2
$\begingroup$

For a field $K$ let denote by $Tr_1(d,K)$ the nilpotent group of all upper triangular $d\times d$-matrices over $K$ with each diagonal entry equal to 1. Let $\mathbb{Q}_p$ the field of $p$-adic numbers and consider $Tr_1(d,\mathbb{Q}_p)$ with the $p$-adic topology (observe that this is a $p$-adic analytic group). Then $Tr_1(d,\mathbb{Q})$ is a dense subgroup of $Tr_1(d,\mathbb{Q}_p)$. The question is: If $H$ is a closed subgroup of $Tr_1(d,\mathbb{Q}_p)$ is it true that $H\cap Tr_1(d,\mathbb{Q})$ is dense in $H$?.

Now $Tr_1(d,\mathbb{Z}_p)$ is an open compact subgroup of $Tr_1(d,\mathbb{Q}_p)$ and in fact it is the pro-$p$ completion of $Tr_1(d,\mathbb{Z})$. It is a fact that for every closed subgroup $H$ of $Tr_1(d,\mathbb{Z}_p)$ we have that $H\cap Tr_1(d,\mathbb{Z})$ is dense in $H$. Using this it is easy to see that for any closed subgroup $H$ of $Tr(d,\mathbb{Q}_p)$ we have that $\overline{H\cap Tr(d,\mathbb{Q})}$ is a an open subgroup of $H$. But I can't prove that it is equal to $H$.

$\endgroup$
3
$\begingroup$

Pick $\lambda \in \mathbb{Z}_p \backslash \mathbb{Q}$ and define

$H = \left\lbrace\begin{pmatrix} 1 & 0 & a \cr 0 & 1 & \lambda a \cr 0 & 0 & 1 \end{pmatrix} : a \in \mathbb{Z}_p\right\rbrace$.

Then $H$ is a closed subgroup of $Tr_1(3,\mathbb{Q}_p)$ but $H \cap Tr_1(3,\mathbb{Q})$ is the trivial group, hence certainly not dense in $H$.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.