1
$\begingroup$

Let $\Delta$ be a Zariski dense finitely generated subgroup of $SL(3,\mathbb{R})$. Assume that $\Delta$ contains no element of finite order. Then, does there exist a finite-order element $A \in SL(3,\mathbb{R})$ such that the group generated by all the elements of $\Delta$ and $A$ contains $\Delta$ as a finite index subgroup? We assume that $A$ is not an identity.

Is there some known conditions for $\Delta$ that the above statement holds?

$\endgroup$
3
  • $\begingroup$ This is already false for Fuchsian subgroups of $SL(2,R)$ (extended by $-I$). If you take holonomy of a generic convex real projective structure on a closed hyperbolic surface, you get $SL(3,R)$ examples. $\endgroup$
    – Misha
    May 7, 2013 at 19:07
  • $\begingroup$ Let $\Delta$ be a holonomy groups of a convex real projective structure on some closed hyperbolic surface. Suppose that there is a finite-order element $A$ that makes the above statement holds. Then must the group generated by $\Delta$ and $A$ be a holonomy group of a convex real projective structure on some hyperbolic orbifold? $\endgroup$
    – kchoi
    May 8, 2013 at 5:09
  • $\begingroup$ Yes, of course. The point is that a finite index extension of a discrete group of projective transformations of a convex domain is still discrete. $\endgroup$
    – Misha
    May 8, 2013 at 5:25

0

Your Answer

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

Browse other questions tagged or ask your own question.