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Consider a Riemannian manifold $(S,g)$ of dimension 2. What can we say about the possibility of an isometric immersion of this surface into $\mathbb{R}^3$?

Of course this is not unique, even up to rigid motions (like in the case of zero Gaussian curvature). How can we find at least one of the possible isometries? Do we have a condition for uniqueness?

If you want to make it simpler, think about a $U \subset \mathbb{R}^2$ with assigned $g_{ij}\neq \delta_{ij}$ (with respect to the standard Cartesian coordinates).

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    $\begingroup$ see here: en.wikipedia.org/wiki/Nash_embedding_theorem $\endgroup$ – Thomas Sep 12 '18 at 17:41
  • $\begingroup$ @Thomas, let's stick to simpler cases than the general one: very docile surfaces and $C^\infty$ maps. I basically need equations that allow me to immerse/embed one of those surfaces... $\endgroup$ – Gaff Sep 12 '18 at 18:31
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    $\begingroup$ Are you asking a local question or a global question? $\endgroup$ – Ted Shifrin Sep 12 '18 at 22:15
  • $\begingroup$ Yes, I am looking for just one map. $\endgroup$ – Gaff Sep 13 '18 at 13:24
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The Cartan-Janet Theorem says (see, for example, Spivak, A Comprehensive Introduction to Differential Geometry, Chapter 11 in volume V) that if you start with a real analytic metric, you can locally isometrically embed the surface in $\Bbb R^3$.

Of course, hoping for a global theorem is hoping too much, as you can't embed isometrically a torus with $g>1$ and constant negative curvature metric in $\Bbb R^3$. Indeed, you can't isometrically embed any compact surface with everywhere negative curvature.

REMARK: A bit of exploration with Google leads me to a new book by Han and Hong called Isometric Embedding of Riemannian Manifolds in Euclidean Spaces. They also discuss the history of the problem. They mention Pogorelov's 1972 example of a $C^{2,1}$ metric on the unit ball so that there's no $C^2$ isometric embedding of any neighborhood of the origin into $\Bbb R^3$.

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  • $\begingroup$ What about compact simply connected subsets of $\mathbb{R}^2$ with a $C^{\infty}$ metric (or, if we need it, analytic)? Can we find a global (even not unique) isometric embedding? Do we need conditions on the curvature? $\endgroup$ – Gaff Sep 13 '18 at 13:56
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By the Cartan-Janet theorem it is possible to locally embed a 2-dimensional Riemannian manifold isometrically into $\mathbb{E}^3$, assuming that the metric is analytic. Note that this is a special case of the Nash embedding theorem. Two-dimensional Riemannian manifolds cannot always be globally immersed or embedded. A counter-example is given by Hilbert's theorem that states that the hyperbolic plane $\mathbb{H}^2$ cannot be immersed in Euclidean three-space.

However, I believe it is not easy to find in general an immersion for a given surface and metric, since this involves solving a system of three non-linear partial differential equations $$\sum_{i=1}^3 \frac{\partial u_i}{\partial x_j}\frac{\partial u_i}{\partial x_k} = g_{jk}(x),$$ where $u_1,u_2,u_3$ are the components of $\mathbb{E}^3$ and $x=(x_1,x_2)$ the coordinates on $U\subset \mathbb{R}^2$.

Also, the isometric embedding is not unique in general. A simple example is to think of a sphere with a dent in it. If you invert the dent (that is, make a bulge of it), you get an isometric surface that is not congruent (i.e. not the same up to isometries of the ambient three-space $\mathbb{E}^3$).

There are however some rigidity results. A well-known one is the theorem of Cohn-Vossen: two isometric ovaloids (compact surfaces with Gauss curvature $K>0$) are congruent, and hence the embedding is essentially unique.

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  • $\begingroup$ I guess that unfortunately we don't know anything about those PDE... $\endgroup$ – Gaff Sep 13 '18 at 14:05
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    $\begingroup$ @Gaff The classic trick to obtain some solutions for a PDE, is to assume extra symmetry. For instance, you might assume that $g_{jk}(x)$ only depends on $x_1$ or $x_2$, or the radial coordinate $r=\sqrt{x_1^2+x_2^2}$. It might give explicit solutions, but these are of course very restrictive cases. $\endgroup$ – Ernie060 Sep 13 '18 at 14:19
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    $\begingroup$ I see. However I think that the equation should read $g_{ij} = \delta_{hk} \frac{\partial u^h}{\partial x^i} \frac{\partial u^k}{\partial x^j}$ with $h,k=1,2,3$ and $i,j=1,2$, i.e., the summation is on the indices of the embedding $u$, not on those of the coordinates $x$. $\endgroup$ – Gaff Sep 13 '18 at 15:19
  • $\begingroup$ True. I edited the answer. Thank you. $\endgroup$ – Ernie060 Sep 13 '18 at 19:57

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