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."

Comments

Read the conjecture...

[CommieLib]

Mmmm...hypothetical donut...

So?

[m0i]

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

Re:So?

[unicron]

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

Re:So?

[jasonditz]

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

Re:So?

[Anonymous Coward]

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.

Re:So?

[Telastyn]

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

Re:So?

[seattle2napa]

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?

Re:So?

[Listen Up]

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.

Re:So?

[elrond2]

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

Re:So?

[GuyMannDude]

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

Re:So?

[doi]

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!" :-)

Re:So?

[sbar]

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? ;)

Re:So?

[richie2000]

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!

Re:So?

[simong_oz]

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 ...

Re:So?

[Reality Master 101]

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.

Re:So?

[King Babar]

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. :-)

Re:So?

[tssm0n0]

"mathematicans do it smoothly and continuously"

That beats my favorite math saying:

What does a constipated mathematician do?

Works it out with a pencil.

Re:So?

[fishexe]

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

Re:So?

[mertner]

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.

Re:So?

[jejones]

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!)

I solved it!

[gerf]

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

Re:I solved it!

[Dr Caleb]

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!".

2 Dimensional Sphere?

[malfunct]

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

Re:2 Dimensional Sphere?

[coult]

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.

Re:2 Dimensional Sphere?

[UberQwerty]

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.

Re:2 Dimensional Sphere?

[exeph]

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.

Re:2 Dimensional Sphere?

[Hard_Code]

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.

Re:2 Dimensional Sphere?

[Anonymous Coward]

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.

Re:2 Dimensional Sphere?

[ThrasherTT]

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.

Re:2 Dimensional Sphere?

[Yobgod Ababua]

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

Re:2 Dimensional Sphere?

[pal]

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

Re:2 Dimensional Sphere?

[BabyDave]

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)

Re:2 Dimensional Sphere?

[nahdude812]

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.

Re:2 Dimensional Sphere?

[muon1183]

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).

Re:2 Dimensional Sphere?

[Serveert]

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?

Re:2 Dimensional Sphere?

[Asprin]

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....

Re:2 Dimensional Sphere?

[DrVeg]

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.

Re:2 Dimensional Sphere?

[PacoSuarez]

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.

Re:2 Dimensional Sphere?

[zeus_tfc]

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

Re:2 Dimensional Sphere?

[barawn]

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.

Re:2 Dimensional Sphere?

[ThrasherTT]

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?

Re:2 Dimensional Sphere?

[Alsee]

I think my brain broke.

BTW, your warranty expired last week.

-

Re:2 Dimensional Sphere?

[Alsee]

Wrap your head around that.

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

-

Re:2 Dimensional Sphere?

[Yobgod Ababua]

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.

Re:2 Dimensional Sphere?

[exeph]

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

Re:2 Dimensional Sphere?

[osu-neko]

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...

Re:2 Dimensional Sphere?

[istewart]

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.

Re:2 Dimensional Sphere?

[ThrasherTT]

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

Re:Though i'm not stupid

[MattRog]

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

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

Ok

[kenp2002]

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.

Re:Ok

[b0r0din]

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.

Re:Ok

[sneakcjj]

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

Re:Ok

[noquarter83]

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.. ..

Re:Ok

[Codifex Maximus]

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.

Re:Ok

[cperciva]

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.

Re:Ok

[fruey]

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.

Re:Ok

[Soul-Burn666]

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...

Re:Ok

[fishbowl]

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

How can it be both?

Re:Ok

[bnenning]

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

Re:Ok

[kalimar]

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.

Re:Ok

[fishbowl]

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.

Re:Ok

[crawling_chaos]

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.

Re:1. PROVE POINCARE 2. ??? 3. PROFIT!

[flynt]

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

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

It's sad that after reading the problem...

[IvyMike]

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

Re:It's sad that after reading the problem...

[IvyMike]

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.

Someone throw cold water on my face

[ekrout]

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.

Re:Someone throw cold water on my face

[zapfie]

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

Re:Someone throw cold water on my face

[luisdom]

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

So?

[i_need_no_nick]

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!

Poincar� Conjecture

[taphu]

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

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

Re:Poincar� Conjecture

[gsfprez]

>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.

Re:Though i'm not stupid

[twistedcubic]

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.

Umm,no

[wiredog]

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.

Mathematical Proof

[SniffleBear]

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.

Re:Mathematical Proof

[selderrr]

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.

Re:Mathematical Proof

[Infernus]

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?

Re:Mathematical Proof

[Tenebrious1]

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

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

Poincar�

[Traa]

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

Poin (POINt) - Ca (CAtastrophic)- Ray

bah, screw it, just call me Henri

Re:Poincar�

[jzs9783]

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

Re:Poincar�

[rsidd]

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.

Re:Poincar�

[Mithal]

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!

Re:Poincar�

[rsidd]

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)

Re:Poincar�

[Abcd1234]

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. :)

Re:Poincar�

[Godwin O'Hitler]

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

Re:1. PROVE POINCARE 2. ??? 3. PROFIT!

[Capt Dan]

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

Re:Though i'm not stupid

[stratjakt]

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?

How about some money ?

[apankrat]

This problem is priced at $1 million if solved.

Re:Though i'm not stupid

[tomzyk]

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?

Re:1. PROVE POINCARE 2. ??? 3. PROFIT!

[br0ck]

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.

Period Three Equals Chaos

[yup2000]

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...

Poincare Conjecture

[jkauzlar]

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.

Re:Poincare Conjecture

[jimmyCarter]

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

Re:Poincare Conjecture

[jkauzlar]

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

Re:Poincare Conjecture

[LadyLucky]

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.

Re:Poincare Conjecture

[Slime-dogg]

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.

Re:Poincare Conjecture

[Viking Coder]

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.

Re:Poincare Conjecture

[Elazro]

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

Re:Poincare Conjecture

[Listen Up]

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

Re:Poincare Conjecture

[opello]

*cough*plajorism*cough*

Re:Poincare Conjecture

[stonecypher]

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?

Re:Poincare Conjecture

[sql*kitten]

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!

Think Quantum &co..

[oliverthered]

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

Re:Think Quantum &co..

[bwt]

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.

Re:Think Quantum &co..

[oliverthered]

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.

Only need an answer for n = 3

[dvdeug]

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.

Re:Only need an answer for n = 3

[Soul-Burn666]

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.

Re:Though i'm not stupid

[Flakeloaf]

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?

FINALLY!!

[paradoxmember]

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

Re:FINALLY!!

[jcoleman]

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.

My proof

[PygmyTrojan]

Every simply connected closed 3-manifold is homeomorphic to the 3-sphere

Well, duh.

Beam me up

[tiredwired]

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?

Re:Beam me up

[adb]

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!

Re:Beam me up

[u38cg]

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.

Re:Beam me up

[fruey]

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.

Re:Beam me up

[panurge]

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.

Re:Beam me up

[Fnkmaster]

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.

Re:Beam me up

[machine of god]

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

Re:Beam me up

[Spunk]

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

That's a good little Tivo.

Re:Beam me up

[Slartibartfarst_UK]

QED, mother fucker.

How terribly American of you. ;-)

Re:Though i'm not stupid

[stratjakt]

> 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?

Poincare Conjecture

[Listen Up]

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.

Poincare Gnomes!

[i_need_no_nick]

Phase 1: Prove Poincare Conjuncture

Phase 2: Collect Clay Math Prize

Phase 3: Profit

Now *there's* a business model!

Re:ANNULUS

[Anonymous Coward]

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.

I thought they did it

[wiredog]

discretely.

Re:I thought they did it

[nevershower]

Only the Computer Science/Math doubles.

Re:I thought they did it

[cperciva]

discretely

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

Re:I thought they did it

[Darby]

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.

Comment removed

[account_deleted]

Comment removed based on user account deletion

2D or 3D?

[richie2000]

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?

Re:2D or 3D?

[EricWright]

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!

Re:2D or 3D?

[richie2000]

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. :-)

Re:2D or 3D?

[EricWright]

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

mmmm.... apples....

Re:2D or 3D?

[Darby]

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.

Re:Though i'm not stupid

[yerricde]

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.

Simplex K?

[Maxwell_E]

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...

video proof

[tgibbs]

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....

having said that...

[Patersmith]

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

Explanation in kindergarten terms

[yerricde]

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.

Re:Explanation in kindergarten terms

[pediddle]

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.

Hiero II of Syracuse

[Theatetus]

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

Re:Hiero II of Syracuse

[Just Some Guy]

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

Re:Though i'm not stupid

[Flakeloaf]

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.

Some smart person: is this right?

[jolshefsky]

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.

Re:Some smart person: is this right?

[jim3e8]

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!

;)

Re:Some smart person: is this right?

[ph43thon]

(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

Re:Some smart person: is this right?

[jolshefsky]

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.

Re:Some smart person: is this right?

[laertes]

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!

Re:Some smart person: is this right?

[spectatorion]

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.

Re:Some smart person: is this right?

[Kragg]

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

Jesus Man!

[kenp2002]

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

Re:Jesus Man!

[dvdeug]

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

Re:Jesus Man!

[kenp2002]

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.

Smash-O-Matic

[pixelpusher220]

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

Re:Smash-O-Matic

[stonecypher]

> Smash-O-Matic

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

Re:Smash-O-Matic

[!!!!!!blameblameblam]

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

I think I understand the problem

[NibbleAbit]

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.

Re:Though i'm not stupid

[MattRog]

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

weird.

[hanwen]

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

Sounds fishy to me.

Re:Someone explain what this sentence means

[Rip!ey]

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.

This is the significance

[Anonymous Coward]

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.

What's "uniform density"?

[yerricde]

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?

Re:What's "uniform density"?

[schmink182]

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

Re:Someone explain what this sentence means

[Big Mark]

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!

Coming from a VERY unexpected corner

[hysterion]

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.

Re:Coming from a VERY unexpected corner

[hanwen]

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

Re:Coming from a VERY unexpected corner

[elBart0]

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"

Re:Coming from a VERY unexpected corner

[call -151]

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!

Re:Coming from a VERY unexpected corner

[delphi125]

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.

The real 1-2-3 steps

[Rayonic]

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) ????????
  

I'm a bit confused.

[fishexe]

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...

Ok...I'll waste a few and explain this to you...

[Anonymous Coward]

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.

abstract and a little background

[call -151]

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

Visualizing a 3-D Sphere

[sisukapalli1]

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

Regarding the remark about primality testing algo

[Anonymous Coward]

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.

Re:Regarding the remark about primality testing al

[jkauzlar]

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%.

Bah.. Math chicks..

[Hydro-X]

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

The answer...

[241comp]

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

= 42

Re:Though i'm not stupid

[navels]

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.

Re:Someone explain what this sentence means

[Old Wolf]

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

Why Eric Weissman rules my world

[Nyarly]

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?

Re:Why Eric Weissman rules my world

[starling]

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.

Re:Why Eric Weissman rules my world

[sqlgeek]

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

Re:Why Eric Weissman rules my world

[Nyarly]

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.

Re:Why Eric Weissman rules my world

[3am]

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.

Re:Why Eric Weissman rules my world

[Cy Guy]

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.

Did I miss something?

[FallLine]

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.

Re:Did I miss something?

[sql*kitten]

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?

Re:Did I miss something?

[FallLine]

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

Re:Did I miss something?

[kevlar]

I think he is running from the Man.

Re:Math Chicks

[3waygeek]

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).

Re:Though i'm not stupid

[Anonymous Coward]

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

Re:Mod parent up!

[Rick the Red]

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

Yeah, I can see applications

[sakusha]

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.

Re:Though i'm not stupid

[spectatorion]

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.

riemann

[dollargonzo]

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.

random walks

[dollargonzo]

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...

Also of note

[lordbad]

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.

Re:Also of note

[sushimonster]

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

MOD THIS UP, forXsake!

[the_real_tigga]

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.

Anagrams! (Someone had to do it)

[Anonymous Coward]

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

Ya' sure it should be called a sphere?

[fireboy1919]

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?

Blows my mind

[Angstrom91]

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.

Re:Mod parent up!

[Trusty Penfold]

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.

James Harris is a well-known sci.math crank.

[rpresser]

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

I think I got it, but one question...

[teamhasnoi]

What AD&D module is this die used in?

Re:James Harris was suckered.

[rpresser]

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.