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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."
I'm not that much into maths, but what will this proof achieve?
have you been defaced today?
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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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!!!!
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?
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.
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!
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!
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)
"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"
That said I have not read the paper, don't have the time right now.
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.
:) Note that IANAM (i am not a mathemetician), so I may not be as precise, or confusing, as a formal proof.
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?
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.