# Quantum Curvature

2004-05-10T18:09:00Z

I was reading an article in Scientific American on loop quantum gravity. It reminded me that I wanted to make a post a while back on quantum curvature, even though the article has little, or nothing to do with quantum curvature.

I was playing around with polydrons back in March. We were trying to make a torus out of equilateral triangles. I realized that the total gaussian curvature of the object must be 0. Since flat faces have no curvature, and since edges have no gaussian curvature (because they bend in only one direction, and don’t form a cup shape) all the curvature must come at the vertexes. I wanted to know how much curvature a vertex has for a given degree.

After a lot of thinking, the answer soon struck me. A tetrahedron has 4 vertexes of degree 3. The total gaussian curvature of a tetrahedron is the same as for a sphere, which is 4π. So each vertex of degree 3 contributes π or (3/3)π curvature.

An octahedron has 6 vertexes of degree 4. So each vertex of degree 4 contributes (4/6)π or (2/3)π amount of curvature.

An icosahedron has 12 vertexes of degree 5. So each vertex of degree 5 contributes (4/12)π of (1/3)π amount of curvature.

6 equilateral triangles form a flat hexagon, so each vertex of degree 6 contributes 0 or (0/3)π amount of curvature.

See the pattern? Let us call (1/3)π one quanta of curvature. Then (by engineering induction) we see that a vertex (in a mesh of equilateral triangles) contributes 6−d quanta of curvature where d is the number of triangles surrounding the vertex. There are a total of 12 quanta of curvature needed to make a sphere. If you put 7 equilateral triangles around a vertex (yes you can do this) that vertex contributes −1 quanta of curvature.

As interesting as this is, tomr managed to just make the torus without the deep mathematical understanding of curvature.

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