[trominos]

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).

[mnemonicsloth]

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.

[Sukotto]

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

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

:-(