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.
Best Answer
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.