3 ms·
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
by Sukotto 17y ago
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"
- Retric 17y agoWhy did this make the cut, the answer seemed obvious.
- jplewicke 17y agoBecause 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.
- trominos 17y agoThat 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).
- mnemonicsloth 17y agoIf 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.
- Sukotto 17y agoIt seems I've had that wrong for years. Crap Unfortunately, HN will not allow me to delete or even edit my previous post. :-(