arithmetic groups VS. Zariski dense discrete subgroups? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T06:19:06Z http://mathoverflow.net/feeds/question/46358 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46358/arithmetic-groups-vs-zariski-dense-discrete-subgroups arithmetic groups VS. Zariski dense discrete subgroups? genshin 2010-11-17T14:40:12Z 2010-11-18T10:39:02Z <p>Assume that \$G\$ is a semi-simple linear algebraic group defined over \$\mathbb{Q}\$, which is \$\mathbb{Q}\$-simple, and that \$G(\mathbb(R)\$ is non-compact, without \$\mathbb{R}\$-factors of rank 1. Then by Margulis's works, the arithmetic subgroups of \$G(\mathbb{R})\$ are the same as discrete lattice in \$G(\mathbb{R})\$. Here a lattice is a discrete subgroup \$\Gamma\$ such that the quotient \$\Gamma\backslash G(\mathbb{R})\$ is of finite volume with respect to the measure deduced from the left Haar measure. In particular, an arithmetic subgroup in \$G(\mathbb{R})\$ is Zariski dense in \$G_\mathbb{R}\$.</p> <p>Conversely, a discrete subgroup \$\Gamma\$ in \$G(\mathbb{R})\$ is given, such that \$\Gamma\$ is also dense in \$G_\mathbb{R}\$ for the Zariski topology, what condition should one impose to make it arithmetic? Shall I assume \$\Gamma\$ to be finitely generated, or stable under certain actions such as \$Aut(\mathbb{R/Q})\$? I feel that such kind of results are more or less available in the literature, bu I'm far from an expert in this field.</p> <p>Many thanks!</p> http://mathoverflow.net/questions/46358/arithmetic-groups-vs-zariski-dense-discrete-subgroups/46364#46364 Answer by Keivan Karai for arithmetic groups VS. Zariski dense discrete subgroups? Keivan Karai 2010-11-17T15:11:29Z 2010-11-17T15:11:29Z <p>an arithmetic subgroup is a lattice. This is in characteristic zero due to Borel and Raghunathan (MR0147566) and in positive characteristic due to Harder and Behr.</p> http://mathoverflow.net/questions/46358/arithmetic-groups-vs-zariski-dense-discrete-subgroups/46468#46468 Answer by Tobias Hartnick for arithmetic groups VS. Zariski dense discrete subgroups? Tobias Hartnick 2010-11-18T10:39:02Z 2010-11-18T10:39:02Z <p>Arbitrary Zariski-dense subgroups in a semisimple group can be very small from a real-analytic point of view. It seems that algebra cannot distinguish between "small" and "large" Zariski-dense subgroups, so most criteria to distinguish between the two have a strong non-algebraic flavour. (Of course one can also characterize arithmetic groups algebraically, but this has even less to do with the line of argument you seem to suggest.) From a dynamical point of view, the key difference between lattices and arbitrary Zariski-dense subgroups is that the former act transitively on the product of the Furstenberg boundary of the ambient Lie group with itself ("double ergodicity"). This is a sort of "largeness" property. There are various ways to capture this property, the most systematic way seems to me the concept of a generalized Weyl group due to Bader and Furman. </p>