3

Usual examples of manifolds that I see are (up to homeomorphism) always a subset of $\mathbb{R}^n$ (ie curves, spheres, klein bottles, etc.). Is this true for every manifold?

1 Answers1

2

It is true that every manifold can be embedded in $\Bbb R^N$ for sufficiently large $N$. This is the famous Whitney embedding theorem. However, most mathematicians do not assume an arbitrary manifold is a subset of some Euclidean space. That is, the choice of embedding may be very important.

In addition, we want something like "the Klein bottle" to be a well-defined object in its own right. If we also want an embedding in $\Bbb R^n$, then this is really some additional structure on top of the Klein bottle. The manifold itself should be thought of as an independent object which happens to embed in Euclidean space.

Some authors (Guillemin & Pollack) skirt this issue by defining a (smooth) manifold to be a subset of Euclidean space. More commonly, one sees the following definitions:

A space $M$ is an $n$-manifold if $M$ is Hausdorff and second countable, and for every $x\in M$ there is some neighborhood $U$ of $x$ in $M$ along with a homeomorphism $\varphi\colon U\to V$, where $V$ is a neighborhood in some $\Bbb R^n$.

Smooth manifolds are a bit trickier to define in generality, but its just a matter of bookkeeping.

pancini
  • 19,216
  • Interesting, however I meant to ask about manifolds in general, not just smooth ones. –  Apr 21 '20 at 17:11
  • This is true for topological manifolds as well. – pancini Apr 21 '20 at 17:12
  • (See here) https://mathoverflow.net/questions/34658/is-there-a-whitney-embedding-theorem-for-non-smooth-manifolds – pancini Apr 21 '20 at 17:15
  • 1
    It's also worth pointing out that it is fairly rare to find a manifold which does not admit a smooth structure, unless you go looking for them. – pancini Apr 21 '20 at 17:17