$K=Q(\sqrt{d} ) , d<0 $, $\Gamma $ an arithmetic subgroup of $G=SU(2,1)(K)$ . Is $\cup\_{g\in G}(g^{-1}\Gamma g)$ dense in G for the complex topology?



density of conjugate of arithmetic subgroup
Since $SU(2,1)(K)$ is dense in $H=SU(2,1)({\mathbb C})$ (as in your 
[previous question](https://mathoverflow.net/questions/95833/density-in-su2-1)), it suffices to consider density in $H$ of the union of $H$-conjugates of your arithmetic subgroup $\Gamma$. Now, I will think of $H$ as a *real* algebraic Lie group. All arithmetic subgroups of $H$ are obtained (up to commensurability and dividing by the finite center) by taking (polynomial) linear representations $\phi: H\to GL(n, {\mathbb R})$ and then taking $\phi^{-1}(GL(n, {\mathbb Z}))$. Since replacing a group with a commensurable one does not affect density, we can as well assume that $\Gamma=\phi^{-1}(GL(n, {\mathbb Z}))$. Now, if you consider characteristic polynomials of the elements of $\phi(\Gamma)$, as suggested by Mrc Plm, then they all have integer coefficients (unlike characteristic polynomials of elements of $\phi(H)$). Thus, the union 
$$
\bigcup\_{h\in H} \phi(h \Gamma h^{-1})
$$
cannot be dense in $\phi(H)$, hence, 
$$
\bigcup\_{h\in H} h \Gamma h^{-1}
$$
cannot be dense in $H$ either. 


Actually, more is true. Let $H$ be a real semisimple Lie group, $\Gamma\subset H$ be a lattice. Then the set 
$$
C:=\bigcup\_{h\in H} h \Gamma h^{-1}
$$
cannot be dense in $H$. The proof in the arithmetic case is explained above. In non-arithmetic case, we can assume that $H$ has rank 1 (by Margulis' arithmeticity theorem and checking that the case of reducible lattices reduces to lattices in simple rank 1 Lie groups). Now, consider a sequence of elements $\gamma\_n\in \Gamma$ and their translation lengths $L\_n=L(\gamma\_n)$ in the associated symmetric space $X=H/K$. I claim that the sequence $L\_n$ (up to a subsequence) is either constant or diverges to infinity. Indeed, if such a sequence is contains no equal elements and is bounded, then, the corresponding sequence of closed geodesics $\beta\_n\subset M=\Gamma\backslash X$ (represented by the elements $\gamma\_n$ and having of length $L\_n$), will subconverge (by Arzela-Ascoli theorem and thick-thin decomposition of $M$) to a closed geodesic, which is impossible. Since the translation lengths of elements of $\Gamma$ are preserved by conjugation via elements of $H$ and since the set of translation lengths of elements of $H$ equals ${\mathbb R}\_+$, it follows that $C$ cannot be dense in $H$. 


