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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.
I'm not that much into maths, but what will this proof achieve?
have you been defaced today?
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.
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.
...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
For super-geeks, here is is a more thorough discussion of the Poincaré Conjecture.
t ml
http://mathworld.wolfram.com/PoincareConjecture.h
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.
I think. Maybe it is a 3-d doughnut. It's been ten years since I studied that stuff at college.
Best Slashdot Co
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
The name is Poincaré (with the litte thingy on the e)
Poin (POINt) - Ca (CAtastrophic)- Ray
bah, screw it, just call me Henri
Mmmmmmmmm... million dollar donut.... gaaaawwwwww...
Sig:
Barbeque is a noun. Not a verb.
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!!!!
This problem is priced at $1 million if solved.
3.243F6A8885A308D313
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.
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.
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.
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.
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"?
wow.. finally.. i can sleep at night!!
Well, duh.
Trying is the first step towards failure.
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?
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.
Phase 2: Collect Clay Math Prize
Phase 3: Profit
Now *there's* a business model!
discretely.
Best Slashdot Co
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
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
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....
How do we use this to take down the RIAA/MPAA?
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
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)
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?
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
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
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.
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)
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
Thanks for modding redundant even though mine was posted 9 minutes before the last link. :(
Thanks,
--
Matt
I've at it since 1998 and spend way to much time at a computer. Given that, it's not hard.
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.
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?
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.
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!
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
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.
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? ]=-
"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.
Wrap your head around that.
I tried, but since the problem is homeomorphic to a sphere my mind always slid off at a point.
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- - You can't take something off the Internet! That's like trying to take pee out of a swimming pool.
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
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?
Actually, all the article says is that they have finally realized that Douglas Adams is right.... the last line of the proof is:
= 42
Full-Featured GPL Web Hosting Control Panel
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?
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
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
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.
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.
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.
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.
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.
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
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!
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.
Don't waste your time increasing his web page hits.
What AD&D module is this die used in?