On ends of finite-volume noncompact manifolds of nonpositive curvature

Abstract

In this paper we confirm a folklore conjecture which suggests that for a complete noncompact manifold $M$ of finite volume with sectional curvature $-1 \leq K \leq 0$, if the universal cover of $M$ is a visibility manifold, then the fundamental group of each end of $M$ is almost nilpotent.

Appendix: A visibility manifold with a finite volume quotient is not necessarily Gromov hyperbolic

In this appendix we construct a complete surface of finite volume with curvature $-1 \leq K <0$, whose universal cover is a visibility manifold but not a Gromov hyperbolic space. This example shows that the argument in the proof of Corollary 3.5 is not adapted to general visibility manifolds.

Let $(S,g)$ be a noncompact surface of constant negative curvature −1 with finite volume. Such a surface has only finitely many cusps. For simplicity we further assume that $S$ has only one cusp $E$, which can be expressed as $\mathbb{S}^{1} \times [0,\infty )$ endowed with the hyperbolic metric $\exp (-2t) \mathrm{d}\tilde{s}^{2}+\mathrm{d}t^{2}$, where $\mathrm{d}\tilde{s}^{2}$ is the flat metric on $\mathbb{S}^{1}$.

Let $h:[0,\infty ) \to \mathbb{R}$ be a smooth function such that

• (1) $h$ is positive and monotonically decreasing,

• (2) $h''/h$ is positive and monotonically decreasing,

• (3) $h=\exp (-t)$ for $0 \leq t \leq 1$,

• (4) $h=\dfrac{1}{t^{2}}$ for $t \geq 3$.

Such a function could be constructed by elementary calculus.

We change the metric on $E=\mathbb{S}^{1} \times [0,\infty )$ to $h^{2}(t) \mathrm{d}\tilde{s}^{2} + \mathrm{d}t^{2}$ and obtain a new smooth metric $g'$ on $S$. Since both $\mathbb{S}^{1}$ and $[0,\infty )$ are complete, their warped product is also complete (e.g., see [5, Lemma 7.2]). We first show that $(S,g')$ is a finite volume surface with bounded nonpositive curvature whose universal cover is a visibility manifold.

Let $S'=S - (\mathbb{S}^{1} \times (3,\infty ))$. $S'$ is a compact surface with boundary. The volume of $S$ with respect with the new metric $g'$ is

$$\begin{aligned} \textrm{Vol}_{g'}(S) =& \textrm{Vol}_{g'}(S')+\textrm{Vol}_{g'}( \mathbb{S}^{1} \times (3,\infty )) \\ =& \textrm{Vol}_{g'}(S') + \int _{0}^{1} \int _{3}^{\infty }h(t) \, \mathrm{d}s \mathrm{d}t \\ =& \textrm{Vol}_{g'}(S') + \int _{3}^{\infty }\dfrac{1}{t^{2}} \, \mathrm{d}t \\ =& \textrm{Vol}_{g'}(S') + \dfrac{1}{3}< \infty . \end{aligned}$$

To see that $(S,g')$ is visible, we check that it satisfies a visibility criterion due to Eberlein and O’Neill [19, Proposition 5.9], which states that if a nonpositively curved manifold $M$ has curvature order $\leq 2$ at a point $p \in M$, i.e., if

$$ \int _{1}^{\infty }|k(\gamma _{\omega}(t))| t \,\mathrm{d}t = \infty \textrm{ for all } \omega \in \mathrm{S}_{p} M, $$

where $\gamma _{\omega}$ is the geodesic ray with initial velocity $\omega$ and $k(\gamma _{\omega}(t))=\min \{|K(\pi )|: \pi \subset \mathrm{T}_{ \gamma _{\omega}(t)} \text{ is a two-dimensional subspace}\}$. Then $M$ is a visibility manifold.

In fact, direct computation gives

$$ K_{g'}(q)= \textstyle\begin{cases} \;\;-1 & \text{for } q \in S \setminus E \\ -\dfrac{h''(t)}{h(t)} & \text{for } q \in \mathbb{S}^{1} \times \{t\} \subset E \end{cases} $$

where $K_{g'}(q)$ is the Gaussian curvature at $q$ with respect with the metric $g'$. By the choice of $h$, $-1 \leq K_{g'}<0$. We have for any $p \in \mathbb{S}^{1} \times \{0\}$ and for any $\omega \in \mathrm{S}_{p} S$,

$$\begin{aligned} \int _{1}^{\infty }|K_{g'}(\gamma _{\omega}(t))| t \,\mathrm{d}t \geq & \int _{1}^{\infty }\dfrac{h''}{h}(t) t \,\mathrm{d}t \\ \geq & \int _{3}^{\infty }(\dfrac{1}{t^{2}})''/(\dfrac{1}{t^{2}}) \cdot t \, \mathrm{d}t \\ =& \int _{3}^{\infty }\dfrac{6}{t} \,\mathrm{d}t = \infty , \end{aligned}$$

hence by the visibility criterion the universal cover of $(S,g')$ is a visibility surface.

It remains to show that $(S,g')$ is not Gromov hyperbolic. Let $\tilde{S}$ be the universal cover of $(S,g')$. Assume now that $\tilde{S}$ is $\delta$-hyperbolic for some $\delta >0$. Let $\gamma$ be a geodesic ray in $S$ converging to the end $E$ and $\tilde{\gamma}$ be a lift of $\gamma$ to $\tilde{S}$. It is not hard to see that a horoball in $\tilde{S}$ centered at $\tilde{\gamma}(\infty )$ is given by $\mathbb{R} \times [0,\infty )$ endowed with the metric $h^{2}(t)\mathrm{d}s^{2}+\mathrm{d}t^{2}$. Denote by $H_{-t}$ the horosphere $\mathbb{R} \times \{t\}$ and by $d_{H}$ the horospherical distance between two points on the same horosphere.

Let $T>3+12 \delta$ be a sufficiently large constant. Set $x_{0}=(0,T)$ and $y_{0}=(2 \delta T^{2},T)$. We have

$$ d_{H}(x_{0},y_{0})=\int _{0}^{2 \delta T^{2}} \dfrac{1}{T^{2}} \, \mathrm{d}s= 2\delta . $$

Where $d_{H}(\cdot , \cdot )$ is the induced distance on the horosphere $H$.

Let $x=(0,T-2\delta )$ and $y=(2 \delta T^{2},T-2\delta )$. The projection of the geodesic segment $\gamma _{x,y}$ onto the horosphere $H_{-T}$ is exactly the horospherical geodesic segment from $x_{0}$ to $y_{0}$. Together with a standard infinitesimal argument, this yields that

$$ d(x,y) > d_{H}(x_{0},y_{0})=2\delta . $$

(40)

On the other hand,

$$ d(x,y) \leq d_{H}(x,y) =\dfrac{2\delta T^{2}}{(T-2\delta )^{2}}< 3 \delta . $$

Set $w=(0,T-8\delta )$ and $z=(2 \delta T^{2},T-8\delta )$, we have

$$ d(x,w)=d(y,z)=6 \delta \geq 2d(x,y) $$

and

$$ d(z,w) \leq d_{H}(z,w)=\int _{0}^{2 \delta T^{2}} \dfrac{1}{(T-8\delta )^{2}} \, \mathrm{d}s= 2\delta \cdot \dfrac{T^{2}}{(T-8\delta )^{2}}. $$

Together with (40) we obtain

$$\begin{aligned} d(z,w)-d(x,y) \leq & 2\delta \cdot \dfrac{T^{2}}{(T-8\delta )^{2}}-2 \delta \\ =& 32 \delta ^{2} \cdot \dfrac{T- 4\delta}{(T-8\delta )^{2}}, \end{aligned}$$

which tends to 0 as $T \to \infty$. This contradicts Lemma 3.4. Therefore $\tilde{S}$ cannot be Gromov hyperbolic. See Fig. 2.

Fig. 2

$d(z,w)$ and $d(x,y)$ can be arbitrarily close

Hence we have constructed a complete visibility surface $S$ with bounded nonpositive curvature and finite volume whose universal cover is not Gromov hyperbolic. Examples in higher dimensions can be obtained similarly using warped products.