Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

As I see it, the core question of topology is to figure out whether a homeomorphism exists between two topological spaces.

To answer this question, one defines various properties of a space such as connectedness, compactness, the fundamental group, betti numbers etc.

However, it seems that these properties can at best be used to distinguish two spaces - i.e. if X is a space with property Q, and Y is a space without property Q, then we can say with certainty that X & Y are not homeomorphic.

My question is this: given two arbitrary spaces, how does one show that they are homeomporphic, without explicitly showing a homeomorphism?

share|improve this question
5  
I'm not sure the premise of this question is valid any more than the core question of group theory is to figure out whether an isomorphism exists between two groups. –  Qiaochu Yuan Oct 17 '09 at 6:37
4  
You might rephrase this question, removing the claim that this is the core question of topology, especially in the light of negative answers below. –  Scott Morrison Oct 17 '09 at 14:02
    
Thanks so much guys, very helpful! –  Tejus Oct 18 '09 at 8:39
    
I agree, the question is too general and not properly phrased. But thanks for the responses, very helpful to see the big picture of what topology is all about. –  Tejus Oct 18 '09 at 8:43
    
What about homotopy? –  Harry Gindi Jan 6 '10 at 7:41

8 Answers 8

up vote 12 down vote accepted

As others have noted, it's hopeless to try to answer this question for general topological spaces. However, there are a few positive results if you assume, say, that X and Y are both simply connected closed manifolds of a given dimension. For example, Freedman showed that if X and Y are oriented and have dimension four, then to check whether they're homeomorphic you just need to compute (i) the bilinear "intersection" forms on H^2(X;Z) and H^2(Y;Z) induced by the cup product; and (ii) a Z/2-valued invariant called the Kirby-Siebenmann invariant. The invariant in (ii) obstructs the existence of a smooth structure, so if you happened to know that both X and Y were smooth manifolds (hence that their Kirby-Siebenmann invariants vanished) you'd just have to look at their intersection forms to determine whether they're homeomorphic (however a great many examples show that this wouldn't suffice to show that they're diffeomorphic).

In higher dimensions, Smale's h-cobordism theorem shows that two simply connected smooth manifolds are diffeomorphic as soon as there is a cobordism between them for which the inclusion of both manifolds is a homotopy equivalence. Checking this criterion can still be subtle, but work of Wall and Barden shows that in the simply-connected 5-dimensional case it suffices to check that there's an isomorphism on second homology H2 which preserves both (i) the second Stiefel-Whitney classes, and (ii) a certain "linking form" on the torsion subgroup of H2.

If you drop the simply-connected assumption, things get rather harder--indeed if n>3 then any finitely presented group is the fundamental group of a closed n-manifold (which can be constructed in a canonical way given a presentation), and Markov (son of the probabilist) showed that the impossibility of algorithmically distinguishing whether two presentations yield the same group translates to the impossibility of algorithmically classifying manifolds. Even assuming you already knew the fundamental groups were isomorphic, there are still complications beyond what happens in the simply-connected case, but these can sometimes be overcome with the s-cobordism theorem.

In a somewhat different direction, in dimension 3 one can represent manifolds by link diagrams, and Kirby showed that two such manifolds are diffeomorphic (which in dimension 3 is equivalent to homeomorphic) iff you can get from one diagram to the other by a sequence of moves of a certain kind. (see Kirby calculus in Wikipedia; similar statements exist in dimension 4). I suppose that one could argue that this isn't an example of what you were looking for, since if one felt like it one could extract diffeomorphisms from the moves in a fairly explicit way, and one can't (AFAIK) just directly extract some invariants from the diagrams which completely determine whether the moves exist.

share|improve this answer
    
Thank you! That does help a lot actually. –  Tejus Oct 18 '09 at 8:42
    
Could you provide a reference that homotopy equivalence which pulls back tangent bundles is a diffeomorphism? Does it somehow follow from the H-cobordism theorem? –  Jason DeVito Nov 16 '09 at 19:05
    
Smale did not show that "two smooth manifolds are diffeomorphic as soon as there is a homotopy equivalence between them which pulls back the tangent bundle on one to the tangent bundle of the other", because this is not true. I think, exotic 7-spheres should give a counterexample. –  Igor Belegradek Feb 28 '10 at 3:42
    
Sorry--in a hasty effort to find a clean statement in the literature without using cobordism language I overlooked some obviously-rather-important parts of the hypothesis of Theorem 7.1 of Smale's "On the structure of manifolds"...namely that the manifolds need to have vanishing cohomology in degrees above around half the dimension (so obviously they can't be closed, among other serious restrictions). I've edited the error –  Mike Usher Mar 1 '10 at 17:39
    
In your comment you sound like the h-cobordism theorem does NOT apply to manifolds that are closed. In fact it applies to closed simply-connected manifolds of dimension >4: two such manifolds are h-cobordant iff they are diffeomorphic. –  Igor Belegradek Mar 1 '10 at 22:41

Among topological spaces simplicial complexes are very nice. But even then we run into problems answering your question. Determining whether two finite simplicial complexes are homeomorphic is an undecidable problem. That means there is no algorithm that can tell you if two finite simplicial complexes are homeomorphic, in finite time. Note that these are particularly nice topological spaces and in general topological spaces can be horrendous. I'm not an expert, but considering this I would say that in general the answer to your question would be: we can't.

share|improve this answer
    
I would guess that there is a semi-algorithm that will tell you reliably if two simplicial complexes are homeomorphic - which is what the question asks for. (Though it probably does this by exhibiting an explicit homeomorphism, which the question rules out!) –  HJRW Jan 6 '10 at 17:43

Like the other responders, I find your question a bit too general to address sensibly. However, I'll give one example of a way to prove two spaces are homeomorphic without providing a homeomorphism: Let M be a connected manifold and f: M --> B a submersion to another manifold. Then any two fibers of f are homeomorpic, but it can be very hard to extract an explicit homemorphism from this data. Rather than requiring that f be a submersion it is enough to require that the critical points of f have codimension 2 in B. For example, this is the easiest way to show that any two smooth hypersurfaces in CP^n of the same degree are homeomorphic.

share|improve this answer

Sometimes we can uniquely categorise a space $X$. We can then find a (preferably) finite list of topological properties, such that any space $Y$ that satisfies these properties must be homeomorphic to $X$. Such characterisations exist for many classical spaces like the Cantor set $C$, the rationals $Q$, the irrationals $P$, the Cantor set minus a point, the real line, the plane $R^2$, the Hilbert cube $I^N$, etc. In that case, in the proof of such theorems, we do show a homeomorphism exists, but once we have this theorem, other mathematicians need not find explicit homeomorphisms any more. I have found such theorems to be quite useful. Of course, only sufficiently nice and/or simple spaces can be characterised in this way, and the reach of such a method is quite limited, as there are far many more spaces than there are such nice lists of properties. But using such theorems, topologists could show that all completely metrisable separable topological linear spaces are homeomorphic, e.g.

share|improve this answer
1  
Aren't there some neat results from Hilbert manifold theory (due to Chapman perhaps) related to your point. I vaguely remember that for Hilbert cube manifolds proper homotopy equivalence implies homeomorphism (perhaps faulty memory)?? –  Tim Porter Feb 27 '10 at 8:00

Maybe this is my algebraic topology bias but I'm not sure there's anything one can say about this question in general--there are just too many topological spaces to try to classify them in any sense.

If you only want to know whether two spaces are homotopy equivalent, you can do a lot better. For example, if X and Y are simply connected CW complexes, you can (in principle) show they are homotopy equivalent without writing down any map from X to Y, by computing k-invariants.

share|improve this answer
3  
What are k-invariants? –  Kevin H. Lin Feb 27 '10 at 8:38
    
I'll describe these in the simplest case (where $\pi_1$ acts trivially on all homotopy groups). In that case for a (nice) space X we have a Postnikov tower $$X \to \cdots P_2 \to P_1 \to P_0$$ Each of the maps $P_{i+1} \to P_i$ is a fibration with fiber $K( \pi_{i+1}X, i+1)$. Under the hypothesis, one can show that each of these is (up to homotopy) a principal bundle and classified by a k-invariant $P_i \to K( \pi_{i+1}X, i+2)$, i.e. a certain cohomology class. This is described in Hatcher's Alg. Top. book. The general case is more complicated. One needs twisted cohomology. –  Chris Schommer-Pries Mar 1 '10 at 13:25
    
Thanks, Chris. In brief, the k-invariants are algebraic data which tell you how to solve the extension problems as you go up the Postnikov tower. You can read a bit about them near the end of section 4.3 in Hatcher's book (p. 412 in the online copy). –  Reid Barton Mar 1 '10 at 18:02

You often need to put some assumptions on your spaces to have a sensible answer to this question; there are just too many terrible spaces out there, and you can write down a host of topological invariants that detect differences between weird spaces but never fully answer the question.

One of the most studied classification attempts is the study of the classification of smooth closed manifolds. This leads to a lot of topics like surgery theory and Morse theory that allows you to give constructive procedures to build any manifold by elementary moves, and so the main question becomes one of extracting invariants. Homotopy type is one invariant, and homology groups can be extracted from it.

And then you're led into questions like the Poincare conjecture, or topological quantum field theories, or the classification of simply-connected 4-manifolds, et cetera, et cetera.

share|improve this answer

The only way I can think of giving a meaningful answer is by listing examples, interpreting this as a "big list" question. I'll give an example of proving homeomorphism to S3 non-constructively.

  • Surgery along a framed link in S3 gives rise to a 3-manifold M, and a presentation for the fundamental group π of M. If π turns out to be the trivial group (you might prove this by the Todd-Coxeter process or something), the Poincare conjecture tells us that M is homeomorphic to S3. To exhibit that homeomorphism might be painful, because the proofs of the Kirby theorem are non-constructive and gives no algorithm to simplify surgery presentations of 3-manifolds.

I'm using the fact that there is a unique 3-manifold with trivial fundamental group (whose proof is non-constructive) and then finding an arbitrarily complicated construction to give you a manifold with those properties (and there are many variations on that theme). These constructions, based on surgery or some other violent operation on the manifold, give no hint of how a homeomorphism might look or how one might try to find one.

share|improve this answer
2  
For 3-manifolds there is an effective algorithm to do what you're talking about. There are standard procedures to construct a triangulation of a 3-manifold obtained by surgery on a link in $S^3$. Then you apply the Rubinstein 3-sphere recognition algorithm to that triangulation, and you're done. It has exponential run-time in the number of tetrahedra in the 3-manifold triangulation and that in turn looks something like the number of crossings in your diagram times a function that measures the size of the surgery coefficients. –  Ryan Budney Jan 6 '10 at 9:23

I would respectful suggest that there are other important problems in topology. One thing that the wikipedia article linked doesn't spend much time on is the "placement problem". Instead of attempting a definition, here are the first two examples that spring to my mind: classify curves in a fixed Riemann surface or classify curves in the three-sphere, each time up to isotopy. The first leads to the study of the mapping class group and perhaps Teichmuller spaces. The second leads one towards knot theory.

Notice that if $\alpha$ and $\beta$ are curves in a surface $S$ then deciding if the pairs $(S, \alpha)$ and $(S, \beta)$ are homeomorphic reduces to the classification of surfaces and so is "easy". The mapping class group still manages to be important, however!

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.