Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have heard that some issues in group theory prevent classifying all manifolds upto homotopy using the fundamental group invariant. Does anyone know what are those issues?

Thanks, K.

share|improve this question
3  
Your question is not quite interesting because the fundamental group focuses on curves - it does not really reflect on 2-cells, 3-cells, etc, as shown in the case where S^2 and a point are both simply connected. A more interesting question would be whether two non-homotopy equivalent spaces can have identical homotopy groups. See mathoverflow.net/questions/3540/… –  Soarer Aug 9 '10 at 13:51
6  
I think that the OP is alluding to the insolvability of the word problem, and the impact that this has on the feasibility of giving an algorithmic classification of higher dimensional manifolds. See my answer below for more details. –  Matt E Aug 9 '10 at 14:35
    
I also read this in the way Soarer does, maybe the OP could clarify. However Matt E makes a good and interesting point. –  BBischof Aug 10 '10 at 2:27

5 Answers 5

This would have belonged as a comment to MattE answer, but I do not have the reputation to leave comments. The reason it is impossible to algorithmically classify topological manifolds (up to homotopy) is, as MattE very clearly explained, that the word problem for groups can be embedded in the question of classifying manifolds. Indeed, suppose that you want to decide if a finitely presented group $G$ is trivial. First, there are several constructions that yield a compact 4-manifold whose fundamental group is isomorphic to $G$; choose your favorite construction and let $M$ be the resulting manifold. Observe that the intersection pairing in homology is computable, so we may just as well assume that we know what this is for our manifold $M$. It turns out that the homotopy type of a simply connected manifold is completely determined by the intersection form on the second integral homology group (and that any intersection form can be realised). So if on the side we construct the simply connected manifold $N$ with intersection form coinciding with the one of $M$, then deciding whether the homotopy types of $M$ and $N$ coincide will decide if the fundamental group of $M$ is trivial or not, i.e. if $G$ is the trivial group or not.

There are some issues on how you compute the intersection pairing and how you find the manifold $N$ given the intersection form, and there are tricks to simplify all these questions, but I will not go into this, in the hope that what I wrote will suffice! Of course, if you want more details, feel free to ask, and I will try to answer!

share|improve this answer
    
Thanks very much for this. –  Matt E Aug 10 '10 at 20:34
    
+1: thanks for this very informative answer. –  Pete L. Clark Aug 10 '10 at 21:31

I think that what you are thinking of is the insolvability of the so-called word problem in group theory. What this means is (among other things) that it is not possible to write an algorithm (i.e. a computer program) which can tell whether two groups, each presented by finitely many generators and relations, are or aren't isomorphic.

If you are given a manifold (presented say by gluing together various open balls), you can get such a presentation of its fundamental group, but (at least when the dimension is 4 or higher --- and I think that 4 is the right bound here) in fact you can get an arbitrary presentation in this way, and so the insolvability of the word problem means that, in practice, you don't have any way to figure out what the fundamental group of your manifold is (even whether or not it is trivial!)

As a consequence, you can't find an algorithm to determine whether two manifolds (of dimension 4 or higher) are homotopic (let along homeomorphic or diffeomorphic), since homotopic manifolds will have isomorphic $\pi_1$s, and so we could use such an algorithm to solve the word problem, by encoding two finitely presented groups as the $\pi_1$ of some manifolds, and then applying our hypothetical manifold algorithm. Following Pete Clark's comment below, I'm no longer sure about the correctness of the preceding statement. What is true is that there is no general algorithm to determine the fundamental group of a manifold (in dimension $\geq 4$), or even to determine if a manifold is simply connected. (Whether this actually prohibits a classification is not clear to me; with luck, someone with more expertise will weigh in.)

For this reason, a lot (although not all) investigations of higher dimensional manifolds restrict attention to simply connected manifolds. Then there is no obstruction to classification coming from issues of the uncomputatibility of $\pi_1$. Of course, as others have noted in their answers, there can be many non-diffeomorphic, non-homeomorphic manifolds, and non-homotopic manifolds of a given dimension, so the problem with $\pi_1$ is far from the only problem when it comes to classifying manifolds. But it is certainly one problem, and you are correct that (at least from the point of view I'm explaining here) it is a problem of group theory.

Added: See Damiano's answer to this question for a precise explanation of the relationship between the word problem for groups and the classification problem for manifolds.

share|improve this answer
    
I am a little confused about your third paragraph (although I have heard similar claims many times, so I must just be missing something). How does an algorithm to classify manifolds up to diffeomorphism lead to an algorithm to test for isomorphism of finitely presented groups? You can find two compact $4$-manifolds with corresponding presentations. If they are diffeomorphic, you know that the groups are isomorphic. But maybe they are not diffeomorphic even though the fund. groups are isomorphic. What then? –  Pete L. Clark Aug 10 '10 at 1:05
    
Dear Pete, This is a good question, to which I don't have a good answer. What does follow from insolvability of the word problem (or, more precisely, from unrecognizability of the trivial group), is that one cannot algorithmically determine if a general $n$-manifold (with $n \geq 4$) is simply connected. But now that you raise the question, I don't see immediately why this prohibits a classification (or at least, an algorithmic recognition procedure); it just means that any such classification can't use "is simply connected" as one of its determining attributes. –  Matt E Aug 10 '10 at 2:17
    
Dear Matt,Thanks for your prompt answer, although I was hoping you would explain why I was missing something silly! I started thinking about this sort of thing after a talk that Bjorn Poonen gave at UGA (google "Poonen Undecidability Everywhere") as a possible attack on the open problem of decidability of isomorphism of varieties over Q-bar: I wanted (momentarily, at least) to use the etale fund. group to reduce to some profinite version of the word problem (not that I know that such a thing exists). But I ran into the same problem as above... –  Pete L. Clark Aug 10 '10 at 3:28

For a better pair of examples take the once punctured torus and a pair of pants. Both have fundamental groups free of rank 2. They are both homotopy equivalent to a figure 8, but as manifolds they are not homeomorphic since one has 3 boundary components and the other has one boundary component.

There are simply connected closed manifolds of dimension 4 and higher that are not homeomorphic. In even dimensions they can be distinguished by their middle dimensional homology intersection forms.

share|improve this answer

Well, there are no reasons to expect that manifolds with equal π1 are homotopy equivalent, and in general, they indeed aren't. Even for closed manifolds of fixed dimension: say, both CP2 and S4 have trivial fundamental group.

share|improve this answer

The following (famous) question is perhaps in the spirit of the OP's question:

If $M$ and $N$ are closed aspherical $n$-manifolds with isomorphic fundamental groups, are $M$ and $N$ homeomorphic? (This is known as the Borel Conjecture.)

Scott Carter's example illustrates the need for the assumption that the manifolds are closed. But the question above suggests that perhaps the fundamental group is sufficient to characterize the topological type of a closed aspherical manifold.

share|improve this answer

Your Answer

 
discard

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