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Jan 30, 2023 at 14:04 vote accept Paul
Jan 29, 2023 at 13:52 answer added Anton Petrunin timeline score: 6
Oct 15, 2018 at 18:33 comment added Paul Yes, indeed I was suspected such a problem. But can I assume a lower bound on the curvature and the injectivity radius, for instance?
Oct 15, 2018 at 18:15 comment added Peter Michor The answer mathoverflow.net/a/124878/26935 shows that it is not possible in general to get a a tubular neighborhood of fixed width.
Oct 15, 2018 at 13:52 comment added Paul Good remark, I have edited the post with $C^\infty$ in fact at least $C^2$ should be good, I just want to preserve curvature...
Oct 15, 2018 at 13:51 history edited Paul CC BY-SA 4.0
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Oct 15, 2018 at 13:25 comment added Piotr Hajlasz There are two very different Nash embedding theorems; One for $C^1$ embeddings and one for $C^\infty$ embeddings. Which one are you interested in?
Oct 15, 2018 at 5:03 history asked Paul CC BY-SA 4.0