## Tuesday, August 29, 2006

### The Poincaré conjecture, I

We've said a bit about the circumstances surrounding the proof of the Poincaré conjecture, but little about the conjecture itself. What is it?

In the most general terms, it is about the question: When is a topological object "the same" as a sphere?

In order to give a mathematical answer to such a question we first have to precisely define at least three terms: "topological object", "the same", and "sphere".

Mathematicians define a topological object as a set of points that satisfy certain axioms. ("Points" themselves are left undefined. The notion of "set" can also be axiomatized, but the intuitive idea is sufficient here; more detail would take us too far afield.) Intuitively, you can think of topological objects as the sort of things studied in high school geometry -- lines, surfaces, solids, etc. Axiomatization of high school ("Euclidean") geometry itself is more complicated than actually taught in high school -- it's more than just Euclid. It was done by David Hilbert in 1899, but details aren't for the faint-of-heart. (Hint: He needed to use 19 axioms.)

The Poincaré conjecture doesn't concern topological objects in full generality. Instead, it's about objects of a certain kind, which mathematicians call "manifolds". I intend to go into more detail about manifolds in another article, but it's not necessary to be more precise at this time. Just keep thinking about lines, surfaces, solids, and higher dimensional analogues. (If you really want to know right now, take a look at this or this.)

In order to add some intuitive content to the notion of manifold, we can consider how "sphere" is defined. The simplest sphere is two points (+1 and -1), sometimes written (a little ostentatiously) as S0. The zero refers to "dimension" -- a point is 0-dimensional. The meaning of "dimension" here is wrapped up in the definition of manifold, but it's just the intuitive thing. An example of a 1-dimensional manifold is a (possibly curved) line. Another example is a circle. A "standard" circle is the set of points in a plane (which itself is 2-dimensional) that satisfy the equation x2 + y2 = r2, where (x,y) is a point in the plane given by x and y coordinates, and r is the radius.

Now, topologically, the size and exact shape of an object is not important. Topology is sometimes called "rubber-sheet geometry" because exact sizes and shapes don't matter. Any transformation of an object on a rubber sheet resulting from stretching or bending is considered to be "the same" object, as long as the sheet isn't ripped. So any figure on a rubber sheet which is "the same" as a standard circle (of radius 1) is just another example of a 1-dimensional "sphere" (a "1-sphere").

When speaking topologically, mathematicians are often deliberately ambiguous. Technically, a circle (on a flat plane) of radius 1, given by the equation above, is a particular manifold, while any stretched, squashed, or twisted version is a different manifold. But topologically they are all "the same" or "equivalent", and any particular instance is referred to simply as a 1-sphere, or S1 for short.

Continuing in the same vein, a "standard" 2-sphere is the set of points (x,y,z) in 3-space defined by the equation x2 + y2 + z2 = r2. Anything that's "the same" as a "standard" 2-sphere is itself usually referred to as a 2-sphere, S2.

One can similarly define n-spheres, Sn, for any integer n≥1. Note that n is the dimension of the object as a manifold, but it is defined in terms of points in a space of dimension n+1. That is, the points themselves each have n+1 coordinates, yet the equation that specifies the points reduces the dimension of the resulting manifold by 1.

Finally, what do we mean by "the same"? As noted, two objects are considered "the same" if there is a transformation that stretches, compresses, or bends the overall space, without tearing it, and carries one object into the other. (That's not quite right, since a manifold need not be defined as embedded in a "Euclidean" space of a particular dimension, but it's close enough, absent a more precise definition of "manifold".)

With those preliminaries out of the way, we can return to the original question: When is a topological object "the same" as a sphere?

For concreteness, consider the case n=2. When is a manifold topologically the same as S2, which is just what we ordinarily think of as a sphere in 3-dimensional space (but a hollow one, like a globe, without the solid interior).

The game is rigged. All the precise definitions of "manifold", "the same", and so forth are arranged so that everything that is topologically the same has the same (topological) properties. So anything that's equivalent to the 2-sphere S2 must have all of its properties. Having each of the properties is a necessary condition for equivalence. So let's list some of them.

• S2 is a 2-dimensional manifold (a surface).
• A sphere is just a single piece, not 2 or more disconnected pieces. Topologists say it is "connected".
• A sphere is finite and bounded. It does not extend infinitely in any direction. Topologists say it is "compact".
• A sphere has no "boundary". That is, there is no "edge", as there is on a 2-dimensional disk, for example.

Unfortunately, it turns out that this set of necessary conditions isn't enough to be sufficient to ensure that a manifold meeting the conditions is topologically equivalent to a 2-sphere. The simplest example of a nonequivalent manifold meeting the conditions above is a (2-)torus, which is the surface of an ordinary 3-dimensional donut. (Ummmmm, donuts!)

How can one visualize the reason this is so? Well, just think of drawing a circle on a torus. In some cases you can imagine the circle contracted to a point without ever leaving the surface or breaking the circle. But in other cases, if the circle goes either around or through the central hole, this is impossible. Because of this circumstance, a torus is not topologically equivalent to a 2-sphere, where a circle can always be contracted to a point.

The condition that any circle drawn on a 2-manifold can be contracted to a point without breaking the circle or leaving the manifold must be met by any manifold that is equivalent to S2. This condition is necessary. The neat thing is that you do get a sufficient set of conditions for equivalence if you add this last condition to the earlier ones. This fact was (implicitly) proven in the mid 1800s by Bernhard Riemann, who gave a simple classification of 2-dimensional manifolds.

In 1904 it occurred to Henri Poincaré to wonder whether an analogous set of conditions applied to 3-dimensional manifolds. (The case of dimension n=1, essentially just closed loops, is fairly trivial.) He conjectured that, with the appropriate definitions, the answer would be yes. That is, if the right conditions are met, a manifold is equivalent to the 3-sphere S3. This is the Poincaré conjecture. At first he even thought, incorrectly, he could prove it.

Others later conjectured that an analogous result holds in any dimension n≥3. In the next installment, we will discuss how to generalize the necessary concepts to higher dimensions, and recount the history of the conjecture up to the present time.

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David Wilson said...

good on ya! not entirely clear to me but a good start, and i thank you for the effort

be well.

9/01/2006 04:31:00 AM
Charles Daney said...

Thanks, David.

Anyone: just let me know what's unclear, and I'll try to clarify things in subsequent articles.

9/01/2006 10:13:00 AM
Anonymous said...

fyi --
S^0 is not a single point, but actually two points...all n-spheres must be embedded in R^n+1.

9/22/2006 12:12:00 PM
Charles Daney said...

About S^0 -- Yes, of course it is two points, not one. My bad. It is the solution (in R^1) of x^2 = 1.

9/25/2006 12:11:00 AM