Timeline for The rigidity of $2$-dim sphere with constant sectional curvature in $\mathbb{R}^n$ for $n> 3$

Current License: CC BY-SA 4.0

10 events

when toggle format	what		by	license	comment
Dec 11, 2023 at 15:50	comment	added	YCor		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!]
Dec 11, 2023 at 15:48	history	edited	YCor		edited tags
Dec 9, 2023 at 22:56	comment	added	Ryan Budney		The question refers to rigidity, i.e. does the group of isometries of $\mathbb R^n$ act transitively on the isometric embeddings of $S^2$ in $\mathbb R^n$. As mentioned, the answer is generally no.
Dec 9, 2023 at 20:49	comment	added	Deane Yang		@WillieWong’s counterexample is still one for your new version of the question.
Dec 9, 2023 at 16:02	comment	added	Daniel Asimov		(Although you don't state explicitly what you mean by "S^2 ⊆ R^n".)
Dec 9, 2023 at 16:00	comment	added	Daniel Asimov		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.
Dec 9, 2023 at 14:09	history	edited	mmaatthh	CC BY-SA 4.0	added 9 characters in body
Dec 9, 2023 at 14:07	comment	added	mmaatthh		@WillieWong, my question should be claified as the follows: Is $𝑓(𝑆^2)$ always $\mathbb{S}^2\subseteq \mathbb{R}^n$ modulo an isometry of $\mathbb{R}^n$?
Dec 8, 2023 at 14:12	comment	added	Willie Wong		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).
Dec 8, 2023 at 13:43	history	asked	mmaatthh	CC BY-SA 4.0