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A Riemannian manifold $(X,g)$ is called Hadamard manifold if it is complete and simply connected and has everywhere non-positive sectional curvature. An example of such a manifold would be the hyperbolic spaces. A group $G$ of isometries of a Hadamard manifold $X$ is elementary if it is discrete and without torsion, fixes a point $x \in X(\infty)=$ ideal boundary of $X$ [1], and such that it is of one of the following two types:

a) $G$ leaves a geodesic $\gamma$ in $X$ emanating from $x$ invariant.

b) $G$ leaves Busemann functions associated to $x$ invariant.

I am interested in the surfaces, which are the quotient of the ideal boundary of the hyperbolic plane by the elementary group of isometries, but I do not have explicit examples in hand.

Question: What are some specific examples of elementary group of isometries in the above particular case, and what would be the nature of the resulting quotient surface, i.e., the existence of cusps, finite area, etc.?

Reference

[1] What is the definition of ideal boundary?

Thank you.

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    $\begingroup$ I suppose the prime examples fixing the point at infinity in the upper half-plane model of $\mathbb{H}^2$ are the infinite cyclic groups generated by the horocyclic translation $z \mapsto z + 1$ (should fall under definition 2, but I'd have to give Busemann functions a proper think to say for sure) and the loxodromic isometry $z \mapsto 2z$, which preserves the geodesic line $\Re z = 0$. In both cases, the quotient surface is topologically an open annulus and has infinite area; in the former it has one cusp. $\endgroup$
    – Robbie Lyman
    Commented Mar 27, 2023 at 18:37
  • $\begingroup$ More to the point, if $\Sigma$ is a finite-area hyperbolic surface (without boundary), which we identify with $\Gamma\backslash\mathbb{H}^2$ for some discrete, torsion-free subgroup $\Gamma \le \operatorname{Isom}(\mathbb{H}^2)$, then $\Gamma$ is never elementary. I won't attempt to prove this. $\endgroup$
    – Robbie Lyman
    Commented Mar 27, 2023 at 18:42

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