What does a flat disk in $\mathbb H^3$ look like?

Question

I am trying to understand hyperbolic space better, in particular $\mathbb H^3$ (assume curvature equal to $-1$). If we use the Beltrami–Klein model, the space can be represented by the open unit ball in $\mathbb R^3$ and geodesics appear as straight lines. An open disk like $$ \{(x,y,0)\in\mathbb R^3: x^2+y^2 < \frac 12\}$$ appears flat, but actually has (I assume) curvature equal to $-1$.

What does a surface look like that is flat in $\mathbb H^3$ ? (e.g. a disk that is flat with respect to the hyperbolic metric).

Answer 1

Even in $\mathbb R^3$, surfaces of curvature $0$ can be a little hard to describe in complete generality (I don't like the word "flat" for subsurfaces, because it clashes with other uses of "flatness" in Riemannian geometry). For example developable ruled surfaces in $\mathbb R^3$ have curvature $0$.

So I'll just give some important examples of curvature $0$ surfaces in $\mathbb H^3$, namely the horospheres. I don't know how to describe them in the Beltrami-Klein model, but they are easy to describe in two other models:

• Horospheres in the Poincare ball model $\mathbb B^3$ are Euclidean spheres inside the Poincare ball and tangent, at a single point, to the boundary sphere at infinity $\mathbb S^2_\infty$.
• Horospheres in the upper half space model $\{(x,y,z) \mid z>0\}$ are of two types: Euclidean spheres in upper half space tangent to the plane $z=0$; and horizontal Euclidean planes of the form $\{(x,y,z_0)\}$ for some constant $z_0>0$.

It's not hard to check that the upper half plane metric restricted to a horizontal Euclidean plane is actually isometric to the ordinary Euclidean plane, and so has Gaussian curvature $0$. For the other types of horospheres, one simple way to verify that their Gaussian curvature equals zero is to produce an isometry of the whole model which takes the given plane to a horizontal Euclidean plane in the upper half space model.