A (Correct) Poincare Proof!? 324
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."
Read the conjecture... (Score:5, Funny)
Umm,no (Score:2)
I think. Maybe it is a 3-d doughnut. It's been ten years since I studied that stuff at college.
So? (Score:2, Interesting)
Re:So? (Score:5, Funny)
Re:So? (Score:5, Funny)
Well, if you give the proof at a big university chances are you'll totally score with some PhD Math chick.
Before you guys all start dissing math chicks, remember that "mathematicans do it smoothly and continuously". I wanted to put that on a bumper sticker and slap it on my car but I went with "My girlfriend can't wrestle but you should see her box" instead.
GMD
I thought they did it (Score:5, Funny)
Re:I thought they did it (Score:5, Funny)
Or, more specifically, in groups, and in fields.
Re:So? (Score:2, Funny)
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? (Score:2)
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? (Score:4, Informative)
Re:So? (Score:5, Insightful)
Pure science is pure science. All great discoveries that have ever existed have been because of small, previously unrelated pure mathematical works (or other pure true sciences), which when done on their own seem to have no superficial meaning to someone such as an engineer or common layman, but pure mathematics is akin to pieces of a grand puzzle. Each piece is intrinsically linked to the whole picture. Looking at each piece will not reveal the puzzle, although solving each piece on its own will. This proof need not prove anything to an engineer, a computer scientist, a ballerina, or the mailman, but to a mathematician and others who understand its significance (among others) this proof advances the pure science of mathematics...and by that the world will eventually be forever changed.
Re:So? (Score:3, Insightful)
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? (Score:3, Informative)
(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:Hiero II of Syracuse (Score:2)
Re:So? (Score:3, Funny)
Archimedes invented the screw pump while taking a bath
Actually, it's a bit more logical than that. He discovered the principal of displacement while taking a bath.
I'm not exactly sure how one would think of "screw pumps" while in the bath. Come to think of it, I'm pretty sure I don't want to know.
How about some money ? (Score:3, Informative)
Think Quantum &co.. (Score:2, Informative)
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
I solved it! (Score:4, Funny)
Re:I solved it! (Score:3, Funny)
before I had a seziure.
Q.E.D. my ass! It shoulda been "Whoo hoooo!".
Ok (Score:5, Funny)
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 (Score:5, Funny)
"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 (Score:5, Informative)
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 (Score:2)
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 (Score:2)
>...Finally, suppose that the universe is finite in volume.
How can it be both?
Re:Ok (Score:2)
riemann (Score:4, Informative)
Why Eric Weissman rules my world (Score:5, Informative)
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?
Re:Ok (Score:3, Funny)
Hehe... gotta go back and get my advanced trigonometry and calculus credits.
It's sad that after reading the problem... (Score:5, Funny)
...the most intelligent thing I can think of is: "Mmmmm...donuts."
Re:It's sad that after reading the problem... (Score:2)
You're probably right. A data point: every time I've attempted to make a Simpsons reference, even if the story has almost no comments when I start to post, my comment is already redundant by the time it shows up. Sorry all.
Someone throw cold water on my face (Score:5, Funny)
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.
Did I miss something? (Score:2)
Back to the drawing board... (Score:5, Funny)
Re:Back to the drawing board... (Score:2)
Poincaré Conjecture (Score:5, Informative)
http://mathworld.wolfram.com/PoincareConjecture.h
Mathematical Proof (Score:5, Funny)
Re:Mathematical Proof (Score:2)
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.
video proof (Score:2)
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....
Poincare Conjecture (Score:5, Informative)
Re:Poincare Conjecture (Score:2)
Oh, I didn't realise it was that simple.
Re:Poincare Conjecture (Score:5, Funny)
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 (Score:2, Informative)
Only need an answer for n = 3 (Score:5, Informative)
Re:Only need an answer for n = 3 (Score:2)
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:Jesus Man! (Score:2)
Re:Jesus Man! (Score:2)
FINALLY!! (Score:5, Funny)
My proof (Score:3, Funny)
Well, duh.
Beam me up (Score:2, Funny)
Re:Beam me up (Score:2)
Re:Beam me up (Score:2, Informative)
Poincare Conjecture (Score:5, Informative)
The Poincaré Conjecture is widely considered the most important unsolved problem in topology. It was first formulated by Henri Poincaré in 1904. In 2000 the Clay Mathematics Institute selected the Poincaré conjecture as one of seven Millennium Prize Problems and offered a $1,000,000 prize for its solution.
The conjecture is that every simply connected compact 3-manifold without boundary is homeomorphic to a 3-sphere. (Loosely speaking, that every 3-dimensional object that has a set of sphere-like properties can be stretched or squeezed until it is a 3-sphere without breaking it).
Analogues of the Poincaré Conjecture in dimensions other than 3 can also be formulated. The difficulty of low-dimensional topology is highlighted by the fact these analogues have now all been proven, while the original 3-dimensional version of Poincaré's conjecture remains unsolved. Its solution is central to the problem of classifying 3-manifolds.
On April 7, 2002 there were reports that the Poincaré conjecture might have been solved by Martin Dunwoody; on April 12 Dunwoody acknowledged a gap in the proof and was attempting to fix it. In October of that year, Sergei Nikitin of Arizona State University announced that he had proved the result.
Poincare Gnomes! (Score:2, Funny)
Phase 2: Collect Clay Math Prize
Phase 3: Profit
Now *there's* a business model!
2D or 3D? (Score:2)
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? (Score:2)
No one cares what goes on inside the sphere... no one can see it!
Re:2D or 3D? (Score:2)
And, if you want to get picky, the surface representation in 3D does care, you need to define the curvature somehow and you can't do that in 2D - without that, it's just another plane. *ponders* Nah, I don't want to get picky. Let's leave it. :-)
Re:2D or 3D? (Score:2)
mmmm.... apples....
having said that... (Score:2, Funny)
How do we use this to take down the RIAA/MPAA?
Some smart person: is this right? (Score:5, Interesting)
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 [wolfram.com].) 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 [wolfram.com] 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.
Smash-O-Matic (Score:2, Funny)
This is the significance (Score:3, Informative)
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.
Coming from a VERY unexpected corner (Score:5, Insightful)
Re:Coming from a VERY unexpected corner (Score:2, Informative)
Also from his CV: "1996-t/n Linux Consultant in Arizona"
Ok...I'll waste a few and explain this to you... (Score:2, Interesting)
"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.
abstract and a little background (Score:3, Informative)
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 [ucdavis.edu]
Visualizing a 3-D Sphere (Score:5, Funny)
Read this a while ago somewhere. Couldn't resist posting it.
S
The answer... (Score:2, Funny)
= 42
Anagrams! (Someone had to do it) (Score:3, Funny)
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
Re:Though i'm not stupid (Score:5, Informative)
http://mathworld.wolfram.com/PoincareConjecture.h
Re:Though i'm not stupid (Score:2, Informative)
Explanation in kindergarten terms (Score:3, Informative)
Is it trying to prove that there can exist a 4-dimensional object that has all points equidistant from a single point in space-time or something?
Assume that you have a sculpture made of Play-Doh® modeling compound, without any holes in it. If the Poincaré conjecture is true, then you can reshape the sculpture into a ball without breaking or joining anything.
Re:Though i'm not stupid (Score:2, Informative)
Probably the simplest layman's explanation I can think of: if something feels like it is the 3-sphere, then it is the 3-sphere.
By "feels like" I mean that it has certain properties which strongly suggest that it is the real thing.
Re:Though i'm not stupid (Score:2, Interesting)
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?
Re:Though i'm not stupid (Score:2)
What's "uniform density"? (Score:2, Funny)
Aye, as most objects of uniform density do :)
Doesn't "uniform density" mean "as opposed to something like swiss cheese"? I was talking about holes as in donut, not holes as in swiss cheese or holes as in IIS. Can a torus have a uniform density?
Re:1. PROVE POINCARE 2. ??? 3. PROFIT! (Score:4, Informative)
http://www.claymath.org/prizeproblems/poincare.
Re:1. PROVE POINCARE 2. ??? 3. PROFIT! (Score:2)
Re:1. PROVE POINCARE 2. ??? 3. PROFIT! (Score:2)
The real 1-2-3 steps (Score:3, Funny)
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) ????????
Re:2 Dimensional Sphere? (Score:2, Informative)
Re:2 Dimensional Sphere? (Score:5, Interesting)
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? (Score:5, Interesting)
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? (Score:2)
BTW, your warranty expired last week.
-
Re:2 Dimensional Sphere? (Score:2)
I tried, but since the problem is homeomorphic to a sphere my mind always slid off at a point.
-
Re:2 Dimensional Sphere? (Score:5, Interesting)
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? (Score:2, Informative)
the "sphere" you know and love, we call the "two sphere."
- pal
Re:2 Dimensional Sphere? (Score:2, Informative)
Re:2 Dimensional Sphere? (Score:2)
This is a bit of an obnoxious distinction to make, and I certainly think they should have phrased it differently. Usually people say three-sphere when they mean a sphere in three dimensions, which is in fact just a surface and thus two-dimensional. Then, we can say n-sphere and have a sphere in n-dimensional space that is an n-1 dimensional object. However, this kind of quibbling tends to have no effect on proof, which is what math on this level concerns itself with anyway.
Yeah, Topologists always were kinda weird....
Re:2 Dimensional Sphere? (Score:2, Informative)
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? (Score:2)
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? (Score:2, Interesting)
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?
Re:2 Dimensional Sphere? (Score:2)
When we say the sphere is two-dimensional, what we mean is roughly that we can describe it using two coordinates. We do this every day, using latitude and longitude to describe positions on the Earth's surface.
To be a little more precise about it, at each point in a 2-dimensional manifold, we can find a small neighbourhood around it which "looks like" part of the plane. (which of course is 2 dimensional)
Re:2 Dimensional Sphere? (Score:2)
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.
Poincaré (Score:2)
Poin (POINt) - Ca (CAtastrophic)- Ray
bah, screw it, just call me Henri
Re:Poincaré (Score:2)
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:Poincaré (Score:2, Interesting)
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:Poincaré (Score:2)
Re:Someone explain what this sentence means (Score:2, Informative)
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!
Re:Someone explain what this sentence means (Score:2)
there should be a $1m prize for taking an apple with a rubber band on it and getting the rubber band to a point
Re:Mod parent up! (Score:2)