# Is $Bw_0B$ a dense open subset of $G$?

Let $G=GL_n(\mathbb{C})$ and $B$ the Borel subgroup of $G$ containing all upper triangular matrices in $G$. Let $w_0$ be the longest word in the Weyl group $W$ of $G$. We have the Bruhat decomposition $G = \cup_{w \in W} BwB$. Is $Bw_0B$ a dense open subset of $G$?

I think that $Bw_0B$ is open because $Bw_0B = \{g \in G: \Delta_{\{n-i+1, \ldots, n\}, \{1,\ldots,i\}}(g) \ne 0, i=1,2,\ldots,n \}$.

How to show that $Bw_0B$ is dense in $G$? Thank you very much.

It is Zariski-open in the irreducible variety $\mathrm{GL}_n(\mathbf{C})$. Every Zariski-open subset of an irreducible variety is dense.
If you consider $G/B$ which is a flag variety and we have Bruhat decomposition and $BwB/B$ is an affine set which is isomorphic to $\mathbb{C}^{l(w)}$ where the dimension is the length of the Weyl group element. Also the $B$ orbits form a stratification by which I mean the closure of a B-orbit is union of other B orbit. If we consider the projection from $G$ to $G/B$ then the set $Bw_0B$ is the preimage of of an open dense set so it is also open and dense.
• This answer does not explain why $Bw_0B$ is open, it simply asserts a more general version of the statement. – David Loeffler May 20 '18 at 8:01