Lattices in SOL - MathOverflow

Lattices in SOL

2011-08-02T02:36:11Z

Consider a semi-direct product $\mathbb{Z}^2\rtimes_A\mathbb{Z}$, where $A\in SL_2(\mathbb{Z})$ and $|Tr(A)|>2$. It is clear that it is isomorphic to a lattice in the 3-dimensional solvable Lie group SOL. To what extent do these examples exhaust lattices in SOL? (i.e., up to a suitable equivalence relation, is every lattice in SOL of this form?)

The question comes from a desire to understand better the Eskin-Fisher-Whyte result on quasi-isometric rigidity of SOL: every finitely generated group quasi-isometric to SOL is virtually a lattice in SOL.

Answer by Igor Rivin for Lattices in SOL

2011-08-02T03:08:53Z

See http://arxiv.org/pdf/1106.4646 the abstract is here:

http://arxiv.org/abs/1106.4646

Answer by Alex Eskin for Lattices in SOL

2011-08-02T17:18:43Z

To add to Igor Rivin's answer: it seems that all the lattices in SOL are isomorphic as abstract groups to $\mathbb{Z}^2\rtimes_A\mathbb{Z}$ for hyperbolic $A\in SL_2(\mathbb{Z})$. If I am reading the paper correctly, it is in Theorem 2.1 of the paper linked in Igor's answer.

I think that this fact can also be easily derived from the following theorem (Corollary 3.5 in Raghunathan's book, which is due to either Auslander or Mostow):

If $G$ is a connected solvable Lie group, and $N$ is its maximum connected (normal) closed nilpotent Lie subgroup, then for any lattice $\Gamma$ in $G$, $\Gamma \cap N$ is a (cocompact) lattice in $N$.

Thus you always have the short exact sequence $$1 \to \Gamma \cap N \to \Gamma \to \Gamma/(\Gamma \cap N) \to 1.$$ The fact that $\Gamma \cap N$ is cocompact in $N$ implies that $\Gamma/(\Gamma \cap N)$ is a discrete subgroup of $G/N$.

If $G = SOL = \mathbb{R}^2 \rtimes \mathbb{R}$, then $N \approx \mathbb{R}^2$, and so $\Gamma \cap N \approx \mathbb{Z}^2$. Also since $G/N \approx \mathbb{R}$, $\Gamma/(\Gamma \cap N) \approx \mathbb{Z}$. So the short exact sequence above reads $$1 \to \mathbb{Z}^2 \to \Gamma \to \mathbb{Z} \to 1.$$ Such a sequence must split, so $\Gamma$ is a semidirect product.

The linked paper does something much more detailed and impressive, sort of like the classification of crystallographic groups.