Return to Answer

First of all, thank you for the adjective "fantastic"(!). The question is actually studied in a paper of mine (link to the MR review). Sorry for talking about my own paper (I have no option, since I do not know if anyone else is interested enough in these questions).

I should add that, in all this, $g\geq 2$. In that paper, 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 first question, there are no in between subgroups of infinite index in $Sp_{2g}(O_K)$ other than $Sp_{2g}(\mathbb{Z})$.

First of all, thank you for the adjective "fantastic"(!). The question is actually studied in a paper of mine (link to the MR review). Sorry for talking about my own paper (I have no option, since I do not know if anyone else is interested enough in these questions).

[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 $Sp_{2g}(\mathbb{Z})$.

First of all, thank you for the adjective "fantastic"(!). The question is actually studied in a paper of mine [see the link to the review

http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=JOUR&pg6=PC&pg7=ALLF&pg8=ET&review_format=html&s4=Venkataramana%2CT%2A&s5=Israel%20J%20Math&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=2&mx-pid=1324463]. Sorry for talking about my own paper (I have no option, since I do not know if anyone else is interested enough in these questions).

I should add that, in all this, $g\geq 2$. In that paper, 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 first question, there are no in between subgroups of infinite index in $Sp_{2g}(O_K)$ other than $Sp_{2g}(\mathbb{Z})$.

First of all, thank you for the adjective "fantastic"(!). The question is actually studied in a paper of mine (link to the MR review). Sorry for talking about my own paper (I have no option, since I do not know if anyone else is interested enough in these questions).

I should add that, in all this, $g\geq 2$. In that paper, 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 first question, there are no in between subgroups of infinite index in $Sp_{2g}(O_K)$ other than $Sp_{2g}(\mathbb{Z})$.

First of all, thank you for the adjective "fansatsic"(!). The question is actually studied in a paper of mine [see the link to the review

http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=JOUR&pg6=PC&pg7=ALLF&pg8=ET&review_format=html&s4=Venkataramana%2CT%2A&s5=Israel%20J%20Math&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=2&mx-pid=1324463]. Sorry for talking about my own paper (I have no option, since I do not know if anyone else is interested enough in these questions).

I should add that, in all this, $g\geq 2$. In that paper, 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 first question, there are no in between subgroups of infinite index in $Sp_{2g}(O_K)$ other than $Sp_{2g}(\mathbb{Z})$.

First of all, thank you for the adjective "fantastic"(!). The question is actually studied in a paper of mine [see the link to the review

http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=JOUR&pg6=PC&pg7=ALLF&pg8=ET&review_format=html&s4=Venkataramana%2CT%2A&s5=Israel%20J%20Math&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=2&mx-pid=1324463]. Sorry for talking about my own paper (I have no option, since I do not know if anyone else is interested enough in these questions).

I should add that, in all this, $g\geq 2$. In that paper, 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 first question, there are no in between subgroups of infinite index in $Sp_{2g}(O_K)$ other than $Sp_{2g}(\mathbb{Z})$.