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Let $F$ be a smooth embedding of the unit circle $\mathbb{S}^{1}$ in $\mathbb{R}^{3}$ such that the image $F(\mathbb{S}^{1})$ is the unknot.

Does there exist a developable surface having $F(\mathbb{S}^{1})$ as boundary?

More formally, can one always find an isometric embedding $G \,\colon V \to \mathbb{R}^{3}$ of some closed subset $V \subset \mathbb{R}^{2}$ such that $G(\partial V) = F(\mathbb{S}^{1})$?

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    Related questions https://math.stackexchange.com/q/634415/242708 and https://math.stackexchange.com/q/2291752/242708 –  May 23 '17 at 10:16
  • According to the answers to the first related question, the behavior of isometric embeddings is very sensitive to the degree of smoothness. Since you are interested in developable surfaces, I assume you want $G$ to be at least $C^2$? –  May 24 '17 at 05:16

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