[math-fun] The punctured torus
It's well known that among the compact orientable surfaces (without boundary), the sphere S^2 can be given a metric of constant curvature K only if K > 0; the torus T^2 can be given one only if K = 0; and the surfaces of genera >= 2 can be given one only if K < 0. I had rarely thought about non-compact surfaces in such terms, but for these one would look for not just any metric of constant curvature but one that is *complete* — meaning that every geodesic curve can be extended indefinitely without "going off the edge" of the surface. So today I was mildly surprised to see that the torus with one puncture (or what's the same topologically, the torus with a disk removed) can be given a complete metric of constant negative curvature (say K = -1). What's more, this can be tiled with exactly two ideal triangles (<https://en.wikipedia.org/wiki/Ideal_triangle>). Which is pretty, since all ideal triangles are congruent. It cannot have a complete metric of constant 0 curvature, even though the unpunctured torus does, easily. Surprisingly, the cylinder (without boundary) can be given a complete metric of constant curvature either if K = 0 or else if K < 0 — in which case it, too, can be tiled with two ideal triangles. (Same goes for the Möbius band.) —Dan
References please? WFL On 6/1/20, Dan Asimov <dasimov@earthlink.net> wrote:
It's well known that among the compact orientable surfaces (without boundary), the sphere S^2 can be given a metric of constant curvature K only if K > 0; the torus T^2 can be given one only if K = 0; and the surfaces of genera >= 2 can be given one only if K < 0.
I had rarely thought about non-compact surfaces in such terms, but for these one would look for not just any metric of constant curvature but one that is *complete* — meaning that every geodesic curve can be extended indefinitely without "going off the edge" of the surface.
So today I was mildly surprised to see that the torus with one puncture (or what's the same topologically, the torus with a disk removed) can be given a complete metric of constant negative curvature (say K = -1). What's more, this can be tiled with exactly two ideal triangles (<https://en.wikipedia.org/wiki/Ideal_triangle>). Which is pretty, since all ideal triangles are congruent. It cannot have a complete metric of constant 0 curvature, even though the unpunctured torus does, easily.
Surprisingly, the cylinder (without boundary) can be given a complete metric of constant curvature either if K = 0 or else if K < 0 — in which case it, too, can be tiled with two ideal triangles. (Same goes for the Möbius band.)
—Dan
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