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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."
Mmmm...hypothetical donut...
If your bitterest enemies are people who hack the heads off civilians, then I would say you're doing something right.
But alas, the space alloted in a regular comments window is insufficient to explain further...
Here is an example with all sorts of definitions you can read.
t ml
http://mathworld.wolfram.com/PoincareConjecture.h
Thanks,
--
Matt
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? ]=-
Here is step 2! You win 1 million dollars for the correct proof from claymath!
h tm
http://www.claymath.org/prizeproblems/poincare.
...the most intelligent thing I can think of is: "Mmmmm...donuts."
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
Well, if you give the proof at a big university chances are you'll totally score with some PhD Math chick.
For super-geeks, here is is a more thorough discussion of the Poincaré Conjecture.
t ml
http://mathworld.wolfram.com/PoincareConjecture.h
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.
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.
PUBLIC SPLIT ON WHETHER BUSH IS A DIVIDER -CNN scrolling banner, 10/15/2004
There's a $1,000,000 award for one thing...
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.
This problem is priced at $1 million if solved.
3.243F6A8885A308D313
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?
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.
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
watch this
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.
wow.. finally.. i can sleep at night!!
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?
Well, duh.
Trying is the first step towards failure.
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.
(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
discretely.
Best Slashdot Co
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.
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.
Slay a dragon... over lunch!
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?
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)
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.
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.
Timeo idiotikOS et dona ferentes
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) ????????
[PowerPoint] is a tool for capitalist presentation
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.
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
Piece by piece:
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.
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
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.
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