# Whitney Embedding Theorem

This is very basic question, but from my previous question I learnt that "Whitney Embedding Theorems states that any smooth n dimensional manifold can be embedded in Euclidean space of dimension at most 2n". Fine, but what about spring? It is 1D manifold that can be embedded at least in 3D, but according to theory it should be 2D. What am I missing?

• By a spring, do you just mean a line segment? Bent up and curved, sure, but in the end just a line segment. No problem with embedding it in the plane (or even in the line). – lulu Jul 13 '15 at 21:06
• In general, connected 1-manifolds are either lines or circles (see, for example, math.brown.edu/~tomg/2110/1-manifolds.pdf). In either case the embedding theorem is clear. – lulu Jul 13 '15 at 21:07

smooth $n$ dimensional manifold can be embedded in Euclidean space of dimension at most $2n$
Whitney's theorem just says that an $n$-dimensional manifold $M$ can be smoothly embedded in $\mathbb{R}^{k}$ for $k=2n$ (and therefore certainly for $k\geq 2n$). Note also that this does not prevent the possibility that a particular $M$ can embed in $\mathbb{R}^k$ for $k<2n$.
What that commenter might have meant is that, given an $n$-dimensional manifold $M$, we can ask what the smallest possible $k$ for which $M$ smoothly embeds in $\mathbb{R}^k$, and that depending on the manifold this $k$ can vary, but it is always at most $2n$ (which is a correct phrasing of the Whitney theorem).
A spring $S$ is the image of a smooth embedding $$\gamma:(0,1)\to\mathbb{R}^3.$$ But then $S$ is diffeomorphic to $(0,1)$ itself, so it can actually be embedded in $\mathbb R$ (and hence also in $\mathbb R^2$).