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Apr 26, 2015 at 10:33 comment added Robert Bryant It's probably worth pointing out, though, that there are constraints: In $n$ dimensions, if you specify $n$ functions $f^1,\ldots, f^n$ and $n$ vector fields $X^1,\ldots, X^n$ and you want to know whether there is a metric $g$ such that $X^i=\nabla^g f^i$, then you will need it to be true that the functions $h^{ij} = \mathrm{d}f^i(X^j)$ satisfy $h^{ij} = h^{ji}$, since, by definition, $$h^{ij} = \mathrm{d}f^i(X^j) = g(\nabla^gf^i,\nabla^gf^j).$$ Of course, you also need the matrix $h = (h^{ij})$ to be positive definite as well as symmetric. With these assumptions, $g$ exists and is unique.
Apr 23, 2015 at 22:27 vote accept Liviu Nicolaescu
Apr 23, 2015 at 21:17 history edited Joonas Ilmavirta CC BY-SA 3.0
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Apr 23, 2015 at 20:40 history answered Joonas Ilmavirta CC BY-SA 3.0