While learning about the rigorous definition of manifolds, my text mentions that any $n$-dimensional manifold can be embedded in $\Bbb{R}^{2n}$, which is called Whitney's embedding theorem. I have attempted to prove this theorem using the rigorous definition of a manifold, but I am stuck. I have only covered tensors and manifolds in my study of differential geometry, so do I have to know more mathematics to prove this theorem? If not, what is the rigorous proof of Whitney's embedding theorem?
[Math] Proof of whitney’s embedding theorem
differential-geometrymanifolds
Best Answer
It's a nontrivial result, and it's probably not something amenable to a proof going back to first principles. In very broad terms, the idea of the proof is:
Show that the manifold $M^n$ embeds into some $\mathbb{R}^N$. This part is straightforward in the compact case: Choose a finite good cover, construct an embedding on each element of the cover (which is trivial), and patch them together using a partition of unity to get a smooth embedding into $\mathbb{R}^N$ for large $N$.
Show that we can take $N = 2n + 1$. The idea here is to use Sard's theorem to construct a nice map onto lower dimension, but there are some tricks involved in controlling the behavior of the resulting map.
Show that we can take $N = 2n$. This is nontrivial, and it uses what's now known as the Whitney trick. Basically, the idea is that you reduce to an embedding with nice singularities, then use the fact that the dimension is high to show that the singularities can be pulled together in pairs through an embedded disk and removed. This trick fails (spectacularly; it's the origin of a lot of the weirdness of $4$-manifold topology) below dimension $5$, but the case $n = 2$ is easy anyway.
You might also be interested in looking up the $h$-cobordism theorem, which contains a lot of the same ideas (including the Whitney trick) and is a more modern result.