[Math] Compactification of Manifolds


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.


Best Answer

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 $M_n\subset int(M_{n+1})$.

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