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All smooth simple closed curves in $\mathbb R^3$ (knotted or not) can be isometrically embedded into $\mathbb R^2$ as a circle of equal arclength.

The situation for knotted spheres seems more complicated because metric surfaces are more rigid. Here I am unsure that if given a smooth knotted sphere in $\mathbb R^4$ if there is a way to embed into $\mathbb R^3$ while preserving its metric structure. So my question is:

Does there exist a nontrivial smooth knotted sphere in $\mathbb R^4$ which can be smoothly isometrically embedded into $\mathbb R^3$? And if so, are there any concrete examples?

In all of the above, the Riemannian metric structures are assumed to be inherited from the ambient space.

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    $\begingroup$ How can a knot be embedded into the plane?? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jul 5, 2020 at 5:27
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    $\begingroup$ There's a slight inconsistency in terminology, but I believe the author at that point in the question is thinking of "circle" as being a synonym for "knot". $\endgroup$
    – Ryan Budney
    Commented Jul 5, 2020 at 6:01
  • $\begingroup$ Can you clarify the metrics you're using in the isometries, what exactly you mean by a round circle, and be specific about what the analogous situation is when you move up in dimension? $\endgroup$
    – guest
    Commented Jul 5, 2020 at 7:39
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    $\begingroup$ Thank you for mentioning it can be done C^1 (via the Nash-Kuiper theorem I presume). However, I am mainly interested in maps with more regularity. Smooth would be ideal, but C^2 would be interesting too. $\endgroup$
    – Christian Bueno
    Commented Jul 7, 2020 at 23:37
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    $\begingroup$ Maybe you are aware of this, but this paper gives an example of a metric on $S^2$ which does not admit a $C^2$ isometric embedding into $\mathbb R^3$: projecteuclid.org/download/pdf_1/euclid.jdg/1214429999. $\endgroup$
    – Justthisguy
    Commented Jul 8, 2020 at 3:51

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A spun knot will give a 2-sphere embedded in $\mathbb{R}^4$ whose intrinsic metric embeds into $\mathbb{R}^3$ isometrically as a surface of revolution.

Take a tangle $T$ in $\mathbb{R}^3_+$ with two ends on $\mathbb{R}^2$, that is an embedding $e: ([0,1],\{ 0, 1\}) \to (\mathbb{R}^3_+,\mathbb{R}^2)$. We may assume that $e$ is an isometry. Spin $T$ about $\mathbb{R}^2$ in $\mathbb{R}^4$ to get a surface $S$.

spun knot If $T$ is knotted then $S$ will be knotted. Then there exists an isometric embedding $e’:([0,1],\{0,1\})\to (\mathbb{R}^2_+,\mathbb{R})$ so that $d(e’(t),\mathbb{R})=d(e(t),\mathbb{R}^2)$ for all $t,\in [0,1]$ and thus the surface of revolution $S’$ of $e’([0,1])$ about $\mathbb{R}$ in $\mathbb{R}^3$ is isometric to $S$.

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