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Problem. There is no embedding $f:S^2\to \mathbf R^3$ such that for all points $p, q\in S^2$, we have $d_{S^2}(p, q)=d_{\mathbf R^3}(f(p), f(q))$.

Here $d_{S^2}$ is the metric on $S^2$ which is induced from the Riemannian metric induced on $S^2$ from $\mathbf R^3$, that is, $d_{S^2}(p, q)$ is the length of the shorter great circular arc joining $p$ and $q$. Here $d_{\mathbf R^3}(f(p), f(q))$ is the Euclidean distance between $f(p)$ and $f(q)$.

In this paper, it is remarked in the introduction that this problem can be solved by noticing that the curvature of $S^2$ does not vanish.

I fail to see how we are making use of curvature to solve this problem.

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Consider $X$, set of four points in $S^2(1)$ : $|pq|=|qr|=|pr| = \frac{\pi}{2}$ and $|qt|=|tr|=\frac{\pi}{4},\ |pt|= \frac{\pi}{2}$. So there is no isometric embedding $f$ from $X$ to $\mathbb{E}^3$.

[Add] In the above we have shown that there is no isometry from $T$ to Euclidean space where $T$ is equilateral triangle of side length $\frac{\pi}{2}$ in a sphere. Hence sphere is not flat.

In the paper, we can doubt that there may be local isometry. Hence it takes a small equilateral triangle $T'$ in a sphere. A little calculation (But idea is same to the above) show that there is no isometry around $T'$. Hence the paper concludes that sphere is not locally flat.

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    $\begingroup$ This is $(0,0,1)$ (North Pole), $(0,1,0)$ ("right" pole?), $(1,0,0)$ ("front" pole?), and $(\sqrt2/2,\sqrt2/2,0)$ (halfway between the last two), if I understand you correctly. $\endgroup$ – Akiva Weinberger Jun 26 '17 at 8:35
  • $\begingroup$ Yes. $X$ is a set of those points. $\endgroup$ – HK Lee Jun 26 '17 at 8:42
  • $\begingroup$ @HKLee This is a very simple solution. Thanks. But can you see how to use curvature in order to solve the problem? $\endgroup$ – caffeinemachine Jun 27 '17 at 4:27
  • $\begingroup$ @caffeinemachine : I add some. $\endgroup$ – HK Lee Jun 27 '17 at 4:59
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Take the equatorial circle $S^1\subseteq S^2$. If $S^2$ is embedded in the strongly isometric sense you described, then $S^1$ is also. But there is no strongly isometric embedding of a circle in Hilbert space since such an embedding would have to be a straight line (dealing with Gaussian curvature is a red herring).

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