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I wonder if someone can clarify whether it is known that every closed, orientable surface (2-manifold) has a smooth isometric immersion in $\mathbb{R}^4$?

This topic has been discussed rather thoroughly in the earlier MO question, "Nash embedding theorem for 2D manifolds." However, I am puzzled by two claims that seem at odds, but are surely not—rather I just don't understand the terminology.

First, Gromov established that every compact surface has a smooth embedding in $\mathbb{R}^5$. This is from his 1986 book, Partial Differential Relations. See the above MO question for a quote.

Second, as user jc said, "Poznyak proved that $\mathbb{R}^4$ works for any compact part of a complete surface." Here is a more precise statement of Poznyak's result, from the book by Jia-Xing Hong, Isometric embedding of Riemannian manifolds in Euclidean spaces (p.40):

Theorem 2.3.1. Let $g$ be a smooth complete metric defined in $\mathbb{R}^2$. Then for any compact set $\Omega$ in $\mathbb{R}^2$, $(\Omega,g)$ admits a smooth isometric immersion in $\mathbb{R}^4$.

Since Poznyak's $\mathbb{R}^4$ result is from 1973, and Gromov's $\mathbb{R}^5$ from 1986, it must be that Poznyak's theorem does not improve Gromov's to $\mathbb{R}^4$. Is this because "a compact part of a complete surface" is not the same as a "compact surface"? Can anyone explain this to me? Thanks!


E.G. Poznyak, "Isometric immersions of two-dimensional Riemannian metrics in Euclidean space." Russian Math. Surveys, 28:4 (1973), pp. 47–77. Uspekhi Mat. Nauk, 28:4 (1973), pp. 47–76.


Answered. Robert Bryant resolved my confusion: "A compact surface without boundary can't be identified with a subset of the plane." It is my understanding now that it remains an open question of whether or not every compact surface (without boundary) has a smooth immersion in $\mathbb{R}^4$: neither Gromov's nor Poznyak's results settle this question, and as far as I know there is no known counterexample.

I'm adding the open-problem tag, because although Robert completely answered my specific question, the question suggested by the title is open.

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    $\begingroup$ It seems that jc's paraphrase of Poznyak's result is inaccurate, since the Theorem 2.3.1 that you quote does not imply what jc wrote. A compact surface without boundary can't be identified with a subset of the plane. If the surface has genus bigger than one, you can isometrically embed any compact subset of its universal cover, but that doesn't seem to help with isometrically embedding the original surface. $\endgroup$ Oct 13, 2011 at 15:01
  • $\begingroup$ @Robert: Thanks so much! I believe that resolves my confusion. And to jc's credit, his was just a tangential comment, and a useful one at that. But you nailed it: "A compact surface without boundary can't be identified with a subset of the plane." That's clear! $\endgroup$ Oct 13, 2011 at 15:21
  • $\begingroup$ Now I'm puzzled; "my" paraphrase was actually taken verbatim from Han and Hong's text before their statement of Poznyak's theorem. Perhaps the meaning of "compact part" is tripping me up??? See the three pages in question here books.google.com/… In any case, the main thrust of my comment was that there is a difference in what can be proved for embeddings versus immersions (which are like "local" embeddings). $\endgroup$
    – j.c.
    Oct 13, 2011 at 18:45
  • $\begingroup$ @jc: Maybe then it is Han & Hong whose paraphrase was inaccurate, but they knew they were following their paraphrase in the next sentence with the formal theorem, so there was little risk of confusion. In any case, I appreciate that you were focused on embed vs. immerse, whereas here I am focused on the difference between a compact set in $\mathbb{R}^2$ vs. a compact surface. Your tip to Poznyak was very useful--thanks! $\endgroup$ Oct 13, 2011 at 18:52
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    $\begingroup$ OK, so what I was confused about was that the paraphrase is not specific about which "compact parts" are allowed -- the corollary makes it clear that it's NOT any compact part, but rather only sufficiently small compact parts; according to their definition of "local": books.google.com/… $\endgroup$
    – j.c.
    Oct 13, 2011 at 19:15

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