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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.
I see no inelegance to this method. One of the steps in the general proof may only work if n>=5. This does not mean that the general proof is invalid.
Essentially, the same method underlies inductive proof (e.g. a general proof that holds for n>s, and a demonstration that n=s combine to n>=s).
Nothing is proven until it is peer reviewed and published in a prestigious journal, and then it must to be out there for some time before it is truly accepted. Also, there may be a mistake that throws the proof off for a few years.
You show great skill at cut & pasting from http://www.claymath.org/prizeproblems/poincare.htm : )
Just kidding. Go ahead, enjoy the cut & paste karma.
Which is exactly what they've done in the paper. They've depicted on how the mesh could possibly collapse.
:-(
They have depicted an 8-gon curve which satisfies the intersection properties, extrapolate using a 2 vertex model and use that to show the possible collapse. They've not depicted the collapse per-se in action tho.
This proof does just d=3 and it's interesting that it's essentially combinatorial. Smale's proof for d>=5 was based on differential topology, a grand and beautiful branch of pure higher math. Freedman's proof for d=4 used Yang-Mills theory developed in particle physics. d=3 looks like essentially a computer scientist's proof.
Disclaimer: I don't understand this stuff in any detail--these remarks are based on looking at the preprint and remembering stuff that I heard in math class long ago. Also, I think I'll wait to hear what the math community says, before believing the problem is really finally solved.
No, one can have a cross (outer) product in any number of dimensions, just as one can have an inner ("dot") product in any number of dimensions. Tensor calculus is the generalisation of ordinary geometric calculus that describes this.
Dan
I don't get it, what is so difficult? Here is one more time for you slow guys:
The n = 1 case of the generalized conjecture is trivial, the n = 2 case is classical, n = 3 remains open, n = 4 was proved by Freedman (1982) (for which he was awarded the 1986 Fields medal), n = 5 by Zeeman (1961), n = 6 by Stallings (1962), and n >= 7 by Smale in 1961. Smale subsequently extended his proof to include n >= 5.
Now what part doesn't make sense? *efg*
i hate pansy republicans
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.
http://www.math.princeton.edu/jfnj/texts_and_graph ics/erratum.txt
-Kevin
No, x/0 is undefined. However, you can do things like
because, when y approaches zero, x/y will obviously become larger. But that is not the same as0*infinity is undefined, however, continuing the example above, I could write:
i.e. in that example, "0*infinity" would be zero.
The problem with infinity is that you can't use it like a number, because it isn't one. Infinity literally means that there is an infinite number of things, e.g. the set of integers is infinite, meaning you can never list all integers because there is always a successor. You'll never "arrive at infinity" when listing integers. This means you can calculate with infinity only with equations that involve sequences and their limits. (Like the above-mentioned lim y->0 which means that y is a sequence of numbers approaching zero, and not y = 0. A suitable sequence might e.g. be y[n] = 1/n with n = 1, 2, .... Obviously, this sequence is approaching zero, but will never be equal to zero.)
Sig (appended to the end of comments I post, 54 chars)
I'll try, but I get WAY out of my league when you talk about anything bigger than n=3.
n represents dimensions. i.e. n=1 is one dimension, n=2 is two dimensions, n=3 is three dimensions, etc.
"simply connected" just means that the boundary surrounding something is connected. For example, in n=2 space (a piece of paper for example), if you drew a line around a bunch of ants, and connected the ends, it would be simply connected. If your line was actually two lines, and weren't connected (you had two groups of ants) then you have a multiply connected boundary.
A manifold (sorry, had to use it) is just an object without a boundary. The earth is a manifold, as is any other n=3 space (3d) object that is connected to itself. In n=3 space, the only way you can have a boundary is to have two different objects, in two different locations.
Homeomorphic just means one object is like another.
They generalize the whole idea to one where all the objects are compact. That just means that the objects "surface" area is as small as it can get for a given internal volume. For example in n=3 space, you can minimize an area (like the material of a balloon) in relation to the volume inside (the helium). Circles are compact for n=2 space, and spheres are compact for n=3 space. BTW, even though I state this as if it were a fact, we don't know about all the compact spaces where n > 2. It would *seem* to make sense that a sphere is the only compact object to 3 space, but stating that as a truth, as of today, isn't possible. Maybe we can do that after they win their million bucks....
So, the whole thing boils down to showing that a compact 3d object is the same as a sphere.
Kord
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The way I'm thinking about the problem, is this. Given the condition that no point on the rubber band can ever break contact with the surface of the object it's wrapped around (sphere or doughnut -- I think that's a torus):
You could move the rubber band towards an arbitrary apex of a sphere until the rubber band condenses to a single point at the apex. This applies to other volumes such as cubes and cones or even a randomly squeezed bit of toothpaste.
On the other hand, this can't be done for a torus when you've stretched a rubber band around the wide way, because dealing with the hole in the doughnut would mean having to break contact with the surface of the dougnut.
They're asking for topological proof that this is the case. Don't ask me to describe simple connectedness in plain English; it's an intuitive thing for me -- someone whose last math course was calculus 101.
What I don't get is why you can't cheat when initially placing the rubber band on the doughnut, and stick it to one side of the hole so that the shrinking process never has to cross the chasm, as it were. Or is that besides the point?
Also, what are the real world applications of this proof?
I'm somewhat familiar with this proofs used in different dimension ranges. It's absolutely necessary to separate out the proof into separate cases because the topology changes wildly with dimension. Roughly speaking in dimensions 4 there is so much room that certain powerful general techniques become possible (essentially, half the dimension of the manifold is more than 2 dimensions away from the full dimension --- so submanifolds of half the dimension cannot be KNOTTED). In dimension 3 and 4 special techniques must be used (and they are different in each case). In dimension 4, a submanifold of half the dimension (i.e, 2) can be knotted in the full manifold, but one can analyze the types of knotting that occurs. Manifolds of dimension 3 need techniques UNIQUE to this dimension (incompressible surfaces, etc.). The case of dimension 3 has been the hardest.
Yes, the proof is only for n=3. Poincare's original conjecture was only for n=3, but was later extended to be for all n. The other cases have been proven already, so this proof takes care of the remaining (original) case.
h tml" is excellent. (And it's where I learned the above stuff about the various cases.
If you're interested in reading up on it a bit, the link in the original post to "http://mathworld.wolfram.com/PoincareConjecture.
Sigh. My id isn't prime. 2 2 2 2 2 3 5 313
He also has some other books on more advanced topics in algebraic topology, in various stages of completion, but I haven't read those yet.
I'll try, but I get WAY out of my league when you talk about anything bigger than n=3.
**You're out of your league already
n represents dimensions [wolfram.com]. i.e. n=1 is one dimension, n=2 is two dimensions, n=3 is three dimensions, etc.
"simply connected" just means that the boundary surrounding something is connected. For example, in n=2 space (a piece of paper for example), if you drew a line around a bunch of ants, and connected the ends, it would be simply connected. If your line was actually two lines, and weren't connected (you had two groups of ants) then you have a multiply connected boundary.
**No, simply connected means that any (n-1)-loop embedded in the topological space is homotopic to a point
A manifold (sorry, had to use it) is just an object without a boundary. The earth is a manifold, as is any other n=3 space (3d) object that is connected to itself. In n=3 space, the only way you can have a boundary is to have two different objects, in two different locations.
**No, a n-manifold is an object such that every point is locally homeomorphic to R^n
Homeomorphic just means one object is like another.
**No. Homeomorphic means that there is a bijective continuous function between two spaces such that its inverse is also continuous.
They generalize the whole idea to one where all the objects are compact. That just means that the objects "surface" area is as small as it can get for a given internal volume. For example in n=3 space, you can minimize an area (like the material of a balloon) in relation to the volume inside (the helium). Circles are compact for n=2 space, and spheres are compact for n=3 space. BTW, even though I state this as if it were a fact, we don't know about all the compact spaces where n > 2. It would *seem* to make sense that a sphere is the only compact object to 3 space, but stating that as a truth, as of today, isn't possible. Maybe we can do that after they win their million bucks....
**No. Compactness means that any open cover of the space has a finite subcover.
No, it means that every loop in the space can be smoothly contracted to a point. This can happen in a plane, for instance. But suppose you punctured the plane with a hole; a loop surrounding the puncture would get "caught" on it if you tried to shrink the loop down to a point. A punctured plane is not simply connected.
An "object"? That's kind of vague. A manifold is, roughly, just an n-dimensional "surface" -- e.g., a 0D point, a 1D line or curve, a 2D plane or sphere, a 3D Euclidean volume, etc. Somewhat more precisely, it's something that "locally looks like" an n-dimensional Euclidean space. (e.g., if you look at a piece of a sphere, it looks like a piece of a 2D Euclidean plane, as far as topology is concerned; it's only when you look at the whole sphere globally at once that you see that it's not a plane.)
Sometimes the definition of "manifold" includes the possibility of a boundary; sometimes they are separated denoted "manifolds-with-boundary".
Pretty vague again. There are many ways an "object" can be "like" another object.
Two topological spaces are "homeomorphic" if one can be continuously deformed into the other (e.g., by stretching, squeezing, and bending, but not cutting, pasting, or tearing.)
(A manifold is a topological space that is locally homeomorphic to an n-dimensional Euclidean space.)
That's completely wrong. The notion of "compactness" exists even when no geometry (and hence no measure of area or volume) is defined. What you are describing is not a compact space but a "minimal surface".
Compactness roughly means "finite", as in the surface of a sphere, compared to an infinite plane. More precisely it means that if you can cover the space with a bunch of (maybe infinitely many) "patches" (open sets or neighborhoods), you can always make do with only finitely many of them in order to cover the space. ("Able to be partrolled by finitely many near-sighted policemen" is how one professor put it.)
Your example is of a 3-manifold with boundary. The Poincare conjecture only considers 3-manifolds without boundary. (Obviously, no 3-manifold with boundary is homeomorphic to a 3-manifold without boundary, since "boundary" is a topologically defined concept.)
Well sorry, but to truly understand this stuff you really do need to have studied a lot of mathematics. I'd say two years minimum of in depth, theory level college mathematics would allow you to read and at least get the gist of most mathematics texts/problems.
The poincare conjecture in the n=3 case is fairly simple to state, it's significance is what is more interesting, and that I cannot remember or find anything useful on at the moment.
Which is not to say you can't have a lot of fun trying to wrap your head around this stuff or other higher level mathematics anyway. Here's a couple general mathematics books with some fun problems in them.
Archimedes Revenge is fairly accessible.
From Here to Infinity By Ian Stewart, that is pretty in depth, but just trying to get the gist could be fun. It has a good chapter on Fermat's Last Theorem
And some of Ian's other books are probably good. Try here
It is not a compact manifold (as you said it is open), therefore Poincare conjecture does not apply.
And of course it is not homeomorphic to the 3-sphere, it is homotopic to the 2-sphere.
A circle is 1-D because topology makes a distinction between a 'circle' (a line figure) and the region bounded by it. The region inside the circle isn't part of the circle, any more than the region outside the circle is. Casual English uses the word 'circle' for the line figure and the region it encloses.
Similarly, if we call a basketball a 'sphere', we are discussing its 2-D surface. If we want to talk about a basketball as a 3-D object, including its internal volume, we must call it a 'ball'. There is such a thing as a 3-D sphere (3-sphere), but it is the surface of a four dimensional 'ball', which, I assure you, is like nothing you've ever seen. Many posters seem to have forgotten this today, and are speaking of 3D spheres, when they mean 2-D spheres enclosing 3-D balls.
2-D spheres have more interesting properties than 3-D balls - which you might not suspect if you think of them (as many do) as mere surfaces (i.e a part or property) of 3-D balls.