A uniform lattice in a locally compact group $G$ is a discrete subgroup $\Gamma\subset G$ such that $G/\Gamma$ is compact.
My question is whether a uniform lattice exists in the group
$$
G={\mathbb R}^2\rtimes SL\_2({\mathbb R}).
$$
This group is unimodular, so an obvious criterion is satisfied. It also admits a lattice (=discrete subgroup of finite covolume), for example the group of integer valued points ${\mathbb Z}^2\rtimes SL\_2({\mathbb Z})$, but this is not cocompact.



Uniform lattice in semidirect product
No, there's no uniform (=cocompact) lattice in $\mathbf{R}^n\rtimes\mathrm{SL}\_n(\mathbf{R})$ for any $n\ge 2$. Up to the action by automorphisms of $\mathrm{GL}\_n(\mathbf{R})$, all lattices are contained in $\mathbf{R}^n\rtimes\mathrm{SL}\_n(\mathbf{Z})$, which is not cocompact.


Indeed, let $\Gamma$ be a lattice. In a connected Lie group, the intersection of a lattice with the amenable radical is a lattice in the amenable radical. So $\Gamma\cap\mathbf{R}^n$ is a lattice in $\mathbf{R}^n$; hence modulo a global automorphism induced by some element of $\mathrm{GL}\_n(\mathbf{R})$, we can assume that $\Gamma\cap\mathbf{R}^n=\mathbf{Z}^n$. The subgroup $\Gamma$ is then contained in the normalizer of $\Gamma\cap\mathbf{R}^n=\mathbf{Z}^n$, and this normalizer is equal to $\mathbf{R}^n\rtimes\mathrm{SL}\_n(\mathbf{Z})$.




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(Edit; added to reflect the comments:) On the other hand, $\mathbf{R}^3\rtimes\mathrm{SL}\_2(\mathbf{R})$ admits cocompact lattices. Indeed, let $q=X^2+Y^2-7Z^2$. Let $G(q)(K)=K^3\rtimes\mathrm{SO}(q)(K)$. Then $q$ is $\mathbf{Q}$-anisotropic; so $G(\mathbf{Z})$ is a cocompact lattice in $G(\mathbf{R})=\mathbf{R}^3\rtimes \mathrm{SO}(q)(\mathbf{R})$. Pulling back, we get a cocompact lattice in $\mathbf{R}^3\rtimes \mathrm{SL}\_2(\mathbf{R})$.


