Surfaces with non-constant negative curvature

Question

Are there any nice models of surfaces with non-constant negative curvature, analogous to the Poincare disk for constant negative curvature. I have found lots of general results and theory but no nice clean models.

Answer 1

If you just want examples for which it's not hard to figure out how the geodesics behave, here's a class of examples with negative and non-constant curvauture in the plane where the geodesics are relatively easy to understand:

Let $a$ and $b$ be smooth functions on $\mathbb{R}$ such that $a(x)+b(y)>0$ for all $x,y\in\mathbb{R}$ and consider the metric $$ g = \bigl(a(x) + b(y)\bigr)(\mathrm{d}x^2 + \mathrm{d}y^2) $$ on $\mathbb{R}^2$. The curvature of this metric is $$ K = \frac{a'(x)^2+b'(y)^2-\bigl(a''(x)+b''(y)\bigr)\bigl(a(x)+b(y)\bigr)} {2\,\bigl(a(x)+b(y)\bigr)^3} $$ It's easy to choose $a$ and $b$ so that $K<0$. For example, $a(x) = x^2+1$ and $b(y) = y^2+1$ gives a complete metric on $\mathbb{R}^2$ that has non-constant negative curvature $K = -4/(x^2{+}y^2{+}2)^{3}<0$.

Note that, taking $a$ (respectively, $b$) to be a constant gives a metric $g$ that has a Killing vector field, namely $\partial/\partial x$ (respectively, $\partial/\partial y$), but, for generic choices of $a$ and $b$, the metric $g$ will have no Killing vector field.

As for geodesics, the good thing about these metrics (called Liouville metrics in the literature) is that their geodesic flows are integrable: Any unit speed geodesic $(x(t),y(t))$ satisfies $$ \bigl(a(x)+b(y)\bigr)\bigl(\dot x^2+\dot y^2\bigr) = 1 \quad\text{and}\quad \bigl(a(x)+b(y)\bigr)\bigl(b(y)\,\dot x^2- a(x)\,\dot y^2\bigr) = c $$ for some constant $c$. (Note that, when either $a$ or $b$ is constant, this second 'first integral' of the geodesic equations specializes to the well-known 'Clairaut integral' for surfaces of revolution.)

In particular, $$ \bigl(b(y)-c)\bigr)\,\dot x^2 - \bigl(a(x)+c)\bigr)\,\dot y^2 = 0, $$ and, assuming that you are in a region when $a(x){+}c$ and $b(y){-}c$ are both positive, $$ \frac{\mathrm{d}x}{\sqrt{a(x)+c}} \pm \frac{\mathrm{d}y}{\sqrt{b(y)-c}}=0, $$ which gives two foliations of this region by geodesics, which can be found by quadrature.

In any case, you will have good qualitative control over these geodesics and can draw some nice pictures.

Added remark (12 May 2020): As an example of what one can do with this more explicit information, you might be interested in this answer of mine to an old question about Riemannian surfaces for which one can compute an explicit distance function.

Answer 2

Minimal surfaces without umbilics. They are nice and have negative Gaussian curvature. The catenoid is a particular example.

Answer 3

You don't even need to settle for a model. You can make such a surface. Take any positive strictly convex smooth function ($f(x) > 0$ and $f''(x) > 0$) and rotate it around the x axis.

The single-sheet hyperboloid, $x^2 + 1 = y^2 + z^2$, has some nice geodesics, but also some messier ones: https://math.stackexchange.com/questions/1601158/how-can-we-find-geodesics-on-a-one-sheet-hyperboloid

You may be able to calculate the geodesics with methods given here: Geodesics on a hyperbolic paraboloid (which is yet another negative curvature surface)

Answer 4

You can play around with this in Mathematica following the suggestions here on Mathematica StackExchange. E.g.

gccolor[{Cos[u] Sech[v], Sin[u] Sech[v], v + v^2 - Tanh[v]},
  {u, 0, 2 \[Pi]}, {v, -2, 3}]

gives the following graph, with different blues for different negative curvatures.