6
$\begingroup$

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.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
7
$\begingroup$

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.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Alex: Why can't you have \Gamma \cap N a Klein bottle group and/or \Gamma / (\Gamma \cap N) the infinite dihedral group? $\endgroup$
    – Daniel Groves
    Aug 2, 2011 at 23:11
  • $\begingroup$ 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. $\endgroup$
    – Alex Eskin
    Aug 3, 2011 at 0:38
  • $\begingroup$ Oh, sorry, I was thinking of lattice in Isom(R^2), etc. Sorry. $\endgroup$
    – Daniel Groves
    Aug 3, 2011 at 1:46
  • $\begingroup$ 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... $\endgroup$
    – Daniel Groves
    Aug 3, 2011 at 1:54
2
$\begingroup$

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

http://arxiv.org/abs/1106.4646

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ @Igor: it may be friendlier to link to the abstract as well as (or instead of) the PDF. $\endgroup$
    – José Figueroa-O'Farrill
    Aug 2, 2011 at 3:43
  • $\begingroup$ True. I didn't look, I was thinking I WAS linking to the abstract, will fix. $\endgroup$
    – Igor Rivin
    Aug 2, 2011 at 13:24
  • $\begingroup$ OK, fixed. (and another five characters) $\endgroup$
    – Igor Rivin
    Aug 2, 2011 at 13:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.