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$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?

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The trace map restricted to SL2(Z) has no dense image in $|x| >2$, but SL2(R) does. I would assume that looking at the coefficients of the characteristic polynomial gives you a negative answers for most variants of this question. – Marc Palm May 7 '12 at 15:21
up vote 3 down vote accepted

Since $SU(2,1)(K)$ is dense in $H=SU(2,1)({\mathbb C})$ (as in your previous question), 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 $H$.

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