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.

This particular variety is irreducible because its coordinate ring is the localization of the ring of polynomial functions on all matrices by the determinant, and is therefore an integral domain.


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.

  • $\begingroup$ This answer does not explain why $Bw_0B$ is open, it simply asserts a more general version of the statement. $\endgroup$ – David Loeffler May 20 '18 at 8:01

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