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stoutfellowThe Gauss-Bonnet Theorem is one of the most unexpected and (to my mind) beautiful results anywhere in mathematics.
We saw before that curvature is one of the key characteristics of a curve. To describe the shape of a surface, though, is rather more difficult, because, in the vicinity of a particular point, it may curve differently in different directions. Consider a torus again, and think of it as lying flat in front of you. At the point closest to you on the outer rim, the torus curves rather gently in the horizontal direction, and more sharply in the vertical direction; it curves away from you in both directions, however. At the nearest point on the inner rim, though, the torus curves away from you horizontally, but towards you vertically.
Let's make this more precise. Suppose you have a normal vector n defined at a point P. Pick any tangent vector v at P. Taken together, n and v define a plane through P, which intersects the surface in a curve. (Most of the time, this curve will be a regular curve; if it isn't, some special things have to be done.) Let C(v) be the curvature of that curve at that point, but multiply it by -1 if the curve bends away from n.
At your point P, then, you have not one but many curvatures, one in each direction. That much data can be useful, but it's rather too much to interpret easily. Can we extract a simpler piece of information?
The most useful piece of information that can be extracted is the "Gaussian curvature", which is computed thus. Find the direction in which the curvature is as large as possible, and the one in which it's as small as possible. (For these purposes, negative values are small.) Those directions turn out to be perpendicular. The Gaussian curvature K(P) is the product of the curvatures in those two directions.
Let's do a couple of examples. Take a sphere of radius r, and use the inward-pointing normal. At any point, all of the curves that arise as described above turn out to be great circles - i.e., circles centered at the center of the sphere, with radius r. Now, the curvature of a circle is the reciprocal of its radius, and all of these circles bend towards n, so the curvature is 1/r in all directions. Therefore the Gaussian curvature at any point of the sphere is 1/r2.
What about a torus? Again, use the normal pointing into the doughnut. At a point on the outer rim, the maximum curvature is the vertical one. (If you slice the torus vertically, you get a circle with radius r.) The minimum curvature is horizontal. (A horizontal slice gives a larger circle; we'll write the radius of that circle as R+r.) The two curvatures, then, are 1/r and 1/(R+r), so the Gaussian curvature there is 1/r(R+r). At a point on the inner rim, the maximum curvature is again vertical, equal to 1/r. The minimum curvature is horizontal, and that circle bends away from n, so it counts negatively: it's -1/(R-r). So the Gaussian curvature at that point is -1/r(R-r).
One thing you should notice is that if the surface always curves toward n (or away from n, for that matter), the Gaussian curvature is positive; if the surface curves toward n in some directions and away from n in others, the Gaussian curvature is negative.
Now here's the deal. If the surface is fairly well-behaved, something interesting happens. "Well-behaved", here, means that it's compact - roughly speaking, it must be finite in area - and it has no boundary. The plane is disqualified, because it isn't compact; the cylinder and the Möbius strip fail too, because they have boundaries. (It's possible to deal with the existence of a boundary, but it makes the formula more complex and uglier.) If you take the total of the Gaussian curvature, all over the surface, the result comes out to 2π times the Euler characteristic of the surface. (How do you take the total Gaussian curvature, when there are infinitely many points involved? Well, technically it's not a sum but an integral; the details are a bit messy.)
Why is this theorem remarkable? Because the Gaussian curvature is defined in terms of calculus, and is sensitive to deformations, but the Euler characteristic is a topological idea, independent of deformations. That they should be linked as they are is startling.
Here's an example that might help show why the total curvature isn't sensitive to deformation. Take a sphere, and push a little dimple into it. Now, in the center of the dimple, the curvature is positive (and greater than the curvature of the original sphere). But further up, near the rim of the dimple, the curvature is negative, and just negative enough to cancel out the increased curvature at the center.
I don't know about you, but I think that's neat!
A-a-and we're done.
Note: "Bonnet" rhymes with "way". Wordplay courtesy of Michael Spivak.
We saw before that curvature is one of the key characteristics of a curve. To describe the shape of a surface, though, is rather more difficult, because, in the vicinity of a particular point, it may curve differently in different directions. Consider a torus again, and think of it as lying flat in front of you. At the point closest to you on the outer rim, the torus curves rather gently in the horizontal direction, and more sharply in the vertical direction; it curves away from you in both directions, however. At the nearest point on the inner rim, though, the torus curves away from you horizontally, but towards you vertically.
Let's make this more precise. Suppose you have a normal vector n defined at a point P. Pick any tangent vector v at P. Taken together, n and v define a plane through P, which intersects the surface in a curve. (Most of the time, this curve will be a regular curve; if it isn't, some special things have to be done.) Let C(v) be the curvature of that curve at that point, but multiply it by -1 if the curve bends away from n.
At your point P, then, you have not one but many curvatures, one in each direction. That much data can be useful, but it's rather too much to interpret easily. Can we extract a simpler piece of information?
The most useful piece of information that can be extracted is the "Gaussian curvature", which is computed thus. Find the direction in which the curvature is as large as possible, and the one in which it's as small as possible. (For these purposes, negative values are small.) Those directions turn out to be perpendicular. The Gaussian curvature K(P) is the product of the curvatures in those two directions.
Let's do a couple of examples. Take a sphere of radius r, and use the inward-pointing normal. At any point, all of the curves that arise as described above turn out to be great circles - i.e., circles centered at the center of the sphere, with radius r. Now, the curvature of a circle is the reciprocal of its radius, and all of these circles bend towards n, so the curvature is 1/r in all directions. Therefore the Gaussian curvature at any point of the sphere is 1/r2.
What about a torus? Again, use the normal pointing into the doughnut. At a point on the outer rim, the maximum curvature is the vertical one. (If you slice the torus vertically, you get a circle with radius r.) The minimum curvature is horizontal. (A horizontal slice gives a larger circle; we'll write the radius of that circle as R+r.) The two curvatures, then, are 1/r and 1/(R+r), so the Gaussian curvature there is 1/r(R+r). At a point on the inner rim, the maximum curvature is again vertical, equal to 1/r. The minimum curvature is horizontal, and that circle bends away from n, so it counts negatively: it's -1/(R-r). So the Gaussian curvature at that point is -1/r(R-r).
One thing you should notice is that if the surface always curves toward n (or away from n, for that matter), the Gaussian curvature is positive; if the surface curves toward n in some directions and away from n in others, the Gaussian curvature is negative.
Now here's the deal. If the surface is fairly well-behaved, something interesting happens. "Well-behaved", here, means that it's compact - roughly speaking, it must be finite in area - and it has no boundary. The plane is disqualified, because it isn't compact; the cylinder and the Möbius strip fail too, because they have boundaries. (It's possible to deal with the existence of a boundary, but it makes the formula more complex and uglier.) If you take the total of the Gaussian curvature, all over the surface, the result comes out to 2π times the Euler characteristic of the surface. (How do you take the total Gaussian curvature, when there are infinitely many points involved? Well, technically it's not a sum but an integral; the details are a bit messy.)
Take your surface, then, and divide it into a bunch of regions R1, R2, ..., Rn. In each region, pick a point: P1, P2, ..., Pn. Take the Gaussian curvature at each of these points, multiply it by the area A of the region, and add them up: K(P1)A(R1) + K(P2)A(R2) + ... + K(Pn)A(Rn). This sum will be close to 2π times the Euler characteristic, and if you take more and more smaller and smaller regions, you can make the sum arbitrarily close to that number.Let's do an example or two. On a sphere of radius r, the curvature at every point is 1/r2. The area of the sphere is 4πr2, so the total curvature is just the product, 4π. (Remember, the Euler characteristic of the sphere is 2.) On the torus, points on the outside have positive curvature, and points on the inside have negative curvature. The region where the curvature is positive is larger than the area where it's negative, but on the other hand the curvatures on the negative part are larger (in absolute value), on average, and they just cancel each other out. (A torus has Euler characteristic 0.)
Why is this theorem remarkable? Because the Gaussian curvature is defined in terms of calculus, and is sensitive to deformations, but the Euler characteristic is a topological idea, independent of deformations. That they should be linked as they are is startling.
Here's an example that might help show why the total curvature isn't sensitive to deformation. Take a sphere, and push a little dimple into it. Now, in the center of the dimple, the curvature is positive (and greater than the curvature of the original sphere). But further up, near the rim of the dimple, the curvature is negative, and just negative enough to cancel out the increased curvature at the center.
I don't know about you, but I think that's neat!
A-a-and we're done.
Note: "Bonnet" rhymes with "way". Wordplay courtesy of Michael Spivak.
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no subject
Date: 2006-01-26 08:01 pm (UTC)no subject
Date: 2006-01-26 10:07 pm (UTC)Yeah. I don't recall being exposed to this in my checkered past. Thank you!
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Date: 2006-01-27 01:58 pm (UTC)