Let $G=\mathrm{SL}_n(\mathbb{R}),\mathrm{Sp}_{2n}(\mathbb{R}),\mathrm{Spin}_n(\mathbb{R})$ be a semi-simple simply connected classical group, $\Gamma\subset G$ a discrete and cocompact subgroup. Then Ratner's orbit closure theorem states that for any unipotent subgroup $U$ of $G$, the closure $\overline{\Gamma U}$ is always of the form $\Gamma H$ for some closed subgroup $H$ of $G$ (see here. Note that Ratner's theorem holds for much more general groups than the $G$ considered here).
Now I would like to consider some particular $U$, which is a maximal abelian unipotent subgroup. Then the closure $\overline{\Gamma U}$ is of the form $\Gamma H$ for some closed subgroup $H$ of $G$.
My question is: what can we say about $H$? Is it true that $H=G$? I guess this is a rather trivial question, yet I failed to find an answer by myself nor on the internet. Any help would be appreciated!
Added: if we drop the assumption 'abelian' on $U$, then it is not difficult to see $H=G$: for example, take $G=\mathrm{SL}_n(\mathbb{R})$ and $U$ the subgroup consisting of unipotent upper triangular matrices. Then take a one-parameter unipotent subgroup $U'=1_n+\mathbb{R}E_{1,2}$ of $U$, then $\overline{\Gamma U'}=\Gamma H'$ where $H'$ contains a subgroup $N$ isomorphic to $\mathrm{SL}_2(\mathbb{R})$. $N$ can not be contained in the upper triangular Borel subgroup of $G$. So a simple argument shows that $N$ and $U$ generates the whole $G$ and so $H$, a subgroup of $G$ containing $U$ and $N$, must be equal to $G$ (this last sentence is too simplified, see the comments below).
PS: this question is also posted on math stackexchange (here)
