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Comment by dmurray | Hacker News Reader

Why can't I do the same thing with the surface of a sphere?

Take two circles, side by side, connected by an infinitesimally small overlap, and put portals all around the circles to their equivalent point in the other circle. Then fold this over and inflate it into a sphere.

If you're suspicious about the discontinuity where they overlap, I think that's a red herring - no one is claiming the sphere is isomorphic to two disjoint surfaces of any shape.

Alternatively, make it a cube. I can definitely fold a single piece of paper into a cube, or equivalently, I can put portals on my piece of paper to give it the topology of a cube. Intuitively that is flat. But the cube isn't one of the 18 forms proposed for the universe. Nor is any of the other millions of 3d shapes I can make with the same process.

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Here's an easy way to test the curvature of these examples. Draw a circle (all points equidistant to a given one) centered at a point on one of the "glue" edges. The amount that the circumference of that circle is short of the expected 2πr is a measure of the curvature at that point.

In the two-discs sphere construction, a circle centered on the disc boundaries will appear half in each disc. But it will be short of the usual circumference due to the shape of the discs. All the curvature has been pushed to these boundaries.

In the cube, a circle centered at a vertex will appear on three sides; its circumference will be three-quarters of the usual value. Note that all the curvature of the cube is at the vertices, not the edges!

For the portal-torus, try drawing a circle anywhere, passing through the portals... it will have the usual circumference, zero curvature everywhere.

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.

Yes!

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.

Nice example! A sphere is pretty tough to see...

One thing that helped me is considering how to deform the edges of your glued circle examples into a flat square. First, map the two sides of the seam two the two corners of the plane. This becomes a sphere by folding it into a triangle and then gluing (or portaling) up neighboring edges. Having the portals on neighboring edges means that they change directions in weird ways.

This is somewhat different than the torus with portals example, where the portals were placed on opposite edges of the plane.

Edit: Found some diagrams!

https://en.wikipedia.org/wiki/Fundamental_polygon#Examples_o...

That's a good response! I had to spend some time to work out what goes wrong here. But I figured it out.

Parallel transport is broken in your model of the sphere. Take this example. Take a vector pointing north in circle 1, and send through the north portal. It should be pointing south when it gets to the other circle. Fine -- north in circle 1 corresponds to south in circle 2.

Now send the north-pointing vector east instead. It is going to point north in circle 2.

So the north vector changes direction depending on which portal it goes through. So parallel transport changes directions. Hence your model of the sphere is not flat.

Thinking about this a bit more, I think I see what you mean - the example where you fold it into a torus is "intuitively flat" because when you leave the portal travelling north, you re-enter travelling north, etc. That property doesn't hold for my cube example.

I wouldn't say the curvature implied by this property is "intuitively" 0 - why not 1, or 360°, or 180°? - but I can see how it's an interesting property that only a few shapes can have.

The cube isn’t flat. The curvature is just concentrated into the corners. You can measure this curvature by adding the interior angles of a triangle that contains a corner.

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