7
$\begingroup$

This question arose out of this stack exchange post. I am wirting a thesis about the $s$-cobordism theorem and Siebenmann's work about end obstructions. Combined they give a quick proof of the uniqueness of the smooth structure for $\mathbb{R}^n$, $n \geq 6$. Roughly Siebenmann's theorem says that for $n \geq 6$ a contractible $n$-manifold $M$ that is simply connected at infinity embeds (smoothly) as the interior of a compact manifold. Since this compact manifold is contractible, by the $s$-cobordism theorem, it is diffeomorphic to the standard $n$-disk $D^n$ (see Minor's Lectures on the $h$-cobordism theorem for example). It follows that $M = \text{int } D^n$ is diffeomorphic to $\mathbb{R}^n$.

The problem is that the case $n = 5$ is not covered. I am aware of Stallings beautifully written On the Piecewise-Linear Structure of Euclidean Space but I am searching for a way to deal with the $n = 5$ case via Siebenmann's end theorem and the proper $s$-cobordism theorem (see link to the mse question). This leads me to the following question, which is interesting in itself

Given a smooth properly embedded codimension 1 submanifold $S \subset \mathbb{R}^{n+1}$, is there a self diffeomorphism $\mathbb{R}^{n+1} \rightarrow \mathbb{R}^{n+1}$ which carries $S$ into a one dimensional bounded region $\mathbb{R}^n \times (-1, 1)$ ?

Now if $M$ is a manifold that is homeomorphic to $\mathbb{R}^5$, the product $M \times \mathbb{R}$ is homeomorphic to $\mathbb{R}^6$, and hence also diffeomorphic. Granted the existence of the diffeomorphism in my question, we could find a diffeomorphism $f : M \times \mathbb{R} \rightarrow \mathbb{R}^6$ that maps $M \times 0$ into $\mathbb{R}^5 \times (-1, 1)$. This would produce a proper $h$-cobordism between $M$ and $\mathbb{R}^5$ by taking the region between $f(M \times 0)$ and $\mathbb{R}^5 \times 1$ in $\mathbb{R}^5 \times \mathbb{R}$. Since $M$ is simply connected, the proper $s$-cobordism theorem applies and shows that $M$ and $\mathbb{R}^5$ are actually diffeomorphic.

New contributor
Lukas is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
  • $\begingroup$ I think you need some assumptions that $S$ is homeomorphic to $\mathbb{R}^n$ or has trivial homology or something like that. There are surfaces in $\mathbb{R}^3$ that separate $\mathbb{R}^3$ into two connected components that can be mapped to each other with a finite translation. I'm trying to think what to google to get a picture. $\endgroup$ – quarague Jul 12 at 9:15
  • $\begingroup$ Essentially take a checkerboard coloring of unit cubes in $\mathbb{R}^3$ and then build a surface that separates the black and white components. One can approximate this surface with a smooth $2$-dimensional manifold. $\endgroup$ – quarague Jul 12 at 9:19
  • $\begingroup$ @quarague Is this supposed to be a counterexample? I don't see how this denies the existence of my desired diffeomorphism. $\endgroup$ – Lukas Jul 12 at 9:30
  • $\begingroup$ I thought something like this would be a counterexample, yes. A unit ball centered anywhere in $\mathbb{R}^3$ contains parts of this surface, so I don't see how a diffeomorphism of $\mathbb{R}^3$ should shrink that to a strip. If you can show that your conjecture holds for surfaces like this I would be much more inclined to believe it. $\endgroup$ – quarague Jul 12 at 10:40
  • $\begingroup$ This en.wikipedia.org/wiki/Regular_skew_apeirohedron was what I was thinking of. You can use a smoothed version of this and I don't see how you squash this to strip in $\mathbb{R}^3$. $\endgroup$ – quarague Jul 12 at 10:45
5
$\begingroup$

For $n\geq 5$, such a diffeomorphism exists iff the complement of $S$ has an unbounded connected component. Indeed, the existence of such a diffeomorphism implies there is an unbounded component of the complement of $S$. Conversely, suppose that the complement of $S$ has an unbounded component. Choose a proper arc $\gamma:[0,\infty)\to\mathbb R^n$ in the complement of $S$ (i.e. going to infinity). Since $n\geq 5$, the sweepout of a generic proper homotopy from $\gamma$ to a standard coordinate ray is properly embedded, hence is an isotopy (this might still work for $n=4$, but it's definitely a bit more delicate, and I haven't thought about it carefully). Using this isotopy on $S$, we reduce to the case where $S$ is disjoint from a standard coordinate ray, which is easy.

A sufficient (but not necessary) condition for the complement of $S$ to have an unbounded component is that $S$ have finitely many components (this implies the complement of $S$ also has finitely many components, hence one must be unbounded).

$\endgroup$
  • $\begingroup$ How do you extend the isotopy to a diffeotopy of $\mathbb{R}^n$ as $[0, \infty)$ is not compact? $\endgroup$ – Lukas Jul 12 at 14:31
  • 1
    $\begingroup$ I think this is not any harder than the compact case; all one needs is that $\gamma:[0,\infty)\times[0,1]\to\mathbb R^n$ is a proper embedding. $\endgroup$ – John Pardon Jul 13 at 9:29
2
$\begingroup$

This may be overkill but you could argue as follows for the special case of $M = \mathbb{R}^n$. A theorem of Kirby (On the set of non-locally flat points of a submanifold of codimension one. Annals, 88, (1968), 281-290) says that an embedding of an $n$-manifold, $n\geq 3$ can't fail to be locally flat at just one point. So take the one-point compactification $M^+$ of $M$, which is a sphere embedded in $S^{n+1} =$ one-point compactification of $\mathbb{R}^{n+1}$. By Kirby's theorem, it is locally flat.

Using local flatness at $p$, choose two balls on either side of $p$ joined by an arc (say a fiber of the normal bundle) that meets $M^+$ transversally in one point. The region between the two balls is a product $S^n \times (0,1)$, and then the region you want is obtained by removing the arc. (To get the picture you're asking about, you'd probably want to remove the inner balls of nested pairs of balls...)

A caveat is that this may be a circular argument from your point of view; Kirby uses a certain amount of machinery that was in the air in the mid-sixties. But I suspect that this machinery doesn't include the result you're trying to prove, since the assumptions on dimension don't match the ones for engulfing or h-cobordism techniques.

$\endgroup$
2
$\begingroup$

As you probably know, Siebenmann's thesis can be extended to dimension 5 (subject to a fundamenal group condition) by using the 4-dimensional work of Freedman. Unfortunately, one loses smoothness in the final conclusion, so your hope for a diffeomorphism would fall short.

An alternative to using Siebenmann's thesis might be to apply his Open Collar Theorem from the later paper "On detecting open collars". The beauty of that approach is that it is valid in the smooth category down to and including dimension 5.

New contributor
CraigG is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$

Your Answer

Lukas is a new contributor. Be nice, and check out our Code of Conduct.

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.