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I am interested in the following question: Can semisimple Lie group of real rank $\geq 2$ contain an abelian lattice?

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    $\begingroup$ To make a short argument: if a locally compact group $G$ has an amenable lattice, then it is amenable as well. If moreover $G$ is (virtually connected) Lie semisimple, this forces $G$ to be compact (and the lattice to be finite). The only facts used are that noncompact semisimple Lie groups are nonamenable (very classical, e.g., they have a discrete nonabelian free subgroup), and that having an amenable lattice $H$ implies $G$ amenable: indeed let $G$ act on a compact convex set $X$; then $H$ fixes a point, and then one can pushforward an invariant probability from $G/H$ to $X$. $\endgroup$ – YCor Mar 27 '19 at 9:57
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No, and it has nothing to do with the higher-rank assumption.

Probably the easiest way to see this is by the Borel Density Theorem, which says that if $G$ is a semisimple Lie group which has no compact factor and it is algebraic, and if $\Gamma$ is a lattice in $G$, then $\Gamma$ is Zariski dense in $G$.

Now, if you have a non-compact semisimple Lie group and a lattice in it, you can mod up the center and get an algebraic group and a lattice in it. Further, the algebraic group is a product of factors and by modding out compact ones you are in the situation described above.

Another easy way to see that a lattice cannot be abelian is by showing that it cannot be amenable. If it was then the enveloping group would be amenable as well, but it is not. Eg it contains a closed free group.

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    $\begingroup$ To complement Uri's answer: if $G$ is an arbitrary locally compact group with an abelian lattice (or more generally a lattice with polynomial growth), then $G$ is amenable. Assume now that $G$ is a Lie group with finitely many components. Then it is known that every lattice in such a group is cocompact. It follows that $G$ has the same growth rate as its lattice, hence has polynomial growth as well. Of course among semisimple real Lie groups, only compact ones have polynomial growth. $\endgroup$ – YCor Jun 22 '17 at 7:31
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No. A lattice in a semisimple Lie group of higher rank has to contain distorted elements of infinite order (a theorem of Lubotzky, Mozes and Raghunathan). On the other hand, every element of infinite order in a finitely generated abelian group is undistorted.

Edit: I missed the assumption that the lattice should be non-uniform (Thanks Yves).

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  • $\begingroup$ Could you please give me references for this theorem? $\endgroup$ – Maria Gerasimova Jun 21 '17 at 21:03
  • $\begingroup$ Lubotzky, Alexander(IL-HEBR-IM); Mozes, Shahar(IL-HEBR-IM); Raghunathan, M. S.(6-TIFR-SM) The word and Riemannian metrics on lattices of semisimple groups. (English summary) Inst. Hautes Études Sci. Publ. Math. No. 91 (2000), 5–53 (2001). $\endgroup$ – Jarek Kędra Jun 21 '17 at 21:09
  • $\begingroup$ For online access: numdam.org/item/PMIHES_2000__91__5_0 $\endgroup$ – Jim Humphreys Jun 21 '17 at 22:58
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    $\begingroup$ Hi, Jarek. You shoot this fly with a cannon! $\endgroup$ – Uri Bader Jun 22 '17 at 5:51
  • $\begingroup$ It's only true in the non-cocompact case. Cocompact lattices in such groups have no distorted element. $\endgroup$ – YCor Jun 22 '17 at 7:02

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