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Let $C$ be a closed, simply connected Riemannian submanifold of $\mathbb{R}^{2}$. Let $F$ be an isometric embedding of the boundary $\partial C$ of $C$ in $\mathbb{R}^{3}$. In particular, assume that $F$ is smoothly isotopic to the trivial embedding $F_{0} \,\colon C \ni (x,y) \mapsto (x,y,0) \in \mathbb{R}^{3}$.

Does there exist an isometric embedding $\tilde{F}$ of $C$ in $\mathbb{R^{3}}$ such that its restriction to $\partial C$ is precisely $F$? If yes, is it unique?

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    Presumably you'd want to impose some additional restriction, e.g., that $\partial C$ is simple connected. Otherwise, if you took $C$ to be an annulus, so that $\partial C$ is a disjoint union of two circles $E_1$ and $E_2$, you could simply choose $F$ so that $d(F(E_1), F(E_2)) > \textrm{diam}(C)$. Then any embedding $\tilde{F} : C \to \Bbb R^3$ extending $F$ satisfies $\textrm{diam}(\tilde{F}(C)) > \textrm{diam}(C)$ and so cannot be isometric. – Travis Willse May 22 '17 at 12:50
  • Good point, thanks –  May 22 '17 at 13:08
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    Take $C$ to be a disk and take $F$ to map $S^1$ onto a nontrivial knot. Unless I am much mistaken, this provides a family of topological (not even geometric) obstructions. – Neal May 22 '17 at 13:15
  • @Neal Thanks, I have edited the question to take care of your counterexample –  May 22 '17 at 20:19
  • I guess this question solves the uniqueness part. –  May 22 '17 at 21:02
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    I guess you need $F$ to not increase chordal distances as well, otherwise one could take $C$ to be a disk and take $F$ to map $S^1$ to a long thin ellipse whose major axis is almost $\pi$. –  May 22 '17 at 21:13
  • @Rahul Yes that resolves the issue, thanks. I got a bit confused, what I truly wanted to ask is this. –  May 23 '17 at 10:23

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