lines in projective spaces [closed]

Question

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 ?

Answer 1

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.