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by Icy0
753 days ago
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I love this! And even more is true -- you can read off the Euler characteristic from adding up how many fractions of a circle are lost over all the points. For the cube, at each vertex you've lost a quarter circle, and there are 8 vertices -- hence the Euler characteristic of a cube is 2. For the two-disk model of the sphere, a similar thing should be true, I think, but I haven't worked it out in detail -- the integral of "circles lost" over the sphere (the support of this integral is the shared boundary of the disks) should be 2 as well. This is the Gauss-Bonnet theorem. |
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For the two-disc sphere, I can't think of an intuitive way to "see" the "circles lost" integral. But here's a different intuitive way to see the total curvature.
Another way to measure the curvature is to look at how much the sum of the interior angles of an n-sided polygon exceeds the usual sum π(n - 2). It's most common to think about triangles, but we can also think about 2-gons... these are usually degenerate shapes with a sum of interior angles of zero.
But on the two disc-sphere, draw two lines, each from the center of one disc to the center of the other disc, passing straight through the glued boundary. These form a 2-gon with sum of interior angles (and also excess over the usual value) equal to twice the angle between the lines. To get the total curvature of the whole sphere, let each of the two interior angles be 2π, for a total of 4π... two circles, the Euler characteristic.