# Compactification of a manifold

This is just a curiosity and the question is really foggy. I'm wondering if there can exist a notion of "minimal smooth compactification" (when I say minimal I think something like adding a finite number of points or at least cells of dimension less than than the dimension of manifold) for a smooth non-compact manifold, in this sense: if the one point compactification of the manifold is smooth and the embedding is smooth, we are done; but what if the one point compactification is singular? Can I embed the manifold in a "minimal" compact manifold of the same dimension?

• Just a remark, it is interesting to look at $\mathbb{T}^2$ times $(0,1)$ which has two boundary components'', but apparently the "minimal" way to embed it in a compact manifold is to include it in $\mathbb{T}^2 \times S^1$ (the other possible compactifications are "singular"). How can we define minimal though? – rpotrie Aug 5 '10 at 9:54
• If you attach finitely many cells to a CW complex with finite dimensional homology groups, then the resulting CW complex still has finite dimensional homology groups. Thus, if you start with a manifold with a non-finitely generated homology group, you cannot "complete" it with finitely many cells. Thus, the complement of a closed set of points in the plane with infinitely many components will do. For instance, you can remove the Cantor set in the $x$-axis from the plane. – damiano Aug 5 '10 at 10:01
• Maybe the question can be changed to: Is it possible to embed any manifold in a compact manifold in order that the image is open and dense? This seems quite "minimal" to me. – rpotrie Aug 5 '10 at 10:09
• You should post that as an answer! – rpotrie Aug 5 '10 at 10:14
• On a slightly different setting, there is a characterization of manifolds that are the interior of a manifold with boundary; This was investigated by Larry Siebenmann and browsing its web page or MathSciNet references should help. – Benoît Kloeckner Aug 5 '10 at 18:03

A "surface of infinite genus" $S$ is an example of a manifold that is not an open subset of a compact manifold.
The reason $S$ cannot be embedded in a compact manifold is straightforward: we can find simple closed curves $a_1 , b_1 , \ldots , a_n , b_n , \ldots$ on $S$ such that, for each positive integer $g$, the curves $a_1 , b_1 , \ldots , a_g , b_g$ form a standard basis for a surface of genus $g$. Thus, considering the product in homology of the classes of these curves, we deduce that they are independent. If $S$ were an open subset of a compact manifold $M$, the same argument would imply that the images of the curves constructed above would also be independent in the homology of $M$. This contradicts the fact that the homology of the compact manifold $M$ is finite dimensional.
Observe that this example is not particularly different from the example of the complement in the complex plane of the integers. Indeed, $S$ can be realized as the double cover of $\mathbb{C}$ branched along the integers.
• Yes, but the point made in the comments about infinitely-generated homology is already in the second sentence, really. The complement o the points (n,0) where n is an integer $can$ be compactified by adding one point for each n, then compactified in the 2-sphere. Do yo want that? – Charles Matthews Aug 5 '10 at 10:19