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Let $S^2$ be the $2$-dimensional sphere with a metric $g$.

Q:

Can we or how to find a smooth map $f:S^2\to \mathbb R^3$, such that

(1) $f$ is diffeomorphic to its image $Im(g)=:M$,

(2) $M$ with the induced metric from $ds^2$ of $\mathbb R^3$, is isometric to $(S^2,g)$, where $ds^2$ is the standard metric on $\mathbb R^3$, i.e. $ds^2=dx^2+dy^2+dz^2$, and the isometric is given by $f^{-1}:M\to S^2$.

I do not know any work about the problem(maybe solved). Any reference about the problem is welcome

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If the Gauss curvature is positive, such an embedding is known to exist, a result of Nirenberg. But if there is a point of zero Gauss curvature, I think that nothing is known. See the book Qing Han, Jia-Xing Hong, Isometric Embedding of Riemannian Manifolds in Euclidean Spaces for the best known results.

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    $\begingroup$ A little clarification: When the Gauss curvature is positive, then it's known as the Weyl problem and was proved independently. by Nirenberg and Pogorelov. For nonnegative Gauss curvature, there is a result by Guan and Li, projecteuclid.org/euclid.jdg/1214454874. If the Gauss curvature is negative somewhere, then nothing is known. $\endgroup$
    – Deane Yang
    Commented May 13, 2020 at 16:26
  • $\begingroup$ Pogorelov showed the following more general result. For a general metric on $S^2$, if the Gaussian curvature is strictly greater than $-\kappa$ for a nonnegative number $\kappa$, then the metric can be isometrically embedded into the simply-connected complete space form of curvature $-\kappa$. $\endgroup$
    – Quarto Bendir
    Commented May 24, 2020 at 21:34

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