_{j=1}^{n-k}a\_{i\_j}w\_{i\_j}$, then $\bar{V}$ is generated by $\bar{x\_i}$'s where $\bar{x\_i}= \sum\_{j=1}^ka\_{i\_j}v\_{i\_j}-\sum\_{j=1}^{n-k}a\_{i\_j}w\_{i\_j}$. Then $V$ and $\bar{V}$ represent two points in $\mathbb P(\bigwedge^n\mathbb C^{2n})$. My guess is that the line joining these two points in $\mathbb P(\bigwedge^n\mathbb C^{2n})$ intersect $Gr(n, \mathbb C^{2n})$ exactly at two points namely at $V$ and $\bar{V}$. Is there way to prove my claim ? If I take the set of all such lines, then what could be the Zariski closure of this set ? 



lines in projective spaces
Since the Grassmannian is an intersection of quadrics, the intersection $Gr \cap L$ is either zero-dimensional of length at most 2, or equals to $L$. So, to prove the claim it is enough to rule out the possibility that $L \subset Gr$.


For this just note that a line on $Gr(n,2n)$ is determined by a flag $U\_{n-1} \subset U\_{n+1}$ and is formed by all $V$ sandwiched between $U\_{n-1}$ and $U\_{n+1}$. So, to prove the claim it is enough to check that $\dim (V \cap \bar{V}) < n - 1$. 


This, however, is not always true. For instance, if $V$ is the span of $v\_1,\dots,v\_{n-1}$, and $v\_n + w\_n$, then the intersection is $(n-1)$-dimensional and the line is contained in the Grassmannian.


