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Let $G\leq \operatorname{SO}_{6}(\mathbb Z)$ be a finite-index normal subgroup, so it's a Zariski dense subgroup of $\operatorname{SO}_{6}(\mathbb C)$; and let $H$ be the subset of $\operatorname{SO}_{6}(\mathbb C)$ given by $$ H=\left\{e_{x,y}= \begin{pmatrix} 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & x & 0 \\ 0 & 0 & 0 & 0 & 0 & y \\ \frac{1}{y} & 0 & 0 & 0 & 0 & 0 \\ 0 & \frac{1}{x} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ \end{pmatrix} \middle| x,y\in\mathbb C^* \right\}. $$

Let $B=B(\mathbb{C})$ be the Borel subgroup which consists of upper triangular matrices.

Is it true that $G\cap Be_{1,1}B$ is necessarily non-empty?

As noticed in the comments, $G\cap H$ is empty for most congruence subgroups $G$.

Here $\operatorname{SO}_{n}(k)$ is defined as $\{A\in\operatorname{GL}_{n}(k)\mid A^TJ_nA=J_n\}$, where $J_n$ is the identity matrix flipped 90 degrees.

My progress: I've been trying to study this subset of $Be_{1,1}B\cap\operatorname{SO}_{6}(\mathbb Z)$: $$ L=\left\{\begin{pmatrix} b_{1} & b_{3} & 1 & a_{2} b_{3}+a_{1} b_{1} & a_{3} b_{1}-a_{2} & -a_{3} b_{3}-a_{1} \\ b_{2} & 0 & 0 & a_{1} b_{2}-b_{3} & a_{3} b_{2}+1 & 0 \\ 0 & -b_{2} & 0 & -a_{2} b_{2}-b_{1} & 0 & a_{3} b_{2}+1 \\ 1 & 0 & 0 & a_{1} & a_{3} & 0 \\ 0 & 1 & 0 & a_{2} & 0 & -a_{3} \\ 0 & 0 & 0 & 1 & 0 & 0 \\ \end{pmatrix} \middle| a_i,b_i\in\mathbb Z\right\}. $$

I noticed that the characteristic polynomials of the matrices in $L$ are of the form $p_{a,b,c}(x)=1+ax+bx^2+cx^3+bx^4+ax^5+x^6$, maybe

$ \{A\in \operatorname{SO}_{6}(\mathbb C)|\exists a,b,c\in\mathbb{C}\;\text{s.t.}\;p_{a,b,c}(x)=p_{\text{char}}(A)(x)\}\subset Be_{1,1}B$

Here:

  • $p_{\text{char}}(A)(x)$ is the characteristic polynomial of $A$.

There is something similar in $\operatorname{SL}_{n}(\mathbb C)$ where instead of $e_{1,1}$ we take $e$ to be the permutation matrix of $(123..n)$ and $G$ a finite index normal subgroup of $\operatorname{SL}_{n}(\mathbb Z)$ and $B$ the upper triangular matrices. Then since $J=\{A\in \operatorname{SL}_{n}(\mathbb C)|disc(A)\neq 0\}$ is Zariski open there is a $g\in J\cap G$ and since every element of $J$ is similar to some $b_1eb_2$ ,$b_i\in B$ , we get $g\in G\cap BeB$.

Here $disc(A)$ is the discriminant of the characteristic polynomial of A.

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    $\begingroup$ You can even take $G$ to be a "generic" Zariski dense 2-generated subgroup in $SO_6({\mathbb C})$. Then the intersection with $H$ will be empty. Here generic means that the two generators are chosen generically. $\endgroup$
    – Misha
    Commented Oct 4, 2019 at 22:54
  • $\begingroup$ @Misha thank you, I edited my question a bit. $\endgroup$
    – Ami
    Commented Oct 4, 2019 at 23:14
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    $\begingroup$ You should look more closely at what's written and revise again: Since $G$ is supposed to have integer matrix entries, if the intersection is nonempty then $x=\pm 1, y=\pm 1$. Hence, $G\cap H=\emptyset$ for most congruence subgroups. I am not sure what happens for double cosets with Borel. $\endgroup$
    – Misha
    Commented Oct 4, 2019 at 23:25
  • $\begingroup$ Since your specific choice of $H$ (and the requirement that the upper-triangular matrices form a Borel subgroup) means that the form of $\operatorname{SO}_6$ matters, you should specify what form you use. $\endgroup$
    – LSpice
    Commented Oct 5, 2019 at 1:34
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    $\begingroup$ Since (passing to a smaller finite index subgroup) you can suppose that $G$ is normal in $\mathrm{SO}_6(\mathbf{Z})$ (and hence normalized by $e_{1,1}$), you can write $G$ instead of $A^{-1}GA$ in your new question. $\endgroup$
    – YCor
    Commented Oct 5, 2019 at 19:37

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