First Clay Millennium Prize goes to Grigoriy Perelman | Hacker News Reader

First Clay Millennium Prize goes to Grigoriy Perelman

claymath.org

77 pointsPanoramix13 comments

The full press release is here: http://www.claymath.org/poincare/millenniumPrizeFull.pdf

After Perelman rejected the Fields medal for this proof, he said about the Clay Millennium prize: "I'm not going to decide whether to accept the prize until it is offered". This was a few years ago.

Now that it is offered I would like to know now if he decided to accept it.

Although this was expected, it is very exciting now that it is official.

Now, we'll have to see whether Dr. Perelman will accept this honor. He astonished people worldwide when he declined the Field Medal: http://www.nytimes.com/2006/08/22/science/22cnd-math.html

I recommend reading "Perfect Rigor - A genius & the mathematical breakthrough of the century" by Masha Gessen. It provides the best possible insights into Perelman's life and work and details why he most probably will not ever accept this prize.

hmmm. http://news.ycombinator.com/item?id=1008102

She certainly isn't a Ph.D., but a journalist who writes stories. I found the book interesting nonetheless, one can read it pretty quickly. It illuminated Perelman's decisions and provided insights into his character. Relating to math, it's certainly only for laymen.

I admire Dr. Perelman for his work (even though I don't completely understand the proof), but most of all I admire him for his integrity. He withdrew from mathematics because he felt dishonesty was tolerated and he didn't want to be part of it. He refers to Shing-Tung Yau of course:

"I can't say I'm outraged. Other people do worse. Of course, there are many mathematicians who are more or less honest. But almost all of them are conformists. They are more or less honest, but they tolerate those who are not honest."

( from "Manifold Destiny", New Yorker, Aug 28, 2006: http://www.newyorker.com/archive/2006/08/28/060828fa_fact2?p... )

Of course, if the good people leave, where does that leave mathematics?

Not mentioned in the title. Perelman proved that the Poincare Conjecture is true.

Imagine a sphere (in our 3 dimensions). If you drip a bit of water on it, the water will run down to a particular point. It goes down to the same point no matter where you drip the water. If you rotate it, the water will still all drip down to a particular point (just not the same one as before).

The above is true for some shapes (eggs, lima beans) and not true for others (Lego bricks)

So the question is: what about a 4 dimensional sphere? If you could somehow drip water on it... would the drips all converge to a particular point?

The answer is "yes"

Why did this make the cut, the answer seemed obvious.

Because time and again, things that the entire mathematical community believe to be obviously true turn out to be completely false, and vice versa. One's intuitions about what is hard and what is easy don't count for much until you've spent some time actually trying to prove it.

That isn't the Poincare Conjecture at all... you're asking, essentially, whether a smooth, connected, finite surface has a single local minimum that's also an absolute minimum. That's not a hard question.

The Poincare Conjecture postulates that if any loop on a "nice" surface can be shrunk to a point, it's topologically equivalent to a sphere. ("Nice" here means connected, finite, and without a boundary -- like a sphere or pyramid, but not a disk or infinite plane.) For instance, if the conjecture is true, a cube is topologically equivalent to a sphere, because if you draw a loop on it you can always shrink it down to a point; but a torus (donut) isn't, because a loop around a vertical cross-section can't be shrunk.

Perelman proved the conjecture for three-dimensional surfaces (which are the boundaries of four-dimensional objects).

If the conjecture is true, a cube is topologically equivalent to a sphere

No, the cube and sphere are homeomorphic regardless. (Pf: Points on S^3 are unit 4-vectors. Projecting onto the unit cube is continuous and invertible)

If the conjecture is true, and if you find yourself on a nice 3-surface, then you can conclude you're on the 3 sphere.

It seems I've had that wrong for years. Crap

Unfortunately, HN will not allow me to delete or even edit my previous post.

:-(

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