Let $g$ be a Riemannian metric in $\mathbb{R}^2$. How can I find a surface in $R^3$ such that their curvature are the same? The shape of the surface in $3$ space is important. I mean I dont like to equip for example the sphere with the metric $g$. Since, I see it as a sphere and its curvature is $1$ every where.
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2$\begingroup$ Possible duplicate of Nash embedding theorem for 2D manifolds $\endgroup$ – j.c. Sep 12 '17 at 19:32
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