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The strongest version of Whitney's embedding theorem says that every smooth real $n$-dimensional manifold $M^n$ (Hausdorff and second-countable) can be embedded in $\mathbb{R}^{2n}$.

This should mean that every smooth curve (i.e. smooth real $1$-dimensional manifold) can be embedded in the real plane $\mathbb{R}^2$. Is this true? Or there is something I am missing? My question comes from the fact that I am struggling to see how this is posible for curves that naturally lives in $\mathbb{R}^3$.

For example, while for the Helix: $$H=\{cos(t),sin(t),t | t\in\mathbb{R}\},$$ I can imagine an embedding in $\mathbb{R}^2$ given by the projection to the second and third coordinate, I have problems imagining something similar for the Trefoil: $$T=\{(sin(t)+2sin(2t), cos(t)-2cos(2t), sin(3t)) | t\in[0,1]\}.$$

Pgatti
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2 Answers2

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A connected smooth $1$-manifold is either a line or a circle. The above Helix is a line. The Trefoil is a circle. Both can be smoothly embedded in the plane.

Amitai Yuval
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Sometimes you want to study the properties of the embedding in the sense that you want to study the embedded manifold together with its surrounding. Notice that knot theory is not equivalent to the classification of closed 1-manifolds!

Every connected 1-manifold (being either an interval or $S^1$) embeds in a standard way into $\mathbb R^2$.

Daniel Valenzuela
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  • Thank you for the precisation, but at the moment I am not interested in knot theory. I choose the example of the Trefoil just because I could not follow the same approach I instinctively used for the Helix. Anyway, as everybody pointed out, using the parametrizations of the curves to build diffeomorphisms with subsets of $\mathbb{R}^2$ is the easiest way. – Pgatti Jan 23 '15 at 08:18