Counting orbits of certain infinitely generated non-sharp discontinuous groups for the anti-de Sitter space

Abstract

Inspired by an example of Guéritaud and Kassel (Geom Topol 21(2):693–840, 2017), we construct a family of infinitely generated discontinuous groups $\Gamma$ for the 3-dimensional anti-de Sitter space $\textrm{AdS}^{3}$. These groups are not necessarily sharp (a kind of “strong” proper discontinuity condition introduced by Kassel and Kobayashi (Adv Math 287:123–236, 2016), and we give its criterion. Moreover, we find upper and lower bounds of the counting $N_{\Gamma }(R)$ of a $\Gamma$-orbit contained in a pseudo-ball B(R) as the radius R tends to infinity. We then find a non-sharp discontinuous group $\Gamma$ for which there exist infinitely many $L^2$-eigenvalues of the Laplacian on the noncompact anti-de Sitter manifold $\Gamma \backslash \textrm{AdS}^{3}$, by applying the method established by Kassel–Kobayashi. We also prove that for any increasing function f, there exists a discontinuous group $\Gamma$ for $\textrm{AdS}^{3}$ such that the counting $N_{\Gamma }(R)$ of a $\Gamma$-orbit is larger than f(R) for a sufficiently large R.

A Proof of the formula (1.3)

The formula (1.3) is probably well known, but we give a proof for the reader’s convenience. It suffices to show the following proposition:

Proposition A.1

Let $\Gamma$ be a discrete group of isometries of a complete Riemannian manifold X. We write B(x; R) for the ball of radius R centered at $x\in X$. Fix a point $x_{0}\in X$. Then,

$$\begin{aligned} \forall x\in X,\ \exists c>0,\ \#(\Gamma x\cap B(x_{0}; R))\le \frac{\textrm{vol}(B(x_{0}; R+c))}{\textrm{vol}(B(x; c))}. \end{aligned}$$

Proof

Let $x\in X$. We denote by $d_{X}(\cdot ,\cdot )$ the distance of X, and set $3c:=\inf \{d_{X}(x,y)\mid y\in \Gamma x\smallsetminus \{x\}\}$. By [4, Ch. IV, Thm. 2.2], the orbit $\Gamma x$ of the discrete group $\Gamma$ is discrete, hence we have $c>0$. By the triangle inequality, the inclusion relation

$$\begin{aligned} \bigcup _{y\in \Gamma x \cap B(x_{0}; R)} B(y; c) \subset B(x_{0}; R+c) \end{aligned}$$

(5.4)

holds. We note that the left hand side of (5.4) is a disjoint union by the definition of c. Since $\textrm{vol}(B(y; c))=\textrm{vol}(B(x; c))$ for any $y\in \Gamma x$, we obtain the desired inequality taking the volumes of the both hand sides of (5.4). $\square$