Uniqueness of compactification of an end of a manifold

Question

Let $M$ be an $n$-dimensional manifold (smooth or topological). I call $\bar{M}$ a compactification of $M$ if it is an $n$-dimensional compact manifold with boundary $\partial \bar{M}$, an $(n-1)$-dimensional manifold, such that $M$ is the interior of $\bar{M}$. I understand that not every manifold $M$ has such a compactification. The obstructions have been discussed in some previous MO questions (22441, 34602). Essentially, no compactification can exist if the ends of the manifold are too "wild".

However, I am more interested in how two compactifications are related to each other, provided they exist. For simplicity, let me assume that $M$ has only one end and a compactification exists. If $\bar{M}_1$ and $\bar{M}_2$ are both compactifications, what can be said about the relationship between them and between $\partial\bar{M}_1$ and $\partial\bar{M}_2$? Does there exist some kind of "universal" compactification $\bar{M}_U$ from which both $\bar{M}_1$ and $\bar{M}_2$ could be obtained? If these questions are easier to answer in the topological category, then I would be quite happy with just that information.

Here's an example. Consider $M=\mathbb{R}^n$ ($n\ge 2$). It has one end. An obvious compactification is to consider it as the interior of a closed ball $\bar{M}_1=\bar{B}^n$, so the boundary is a sphere $\partial\bar{M}_1=S^{(n-1)}$. However, I can apply a homotopy to $\bar{B}^n$ which maps the boundary into itself and restricts to a homoemorphism on the interior. Composing this operation with the embedding of $M$ in $\bar{M}_1$ I get a different compactification $\bar{M}_2$. By construction, there is a map $\bar{M}_1 \to \bar{M}_2$, which is a homeomorphism between the interiors but only a homotopy between the boundaries. In particular, the homotopy could blow up a point on $\partial\bar{M}_1$ to a closed set with open interior. Then two curves that had the same end point on the boundary of $\bar{M}_1$ could be mapped to curves with distinct end points on the boundary of $\bar{M}_2$. Based on this example, I would naively guess that $\mathbb{R}^n$ can be compactified by adding an $S^{(n-1)}$ boundary and all other compactifications can be obtained by applying a homotopies to any one element of this class of compactifications. So this compactification could be considered universal. How far is this intuition from reality?

I'm aware of resources like L. Siebenmann's thesis and the book Ends of Complexes by Ranicki and Hughes. Unfortunately, I don't have enough algebraic and topological background to immediately see whether they contain the answer or how to dig it out if they do. So precise suggestions for where to look would also be helpful!

Answer 1

Suppose $\overline{M}_i$, $i=0,1$, are compact smooth manifolds with boundary whose interiors are diffeomorphic: let $\psi$ be such a diffeomorphism, and $M$ for either interior (identified via $\psi$). For both manifolds one can find a smooth collar $c_i : \partial \overline{M}_i \times [0, 1) \hookrightarrow \overline{M}_i$, as they are compact. By shrinking $c_0$ (and re-parametrising) we can suppose it has image inside that of $c_1$ (restricted to the interior, compared via $\psi$), and then we obtain an embedding $$\partial \overline{M}_0 \times [\tfrac{1}{2}, 1) \hookrightarrow \partial \overline{M}_1 \times (0, 1) \subset \overline{M}_1 \times [0, 1).$$ The complement of this embedding has two components, precisely one of which is compact and is a cobordism $W : \partial \overline{M}_0 \leadsto \partial \overline{M}_1$.

This is an $h$-cobordism, as each of the $\partial \overline{M}_i \times [0, 1)$ and $W$ are homotopy equivalent to the space $$\mathcal{E}(M) := \mathrm{holim}_{K \subset M \text{compact}} M \setminus K$$ as there are evident cofinal subdiagrams indexed by $K = M \setminus c_i(\partial \overline{M}_i \times [0, \epsilon))$ for either $i$, which are homotopically constant.

Thus, the two boundaries $\partial \overline{M}_i$ are $h$-cobordant (so diffeomorphic if they are simply-connected). If the boundaries are not simply-connected, then I imagine that this $h$-cobordism can be non-trivial, as in Benoît Kloeckner's discussion.

Answer 2

Let us consider smooth object, I am more familiar with them. First you have to precise the definitions. In particular, "$M$ is the interior of $\bar M$" is not clear; I guess you mean that $M$ is diffeomorphic to the interior of $\bar M$. But then, you should precise whether the compactification is the data of solely $\bar M$, or if it is the data of both $\bar M$ and a smooth embedding $M\to\bar M$. The notion of isomorphism is then to be precised (most naturally, existence of a diffeomorphism $\bar M_1\to\bar M_2$ in the first case; in the second case you have the choice between: existence of diffeomorphisms $\bar M_1\to \bar M_2$ and $M \to M$ making the obvious square-shaped diagram commutative, or existence of a diffeomorphism $\bar M_1\to\bar M_2$ making the obvious triangular-shaped diagram commutative). I'll assume we are in the first case.

Anyway, probably the most important reference to start investigating your question is Milnor's "Two complexes which are homeomorphic but combinatorially distinct", Ann. of Math. (2) 74 (1961), pp. 575--590, where it is shown that the manifolds with boundary $L_{7,1}\times \bar B^5$ and $L_{7,2}\times \bar B^5$ are not diffeomorphic but have diffeomorphic interiors. Such an example does not exist in dimension $3$ (Edwards, "Concentricity in $3$-manifolds", Trans.AMS 113 (1964), pp. 406--423).

Moreover, Barden, Mazur and Stallings have found a $h$-cobordism $(\bar W,M,M')$, where $M$ is diffeomorphic to $M'$ and the interior of $\bar W$ diffeomorphic to $M\times ]0,1[$, but $\bar W$ is not diffeomorphic to $M\times [0,1]$ (see Milnor's "Whitehead torsion" in the Bull. Amer. Math. Soc. 72 (1966) pp.358--426 if I believe my old notes). In particular, knowing the interior and boundary of a manifold is not sufficient to knowing the manifold itself.

Concerning the compactifications of $\mathbb{R}^n$, it is a consequence of the $h$-cobordism theorem that the only compact manifold with boundary having an interior diffeomorphic to $\mathbb{R}^n$, where $n\geqslant 6$, is the closed ball (up to diffeomorphism). For $n=5$, adding Freedman's proof of the topological $4$-dimensional Poincaré conjecture you get the same result up to homeomorphism. For $n=4$, the question is open as far as I know.