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8.16.2006

The Poincare conjecture 

There's an article in today's New York Times about the Poincare conjecture, and how it was solved (or at least thought to have been solved) by Grisha Perelman.

For the non-geometers in the audience, the Poincare conjecture asserts that if X is a smooth orientable 3-dimensional space that has the property that any closed non-self-intersecting loop can be shrunk to a point, then X is a 3-sphere. (To see that this is a nontrivial statement, think about drawing a loop on the surface of a donut.) Such a space is said to be "simply connected".

The basic technique that Perelman uses is Ricci curvature flow. Curvature is fairly easy to explain for 2-dimensional spaces sitting in our ordinary 3-space. Pick up a ball and pick any point on it. Draw a curve on the ball going through the point. The curve bends in a certain direction. Now draw another curve going through the point, but perpendicular to the first curve. The second curve bends in the same direction as the first one. The curvature of the sphere at that point is positive; in fact, the curvature of a sphere at any point is positive.

Now pick up a Pringle, and pick a point on it. If you draw two curves through the point, they'll bend in opposite directions. The curvature of the Pringle at any point is negative.

The neat thing is that the curvature of a surface, which is a local property of each point on the surface, tells you something about the global topology of the surface. If you have a closed, orientable 2-dimensional surface and its average curvature is positive, it must be topologically equivalent (smoothly deformable) to a sphere. ("Average" here means "integrate a 2-form and divide by a constant", but that's a mouthful.) If its average curvature is zero, it must be topologically equivalent to a donut with one hole. (Pick up a donut; the points on the outside rim have positive curvature, while the points on the inside rim have negative curvature.) If its average curvature is negative, it must be topologically equivalent to a donut with more than one hole. This is essentially the Gauss-Bonnet theorem.

Sadly, the previous paragraph's formulation of curvature is insufficient for more general situations -- specifically, we need a notion of curvature for higher-dimensional manifolds, and this notion of curvature should be intrinsic to the manifold instead of depending on the way it is situated in an ambient space. The modern formula of curvature is as a (1,3)-tensor, meaning that it is a function which takes three tangent vectors and spits out another tangent vector. Here's how it works: pick a point on the manifold, pick two tangent vectors, and draw a little square anchored at the point using the two vectors. Take a third vector, and transport it around the square. The curvature tensor tells you how much the third vector changes when you transport it. To see that this is not trivial, think about sitting at the Equator of the Earth carrying a vector that points north. Walk east around the Equator 90 degrees, then walk up to the North Pole, and finally walk back down to your starting point, all the while not changing the direction of the arrow.

The curvature tensor is a mouthful, so Ricci invented a contraction of it called the Ricci tensor. It's a (0,2) tensor, so it takes as input two tangent vectors and spits out a number. You lose some information this way, but it's a more managable object to deal with.

Oh, I need to tell you about the metric tensor on a general manifold. It too is a (0,2) tensor, meaning that you take two tangent vectors at a point, and it gives you a number. In our ordinary geometry this number would just be the product of the lengths of the vectors times the cosine of the angle between them. In abstract differential geometry, we need the metric tensor to define the notions of "length" and "angle" on our manifold, and in fact we need to have the metric tensor first in order to properly define the curvature tensor, but whatever.

Okay, so now we can sort of say what Ricci flow is. The Ricci flow simply causes the metric to change with time, and the rate of change of the metric tensor is given by the (negative of the) Ricci tensor. They're both (0,2)-tensors, so this makes sense mathematically. However, as the metric tensor changes, so does the Ricci tensor, so the Ricci flow equation becomes highly nonlinear and difficult to solve. In our previous discussion of 2-dimensional surfaces, what would happen is that the positive-curvature regions would shrink and the negative-curvature regions would grow. If you took a sphere, squished it so it had more of a dumbbell shape and then ran the Ricci flow, the thin neck of the dumbell would grow, while the round ends would shrink. Eventually the kinks would smooth out and the dumbbell would turn into a sphere, then start shrinking to a point.

Naturally, the Ricci flow is a bit more complicated in higher dimensions, but what Perelman essentially does is take the simply connected 3-manifold X, run the Ricci flow on it, perform surgery on the singularities that appear and resume the Ricci flow. Only finitely many surgeries are required in any finite time interval, and Perelman ends up studying the limit as t goes to infinity, then infers from the geometry of the limiting manifold that the original X was topologically a 3-sphere.

Comments:

Mmmmm, donuts. And Pringles.
 


Your theory of a donut-shaped universe is intriguing, Homer. I may have to steal it.

The Pringles idea was itself stolen from Tom Weston.
 


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