Slashdot Mirror
The Poincaré Conjecture has Been Proved
Martin Dunwoody, a famous mathematician who works in the field of topology has a preprint that provides a proof of the Poincaré conjecture. This was one of the seven Clay Mathematics Institute millenium prize problems (reported on Slashdot here). The solution to each of the problems carries a monetary reward of 1 million dollars. However there are a number of conditions that still need to be met for the prize to be awarded in the case of the Poincaré conjecture.
You show great skill at cut & pasting from http://www.claymath.org/prizeproblems/poincare.htm : )
Just kidding. Go ahead, enjoy the cut & paste karma.
Wow. I wish I could highlight a section of your post to point out as being wrong as you did the the grandparent. Unforutnately, I can't. You're wrong all throughout.
First, how do you show something is proven? Well, you give a proof. How do I know the proof is correct? I work through all the steps... But what if I mess up and sneeze and my thinking gets confused and I accept something that isn't true? It could happen. Well, I'll just push it through a formal logic computer program that checks it.
But what if the computer has a glitch and a 0 or a 1 gets accepted. Or worse, I made the error while programming the formal logic system. Or more subtly, the compiler or hardware.
Basically, it's like this, proofs are as much a social event as a mathematical cedrtainty. Proofs are presented, and believed, and then refuted. Mathematical proof is a social process carried on by mathamaticians, and you can't forget that. I'm sure that I've proved things incorrectly before, and believed them. Just because nobody's found an error in a published and accepted proof doesn't mean one doesn't exist. If you think that humans can do ANYTHING with probability 1, you're sorely mistaken and are seeing the world in too convenient terms.
Sorry to burst your bubble, but there's a lot of thinking in this. Peer review does not imply flawlessness.
That being said, Martin Dunwoody is a remarkable researcher and this work relies on important, ground-breaking work of Abby Thompson and Hyam Rubenstein, and this preprint sounds very promising!
It's psychosomatic. You need a lobotomy. I'll get a saw.
Surely it should read:
The conjecture that every *compact* simply connected 3-manifold is homeomorphic to the 3-sphere,
Normal euclidean space R^3 is simply connected,
and definitely NOT homeomorphic to to the
3-sphere !!
(That they are not homeomorphic can be proved by
comparing their homotopy or homology groups).
Liam.
Maybe we should give these problems to the people at the next ACM International Programming Contest.
4 (Insightful)!? Almost every statement in this post is incorrect.
The description of simply connected is a description of connectedness. Simply connected means your space doesn't have holes in it, in addition to being connected. This is required, since there are obviously 2-D surfaces (think of donuts) that are connected, yet not homeomorphic to a 2-sphere.
A manifold is a space that is locally homeomorphic to Euclidean space. i.e. if you take a very small piece of the space near a point, it looks like a small piece of R^n. A figure 8 curve is an example of a 1-dimensional space that is not a manifold.
Homeomorphic means that there exists a bicontinuous (continuous in both directions) one-one correspondence between the spaces.
Compactness has precisely nothing to do with surface areas and volumes. If an objects surface area is as small as it can get wrt its volume, it's a sphere, and this has been known for a long time. Secondly, circles are 1-D, not 2-D.
Intuitively the notion of compactness corresponds to being `finite'. In R^n, a set is compact if it is closed (i.e. contains its boundary) and bounded (doesn't stretch off to infinity). The general definition of compactness is more hairy: one way of stating it is that every infinite sequence in the set has a convergent subsequence (note that the limit also has to be in the set).
What the Poincare conjecture states, roughly, is that any closed bounded d-dimensional object in R^n that doesn't have any holes in it (this makes it homotopy equivalent to a d-sphere) is actually homeomorphic to a d-sphere. (Note: it's non-trivial to prove that a compact d-dimensional manifold can actually be embedded in R^n for some n).
Be careful how you phrase that last sentence - your carefree use of the word "obvious" in reference to math calls to mind an old joke:
Two mathematicians were talking one day about some recent work they'd done. One described a proof to the other but quickly glossed over a complicated step. The second one said, "Wait a minute - you didn't prove your last assertion." The reply: "It's obvious."
So the second mathematician wordlessly took a piece of chalk, went to the nearby blackboard, and began to fill it with long statements full of obscure symbols. Nearly half an hour later, he stopped writing, turned around, and said, "You're right. It is obvious."