Thanks - I can see the weaknesses in my argument. However, is there anything wrong with this; PC is true because from the 3-sphere, a compact, simply connected manifold, through only operations which preserve homeomorphicity, we can attain every other compact, simply-connected 3-manifold. 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 -----Original Message----- From: Andy Latto [mailto:Andy.Latto@gensym.com] Sent: 14 October 2003 19:00 To: 'perry@globalnet.co.uk' Subject: RE: [math-fun] Poincaré's Conjecture

-----Original Message----- From: Jon Perry [mailto:perry@globalnet.co.uk] Sent: Tuesday, October 14, 2003 7:00 AM To: math-fun Subject: RE: [math-fun] Poincaré's Conjecture

is this clear, or do I require further explaning?

Apparently, Dan Asimov's post, which I found completely clear, requires more explanation. Let me try to make it clearer. You think you have a new way of proving facts about topology. You are wrong. Your techniques claim to prove that all simply connected manifolds of a given dimension are homeomorphic. This is a false statement. Even if you discover that you have used compactness in your "proof", and weaken your claim to "all simply connected compact manifolds of a given dimension are homeomorphic", this is still false, since S2 x S2 and S4 are simply connected compact 4-manifolds that are not homemorphic. So I am absolutely certain, as is Dan, and everyone else on this mailing list with the slightest understanding of topology, is certain that your proof technique is seriously flawed. So the process of trying to understand it from your vague hints would not be a useful one, since it is certain to result, if you are able to make what you are saying sufficiently precise to be comprehensible, in discovering that what you are saying is in error. So what you should be saying isn't "I have a proof of a result that those who have studied topology intensely for decades have not been able to prove" (because you clearly don't), but rather "I must be confused in some way about the basic concepts of topology; I have what seems to me to be a simple argument, but it produces a contradiction; can you help me find the flaw?". You might get more useful feedback with that sort of attitude; otherwise, people will just think you're a crank, who is producing ravings, rather than mathematics. If you want to make your argument clear to others, the first step is to make it clear to yourself. Here are a few steps that may help in this process: 1. You use the term 'solid'. Mathematics deals only with precisely defined terms. This is not a standard technical term in topology. So you need to define it precisely. It would probably also be useful for you to prove some elementary theorems about "solid" spaces, using your definition. What are the properties of solid spaces that guarantee that the sections of a simply connected manifold are solid? What are the properties of solid spaces that guarantee that a space with "solid" projections must be homeomorphic to a sphere? Prove that all of these properties hold, directly from your definition of "solid". 2. Be careful with quantifiers. Are you saying that in every projection of a simply connected space to a lower dimension, every projection must be solid? Or are you just saying that every simply connected space has at least one projection in which every projection is solid? In either case, can you prove this? 3. Again, define your terms, and be careful when you use pre-existing terms. A projection, say from (x,y,z) space to (x,y) space, is a map that is many-1, that maps every point of the form (x,y,z) to the point (x,y). So the projection of a geomtric sphere, for example, would be a closed disk. I'm guessing that isn't what you mean when you say projection, that instead, you are looking at the manifolds (x,y,t) for each value of t. These would be circles (and notice that the circle is *not* simply connected, even though the sphere is), and in 2 cases consist not of a manifold 1 dimension lower at all, but of a single point. So if you are looking at these lower-dimensional manifolds, you need to deal with the fact that they need not be simply connected, and need not be manifolds at all. 4. The first step in proving something is to understand the question. The Poincare conjecture states that (A)Any compact, simply-connected 3-manifold is homeomorphic to the sphere. Since the statement (B)Any simply-connected 3-manifold is homeomorphic to the sphere is false (consider R3), any attempt to prove this statement is obviouly completely futile, and doomed to failure. So any proof of the Poincare conjecture that has a hope of being correct must make use of compactness. But you can't make use of compactness without knowing what compactness is. And I don't mean a vague idea; I mean a completely and absolutely precise understanding. So if you can't define compactness, you have no hope of proving the Poincare conjecture, and it would obviously be a complete waste of time for anyone to look at anything that you thought was a proof. Can you define compactness for me, and show me exactly where you use compactness in your proof? Andy Latto andy.latto@pobox.com

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 -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com]On Behalf Of Jon Perry Sent: 13 October 2003 13:19 To: math-fun Subject: RE: [math-fun] Poincaré's Conjecture

Consider a circle that varies in radius from 0 to R over time. Placing this circle at the start of a 3-d space and extending over the z-axis produces a sphere.

So, in 3d space we have 3 axis-dependent projections for a space. Similarly in n-d space.

Reversing the process and taking the projections for the space, simply connected spaces consist only of 'solid' projection, and hence are homeomorphic. Other spaces have various but definitely different projective spaces, for example the cylinder with a hole consists of a circle with a hole when viewed down the tube, and a widening than shrinking rectangle that splits when viewed from either perdendicular. 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 -----Original Message----- From: math-fun-bounces@mailman.xmission.com

[mailto:math-fun-bounces@mailman.xmission.com]On Behalf Of asimovd@aol.com Sent: 13 October 2003 08:31 To: math-fun@mailman.xmission.com Subject: Re: [math-fun] Poincaré's Conjecture Jon Perry writes: << ... 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.

...

It really would be better if you devoted some effort toward making your posts comprehensible. Here you don't say what kind of "equivalent" you are talking about, so your assertion conveys zero information to anyone reading it. If you mean topologically equivalent, then this assertion is false, since (e.g.) each of the 4-manifolds S^4 and S^2 x S^2 is simply-connected but they are not homeomorphic. You also don't say what kind of manifolds you are considering (e.g., compact). If noncompact manifolds are covered by your assertion, then R^2 and S^2 provide a simpler counter-example of two simply-connected manifolds of ther same dimension that are non-homeomorphic. --Dan Asimov