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Consider a smooth Nash embedding, $f$, of a Riemannian manifold $Σ$ into Euclidean space $\mathbb R^n$. To what extent is this embedding not unique?

It is clear that the set of all such embeddings contains, at least, all the embeddings obtained by action of the Euclidean group $E(n)$ on $f$, but is there a way of understanding the entire set of embeddings?

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    $\begingroup$ My understanding is that as the co-dimension grows the embeddings get an increasing amount of local freedom. For example, the Nash embeddings of $S^1$ in $\mathbb R^3$ are simply the smooth embeddings having a prescribed global length, parametrrized at constant speed. This has the same homotopy-type as the space of smooth embedings of $S^1$ in $\mathbb R^3$, which is an infinite-dimensional manifold. $\endgroup$
    – Ryan Budney
    Commented Jul 22, 2022 at 17:45
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    $\begingroup$ It's not clear what you mean by 'the' Nash embedding or what you mean by 'equivalent embeddings', since you clearly don't mean 'rigidly equivalent embeddings'. What is the equivalence relation on embeddings that you want to consider if it's not 'differing by rigid motion'? Also, you should bear in mind that the theory is very different if you consider, say, $C^1$ embeddings vs. smooth embeddings. $\endgroup$
    – Robert Bryant
    Commented Jul 22, 2022 at 17:53
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    $\begingroup$ My understanding is that there is an $h$-principle identifying the homotopy type of the space of isometric immersions. In the $C^1$ case, I think it is that the inclusion of isometric immersions into immersions is a homotopy equivalence. In the smoother case, there is some normal bundle data, checking that second fundamental form works out, but ultimately it is just bundle data over the space of immersions. $\endgroup$
    – Ben Wieland
    Commented Jul 22, 2022 at 18:21
  • $\begingroup$ As for embeddings, people usually just say that in high enough codimension there is wiggle room to make a generic immersion be an embedding. But Gromov says that something like the Haefliger functor captures the homotopy type in the metastable range. Surely this generalizes to the full embedding calculus and the final answer is that in codimension at least 3 the space of isometric embeddings has the homotopy type of the fiber product of the space of isometric immersions and the space of embeddings, fibered over the space of immersions. $\endgroup$
    – Ben Wieland
    Commented Jul 22, 2022 at 18:24
  • $\begingroup$ @RobertBryant What I mean by equivalent is: the set of all smooth isometric embeddings of $\Sigma$ into $\mathbb{R}^n$. $\endgroup$
    – dennis
    Commented Jul 22, 2022 at 20:23

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