Subgroups of $Sp(2n,\mathbb{R})$ between $Sp(2n,\mathbb{Z})$ and some arithmetic group

The fantastic answers to my previous question Subgroups of $SL_2(\mathbb R)$ which contain $SL_2(\mathbb Z)$ as a finite index subgroup led me to the following question.

Let $O_K$ be the ring of integers of $K= \mathbb{Q}(\sqrt{p})$, where $p$ is a prime number.

What are the subgroups $\Gamma \subset \mathrm{Sp}(2g,O_K)$ of infinite index which contain $\mathrm{Sp}(2g,\mathbb Z)$? Are there any?

[I should add that, in all this, $g\geq 2$]. In the paper referred to above, what is proved is that any intermediate subgroup either has finite index in $Sp_{2g}(O_K)$ or else contains $Sp_{2g}(\mathbb{Z})$ as a finite index subgroup. In particular, since $Sp_{2g}(\mathbb{Z})$ is a maximal discrete subgroup of $Sp_{2g}(\mathbb{R})$ by the answers to your previous question, there are no in-between subgroups of infinite index in $Sp_{2g}(O_K)$ other than $Sp_{2g}(\mathbb{Z})$.
• Thank you again! I have three questions if that's ok. 1) Is it important in your paper that $K$ is a real quadratic field? Or could $K$ be any number field? 2) Is it important that $O_K$ is a maximal order in $K$? What if we take any order $O$ and ask a similar question? 3) What if we look at subgroups between $Sp(2g,O_K)$ and some finite index subgroup of $Sp(2g,\mathbb Z)$? Sorry for the many questions; I'm only now discovering these fantastic results. – Honing Oct 13 '16 at 21:05
• @Honing. Thank you ! Yes, in the paper, it was important that it was (usually real) quadratic extension. In general, it seems hard to extend the results of that paper to other extensions. Your question 3 (for real quadratic $K$) is that any intermediate group either has finite index in the larger group or else contains the smaller group as a finite index subgroup (this is weaker than what you have asked for $Sp_{2g}(\mathbb{Z})$). Yes to Question 3, in this weaker sense, implies yes to question 2. – Venkataramana Oct 14 '16 at 0:15