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A (Correct) Poincare Proof!?

aphyscher writes "About a year ago, there was an announcement that M.J. Dunwoody had proved the (in)famous Poincare conjecture. His paper turned out to have a slight problem, and so it remained unsolved... until perhaps now! Sergey Nikitin has posted a preprint of what may perhaps be an actual proof."

318 comments

  1. 1. PROVE POINCARE 2. ??? 3. PROFIT! by Subject+Line+Troll · · Score: 0, Funny
  2. Though i'm not stupid by Squareball · · Score: 0

    can any one just go ahead and explain it in layman's terms? I mean, I haven't read it yet but I can just tell i'm not going to get it ;) hehe j/k

    1. Re:Though i'm not stupid by Alexis+Morissette · · Score: 0, Redundant

      I didn't understand it fully, but from what I can tell, it has something to do with rubber bands and doughnuts.

      --
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      This
    2. Re:Though i'm not stupid by MattRog · · Score: 5, Informative

      Here is an example with all sorts of definitions you can read.

      http://mathworld.wolfram.com/PoincareConjecture.ht ml

      --

      Thanks,
      --
      Matt
    3. Re:Though i'm not stupid by twistedcubic · · Score: 2, Informative


      Probably the simplest layman's explanation I can think of: if something feels like it is the 3-sphere, then it is the 3-sphere.
      By "feels like" I mean that it has certain properties which strongly suggest that it is the real thing.

    4. Re:Though i'm not stupid by stratjakt · · Score: 2, Interesting

      I took a bunch of math, but am no super-whiz. I think it's something like this.

      A simply connected object is homeomorphic to a sphere in 3-space. Ie; An egg-shaped object is a sphere that's been stretched, and can be a sphere again by compressing along one of the axes. A doughnut (properly called a torus) isnt

      This is true in 2 and 3 space. An ellipse is a stretched circle, an egg is a stretched out sphere.

      Poincare's conjecture extends this into n-space. So a 'simply connected' n-dimensional object should be homeomorphic to an n-dimensional sphere.

      At least I think?

      --
      I don't need no instructions to know how to rock!!!!
    5. Re:Though i'm not stupid by tomzyk · · Score: 1

      oh well, THAT answered all of my questions... [/sarcasm]

      Um. Can anyone actually explain this in, like, "plain english"? Is it trying to prove that there can exist a 4-dimensional object that has all points equidistant from a single point in space-time or something?

      --
      Karma: NaN
    6. Re:Though i'm not stupid by Flakeloaf · · Score: 2

      So in other words, what you're saying is that the Poincaré conjecture is the supposition that any n-dimensional solid object of uniform density can be deformed by some reversible mathematical translation into an n-dimensional sphere?

      --

      Am I the only one who heard Roxette to sing "I'm gonna get blitzed for some sex"?

    7. Re:Though i'm not stupid by stratjakt · · Score: 1

      > So in other words, what you're saying is that the Poincaré conjecture is the supposition that any n-dimensional solid object of uniform density can be deformed by some reversible mathematical translation into an n-dimensional sphere

      Holy shit! I said that?

      --
      I don't need no instructions to know how to rock!!!!
    8. Re:Though i'm not stupid by Anonymous Coward · · Score: 0

      No. It has not been proved in 3 space. It has been proved for a superior number of spaces because they are combination of lower number of dimensions.
      (whatever that is. And I invoke the "For god's sake jim, I'm a physicist, not a mathematician" argument)

    9. Re:Though i'm not stupid by yerricde · · Score: 1

      what you're saying is that the Poincaré conjecture is the supposition that any n-dimensional solid object of uniform density can be deformed by some reversible mathematical translation into an n-dimensional sphere?

      True, provided that the solid object doesn't have any holes in it. It's already been proven for n = 1, n = 2, and n > 3; the $1 million prize is for proving the conjecture for n = 3, getting your proof published, and defending the proof for two years.

      --
      Will I retire or break 10K?
    10. Re:Though i'm not stupid by Flakeloaf · · Score: 1

      True, provided that the solid object doesn't have any holes in it.

      Aye, as most objects of uniform density do :)

      Now according to the rules of social karma, the penalty for pointing out someone else's misstatement is to make a doubly foolish error of my own.

      "I have a penchant for bowlegged women who can waterski and sport tattoos of the feet of Russian historical figures."

      There we go.

      --

      Am I the only one who heard Roxette to sing "I'm gonna get blitzed for some sex"?

    11. Re:Though i'm not stupid by MattRog · · Score: 2, Informative

      Thanks for modding redundant even though mine was posted 9 minutes before the last link. :(

      --

      Thanks,
      --
      Matt
    12. Re:Though i'm not stupid by navels · · Score: 1

      The extension into higher dimensions is that a simply-connected, closed n-manifold with the same homotopy groups as the n-sphere is homeomorphic to the n-sphere. For a 3-manifold, you don't have to say the bit about homotopy groups . . . that follows from the simply-connected assumption and Poincare duality.

    13. Re:Though i'm not stupid by Anonymous Coward · · Score: 1, Funny

      Dang. I have moderator points, but I don't see the option to mod this post "Whiney".

    14. Re:Though i'm not stupid by spectatorion · · Score: 1

      this actually has nothing to do with "density," which is a concept from physics more than mathematics. the poincare says that any 3-manifold (think some "solid" embedded in 4-dimensional space - contrast with a 2-manifold like the surface of a sphere or a torus, or the surface any polyhedron) that has no holes, or other undesirable properties, is "homeomorphic" to the set of points in 4-dimensional space that are unit distance from the origin (aka the 3-sphere). homeomorphic roughly means that one space can be continuously deformed into another in a way that is continuously invertible. (taking a square to a circle is homeomorphism, shrinking a square to a point is not because it can't be undone). as you can see, density really has nothing to do it.

    15. Re:Though i'm not stupid by Anonymous Coward · · Score: 0

      Also known as "The Duck Principle"

      If it walks like a duck and quacks like a duck, it's probably a duck.

      BTW: Keep this in mind when dealing with the ladies, fellas.

  3. Read the conjecture... by CommieLib · · Score: 5, Funny

    Mmmm...hypothetical donut...

    --
    If your bitterest enemies are people who hack the heads off civilians, then I would say you're doing something right.
    1. Re:Read the conjecture... by Anonymous Coward · · Score: 0

      "Why shouldn't poor people be allowed to send their children to private schools on the money they pay to public schools? "

      Because they're poor, and most private schools cost a lot more than most public schools. Giving people credits or vouchers based on their paid taxes is reasonable though.

      Oh wait, your nick is "CommieLib," you must be a troll, a 19 y.o. and know everything, or brain damaged. Maybe all three!

      Get a haircut, stop smoking ganja and rolling on X, and maybe some rational thought might start to return! :)

  4. So? by m0i · · Score: 2, Interesting

    I'm not that much into maths, but what will this proof achieve?

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    have you been defaced today?
    1. Re:So? by unicron · · Score: 1

      That yes, ants can be trained to assort tiny screws in space.

      --
      Finally, math books without any of that base 6 crap in them.
    2. Re:So? by jasonditz · · Score: 5, Funny

      Well, if you give the proof at a big university chances are you'll totally score with some PhD Math chick.

    3. Re:So? by Anonymous Coward · · Score: 1, Funny

      It will allow industry analysts to work out just how much money the RIAA looses each second because of MP3s.

      Oh, and it will also predict when Linux 3.0.0 is due for release.

    4. Re:So? by Anonymous Coward · · Score: 0

      1. Prove Poincare conjecture
      2. ???
      3. PROFIT!!!

    5. Re:So? by Telastyn · · Score: 4, Informative

      There's a $1,000,000 award for one thing...

    6. Re:So? by seattle2napa · · Score: 1
      Let's look at the options:

      a. World peace

      b. End of all anti-MS posts on slashdot

      c. Modicum of fame for the author

      d. Riches beyond compare!

      e. None of the above.

      Well, what do you think?

    7. Re:So? by Listen+Up · · Score: 5, Insightful


      Pure science is pure science. All great discoveries that have ever existed have been because of small, previously unrelated pure mathematical works (or other pure true sciences), which when done on their own seem to have no superficial meaning to someone such as an engineer or common layman, but pure mathematics is akin to pieces of a grand puzzle. Each piece is intrinsically linked to the whole picture. Looking at each piece will not reveal the puzzle, although solving each piece on its own will. This proof need not prove anything to an engineer, a computer scientist, a ballerina, or the mailman, but to a mathematician and others who understand its significance (among others) this proof advances the pure science of mathematics...and by that the world will eventually be forever changed.

    8. Re:So? by Anonymous Coward · · Score: 0

      yep, cos there's nothing hotter than math chicks. hoo-boy, them's some fine-ass bitches.

    9. Re:So? by Anonymous Coward · · Score: 0

      I know neither mathematics nor physics but I can offer a guess.
      Regarding only spatial dimensions, the Universe is a 3-manifold. I don't think that the coiled dimensions of string theory would disqualify it from being a 10-manifold. If the conjecture was true then the Universe CAN collapse to a point because it is simply connected. If the conjecture was false, in addition to the universe being multiply connected, then it would be IMPOSSIBLE for the universe to completely collapse into a point.

    10. Re:So? by elrond2 · · Score: 1

      with the phd math chick on the side, sounds like a good prize.

    11. Re:So? by GuyMannDude · · Score: 5, Funny

      Well, if you give the proof at a big university chances are you'll totally score with some PhD Math chick.

      Before you guys all start dissing math chicks, remember that "mathematicans do it smoothly and continuously". I wanted to put that on a bumper sticker and slap it on my car but I went with "My girlfriend can't wrestle but you should see her box" instead.

      GMD

    12. Re:So? by Anonymous Coward · · Score: 0

      I know neither mathematics nor physics but I can offer a guess.
      Regarding only spatial dimensions, the Universe is a 3-manifold. I don't think that the coiled dimensions of string theory would disqualify it from being a 10-manifold. If the conjecture was true then the Universe CAN collapse to a point because it is simply connected. If the conjecture was false, in addition to the universe being multiply connected, then it would be IMPOSSIBLE for the universe to completely collapse into a point.

    13. Re:So? by Anonymous Coward · · Score: 0

      Well i know a few who are wat hot at least.

    14. Re:So? by doi · · Score: 3, Insightful
      All great discoveries that have ever existed have been because of small, previously unrelated pure mathematical works

      Unless I'm mistaken, Archimedes invented the screw pump while taking a bath, and wasn't thinking about the intricacies of helical structures before then. Certainly the mathematics of the time weren't sufficient to fully describe that structure either...it was a purely practical device for a purely practical application, and definitely WAS one of the great discoveries of all time.

      Not to mention the discovery of the word "Eureka!" :-)

      --
      A man's reach must exceed his grasp, or what's an erection for?
    15. Re:So? by Anonymous Coward · · Score: 0

      With 3 Eyes and 1 Nipple.

    16. Re:So? by Anonymous Coward · · Score: 0

      In other words: It doesn't have any practical relevance at all.

    17. Re:So? by Anonymous Coward · · Score: 0

      Which box would that be, then?

    18. Re:So? by sbar · · Score: 1

      Watch out! The phrase Simply connected in topology means that every two paths between any two points can be deformed into eachother in a continuous manner. E.g. Every n-sphere is simply connected, but the torus is not.
      So how do you know that the Universe is simply connected? ;)

    19. Re:So? by Anonymous Coward · · Score: 0

      ive seen your girlfriend's box. it is nice

    20. Re:So? by richie2000 · · Score: 3, Informative
      Archimedes invented the screw pump while taking a bath

      (Note to reader: I'll ignore the obvious troll potential in that statement and go for the semi-serious approach that tapers out at the end) IIRC, he noticed the displacement of a fluid when a body is submerged in it. This lead to displacement of a goldsmith's head since it provided him with a method to test the density (and hence deduce the proportions of the different metals) of a newly manufactured golden crown for the King (whose name I have conveniently forgotten, let's hope no one knows who George Bush was two thousand years from now, but everyone has heard of Stephen Hawking).

      Little known conjecture: If Alexander Graham Bell had been alive at the time, Archie would have forgotten the whole thing when he had to climb out of the bath to answer the phone. Let's decapitate telemarketers!

      --
      Money for nothing, pix for free
    21. Re:So? by simong_oz · · Score: 2, Funny

      remember that "mathematicans do it smoothly and continuously"

      Heh, I got that beat hands down - I'm a tribologist.

      Tribology = study of friction, wear and lubrication

      What's more, my specialisation is biotribology (lubrication mostly) - tribology applied to biological systems. I'm sure you can see where this is headed ...

      --
      "Because it's there." - George Mallory, when asked why he wanted to climb Mt Everest, March 18, 1923 (New York Times)
    22. Re:So? by Reality+Master+101 · · Score: 3, Funny

      Archimedes invented the screw pump while taking a bath

      Actually, it's a bit more logical than that. He discovered the principal of displacement while taking a bath.

      I'm not exactly sure how one would think of "screw pumps" while in the bath. Come to think of it, I'm pretty sure I don't want to know.

      --
      Sometimes it's best to just let stupid people be stupid.
    23. Re:So? by King+Babar · · Score: 2
      Before you guys all start dissing math chicks, remember that "mathematicans do it smoothly and continuously".

      OK, but mathematicians can usually only prove things about what we might "bodily functions" when they consider what goes on in tiny local neighborhoods that are only frequented by other mathematicians. :-)

      --

      Babar

    24. Re:So? by Anonymous Coward · · Score: 0

      The top tribologist at NASA quit the space shuttle program to market a water based lubricant he invented as Astroglide. It's the best lube ever.

    25. Re:So? by tssm0n0 · · Score: 1

      "mathematicans do it smoothly and continuously"

      That beats my favorite math saying:

      What does a constipated mathematician do?

      Works it out with a pencil.

    26. Re:So? by fishexe · · Score: 1

      And I'm stuck in this 95% male tech college.
      Dammit, dammit, dammit!!!!!

      --
      "I don't care about the Constitution!" --Bill O'Reilly, November 17, 2009
    27. Re:So? by Anonymous Coward · · Score: 0
      with the phd math chick on the side, sounds like a good prize.


      You are forgetting one of the fundamental mathematical laws:
      Brains + Beauty = Constant Z

      (Z is the constant that ensures all women are insane)
    28. Re:So? by Anonymous Coward · · Score: 0

      penicilliin-poor experimental controls

      radioactivity-camera film coincidally stored next to readioactive rocks

      transistor-np junctions of large crystals were utilized in radios for years before anyone had a clue as to what was happening.

      vulcanized rubber-mr goodyear and his gum, etc

      teflon - hey..what's this crap at the bottom of my flask?

      lifting properties of an airplane wing - it's the boernelli princliple, no it isn't, yes it is!

      boat propellors - you look at propellors from the past and look at modern ones, you'll know it's been trial and error all the way baby.

      smallpox vaccine - hey, those milkmaids aren't getting sick! let's try to be like milkmaids for awhile and see what happens.

      Look, I don't mean to discredit these discoveries, they've all been great. But to attribute to genius that which comes from somewhere else, is to overlook important mechanisms that play a critical role in the development of technology.

    29. Re:So? by Anonymous Coward · · Score: 0

      "And I'm stuck in this 95% male tech college."

      You would probably go crazy at ASU (Nikitin's college, mine also :-)

      Not only is it SWARMING with highly attractive (by American male standards) females, it's in Tempe Arizona (Warm all year -- they dress "appropriately"). And, yes, this includes math, engineering, and physics students.

    30. Re:So? by Anonymous Coward · · Score: 0

      Economists do it with models.

    31. Re:So? by mertner · · Score: 1

      In many cases, it is necessary to use a conjecture to prove something else. Once a fundamental conjecture is proven to be true, a whole range of other mathemathical conjectures based on it may all of a sudden seem become real proofs!

      Conversely, disproving a conjecture widely assumed to be true could invalidate a lot of work - so getting the proof right is definitely important.

      --
      -- As long as the answer is right, who cares if the question is wrong?
    32. Re:So? by Anonymous Coward · · Score: 0

      Assume a block on a pendulum...
      Hit it with infinite force over zero time...
      So does that mean foreplay comes afterwards?

    33. Re:So? by jejones · · Score: 2

      Well, it depends. Analysts do it continuously. Algebraists do it in groups. Topologists...I don't want to think about that. (Strong deformation retract? Ouch!)

    34. Re:So? by Anonymous Coward · · Score: 0
      What does a constipated mathematician do?

      Works it out with a pencil.

      And what does a constipated tenured mathematician do?

      He has his grad student work it out with a pencil for him...

    35. Re:So? by Anonymous Coward · · Score: 0

      Tribadism?

  5. I solved it! by gerf · · Score: 4, Funny

    But alas, the space alloted in a regular comments window is insufficient to explain further...

    1. Re:I solved it! by Dr+Caleb · · Score: 3, Funny
      I got to:
      and a ? @K + K has less generators
      than M: Hence, by the assumption of mathematical induction
      a ? @K + K @(1 2 3 4 5) and consequently
      M @(1 2 3 4 5):
      The proof is completed. Q.E.D.

      before I had a seziure.

      Q.E.D. my ass! It shoulda been "Whoo hoooo!".

      --
      "History doesn't repeat itself, but it does rhyme." Mark Twain
    2. Re:I solved it! by Anonymous Coward · · Score: 0
      But alas, the space alloted in a regular comments window is insufficient to explain further...


      Nahh, the real reason is that the HTML standards bodies allowed real math typsetting to be castrated in about 1993. Being able to really set equations in HTML was going to be soooo cool. No, so what did we get more flippin' detailed formatting support.
    3. Re:I solved it! by Lil'wombat · · Score: 0, Redundant

      I solved it too! But the lameness filter prevents me from posting.

      --

      Truth: If it's not one thing, it's another

    4. Re:I solved it! by Anonymous Coward · · Score: 0

      I figured out a perl one-liner to illustrate the proof but it won't get past the lameness filter...

  6. 2 Dimensional Sphere? by malfunct · · Score: 1

    Some explain to me what this is. Is it different than a circle? Doesn't a sphere in its basic definition mean 3 dimensions?

    --

    "You can now flame me, I am full of love,"

    1. Re:2 Dimensional Sphere? by coult · · Score: 2, Informative

      A 2 dimensional sphere is one where the sphere is "locally" the same as a flat 2-dimensional plane. That's what we call a sphere. A 3-dimensional sphere is really a four-dimensional object, because it consists of those points in four dimensional euclidean space that are equidistant from the center of the 3 dimensional sphere. So, a 3-dimensional sphere looks "locally" like flat 3-dimensional space. Its hard to visualize.

      --

      All is Number -Pythagoras.

    2. Re:2 Dimensional Sphere? by Anonymous Coward · · Score: 0

      I don't know you, but I am so attracted to you right now, it's hard to describe.

    3. Re:2 Dimensional Sphere? by UberQwerty · · Score: 5, Interesting

      A "two-dimensional" sphere is an ambiguous thing to say. The article could have meant several things. Let me start from the top.

      Doesn't a sphere in its basic definition mean 3 dimensions?

      No. Strictly speaking, a sphere is "the set of all points an equal distance from a particular point." When we say sphere without saying how many dimensions we're working in, people tend to assume we're working in the standard three dimansions.
      A sphere in one dimension is the two points the same distance away from the sphere's center in either direction.
      A sphere in two dimensions is a circle (just the curve around the outside - not the middle area)
      A sphere in three dimensions is a hollow ball.
      A sphere in four dimensions can't be pictured.

      However, a sphere in two dimensions itself is only a one-dimensional thing. It's a bent line. It has length, but no area.

      Likewise, a sphere in three demensions is a two-dimensional sphere. It's only the shell - it has (surface) area, but no volume.

      Without proving it, I can see that a sphere in four dimensions (commonly called a hypersphere) will be three-dimensional. So, when the article mentions a three-dimensional sphere, they really mean a sphere in four dimensions.

      This is a bit of an obnoxious distinction to make, and I certainly think they should have phrased it differently. Usually people say three-sphere when they mean a sphere in three dimensions, which is in fact just a surface and thus two-dimensional. Then, we can say n-sphere and have a sphere in n-dimensional space that is an n-1 dimensional object. However, this kind of quibbling tends to have no effect on proof, which is what math on this level concerns itself with anyway.

      --


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    4. Re:2 Dimensional Sphere? by exeph · · Score: 1

      If you want to get an intuitive grasp of what a 3-sphere is...

      A mathematical sphere of n-dimension is like a mathematical ball (a ball is like a solid sphere) of n dimension where the edges have been glued together. For example, a 2-ball is a disc, and if you glue all the edge of a disc together at one point you get a 2-sphere, which is like a hollow baseball. So to get a 3-sphere, you start with a 3-ball (solid baseball) and glue all the outside edge together at a single point.

    5. Re:2 Dimensional Sphere? by Hard_Code · · Score: 5, Interesting

      Or in other words:

      A 3-d (in layman's terms) sphere casts a 2-d 'shadow' (a circle).

      A 4-d sphere casts a 3-d 'shadow' (a normal sphere)

      Wrap your head around that.

      --

      It's 10 PM. Do you know if you're un-American?
    6. Re:2 Dimensional Sphere? by Anonymous Coward · · Score: 0

      Wouldn't the three dimensional shadow of a fourth dimensional sphere look and act like a solid sphere that gradually becomes hollow and then fills? It would oscillate in a sinusoidal curve between those two states filing from the outside and emptying from the inside. This is all just how I think it would act. I might be totally wrong.

    7. Re:2 Dimensional Sphere? by Anonymous Coward · · Score: 1, Interesting

      A 4-dimensional shape can be pictured by taking time as the fourth dimension.

      Imagine a particle appearing in front of you, gradually growing into a large sphere, and then gradually shrinking back to a point before disappearing completely.

      You will have seen a sphere in 4 dimensions.

    8. Re:2 Dimensional Sphere? by ThrasherTT · · Score: 1

      Wouldn't your "three-sphere, two-dimensional" object mean that the 2D "space" that the three-sphere exists in must be curved into a 3rd dimension, and thus would be 3 dimensional?

      I don't see how one can claim that an object, being defined as a collection of points, each of which having 3 components (x, y, z), could be considered two dimensional.

      --

      All Your Memory Are Belong To Java
    9. Re:2 Dimensional Sphere? by Yobgod+Ababua · · Score: 2

      You can always assign extra components to an object by simply measuring it in additional dimensions. If you specify the time your sphere existed at, giving the points x,y,z and t components, it doesn't make it a 4-dimensional object.

      The dimension of an object is the -minimal- number of components neccessary to differentiate all of the points. In the case of the sphere, you can specify all of the points involved with only latitude and longitude: two dimensions. Similarly, with the circle, you can specify any point with only one dimension, the length along the circle from a defined origin.

      Don't get confused between the dimension of an object and the dimension of the space it exists in. (The minimal space needed to contain the sphere, as you note, is three-dimensional-space).

      Mmmm, topology

    10. Re:2 Dimensional Sphere? by pal · · Score: 2, Informative

      this is not quite right. when people say "three sphere," they mean the same object that you call a sphere in four dimensional space. however, it doesn't necessarily need to be embedded in any four (or higher) dimensional space. the important part is that viewed in isolation, it's got three dimensions of its own.

      the "sphere" you know and love, we call the "two sphere."

      - pal

    11. Re:2 Dimensional Sphere? by BabyDave · · Score: 2

      When we say the sphere is two-dimensional, what we mean is roughly that we can describe it using two coordinates. We do this every day, using latitude and longitude to describe positions on the Earth's surface.

      To be a little more precise about it, at each point in a 2-dimensional manifold, we can find a small neighbourhood around it which "looks like" part of the plane. (which of course is 2 dimensional)

    12. Re:2 Dimensional Sphere? by nahdude812 · · Score: 5, Interesting

      ooh, I love playing with this kind of stuff in my head. Ok, how about this.

      A 1-d line segment (a non-infinite line) casts a 0-d or 1-d shadow in a 1-d or above world: If the "light source" is cast straight down from the end of the line, it will cast a 0-d shadow [a point]. If cast from anywhere else, it casts a 1-dimensional shadow. That is to say that if it is cast from OUTSIDE of the first dimension, it will cast a 1st dimensional shadow [a distorted line segment], given that you have at least a 1-dimensional surface on which to cast the shadow.

      So a 2-d circle casts a 2-d or 1-d shadow. If cast from within the same two dimensions, it casts a line segment shadow, if cast from the third dimension, it casts a 2-dimensional shadow [a distorted circle], given that you have at least a 2-dimensional surface on which to cast te shadow. As far as I am aware, you cannot cast a 0-d shadow off of a 2-d object unless you cast that shadow from the 0th dimension, see my expansion on this in the 3d world below.

      So a 3-d circle casts a 3-d or 2-d shadow. If you cast the shadow from within the same 3 dimensions, you get a 2-dimensional shadow. If you cast it from the 4th dimension, you could get a 3-dimensional shadow given that you have at least a 3-dimensional "surface" on which to cast the shadow. Unless you are casting the shadow from the 2nd dimension (a planar light source), you cannot get a 1-dimensional shadow, and unless you are casting the source from the 1st dimension [a line light source, similar to a laser], you cannot get a 0-dimensional [point] shadow.

      So we derive a formula:
      shadowdimension = lesser([dimension_of_light-1], [dimensions_of_object],[dimension_of_surface])

      Meaning that if we work with a 16th dimensional sphere, and we cast a shadow from the 8th dimension, we will get a 7th dimensional shadow so long as we have an 7th dimensional "surface" on which to cast. The same 16th dimensional sphere with a 23rd dimensional light source would cast a 16th dimensional shadow so long as we have at least a 16th dimensional "surface". And no matter how many dimensions our object and light source are, we can only get a 5th dimensional shadow if we only have a 5th dimensional "surface."

      Did I do that right? I think my brain broke.

    13. Re:2 Dimensional Sphere? by muon1183 · · Score: 2, Informative

      In the mathematical sense, the dimensionality of an object refers to how many dimensions the object itself has, not the dimensionality of the ambient space. A 2-sphere, denoted S2, lives in R3. However, if you examine the surface itself, it is a 2 dimensional surface (as in, a basis for the points on the 2 sphere has only 2 elements). When doing math, dimensionality is a property of the object, not of the ambient space. When you think about this, it makes sense, since there are plenty of examples of 2 dimensional objects which cannot be found in less than 4 examples, the most common of which is the klein-bottle (think of a torus, except it intersects itself if represented in 3 dimensions, since it's made out of a mobius strip instead of an anulus).

      --

      There's no sig like SIGSEG
    14. Re:2 Dimensional Sphere? by Serveert · · Score: 1

      So to get a 3-sphere, you start with a 3-ball (solid baseball) and glue all the outside edge together at a single point.

      I was with you up until this sentence. Maybe I'm being dumb here but I can't visualize this. Could you elaborate? With a disc you have edges to glue together so it's easy to think of glueing the edges together to form a 2-sphere. But with a 3-ball(solid baseball), you don't have such edges.. well you have the ouside of the 3-ball. So do you peel all the edges off like an onion and connect the resulting pieces at one point so you have a bunch of connected 2-balls ranging from small(center of 2-ball) to large(edges of 2-ball) connected at one point?

      --
      2 years and no mod points. Join reddit. Because openness is good.
    15. Re:2 Dimensional Sphere? by Asprin · · Score: 2


      This is a bit of an obnoxious distinction to make, and I certainly think they should have phrased it differently. Usually people say three-sphere when they mean a sphere in three dimensions, which is in fact just a surface and thus two-dimensional. Then, we can say n-sphere and have a sphere in n-dimensional space that is an n-1 dimensional object. However, this kind of quibbling tends to have no effect on proof, which is what math on this level concerns itself with anyway.

      Yeah, Topologists always were kinda weird....

      --
      "Lawyers are for sucks."
      - Doug McKenzie
    16. Re:2 Dimensional Sphere? by DrVeg · · Score: 2, Informative

      This is partially correct and partially misleading. The part that is misleading is to think of a n-sphere as necessarily being embedded in some other space. I think that is where UberQwerty gets confused.
      A topological space is a set of points with the notion of a neighborhood of each point. If every point in the space has a neighborhood that looks like (homeomorphic) familiar 3-space then it is a 3-manifold. Similarly for n-manifolds. (Example: the ordinary hollow sphere is a 2 manifold because little sections of it look like a plane.)
      Our ordinary experience (excluding relativity, string theory, etc.) says we live in a universe that is a 3-manifold.
      Poincare says that a 3-manifold that is simply connected (e.g., able to draw a curve between any two points without going out of the space) and closed (any sequence of points that tend to a limit have that limit in the manifold) is actually topologically eqivalent to the set of points in 4-space equidistant from a given point.
      So thinking of the apparent 3-dimensional universe, it doesn't have "holes" or weird twists like you can do in 2 dimensions on a Mobius band.

    17. Re:2 Dimensional Sphere? by PacoSuarez · · Score: 1
      A sphere in one dimension is the two points the same distance away from the sphere's center in either direction. A sphere in two dimensions is a circle (just the curve around the outside - not the middle area)
      That's not right. A 1-sphere is a circle. A 2-sphere is a hollow ball. We measure the dimensionality of the sphere itself, not the space in which it is embedded, because in topology you can define a sphere without a space in which to embed it. For instance, you can define a n-sphere to be the compactification of R^n by adding a single point at infinity.
    18. Re:2 Dimensional Sphere? by zeus_tfc · · Score: 1

      However, a sphere in two dimensions itself is only a one-dimensional thing. It's a bent line. It has length, but no area.

      Likewise, a sphere in three demensions is a two-dimensional sphere. It's only the shell - it has (surface) area, but no volume.


      Although, following this logic, one cannot view a complete four dimensional sphere in three dimensions, even though, as you say, it is a three dimensional object. Since it is curved through the fourth dimension we can only view a cross section of it at any one time.

      Argh! Too much... head hurts...
      Too much theoretical math for this simple engineer.
      Ouch

      --
      "...At the end of the day"..."when everyone goes home, you're stuck with yourself." RIP Layne Staley
    19. Re:2 Dimensional Sphere? by barawn · · Score: 2

      He said a sphere in 1 dimension, not a sphere of one dimension. The "in" there implied embedding. A 1-sphere is a sphere in 2 dimensions, with dimension 1. A "sphere in 1 dimension" is a 0-sphere.

      Yes, you don't need to embed the sphere in a space, but if you're going to explain it to someone, it can help.

    20. Re:2 Dimensional Sphere? by ThrasherTT · · Score: 1

      I understand what you are saying, but isn't each point on a sphere defined by three values: Latitude, Longitude, Radius? Or do we ignore the radius component since it is constant by definition of a sphere?

      --

      All Your Memory Are Belong To Java
    21. Re:2 Dimensional Sphere? by Alsee · · Score: 2

      I think my brain broke.

      BTW, your warranty expired last week.

      -

      --
      - - You can't take something off the Internet! That's like trying to take pee out of a swimming pool.
    22. Re:2 Dimensional Sphere? by Alsee · · Score: 2

      Wrap your head around that.

      I tried, but since the problem is homeomorphic to a sphere my mind always slid off at a point.

      -

      --
      - - You can't take something off the Internet! That's like trying to take pee out of a swimming pool.
    23. Re:2 Dimensional Sphere? by Yobgod+Ababua · · Score: 2, Interesting

      You do need the radius to describe each point's absolute position in 3-space, but you don't need it to describe a point's position relative to the other points on the surface.

      In other words, because the radius is the same for every point on the sphere, we can ignore it in the same way that we ignore the time on every point on the sphere, or the 5th dimensional position of every point on the sphere, because they are constant.

      A classical plane (which most people accept as a 2-dimensional object) could be defined in 3-dimension space by something like "z=5". Every point has an x,y and z component, but because the z is constant for every point, it can be ignored, leaving the plane as a 2-dimensional x,y specified object. A sphere is just a type of plane described with polar coordinates rather than cartesian.

      Does that help? :) Note that IANAM (i am not a mathemetician), so I may not be as precise, or confusing, as a formal proof.

    24. Re:2 Dimensional Sphere? by exeph · · Score: 1

      the "edge" of the 3-ball is the outside layer(which is a 2-sphere). in a 3-sphere, you have a 3-ball where every point on this outside layer is the same as every other point. you can't picture it in 3-d, but a 3-sphere is kind of like a 3-ball that you can fly around inside and when you get to the edge you're at ALL the edge at the same time

    25. Re:2 Dimensional Sphere? by Anonymous Coward · · Score: 0

      In case this is relevant, in Doom 3 there will be 8 light sources at any given point. Thus, er, wait. No.

    26. Re:2 Dimensional Sphere? by osu-neko · · Score: 1
      Maybe I'm being dumb here but I can't visualize this.

      If you say you can visualize this, I believe you'd either be a liar or a god...

      --
      "Convictions are more dangerous enemies of truth than lies."
    27. Re:2 Dimensional Sphere? by istewart · · Score: 1

      I'm going to reply to this with my highschool-precalculus-student's two cents worth.

      Below, a poster made a comment about the universe folding in on itself in the shape of a hypersphere. Going off the above definition that a sphere is defined as the set of points being equidistant from a given central point, and considering the examples already given of the representations of 1, 2 and 3-D spheres, the following popped into my head. If the fourth dimension is what we perceive as time, then a 4-D sphere would be all points in 3 dimensions that exist around a certain point in 4-D space, which can also be expressed as a certain point in time. Since all points in 3-D space are represented in a hypersphere, our universe could conceivably fit the definition of a hypersphere.

      It sounded good when I first thought of it, anyway.

    28. Re:2 Dimensional Sphere? by ThrasherTT · · Score: 1

      Yes, that does help. Thanks for clearing that up for me (and thanks for not making it a formal proof!).

      --

      All Your Memory Are Belong To Java
  7. Ok by kenp2002 · · Score: 5, Funny

    Ok so rad the pre-writeup on this and I can say this: WAY OVER MY HEAD! I understood about
    1.05E-60% of that. Holy cow. There is proof that higher education still turns out some bright people. I wish I knew what the hell all that was about, it "looks" cool. You could use that as a prop in a movie for some secret formula or something.

    --
    -=[ Who Is John Galt? ]=-
    1. Re:Ok by b0r0din · · Score: 5, Funny
      It's very simple, as the problem asks.

      "Consider a compact 3-dimensional manifold V without boundary. Is it possible that the fundamental group of V could be trivial, even though V is not homeomorphic to the 3-dimensional sphere?"

      What he's saying is, the...er...well, he means that the, uh...


      I fucking HATE french people.

    2. Re:Ok by sneakcjj · · Score: 1

      People doing this stuff are probably genius level, not just taught :).

    3. Re:Ok by noquarter83 · · Score: 1

      This is the sort of thing I'm studying at school (not this EXACT problem, but high-level math.)

      Not to say that I understand this problem or its solution, but I have professors who probably would. And holy crap, the things you hear the guys who do this sort of stuff discussing on their coffee breaks.. ..

    4. Re:Ok by Codifex+Maximus · · Score: 3, Funny

      Ok, I got to page two and hadda stop. He started in with the big sigma and I lost it.

      Hehe... gotta go back and get my advanced trigonometry and calculus credits.

      --
      Codifex Maximus ~ In search of... a shorter sig.
    5. Re:Ok by fruey · · Score: 0, Troll
      I fucking HATE french people.

      I hope that's a joke.

      --
      Conversion Rate Optimisation French / English consultant
    6. Re:Ok by Anonymous Coward · · Score: 0

      Yeah, one of those obvious types, like Seinfeld. Doesn't everyone hate the french?

    7. Re:Ok by cperciva · · Score: 5, Informative

      What he's saying is, the...er...well, he means that the, uh...

      Look at it this way:
      Suppose the universe doesn't have any "edges" -- you can keep on going forever in a straight line without "falling off the edge of the world". Suppose further than there aren't any "wormholes" -- that given two paths between a pair of points, you can continuously deform one into the other. Finally, suppose that the universe is finite in volume.

      Now, the first and third conditions above imply that the universe "folds in on itself". Add in the "no wormholes" condition, and Poincare's conjecture/theorem, and you find that there is only one possible way that it can fold in on itself -- as a hypersphere.

      At least, that's the best explanation I can provide without any formal background in topology or astrophysics.

    8. Re:Ok by fruey · · Score: 1
      Yeah, one of those obvious types, like Seinfeld. Doesn't everyone hate the french?

      Not as much as they hate the British... but anyway, I believe you are only qualified to say you hate a nation of 60 million if you've been there a long time and met more that you dislike than like.

      Of course, I'm married to a French-Moroccan and have a number of great friends in France, so maybe I'm biased. But I reckon that the average young French person (forget their bullshit government) is amongst the cooler peoples in Europe. They don't give a shit about petty authority, smoke good dope, and have great wine.

      --
      Conversion Rate Optimisation French / English consultant
    9. Re:Ok by Soul-Burn666 · · Score: 2

      Doesn't the "without boundry" mean that's it's an "open section"?

      Like, In 2d, x^2+y^2=1 is closed cause it has the border, meaning, a circle around a dot on the border will always be on the 2 sides of the border, however small, while x^2+y^21 is open because every cirlce around any specific dot inside it can be made small enough to be wholely in the area.

      That's studied in basic Calcalus courses...

      --
      ^_^
    10. Re:Ok by fishbowl · · Score: 2

      >Suppose the universe doesn't have any "edges"...
      >...Finally, suppose that the universe is finite in volume.

      How can it be both?

      --
      -fb Everything not expressly forbidden is now mandatory.
    11. Re:Ok by bnenning · · Score: 2

      A sphere has no edges but has a finite area. Just bump it up a dimension.

      --
      How to solve most of our problems: 1.Lots of nuclear plants. 2.Cure aging.
    12. Re:Ok by kalimar · · Score: 1
      Doesn't the "without boundry" mean that's it's an "open section"?
      If I'm understanding you correctly, then no.
      The surface of a sphere (ball) has no boundary. You can keep going in the same direction and never reach an edge. However, it's not "open" per se. The sphere (ball) surrounds a finite space. It's been a while since topology was in the forefront of my mind.
      Another example:

      Take a mobius strip. It is a closed loop. However in one direction it is without boundary.

    13. Re:Ok by fishbowl · · Score: 1

      Ah. Okay. I'm still struggling with what's meant by a Hilbert Space. Actually, I'm still struggling with the plane, r-theta coords, and I totally suck at arithmetic. :-)

      But I'm still going to keep taking maths and eventually get up to number theory and scientific computing. It might take 10 more years but I'm working on it.

      --
      -fb Everything not expressly forbidden is now mandatory.
    14. Re:Ok by crawling_chaos · · Score: 1
      Hehe... gotta go back and get my advanced trigonometry and calculus credits.

      I'm afraid, you'll need more than that. Algebraic Topology makes my head hurt.

      --
      You can only drink 30 or 40 glasses of beer a day, no matter how rich you are.
      -- Colonel Adolphus Busch
    15. Re:Ok by Anonymous Coward · · Score: 0

      I hope it's a joke how they hate Americans. SERIOUSLY hope so.

  8. Re:1. PROVE POINCARE 2. ??? 3. PROFIT! by flynt · · Score: 4, Informative

    Here is step 2! You win 1 million dollars for the correct proof from claymath!

    http://www.claymath.org/prizeproblems/poincare.h tm

  9. It's sad that after reading the problem... by IvyMike · · Score: 5, Funny

    ...the most intelligent thing I can think of is: "Mmmmm...donuts."

    1. Re:It's sad that after reading the problem... by Anonymous Coward · · Score: 0

      It's okay. You still got a +3 Funny, even though someone else already made the same joke and got a +4 Funny.

      I have a theory that all simpsons jokes get modded funny. I'm working on an inductive proof but I need some help. It's an inductive proof but all I have so far is the inductive step:

      Q.E.D.

      Anybody want to help with the basecase?

    2. Re:It's sad that after reading the problem... by Anonymous Coward · · Score: 0

      3) PROFIT!!!

    3. Re:It's sad that after reading the problem... by IvyMike · · Score: 2

      You're probably right. A data point: every time I've attempted to make a Simpsons reference, even if the story has almost no comments when I start to post, my comment is already redundant by the time it shows up. Sorry all.

  10. Someone throw cold water on my face by ekrout · · Score: 5, Funny

    A.) There's now a correct proof of the Poincare problem!
    B.) Jon Katz no longer posts to Slashdot!
    C.) Chris D. starts his own gaming company; plans to fill-in Part 2 of the traditional Steps 1, 2, & 3 to Profit!
    D.) Microsoft is now the largest paid advertiser on Slashdot.org, the be-all-end-all for all Open-Source/Free-Software news

    My brain needs a reboot.

    --

    If you celebrate Xmas, befriend me (538
    1. Re:Someone throw cold water on my face by I'm+not+a+script · · Score: 0, Offtopic

      Microsoft
      Visual Studio.net

      See what develops.

      Try it now.
      Free online hosted session.


      Click here.

      Slashdot is proud to be associated with Microsoft.
      Together, for better software.

      --
      kthx
    2. Re:Someone throw cold water on my face by zapfie · · Score: 1

      Microsoft buys adspace on Doubleclick. Slashdot, among other websites, uses Doubleclick for some ads.

      --
      slashdot!=valid HTML
    3. Re:Someone throw cold water on my face by luisdom · · Score: 1

      And guess what... Nostradamus predicted it! Apocalypse is coming!

  11. So? by i_need_no_nick · · Score: 1
    What's the significance of this proof? I'm no maths guy, so forgive me if it's something earth-shattering in it's importance.

    Is it just because it was a thorny problem, a bit of knowledge for the sake of knowledge?

    Or is it just that the guy is now eligible for the Clay maths prize?

    --

    Must go buy a lottery ticket, front page on /. here I come!

  12. Poincar� Conjecture by taphu · · Score: 5, Informative

    For super-geeks, here is is a more thorough discussion of the Poincaré Conjecture.

    http://mathworld.wolfram.com/PoincareConjecture.ht ml

    1. Re:Poincar� Conjecture by Anonymous Coward · · Score: 0

      Thanks, it all makes perfect sense now.

    2. Re:Poincar� Conjecture by gsfprez · · Score: 1

      >For super-geeks, here is is a more thorough discussion of the Poincaré Conjecture.

      i had to click on every single link of every single link of that explanation page...

      to discover that i am a stupid ass-hat.

      screw the frog math guy that came up with this.. and the frenchman he rode in on.

      --
      guns kill people like spoons make Rosie O'Donnell fat.
    3. Re:Poincar� Conjecture by Anonymous Coward · · Score: 0

      In my opinion this is a better explanation of the conjecture, in part because it was written by an actual mathematician.

  13. Umm,no by wiredog · · Score: 2
    Hypothetical hyper-doughnut. It's a 4 (or more) dimensional object.

    I think. Maybe it is a 3-d doughnut. It's been ten years since I studied that stuff at college.

  14. Mathematical Proof by SniffleBear · · Score: 5, Funny

    This is just mathematical proof that you can wrap a rubberband around an apple. I think the rest of us would be satisfied by a videotape instead.

    1. Re:Mathematical Proof by selderrr · · Score: 2

      nonono ! It's *four* dimensional d00d !

      That's way cooler : you wrap a rubberband around an apple faster than a jury can see it. The fastest math geek gets the $1M and the chicks.

      For the tech-savvy : they're using tiBooks to wrap the band around. Imacs turned out to be a pain when the band reaches the power chord.

    2. Re:Mathematical Proof by Infernus · · Score: 1

      Ah, but a videotape won't prove that all rubberbands or topologically-rubberband-like-objects can wrap around all topologically-apple-like-objects, thus the videotape would be completely worthless, for though the rubberband in the video went around the apple, how will you be SURE the rubberband in your pocket will go around the apple in your lunchbag without the mathematical proof?

    3. Re:Mathematical Proof by Tenebrious1 · · Score: 1, Offtopic



      ... I think the rest of us would be satisfied by a videotape instead.

      As long as it's not quicktime or other proprietary format.

      --
      -- If god wanted me to have a sig, he'd have given me a sense of humor.
  15. Poincar� by Traa · · Score: 2

    The name is Poincaré (with the litte thingy on the e)

    Poin (POINt) - Ca (CAtastrophic)- Ray

    bah, screw it, just call me Henri

    1. Re:Poincar� by jzs9783 · · Score: 1

      I like to call that an "accent" Ack (ACknowledge) - Cent (SENTence)

    2. Re:Poincar� by rsidd · · Score: 2
      Poin (POINt) - Ca (CAtastrophic)- Ray

      Uh, it's poin as in french (roughly poan or pwan with the n a nasal ending, not hard);
      ca somewhere between ca (car) and cu (cuff); re as in ray, as you say.

      bah, screw it, just call me Henri

      pronounced, roughly, ahn-ree or on-ree, again the n is a nasal sound not a hard consonant.

      And btw, if you're not on a French azerty keyboard, feel free to leave out the accents. Most French people I know on qwerty keyboards drop the accents, in fact. As they generally do with capital letters too.

    3. Re:Poincar� by Mithal · · Score: 2, Interesting
      May I suggest that you talk your French friends into trying the "French Canadian" keyboard? [Available in both Windows and Linux]

      A little hard to learn (as it relocates a few key characters like '\' and '~', but it is a QWERTY based keyboard layout that allows me to use all the french accents without any problem... including the capital letters. 'ÀÈÏÔÇ'

      As for Poincaré, I would say: "Pwain Ca (CAtastrophic) Ray", as you did, but I would roll the "R"... But as this sound doesn't exist in english, I suppose it's hopeless to try to teach it here!

    4. Re:Poincar� by rsidd · · Score: 1
      May I suggest that you talk your French friends into trying the "French Canadian" keyboard?

      I suspect it's even harder to find in France than a US qwerty keyboard. I think they're happy the way they are....

      For myself, when I use vim I can use its built-in digraph support, otherwise I just drop accents too.

      As for Poincaré, I would say: "Pwain Ca (CAtastrophic) Ray", as you did,

      Are you Quebecois? The French would not say "pwain" to rhyme with "Twain" anyway -- it would be closer to a rhyme with "won" but not quite that either. And nobody in France pronounces "ca" as in "catastrophic" even in English words -- or perhaps you mean "catastrophique"? (cah-tah-strofeek, with both t's soft)

    5. Re:Poincar� by Abcd1234 · · Score: 2

      Although, it's not really a "rolled" R sound, is it? At least not what I think of as a real rolled R, using the tongue. As I understand it, it's more of a back-of-the-throat sound... but then again, what do I know. :)

    6. Re:Poincar� by Godwin+O'Hitler · · Score: 1

      Back of the tongue actually, like dogs when they trap criminals in corners and say grrrrrrr

      --
      No, your children are not the special ones. Nor are your pets.
  16. Re:1. PROVE POINCARE 2. ??? 3. PROFIT! by Capt+Dan · · Score: 2

    Mmmmmmmmm... million dollar donut.... gaaaawwwwww...

    --
    Sig:
    Barbeque is a noun. Not a verb.
  17. How about some money ? by apankrat · · Score: 3, Informative

    This problem is priced at $1 million if solved.

    --
    3.243F6A8885A308D313
  18. ANNULUS by Anonymous Coward · · Score: 0

    Is the proper name for a doughnut shape.

    Just thought you might find it amusing.

    1. Re:ANNULUS by Anonymous Coward · · Score: 1, Informative

      No, the proper name for a donut shape is 'torus'. An annulus is the figure bounded by and containing the area between two concentric circles.

    2. Re:ANNULUS by Anonymous Coward · · Score: 0
      No, the proper name for a donut shape is 'torus'. An annulus is the figure bounded by and containing the area between two concentric circles.


      No, that's not...oh...nevermind...I was thinking of analingus.

    3. Re:ANNULUS by Anonymous Coward · · Score: 0

      It looks a bit like an ass, though.. got that pinched oval shape.

  19. Re:1. PROVE POINCARE 2. ??? 3. PROFIT! by br0ck · · Score: 2

    Unfortunately the money may take a while in coming. The rules state, "a proposed solution must be published in a refereed mathematics journal of world wide repute, and it must also have have general acceptance in the mathematics community two years after." After this a committe is formed to determine whether the money should be awarded.

  20. Period Three Equals Chaos by yup2000 · · Score: 1
    I did a introductory paper on chaos theory once....
    glad to see that Poincare did other extremely cool stuff as well... even if it did take us a hundered years to prove it to ourselves... :)

    hrm, I've really got to get off of this chaos=order kick... its affecting my life in very unpredictable ways... er, so that was sort of randomly normal... er...

  21. MOD PARENT UP by Anonymous Coward · · Score: 0

    By far one of the few intelligent things being posted for this article.

  22. Poincare Conjecture by jkauzlar · · Score: 5, Informative

    Simpy put, the poincare conjecture implies that every compact n-manifold is homotopy-equivalent to the n-sphere iff it is homeomorphic to the n-sphere. (From Wolfram's MathWorld) Actually I don't know what that means, but having read and studied a bit about math, I can offer some explanation on the importance of such a proof. When a proof attempts to show that two algebraic structures are equal, as does this conjecture, it allows mathematicians the freedom to look at a problem in two ways instead of one. At last, a compact n-manifold problem can be safely regarded as an n-sphere problem and all the rules regarding n-spheres can be applied to certian n-manifolds. On another topic, these long-standing, but near-universally-believed-to-be-true conjectures are often assumed to be true in order to prove other theorems. i.e. a ground-breaking new primality testing algorithm ASSUMES the truth of the unproven Reimann Hypothesis. So, future encryption keys may rely on unstable hypotheses for their unbreakability.

    1. Re:Poincare Conjecture by jimmyCarter · · Score: 1

      Very impressive, but tell me what 7 x 10 yields and you'll have me convinced of your math knowledge.

      --

      -- jimmycarter
    2. Re:Poincare Conjecture by jkauzlar · · Score: 2, Informative

      Very impressive, but tell me what 7 x 10 yields and you'll have me convinced of your math knowledge. 70

    3. Re:Poincare Conjecture by LadyLucky · · Score: 2
      Simpy put, the poincare conjecture implies that every compact n-manifold is homotopy-equivalent to the n-sphere iff it is homeomorphic to the n-sphere

      Oh, I didn't realise it was that simple.

      --
      dominionrd.blogspot.com - Restaurants on
    4. Re:Poincare Conjecture by Slime-dogg · · Score: 1

      Simpy put, the poincare conjecture implies that every compact n-manifold is homotopy-equivalent to the n-sphere iff it is homeomorphic to th...

      Oh, you said "Simpy" instead of "Simply." I get it now.

      --
      You need to restart your computer. Hold down the Power button for several seconds or press the Restart button.
    5. Re:Poincare Conjecture by Viking+Coder · · Score: 5, Funny

      Frink: Well, it should be obvious to even the most dim-witted individual who holds an advanced degree in hyperbolic topology, n'gee, that Homer Simpson has stumbled into...[the lights go off] the third dimension.

      Lisa: [turning the lights back on] Sorry.

      Frink: [drawing on a blackboard] Here is an ordinary square --

      Wiggum: Whoa, whoa -- slow down, egghead!

      Frink: -- but suppose we extend the square beyond the two dimensions of our universe - along the hypothetical Z axis, there.

      Everyone: [gasps]

      Frink: This forms a three-dimensional object known as a "cube", or a "Frinkahedron" in honor of its discoverer, n'hey, n'hey.

      --
      Education is the silver bullet.
    6. Re:Poincare Conjecture by rocksh · · Score: 0

      It is intresting to note that Nikitin's paper has no reference of Dunwoody work at all, most references are dated to 1920s-1960s. Is it just my observation or general feeling about the spirit in the field of mathematics? If so it is not surprizing that Noebel has not mentioned mathematitians in his will and we don't have Noebel Prize in Mathematics. So, who got $1M prize or this publication @slashdot is just marketing campain with the goal of getting one?

      --
      >
    7. Re:Poincare Conjecture by Anonymous Coward · · Score: 0
      ...it allows mathematicians the freedom to look at a problem in two ways instead of one.

      Which is of course why the proofs are important in general. It'd be incredibly painful to write stuff like "Assuming Pythagoras' Conjecture is correct..." almost every time when doing geometry if there would be no proof for Pythagoras' Theorem, or "Assuming 1 + 1 = 2..." every time when doing anything at all if there would be nothing holy so to speak...

    8. Re:Poincare Conjecture by Elazro · · Score: 1

      That's a well put post, except for one glaring error:

      On another topic, these long-standing, but
      near-universally-believed-to-be-true conjectures
      are often assumed to be true in order to prove
      other theorems. i.e. a ground-breaking new
      primality testing algorithm ASSUMES the truth of
      the unproven Reimann Hypothesis.

      A conjecture can not be used to prove a theorem (except in a strict mathematical sense which I'll describe in a bit). A proof is extremely rigorous, built up of elements of logic, axioms, and other (previously proven) theorems. Thus, while your hypothetical new primality testing algorithm can be very useful, it can not be PROVEN to be effective or correct until the Riemann Hypothesis is shown to be correct. Mathematics is extremely strict about this. In the (empirical) sciences and law a preponderance of evidence usually serves as proof, but absolutely not in math. So while your new primality testing algorithm may wow generations of computer scientists with its seemingly unerring accuracy, it still won't be proven, and might become completely useless for primes over a googleplex (10^(10^100)) or something. (Of course in that case, you'll end up proving the contrapositive of "If the Riemann hypothesis is true, then my algorithm determines primes.", and thus you'll have disproven the Riemann Hypothesis. Which may piss some people off)

      On the other hand, it is not uncommon for a mathematician to prove a link between two conjectures - in fact, this is the story of how Fermat's last theorem was proved. You must note that FLT was a horrible name, and still is. FLT was, until 1993, merely a conjecture. In 1986 Andrew Wiles, who had been mildly obsessed with this FLT-conjecture throughout his whole life, came across a proof by Ken Ribet that linked FLT to the Taniyama-Shimura Conjecture, a complex mathematical statement which says nothing at all about FLT. Ribet actually proved a theorem - "If the Taniyama-Shimura Conjecture is true, then Fermat's Last Theorem is also true." . Andrew Wiles took one look Ribets theorem and got to work proving the Taniyama-Shimura conjecture, which he finished seven years later.

      -matt

    9. Re:Poincare Conjecture by Listen+Up · · Score: 1


      Would you care to explain what exactly you are trying to say? I would love to answer your questions if I could. Thanks.

    10. Re:Poincare Conjecture by opello · · Score: 1

      *cough*plajorism*cough*

    11. Re:Poincare Conjecture by Anonymous Coward · · Score: 0

      This is mostly correct, except that if you write a proof that relies on the Riemann hypothesis being true it will still be accepted by mathematics journals. In general this is not true. For instance, if you attempt to use the (unproven)Hodge conjecture in a paper, it will most likely not be accepted. This is not the case with use of the Riemann hypothesis. Incidentally, I believe that the prime number algorithm in question used the extended Riemann hypothesis which is a generalization of the Riemann hypothesis.

    12. Re:Poincare Conjecture by stonecypher · · Score: 1

      Simpy put, the poincare conjecture implies that every compact n-manifold is homotopy-equivalent to the n-sphere iff it is homeomorphic to the n-sphere.

      You don't really have a very good grasp of the word "simply", do you?

      --
      StoneCypher is Full of BS
    13. Re:Poincare Conjecture by sql*kitten · · Score: 2

      The conjecture is that every simply connected compact 3-manifold without boundary is homeomorphic to a 3-sphere. (Loosely speaking, that every 3-dimensional object that has a set of sphere-like properties can be stretched or squeezed until it is a 3-sphere without breaking it).

      I'm sphere-like but I ain't no homeomorph, buddy!

  23. Think Quantum &co.. by oliverthered · · Score: 2, Informative

    This has a lot of implications for anything in 4d space.
    Basicly before the proof you couldn't be sure what limits existed, now an extra limit has been placed on 4d environments.

    The proof may also point the way to other proofs

    --
    thank God the internet isn't a human right.
    1. Re:Think Quantum &co.. by bwt · · Score: 2

      Actually it is a major piece in the classification 3 dimensional worlds (mathematicians call them "manifolds"). A manifold is the higher dimension analog of a surface. 3-manifolds in particular have direct relevence to virtual reality in CS. We've all seen 2-D games where you go off the top and come up from the bottom and similar with the right and left side. You probably never thought of the fact that such a game is "topologically" played on a surface that is a torus. Variations to this can exist -- imagine that moving off the top brings you in from the bottom but reflected so if you go off the top right, you come in the bottow left (right-left wraps as before). Now you are playing on a "Klein bottle", which is a surface that cannot be embedded in the normal 3-D world. The classification of all surfaces is a standard problem in first year graduate topology and/or algebraic topology.

      One approach is to look at the different ways you can draw a loop, where two loops are equivalent if you can make a movie where one morphs into the other continuously. It turns out that the structure of the fundamentally different ways of putting loops in surfaces completely classifies surfaces. For example, if all loops are equivalent then the surface is a sphere.

      For 3-D "worlds" there are many more possibilities. It is actually a long standing problem to identify what all the possibilites are. The Poincare conjecture, which may now be solved, completes a major piece of that puzzle. It may even complete it all (I don't recall). The obvious thing to do with 3-manifolds is to look at how you can embed spheres, but this approach has failed miserably, because nobody could prove that if any sphere embeddings can be morphed into any other then the "world" is the 3-sphere.

    2. Re:Think Quantum &co.. by oliverthered · · Score: 1

      Cheers, After reading the rest of the comments it was apparent that the first part of the Poincare conjecture was proved and the second part had already been proved.

      I very much a layman when it somes to topology, but it was still fairly obvious that the proof is very important.

      --
      thank God the internet isn't a human right.
  24. Only need an answer for n = 3 by dvdeug · · Score: 5, Informative

    I don't know which is worse; a problem like the Poincare problem, which has been definitively solved for 1-manifolds, 2-manifolds, and n-manifolds where n > 3, leaving only one little hole; or something like Femat's Last Theorem, which was solved for everything up to n equals a million billion and most numbers beyond that, before someone finally come up with a definitive proof.

    1. Re:Only need an answer for n = 3 by Soul-Burn666 · · Score: 2

      In 2d, an area A is "simply connected" means that for every closed _line_ which is wholely inside the area, the area inside the line is wholely inside A and the line is it's border.

      In 3d, a volume A is "simply connected" means that for every closed line which is wholely inside the volume, there exists a surface that is wholely inside A.

      --
      ^_^
  25. I would post something interesting... by Anonymous Coward · · Score: 0

    ... but my head just exploded.

    I was just telling my buddy on EverQuest that these math guys have too much time on their hands.

  26. FINALLY!! by paradoxmember · · Score: 5, Funny

    wow.. finally.. i can sleep at night!!

    1. Re:FINALLY!! by Anonymous Coward · · Score: 0

      Me too. I've had a rubber band wrapped around my head for 2 years trying to prove this stupid theory.

    2. Re:FINALLY!! by jcoleman · · Score: 2

      How is it that a sophomoric comment like "wow I can sleep at night" gets modded to '5 Funny' but the reply which was much more clever is left at 0? I bet you people think the movie "The Sweetest Thing" was funny too. Geez.

  27. My proof by PygmyTrojan · · Score: 3, Funny
    Every simply connected closed 3-manifold is homeomorphic to the 3-sphere

    Well, duh.

    --

    Trying is the first step towards failure.

  28. Beam me up by tiredwired · · Score: 2, Funny

    Finally I can complete the warp engine. We shall fly through space like a rubber band flung from the surface of a sphere. Evil donuts beware. Why do Brits say maths instead of just math?

    1. Re:Beam me up by adb · · Score: 2

      Because they think there's more than one, duh. Little do they know that the One True Math will damn them to the firey pit of Sociology for their blasphemy!

    2. Re:Beam me up by u38cg · · Score: 2, Informative

      We used to treat 'mathematics' as a plural, like you still sometimes hear data treated ("the data were tested..."). When lazy schoolboys and Cambridge ugrads abbreviated it to maths, they kept their plural, as changing it to a singular (at that time) would have felt odd. Now, we treat it as singular, but continue to call it maths. Obviously. One sheep, two sheeps, fish.

      --
      [FUCK BETA]
    3. Re:Beam me up by Anonymous Coward · · Score: 0
      Why do Brits say maths instead of just math?


      Could it be because we speak English (or at least most of us do)!

    4. Re:Beam me up by Anonymous Coward · · Score: 0

      Dialectic Idiosincratic Feature

      Ie: There are differences in the english language when you go to different places around the world.

    5. Re:Beam me up by Anonymous Coward · · Score: 0

      Why do americans say math instead of maths?

      Maths is a short for Mathematics.

      The etymology of the word is from Greek to Latin, where the singular is Mathematics and the plural is Mathematica.

      Although this may be flamebait, you should note that the language is called English not american, therefore, since it is our language, the English are correct and the americans are wrong.

      The accepted de facto standard of English usage is the Oxford English Dictionary. Which has this to say about the use of Math in this context;
      "Math - U.S. colloq."

    6. Re:Beam me up by fruey · · Score: 1
      It doesn't matter whether you treat mathematics as a plural. That's why we say maths in Britain. Because it feels right, because the abbreviation keeps the trailing s.

      In Latin it wasn't plural either (mathematica?) nor in Greek... but usually when abbreviating a long word that ends in s, you still keep the s at the end in the abbreviation.

      --
      Conversion Rate Optimisation French / English consultant
    7. Re:Beam me up by panurge · · Score: 1
      Why do Brits say maths instead of math? Because it's short for mathematics. Whether mathematics actually is plural or not is irrelevant, it's a syntactic usage rather than grammatical.

      Just as utterly off-topic, what irritates me is marketoids referring to vehicles without the definite article, i.e instead of saying "The klutzsuv gets five milles to the gallon and can climb a kerb without the wheels falling off", they say "Klutzsuv....". They're trying to suggest the thing has a personality and so has a proper name, when it's just a metal thing and lots of them have the same name on the back end. Oh well, I can either post pointlessly to this discussion or moderate from a position of total ignorance. I've made my choice.

      --
      Panurge has posted for the last time. Thanks for the positive moderations.
    8. Re:Beam me up by Fnkmaster · · Score: 1

      Or you could note that "mathematics" is NOT plural in Latin, and therefore when shortening/abbreviating it, keeping the s makes little sense, since it was never truly plural to begin with. QED, mother fucker.

    9. Re:Beam me up by machine+of+god · · Score: 1

      In the romance languages when you say math it's plural.

    10. Re:Beam me up by Spunk · · Score: 1

      Do they really? Tivo eats my commercials so I don't see them anymore.

      That's a good little Tivo.

    11. Re:Beam me up by Anonymous Coward · · Score: 0

      Are you studying mathematics, or are you studying mathematic?

    12. Re:Beam me up by Slartibartfarst_UK · · Score: 1
      QED, mother fucker.

      How terribly American of you. ;-)

  29. Of course by Anonymous Coward · · Score: 0

    All the answers can be found at http://quiz.ravenblack.net/blood.pl?3634734612

  30. Poincare Conjecture by Listen+Up · · Score: 5, Informative


    The Poincaré Conjecture is widely considered the most important unsolved problem in topology. It was first formulated by Henri Poincaré in 1904. In 2000 the Clay Mathematics Institute selected the Poincaré conjecture as one of seven Millennium Prize Problems and offered a $1,000,000 prize for its solution.

    The conjecture is that every simply connected compact 3-manifold without boundary is homeomorphic to a 3-sphere. (Loosely speaking, that every 3-dimensional object that has a set of sphere-like properties can be stretched or squeezed until it is a 3-sphere without breaking it).

    Analogues of the Poincaré Conjecture in dimensions other than 3 can also be formulated. The difficulty of low-dimensional topology is highlighted by the fact these analogues have now all been proven, while the original 3-dimensional version of Poincaré's conjecture remains unsolved. Its solution is central to the problem of classifying 3-manifolds.

    On April 7, 2002 there were reports that the Poincaré conjecture might have been solved by Martin Dunwoody; on April 12 Dunwoody acknowledged a gap in the proof and was attempting to fix it. In October of that year, Sergei Nikitin of Arizona State University announced that he had proved the result.

  31. 4D Sphere by Anonymous Coward · · Score: 0

    Sure, it can be pictured:

    1) Look at it.
    2) Go away for a while.
    3) Come back, look at it again.
    4) Go away for a while.
    5) Come back, look at it again.

    Congrats, you have successfully viewed a sphere in four dimensions. Length, width, height, time.

  32. I won't believe the proof until... by Anonymous Coward · · Score: 0

    If James Harris (JSH) gives his blessing then maybe I'll believe it.


    NOT!

    1. Re:I won't believe the proof until... by JamesHarris · · Score: 0, Offtopic

      Sigh. Someone flamed me, and did it anonymously. If you followed the flame link and are curious about what mathematical work could generate so much heat, then look at http://groups.msn.com/AmateurMath, as I'm one of the most well known people in math circles that most of you have probably never heard of. Oh, and after you look around you should check out a Java implementation of THE prime counting function at http://groups.msn.com/AmateurMath/page.msnw?&pss=k . (Yes I'm trying to make lemonade out of a lemon.) James Harris

  33. Micro$oft ads on Slashdot?! by Anonymous Coward · · Score: 0

    Of course I know this is offtopic, but I kinda doubt the subject will come up...

    What's with the Micro$oft ads on Slashdot? Just saw one for .NET Developer Studio and I had to do a double take to make sure of what website I was on.
    Is this just the first step?

  34. Poincare Gnomes! by i_need_no_nick · · Score: 2, Funny
    Phase 1: Prove Poincare Conjuncture

    Phase 2: Collect Clay Math Prize

    Phase 3: Profit

    Now *there's* a business model!

  35. Sadly... by Anonymous Coward · · Score: 0

    Who stands a better chance at proving/refuting the hypothesis: mathematicians pushing chalk or an NSA supercomputer using brute force? After all We're already being searched at airports, now mathematicians can't carry a protractor or a compass without being looked as being suspicious. When will terrorists learn that attacking math problems never solves anything. Wait, maybe it does... Can someone explain exactly what this is and what it means in very small words? And what is n cm? the length of the piece of string is that: I'm not sure it's a good idea to be publicizing this. know that the Poincare conjecture problem is really hard, and that there's no shame at all in all of these Clergy being completely at a loss. But if the public catches wind of such a perceived failure, they may lose faith. A) You can't predict prime numbers. D)If we know how he predicted prime numbers. B) That guy predicted prime numbers. C) Alot of money goes to whoever proves how the hell he predicted prime numbers.

  36. front-page summary by Anonymous Coward · · Score: 0, Funny

    And now your Slashdot front page summary.

    Postmodern Computer Science: My cat's breath smells like cat food.

    Financial Institutions Balk at MS Licensing: I'm going to eat chocolate till I barf!

    Canada to Launch Countrywide Virtual Supercomputer: What's a diorama?

    Satellite Radio in Fiscal Trouble: Was President Lincoln okay?

    Portable CD-RW/DVD Player: I dress myself!

    Hardware Manufacturing in China's 'Hot Zone': Lisa's bad dancing makes my feet sad.

    Congress Members Oppose GPL for Government Research: Then the doctor told me that BOTH my eyes were lazy.

    Music and the Internet Reprise: Everybody's hugging!

    and finally,

    A (Complete) Poincare Proof!?: When I grow up, I'm going to Bovine University.

  37. I thought they did it by wiredog · · Score: 5, Funny

    discretely.

    1. Re:I thought they did it by nevershower · · Score: 1

      Only the Computer Science/Math doubles.

      --
      Look, ma! I'm a karma whore
    2. Re:I thought they did it by cperciva · · Score: 5, Funny

      discretely

      Or, more specifically, in groups, and in fields.

    3. Re:I thought they did it by Darby · · Score: 2

      discretely.

      Might not be possible since they can't tell the difference between their ass and a hole in the ground.
      They can, however, tell the difference between their ass and *two* holes in the ground.

  38. Comment removed by account_deleted · · Score: 1

    Comment removed based on user account deletion

  39. Ahh haa! by Anonymous Coward · · Score: 0

    I have zee MICROFILM!

  40. 2D or 3D? by richie2000 · · Score: 2
    two dimensional sphere

    Exqueeze moi? That one made me do a double-take. Or maybe those two made me do a triple-take, I'm not sure... Maybe if you look at a coloured circle using red-green glasses?

    --
    Money for nothing, pix for free
    1. Re:2D or 3D? by EricWright · · Score: 2

      A manifold is just an n-dimensional surface. The surface of a sphere is completly defined by two dimensions, theta and phi, latitude and longitude, etc.; hence, a 2D sphere.

      No one cares what goes on inside the sphere... no one can see it!

    2. Re:2D or 3D? by richie2000 · · Score: 2
      Ah, that explains it. But you're wrong - the worm cares about the innards of the apple. And Woz, of course. ;-)

      And, if you want to get picky, the surface representation in 3D does care, you need to define the curvature somehow and you can't do that in 2D - without that, it's just another plane. *ponders* Nah, I don't want to get picky. Let's leave it. :-)

      --
      Money for nothing, pix for free
    3. Re:2D or 3D? by EricWright · · Score: 2

      Well, curvature depends on the radius, which is a constant... for that manifold anyway...

      mmmm.... apples....

    4. Re:2D or 3D? by Darby · · Score: 2

      And, if you want to get picky, the surface representation in 3D does care, you need to define the curvature somehow and you can't do that in 2D - without that, it's just another plane.

      The sphere is just the surface. A Ball is the sphere and its interior
      Just like a circle is the "edge" and a disk is a circle and the interior.

      Remove one point from a circle and you can flatten it out into a line. So it's 1 dimensional, but it lives in 2 space.
      A circle has an arc *length*, but zero area.
      In 2+ space, it surrounds an area though.

      A sphere has a surface area, but zero volume, although it surrounds a volume of 3 space.

      If you look at just the sphere by itself, without embedding it in 3 space, there isn't even an interior to worry about. It is by itself, a 2 dimensional object.

  41. My favorite way to prove God by kenp2002 · · Score: 0, Offtopic

    It's simple

    N=N+1

    Therefore God must exist.

    --
    -=[ Who Is John Galt? ]=-
    1. Re:My favorite way to prove God by Anonymous Coward · · Score: 0

      Let me be the first to ask: WTF??

      I thought that was a proof against god? Or is that the inverse?
      N!=N+1

  42. just shut up by Anonymous Coward · · Score: 0

    no text

  43. Simplex K? by Maxwell_E · · Score: 1

    See, he's got it all wrong already. It's the Circle K, and strange things are afoot there.

    Seriously, there was more greek in that thing than the local sammitch shop. Mmm. Gyro...

  44. Re:I hate your fucking sig by Anonymous Coward · · Score: 0

    "Don't do what Donny Dont does" is a Simpsons reference from when Bart tries to get his knife after joining the Boy Scouts. I thought it was funny.

  45. Mod parent up! by Anonymous Coward · · Score: 0

    Dear God, moderator, don't you understand a joke when you see one?

    1. Re:Mod parent up! by Rick+the+Red · · Score: 2

      Haven't you read the FAQ? You don't get mod points if you have a sense of humor.

      --
      If all this should have a reason, we would be the last to know.
    2. Re:Mod parent up! by Trusty+Penfold · · Score: 1

      I went through moderator indoctrination once; it's easy.

      Stephen King : Troll
      John Carmack : Informative.
      Goatse : Troll
      "Do we like the RIAA today?" : Funny.
      Microsoft : Troll
      Linux : Insightful interesting information.

  46. video proof by tgibbs · · Score: 2
    This is just mathematical proof that you can wrap a rubberband around an apple. I think the rest of us would be satisfied by a videotape instead.

    It's more like a proof that the rubber band will pop off the apple if you nudge it a bit. Oh, yeah, and the apple has to be 4-dimensional (and no, you can't count time as one of them). Good luck making that video....

  47. having said that... by Patersmith · · Score: 2, Funny


    How do we use this to take down the RIAA/MPAA?

    1. Re:having said that... by Anonymous Coward · · Score: 0

      this has to be the funniest response...Im still giggling... someone please mod this one up to Funny:5.. rofl!

  48. Explanation in kindergarten terms by yerricde · · Score: 3, Informative

    Is it trying to prove that there can exist a 4-dimensional object that has all points equidistant from a single point in space-time or something?

    Assume that you have a sculpture made of Play-Doh® modeling compound, without any holes in it. If the Poincaré conjecture is true, then you can reshape the sculpture into a ball without breaking or joining anything.

    --
    Will I retire or break 10K?
    1. Re:Explanation in kindergarten terms by pediddle · · Score: 1

      Also, if you have other sculptures with holes in them, you cannot reshape them into a sphere without breaking or joining anything.

      Now, for the purposes of this problem, we only actually care about the surface of the Play-Doh ball. The surface has 2 dimensions, making the Play-Doh ball a 2-sphere (a 2-manifold). It's pretty obvious that the conjecture is true for Play-Doh, and proving that the conjecture is true for 2-manifolds has already been done. Immagine, if you can, that the Play-Doh's surface is actually in three dimensions, and the Play-Doh itself is actually 4 dimensions. Then you get an idea of the problem.

  49. Someone explain what this sentence means by Anonymous Coward · · Score: 0

    "If we stretch a rubber band around the surface of an apple, then we can shrink it down to a point by moving it slowly, without tearing it and without allowing it to leave the surface."

    How is the band being shrunk? How can it be shrunk to a single point yet still be wrapped around the apple? Why can't science types properly describe something in english? Science is about descriptions!

    1. Re:Someone explain what this sentence means by Rip!ey · · Score: 1

      How is the band being shrunk? How can it be shrunk to a single point yet still be wrapped around the apple? Why can't science types properly describe something in english? Science is about descriptions!

      Consider the earth as the apple.

      If we all stood on the equator holding hands to form a human chain (the rubber band), and we all walked (or swam) at 90 degrees to the equator (same direction, north or south), we will all end up at either the north or south poles.

      The north or south poles represent the single point.

    2. Re:Someone explain what this sentence means by Big+Mark · · Score: 2, Informative

      If you imagine that the Earth was a perfect sphere (it's not, but just for the sake of argument let's say it is) and that the equator was the rubber band. See how it slices the Earth into two bits?

      Start sliding the equator up towards the geographical north pole.

      Keep sliding. See how the total length of the "equator" has shrunk? See how there is one slice of the Earth that's bigger than the other? Imagine taking the top off of a boiled egg, if that helps... Slide some more.

      Stop right there! You're just about to reach the north pole. Push it perfectly onto the north pole...

      See? It is still on the Earth, but the "slice" of the Earth formed by the "equator" here is so thin that the "equator" now has zero length, and the second slice has no volume.

      This, of course, requires a degree of perfection mere humans could neverachieve. I'm talking perfect perfection here. Not one merest of iota away from where it should be. Hey, it is theoretical...

      Move the "equator" back anywhere near where it should be...

      and it gets a non-zero length again. Push it even slightly further than the north pole...

      And it is no longer on the Earth.

      Congratulations on pinging the "equator" at the Sun, by the way. You've just annoyed every geographer on the planet. It looks like you've hurt the Sun as well... oh dear, it's going supernova! We're all going to die!

    3. Re:Someone explain what this sentence means by Anonymous Coward · · Score: 0

      You win the prize for the most sensible and clear explanation here - thanks

    4. Re:Someone explain what this sentence means by Old+Wolf · · Score: 2

      The rubber band would shoot off the apple as soon as you had moved it a few millimetres..

      there should be a $1m prize for taking an apple with a rubber band on it and getting the rubber band to a point

  50. Hiero II of Syracuse by Theatetus · · Score: 1

    Though that story seems to be apocryphal, especially given that much earlier metallurgists than Archimedes already knew ways to test the purity of metal.

    --
    All's true that is mistrusted
    1. Re:Hiero II of Syracuse by Just+Some+Guy · · Score: 2

      Not that I'm disagreeing, but were those early metallurgists able to nondestructively test metals? IIRC, that was the alleged advancement.

      --
      Dewey, what part of this looks like authorities should be involved?
  51. Some smart person: is this right? by jolshefsky · · Score: 5, Interesting
    Let's see if I've got this right. A manifold is just a way of describing some thingy in a some specific N-dimensional space. You would say that a Dixie cup is a manifold that is homeomorphic to a sphere because if the Dixie cup were maleable, you could stretch and push the outer surface of it to form a sphere.

    You could take a magical rubber band and stretch it around a sphere and then slide the rubber band along the surface. As you work your way around, you'd find that the length of the rubber band varied along the surface. The important thing is that you can slide the rubber band so its length is essentially zero--a.k.a. a single point.

    Poincare (Poincaré really but thanks a lot Slashdot for not letting me ...) speculated that if you had any simply connected closed 3-manifold is homeomorphic to the 3-sphere (which I'm just parroting back from the Mathworld site at Wolfram.) The theory was later expanded to include the equivalent in N dimensions. In other words, if you take something that only has an "outside" and no holes, you could mash it into the shape of a sphere, or slide the rubber band right off it no matter what.

    The other side of the story is things like a torus, or for a tastier example, donut. It's not "simply connected closed 3-manifold" because if you put a rubber band around the "meat" of the donut through the hollow middle and never be able to get the rubber band off without breaking the donut or the rubber band. Yum.

    The thing that hasn't yet been proven is whether this is true for 3-dimensions as originally speculated. The Mathworld site at Wolfram says that it's been proven for N=1, 2, 4, 5, 6, and >=7, but not for 3. I don't know why ... I mean, can't you just define a point that is in the center of a given manifold then make a sphere that is the average distance from all points on the surface and define a new surface that is half-way between the two surfaces, and repeat forever to show that you really get a sphere ... for a torus, for instance, you'd get a point, but for a cube you'd get a finite sized sphere ... same for a Dixie cup, except it'd be really small.

    --
    --- Jason Olshefsky

    Karma: Poser (mostly affected by adding this line long after everyone else did)

    1. Re:Some smart person: is this right? by jim3e8 · · Score: 1
      it's been proven for N=1, 2, 4, 5, 6, and >=7, but not for 3. I don't know why ... I mean, can't you just define a point that is in the center of a given manifold then make a sphere that is the average distance from all points on the surface and define a new surface that is half-way between the two surfaces, and repeat forever to show that you really get a sphere...

      Holy shit, you've done it! Give the guy his million bucks!

      ;)

    2. Re:Some smart person: is this right? by Anonymous Coward · · Score: 0

      actually, a dixie cup is _not_ homeomorphic to a sphere: it doesn't enclose any space. that is, to morph the dixie cup into a sphere, at some point you'd have to sew the the opening shut.

    3. Re:Some smart person: is this right? by ph43thon · · Score: 1

      (I was inclined to say a dixie cup would be homeomorphic to a sphere with a point removed) But.. A dixie cup is different than a sphere with a point removed.. A sphere with a point removed is homeomorphic to a dixie cup with no boundary on the "lip" (or top circular edge of the cup) A dixie cup would be homeomorphic to a square with its boundary included. p

    4. Re:Some smart person: is this right? by jolshefsky · · Score: 1

      Actually I was thinking of a Dixie cup as an actual object rather than a surface defined by the shape of a Dixie cup ... get it? I meant a Dixie cup with thickness. Perhaps a clay cup would have been a better example to emphasize the thickness of the cup shape. I just wanted to use the Dixie trademark as much as I could.

      --
      --- Jason Olshefsky

      Karma: Poser (mostly affected by adding this line long after everyone else did)

    5. Re:Some smart person: is this right? by laertes · · Score: 1
      don't know why ... I mean, can't you just define a point that is in the center of a given manifold then make a sphere that is the average distance from all points on the surface and define a new surface that is half-way between the two surfaces, and repeat forever to show that you really get a sphere

      By God Man! You've just proven the most important conjecture in topology! You get the Clay Institute Prize!

      --

      Yes, I'm still a junky. Are you still a bitch?
    6. Re:Some smart person: is this right? by spectatorion · · Score: 1

      not quite. it's the surface of the sphere (or torus) that we call a 2-manifold. when we say sphere or torus, we really mean the surfaces of these things. the surface of a bowl or cup (without handle) that has some thickness is homeomorphic to a sphere since you can just push in the sphere to get the bowl/cup. and actually, a manifold is not some thingy in n-dimensional space, but rather some thingy that locally looks like n-dimensional space (in some more precise way). the surface of a sphere or a torus locally looks like a plane (so these are 2-manifolds), while a cone does not look like a plane everywhere--namely at its vertex, and also where its base disk intersects its sides, it is not smooth like n-space. thus a cone is not a manifold.

    7. Re:Some smart person: is this right? by Kragg · · Score: 2

      Sounds like the easier proof to the 2-dimensional manifold (in 3d space) rather than the 3d manifold in 4d sapce to me.

      --
      If you can't see this, click here to enable sigs.
  52. Jesus Man! by kenp2002 · · Score: 1, Offtopic

    Do you realize you are bordering on 1000 posts!? Your a maniac! :)

    --
    -=[ Who Is John Galt? ]=-
    1. Re:Jesus Man! by dvdeug · · Score: 2

      I've at it since 1998 and spend way to much time at a computer. Given that, it's not hard.

    2. Re:Jesus Man! by elocutio · · Score: 0, Offtopic

      Do you realize that you posted, like, 23 times, TODAY!? You're even more maniacal! :D

    3. Re:Jesus Man! by kenp2002 · · Score: 2

      Hahahahha you don't even know. You should see my Karma. It's freaky. All I do is patrol /. stirring up trouble for trouble's sake. my theory here is no matter how good or bad my posts are (I some days just write something that will get people talking just for the sake of getting them talking.) I can be comforted in knowing it will force people to think. I am happy with that.

      --
      -=[ Who Is John Galt? ]=-
  53. Eureka? by Anonymous Coward · · Score: 0

    Hadn't you heard? They recently discovered that "eureka" actually means "oh my, this bathwater is too hot".

  54. Smash-O-Matic by pixelpusher220 · · Score: 2, Funny

    Gallagher could reduce both an apple and a donut to a point...with just one swing!

    --
    People in cars cause accidents....accidents in cars cause people :-D
    1. Re:Smash-O-Matic by stonecypher · · Score: 1

      > Smash-O-Matic

      Is that anything like sledge-o-matic? :P

      --
      StoneCypher is Full of BS
    2. Re:Smash-O-Matic by !!!!!!blameblameblam · · Score: 1

      C'mon now, it's Sledge-O-Matic... :)

  55. I think I understand the problem by NibbleAbit · · Score: 1, Informative
    I may be wrong, but I think they are trying to prove that you can map a smooth shape into a sphere using the same number of transformation equations as there are dimensions.

    For example. You can draw a smooth shape (no sharp corners, no intersections) on a piece of paper. Assume you can come up with some equation that defines that shape. Then there exists not more than 2 equations that will transform (map) your original equation into a circle. Now the problem is prooving this in 3 dimensions. You start with a blob (actually just the surface of the blob), the blob has no holes and no sharp angles and doesn't intersect with itself (just your standard blob). There exists not more than 3 equations that will map your original blob defining equation into a sphere.

    I'm no mathematition, but thats how I read the description. I think where this would be useful is in manipulating very complex shapes. You start with a shape, find the transformation equations. Manipulate a shpere (easy), then apply the inverse maping back to the original, and you get the result.

    Like I said, I may be way off on this, but I did pass high school.

  56. weird. by hanwen · · Score: 1

    This guy doesn't have any background in algebraic topology. All his papers are on control theory.

    Sounds fishy to me.

    --

    Han-Wen Nienhuys -- LilyPond

    1. Re:weird. by Anonymous Coward · · Score: 0

      You can't judge someone on their past work.

  57. This is the significance by Anonymous Coward · · Score: 3, Informative

    Here is the importance of this conjecture. It's really about a 3-dimensional subset of 4-dimensional space, but think of the usual 2- in 3- situation if it helps.

    Basically, if somebody gives you a twisted and knotted object, you would like to be able to say whether its really just a twisted and knotted sack (the sphere). It could in principle have any weird basic shape, you can't identify it when it's all twisted up. Showing the object is just a sphere, would require you to try to unknot it and smooth it out and say: look, I told you it was really just an ordinary sack.

    In pathological cases it can be really hard to figure out how to undo all the knotting and twisting possible, and the case of dimension 3- in 4- was the only one still unkown.

    So what you would like to do is not have to provide instructions for untangling the object, but rather just put it into a CAT scan, map its shape and perform some kind of calculation to verify it must be a sphere.

    This is the homotopic equivalence of the conjecture. You can calculate the homotopy groups of the sphere. You can also calculate the homotopy groups of the weird twisted object. If they are equivalent, you don't have to go to the effort of unknotting the shape. Before you didn't know for sure it must just be a sack, but now you do!

    Actually, that's assuming the proof holds up. Don't rush to judge these things so fast.

  58. What's "uniform density"? by yerricde · · Score: 2, Funny

    Aye, as most objects of uniform density do :)

    Doesn't "uniform density" mean "as opposed to something like swiss cheese"? I was talking about holes as in donut, not holes as in swiss cheese or holes as in IIS. Can a torus have a uniform density?

    --
    Will I retire or break 10K?
    1. Re:What's "uniform density"? by schmink182 · · Score: 1

      Yes, a torus can have uniform density, so long as wherever there is matter, it is of the set density.

  59. Re: Your sig by CommieLib · · Score: 0, Offtopic

    What I mean is a maximum limitation to be eligble for the voucher (perhaps I need to revise my sig), i.e., you cannot qualify for a voucher unless you can demonstrate need.
    The purpose of the argument is to demonstrate that the real objection is not about quality of education, it's about the powerful in this country staying powerful by limiting the choices of the less powerful.

    --
    If your bitterest enemies are people who hack the heads off civilians, then I would say you're doing something right.
  60. Re:I hate your fucking sig by Anonymous Coward · · Score: 0

    I've read your stupid god damned sig about 50 times over the last several months... remove it from public display

    Remove it yourself. Slashdot has an option to turn SIGs off. Preferences/Comments/Disable Sigs.

  61. Coming from a VERY unexpected corner by hysterion · · Score: 5, Insightful

    A glance at Nikitin's publication list will show that he works in Control Theory, and never published in Topology or Geometry journals before... It's a bit as if a statistician announced a proof of Fermat, with a (by math standards) surprisingly short and elementary proof. Hats off if it's right, anyway I guess any mistake would be found pretty soon.

    1. Re:Coming from a VERY unexpected corner by hanwen · · Score: 1


      yeah, and I could actually understand the first 4 pages of the paper, which leads me to think that it is flawed.

      --

      Han-Wen Nienhuys -- LilyPond

    2. Re:Coming from a VERY unexpected corner by elBart0 · · Score: 2, Informative

      It's also interesting to note from his CV, that he's only an Associate Professor. ASU might want to make sure this guy's on a tenure track (if he wasn't before, I'm sure he is now.)

      Also from his CV: "1996-t/n Linux Consultant in Arizona"

      --
      09 F9 11 02 9D 74 E3 5B D8 41 56 C5 63 56 88 C0
    3. Re:Coming from a VERY unexpected corner by call+-151 · · Score: 2
      Associate professor usually means recently tenured or not super-productive after getting tenure a while ago. Here is a (generally correct) classification of the animals you might see in the US-version of a typical academic zoo:
      • Professor/Full Professor: tenured and promoted to generally the highest rank, though some institutions have "Distinguished Professor"
      • Associate Professor: promoted from assistant professor, usually with tenure
      • Assistant Professor: untenured faculty member who is on the tenure-track and will probably be considered for tenure after 5 years or so from starting
      • Lecturer: someone whose job expectations are usually entirely from teaching with no research expectations. Can be tenured or (usually) not, depending upon institution.
      • Visiting Professor: researcher visiting from somewhere else, not tenure-track.
      • Instructor: vague title usually non-researcher
      • Postdoc: non-tenure track position, limited term, for someone with a PhD, usually entirely a research job, though there are sometime some teaching obligations.

      The tenure decision is made usually in the fifth or so year and in principle involves research, teaching and possibly departmental/university service as criteria for awarding tenure. Almost always, tenure is awarded simultaneously with promotion from assistant to associate. Usually, consideration for promotion from associate to full is after about six years of being associate professor and being productive in research. There are exceptions- some bigshots are hired with tenure or a higher-rank upon arrival, and sometimes people move institutions with all kinds of complicated arrangements.

      Hope that helps!
      --
      It's psychosomatic. You need a lobotomy. I'll get a saw.
    4. Re:Coming from a VERY unexpected corner by delphi125 · · Score: 2

      Although the initial proof of anything in mathematics may be quite long, shorter proofs will be found. Most long-standing proofs are relatively short, although they may of course rest on other 'lower level' theorems. Or to put it another way, once a mathematician has accepted Peano's axioms, they are not required in every proof concerning integers.

  62. The real 1-2-3 steps by Rayonic · · Score: 3, Funny
    From what I've read of these scientific papers, I've been able to divine the actual list:

    • Step 1) Prove that it is possible that a fundamental group of 3-dimensional manifolds (V) could be trivial, even though V is not homeomorphic to the 3-dimensional sphere.

      Step 2) ??????

      Step 3) ????????
  63. Great news... but it hard to believe by Anonymous Coward · · Score: 0

    There are proofs for all dimensions but the 4th. There were 16 wrong demonstrations of the poincare conjucture for the 4th dimension. So for me, it is just too hard to belive. I won't be happy until I see the Clay Mathematics Institute give him the prize. http://www.claymath.org

  64. I'm a bit confused. by fishexe · · Score: 1

    The explanation of the problem linked in the posting said you can compress an apple down to a point by moving a rubber band around it? I know for a fact there is no way to compress an apple that small without breaking it. Trust me, I've tried.

    Donuts, on the other hand...

    --
    "I don't care about the Constitution!" --Bill O'Reilly, November 17, 2009
  65. Ok...I'll waste a few and explain this to you... by Anonymous Coward · · Score: 2, Interesting
    From Mathworld...

    "The conjecture that every simply connected closed 3-manifold is homeomorphic to the 3-sphere"

    A manifold is simply a surface, like the surface of a peice of paper. There are different types of manifolds ( topological, smooth,...), but for the near term that's not important.

    A 3-manifold is simply a manifold that has a surface of three dimensions.

    A simply connected manifold is a surface on which any loop you place one the surface can be continuously deformed to a point. What that means is that when you place a rubber band on the surface you can squench the rubber band down to a point without having to make it lose contact with the surface. For example you can do this for a soccer ball. But you can't for a dount. So a soccer ball is simply connected while a donut is not.

    To explain the term closed requires a bit of work. When one studies this kind of thing one covers manifolds with smaller sets of points that look just like the normal Euclidian balls, ie all points with a radius less than R say. These are open sets. [Experts only: Yeah I can define another topology but I am trying to explain things here! ] The complement of a set A which is a subset of a set B is the set of all points in B that are not in A. A closed set is a set that has a complement that is open.

    Two manifolds are homeomorphic if they can be [continuously] deformed in to one another.

    Finally a 3-sphere is simply the set of points in, a 4 dimensional space (x,y,z,t) that are equidistant from the origin, (0,0,0,0).

    So that should be it...now you know what this drek...

    "The conjecture that every simply connected closed 3-manifold is homeomorphic to the 3-sphere"

    ...means.

    That said I have not read the paper, don't have the time right now.

  66. abstract and a little background by call+-151 · · Score: 3, Informative

    The preprint is posted on the arXiv.org web site, which is exactly that, a place to put preprints. Preprints that appear there have not been subject to peer review, so at this point, this is an annoucement of a result, which is very different than a number of mathematicians with the appropriate background agreeing that this is a proof.

    The abstract from the arXiv is:
    This paper proves that any simply connected closed three dimensional stellar manifold is stellar equivalent to the three dimensional sphere.


    and the intro of the paper says that "Since every 3-dimensional manifold can be triangulated and any two stellar equivalent manifolds are PL homeomorphic, our result does imply the famous Poincare conjecture."

    There is a nice front end to the math part of the arXiv from UC-Davis at this link

    --
    It's psychosomatic. You need a lobotomy. I'll get a saw.
  67. Visualizing a 3-D Sphere by sisukapalli1 · · Score: 5, Funny

    A mathematician was once asked about how he could visualize a 3-D Sphere. His response was, "Simple! First visualize an n-D Sphere and then set n to 3".

    Read this a while ago somewhere. Couldn't resist posting it.

    S

    1. Re:Visualizing a 3-D Sphere by Anonymous Coward · · Score: 0

      That's cute. You might want to try that for visualizing an infinite dimensional sphere. Personally I think it actually works, but my definition of "visualize" may be different from yours.

  68. Math Chicks by hymie · · Score: 0, Offtopic

    Danica McKellar, who played Winnie Cooper on The Wonder Years, has a degree in mathematics from UCLA.

    1. Re:Math Chicks by 3waygeek · · Score: 1, Offtopic

      Also, Teri Hatcher, IIRC, majored in math before going into acting. A natural choice, since her folks are both geeks (dad's a physicist, and mom was a programmer).

  69. Regarding the remark about primality testing algo by Anonymous Coward · · Score: 1, Informative

    If I recall correctly, assuming that the Reimann Hypothesis is true results in an algorithm that runs *faster* (has a lower polynomial running time), but they also provided an algorithm that is polynomial time *without* assuming Reimann hypothesis is true.

    In other words, primality testing is in P -- unconditionally.

    A.

  70. Unfortunately, this is not a correct proof by Anonymous Coward · · Score: 0, Troll


    In his paper he claims to prove that a simply connected closed 3 dimensional manifold is stellar equivalent to the 3-sphere. However, his definition of page 11 of simply connected is incorrect. Using his "definition" it is trivial to prove the Poincare conjecture in every dimension. (Even though it is known to be false in higher dimensions.)

    Nice try. But I guess we'll have to wait for the next incorrect proof.

    1. Re:Unfortunately, this is not a correct proof by Anonymous Coward · · Score: 0

      I don't agree; Nikitin's definition of simply connected is consistent with the definition I used in anlysis. That is, for example in R^2, a set A is simply connected iff all closed curves in A are homotopic to a point in A.

  71. Re:Regarding the remark about primality testing al by jkauzlar · · Score: 1

    I don't remember if it is in polynomial time, but the one that assumes the Riemann Hypothesis is the first algorithm to give a 100% probability of primality. Previously they fell around a measly 99%.

  72. Bah.. Math chicks.. by Hydro-X · · Score: 1

    Everyone knows engineering chicks have significant figures. Plus, no couple enjoy a better moment.

  73. Re:N Dimensional Sphere? by noshellswill · · Score: 0

    Actually, it seems even a line may cast an (N-1)_dim shadow onto an N_dim surface.

  74. manifolds by orion67 · · Score: 0

    oh yeah? My Edelbrock high-rise manifold on my '68 Camaro can beat the shit out of your "3-dimensional manifold V", plus it comes with a bumper sticker!

  75. The answer... by 241comp · · Score: 2, Funny

    Actually, all the article says is that they have finally realized that Douglas Adams is right.... the last line of the proof is:

    = 42

  76. Why Eric Weissman rules my world by Nyarly · · Score: 5, Informative
    What he's saying is, the...er...well, he means that the, uh...

    Piece by piece:

    • Consider a compact 3-dimensional manifold V without boundary.
      • Basically, V is a set of points with a whole bunch of properties. Among them are the fact that the points are "smooth," as defined by a funky neighborhoods deal, but it's roughly analogous to the definition of a continuous function. (I believe that the points that satisfy a continuous function would be a topological space - the first part of being a compact manifold).
      • Compactness is harder to grasp. Essentially you couldn't find a infinite set of open sets whose union is the set of points in the manifest, from which a finite number of sets could be taken whose union would also be the original set. Almost kinda like saying that there aren't discontinuities in the manifold - there aren't gaps.
      • Without boundary means that it doesn't include it's own boardary - like the open ball, where it's every point inside a certain radius - the surface is the sphere at that radius.
      • 3-dimensional means that you could refer to any point in the manifold with as few as three values. Unless the manifold set is a space, is exists in 4 dimensions. Think about a sphere - you can refer to any point on a sphere with 2 coordinates, like latitude and longitude, but it exists in 3 dimensions.
    • Is it possible that the fundamental group of V could be trivial
      By which he means: there is one equivalence set of loops through the manifold. Every possible loop (A path that returns to its beginning point. Duh.) in the manifold belongs to one set of loops that are pretty much the same - you could push any one of them around and get any other one. A sphere has this quality - any loop you draw on the surface of a 2-sphere (the one that exists in three dimensions), but a torus doesn't - there are the loops that are equivalent to the loop around the outside of the torus, and the ones that run through the hole.
    • even though V is not homeomorphic to the 3-dimensional sphere?" Trans: Even though you can't finagle necessarily finagle it into a 3-sphere, even if I let you do it in higher dimensions.

    What he's asking is, is it possible there could be an object in 4-dimensions, that has some kind of tangle in it such that you can't make it into a hypersphere, but isn't so mangled that there are fundamentally different ways to "draw lines" on it. The suggestion is fairly reasonable, I think.

    For instance, any loop drawn on a plane is homeomorphic to a circle. You're options in connected finite 1-d manifolds amount to line segments, or cirles. But when you upgrade to 2-d, suddenly you've got holes. You can't have holes in 1-d and still be connected, but in 2-d you get donuts and dresses and honeycombs and whatnot. But the thing about a hole is that it means that your fundamental group have more than one set in it, and without a hole, you wind up being a sphere.

    So why shouldn't there be another characteristic of 3-d objects, one which allows for more than one kind of simply connected, non-contractable manifold?

    --
    IP is just rude.
    Is there any torture so subl
    1. Re:Why Eric Weissman rules my world by starling · · Score: 2

      the points are "smooth,"

      Sorry, you lost me there. Wouldn't they be pointy?

      Compactness is harder to grasp

      Nope, can't see it myself. Compact things are *easier* to grasp, surely.

      Thanks for trying to explain though.

    2. Re:Why Eric Weissman rules my world by sqlgeek · · Score: 1
      Or lets simplify this much further.



      "Consider a compact 3-dimensional manifold V without boundary. Is it possible that the fundamental group of V could be trivial, even though V is not homeomorphic to the 3-dimensional sphere?"



      A compact 3-dimensional manfold without boundary is a smooth 3-dimensional space (manifold=smooth space, i.e. there aren't any wierd kinks that can't be ironed out) without a boundary (i.e. you cannot fall off the edge of the universe) that is "compact". So the only hard word here is compact. In my opinion the most readily understood definition of compact is that every convergent sequence of points in the space ends in a point in the space. In other words, there are no points that you can get arbitrarily close to but never reach quite reach. For the positive real numbers such a point would be zero. The only such 1-dimensional space is the circle. The 2-dimensional spaces of this sort are the sphere, the torus (surface of a donut) and all of the possible siamese twin, triple, quadruple, etc. donuts.



      Next "Is it possible that the fundamental group of V could be trivial". This means simply that any loop you could draw in this space can be contracted to a single point. In 2-dimensions the only such space is the sphere, on a donut you can draw a loop around the donut that cannot be contracted to a single point without jumping out of the donut to do it.



      And finally "even though V is not homeomorphic to the 3-dimensional sphere?" simply means that we can't smoothly deform ("is not homeomorphic" = "we can't smoothly deform") any such space into the 3-sphere. What's a 3-sphere? It's like a circle or 2-sphere (the familiar sphere), except a dimension bigger. It's like a solid ball, except when you get to the edge it's only all the same single point. This is analogous to taking a line segment and fusing the ends together to get a circle or taking a solid 2-dimensional disk, gathering up the edge of it like a bag and fusing the edge of it all together to a single point to form a sphere.



      Cheers,
      Scott

    3. Re:Why Eric Weissman rules my world by Nyarly · · Score: 1
      So the only hard word here is compact. In my opinion the most readily understood definition of compact is that every convergent sequence of points in the space ends in a point in the space. In other words, there are no points that you can get arbitrarily close to but never reach quite reach.

      Yeah, I found an excellent explaination of compactness in sci.physics.research about 10 minutes after I posted. However, the sequence of points thing only works for manifolds, not all topological spaces. So it's cogent here but not always.

      "Is it possible that the fundamental group of V could be trivial". This means simply that any loop you could draw in this space can be contracted to a single point.

      Actually, this isn't necessarily so - or if it is, I don't know of the theorem that demonstrates it. I certainly can't produce a 2-dimensional counter example. My understanding was that a trivial fundamental group essentially contained only one equivalence class on homotopy. So, can you define a manifold such that there's only one homotopic equivalence class of loops, but which doesn't include a single point?

      Oh, duh. Of course not - the manifold necessarily must include points in order for there to be loops at all (unless you consider the null manifold to vacuously satisfy the conditions of a trivial fundamental group), and any basepoint is itself sufficient to be a loop. So, since at least one set in the fundamental group includes the single-point loop, all loops have to be homotopic to the point. Okay.

      Honestly, I'm only posting this in the hopes that others may be enlightened by my own folly. If anyone cares.

      Now if I can just wrap my head around why the n-sphere isn't contractable...something about neighboring paths.

      --
      IP is just rude.
      Is there any torture so subl
    4. Re:Why Eric Weissman rules my world by 3am · · Score: 1

      i no longer have mod points, but i would like to offer my congratulations on some very informative posts. you really raised the level of discussion here.

      Are you familiar enough with the research Nikitin and Dunwoody to judge if Nikitin's work looks solid? Unfortunately my knowledge of topology doesn't range much further than the basics (ie, i understand what you mean when you talk about the equivalence classes of paths under homotopy, manifolds, compactness), and Nikitin doesn't seem to mention the Dunwoody's work, and whether he fixed particular holes in that approach or developed an entirely new one.

      --

      A: None. The Universe spins the bulb, and the Zen master merely stays out of the way.
    5. Re:Why Eric Weissman rules my world by Cy+Guy · · Score: 1

      Think about a sphere - you can refer to any point on a sphere with 2 coordinates, like latitude and longitude, but it exists in 3 dimensions.

      As a non-Math Geek, I just want to thank you for providing the single most accessible statement I have read on multidimensional space since reading Flatland.

  77. Did I miss something? by FallLine · · Score: 2

    Did Katz really stop posting to slashdot? Permanently? Did he say why? I haven't seen him for awhile, but I'm not sure if this is because he's on sabbatical or what have you. Please do tell! I rather detest his drivel.

    1. Re:Did I miss something? by sql*kitten · · Score: 1

      Did Katz really stop posting to slashdot? Permanently? Did he say why? I haven't seen him for awhile, but I'm not sure if this is because he's on sabbatical or what have you. Please do tell! I rather detest his drivel.

      Well, you can filter him out in your user preferences, maybe that's what you did?

    2. Re:Did I miss something? by FallLine · · Score: 2

      Nope, I didn't filter him out. Also if you look at www.slashdot.ort/~jonkatz he clearly hasn't posted any comments in awhile. Maybe he posted stories though? SHrug

    3. Re:Did I miss something? by Anonymous Coward · · Score: 0

      I blame the man!

    4. Re:Did I miss something? by kevlar · · Score: 1

      I think he is running from the Man.

  78. SAY WHAT??? by Anonymous Coward · · Score: 0

    You can calculate the homotopy groups of the sphere.

    If you could produce an algorithm for determining the homotopy groups of spheres, you'd be in possession of a Fields medal.

    1. Re:SAY WHAT??? by Anonymous Coward · · Score: 0

      did i miss something? isn't the homotopy group of a sphere the trivial group?

  79. Yeah, I can see applications by sakusha · · Score: 2

    Once someone mentioned Nitikin's specialty is Control Theory, it all snaps into place. Yep, there are realworld applications here. I can almost grasp the networking and communications implications, the ability to deal with n-dimensional spaces as simpler topological states. Ah, if only I'd done that two more years of calculus class. But those applications have little to do with the proof itself, more the implications of the topological network system he uses.
    Now let us know when he wins the Clay Mathematics Prize, wow a megabuck, that's better than a Nobel.

  80. Re: Your sig by Luyseyal · · Score: 0, Offtopic
    Does "solid education system" reduce to "all schools are good"? If so, then many voucher advocates would agree with you. But they would disagree when you tell them children in failing schools right now should be forced to stay there until the district or state gets its proverbial shit together.

    It takes a long time to turn around a failing school. Having a solid educational foundation is critical for elementary school children in setting the tone for their future in education. Those kids don't have 5 years to wait on the district. They need a solution now.

    That said, vouchers are only one part in a total school system solution. Here are some things I thought of off the top of my head. (FWIW, I'm a single parent of a 6 year old who attends a National Blue Ribbon public elementary school.)

    • Teachers should have to meet tougher requirements to teach in low-performing schools
    • They should make more money than teachers at better schools
    • Classroom sizes should be lower in low-performing schools
    • Teachers should be encouraged to send their own children to their school through incentives
    • Those who relocate into the school's neighborhood should have their moving expenses paid for
    • All teachers should be required to publish syllabi for their classes including a list of weekly assignments. These documents must be signed and returned by the parent[s].
    • A myriad of volunteer opportunities should be extended to parents: reading aloud to the kids, encouraging parents to send snacks to all the kids in the classroom, having regular plays or musicals with parent participation, helping with field trips, etc.
    • Regular PTA meetings / potlucks
    • Regular dialog sessions between parents and administration
    • And a bjillion more... as I said, vouchers are only part of the immediate solution. Most of these other things take time to develop.

    As we all know, good schools are only 50% of education. The other 50% is a good home. Districts can't solve those problems, but they can help by encouraging parents who do care to participate in ways that help more children than just their own.

    Cheers,
    -l

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    Help cure AIDS, cancer, and more. Donate your unused computer time to worldcommunitygrid.org. Join Team Slashdot!
  81. Re: Your sig by Anonymous Coward · · Score: 0

    "The purpose of the argument is to demonstrate that the real objection is not about quality of education, it's about the powerful in this country staying powerful by limiting the choices of the less powerful."

    Do you know who gives a person power? Himself.

    If you are equating money==power here, you've just become logic-poisoned by the anti-capitalists.

    There is no vast right-wing conspiracy. Trust me. >:D

  82. Re: Your sig by Anonymous Coward · · Score: 0

    I was waiting to argue with you, on thing such as "Those kids don't have 5 years to wait on the district."

    But then you said: "As we all know, good schools are only 50% of education. The other 50% is a good home."

    And then you got the point, so there was no reason for me to argue.

    Morons beget morons. Bad public schools are bad because in agregate, the the administrators, the students, the parents, and the administrators are all mentally and/or physically lazy. The only ones who can make a difference in raising and educating better kids are smart, patient teachers, reasonably unbureaucratic administrators, and parents who aren't busy sending their kids down the street to buy them crack.

    Remember Twain: I never let schooling get in the way of my education.

  83. riemann by dollargonzo · · Score: 4, Informative

    this is the property of a non-euclidean riemann geometry. suppose that you had a front yard, and you wanted to put a fence around it, to show it was yours. the yard is 2D, so the bigger the yard, the bigger the fence. however, since the flat surface of the earth curves and folds on itself as a sphere, you can own a yard the size of the earth and NOT need a fence, since there are no edges. the same thing applies here.

    --
    BSD is for people who love UNIX. Linux is for those who hate Microsoft.
  84. random walks by dollargonzo · · Score: 2

    what i find really interesting is that the solutions of these topology problems have a characteristic very similar to the mathematics of random walks, both self-avoiding and regular:

    there exists some phenomena that is dimensionally dependent. there exists some critical dimension for which the case is simple and hence reduces, and much easier to prove. the 2d case is classical, and often yields elegant results and rational numbers.

    however, the MOST interesting thing, is that in all 3 problems (random walks, self-avoiding walks, and the pointcare conjecture) the d=3 case is the most mind-boggling. it yields VERY ugly mathematics, and ends up being much harder to prove for. just a thought...

    --
    BSD is for people who love UNIX. Linux is for those who hate Microsoft.
  85. Also of note by lordbad · · Score: 1

    If memory serves correctly, Nikitin was also one of the founding members of the blackdown port of Java to linux.

    I also had a calculus class (95 or 96) with professor Nikitin at ASU, we had many conversations about programming and linux in particular. So not only is he a genius mathmetitian, but also a complete linux geek like the rest of us.

    1. Re:Also of note by sushimonster · · Score: 1

      also know dr nikitin, if anyone has the mind to to validate this proof it is him, you are very correct on noting his linux & blackdown work, he is truely a unique geek/academic/linux/machine langauage creator the practical application of this work is subtle yet omnious in potential change to current technologies such as software engines that drive video, security, network modulators, the more you think about it a $1mil is less than the possible value

      --
      if change is constant, why is it change
  86. MOD THIS UP, forXsake! by the_real_tigga · · Score: 1

    In case this is relevant, in Doom 3 there will be 8 light sources at any given point. Thus, er, wait. No.

    Oh, yes, and the parent, too. Okay, 5, Interesting will suffice.

    --
    my .sig is better than yours.
  87. Anagrams! (Someone had to do it) by Anonymous Coward · · Score: 3, Funny

    The Poincare Conjecture and the issues surrounding it can be described using nothing but anagrams of the famous mathematicians name.

    IE NO CRAP

    Poincare was A NICE PRO by the standards of the time. I wish I had A COIN PER attempt to prove his theorem! Believe me, its NO PI RACE

    I'd ususally begin with a topological approach.
    Take a tennis ball and try to ARC ONE PI around the circumference, then PAIR ONCE.

    Getting too hard, need to go home to use super-computer.

    I OPEN CAR and drive home. ARE I PC ON? Click on PEAR ICON to load fruity maths app.

    Finally prove the theorem!

    I RAP ONCE and then REAP COIN.

    Thats all, I NO RECAP

    Sorry, someone had to do it!

    I. PORN ACE

  88. Ya' sure it should be called a sphere? by fireboy1919 · · Score: 2

    I've been studying EE stuff for a long time now.

    Very frequently we use K-maps, which can take any number of dimensions. 1-d Kmaps exist in things we call LINES. 2-D K-maps exist in things we call SQUARES. 3-D K-maps exist in things we call CUBES, and higher than that, we call the things the K-maps are in HYPERCUBES.

    Extending this terminology...shouldn't a 4-d object with all surface points equidistant from a single point be called a hypersphere to distinguish it from normal usage?

    --
    Mod me down and I will become more powerful than you can possibly imagine!
  89. Blows my mind by Angstrom91 · · Score: 2, Interesting

    So, theoretically, the shape and structure and position if you neglect the uncertainty principle could be mathematically defined.. and we could all be "shadows" of a 4th dimensional world? Makes me wonder what that does to free will.

  90. Re:Ok...I'll waste a few and explain this to you.. by Anonymous Coward · · Score: 0

    Ummmmm..... If you think that a manifold is simply a surface then I can see why you haven't read the paper yet. You should probably be able to understand basic point-set topology before attempting to read proofs of the Poincare conjecture.

  91. James Harris is a well-known sci.math crank. by rpresser · · Score: 2, Insightful

    Don't waste your time increasing his web page hits.

    1. Re:James Harris is a well-known sci.math crank. by JamesHarris · · Score: 0, Offtopic

      Flamed again. I guess I'm wearing a sign "Flame me on SlashDot" or something. Um, I'm less than interested in dealing with mad mathematicians on SlashDot, any advice? Can anyone offer some help? Or is flaming standard operations here?

  92. I think I got it, but one question... by teamhasnoi · · Score: 2

    What AD&D module is this die used in?

  93. Fuck you, E r i c. by Anonymous Coward · · Score: 0

    Fuck you, E r i c.

  94. Re:James Harris was suckered. by JamesHarris · · Score: 0

    I notice that rpresser got a score of 2 for his reply where he put in the subject line that I'm a "crank", and I'm now motivated to see how this discussion area works, so I've noticed a few things.

    1) People who are familiar with it all can pull in an outsider with an anonymous post by making that post and putting up a link, as was done with the original post that drew me here as it was posted on the sci.math newsgroup, which is how I knew about it.

    2) The person they're drawing in, if they rush in and make a post as I did, is unlikely to have an understanding of the system, so they're unlikely to make an effective reply (like my problems with html formatting).

    3) Then someone else can come in and clean up, like rpresser and score some points at the sucker's expense, as he got 2 points even though he was off-topic.

    But isn't that the point of these pages anyway? Get as much action as possible, no matter how it's done?

  95. Re:James Harris was suckered. by rpresser · · Score: 1

    Moderation points are entirely the decision of other people. It's true that potentially everyone can moderate, but in fact not everyone does. (I never have; I'm a relative newcomer, and I've not yet been offered the chance.) I didn't post what I posted hoping to be moderated up; I posted it because it came to mind when I read your ... um ... post.

    The point of these pages varies from person to person, but personally I don't give a damn about whether what I post is modded or not. I read slashdot to be informed; I'm not well known either on Slashdot or elsewhere, and I don't care if I ever am.

    You, on the other hand, are well known. Celebrity always attracts attention; if the attention is uniformly negative, perhaps that says something about you or your contributions.

  96. Re:James Harris was suckered. by JamesHarris · · Score: 0

    And it's people like you who end up helping to break a system. You called me a "crank" which is derogatory, but you were rewarded by the SlashDot system. And here you get another point which makes this look like loser-ville. As for those who might be deluded about the how big SlashDot is, just consider that when I first saw that link on sci.math and clicked on it, soon after it was posted, I couldn't even reach the post as SlashDot itself was unavailable. My Internet activity has created reactions from organizations as diverse as MSN and Google, as I pull and push the Internet, so you might want to consider that when you consider celebrity. And also consider that as far as I'm concerned SlashDot replies more often than not attract people who try to win at someone else's expense, so I *quit* reading them.

  97. Re:Delusion is a terrible thing by JamesHarris · · Score: 0

    I said that about SlashDot, MSN and Google, then I went to check Google statistics and saw no mention of me! Mein Gott!!! I'm cuckoo! Macht's nicht. Ich will nicht geschwunden. And no, ich nicht spreche deutsch. I just know a little bit that I like to throw out when I'm dealing with a certain type of people.