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Let $[0,1]^{k}$ be the $k$-times Cartesian product of the closed interval $[0,1]$. Let $\Omega$ be a contractible, nonempty, and closed region in $[0,1]^k$. Here a region in $[0,1]^k$ refers to a manifold of dimension $k$ that are contained in $[0,1]^k$ as a subset. And a closed region means that the manifold is closed as a topological space.

Question. Whether could it be proved that $\Omega$ is homeomorphic to $[0,1]^k$? Or could we add more conditions on $\Omega$ to guarantee the conclusion that $\Omega$ is homeomorphic to $[0,1]^k$?

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In Knotted contractible $4$-manifolds in $S^4$, Lickorish describes embeddings of contractible $4$-manifolds into $S^4$. His paper is concerned with producing these embeddings such that the fundamental group of the complement is an arbitrary balanced group $G$, but this does not matter for your question.

By Lickorish, we have a contractible $4$-manifold $\Omega$ that embeds into $S^4$. Also $\Omega$ is not homeomorphic to the $4$-ball $B^4$ (and thus not homeomorphic to $[0,1]^4$). So $\Omega$ embeds into $S^4\setminus\{\text{point}\}$, which is homeomorphic to $\mathbb{R}^4$, and $\mathbb{R}^4$ is homeomorphic to $(0,1)^4$. Therefore $\Omega$ also embeds into $[0,1]^4$.

The answer to your first question is no, we cannot conclude that $\Omega$ is homeomorphic to $[0,1]^k$. I am not sure about what additional conditions would have to be assumed to guarantee that $\Omega$ is homeomorphic to $[0,1]^k$.

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