# Bi-asymptotic geodesics in Visibility manifolds

I'm thinking about some properties of geodesics in visibility spaces. Here I give some definitions:

A Riemannian manifold $(M,g)$ with riemannian distance $d$ is said to be a visibility manifold if given $p\in M$ and $\varepsilon>0$ there exists $r=r(p,\varepsilon)>0$ with this property: if $\sigma:[a,b]\longrightarrow M$ is a geodesic segment such that $d(p,\sigma)\geqslant r$, then $\sphericalangle_p(\sigma(a),\sigma(b))\leqslant\varepsilon$, where $\sphericalangle_p(\sigma(a),\sigma(b))$ is the angle between the tangent vectors of the minimizing geodesics joining $p$ to $\sigma(a)$ and to $\sigma(b)$.

Nice models for visibility manifolds includes negatively-curved spaces, such the hyperbolic plane.

We say that two geodesics $\gamma,\beta:\mathbb{R}\longrightarrow M$ are bi-asymptotic if there exists $C>0$ such that $d(\gamma(t),\beta(t))<C$ for all $t\in\mathbb{R}$.

Consider now non-conpact manifolds. My question is:

If $M$ is a visibility manifold, is it true that there is no bi-asymptotic geodesics besides coincident pairs of geodesics?

I see this happening in negatively curved spaces, but I have no clue if this is a more general property or not.

Any help will be really appreciated. Thanks a lot!

• By these definitions, any compact manifold is a visibility manifold (just take $r$ larger than the diameter of the manifold), but any two geodesics in a compact manifold are bi-asymptotic. Jan 13, 2015 at 19:17
• Yes, you are completely right. I have wrote (and after I deleted it) the non-compact condition, but I thought it was perhaps unnecessary to comment. I will put this there to make more sense to the question. Thanks again! Jan 13, 2015 at 19:21
• I think if $\mathbb{H}^2$ is the hyperbolic plane, then $\mathbb{H}^2 \times \mathbb{S}^1$ is still a counterexample. So it seems a stronger condition than non-compact is needed. (Maybe assuming $M$ is a Hadamard manifold, or that it has a nice visual boundary?) Jan 13, 2015 at 19:36
• I have in addition the abscense of conjugate points, if it helps... Jan 13, 2015 at 19:50
• $\mathbb{H}^2 \times \mathbb{R}$ isn't a counterexample; fix a geodesic $\gamma$ of $\mathbb{H}^2$; then the geodesics $\sigma_\lambda = (\gamma(t), \lambda)$ won't satisfy the visibility condition, since $\sigma_\lambda$ and $\sigma_\mu$ are a distance $|\lambda - \mu|$, which can be very large, but the angle from $\sigma_\lambda(0)$ to points very far down $\sigma_\mu$ approaches $\pi$. Jan 13, 2015 at 20:02

Even under the stronger hypothesis that $M$ be simply connected of nonpositive curvature and a visibility manifold, there can exist pairs of bi-asymptotic geodesics.
To construct a counterexample, let $N$ be a negatively curved surface with a single cusp. Chop off the cusp at some finite point and smooth it down so that it ends in a flat cylinder; this gives a nonpositively curved manifold $N'$ with a boundary $\mathbb{S}^1$, such that a neighborhood of this boundary is isometric to a flat cylinder. Let $M$ be the universal cover of the manifold obtained from gluing $N'$ to a copy of its reflection along the flat cylinders.
$M$ is a visibility manifold by the theorem on the first page of [1] (thanks for the reference!) since clearly it is not flat (and hence contains no isometric $\mathbb{R}^2$). But $M$ has lots of pairs of bi-asymptotic geodesics; $M$ even has many flat strips (isometric copies of $\mathbb{R} \times [0,\epsilon]$).