Re: [math-fun] Poincaré's Conjecture
John writes: << The problem is a subtle one, and there have been quite a few experienced topologists who've thought they'd proved it, but had to retract (in some cases, quite embarrassingly).
When I was a graduae student, I thought I might haved proved it until I mentioned this to Stephen Smale (who had won a Fields medal in 1966 for proving the n-dimensional Poincaré conjecture for n >= 5), who told me he, too, came up with the same erroneous proof as a grad student. (Alas, this did not foreshadow my ever winning a Fields medal.) The "proof" goes as follows: Suppose you have a simply-connected 3-manifold. RMove the interiors of two disjoint 3-balls, getting a manifold with two two-spheres as its boundary. A simple argument shows that it admits a nowhere-zero vector field that is inward on one boundary component and outward on the other one. Clearly the only possibility for any trajectory of the vector field is an interval connecting the two boundary components. A simple argument then shows this manifold has the structure of the cartesian product S^2 x [0,1] -- which shows the original manifold was in fact S^3. --Dan
At 02:00 PM 10/8/03 -0400, you wrote:
The "proof" goes as follows: Suppose you have a simply-connected 3-manifold. RMove the interiors of two disjoint 3-balls, getting a manifold with two two-spheres as its boundary. A simple argument shows that it admits a nowhere-zero vector field that is inward on one boundary component and outward on the other one. Clearly the only possibility for any trajectory of the vector field is an interval connecting the two boundary components. A simple argument then shows this manifold has the structure of the cartesian product S^2 x [0,1] -- which shows the original manifold was in fact S^3.
I'm feeling dumb for not spotting the flaw right away. But: assuming that you haven't cheated by hiding the flaw under either of your "simple arguments", is the problem that there might be more than one way to sew the two caps back onto the S^2 x [0,1] cylinder? Jon Perry hasn't shown us much of his proof, but a hint he dropped prompts me to make a guess of what he's thinking. I conjecture that Jon has developed some kind of description, signature, or characterization of 3-manifolds, and can prove that simply-connectedness imposes a constraint that only one such signature meets. If this is the approach, the place to look for a flaw is the necessary lemma that two manifolds with the same signature are homeomorphic. -A
A good external criticism of this argument is that it applies equally to all 3-manifolds. It doesn't mention or use the condition that the manifold is simply-connected, so if there weren't a flaw it would show that the 3-sphere is the only 3-manifold (and that 2 + 2 = 3, etc.) Internally, the problem is that not all trajectories have to enter or exit through the boundary. For instance, there may be circular trajectories of the vector field. There can also be much more complicated phenomena---e.g. orbits whose closure is an "exceptional minimal set" whose cross-sections are Cantor sets. However, this schema is useful in 2 dimensions. Smale used this picture to show that the group of diffeomorphisms of a disk fixed on the boundary is contractible. There's some discussion of this stuff in my book "Three-dimensional geometry and Topology". Bill Thurston wpt4@cornell.edu or wpthurston@mac.com On Thursday, October 9, 2003, at 04:18 AM, Allan C. Wechsler wrote:
At 02:00 PM 10/8/03 -0400, you wrote:
The "proof" goes as follows: Suppose you have a simply-connected 3-manifold. RMove the interiors of two disjoint 3-balls, getting a manifold with two two-spheres as its boundary. A simple argument shows that it admits a nowhere-zero vector field that is inward on one boundary component and outward on the other one. Clearly the only possibility for any trajectory of the vector field is an interval connecting the two boundary components. A simple argument then shows this manifold has the structure of the cartesian product S^2 x [0,1] -- which shows the original manifold was in fact S^3.
I'm feeling dumb for not spotting the flaw right away. But: assuming that you haven't cheated by hiding the flaw under either of your "simple arguments", is the problem that there might be more than one way to sew the two caps back onto the S^2 x [0,1] cylinder?
Jon Perry hasn't shown us much of his proof, but a hint he dropped prompts me to make a guess of what he's thinking. I conjecture that Jon has developed some kind of description, signature, or characterization of 3-manifolds, and can prove that simply-connectedness imposes a constraint that only one such signature meets. If this is the approach, the place to look for a flaw is the necessary lemma that two manifolds with the same signature are homeomorphic.
-A
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At 06:44 AM 10/9/03 -0400, you wrote:
A good external criticism of this argument is that it applies equally to all 3-manifolds. It doesn't mention or use the condition that the manifold is simply-connected, so if there weren't a flaw it would show that the 3-sphere is the only 3-manifold (and that 2 + 2 = 3, etc.)
I don't think Dan Asimov would make an error that blatant. I had presumed that the assumption of simple connectivity was discharged behind one or both of the two fig-leaves in his synopsis, marked by the two occurrences of the phrase "a simple argument shows ...".
Internally, the problem is that not all trajectories have to enter or exit through the boundary. For instance, there may be circular trajectories of the vector field.
If this were the problem, it would be hidden in Dan's second "simple argument", which purports to demonstrate that the structure of the altered manifold is S^2 x [0,1]. As I stated before, Dan's presentation seems to imply that the flaw is in the open, not in one of the omitted simple arguments. -A
'Jon Perry hasn't shown us much of his proof, but a hint he dropped prompts me to make a guess of what he's thinking. I conjecture that Jon has developed some kind of description, signature, or characterization of 3-manifolds, and can prove that simply-connectedness imposes a constraint that only one such signature meets. If this is the approach, the place to look for a flaw is the necessary lemma that two manifolds with the same signature are homeomorphic.' This is close. I have developed a schema for characterizing ANY x-manifold, and in my scheme simply connected spaces fall into exactly one category, and hence are equivalent. As the maths is new, I haven't yet explored all the necessary bits that form a full proof, but I don't see this as being as hard as the original conjecture seems. I would spill the beans, but with $1million up for grabs, my sense of need for privacy is still winning. Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com
At 01:36 PM 10/9/03 +0100, you wrote:
This is close. I have developed a schema for characterizing ANY x-manifold, and in my scheme simply connected spaces fall into exactly one category, and hence are equivalent.
This is exactly where your approach is vulnerable. It depends on a theorem, which you have almost stated explicitly, but not quite: If space S and space T have the same characterization, then S and T are homeomorphic. You must prove that, for your particular characterization scheme, or the approach is worthless. And you have to be prepared for deep skepticism from topologists. Topologists have an enormous arsenal of such characterizations, and every single one suffers from one of two problems: either it does not always assign the same character to the same space, or it sometimes assigns the same character to different spaces. Good luck with yours. But you can't dodge the responsibility for proving that the characterization 'works'; that's the whole game. -A
participants (4)
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Allan C. Wechsler -
asimovd＠aol.com -
Jon Perry -
William P. Thurston