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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.

Alex M.
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    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. – alvarezpaiva Jan 19 '14 at 18:54
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    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. – Bill Johnson Jan 19 '14 at 19:43
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    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. – Bill Johnson Jan 19 '14 at 19:44
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    @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. – Alex M. Jan 19 '14 at 20:06
  • @alvarezpaiva: I might be superficial here, but aren't the two concepts identical (through the Myers-Steenrod theorem)? – Alex M. Jan 19 '14 at 20:07
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    @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. – Anton Petrunin Jan 19 '14 at 20:34
  • Interesting. What is the definition of "isometric embedding" used in differential geometry? And how can someone reading the question realize that it does not mean a global isometric embedding? – Gerald Edgar Jan 19 '14 at 23:11
  • @GeraldEdgar, it is a legth-preserving embedding. It is a historical term, when it was invented (Gauss?) people probably thought that grobal isometric embedding are too boring. (I also hate that isometric embeddings are not isometric...) – Anton Petrunin Jan 19 '14 at 23:31
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    re: the notion of isometric embedding, see http://math.stackexchange.com/questions/87503/isometric-embedding – j.c. Jan 20 '14 at 06:50
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    Alex M, the unit sphere in Euclidean space has two natural induced "metrics": the induced distance (aka chordal metric) and the distance induced by the induced Riemannian metric. They are quite different. For connected manifolds, one is the length metric associated to the other. You should probably clarify the question by explicitely writing what kind of embedding you are looking for (probably length-preserving). Also, your first paragraph seems to obfuscate this point a little further, as length preserving embeddings into Banach space seem unusual. – Benoît Kloeckner Jan 20 '14 at 10:59
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    @AlexM. Benoit already answered your comment (why the induced metric and the induced length metric are not the same). I just wanted to add that while the question of when a metric space can be embedded in a Hilbert space was settled by Frechet, Schoenberg, and Menger, the existence of length preserving embeddings between length spaces and Banach space has not been studied much. Sample problem: does every compact Finsler surface admit an isometric embedding into $L_1$? – alvarezpaiva Jan 20 '14 at 14:44
  • @Bill Johnson's bibliographic indication (Wells and Williams, "Embeddings and extensions in analysis", Springer Ergebnisse 84) answers my problem, I guess, even though the characterization of those metric spaces that can be embedded in Hilbert spaces from this book is quite technical and not very enlightening. – Alex M. Jan 20 '14 at 18:19
  • @alvarezpaiva mentions some results of Frechet and Menger. Are they along the same lines as those of Schoenberg mentioned by Bill Johnson? – Alex M. Jan 20 '14 at 18:20
  • @AlexM. Perhaps you should read some papers that use Shoenberg's theorem. I suggest the papers of Aharoni, Maurey and Mitiagin that use it to classify the Banach spaces that uniformly embed into a Hilbert space and the one by Randrianarivony that classifies the Banach spaces that coarsely embed into a Hilbert space. We do not have analogous classifications of Banach spaces that uniformly or coarsely embed into $L_p$ for other $p$ simply because there is no replacement for Schoenberg's theorem. – Bill Johnson Jan 20 '14 at 19:00
  • @Bill Johnson: I see that the proof found in Wells and Williams depends on Zorn's lemma. Would you happen to know whether this is just for convenience, or a fundamental aspect of this fact? (Lemma 2.3, page 3). – Alex M. Jan 24 '14 at 19:07

2 Answers2

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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?).

Anton Petrunin
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  • 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." – Anton Petrunin Jan 19 '14 at 19:06
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The answer was given in the papers of I. Schoenberg and von Neumann, MR1501980, MR1503439, MR0004644.

Alexandre Eremenko
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    As well as Bill Johnson, you do not read the question to the end. – Anton Petrunin Jan 19 '14 at 23:07
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    I (like Bill and Alexandre) read "isometric embedding" as what it says, not something esoteric known only to specialists. – Gerald Edgar Jan 19 '14 at 23:18
  • @GeraldEdgar, good for you, but you should not do it in diffgeometry :) – Anton Petrunin Jan 19 '14 at 23:20
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    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? – Bill Johnson Jan 19 '14 at 23:35
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    @BillJohnson, you have to read further, once you see Nash's theorem you know it. – Anton Petrunin Jan 19 '14 at 23:46
  • @AntonPetrunin: It certainly doesn't help that the title refers to "isometric embeddings of metric spaces", whereas the question is about something rather different from what that phrase usually means. – Mark Meckes Jan 20 '14 at 10:21
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    @MarkMeckes: Sure, the question should be formulated better. (BTW now I see that OP did not understand what he was asking.) – Anton Petrunin Jan 20 '14 at 20:41