Visualizing hyperbolic metric of punctured sphere

Question

Uniformization of the 3-punctured sphere generates a "pants" configuration with three legs narrowing down to cusps. This is supposed to have a metric of constant negative curvature, and I can see this in the cusps, and also in the saddle regions where they join, but I am at a loss to understand how the curvature can remain negative in the middle region. Simplistic depictions (e.g. on the Wikipedia pair-of-pants page) clearly do NOT show negative curvature throughout. As another example, Hubbard Fig. 3.5.1 (ref below) shows a surface which does have negative curvature throughout, and has two cusps coming in; however, it has a finite-size hole going out, and that hole is clearly going to have to grow, rather than shrink, to maintain negative curvature.

In short, I can't understand how a surface can start at zero radius, then traverse a section where size is growing, and then contract back to zero size, without encountering a section where curvature is positive. I can calculate that it is happens using the uniformizing metric, but I would really appreciate some guidance in how to visualize or understand it property. Even if I think of embeddings in 3 space with self-intersections, I can't seem to construct a visualization that works.

ref: Hubbard, John H., "Teichmuller Theory", volume 1, Matrix Editions, 2006.

Answer 1

I think trying to embed such things curvature-preserving into 3-dimensional space is just the wrong way to think about them. For example iis reallyhard to visualize the flat metric on a torus like that. However, it is much easier by taking just a flat square and identifying opposing sides. This gives us the flat metric on a torus.

Let us apply the same idea to the three punctured sphere. Now we do not want curvature zero, but constant curvature minus 1. So we should not take a flat square, but some (ideal) polygon in two dimensional hyperbolic space. Take 6 ideal triangles with angles 60°,60° and 0° (so the vertices with 0°are ideal) glue them together to get an ideal hexagon with angles 0°,120°,0°,120°,0° and 120° and glue the edges next to one ideal vertex together.

To vertify that the result really has constant curvature minus -1, we have to show that everypoint has a small neighborhood that isometrically embeds into two dimensional hyperbolic space. Let us first go back to the toy example of a flat torus. Why does that argument work there. It is clear for the points in the interior of a square. For the points on the edges it also works, since we glue always glue two edge together. And for the cornerpoint it also works, since there 4 corners with 90° each meet.

And now the same argument works in our hyperbolic setting; in the interior everything is ok, we identify one edge with exactly one other edge and at the corner exactly three corners with 120° each meet.

It might be a nice exercise to relate this construction to the picture in the comment above by მამუკა ჯიბლაძე.