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Let $B^n$ denote the unit ball in $\mathbb{R}^n$ (wrt the standard euclidean metric) and $\bar{B}^n$ denote the unit closed ball. Suppose that $\Sigma$ is a a smooth embedded hypersurface with boundary in $\bar{B}^4$ which is topologically $\bar{B}^3$ and so that $\Sigma$ meets $\partial \bar{B}^4$ transversely and $\partial \Sigma\subset \partial \bar{B}^4$. It should follow from known results that $B^4\backslash \Sigma=U_+\cup U_-$ where $U_\pm$ are open domains homeomorphic to $B^4$. Is it also known whether they are diffeomorphic, or is this equivalent to the smooth 4D Schoenflies problem (it is clearly weaker)?

I'm not a topologist so I apologize if this is a silly question.

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I think this is equivalent to the smooth Schoenflies conjecture; the executive summary is that this is true because smooth balls (in any dimension) are isotopic to standard ones.

Here are some details, starting with some preliminary remarks. Your $\bar{B}^3$ is diffeomorphic to a closed 3-ball, so I'll take that as given. Its boundary is a 2-sphere in $\partial \bar{B}^4$, and so is isotopic to the standard 2-sphere in $S^3$ by the 3D Schoenflies theorem (Alexander's theorem). Also, the standard result about balls in a manifold being standard applies to a codimension $0$ ball $\bar{B}^n$ embedded in an $n$-manifold in a non-proper (or neat) way. In other words the boundary of $\bar{B}^n$ is divided into $\bar{B}^{n-1}_\pm$ with $\bar{B}^n \cap \partial M = \bar{B}^{n-1}_-$ lying in the boundary of $M$, and $\bar{B}^{n-1}_+$ properly embedded. You really should worry a bit about corners here, but I'm going to ignore this (and in what follows).

Suppose first that we have $\bar{B}^4$ as described and that the Schoenflies conjecture holds. By the remarks above, do a preliminary isotopy so the boundary is standard. Now add a 4-ball $B_0^4$, divided into balls $U^0_\pm$, where $\bar{U}^0_+ \cap \bar{U}^0_-$ is a standard $\bar{B}_0^3$. Then $\bar{B}_0^3 \cup \bar{B}^3$ is an embedded sphere, and assuming the Schoenflies theorem, bounds balls on either side. Each of these balls is of the form $U^0_\pm \cup U_\pm$. But since we know that $U^0_\pm$ is a ball, its embedding is standard, so it follows that $U_\pm$ is a ball as well.

The converse is also straightforward. Given $S^3 \subset S^4$, dividing $S^4$ into regions $A_+ \cup A_-$, remove a small ball $B^4_0$ centered at a point of $S^3$. The complement of $B^4_0$ is a ball divided into regions $U_\pm$ as in your description, and by hypothesis each of $U_\pm$ is a ball. It follows that $A_\pm$ are balls as well.

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  • $\begingroup$ Thanks for your example of the converse -- it was what I was missing. $\endgroup$ – foliations Jun 23 '15 at 21:08
  • $\begingroup$ Funny--I thought that was the easier direction and almost didn't mention it! $\endgroup$ – Danny Ruberman Jun 24 '15 at 0:38
  • $\begingroup$ Do you mind specifying what you mean by the "standard result about balls in a manifold being standard"? $\endgroup$ – Kyle Hayden Jul 16 '15 at 18:45
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    $\begingroup$ It is a basic theorem of differential topology that if $M^n$ is a connected manifold, then if $k <n$, any two embeddings of the closed ball $\bar{B}^k$ in $M$ are isotopic. If $k=n$, and $M$ is orientable then the same is true if the embeddings are both orientation preserving or both orientation reversing. If $M$ is non-orientable, then it doesn't matter about the orientations. I think this result is due to Palais; you can find a proof in Hirsch's book on Differential topology. See also en.wikipedia.org/wiki/Disc_theorem. $\endgroup$ – Danny Ruberman Jul 16 '15 at 19:40

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