# Groups quasi-isometric to reducible nonuniform lattices

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".

• If you replace "semisimple Lie group" by "automorphism group of a product of trees", very exotic phenomena may occur; e.g. $F_2\times F_2$ is a cocompact lattice, and so are the Burger-Mozes groups (which are simple). Clearly they are not commensurable. (BTW, this example also shows that linearity is not a q.-i. invariant). – Alain Valette Mar 3 '12 at 20:11
• Alan makes a very good point: You cannot allow more than one factor in your Lie group which is locally isomorphic to $SL(2, {\mathbb R})$. Otherwise, your lattice $\Gamma$ will have two factors commensurable to $F_2$ and one can hardly make any conclusions. However, once you exclude (more than two) $SL(2, {\mathbb R})$ factors and higher rank factors, the Burger-Moses phenomenon does not occur and you get QI rigidity for $\Gamma$. – Misha Mar 3 '12 at 21:26
• Thank you, Alain and Misha. From what you say it seems even in the presence of two $SL(2,\mathbb R)$ factors one still can draw some conclusions; after all not all groups are lattices in the product of trees. – Igor Belegradek Mar 4 '12 at 5:31

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.

• I do not see how Lemma 10.3 implies that the factors are preserved. Proposition 10.1 does imply that, but it uses irreducibility. – Igor Belegradek Mar 4 '12 at 17:21
• The way I read Lemma 10.3 is as follows: Suppose $G = G_1 \times G_2$. If two points $x$ and $y$ in the thick part of $G/Gamma$ have the same projection to $G_1$, then (up to a bounded error) their images have the same projection to $G_1$. Is this enough for factor preserving? – Alex Eskin Mar 5 '12 at 0:01
• I looked over most of the paper. It seems that Lemma 10.3 is not needed; in fact the proof of the main theorem 0.2 carries over to the case where all G_i have higher rank with virtually no modifications. (The stuff in section 10 is not used in the proof of theorem 0.2). I am not sure what happens if you have a product with both rank one and higher rank factors. That case seems open, but I think it should be doable. (You should also ask Kevin Wortman). – Alex Eskin Mar 5 '12 at 13:55
• Many thanks! I just wanted to know the answer, as the issue came up in the paper I am writing, but I am not planning to work on this further. Sounds like a good project for a student. – Igor Belegradek Mar 5 '12 at 16:22

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

 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 . 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

 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 ), so  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  using the fact that "higher-dimensional" exponential divergence is now known for non-uniform lattices of higher rank.