# Stabilizer of a lattice $\Gamma \subset G$ in the group of automorphisms $\operatorname{Aut}(G)$ is always discrete?

$$\DeclareMathOperator\Aut{Aut}$$Let $$G$$ be a (simply) connected Lie group whose semisimple part has no compact factors, and let $$\Gamma$$ be a lattice (uniform?) in $$G$$. Is it true that the stabilizer $$\operatorname{Stab}_{\Aut(G)} (\Gamma)$$ of $$\Gamma$$ in the group of automorphisms $$\Aut(G)$$ is always discrete? In some special cases this is true (R. Mosak, M. Moskowitz. Stabilizers of Lattices in Lie Groups, Journal of Lie Theory Volume 4 (1994) 1–16).

• You have to assume that $G$ has discrete center, otherwise the claim is clearly false. For semisimple groups the claim holds due to Zariski density of lattices. Commented Jul 10 at 13:27
• We are talking about arbitrary Lie groups. For Abelian Lie groups, the statement I indicated is true (we are not talking about the normalizer of a lattice in a Lie group, but about the stabilizer of this lattice in an automorphism group of this Lie group!). For semisimple Lie groups, the statement I indicated is almost obvious. Commented Jul 10 at 15:08
• Oh, I see, I misread the question. Commented Jul 10 at 15:24

The Lie group $$G$$ is the universal covering for the group $$E_2$$ of orientation-preserving motion of a two-dimensional Euclidean plane. This Lie group is three-dimensional, solvable and has the form $${\bf R} \cdot {\bf R}^2$$. The lattice $$\Gamma = {\bf Z} \cdot {\bf Z}^2$$. Here the stabilizer is isomorphic to $${\bf R}^2$$ and therefore non-discrete.
• Ah, also $\mathbf{Z}^n$ in $\mathrm{Isom}^+(\mathbf{R}^n)=\mathbf{R}^n\rtimes\mathrm{SO}(n)$ works, for the same reason ($n\ge 2$).