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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.

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up vote 6 down vote accepted

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.

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Alex: Why can't you have \Gamma \cap N a Klein bottle group and/or \Gamma / (\Gamma \cap N) the infinite dihedral group? – Daniel Groves Aug 2 '11 at 23:11
Daniel: I am probably missing the point. I was thinking N is R^2 (where the group operation is vector addition), and \Gamma \cap N is a lattice in R^2, so it must be Z^2. Also \Gamma/(\Gamma \cap N) is a lattice in R, so it should be Z. – Alex Eskin Aug 3 '11 at 0:38
Oh, sorry, I was thinking of lattice in Isom(R^2), etc. Sorry. – Daniel Groves Aug 3 '11 at 1:46
The reason I was confused is that 3-manifolds that admit SOL geometry needn't have fundamental groups which are lattices in SOL (since SOL is only the connected component of 1 in Isom(SOL)). So I was thinking of these SOL 3-manifolds, which don't have the form you claim (but have a finite-index subgroup which does). So, I'm afraid I didn't read your answer carefully enough... – Daniel Groves Aug 3 '11 at 1:54
That is nice to know, thanks! – Alex Eskin Aug 3 '11 at 2:10

See the abstract is here:

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@Igor: it may be friendlier to link to the abstract as well as (or instead of) the PDF. – José Figueroa-O'Farrill Aug 2 '11 at 3:43
True. I didn't look, I was thinking I WAS linking to the abstract, will fix. – Igor Rivin Aug 2 '11 at 13:24
OK, fixed. (and another five characters) – Igor Rivin Aug 2 '11 at 13:27

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