Submanifold of dimension k in $\mathbb{R}^n$

Question

If $\Sigma \subset \mathbb{R}^n$ is a submanifold of dimension k, Why can't we just say it's a subset of $\mathbb{R}^k$ or is that usually implied, or does n have to be atleast k+1?

Answer 1

The sphere $S^2$ is a submanifold of $\mathbb{R}^3$ of dimension 2, but it is not a subset of $\mathbb{R}^2$.

Answer 2

You are mixing up the concepts. You can define them as follows and check the differences between them:

A set $A$ is a subset of a set $B$, or equivalently $B$ is a superset of $A$, if $A$ is "contained" inside $B$, that is, all elements of $A$ are also elements of $B$. $A$ and $B$ may coincide.

A submanifold $\Sigma$ of a manifold $M$ is a subset $S$ which itself has the structure of a manifold, and for which the inclusion map $S\to M$ satisfies certain properties.

In other words, any submanifold $\Sigma$ of $\mathbb{R}^n$ is a subset in which you can also define a structure of manifold ($k$-dimensional in your question). But not any subset of $\mathbb{R}^n$ is a submanifold.

Depending on which certain properties you use in the definition of submanifold, you can find different counterexamples of subset not being submanifolds.

Answer 3

In general, a submanifold $\Sigma$ of $\mathbb{R}^n$ does not have to lie in any smaller $\mathbb{R}^k$ (see my caveat later in the post). For example, we generally consider the unit circle $\mathbb{T}$ to be a subset (and a submanifold of dimension 1) of the plane $\mathbb{R}^2$. Then we can consider the torus $\mathbb{T}^2$ as a subset (and submanifold of dimension 2) of $\mathbb{R}^4$. However, this definition of the torus does not allow us to view it as a subset of $\mathbb{R}^3$ or $\mathbb{R}^2$ via the natural embedding of these spaces into $\mathbb{R}^4$ (i.e., $(x_1,x_2,x_3) \mapsto (x_1,x_2,x_3,0)$ or $(x_1,x_2) \mapsto (x_1,x_2,0,0)$). This is because it contains points with nonzero coordinates in each orthogonal direction (e.g., $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right) \in \mathbb{T}^2$.

That being said, there is a result called the Whitney embedding theorem, which states that any manifold of dimension $k$ can be smoothly embedded in $\mathbb{R}^{2k}$. What this means is that if you have an embedded submanifold $\Sigma$ of dimension $k$ inside $\mathbb{R}^n$, then (if you want) you can realize it (via a diffeomorphism) as an embedded submanifold of $\mathbb{R}^{2k}$ instead. In general, the power $2k$ is optimal, although if you know certain facts about your manifold you can sometimes do better. Also, if $k$ is not a power of $2$ then you can embed in $\mathbb{R}^{2k-1}$. To see an example where you really do need $2k$, consider that the Klein bottle is a smooth $2$-manifold and cannot be embedded in $\mathbb{R}^3$.