Let $Lie(B)$ be the subalgebra of upper triangular matrices in $M_{n \times n}(\mathbb{C})$. Is the set $\{M[x] \text{ s.t. } [x] \in Gr_k^n(\mathbb{C}) \subset \mathbb{P}(\wedge^k \mathbb{C^n}) , M \in Lie(B), Mx \neq 0, \}\ \subset Gr_k^n(\mathbb{C})\subset P(\wedge^k(\mathbb{C^n}))$ a closed subset?
The obvious thing to do is to show that this is the closure of $B[x]$. It is more or less obvious that the above set is in the closure of $B[x]$. But the other direction has been running me into trouble.
If you are given $[y] \in \overline{B[x]}$ and a sequence $b_i [x]$ converging to $[x]$, with the absolute value of the largest eigenvalue of $\wedge^k b_i=1$, then one can pick out a convergent subsequence, still denoted $\wedge^k b_i$, converging to some matrix $M \in End(\wedge^k \mathbb{C})$. $M= \wedge^k N$ for some $N \in End( \mathbb{C})$ because the set of matrices that can be written like this are closed(this is the theorem of the plucker embedding).
What I am not sure about is why $N$ needs to be in $\overline{B}=Lie(B)$.