Let $M$ be a compact hyperbolic manifold, and $\tilde M = H^n$ the universal covering. Now let $\Gamma$ be the group of Decktransformations. So we have $\tilde M / \Gamma = M$.
My question: Is it possible, that all $g \in \Gamma$ have one common fixpoint at infinity?
I know that all $g \in \Gamma$ are of hyperbolic type, so they have two fixpoints at infinity. If we look at this in the Poincaré-ball model of the hyperbolic space, we have or each $g \in \Gamma$ two points $p_1$, $p_2 \in \partial B^n$ with $g(p_1) = p_1$ and $g(p_2)=p_2$. So could there exist a point $p \in \partial B^n$ such that $g(p) = p$ for all $g \in \Gamma$?
I think the answer is no, but why?
I think that it is straightforward to show to show that if two hyperbolics have exactly one common fixed point, then the subgroup they generate is not discrete; actually, that is also the case if one of them is a parabolic. Don't take my word for it and check that this is true!
I think you can conclude from there?