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Apparently, there is the following fact:

The set of homeomorphism classes of connected manifolds has the same cardinality $c$ as that of $\mathbb R$.

I find it to be interesting; but would be happier to see a proof, and would be grateful for a reference somewhere.

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Already in dimension 3 there are continuously many non-homeomorphic manifolds (in fact even contractible open subsets of $\mathbf{R}^3$, see my answer here mathoverflow.net/questions/4155/…). –  algori May 17 '10 at 2:05

3 Answers 3

up vote 12 down vote accepted

Upper bound (assuming the manifolds are second countable): every manifold admits a complete metric, and the "set" of isometry classes of complete separable metric spaces is of cardinality continuum. Indeed, every such a space is a completion of a countable metric space, and there is only continuum of metrics on a countable set.

Lower bound: there is a continuum of non-isomorphic fundamental groups of manifolds. E.g. any countable sum of cyclic groups is a first homology group of a connected manifold: just take a connected sum of an appropriate countable collection of lens spaces.

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To be more explicit in the second part: You can encode an infinite binary string as a subset $S$ of the prime numbers. Then you can take a connected sum of the lens spaces $L(p,1)$ for each $p \in S$. You can then use the fundamental group or first homology as an invariant to recover $S$, or even Milnor's theorem that 3-manifolds uniquely factor with respect to connected sum. Technically speaking Milnor's theorem is in the closed category, but it applies to this case. I think that there are only countably many compact manifolds. –  Greg Kuperberg May 16 '10 at 22:20
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Greg, your last statement is correct. Any compact manifold can be triangulated with finitely many simplexes, and there are only countably many ways to glue these guys together. –  Kevin Ventullo May 17 '10 at 1:54
    
Kevin, are you talking about the smooth case? Aren't there non-triangulable topological manifolds? –  Sergei Ivanov May 17 '10 at 8:40

There are countably many compact topological manifolds, as was shown by Cheeger-Kister in "Counting topological manifolds", Topology 9 1970 149--151. The proof depends on results of Edwards-Kirby of deformation of homeomorphisms. A sketch of the proof is here.

Incidentally, a good discussion of counting non-metrizable manifolds is in Gauld's paper, section 3.

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I am worried that this may not be true in TOP (the topological category). Let us work in PL (piece-wise linear) instead. Every PL manifold can be expressed as a locally finite simplical complex, with at most a countable number of vertices. Thus there are at most $2^N$ of these. (Here $N$ is the natural numbers.) I think I will leave the lower bound as an exercise -- I have a way of doing this by encoding binary sequences using generalized Heegaard splittings of three-manifolds, but that is a hack. I am sure that there is a more beautiful way to give the lower bound just using non-compact surfaces.

I looked, but could not find a reference.

EDIT: Just in case the above is too brief here is an "easier" exercise. The number of simple locally finite graphs on at most a countable number of vertices is again $2^N$. [Hint: consider the adjacency matrix.] [In fact, the locally finite hypothesis is not necessary in this case.]

EDIT2: After thinking a bit more about my three-manifold examples, I realized the statement can be simplified a bit. You can embedd the set of binary sequences into the set of homeomorphism classes of submanifolds of $S^3$ as follows: Choose your two favorite distinct hyperbolic links in $S^3$, each with two components. Lets call these links $L_0$ and $L_1$. Given a binary sequence $s \colon N \to \{0,1\}$ form a three manifold $M_s$ by gluing copies of $L_0$ and $L_1$ as instructed by $s$. Note that $M_s$ embeds in the three-sphere; the complement looks a bit like Antoine's necklace. Finally, $M_s$ determines $s$ by the uniqueness of the JSJ decomposition. As a remark - it is also possible to do this with hyperbolic manifolds (thus having trivial JSJ decomposition) again embedding in the three-sphere.

Ah - following the link that algori provides reminds me that there are (up to homeomorphism) uncountably many Whitehead manifolds - open, contractable submanifolds of $S^3$. The point of all of these examples is that it is "easy" to encode information in the end of a three-manifold. I'll leave my examples here, as they are a useful warm-up to understanding the Whitehead manifolds.

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