Fixed directions and Zariski density of hyperbolic groups

Question

It is a fact that if $\Lambda$ is a nonelementary subgroup of ${\rm PSL_2}(\mathbb{C})$ which contains an hyperbolic transformation and moreover ${\rm tr}(g)\in\mathbb{R}/\pm 1$ for all $g\in\Lambda$ then in fact $\Lambda$ is a Fuchsian subgroup, i.e. contained in a conjugate of ${\rm SL}_2(\mathbb{R})$. The proof I am aware of for this is rather unenligthening since it proceeds by simple algebraic manipulations on $2\times 2$ matrices.

Is there a more geometric proof of this? The main motivation for asking this is the following question, which is a tentative generalization of the above fact to higher dimensions: can a (torsion-free, say) nonelementary subgroup $\Lambda$ of ${\rm SO}(n,1)^\circ$ ($n$ odd) satisfy the condition that any of its semisimple elements has an holonomy (i.e. rotational part) which fixes a subspace of codimension $\ge 2$ in $\mathbb{R}^n$ and be Zariski-dense (if no, the proof would likely proceed by showing that it must preserve a totally geodesic subspace)?

For lattices there is enough holonomies so that the condition above is never satisfied: actually the holonomies becomes equidistributed in ${\rm SO}(n)$ as one goes through the hyperbolic conjugacy classes, by a result of P. Sarnak and M. Wakayama.

Edit: As remarked in Misha's answer the question was carelessly phrased for even-dimensional hyperbolic spaces; I added the oddness condition in the original formulation, for an even dimension $n$ the question becomes 'can a Zariski-dense subgroup of ${\rm SO}(n,1)$ contain only elements which fix a subspace of dimension $\ge 3$ in $\mathbb{R}^{n+1}$?'

Answer 1

I will start with few observations; along the way, I will correct your question. Let $g\in SO(n,1)$ be an orientation-preserving loxodromic element. Let $U_g\in SO(n)$ denote the rotational part of $g$. Then $U_g$ always has a nonzero fixed vector in ${\mathbb R}^n$. Furthermore, if $n$ is even then it follows that the dimension of the fixed-point set of $U_g$ is $\ge 2$. Define $\delta_n=1$ if $n$ is odd and $\delta_n=2$ if $n$ is even. Let's call $g$ "strictly loxodromic" if dimension of the fixed point set of $U_g$ is exactly $\delta_n$. Clearly, $SO(n,1)$ contains orientation-preserving strictly loxodromic elements.

Let $\rho=\rho_g: {\mathbb Z}\to SO(n,1)$ denote the representation sending the generator to loxodromic $g$. Then dimension of the fixed-point set of $U_g$ is nothing but $b^0_\rho= dim H^0({\mathbb Z}; R^{n,1}_\rho)$. Now, it is a general fact that $$ \{\rho: {\mathbb Z}\to SO(n,1): b^0_\rho\ge m\} $$ is an algebraic subvariety of $Hom({\mathbb Z}, SO(n,1))=SO(n,1)$. The same holds if you consider $$ dim H^i(\Gamma, V_\rho) $$ where $G$ is an algebraic group, $\Gamma$ is a finitely-generated group and $V$ is a finite-dimensional linear algebraic representation of $G$ (just $Hom(\Gamma, G)$ will be more complicated in this case). In our setting, this is especially easy to see since $$ b^0_{\rho_g}= dim Ker (g-I), $$ where I regard $g$ as a matrix in $SO(n,1)$.

In particular, $$ \{g\in SO(n,1): b^0_{\rho_g}\ge \delta_n+1\} $$ is a (proper) algebraic subvariety. In particular, if $\Lambda < SO(n,1)$ is a Zariski dense torsion-free subgroup, then it cannot consist entirely of elements with $b^0_{\rho_g}\ge \delta_n+1$. In other words, such $\Lambda$ always contains a strictly loxodromic element, thus, answering the corrected version of your question.

Edit: As for your request for a geometric proof of exietence of loxodromic elements, see for instance Lemma 3.24 here. Lastly, the existence of axial isometries of the symmetric space $G/K$ in a Zariski dense subgroup of a (real) semisimple Lie group $G$ is the first step in the proof of the Tits' alternative. Any proof of TA you find (and there are many by now) will contain an explanation generalizing the "unenlightening matrix manipulation".