It is known that for any locally compact Hausdorff space X, we can define a Hausdorff one-point compactification containing X. In the case of the (differentiable) manifold $\mathbb R^n$ this one-point compactification turns out to be (homeomorphic to) $\mathbb S^n$, which is again a (differentiable) manifold.

This leads to the following question:

What does the picture look like in the general case for compactifications of an arbitrary manifold $M$?

Although the one-point compactification of $M$ is not a manifold in general (e.g. $\mathbb R^n - 0$); is it possible to view every manifold as an open (dense?) subset of a compact manifold by taking some other kind of compactification? In the differentiable case? In the $C^0$-case?

I had thought along the following lines at first: By the Whitney embedding theorem, every manifold $M$ can be thought of as a closed submanifold of $\mathbb R^n$ for some $n$. And by embedding $\mathbb R^n$ into $\mathbb S^n$, we can think of $M$ as an embedded submanifold of a compact manifold. But I guess taking the closure of $M$ in $\mathbb S^n$ will not in general leave us with a manifold anymore (?), so this does not answer my question...

Has this been looked into?

Thanks for any thoughts.



1 Answer 1


Just to take this question from the "unanswered" list. (It was actually answered in comments.)

(1). The simplest example of a manifold which is not homeomorphic to an open subset of a compact manifold is an infinite disjoint union of circles. [Edit: I stand corrected: the simplest example is ${\mathbb N}$ with discrete topology. I was only thinking about manifolds of positive dimension.]

(2). If you want a connected example, then it first appears in dimension 2: If a connected surface $S$ has infinite genus then it is not homeomorphic to an open subset of a compact surface. This is intuitively clear, but I will nevertheless give a proof which works in all dimensions.

Since $S$ has infinite genus, the image of the natural map $$ \phi: H^1_c(S; {\mathbb R})\to H^1(S; {\mathbb R}) $$ has infinite rank (each "handle" in $S$ contributes a 2-dimensional subspace). Suppose that $S\to T$ is an open embedding of $S$ to a compact surface $T$. Then we have the commutative diagram $$ \begin{array}{ccc} H^1_c(S; {\mathbb R}) & \stackrel{\phi}{\to} & H^1(S; {\mathbb R})\\ \psi\downarrow & ~ & \eta\uparrow \\ H^1_c(T; {\mathbb R}) & \stackrel{\cong}{\to} & H^1(T; {\mathbb R}) \end{array} $$
(Note that the induced maps of ordinary and of compactly supported cohomology groups go in opposite directions, this is what used in the proof.) Since $H^1(T, {\mathbb R})$ is finite-dimensional, the image of $\eta\circ \psi$ is also finite-dimensional, which is a contradiction.

Edit. There are less trivial examples in dimension 3. Haken proved that a certain open contractible 3-manifold does not embed in any compact 3-manifold:

W. Haken, Some results on surfaces in 3-manifolds, Studies in Modern Topology, Math. Assoc. Amer. (distributed by Prentice-Hall, Englewood Cliffs, N. J.), 1968, 39-98.

This result was generalized in

R. Messer, A. Wright, Embedding open 3-manifolds in compact 3-manifolds. Pacific J. Math. 82 (1979), no. 1, 163–177.

who found necessary and sufficient conditions for embedding in compact 3-manifolds of open 3-manifolds of the form $$ \bigcup_{n \in {\mathbb N}} M_n, $$ where for all $n$, $M_n$ is a compact submanifold with toral boundary and $M_n\subset int(M_{n+1})$.

  • $\begingroup$ For (1), I'm pretty sure that the disjoint union of infinitely many singletons is "simpler" (for any notion of "simple"). $\endgroup$ Commented Oct 22, 2018 at 9:16

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