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The Poincaré Conjecture has Been Proved 307

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.
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The Poincaré Conjecture has Been Proved

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  • Wierd Problem (Score:3, Interesting)

    by KagatoLNX ( 141673 ) <`kagato' `at' `'> on Sunday April 07, 2002 @05:51AM (#3298337) Homepage
    If you follow the link to the description of the problem, it gets really wierd. Apparently this is one of those problems where you have to prove it for 1=7} but no one ever managed n=3 (which was the original, non-generalized conjecture anyways). Funny that this guy just had to fill in the last blank.
    • seems rather an inelegant way to make a general proof; general proof for n>=5, the seperate proofs for n=1, n=2, n=3 and n=4. Does this new proof just do n=3, or is it a _nice_ general proof?
      • Re:Wierd Problem (Score:2, Informative)

        by leviramsey ( 248057 )
        seems rather an inelegant way to make a general proof; general proof for n>=5, the seperate proofs for n=1, n=2, n=3 and n=4. Does this new proof just do n=3, or is it a _nice_ general proof?

        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).

        • This is an inductive proof. Anyone who finds the technique inelegant better worry, since pretty much the entire field of Computer Science is held up by similiar "inelegant" logic.

          Come to think of it, most of Abstract Algebra is as well. Someone want to help me out here? I only audited AA at eight in the morning and dated a math major. I wasn't paying much attention to math on either occasion.

      • Without reading the preprint, I cannot say (not that I could understand it anyway :) ). But it wouldn't surprise me if the proof was just for 3.

        R^3 is kind of a magical place. R^2 might not have enough wiggling room, but R^4 might have too much. There exists a cross product in only R^3.
        • Re:Wierd Problem (Score:2, Informative)

          by avsed ( 168886 )
          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.

          • Uh... sort of. First of all, the cross product isn't just any outer product, it's an antisymmetrised one, more of a wedge product than a tensor product. Anyhow, you can certainly do similar things in other dimensions, but the objects you get won't all be "of the same type", in a loose way of speaking. Somewhat more precisely, only in 3 dimensions do we have that the spaces of 1-forms and 2-forms have the same dimension, so that if a and b are one forms, (a /\ b) can be somehow identified with another one form. Even in 3 dimensions there really isn't a canonical way to do this, which is why cross products end up needing an orientation (the whole right-hand rule thing, which is rather arbitrary).
        • Re:Wierd Problem (Score:2, Informative)

          by caffeined ( 150240 )
          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.

          If you're interested in reading up on it a bit, the link in the original post to " tml" is excellent. (And it's where I learned the above stuff about the various cases.

          • That site doesn't define 3-manifold and 3-sphere. Is it a 3D sphere in 4D or a 2D sphepe in 3D? It it's the former, then the result hardly has any practical use, since we live in 3D and mostly deal with 2D manifolds :-)
        • If by cross product you mean a bilinear map (a,b)-> a x b on some vector space where the product is orthogonal to the arguments and |a x b|^2+(a.b)^2=|a|^2|b|^2 then you may be surprised to hear that there are many cross products in dimension 7. They come from the octonions, the non-associative 8D generalisation of the quaternions. See here [].
          • Sure, and by the Celebrated Theorem of Froebenius, there is no such thing in a higher dimension. :)
      • Re:Wierd Problem (Score:2, Informative)

        by splorf ( 569185 )
        Here's [] some further info on the Poincare conjecture.

        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.

      • 3 dimensional spaces are exceptionally strange. there are a large number of theorems that have trivial proofs for 1 or 2 dimensional spaces, and work out very easily for 4 or more, but for some reason, 3-space is highly irregular.

        My knowledge of topology isn't terribly deep, so you'll have to ask others for more info.
    • You can prove anything :-)
    • Lots of problems are like this. Namely, every NP-Complete problem is like this.

      You can prove that a problem is NP-Complete by restating the problem as a polynomial-time manipulation of another problem. Sort of like "if the red marbles in this problem are considered the nodes of the tree in this other problem, then I can say this because I know something about red marbles."

      In this case, n > 3 is simply a problem that can be stated as a manipulation of n = 3.
  • by squidinkcalligraphy ( 558677 ) on Sunday April 07, 2002 @05:51AM (#3298339)
    The Poincaré Conjecture proved, and microsoft ads on slashdot
  • by Anonymous Coward
    so we finally have mathematical proof that a teacup is a donut for every teacup in the known (Euclidean) universe
  • by Anonymous Coward
    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 [].
  • Here's the proof:

    assume a, b, c such that: a + b = c

    then 5a + 5b = 5c
    and 4c = 4a + 4b

    adding the two: 5a + 5b + 4c = 4a + 4b + 5c

    shifting some terms around: 5a + 5b - 5c = 4a + 4b - 4c

    simplifying: 5 (a + b - c) = 4 (a + b - c)

    dividing by the common factor (a + b - c): 5 = 4


    • Always nice to see how division by zero can be masked ;-)
    • If a + b = c

      then (a + b - c) = 0

      so 5*0 = 4*0

      someone had to do it ....

    • Theorem: All horses are the same color.

      Proof: By induction. First consider the case of one horse. Clearly, one horse is the same color as itself. Now suppose any set of k horses is the same color. If we take a set of k+1 horses, there are k ways to create sets of k horses, all of which must be the same color under the inductive hypothesis, and all of which contain common horses. Therefore any set of k+1 horses are the same color. Therefore all horses are the same color, by induction.

      • 1) Cows have an even number of legs.
        2) Cows have forelegs and two back legs, equalling six legs.
        3) Six is an odd amount of legs for a cow.
        4) By 1 and 3 cows have both an even number of legs and an odd number of legs.
        5) The only number that is both odd and even is infinity.

        Cows have an infinite number of legs. QED.
  • by LadyLucky ( 546115 ) on Sunday April 07, 2002 @06:32AM (#3298433) Homepage
    could accomplish such a thing :-)
  • by call -151 ( 230520 ) on Sunday April 07, 2002 @07:14AM (#3298528) Homepage
    Normally it take a while for a proof to be verified- a better title would be `A Proof has been announced for the Poincare Conjecture.' The Poincare conjecture has attracted a great deal of attention and lots of remarkable, deep work, but it has also had its fair share of proofs which fell apart under serious scrutiny. Most notably, Colin Rourke and a co-author I can't remember had claimed a proof of the Poincare conjecture in 1987 which took something like a year-plus before the mistakes were found, and took a great deal of energy by a number of mathematicians to find the errors.

    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!

    • Well, actually, it's a beautifully simple, short-and-sweet, easy-to-follow 6 page proof. Most students of topology can easily follow it (well, pretty easily anyway).
      I highly doubt that any errors will pop up at all simply because the proof is elementary. (note to non-mathematicians... elementary and simple or easy are two very different things in math).
      And it's only 6 pages!
  • by Anonymous Coward on Sunday April 07, 2002 @07:42AM (#3298579)
    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).

    • Mods, I think the parent may be a gem lost in the abyss of anonymity. Do what you think is right.
    • Here's the better proof that R^3 and S^3 are not homeomorphic. Here it goes:

      S^3 is compact and R^3 is not.
  • Maybe it's time for me to make a fourth attempt at reading Harper and Greenberg's Algebraic Topology: A First Course. I think I got three quarters of the way through last time...

    Can anyone recommend any other books on algebraic topology?


    • by Tityrus ( 547161 )
      Alan Hatcher's "Algebraic Topology" is, besides being freely available on his homepage [], one of the best & most elegant textbooks I've ever come across.

      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 agree that Hatcher's book is good, although I'm just beginning to learn this stuff. We're using that as the text in Math 263 at Chicago - taught by J.P. May. May has a book as well, "A Concise Course in Algebraic Topology," which is highly categorical in its perspective. I'm struggling to figure out limits and colimits, so I'm afraid I haven't made it past the second chapter yet, but there seems to be a lot of good stuff in that book.

        At a more basic level, Munkres "Topology" is good for point-set stuff, but also has some algebraic topology.

        It isn't about algebraic topology, but I very highly recommend Milnor's "Topology from the Differentiable Viewpoint" and his book on Morse Theory. Guillemin & Pollack also give a very good treatment of differential topology. And Thurston's book on three-dimensional geometry and topology is awesome, but I think I would have had a very hard time getting through much of it without the class I took on it in the fall.
  • English please! (Score:2, Interesting)

    by prestwich ( 123353 )
    Can someone explain what the hell this problem is about in English please? (Preferably avoiding the word manifold).
    • Re:English please! (Score:2, Informative)

      by jso888 ( 114340 )
      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?
  • by the eric conspiracy ( 20178 ) on Sunday April 07, 2002 @09:38AM (#3298774)

    Maybe we should give these problems to the people at the next ACM International Programming Contest.

  • by Anonymous Coward on Sunday April 07, 2002 @10:49AM (#3298949)
    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.
  • I can appreciate that it is very interesting to mathematics folks. thats easy. no one knows what I'm talking about when I mention quantumn physics (I'm not a physicist but I can wrap my head around what I read). Mathematics however just befuddles me to no end. Could several of you math junkies point me in the direction of a good starter text on Mathematics? Something I can pick up at Barnes and Noble. Not the Knuth of Mathematics either. Knuth's titles are enough to make my toes curl and my brain fry. Just a layman's intro to Math will do. I'll ask again when I've figured out the first one.
    • by Anonymous Coward
      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's only been a couple years since I took geometric topology; I shouldn't have forgotten this much, this fast.

    Isn't a sphere with a bubble in it (say, A = {x in R^n: 1/2 < d(x,0) < 3/2}) a 3-manifold? It's an open subset of 3-space.

    Isn't that set A simply connected? You can deformation retract it down to S^2, which is simply connected.

    And yet, even if the fundamental group pi_1(A) = 0, the higher homotopy groups aren't trivial: pi_2(A) isn't zero, so A can't be homeomorphic to a 3-sphere.

    So why isn't this a counterexample to the Poincare conjecture?
    • Isn't a sphere with a bubble in it (say, A = {x in R^n: 1/2

      Sure, but I think the whole point is to prove that a compact 3d shape, that is, the one with the greatest surface area in relation to its internal volume, is a sphere.

      My question is, for the n>3 cases, were they basically doing geometry on hyperspheres? That's one thing I've never been able to wrap my head around.

    • by snarkh ( 118018 )
      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.
    • Basically my problems were:

      The manifold needs to be compact for the conjecture to apply.

      I was thinking of the "3-sphere" as B^3, not S^3.

      Thanks, everyone.
  • that can be made because of this potential breakthrough?

    Just curious, or whether it is just an annoying abstract problem that was solved?


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