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Let $f:S^{n-1} \rightarrow S^n$ be a topological embedding and let $A_f$ and $B_f$ be the components of $S^n \setminus f(S^{n-1})$. If $\overline{A}_f$ and $\overline{B}_f$ are manifolds with boundary $f(S^{n-1})$, then the locally flat Schoenflies conjecture (proved by Mazur and Brown) says that $\overline{A}_f$ and $\overline{B}_f$ are both topological discs. The classical example of the Alexander horned sphere shows that the assumption that $\overline{A}_f$ and $\overline{B}_f$ are manifolds is necessary. Indeed, letting $\alpha:S^2 \rightarrow S^3$ be the Alexander horned sphere, it turns out that one of the components (say, $A_{\alpha}$) is not even simply connected. However, $\overline{B}_{\alpha}$ is a manifold. Using the Schoenflies conjecture, this implies that if we attach a collar neighborhood $\alpha(S^2) \times [0,1]$ to $\overline{A}_{\alpha}$, then we get a $3$-ball which deformation retracts to $\overline{A}_{\alpha}$. In particular, $\overline{A}_{\alpha}$ is contractible.

Question: Does there exist a topological embedding $f^{n-1} \rightarrow S^n$ (preferably with $n=3$) such that neither $\overline{A}_f$ nor $\overline{B}_f$ is contractible?

There definitely exist such embeddings where neither $\overline{A}_f$ nor $\overline{B}_f$ are manifolds. For example, Bing proved that the double of the Alexander horned ball $\overline{A}_{\alpha}$ along its "boundary" $\alpha(S^2)$ equals $S^3$. This means that the above trick will not work in general (but of course we already know that $\overline{A}_{\alpha}$ is contractible, so Bing's theorem does not answer our question).

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    $\begingroup$ Your argument works in general: en.wikipedia.org/wiki/Crumpled_cube I think this was known earlier though. $\endgroup$
    – Ian Agol
    Commented Apr 13, 2015 at 5:17
  • $\begingroup$ @IanAgol : The theorem of Lininger referred to in the wikipedia article is very cool! You should post an answer with that reference so I can accept it (though I guess it only gives the $3$-dimensional case; my suspicion is that no one will respond with a more general result). $\endgroup$
    – Sarah
    Commented Apr 13, 2015 at 5:44
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    $\begingroup$ Actually, by searching on mathscinet I found Daverman, Robert J. Every crumpled n-cube is a closed n-cell-complement. Michigan Math. J. 24 (1977), no. 2, 225–241. which proves the same result for dimensions at least $5$ and Daverman, Robert J. Each crumpled 4-cube is a closed 4-cell-complement. Topology Appl. 26 (1987), no. 2, 107–113. which proves it in dimension $4$. So it always holds. I'd be interested in an easier proof! $\endgroup$
    – Sarah
    Commented Apr 13, 2015 at 5:57
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    $\begingroup$ Ok, looks like you've found a complete answer to your question. You can post it and check the answer yourself. This could be useful to others searching for the same question who aren't familiar with the odd term "crumpled cube". $\endgroup$
    – Ian Agol
    Commented Apr 13, 2015 at 15:43
  • $\begingroup$ @IanAgol : I did (and I also found what I believe is the original reference for the contractibility, a paper of Bing). $\endgroup$
    – Sarah
    Commented Apr 14, 2015 at 0:20

1 Answer 1

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I want to thank Ian Agol for his comments which helped point me in the right direction. In particular, he told me that if $f:S^{n-1} \rightarrow S^n$ is a topological embedding, then the closures of the components of $S^n \setminus f(S^{n-1})$ are known as crumpled $n$-cubes.

It turns out that Bing originally proved that crumpled $n$-cubes are contractible. See Theorem 4 of his paper

Bing, R. H. Retractions onto spheres. Amer. Math. Monthly 71 1964 481–484.

which proves that a crumpled $n$-cube in $\mathbb{R}^n$ is a retract of $\mathbb{R}^n$.

It also turns out that the technique I described in the question for proving that crumpled $n$-cubes are deformation retracts of standard $n$-discs works in general: if $C$ is a crumpled $n$-cube and $M$ is obtained by gluing a standard $n$-disc to the "boundary" of $C$, then $M$ is homeomorphic to $S^n$. This was proved for $n=3$ in

Lininger, Lloyd L. Some results on crumpled cubes. Trans. Amer. Math. Soc. 118 1965 534–549.

for $n \geq 5$ in

Daverman, Robert J. Every crumpled n-cube is a closed n-cell-complement. Michigan Math. J. 24 (1977), no. 2, 225–241

and finally for $n=4$ in

Daverman, Robert J. Each crumpled 4-cube is a closed 4-cell-complement. Topology Appl. 26 (1987), no. 2, 107–113.

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