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If there is a smooth isometric embedding $f: (S^2, g)\rightarrow \mathbb{R}^n$, where $(S^2, g)$ is a sphere with Riemannian metric such that the corresponding sectional curvature is equal to $1$, and $n\geq 4$. Is Is $f(S^2)$ always $\mathbb{S}^2\subseteq \mathbb{R}^n$ modulo an isometry of $\mathbb{R}^n$?

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    $\begingroup$ If you mean $\mathbb{R}^3\subseteq \mathbb{R}^n$ as a vector subspace, then the answer is no. There are many isometric embeddings of $\mathbb{R}^3$ into $\mathbb{R}^4$ that is not extrinsically flat (just roll it up like a fruit roll up). $\endgroup$
    – Willie Wong
    Commented Dec 8, 2023 at 14:12
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    $\begingroup$ Since R^1 smoothly embeds isometrically in an arbitrarily small neighborhood of R^2, R^3 does the same in R^6. So if we first place S^2 in R^3 as the usual unit sphere, and then embed R^3 in an arbitrarily small neighborhood of R^6, this provides a counterexample to your question. $\endgroup$
    – Daniel Asimov
    Commented Dec 9, 2023 at 16:00
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    $\begingroup$ (Although you don't state explicitly what you mean by "S^2 ⊆ R^n".) $\endgroup$
    – Daniel Asimov
    Commented Dec 9, 2023 at 16:02
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    $\begingroup$ @WillieWong’s counterexample is still one for your new version of the question. $\endgroup$
    – Deane Yang
    Commented Dec 9, 2023 at 20:49
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    $\begingroup$ I guess "isometric embedding" is to be interpreted in the Riemannian sense, i.e., infinitesimally isometric, and not in the stronger metric sense (distance-preserving). [The standard embedding in $\mathbf{R}^3$ is not distance-preserving!] $\endgroup$
    – YCor
    Commented Dec 11, 2023 at 15:50

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