Classification problem for non-compact manifolds

Background

It is well-known that the compact two-dimensional manifolds are completely classified (by their orientability and their Euler characteristic).

I'm also under the impression that there is also a classification for compact three-dimensional manifolds coming from the proof of the Geometrization Conjecture and related work.

Unfortunately for $n\ge4$ no similar classification is possible because it can be shown that it is at least as hard as the word problem for groups. Thus for higher-dimensional manifolds we instead focus on classifying all the simply-connected compact manifolds.

My question

Why in these "classification problems" are we only considering compact manifolds? Is there an easy reason why we restrict ourselves to the classification of compact manifolds? Does a classification of general (not necessarily compact) manifolds follow easily from a classification of compact manifolds?

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People don't stick to only simply-connected manifolds. 4-manifolds having amenable fundamental group is a popular class, for example. Basically, any class where some aspects of the fundamental group aren't out of control are okay. From the point of view of classification you want to restrict to a class of groups where the isomorphism problem is tractible. – Ryan Budney Nov 6 2009 at 3:01
And you of course you can do things with manifolds by avoiding direct discussion of the fundamental group. The classification of 3-manifolds in essence says that 3-manifolds are classified by their fundamental groups modulo the issue of lens space summands. But that's not how the proof goes. Similarly, Rubinsteins 3-sphere recognition algorithm does not discuss the fundamental group, even though the 3-sphere is the only compact boundaryless 3-manifold with a trivial fundamental group. Presumably there could be an analogue of Rubinstein's algorithm for triangulated 4-manifolds. – Ryan Budney Nov 6 2009 at 3:56
I think, that the first 10 lines of Maillot's article give an answer to your quesiton. Here is a citation: "For open 3-manifolds, by contrast, there is not even a conjectural description of a general 3-manifold in terms of geometric ones". arxiv.org/PS_cache/arxiv/pdf/0802/0802.1438v2.pdf – Dmitri Jan 12 2010 at 9:05

I think it's because it's too hard when the manifolds are noncompact.

Closed orientable surfaces are classified nicely by their Euler characteristic, but it's not so clear how to classify noncompact surfaces. For instance, there are things like the sphere minus a Cantor set, or surfaces of genus g minus a Cantor set, or infinite genus things like the following.

Take the infinite graph obtained by looking at the union of the edges in the usual tiling of R^2 by squares. Take a 3-dimensional neighborhood of this and call the boundary S. Do the same thing with the graph obtained by taking the union of edges in the tiling of R^3 by cubes to get a surface T.

These aren't homeomorphic, but you need to do some work to differentiate them.

In higher dimensions the situation is even worse.

Some semi-related awesomeness: A large surface is a space that is a surface, except that you don't require it to be second countable. It's a theorem that there are precisely 2^{\aleph_1} connected large surfaces. I forget who proved this, but if anyone can find it in mathscinet, I would love to know the reference (I'd be your best friend, et cetera).

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FYI, non-compact surfaces have been classified. # On the Classification of Noncompact Surfaces # Ian Richards # Transactions of the American Mathematical Society, Vol. 106, No. 2 (Feb., 1963), pp. 259-269 – Ryan Budney Nov 4 2009 at 23:52
Oh, thanks. I knew that at some point. – Richard Kent Nov 5 2009 at 0:21
And regarding non-compact 3-manifolds there has been some progress on this front. For instance, the arXiv papers of Sylvain Maillot, and all the recent work on tameness of non-compact hyperbolic 3-manifolds. There's likely quite a ways to go, I don't want to minimize what hasn't been done. – Ryan Budney Nov 18 2009 at 18:58
@Richard Kent: I think you are wrong concerning the two surfaces you describe in your second paragraph. See my answer below (it does not fit in a comment). – Benoît Kloeckner Apr 13 2010 at 15:42
@Benoit: Oops, you're right. I think I had misunderstood the statement of Kerekjarto's Theorem, which says that they should be homeomorphic (and can be found in the paper by Richards that Ryan cites). – Richard Kent Apr 13 2010 at 16:04

Especially if you're looking at smooth manifolds, there's some weird stuff that happens in the noncompact case. The most famous example is the existence of infinitely many manifolds that are homeomorphic but not diffeomorphic to R^4.

I don't know that similarly weird things happen just for topological manifolds, but I wouldn't count it out.

I guess it's worth mentioning, though, that plenty of utterly crazy things happen even for compact topological 4-manifolds. For instance, you have the E8 manifold, which isn't triangulable, and on the other end you have manifolds that admit way too many piecewise linear structures. This weirdness disappears for compact manifolds of higher dimension (fortunately), but for non-compact manifold you have to deal with stuff like R x E8, which I suspect isn't much nicer.

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I don't know about E8xR, but crossing with R sometimes makes things better. For instance, the Cannon-Edwards Theorem asserts that the double suspension of a homology sphere is homeomorphic to a sphere. It follows that although the cone C on a homology sphere is not a manifold, Cx(R^2) is a manifold, remarkably. On the the other hand, this gives a way of building "bad" triangulations of manifolds. – HW Nov 4 2009 at 23:35

I don't think there is an easy reduction to the compact case. For instance, the Whitehead manifold is a non-compact, contractible 3-manifold which is not homeomorphic to real 3-space. This suggests that the non-compact situation in 3 dimensions is much worse than the compact one (which, as you point out, is now fairly well understood, after a LOT of work).

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I completely agree with this point of view – Dmitri Jan 12 2010 at 10:58

In a sense you more or less know what non-compact manifolds must look like provided you have a classification of the compact manifolds. The starting point is the basic observation (Whitney) that a non-compact manifold has a proper function $f : M \to R$. So the preimage of the intervals $[-n,n]$ for $n=1,2,3,\cdots$ form a nested family of compact submanifolds of $M$ that exhaust the manifold -- provided $f$ is transverse to the integers, which can be accomplished.

So understanding $M$ boils down to seeing how the features of this family of submanifolds "pile up", and putting some sort of reasonable ideal boundary on $M$ -- since "ideal boundary" is a lower-dimensional phenomenon, in principle you might be inclined to think this is reasonable.

Of course I'm being pretty vague but it sounds like you were looking for something like this?

On the other side of things, because of the above non-compact manifolds are decidedly less combinatorial objects. You don't have finite, combinatorial descriptions. Larry Siebenmann shows people the example of the smooth structure on $I \times \mathbb R^4$ such that the projection map $I \times \mathbb R^4 \to I$ is a smooth submersion, but for which the fibers a pairwise non-diffeomorphic $\mathbb R^4$'s.

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Hmm... When I read the question, I assumed the author meant "compact manifolds without boundary". This would seem to make your approach impossible. I wonder, does it? Is there a way to fix this? – Ilya Grigoriev Dec 13 2009 at 0:42
Here's what I meant. I'm used to thinking of non-compact manifolds as being very closely related to compact manifolds with boundary, as your answer seems to confirm. However, all the classification problems I'm familiar with classify compact manifolds without boundary. So, I think that the interesting question is: "Is there an easy reason why we restrict ourselves to the classification of compact manifolds w/o bdry? Does a classification ... compact manifolds w/o bdry?". I'm not sure if that's what the original poster meant; perhaps I should ask this as a separate question. – Ilya Grigoriev Jan 10 2010 at 5:55
The answer to that is probably no, as Henry Witon's answer seems to suggest. – Ilya Grigoriev Jan 10 2010 at 5:55
There's a sense in which classifying manifolds with boundary is easier than classifying them without. Think of the historical precedent -- Thurston proved geometrization for 3-manifolds with incompressible boundary before geometrization was proven in general. A manifold with boundary is stratified, so in some sense it's less than a proper manifold. More strictly, there's issues. In general 3-manifolds with boundary have a more complicated decomposition than closed 3-manifolds as there's the additional Bonahon "compression body" decomposition. – Ryan Budney Jan 11 2010 at 22:58
Similarly, 3-dimensional orbifolds had geometrization (provided the orbifold locus was non-empty) before Perelman. Stratifications give you something to work with, so in a sense, orbifolds are simpler than manifolds even if their definition appears more general. – Ryan Budney Jan 11 2010 at 23:00
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I just wanted to mention that the fact that the word problem is "hard" for groups doesn't mean that there can be no classification for manifolds of dimension $\geq$ 4.

The problem is that the word "hard" here means "there is no algorithm to solve it". There certainly could be (though I'd doubt it) a relatively short list of invariants for n-manifolds which classifies them up to diffeomorphism. The problem then becomes determining, for two specific manifolds $M$ and $N$ whether or not their lists of invariants are isomorphic/equal, which could be "hard".

For a somewhat trivial example: in dimension 2, compact manifolds are classified up to diffeomorphism by their first homology group. This is a true (and useful) statement regardless of how "easy" it is to determine if $H_1(M)$ and $H_1(N)$ are isomorphic.

Further, if one wants to, say, apply the Heawood conjecture for compact surfaces, and one only knows $H_1(M)$, one needs only demonstrate it's NOT isomorphic to 0 or $\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$, not isolate precisely which group it IS isomorphic to.

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If classification simply means "a short list of invariants for n-manifolds which [distinguishes] them up to diffeomorphism", then we already have such a list -- the diffeomorphism type of the manifold is the unique invariant you're after. The problem with this "classification" is we don't know how to compute it. So what does "classification" mean? I think for many people (most?) a classification requires computability starting from some kind of reasonable prescription for the type of objects you're interested in -- triangulations or surgury typically for manifolds. – Ryan Budney Nov 15 2009 at 22:33
I disagree. The Seifert - VanKampen theorem doesn't compute the fundamental group, it computes one presentation for it from the ingredient presentations and amalgamation maps. I think the more you stare at impossibly ugly presentations the more you'll be swayed by this point of view. I'd rephrase what you've been saying as something like this "given an oracle for the isomorphism problem between finitely presented groups, is there an algorithm to classify compact n-manifolds?" Off the top of my head I don't know the answer to that, but I suspect the answer is no. – Ryan Budney Nov 16 2009 at 5:01
Rereading your post I think you're blurring the line between "a presentation for a group" and "a group". The former is readily computable for manifolds via Seifert - VanKampen, the latter isn't. – Ryan Budney Nov 16 2009 at 5:19
There's a big gap between your two examples -- given two triangulated 4-manifolds, determining whether or not their intersection forms are isomorphic, there's a polynomial-time algorithm for that. The isomorphism problem for group presentations isn't so easy. I take a "classification" to mean: you have a collection C with an equivalence relation ~ on it, and you want a list of one representative from each object of C/~ and a decision procedure to determine, given an object c from C, which representative it is isomorphic to. What you call classification I'd call an equivalence. – Ryan Budney Nov 16 2009 at 18:33
Small follow-up. There is an algorithm to determine if simply-connected manifolds are homeomorphic. Nabutovsky and Weinberger Contemp. Math. 1999 245-250. – Ryan Budney Jan 13 2010 at 1:24

Complementing Ryan's answer: as shown by McMillan (Transactions AMS 102, 373-382) there is a continuum of pairwise nonhomeomorphic contractible open subsets of $\mathbf{R}^3$. So classifying noncompact manifolds in general is probably hopeless and some restrictions are necessary (e.g. one can consider only the interiors of compact manifolds with boundary that satisfy some conditions on the fundamental groups etc).

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This answer complement the answers of Henry and Algori. I think, it is worth to strees, that a classification of open manifolds does not follow from a classification of compact manifolds. Open surfaces were classified, but open 3-folds are not, their classification does not follow from the classification of compact ones at least at the present time. In particular, {\it prime decomposition}, or Kneser's theorem (http://en.wikipedia.org/wiki/Prime_decomposition_(3-manifold)) does not hold for non-compact 3-manfiolds. There is a constructiion due to Scott (http://plms.oxfordjournals.org/cgi/pdf_extract/s3-34/2/303) of an example of a simply connected 3-fold that can not be a connected sum of finite or infinite number of prime manifolds.

Open 3-manifold are actively studdied now. Let me give two citations that confirm further that our knowlage of compact 3-manfiolds is not sufficient for understending of non-compact ones.

1) Ricci flow on open 3-manifolds and positive scalar curvature. Laurent Bessi`eres, G´erard Besson and Sylvain Maillot. http://arxiv.org/PS_cache/arxiv/pdf/1001/1001.1458v1.pdf

Thanks to G. Perelman’s proof of W. Thurston’s Geometrisation Conjecture, the topological structure of compact 3-manifolds is now well understood in terms of the canonical geometric decomposition. The first step of this decomposition, which goes back to H. Kneser, consists in splitting such a manifold as a connected sum of prime 3-manifolds, i.e. 3-manifolds which are not nontrivial connected sums themselves. It has been known since early work of J. H. C.Whitehead [Whi35] that the topology of open 3-manifolds is much more complicated. Directly relevant to the present paper are counterexamples of P. Scott [ST89] and the third author [Mai08] which show that Kneser’s theorem fails to generalise to open manifolds, even if one allows infinite connected sums.

The refference for the article of Maillot is the followning.

2) Some open 3-manifolds and 3-orbifolds without locally finite canonical decompositions. http://arxiv.org/PS_cache/arxiv/pdf/0802/0802.1438v2.pdf

Here is the citation

Much of the theory of compact 3-manifolds relies on decompositions into canonical pieces, in particular the Kneser-Milnor prime decomposition [12, 16], and the Jaco-Shalen-Johannson characteristic splitting [10, 11]. These have led to important developments in group theory [22, 7, 9, 24], and form the background of W. Thurston’s geometrization conjecture, which has recently been proved by G. Perelman [19, 20, 21].

For open 3-manifolds, by contrast, there is not even a conjectural description of a general 3-manifold in terms of geometric ones. Such a description would be all the more useful that noncompact hyperbolic 3-manifolds are now increasingly well-understood, thanks in particular to the recent proofs of the ending lamination conjecture [17, 4] and the tameness conjecture [5, 1]. The goal of this paper is to present a series of examples which show that naive generalizations to open 3-manifolds of the canonical decomposition theorems of compact 3-manifold theory are false.

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 Sorry to comment on such an old post, but is the situation of non-compact 3-manifolds, with f.g. fundamental group, now easier to handle thanks to the proof of the ending lamination conjecture? It seems Minsky et al. have managed to come up with a "rough" classification of f.g. Kleinian groups, but how does this translate to the manifolds? – Steve D Aug 29 at 6:49

As noted in the comment of Ryan Budney to Richard Kent answer, the non-compact surfaces have been classified (in a very nice way, hence my need to share the details).

I have no access to Jstor right now, so I rely on my memory and, well, you may want to check all this.

Ian Richards theorem says that non-compact surfaces (without boundary) are classified by their orientablility, their genus (possibly infinite) and a triple of spaces, each one embedded in the preceding, that are:

1. the space of its ends,

2. the space of its ends with genus,

3. the space of its unorientable ends.

The space of ends is constructed by taking an increasing sequence of compact subset that cover a topological space $T$, and looking at the connected components of their complements. An end of $T$ is an infinite, decreasing sequence of such connected components. The point is that you can do that in a way that does not substancially depend on the sequence of compact you chose.

For example, $\mathbb{R}^n$ has only one end provided $n\ge2$ (look at the sequence of balls of integer radius and centered at some point), while $\mathbb{R}$ has two ends. The space of ends of a regular tree is a Cantor set.

An end is said to have genus if the connected components that define it all have genus (they never reduce to annuli). An end is said to be unorientable if the connected components that define it all are unorientable.

Now, consider the surface $S^n$ ($n=1,2$ or $3$ defined as the boundary of a tubular neighborhood of the usual embedding of the usual Cayley graph of $\mathbb{Z}^n$ into $\mathbb{R}^3$ (for $n=1$ you get a cylinder; for $n=2$ some sort of grid; for $n=3$ it is sometimes called a jungle gym). $S^2$ and $S^3$ are the surfaces described by Richard Kent in his third paragraph. These two surfaces have exactly one end which is orientable but has genus. Therefore they all are homeomorphic. This is a pretty incredible result in my opinion. The most simple presentation of this surface is called the Loch-Ness monster: it is constructed by adding to a plane a sequence of handles placed in a row.

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Is 'Lock-Ness' a pun (presumably on the name of someone called 'Lock') or is it a mis-spelling of 'Loch Ness'? – HW Jan 10 2012 at 12:50
@ HW: it was a mis-spelling, thanks for noticing. – Benoît Kloeckner Jan 11 2012 at 11:41
I think for $n=1$ you get a cylinder only when you embed the graph into $\mathbb{R}^3$ instead of $\mathbb{R}^n=\mathbb{R}$. Also, $S^2$ and $S^3$ should lie in $\mathbb{R}^n$. – Junyan Xu Jan 31 2012 at 15:09
@Junyan Xu: you're half right, in fact if you want surfaces from tubular neighborhoods, you need to restrict to $\mathbb{R}^3$. Now corrected, thanks for noticing. – Benoît Kloeckner Feb 1 2012 at 5:38

To illustrate both sides of all of the above comments (non-compact builds on compact, vs. non-compact is much harder than compact), I point out that ALL connected, separable, metric 2-manifolds (with or without boundary) meaning all connected, separable, metric spaces in which every point has an open neighborhood homeomorphic to the closed 2-disk have been classified. See Brown, Edward M.; Messer, Robert: The classification of two-dimensional manifolds. Trans. Amer. Math. Soc. 255 (1979), 377–402. MR0542887

The point is that the analysis is rather delicate and has some surprises, showing that the non-compact case does not follow easily from the compact. Given the examples in dimension 3 by McMillan mentioned by algori, one of the surprises is that it could be done at all.

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Bits and pieces are coming back to me. There are uncountably many open 3-manifolds (non-compact and no boundary) that embed in no compact 3-manifold (suggested by Kister and McMillan and proven by Haken). See reviews MR0144322 and MR0482767. This was generalized to n dimensions in the unpublished thesis of R. Sternfeld. See genealogy.math.ndsu.nodak.edu/id.php?id=9614 – Matt Brin Jan 12 2012 at 5:00