Sunday, September 19, 2004

Poincare Conjecture Proved?
Last week, it was reported that the famous Poincare Conjecture might finally be proved. The Poincare conjecture is one of the Seven Millenium problems. A media-shy Russian mathematician, Grigori Perelman is said to have proven a more general case of the conjecture. Here is Wolfram's MathWorld discussing the news. This is also a highly mathematical description of the problem, so for the rest of us, here is a layman's version about what the conjecture is about. More links here, and here.
The conjecture is about topology, or the study of shapes. It is the study of the properties of an object that do not change under continuous deformations (such as stretching and bending (but not tearing)). Just as there is only one way to bend a two-dimensional plane into a shape without holes -- the sphere -- there is likewise only one way to bend three-dimensional space into a shape that has no holes. If we stretch a rubber band around the surface of an apple, then we can shrink it down to a point by moving it slowly, without tearing it and without allowing it to leave the surface. On the other hand, if we imagine that the same rubber band has somehow been stretched in the appropriate direction around a doughnut, then there is no way of shrinking it to a point without breaking either the rubber band or the doughnut. We say the surface of the apple is "simply connected," but that the surface of the doughnut is not.
Poincaré, almost a hundred years ago, knew that a two dimensional sphere is essentially characterized by this property of simple connectivity, and asked the corresponding question for the three dimensional sphere (the set of points in four dimensional space at unit distance from the origin). This question turned out to be extraordinarily difficult, and mathematicians have been struggling with it ever since.
Though abstract, the conjecture has powerful practical implications - apparently from the study of DNA knot formation and protein folding to the shape of the universe.

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