Question regarding Nash-Kuiper embedding theorem

264 Views Asked by At

In Wikipedia description of Nash-Kuiper theorem, it says:

Let $(M,g)$ be a Riemannian manifold and $f: M^{m} \rightarrow \mathbb{R}^n$ a short $C^{\infty}$-embedding into Euclidean space $\mathbb{R}^n$, where $n \ge m+1$. Then for arbitrary $\epsilon > 0$ there is embedding $f_\epsilon: M^m \rightarrow \mathbb{R}^n$ which is

i) in class $C^1$ ii) isometric iii) $\epsilon$-close to $f$.

I am not sure what $M^m$ is implying. Is it saying that $M$ is m-dimensional? Or is it using m-tuple copies of each point of $M$?

1

There are 1 best solutions below

0
On

The first interpretation is correct: $M$ is $m$-dimensional. The assumption $n\ge m+1$ leaves an extra dimension for "corrugations" introduced by Kuiper (Nash's proof needed two extra dimensions).

The second interpretation would not make much sense, by the way: even when $M$ is a surface, embedding its $m$th power would take at least $2m$ dimensions, making the assumption $n\ge m+1$ redundant.