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Suppose $M$ is a compact smooth manifold with boundary, which is homeomorphic to the compact ball $\mathbb{B}^d\subset \mathbb{R}^d$. Must $M$ be diffeomorphic to $\mathbb{B}^d$ or are there exotic smooth structures?

I suspect that the answer is well known or might follow from a simple argument that experts (which I am certainly not) have up their sleeves.

A natural candidate in $d=4$ would be to take $M$ to be the compact unit-ball in an exotic $\mathbb{R}^4$, but I don't know how to check whether $M$ is exotic (or even whether it has a smooth boundary to be honest).

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    $\begingroup$ This is equivalent of smooth Poincare conjecture. $\endgroup$ Jul 27, 2020 at 15:29
  • $\begingroup$ Would you care to elaborate, @AnubhavMukherjee? Sure, the double of an exotic ball is a sphere, but can one say something about whether it is exotic or not? Somewhat related: I have recently learned about Mazur manifolds, which have standard $S^4$ as double. $\endgroup$
    – Jan Bohr
    Jul 27, 2020 at 15:50
  • $\begingroup$ If you have an exotic sphere, then if you take out a standard ball, then the remaining is an exotic sphere. It is not true that if one glue two copies of exotic mazur manifold then that will give rise an exotic sphere. mathoverflow.net/questions/357200/… . I think with assuming some string conjecture one can prove using Cerf's theorem that an exotic 4 ball implies existenc of exotic sphere. $\endgroup$ Jul 27, 2020 at 20:17
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    $\begingroup$ @AnubhavMukherjee What do you mean by "the remaining is an exotic sphere", since the remaining is either non-compact, or has boundary. Did you mean "the remaining is an exotic ball"? If so, I don't think this is true: I thought it's possible to glue two copies of standard $B^7$'s together via an exotic diffeo (i.e., an orientation preserving diffeo which is not isotopic to Id) to make exotic $S^7$. Maybe I'm misremembering? $\endgroup$ Jul 28, 2020 at 0:10
  • $\begingroup$ Sorry, it's typo and here you can't edit comments. In 4 dim, complement of a standard ball in a exotic sphere would be an exotic ball. Because you can always extend any boundary diffeomorphism of sphere into 4 ball (Cerf's theorem). Am I saying something wrong? $\endgroup$ Jul 28, 2020 at 2:17

2 Answers 2

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In dimension 4, one has the following equivalence:

The set $\mathcal B_4$ of compact smooth manifolds homeomorphic to $B^4$, considered up to oriented diffeomorphism, is in canonical bijection with the set $\mathcal S_4$ of compact smooth manifolds homeomorphic to $S^4$, considered up to oriented diffeomorphism.

Proof. We construct maps both ways, which will be clearly inverse.

$C: \mathcal B_4 \to \mathcal S_4$. Pick $B \in \mathcal B_4$. By the resolution of the 3D Poincare conjecture, there is some oriented diffeomorphism $\varphi: \partial B \to S^3$. One may then "cap off the boundary": define $$C(B) = B \cup_\varphi B^4,$$ defined as $B \sqcup B^4$, modulo identifying $\partial B \cong S^3$ via the diffeomorphism $\varphi$.

$C$ is well-defined by Cerf's theorem that $\pi_0 \text{Diff}^+(S^3) = 1$: all diffeomorphisms are the same up to isotopy. Isotoping $\varphi$ above does not change the diffeomorphism type; so $C$ is a set map.

Conversely one has $D: \mathcal S_4 \to \mathcal B_4$, with $D(S)$ given by deleting the interior of some oriented embedding $\iota: B^4 \to S$. Oriented embeddings of balls into a connected manifold are unique up to isotopy (Palais, but straightforward); this is true in all dimensions.

Clearly $D(C(B)) = B$ (delete the ball you glued) and $C(D(S)) = S$ (glue in the ball you deleted). Therefore $C,D$ are inverse bijections.


However one has the following for $n \geq 6$.

The set $\mathcal B_n$ is trivial.

Proof: Pick $B \in \mathcal B_n$. Delete a standard ball from its interior; then we are provided with a compact manifold $W$, with $\partial W = S^{n-1} \sqcup \partial B$, so that $W \cup_{S^{n-1}} B^n = B$.

$W$ is an h-cobordism (algebra). Therefore by the h-cobordism theorem there is a diffeomorphism $W \cong S^{n-1} \times [0,1]$, which sends $S^{n-1}$ to $S^{n-1} \times \{0\}$ by the identity map. Therefore $W \cup_{S^{n-1}} B^n \cong B^n$. Therefore $B \cong B^n$.


Of course there are no exotic n-balls n = 1,2,3. For n=5 I do not know. I think the answer is that $\partial: \mathcal B_5 \to \mathcal S_4$ is an injection (maybe bijection?) but I don't know off the top of my head. Some googling or looking on Math Overflow should help. Obviously "exotic balls" will get you nowhere but some buzzwords like diffeomorphism might.

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  • $\begingroup$ In your last paragraph, should $\mathcal S_4$ be $\mathcal S_5$? If so, it may be worth noting that Smale showed $\mathcal S_5$ has only a single element in it. $\endgroup$ Jul 28, 2020 at 0:06
  • $\begingroup$ Thank you very much! Do you have a reference for the fact that there are 'of course no exotic $3$-balls'? I have read of Moise's theorem that states the analogue for closed three-folds, but your answer makes me very cautious to jump to the boundary case right away (after all there is an exotic $S^7$, but you claim there is no exotic $B^7$.) $\endgroup$
    – Jan Bohr
    Jul 28, 2020 at 10:00
  • $\begingroup$ For n=5, the problem is equivalent to the smooth Poincare conjecture in dimension 4, see below. $\endgroup$ Aug 3, 2020 at 20:23
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This is an addendum regarding dimensions 4 and 5: While the (smooth) h-cobordism argument for 5-dimensional cobordisms no longer works, Milnor proves in his "h-cobordism" book (pages 110-111, Proposition C) that if $M$ is a 5-dimensional smooth compact manifold homeomorphic to $D^5$, whose boundary is diffeomorphic to $S^4$, then $M$ is diffeomorphic to $D^5$. According to theorems of Milnor-Kervaire and Wall, if $M^4$ is an exotic $S^4$, then it bounds a smooth contractible 5-manifold $W$. The manifold $W$ then will have to be homeomorphic to $D^5$ (by the topological h-cobordism theorem for 5-dimensional h-cobordisms, due to Freedman). Thus, the existence of an exotic $D^5$ is equivalent to the existence of an exotic $S^4$.

In dimension 4 one would need a bit more: If $W$ is a smooth 4-manifold homeomorphic to $D^4$, then its double $DW$ is homeomorphic to $S^4$. If $DW$ is diffeomorphic to $S^4$, assuming, in addition, smooth Schoenflies conjecture in dimension 4, one would obtain that $W$ is diffeomorphic to $D^4$. If smooth Schoenflies conjecture fails, then you would obtain a smooth submanifold inside $S^4$ which is homeomorphic but not diffeomorphic to $D^4$. (The same deal if the smooth Poincare conjecture fails in dimension 4.) So, "in essence," in dimension 4, the problem is equivalent to Poincare+Schoenflies.

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  • $\begingroup$ There is some friction between your second paragraph and the claim of @user812049 that in dimension four the problem is equivalent to the smooth Poincaré conjecture (without appealing to Schoenflies). Do you see a problem with their reasoning? (Or do I misunderstand what you have said?) $\endgroup$
    – Jan Bohr
    Aug 4, 2020 at 8:08
  • $\begingroup$ Oh, you are right, I was not thinking clearly. The point s that 4D smooth Poincare conjecture implies Schoenflies. $\endgroup$ Aug 7, 2020 at 18:48

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