# Existence of lattices in reductive Lie groups

What is known about existence of lattices in reductive Lie groups? The best results I know about existence of lattices in connected Lie groups are either about semisimple groups or nilpotent groups but never about "mixed" cases like reductive groups.

Let me be a bit more precise. Let G be a reductive Lie group. For me, this means that G is a connected Lie group, and the Lie algebra of G is abelian + semisimple (direct sum as Lie algebras).

Question: Does every such G have a lattice? If not, what is a counterexample?

I am actually only interested in the following restricted case where additionally

1. the semisimple Levi factor of G is dense but not closed in G,
2. the Lie algebra of the semisimple Levi factor consists entirely of real rank one simple summands (at least two summands) of the type $\mathfrak{su}(n,1)$ where $n \geq 1$ (and $n$ may vary with each summand).

The following group G is such an example, and I do not know if it has a lattice. Let $H$ be the universal covering group of $SL(2,\mathbb R) \approx SU(1,1)$, let $z$ be a generator of the center of $H$, let $\alpha$ be an irrational number, and consider the following discrete central subgroup $D$ of $H \times H \times \mathbb R$: $$D = \{ ( z^m , z^n , - m - \alpha n) |\ m,n \text{ integers} \}.$$ Let G be the quotient group $(H \times H \times \mathbb R)/D$.

• Let $\Gamma\subset H$ be a uniform lattice containing the center, then it seems to me that $\Sigma=(\Gamma\times\Gamma\times \{1\})D$ is a uniform lattice in $G$. – user1688 Mar 31 '16 at 16:25
• Your example $(H \times H \times \mathbb{R})/D$ does have a (noncocompact) lattice. $SL(2,\mathbb{R})$ has a noncocompact lattice $\Gamma$ that is a free group, so $\Gamma$ lifts to a subgroup $\widetilde\Gamma$ of $H$. Then $(\widetilde\Gamma \times \widetilde\Gamma \times \mathbb{Z})D$ is a lattice in $H \times H \times \mathbb{R}$, and it obviously contains $D$, so it factors down to a lattice in $(H \times H \times \mathbb{R})/D$. – Dave Witte Morris Apr 1 '16 at 1:26
• @Anton, your subgroup contains a dense subgroup of $1 \times 1 \times \mathbb{R}$ and is therefore not a lattice. – Dave Witte Morris Apr 1 '16 at 1:27
• The special case of interest to the OP can be restated as the following question: Is there a lattice in $SU(m,1) \times SU(n,1)$ that lifts to a subgroup of the universal cover? I do not know the answer to this cohomology question. – Dave Witte Morris Apr 1 '16 at 3:19
• A comment on terminology here is that "reductive Lie group" doesn't seem like useful terminology, since a Lie group with an abelian Lie algebra might not be at all what most people think of as "reductive" in the Borel-Tits sense for algebraic groups. Borel and others did attempt to pin down what might be meant by a "real reductive Lie group", but not just in terms of the Lie algebra. – Jim Humphreys Apr 1 '16 at 13:54

My recollection is that not every reductive group has a lattice, and an example is provided by a slight modification of your construction. (I think this is essentially in an old paper of Starkov.) Let $H$ be a simple Lie group of large real rank (maybe greater than 1 is enough) whose center is $\mathbb{Z}$, choose an irrational $\alpha$, let $$D = \{(m, \alpha m + n) \mid m,n \in \mathbb{Z} \} \subset \mathbb{Z} \times \mathbb{R} ,$$ and let $G = (H \times \mathbb{R})/D$. Then $G$ does not have a lattice.
Suppose $\Lambda$ is a lattice in $G$. Let $\overline H = H / \mathbb{Z}$ and let $\widetilde\Lambda$ be the inverse image of $\Lambda$ in $H \times \mathbb{R}$, so $\widetilde\Lambda$ is a lattice in $H \times \mathbb{R}$ that contains $D$. A well-known theorem of Auslander implies that the image $\overline\Lambda$ of $\Lambda$ in $\overline H$ is a lattice, and my recollection (an expert on cohomology of arithmetic groups can correct me) is that a theorem of Borel says that higher real rank implies the restriction $H^2(\overline H ; \mathbb{R}) \to H^2(\overline\Lambda ; \mathbb{R})$ is injective. Applying this to the cocycle corresponding to the extension $1 \to \mathbb{Z} \to H \to \overline H \to 1$ tells us that $[\widetilde\Lambda, \widetilde \Lambda]$ contains a finite-index subgroup of $\mathbb{Z} \times 0$. So $\widetilde\Lambda \supseteq [\widetilde\Lambda, \widetilde \Lambda] D \supseteq (k\mathbb{Z} \times 0)D$, which contains a dense subgroup of $1 \times \mathbb{R}$. This contradicts the fact that $\widetilde\Lambda$ is discrete.