Here's an excerpt describing the Poincare conjecture,"Poincaré conjecture states that every simply connected closed three-manifold is homeomorphic to the three-sphere".The problem is that smooth three-manifolds (octonions) have only a single point projection onto a Euclidean space (the space of the three-sphere)! That's basically the same as saying Fermions have only position and momentum, all the other dimensions are orthogonal/rolled up. BTW, simply connected doesn't mean there are no knots (since there are tons of knots in octonions), but it does mean all the knots must unwind at the single point in Euclidean space.Unfortunately, there is no way to be homeomorphic to a single point. So I's say the Poincare conjecture is neither right or wrong, it's just a stupid question.

http://en.wikipedia.org/wiki/Poincaré_c ... 91QuoteThe Poincaré conjecture is one of the most important questions in topology. It is one of the seven Millennium Prize Problems, for which the Clay Mathematics Institute is offering a $1,000,000 prize for the first correct solution. does Mr.N still feel its a stupid question.....im no math geek/quant/elite#...but after reading those articles i feel watever that conjecture is....its def not a stupid question..

Okay if the question ain't stupid, the answer "N/A" makes it seem so.

QuoteOriginally posted by: NHere's an excerpt describing the Poincare conjecture,"Poincaré conjecture states that every simply connected closed three-manifold is homeomorphic to the three-sphere".The problem is that smooth three-manifolds (octonions) have only a single point projection onto a Euclidean space (the space of the three-sphere)! That's basically the same as saying Fermions have only position and momentum, all the other dimensions are orthogonal/rolled up. BTW, simply connected doesn't mean there are no knots (since there are tons of knots in octonions), but it does mean all the knots must unwind at the single point in Euclidean space.Unfortunately, there is no way to be homeomorphic to a single point. So I's say the Poincare conjecture is neither right or wrong, it's just a stupid question.Hm, first of all it's a manifold so by definition it has a set of maps, where the manifold is covered by all the maps and each map has a homeomorphism onto an open set of euclidian space(basically it's a disco). So how is it a single-point projection?And three-sphere isn't Euclidean Space because you can't enter the local coordinates with the help of one map(e.g. north and south poles have problem) so three-sphere isn't homeomorphic to Euclidean space. I don't get your argument...

Hm, first of all it's a manifold so by definition it has a set of maps, where the manifold is covered by all the maps and each map has a homeomorphism onto an open set of euclidian space(basically it's a disco). So how is it a single-point projection?Requiring n-manifolds to be smooth is a big time constraint. For physicists, smooth n-manifolds are "phase-space". You can't have knots with one-manifolds since there are so few degrees of freedom; all one-manifolds are simply connected and homeomorphic to the one-sphere. (field - complex numbers)Two manifolds have exactly the right number of degrees of freedom for knots, but those knots cannot be unwound. There is only one (among a finite number) of two-manifolds that is simply connected and homeomorphic to the two-sphere, but it's also isomorphic to a one-manifold - so who cares about it? Those non-simply connected smooth manifolds map back to Euclidean space at points on a lattice (as one would expect with elliptic curves). Two-manifolds are unique in that they are the only torsion-free manifolds. (field - quarternions)Three-manifolds that are smooth have just the right number of degrees of freedom to both have knots and also unwind them to stay in Euclidean space. This can only happen at the single point since the space is the direct product of two quaterions. (field - octonions) And of course, we know the algebra of octonions is so limited that there can be only one element in its group, the identity. What's key is that local coordinates don't mean squat unless the global map is correct.

Suppose you are a genius like Poincare and invent a new subject (topology). You bring in some objects of study that have remarkable applications (manifolds). You also invent - or massively popularise - some tools to study and distinguish these objects (homology, the fundamental group).If you invent / discover an invariant - you are obliged to show that it is useful. One of your first questions should be "How good are my tools at distinguishing these things?" i.e. to what extent does the fundamental group determine the space. This is still very much a field of active research (I did my PhD in it).The Poincare Conjecture asks the most simple version of the question. Is the invariant - the fundamental group - sufficient to determine whether the space is trivial?I would say that the question was not irrelevant, although I consider Pure maths as relevant. I will admit that the answer "yes" was a lot less interesting than the answer "no".

QuoteOriginally posted by: timedsSuppose you are a genius like Poincare and invent a new subject (topology). You bring in some objects of study that have remarkable applications (manifolds). You also invent - or massively popularise - some tools to study and distinguish these objects (homology, the fundamental group).If you invent / discover an invariant - you are obliged to show that it is useful. One of your first questions should be "How good are my tools at distinguishing these things?" i.e. to what extent does the fundamental group determine the space. This is still very much a field of active research (I did my PhD in it).The Poincare Conjecture asks the most simple version of the question. Is the invariant - the fundamental group - sufficient to determine whether the space is trivial?I would say that the question was not irrelevant, although I consider Pure maths as relevant. I will admit that the answer "yes" was a lot less interesting than the answer "no".Here's the real question - Can homotopy groups have any relationship (at all) to two and three manifolds? If not, the Poincare conjecture is obviously a pretty stupid idea. If yes, the solution to Navier Stokes and other nonlinear PDEs would be trivial. right?

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