We know that there are compact manifolds with diffeomorphic interiors but their boundaries are not homeomorphic (see the question Manifolds with homeomorphic interiors).

My question is about a very special case:

Assume $D$ is a bounded open set in $R^n$ with smooth boundary. If $D$ is diffeomorphic to the unit ball, is $\bar D$, the closure of $D$, diffeomorphic to the closed unit ball?


2 Answers 2


The case $n=4$ is open as far as I know.

The case $n=3$ follows since $\mathbb R^3$ is irreducible, so it contains no fake 3-disk, i.e. $\bar D$ must be the standard disk.

The case $n=5$ is equivalent to the smooth $4$-dimensional Poincare conjecture (which is still open). Here is why:

  1. Any homotopy $4$-sphere embeds smoothly into $\mathbb R^5$ (Sketch: homology $4$-sphere bounds a contractible smooth manifold $C$ [Kervaire, "Smooth homology spheres and their fundamental groups", Theorem 3]. In our case $\partial C$ is simply-connected, so attaching a collar on the boundary one gets a contractible $5$-manifold that is simply-connected at infinity, and hence it is diffeomorphic to $\mathbb R^5$ by a result of Stallings).

  2. Any embedding of the standard $4$-sphere into $\mathbb R^5$ bounds a standard disk, see [Smale, "Differentiable and Combinatorial Structures on Manifolds", Corollary 1.3]. What Smale actually states is that any embedded $S^{n-1}$ in $\mathbb R^n$ bounds a standard disk unless $n=4$ or $7$. This was before he proved the h-cobodorsm theorem hence he excludes $7$.

Finally, as mentioned in comments if $n>5$, then $\bar D$ is diffeomorphic to the standard disk by the h-cobordism theorem (sketch: since by assumption $D$ is simply-connected at infinity, $\partial D$ is a homotopy sphere and $\bar D$ is a contractible smooth manifold, so removing a small ball in its interior results in h-cobordism between then standard sphere and the embedded one. Proving that this is an h-cobordism involves standard excision considerations in homology).

  • 2
    $\begingroup$ Thank you for your answer and reference. The case $n=3$ is a classical 3-dimensional topology fact, that doesn't require the Poincaré conjecture (that is, $S^3$ is irreducible), see for ins Hatcher, Notes on Basic 3-Manifold Topology, at the very beginning. $\endgroup$ Feb 23, 2014 at 15:15
  • $\begingroup$ @DanieleZuddas: Oh, indeed, I was not thinking clearly. I will edit. $\endgroup$ Feb 23, 2014 at 15:18
  • $\begingroup$ Can we relax the condition that do not require D us embedded in $R^n$? $\endgroup$
    – J. GE
    Feb 23, 2014 at 15:40
  • $\begingroup$ @J.GE at least, you should request that $D$ is embedded outside of a compact subset of it, and that no points of the critical subset overlap with its embedded complement. $\endgroup$ Feb 23, 2014 at 16:42
  • 1
    $\begingroup$ @J.GE: having an embedding is not important: compact smooth $n$-manifold whose interior is an open disk is diffeomorphic to $D^n$, if $n\neq 4,5$. $\endgroup$ Feb 23, 2014 at 20:29

First, your assumption imply that $\bar D$ is a compact smooth manifold with boundary a topological sphere (because is a simply connected homology sphere). So, $\bar D$ is a topological $n$-ball, by the generalized Schoenflies theorem. Now, this reduces to the smooth Schoenflies problem (still open in $\Bbb R^4$). Also, I don't know whether an exotic $(n-1)$-sphere can be smoothly embedded in $\Bbb R^n$.

  • 5
    $\begingroup$ Exotic $n$-spheres with $n>4$ cannot be smoothly embedded into $\mathbb R^{n+1}$ by the h-cobordism theorem (the sphere bounds a contractible smooth manifold, so removing a small ball in its interior results in h-cobordism between then standard sphere and the embedded one). I think the same result is true if $n=4$ but the proof must be harder, and I cannot remember it at the moment. $\endgroup$ Feb 23, 2014 at 4:48
  • $\begingroup$ In the last sentence by a "result" I meant that a homotopy $4$-sphere that embeds into $\mathbb R^5$ bounds the standard ball. $\endgroup$ Feb 23, 2014 at 4:57
  • $\begingroup$ @IgorBelegradek thank you for pointing out this... Actually the h-cobordism theorem doesn't hold in this dimension, I think. $\endgroup$ Feb 23, 2014 at 5:27
  • $\begingroup$ What I mean is that (if memory serves) a different argument takes care of smooth Schoenflies in $\mathbb R^5$. $\endgroup$ Feb 23, 2014 at 5:31
  • 2
    $\begingroup$ @Misha: by assumption $D$ is a disk, so it is simply-connected at infinity. $\endgroup$ Feb 23, 2014 at 5:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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