2
$\begingroup$

What can be said about a discrete finitely generated subgroup $G$ of $PSL(2,\mathbb C)$ whose nontrivial elements are parabolic? If $G$ is geometrically finite, one can show that $G$ must be elementary so the real question is: Can $G$ be geometrically infinite?

$\endgroup$
4
  • 5
    $\begingroup$ Igor: Such $G$ has to be elementary even if it is not discrete and not finitely generated. You can see this for instance by looking at the Zariski closure of $G$ or by looking at the group generated by high powers of two elements of $G$ which have distinct fixed points. $\endgroup$
    – Misha
    Commented Apr 29, 2013 at 20:54
  • $\begingroup$ Thanks, Misha! I do not understand what you mean by the Zariski closure argument (what's is special about the Zarisky closure of a purely parabolic group?), but the other one I get: the group generated by high powers of parabolic elements with disjoint fixed points at infinity is geometrically finite (it visibly has a fundamental polyhedron with 4 faces), and hence it contains a hyperbolic element. $\endgroup$ Commented Apr 29, 2013 at 22:11
  • 2
    $\begingroup$ Igor: The equation $tr(A)^2=4$ is preserved by taking Zariski closure. Thus, Zariski closure is still purely parabolic. Now, you either looks at the list of algebraic subgroups, or use structure properties of algebraic groups (Levi decomposition etc); alternatively, use Lie-Kolchin theorem on subgroups consisting of unipotent elements. $\endgroup$
    – Misha
    Commented Apr 29, 2013 at 23:50
  • $\begingroup$ The original question was actually a toy case of the same question for purely parabolic isometry groups of Hadamard manifolds. Neither argument extends to this setting, as far as I can see. $\endgroup$ Commented Apr 30, 2013 at 3:48

1 Answer 1

2
$\begingroup$

Suppose that $G$ is a (not necessarily discrete) nonelementary group of isometries of the hyperbolic $n$-space. Then pairs of fixed points of loxodromic elements of $G$ are dense in $L(G)\times L(G)$, where $L(G)$ is the limit set of $G$. See Lemma 3.24 in "Hyperbolic Manifolds and Discrete Groups". I am quite sure that this is also proven in Maskit's book and in Beardon's book. The same argument works for arbitrary convergence groups acting on compacts, e.g. on ideal boundaries of proper geodesic Gromov-hyperbolic spaces.

Just as a curiosity: There are 2-generated infinite purely elliptic (nondiscrete, of course) isometry groups of the hyperbolic $4$-space, but not of the hyperbolic 3-space. The existence is a cute application of the Tits alternative for $SU(2)$.

$\endgroup$
3
  • $\begingroup$ I see. This should also extend to the setting of Karlsson-Noskov paper on isometry groups of spaces with contractive bordifications, such as e.g. visibility spaces. $\endgroup$ Commented Apr 30, 2013 at 12:47
  • $\begingroup$ Yes, It does extend to the Karlsson-Noskov setting. $\endgroup$ Commented Apr 30, 2013 at 14:53
  • 1
    $\begingroup$ Igor: There is also a weak convergence property for group actions on boundaries of CAT(0) spaces (something about $\pi/2$-balls in Tits metric) due to Papasoglu (and maybe Swenson), but I forgot the details. In any case, you should probably ask the question you are actually interested in and state what is known. $\endgroup$
    – Misha
    Commented Apr 30, 2013 at 15:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .