Let $G$ be a complex Chevalley group (not necessarily adjoint type) with $\operatorname{\mathbb{C}-rank}\geq2$ 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 permutation matrix of $(123..n)$ when $B$ is the upper triangular matrices and $T$ the diagonal matrices. Then since $J=\{A\in \operatorname{SL}_{n}(\mathbb C)|\operatorname{disc}(A)\neq 0\}$ 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.

When $G=\operatorname{SO}_{2n}(\mathbb C)=\{A\in\operatorname{SL}_{2n}(\mathbb C)\mid A^TJ_{2n}A=J_{2n}\}$ where $J_{2n}$ is the identity matrix flipped 90 degrees and the index order is $1,2...n,-n,...,-1$. We get that $w$ is the permutation matrix of $((n-1)(n-2)...1(1-n)(2-n)...(-1))(n(-n))$ when $B$ is the upper triangular matrices and $T$ the diagonal matrices. I'm thinking of looking for an open set inside $\{A\in \operatorname{SO}_{2n}(\mathbb C)|A $ is conjugate to $C(p_1)\oplus C(p_2) $ in $\operatorname{SL}_{2n}(\mathbb C),p_1|p_2,p_i\in\mathbb{C}[x] $ monic and $deg(p_1)=2 \}$.

$C(p)$ is the companion matrix of the monic polynomial $p$.

In general $G$ where the Coxeter number $h$ is even there is a Coxeter element $w_c$ such that $w_c^{h/2}=w_0$ the longest root. So the open set $Bw_0B=\prod_{i=1}^{h/2}Bw_cB$ may say that $Bw_cB$ is open?