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Here a hexagonal 2-torus TH means any Riemannian flat torus obtained by identifying opposite edges of a regular hexagon in the plane.

It's easy to see that TH can be smoothly (C) isometrically embedded in ℝ6, and for that matter (if suitably scaled) in the unit sphere S5, and also in ℂℙ2.

But is it known whether TH can be smoothly isometrically embedded in ℝ5 or ℝ4 ?

(Also of interest is whether TH can be smoothly isometrically immersed in ℝ5 or ℝ4.)

Edit: Thanks to the Gromov theorem mentioned by Michael Albanese, we know that TH isometrically embeds smoothly in ℝ5, so the question remaining is whether it embeds in ℝ4. Pace Ian Agol's comment, I am waiting to confirm whether TH actually has a smooth isometric embedding in S^3 (up to scaling) ... and hence in ℝ4.

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    $\begingroup$ Gromov showed that every compact Riemannian two-manifold admits a smooth isometric embedding into $\mathbb{R}^5$. See Gromov's Partial Differential Relations, pages 298 - 303. This is also mentioned in one of the answers to this question which may have some other helpful information. $\endgroup$ Commented Oct 29 at 0:45
  • $\begingroup$ I think you can use the principal fibration $U(1)\to SU(2) \to \mathbb{CP}^1$ (Hopf fibration) to isometrically immerse any flat 2-torus in the 3-sphere, and hence into $\mathbb{R}^4$. Any flat 2-torus fibers over the circle as a Riemannian submersion. Rescale $S^3$ so the Hopf fibers have the length of the fibers of the torus, and choose a curve in $\mathbb{CP}^1\cong S^2$ so that the length is the length of the base of the torus fibration and the holonomy agrees with the torus holonomy. Then the preimage of this curve gives the desired torus. I haven’t thought this through carefully though. $\endgroup$
    – Ian Agol
    Commented Oct 29 at 5:21
  • $\begingroup$ Ian — I had thought that result applied to conformally equivalent Riemann surfaces (rather than isometry classes up to scaling). $\endgroup$ Commented Oct 29 at 15:38
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    $\begingroup$ Namely, Hopf tori in S^3, Ulrich Pinkall, Inventiones Mathematicae, v. 81, 1985. $\endgroup$ Commented Oct 29 at 18:41
  • $\begingroup$ @DanielAsimov that reference seems to do it $\endgroup$
    – Ian Agol
    Commented Oct 29 at 21:56

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