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.


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.

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    $\begingroup$ I agree with Henrik very much. Just a side comment: after drawing a lot of pictures about the local geometry in this style, you might also try to fashion a global topological picture of how these local pictures glue together. This can be done most easily with fabric, or some other very flexible material. Bill Thurston was a true genius at imagining these things in his mind, but a lot of the "large scale" understanding of the negative curvature of surfaces of negative euler characteristic can be gained in this way. I'm very far from the right person to ask about this, but here I am anyway.. $\endgroup$ – Andy Sanders Jan 21 '20 at 21:35
  • $\begingroup$ But, you can embed a hyperbolic cylinder and visualize it pretty easily. And, the 3-punctured sphere is just two cylinders coming in, one going out, yet it's behaving very differently from a cylinder, because a hyperbolic cylinder can never grow and then shrink again. I guess I just have to dig into the model and see where this intuition from cylinders is going wrong. $\endgroup$ – William Nelson Jan 22 '20 at 5:20

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