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Oct 13, 2017 at 20:31 vote accept Ivo Terek
Oct 13, 2017 at 20:30 answer added Ivo Terek timeline score: 0
Apr 9, 2017 at 10:33 comment added Robert Bryant The difficulty is that the group of symmetries of the 'metric' $\mathrm{d}x^2$ on $\mathbb{R}^2$ is the set of transformations of the form $$\phi(x,y) = \bigl(\pm x + c, f(x,y)\bigr).$$ where $c$ is a constant and $f_y(x,y)$ is nowhere vanishing. In particular, any curve of the form $\bigl(s,g(s)\bigr)$ can be transformed into $(s,0)$ by the symmetry $\phi(x,y) = \bigl(x,y-g(x)\bigr)$. Thus, all curves on which $\mathrm{d}x$ is nonvanishing are locally equivalent under the symmetry group, so there is no local invariant that can distinguish them.
Apr 9, 2017 at 4:27 history edited Ivo Terek CC BY-SA 3.0
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Apr 9, 2017 at 4:10 history asked Ivo Terek CC BY-SA 3.0