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I perfectly understand that the Milnor-Schwarz lemma tells me that cocompact lattice in semisimple Lie groups of higher rank are not hyperbolic (in the sense of Gromov).

But do there exist noncocompact lattices in higher rank semisimple Lie groups which are hyperbolic? (wikipedia is saying "no" without any reference, but I do not trust that article since it contains mistakes...)

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up vote 4 down vote accepted

If your lattice is not irreducible, it contains $\mathbf{Z}^2$ and you're done. Otherwise, Margulis' arithmeticity and Lubotzky-Mozes-Raghunathan imply, for a non-cocompact lattice, that it contains exponentially distorted elements and this is not possible for a hyperbolic group. Edit: Alternatively, in the irreducible case, you can use Kazhdan-Margulis's theorem that quotients of the lattices by non-central subgroups are finite. In contrast, non-elementary hyperbolic groups have plenty of quotients by infinite subgroups that are infinite. Anyway, both approaches make use of pretty non-trivial results.

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You asked a related question in another post, and you erased the question while I was posting an answer. So I post it here. The question was complementary to the one above so I think it's relevant to include the answer: why are non-uniform lattices in rank 1 symmetric spaces of noncompact type not hyperbolic except in the case of the hyperbolic plane? Here is my answer.

I think it's a result of Garland and Raghunathan (Annals 1970 "Fundamental domains..."). They show that given such a lattice, for some point $\omega$ at infinity, the stabilizer of $\omega$ in the lattice acts cocompactly on each horosphere based at $\omega$. This horosphere is modeled on a $(n-1)$-dimensional simply connected nilpotent Lie group, where $n$ is the real dimension of the rank 1 symmetric space of non-compact type. Thus the nonuniform lattice contains a f.g. nilpotent group of Hirsch length $n-1$. This is possible in a hyperbolic group only if $n\le 2$. (Note: if $n\ge 3$, it follows that the lattice contains a free abelian group of rank 2.) More precise results about the structure of these lattices were formalized into the concept of relatively hyperbolic groups, see Gromov, Farb, etc. They indicate that intuitively, these "peripheral" subgroups are the only obstruction to hyperbolicity.

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Thanks for answer. The reason why I erased the question was that I had figured out the following argument, which seems to be very similar to yours: Let $\Gamma$ be any such lattice. Pick a cusp of $\Gamma$ and fix an Iwasawa decomposition $G=NAK$ of the isometry group $G$ of the space. Using the Garland-Raghunathan result on fundamental domains of Siegel set type, we see that $\Gamma \cap NAM = \Gamma\cap NM$ must be discrete. If $\dim NM > 1$, this implies that $\Gamma$ contains a $\mathbb Z^2$ and hence $\Gamma$ cannot be hyperbolic. Now, $\dim NM = 1$ if and only if $G=SL(2,\mathbb R)$. – Jonas Schulz Nov 6 '12 at 10:00

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