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Jan 13, 2023 at 8:45 comment added Alex M. $\Xi _\varepsilon$ is not an isometry, though, so strictly speaking this looks like half an answer to the OP's question. I believe that the OP was asking about the existence of a Nash isometric embedding theorem but under the weaker requirement of the target space being infinite-dimensional (which, hopefully, would make the proof easier and shorter).
Sep 10, 2018 at 9:09 history edited Liviu Nicolaescu CC BY-SA 4.0
added 178 characters in body
S Sep 9, 2018 at 22:00 history suggested AccidentalFourierTransform CC BY-SA 4.0
fix some typos
Sep 9, 2018 at 19:45 review Suggested edits
S Sep 9, 2018 at 22:00
Sep 9, 2018 at 15:45 comment added Ho Man-Ho Thank you Prof. Nicolaescu. To me this is a very amazing result in the sense that no conditions have to be imposed on the metric $g$.
Sep 9, 2018 at 15:39 vote accept Ho Man-Ho
Sep 9, 2018 at 15:03 history answered Liviu Nicolaescu CC BY-SA 4.0