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


*

*the semisimple Levi factor of G is dense but not closed in G,

*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 $(H \times H \times \mathbb R)/D$.
 A: 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.
