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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".

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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

2 Answers 2

up vote 5 down vote accepted

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.

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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

[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.

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Thank you, this is very helpful. –  Igor Belegradek Mar 4 '12 at 5:24

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