Groups quasi-isometric to reducible nonuniform lattices - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T08:04:20Z http://mathoverflow.net/feeds/question/90130 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90130/groups-quasi-isometric-to-reducible-nonuniform-lattices Groups quasi-isometric to reducible nonuniform lattices Igor Belegradek 2012-03-03T17:50:17Z 2012-03-04T03:37:08Z <p>It is known that a finitely group $G$ is quasi-isometric to a nonuniform <i> irreducible</i> lattice $\Lambda$ in a semisimple Lie group if and only if $G$ and $\Lambda$ are commensurable (see references in <a href="http://www.mathjournals.org/mrl/1997-004-005/1997-004-005-008.pdf" rel="nofollow"> this survey</a> of Farb). </p> <p><b> Question.</b> What is known about groups quasi-isometric to <i>reducible</i> nonuniform lattices in semisimple Lie groups? </p> <p>As usual in this business "semisimple" means "noncompact, connected, semisimple, with finite center".</p> http://mathoverflow.net/questions/90130/groups-quasi-isometric-to-reducible-nonuniform-lattices/90145#90145 Answer by Misha for Groups quasi-isometric to reducible nonuniform lattices Misha 2012-03-03T20:33:26Z 2012-03-03T20:33:26Z <p>Igor, I think it is still (mostly) unknown. Suppose that $\Gamma$ is a product of non-uniform irreducible lattices $\Gamma_i$. If all factors $\Gamma_i$ are lattices in rank 1 Lie groups then quasi-isometries preserve the product structure according to our paper </p> <p>[1] Kapovich, Kleiner, Leeb, Quasi-isometries and the de Rham decomposition, Topology 37 (1998), no. 6, 1193–1211. </p> <p>The reason is that in this case each $\Gamma_i$ contains quasi-geodesics with exponential divergence, so it is of Type I in the sense of [1]. Once you know this, you are in business because the factors $\Gamma_i$ are QI rigid. However, if you allow factors which are non-uniform lattices of rank $\ge 2$, then, conjecturally, they have linear divergence. Special cases of this conjecture are proven in </p> <p>[2] Drutu, Mozes, Sapir, Divergence in lattices in semisimple Lie groups and graphs of groups. Trans. Amer. Math. Soc. 362 (2010), no. 5, 2451–2505.</p> <p>Thus, such non-uniform lattices (at least conjecturally) are of neither type I nor II (in the sense of [1]), so [1] does not apply and, at this point (I think) no other technique is available to handle quasi-isometries of products. However, you should check with Kevin Wortman, since in his work on S-arithmetic lattices and lattices in algebraic groups over functional fields he had to handle similar issues. Thus, there is a chance that QI rigidity for reducible lattices is implicit in his work. </p> <p>Another possible approach would be to generalize [1] using the fact that "higher-dimensional" exponential divergence is now known for non-uniform lattices of higher rank. </p> http://mathoverflow.net/questions/90130/groups-quasi-isometric-to-reducible-nonuniform-lattices/90179#90179 Answer by Alex Eskin for Groups quasi-isometric to reducible nonuniform lattices Alex Eskin 2012-03-04T03:20:22Z 2012-03-04T03:37:08Z <p>Here is a partial answer: Suppose $\Gamma = \Gamma_1 \times \dots \times \Gamma_n$ and all the $\Gamma_i$ are irreducible lattices in $G_i$, where each $G_i$ has real rank at least two. </p> <p>It has been a long time, and I do not remember all the details, but I think it may be true that any quasi-isometry from a product of such lattices $\Gamma_1 \times \dots \times \Gamma_n$ to itself preserves the factors (up to permutation). I am looking at Lemma 10.3 of my paper in JAMS from 1998 <a href="http://www.math.uchicago.edu/~eskin/sl3z.ps" rel="nofollow">http://www.math.uchicago.edu/~eskin/sl3z.ps</a>. It is stated for irreducible lattices, but that does seem to be used in the proof. Of course I could be missing something.</p> <p>If self quasi-isometries are indeed factor preserving, then one has the same classification as for irreducible lattices.</p> <p>One more comment: the reason my proof fails when you have a factor $\Gamma_i$ in a real rank one group $G_i$ is that I quote Lubotsky-Mozes-Raghunathan which does not work in that case. </p>