Groups quasi-isometric to reducible nonuniform lattices - MathOverflow

Groups quasi-isometric to reducible nonuniform lattices

2012-03-03T17:50:17Z

It is known that a finitely group $G$ is quasi-isometric to a nonuniform irreducible lattice $\Lambda$ in a semisimple Lie group if and only if $G$ and $\Lambda$ are commensurable (see references in this survey of Farb).

Question. What is known about groups quasi-isometric to reducible nonuniform lattices in semisimple Lie groups?

As usual in this business "semisimple" means "noncompact, connected, semisimple, with finite center".

Answer by Misha for Groups quasi-isometric to reducible nonuniform lattices

2012-03-03T20:33:26Z

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

[1] Kapovich, Kleiner, Leeb, Quasi-isometries and the de Rham decomposition, Topology 37 (1998), no. 6, 1193–1211.

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

[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.

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.

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.

Answer by Alex Eskin for Groups quasi-isometric to reducible nonuniform lattices

2012-03-04T03:20:22Z

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.

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 http://www.math.uchicago.edu/~eskin/sl3z.ps. It is stated for irreducible lattices, but that does seem to be used in the proof. Of course I could be missing something.

If self quasi-isometries are indeed factor preserving, then one has the same classification as for irreducible lattices.

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.