I've seen stuff like Whitney's embedding theorem and Nash embedding theorem (I don't actually understand their mathematical statement yet) from what I've seen those theorems say you can only embed a n-dimensional manifold in a 2n dimensional euclidean space, maybe 2n – 1 if you're lucky. But you can embed $S^1$ in $\mathbb{R}^2$ and you can embed $S^2$ in $\mathbb{R}^3$ you can embed any 2D manifold in $\mathbb{R}^3$ so why doesn't it generalize to arbitrary dimensions? does that only work with $S^n$? do non-constant curvature manifolds just have too much information like different types of curvature and stuff so that it can't just be embedded into a Euclidean space who's dimension is just 1 higher than the manifold's dimension?
Why can't we isometrically embed an n-dimensional manifolds in $\mathbb{R}^{n+1}$
I'm assuming I need a "isometric embedding" here, I think that means the embedding preserves distances.
my background: I only know linear algebra and multivariable calculus (and some very basic topology) and I haven't actually studied differential geometry and differential topology yet
Best Answer
I find your question somewhat unclear. Perhaps you are asking two separate questions?
The first question is a purely topological one. Can a manifold be embedded into Euclidean space as a hypersurface? Here, there is no notion of distance, and you're allowed to stretch or twist the manifold as much as you want. The Whitney embedding theorem says you can embed an $n$-manifold into Euclidean $2n$-space, but you're asking for something better.
Even this question is vague. Do you want to restrict to closed (compact) manifolds? Based on the examples you cite, it appears so. Do you want to restrict to orientable manifolds? That's a reasonable assumption, since a closed non-orientable manifold, such as the Klein bottle or real projective space, can never be embedded as a hypersurface. Do you want to restrict to smooth embeddings? I suggest that we do, since the situation becomes a lot more subtle yet complicated if you assume less. At that point, your question is a reasonable one, and indeed not so easy to come up with counterexamples. One simple reason is that it requires a topologically nontrivial example that has dimension 3 or more. I'm pretty sure that the complex projective plane (which is 4-dimensional) cannot be embedded into Euclidean 5-space. I'll let others cite other counterexamples and explain why.
Your second question appears to be: Given a Riemannian manifold, can it be embedded isometrically as a hypersurface in Euclidean space? Here, there is a well-defined distance function between two points, both on the Riemannian manifold and in Euclidean space. Isometric means that, given any two points on the hypersurface, the two distance functions agree. Here, there are two major theorems. Nash proved that any closed Riemannian manifold can be embedded smoothly and isometrically into Euclidean space of sufficiently high dimension. On the other hand, Nash and Kuiper proved that if a Riemannian manifold can be embedded as a hypersurface (this is the topological assumption discussed above), then there is a $C^1$ isometric embedding. Not only that, you can do this so that the hypersurface lies inside a ball of arbitrarily small radius. For example, you can embed both the standard unit sphere as well as the flat torus isometrically (distance preserving) into a ball of radius 1/10. This is amazing and obviously means the embedding is very wrinkly. You can see beautiful examples of this here.
Notice that since you can't differentiate a $C^1$ embedding map twice, there is no reasonable notion of curvature for such an embedding. So we usually assume at least $C^2$. Once you can define curvature, things get much more restrictive. In particular, it's easy to prove that any isometrically embedded closed hypersurface must have at least one point with positive curvature. This is easy to see intuitively. Just find the smallest sphere that contains the hypersurface. It has to touch the hypersurface somewhere, and at that point the hypersurface has to be more positively curved than the sphere. So a closed Riemannian manifold with negative curvature cannot be embedded smoothly and isometrically as a hypersurface, even if it can be embedded topologically (i.e., without preserving the distance function). The simplest examples are surfaces with genus greater than 1. These all have Riemannian metrics with negative curvature. With such a metric, they cannot be embedded smoothly isometrically as surfaces in Euclidean space (but can be embedded as $C^1$ isometric maps).
Some quick comments about constant curvature: Any closed orientable 2-manifold has a Riemannian metric with constant curvature, which is positive for a sphere, zero for a torus, and negative for a surface of genus greater than 1. There is a way to define constant curvature for higher dimensional manifolds. There is a similar story as for surfaces.
A lot more can be said about both questions, but I'll stop here. You'll have plenty of time going forward to learn more.