# Is the hyperbolic paraboloid a model for hyperbolic plane?

I have been doing some outreach work on conveying notions of hyperbolic space to people with limited math backgrounds. One idea I like to use is to talk about the hyperbolic plane as being like the opposite of the sphere. For instance, if I start with a circle (which is flat), then I shrink the boundary down to a point, it will bulge out into a sphere. So do the opposite: start with a circle and expand the boundary out to something infinite. This can be used to give some intuition about the classical models for the hyperbolic plane (upper half-plane, Poincaré disk, Klein disk, hyperboloid). In each of those models, one can think of it as having a boundary that is a circle of infinite radius, in different ways.

An audience member asked me an interesting question: why don't we use the hyperbolic paraboloid (familiar from calc 3) as a model for the hyperbolic plane? It captures this notion of a boundary needing to bulge out in different directions as one moves away from the center, like a Pringles potato chip or like choral. I'm wondering if anyone has ever made that idea precise, or if there is some reason why that can't be done. Is there a formula for a metric on the hyperbolic paraboloid that makes it a nice model for the hyperbolic plane? Is there some useful geometric interpretation for this? Are there interesting applications?

• Well, the hyperbolic paraboloid has non-constant curvature, when you consider the metric induced from $\Bbb R^3$. So you would have to find a motivation for a new metric there. And since that paraboloid is diffeomorphic to $\Bbb R^2$, you might as well work in the plane (which does not seem too interesting, since we already have Poincaré's half-plane). – Ivo Terek Jul 20 '18 at 19:05
• I think that's what stopped me from having a good answer. On the other hand, I like the idea that the "center" of the hyperbolic paraboloid might, in some sense, capture what it "looks like" to stand a point in the hyperbolic plane and look around. But I'm not sure there's anything there that can be made rigorous constructively. – j0equ1nn Jul 20 '18 at 19:07
• Well, if $X(u,v)=(u,v,u^2-v^2)$, then $K(X(0,0))= -4$, so the curvature of the hyperbolic paraboloid remains negative in a neighborhood of the "center" by continuity. It is an hyperbolic point in the sense that the surface locally intersect the half-spaces determined by tangent plane at the "center". I don't know how further can we go. The curvature of the paraboloid being non-constant is a great problem. – Ivo Terek Jul 20 '18 at 19:17