Riemannian Manifolds: Conformally Equivalent But Not Hyperbolic

Nov 25, 2025 by Andrew McMorgan 64 views

Hey Plastik Magazine readers! Ever wondered about the wild world of Riemannian manifolds? Today, we're diving deep into a fascinating question that pops up in differential geometry, metric geometry, complex geometry, and good ol' Riemannian geometry. Specifically, we're asking: Are there any Riemannian manifolds $(M, g)$ that are conformally equivalent to a domain $D$ but aren't Gromov hyperbolic?

Diving into the Question

Okay, let's break this down. First off, a Riemannian manifold $(M, g)$ is essentially a smooth manifold $M$ equipped with a Riemannian metric $g$ , which allows us to measure distances and angles locally. Think of it as a curved space where you can do calculus.

Now, what does it mean for $(M, g)$ to be conformally equivalent to a domain $D$ ? Well, it means there's a smooth map $f: M \rightarrow D$ that preserves angles. More formally, there exists a positive smooth function $\lambda: M \rightarrow \mathbb{R}$ such that $f^*(g_D) = \lambda g$ , where $g_D$ is the metric on $D$ . In simpler terms, $(M, g)$ looks like $D$ but with some local scaling applied.

Finally, Gromov hyperbolicity is a property that captures the idea of negative curvature in a large-scale sense. A Gromov hyperbolic space behaves a lot like a hyperbolic space – triangles are thin, geodesics diverge exponentially, and so on. It's a crucial concept in geometric group theory and related areas.

So, the core question is: Can we find a Riemannian manifold that looks like a nice domain (conformally equivalent) but doesn't have this "negative curvature" property (Gromov hyperbolicity)?

The Denjoy Domain Example

The question mentions a crucial hint: Denjoy domains. A Denjoy domain is typically constructed by removing a Cantor set from a circle in the complex plane. These domains have some funky properties. In particular, there exist Denjoy domains that are not Gromov hyperbolic. This non-hyperbolicity arises from the presence of these Cantor sets, which create "bottlenecks" that prevent the space from behaving like a hyperbolic space.

However, the universal cover of such a Denjoy domain can be conformally equivalent to the unit disk. The unit disk, equipped with the Poincaré metric, is Gromov hyperbolic. This is a key observation. It suggests that conformal equivalence to a nice domain doesn't automatically imply Gromov hyperbolicity.

Why This Matters

This distinction is important because it highlights that conformal geometry and large-scale geometry (as captured by Gromov hyperbolicity) are different beasts. Conformal equivalence focuses on local angles and scaling, while Gromov hyperbolicity is concerned with the global behavior of geodesics and triangles. Just because a space looks locally like a hyperbolic space doesn't mean it behaves like one on a large scale.

Let's Dig Deeper: Constructing Such Manifolds

So, how can we explicitly construct such manifolds? Here's a potential approach:

Start with a Denjoy domain $D$ that is not Gromov hyperbolic but whose universal cover $\tilde{D}$ is conformally equivalent to the unit disk $\mathbb{D}$ . Let $\phi: \tilde{D} \rightarrow \mathbb{D}$ be the conformal equivalence.
Equip $\mathbb{D}$ with the Poincare metric $g_{\mathbb{D}}$ , which is famously Gromov hyperbolic.
Consider the Riemannian manifold $(\tilde{D}, \phi^* g_{\mathbb{D}})$ . This manifold is conformally equivalent to $\mathbb{D}$ by construction.
Note that $(\tilde{D}, \phi^* g_{\mathbb{D}})$ is not necessarily Gromov hyperbolic. While $\mathbb{D}$ is Gromov hyperbolic, the pullback metric might not inherit this property on $\tilde{D}$ because the covering map $\pi : \tilde{D} \rightarrow D$ can "distort" the geometry in a way that destroys hyperbolicity. Think of it like flattening a curved surface – you change the geometric properties.
Now, project the metric (if possible). This is the tricky part. We want to find a metric $g$ on $D$ such that $\pi^* g = \phi^* g_{\mathbb{D}}$ . If we can find such a $g$ , then $(D, g)$ would be conformally equivalent to $\mathbb{D}$ (and thus conformally equivalent to a Gromov hyperbolic space) but $D$ itself is not Gromov hyperbolic.

Challenges and Considerations

The main challenge here is ensuring the existence and smoothness of the metric $g$ on $D$ . The projection of a metric from the universal cover to the base space is not always guaranteed to produce a well-behaved metric. You need to carefully consider the covering map and the properties of the original metric.

Also, it's essential to remember that conformal equivalence preserves angles but not necessarily distances or areas. This means that while $(D, g)$ and $\mathbb{D}$ "look the same" locally in terms of angles, their global geometric properties can be drastically different.

More Examples and Further Explorations

While the Denjoy domain provides a concrete starting point, there might be other examples of Riemannian manifolds that satisfy the given conditions. For instance, consider:

Manifolds with specific topological properties: Certain topological obstructions might prevent a manifold from being Gromov hyperbolic, even if it's conformally equivalent to a nice domain. Think about manifolds with "too much" topology, like high-genus surfaces with complicated mapping class group actions.
Manifolds constructed using quasi-conformal maps: Quasi-conformal maps are generalizations of conformal maps that allow for some bounded distortion of angles. These maps can be used to construct manifolds with controlled geometric properties. By carefully crafting a quasi-conformal map, you might be able to create a manifold that is conformally equivalent to a domain but not Gromov hyperbolic.

The Role of Curvature

Another avenue to explore is the role of curvature. Gromov hyperbolicity is often associated with negative curvature (or, more precisely, sectional curvatures bounded above by a negative constant). However, conformal transformations can change the curvature of a Riemannian manifold. So, even if a manifold is conformally equivalent to a negatively curved space, its own curvature might not be negative enough to ensure Gromov hyperbolicity.

Conclusion: A World of Geometric Nuances

So, to answer the original question: Yes, there are examples of Riemannian manifolds that are conformally equivalent to a domain but are not Gromov hyperbolic. The Denjoy domain provides a concrete illustration of this phenomenon. This highlights the subtle interplay between conformal geometry, large-scale geometry, and the topological properties of manifolds.

For you guys out there who are passionate about diving deep into the heart of mathematics, I hope this exploration into Riemannian manifolds has been both enlightening and inspiring. The intersection of differential geometry, metric geometry, and complex analysis offers a rich playground for discovering fascinating and unexpected results. Keep exploring, keep questioning, and keep pushing the boundaries of our understanding. Peace out, Plastik Magazine crew!