A wild embedding of $\mathbb{S}^1$ into $\mathbb{R}^3$

Question

Can one construct an embedding of $\mathbb{S}^1$ into $\mathbb{R}^3$ so that every orthogonal projection onto a two dimensional plane is a unit disc?

It is easy to construct an embedding of $\mathbb{S}^1$ into $\mathbb{R}^3$ so that one orthogonal projection is a unit disc: take a Peano-type curve $\gamma=(\gamma_1,\gamma_2):[0,1]\to\mathbb{R}^2$ that fills the unit disc and define $\Gamma(t)=(\gamma_1(t),\gamma_2(t),t):[0,1]\to\mathbb{R}^3$. Clearly $\Gamma$ is one-to-one so it is an embedding and its projection onto the first two coordinates fills the disc. It remains now to "glue" the ends to make it an embedding of $\mathbb{S}^1$.

Anton Petrunin · Accepted Answer · 2018-12-04T23:41:35.770

14

No.

Assume it is possible, that is, there is an embedding $f\colon\mathbb{S}^1\to \mathbb R^3$ such that any projection of $f(\mathbb{S}^1)$ is a unit disc.

By this answer, the convex hull of the image $f(\mathbb{S}^1)$ is a unit ball. Further note that every extreme point lies in the image; that is, $f(\mathbb{S}^1)\supset\mathbb{S}^2$ --- a contradiction.

P.S. There are embeddings $f\colon\mathbb{S}^1\to \mathbb R^3$ such that any projection of $f(\mathbb{S}^1)$ contains a unit disc. Moreover one can assume that the convex hull $W=\mathop{\rm Conv}f(\mathbb{S}^1)$ is arbitrary close to the unit ball. (Typically, the set of extreme points of $W$ is a Cantor set.) To construct $f$, modify a space filling curve by making it injective, but still intersecting all the lines pass thru the unit ball.

edited Dec 04 '18 at 23:41

answered Dec 04 '18 at 04:57

Anton Petrunin

43,739

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Add the link to this question https://mathoverflow.net/questions/39127/is-the-sphere-the-only-surface-all-of-whose-projections-are-circles-or-can-we – Taras Banakh Dec 04 '18 at 08:08
@TarasBanakh Thank you for the link. It greatly complements Anton's answer. – Piotr Hajlasz Dec 04 '18 at 12:59
@PiotrHajlasz You are welcome. – Taras Banakh Dec 04 '18 at 13:25
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This answer is (was) completely impossible to understand if one does not click the link "is a unit ball", I spent 5 minutes puzzled before clicking... I rephrased to emphasize that this is the main point, not just "note that". – YCor Dec 04 '18 at 21:14
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What if we only ask for every projection to contain a disk? i.e. have nonempty interior? – Wojowu Dec 04 '18 at 21:29
@Wojowu see the postscript. – Anton Petrunin Dec 06 '18 at 21:40

A wild embedding of $\mathbb{S}^1$ into $\mathbb{R}^3$

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