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There are plenty of isometric embeddings of metric spaces in Banach spaces. Nevertheless, I have been unable to find any significant result on isometric embeddings into Hilbert spaces. My question is: how can one recognize those metric spaces that are isometrically embeddable into Hilbert spaces?

Later edit: I have removed two paragraphs from my original question, which created a lot of confusion among those who answered it. I take responsibility for mixing "metric spaces" isometries and "differential geometry" isometries. I apologize.

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    $\begingroup$ The isometric embedding theorems in Banach spaces are of a very different nature: they are distance preserving maps. Nash theorem is about length preserving maps. $\endgroup$
    – alvarezpaiva
    Jan 19, 2014 at 18:54
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    $\begingroup$ Schoenberg's 1935 and 1938 papers address your question. If you do not want to go to the original papers, read about it e.g. Wells and Williams, Embeddings and extensions in analysis, Springer Ergebnisse 84. $\endgroup$
    – Bill Johnson
    Jan 19, 2014 at 19:43
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    $\begingroup$ This question appears to be off-topic because it is about a topic that has been well covered on MO and the OP could have found the answer easily with a Google search. $\endgroup$
    – Bill Johnson
    Jan 19, 2014 at 19:44
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    $\begingroup$ @Bill Johnson: I'm sorry if things are as you say. I assure you that I have first searched the whole web, not just MO, for the answer, before asking my question. I have found only one slightly related question on MO, migrated on SE. $\endgroup$
    – Alex M.
    Jan 19, 2014 at 20:06
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    $\begingroup$ @BillJohnson I am sure you did not read the question to the end. Schoenberg's results (which are by the way totally trivial) are about GLOBAL isometric embedding and OP asks about isometric embeddings as it is usually defined in diff. geometry. $\endgroup$
    – Anton Petrunin
    Jan 19, 2014 at 20:34

2 Answers 2

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The answer was given in the papers of I. Schoenberg and von Neumann, MR1501980, MR1503439, MR0004644.

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    $\begingroup$ As well as Bill Johnson, you do not read the question to the end. $\endgroup$
    – Anton Petrunin
    Jan 19, 2014 at 23:07
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    $\begingroup$ I (like Bill and Alexandre) read "isometric embedding" as what it says, not something esoteric known only to specialists. $\endgroup$
    – Gerald Edgar
    Jan 19, 2014 at 23:18
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    $\begingroup$ Look at the first sentence in the OP's post and the tags, Anton. How is one supposed to know that isometric embedding does not have its usual meaning in metric geometry and metric spaces? $\endgroup$
    – Bill Johnson
    Jan 19, 2014 at 23:35
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    $\begingroup$ @BillJohnson, you have to read further, once you see Nash's theorem you know it. $\endgroup$
    – Anton Petrunin
    Jan 19, 2014 at 23:46
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    $\begingroup$ @MarkMeckes: Sure, the question should be formulated better. (BTW now I see that OP did not understand what he was asking.) $\endgroup$
    – Anton Petrunin
    Jan 20, 2014 at 20:41
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This is the answer to the original question (Not the one which is posted now).

Look in two papers, mine and the paper of Enrico Le Donne.

You are looking for spaces which admit length-preserving embedding into Hilbert space. In my paper I prove that a compact length spaces which (roughly) admit a length-preserving map into Euclidean $m$-space has to be inverse limits of $m$-dimensional polyhedral spaces.

The infinite dimensional case is easier; it can be done along the same lines; in this case the dimension of polyhedral spaces will go to infinity. It seems that if a compact space admits a length-preserving map into infinite dimensional Hilbert space then it can be perturbed into length-preserving embedding. Enrico considers length-preserving embedding in finite dimensional case (which is much harder).

About the last question. Nash's theorem can be indeed simplified considerably if the target space is infinite dimensional. You can apply the same procedure as in the proof of Nash--Kuiper theorem, but since the dimension is infinite, you can use a parallel frame each time. This way you get a sequence of maps $f_1,f_2,\dots$ which converges to isometric embedding and such that the $k$-th coordinate of $f_n$ is the same for all large $n$. This way you can get better regularity of the limit --- instead of $C^1$, you should get $C^\infty$ (am I right?).

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  • $\begingroup$ P.S. An analog of Nash's theorem for Finsler manifolds is given in the paper of Burago and Ivanov, "Isometric embeddings of Finsler manifolds." $\endgroup$
    – Anton Petrunin
    Jan 19, 2014 at 19:06

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