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Wikipedia states the hypothesis of the Nash-Kuiper theorem as follows:

Let $(M,g)$ be an $m$-dimensional Riemannian manifold and $f: M \to \mathbb{R}^n$ a short $\mathcal{C}^\infty$-embedding (or immersion) into Euclidean space $\mathbb{R}^n$, where $n ≥ m+1$.

We usually either talk about (a) "embeddings" between differentiable manifolds or (b) "isometric embeddings" between Riemannian manifolds. So if someone talks about just an "embedding" between Riemannian manifolds, is it implied that this embedding is isometric, or does the embedding only apply to the underlying differentiable manifold structure and not to the metric structure?

I assume that $f$ isn't required to be isometric, because if it is then the Nash-Kuiper theorem seems totally trivial, because you can just let $f_\epsilon$ be $f$ itself.

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No, the main point about the theorem is to prove the existences of an isometric embedding in the first place (which - as explained on the Wiki page, is a consequence of the theorem in question together with the Whitney embedding theorem, which is ignoring the metric structure). So the assumption in the Nash $C^1$ theorem is that you have an embedding in the sense of the underlying differentiable manifold structure.

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  • $\begingroup$ I don't quite understand that sentence in the Wikipedia article ("In particular, as follows from the Whitney embedding theorem, any $m$-dimensional Riemannian manifold admits an isometric $\mathcal{C}^1$-embedding into an arbitrarily small neighborhood in $2m$-dimensional Euclidean space.") Are they using the stronger version or the weak version of the Whitney embedding theorem? It doesn't seem like either version quite applies here, because the strong version only applies to embeddings into all of $\mathbb{R}^{2n}$, not just ... $\endgroup$
    – tparker
    Feb 19, 2022 at 19:00
  • $\begingroup$ ... arbitrarily small neighborhoods, and the weak version only applies to embeddings into manifolds of dimension greater than $2m$. $\endgroup$
    – tparker
    Feb 19, 2022 at 19:01
  • $\begingroup$ I agree that the existence of a short map for a non compact Riemannian manifold is unclear here. $\endgroup$
    – Deane
    Feb 19, 2022 at 19:31

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