Skip to main content
1 of 9
Ami
  • 71
  • 2

Do these Zariski-dense subgroups of complex Chevalley group have non-empty intersection with this cell?

Let $G$ be a complex Chevalley group (not necessarily adjoint type) with $\mathbb{C}-rank\geq2$ and let $H$ be a normal subgroup of $G(\mathbb Z)$ with a finite index (so $H$ is Zariski dense in $G$).

Let $T$ a maximal torus in $G$ and $B$ a Borel subgroup containing $T$, let $w_{\alpha_1},...,w_{\alpha_n}$ represent the simple reflections of the Weyl group in $G$. From Steinberg's 1967-68 Yale lectures $w_α(t)=x_α(t)x_{−α}(−t^{−1})x_α(t),w_α:=w_α(1)$ where $x_α(t)$ is the root element in $G$.

Denote $w:=w_{\alpha_1}\cdots w_{\alpha_n}$ now

is it true that $H∩BwB$ is necessarily non-empty?

When $G=\operatorname{SL}_{n}(\mathbb C)$ we get that $w$ is the permitation matrix of $(123..n)$ when $B$ is the upper triangular matrices and $T$ the diagonal matrices. Then since $J=\{A\in \operatorname{SL}_{n}(\mathbb C)|disc(A)\neq 0\}$ is Zariski open there is a $g\in J\cap H$ and since every element of $J(\mathbb Z)$ is similar in $\operatorname{SL}_{n}(\mathbb Z)$ to some $b_1wb_2$ ,$b_i\in B$ and $H$ is normal, we get $g\in H\cap BwB$.

Here $disc(A)$ is the discriminant of the characteristic polynomial of A, when $disc(A)\neq0$ that mean that $A$ is similar to a companion matrix since all the roots of the characteristic polynomial of A are distinct.

Ami
  • 71
  • 2