Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

$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?

share|improve this question
1  
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. –  plusepsilon.de May 7 '12 at 15:21

1 Answer 1

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 $H$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.