# Gauss-Bonnet theorem for spheres that almost look like a torus

Imagine a torus and a flat disk fitting in the middle of its "hole" (a doughnut with a membrane in the middle). Cut the torus at its inner equator, duplicate the disk, move the two copies away from each other slightly, widen the cut appropriately and join the two flat disks with the sliced torus (anyway you like).

You get a surface $M$ homeomorphic to the sphere - thus with integral curvature $\int_S\kappa = 4\pi$ - , but with integral curvature equal to that of the torus $\int_T\kappa = 0$ plus a contribution from the "regions of agglutination" where the two disks and the sliced torus are glued together (the disks by themselves having zero curvature).

Is it simply a consequence of the Gauss-Bonnet theorem that however smoothly or abruptly you glue the two disks and the sliced torus together the integral curvature in the "region of agglutination" has to be $4\pi$?

Or is there a mistake in my description of the surface or in my understanding of the Gauss-Bonnet theorem?

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First, note that the integral over the torus is $0$, not $2\pi$ (since the Euler characteristic of a torus is $0$). I suppose this just makes the result even more surprising.

There are (at least) two things happening which add up to the net increase of $4\pi$ to the total curvature.

First note that you're cutting out exactly the points of largest negative curvature on the torus: If you imagine revolving a circle around an axis to make the torus, then the outside semicircle rotates to the points with non-negative curvature while the inside semicircle rotations to points with non-positive curvature. So, in some sense, you've cut out the largest contributer of negative curvature.

Of course, you may object that you're really cutting out a very tiny portion, so it shouldn't affect the answer much, but here is where the other part comes into play.

In order to connect the top half of the torus to the top disc in a smooth way, you have to bend the torus quite bit. More specifically, imagine rotation $(x-2)^2 + y^2$ around the $z$-axis to get the torus. And just focus on the semi-semicircle in the top left portion of this circle. Near the equator, the tangent line is a very large positive number. In order to connect up with the disc (horizontal tangent), you must bend the graph so this very large number becomes $0$. That introduces a ton of second derivative (since you're bending the tangent vectors a lot).

More importantly, if you rotate that portion of the graph around, the part near the equator used to be contributing negative curvature, but is now contributing a lot of positive curvature (because it's bent so much). That's where you really get the $4\pi$ from.

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I think it's also worth emphasizing that the first part is actually quite negligible, since it's enough to cut out arbitrarily small area of negative curvature. It is the second part (i.e. bending the torus inwards) that accounts for the difference. – Marek Jan 28 '13 at 21:15
@Jason: Thanks a lot. I am glad you guessed what I was asking for without a picture. – Hans Stricker Jan 28 '13 at 22:21
@Marek: How do I get a "feeling" for what you claim? How is it that the second part does "more" account for the difference than the first part? – Hans Stricker Jan 28 '13 at 22:24
@Hans: The curvature at the inner equator is $\kappa = \frac{-1}{r(R-r)}$ and close to that number at nearby points. If you throw away points in an $\varepsilon$-neighborhood of the inner equator (roughly a cylinder of radius $(R-r)$ and height $\varepsilon$), you'll remove something of order $\kappa \cdot 2\pi (R-r) \varepsilon = \frac{-2\pi\varepsilon}{r}$, which you can make as close to $0$ as you wish. – Martin Jan 29 '13 at 4:49