Given a smooth function $f : \mathbb{R}^2 \setminus \{0\} \to \mathbb{R}$, there exists some smooth vector field $F : \mathbb{R}^2 \setminus \{0\} \to \mathbb{R}^2$ such that $f = \mathrm{curl}\,F$. In other words, it is possible to find smooth functions $F_1, F_2 : \mathbb{R}^2 \setminus \{0\} \to \mathbb{R}$ such that $$ f = \frac{\partial F_2}{\partial x_1} - \frac{\partial F_1}{\partial x_2}. $$ One way to prove this is to show that the (de Rham) cohomology group $H^2(\mathbb{R}^2 \setminus \{0\})$ is trivial, but this usually involves additional machinery (e.g., the Mayer–Vietoris sequence). Does anyone know of an elementary proof of this result?
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@RossMillikan The function $f$ is defined over $2$-dimensional space, so there isn't a way of taking the divergence. – Frank Nov 07 '22 at 03:13
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Yes, choose $F_1 = 0 $. Then use the elementary fact that every continuous one variable function has an antiderivative. – Ninad Munshi Nov 07 '22 at 03:26
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@NinadMunshi I don't think that works because of the puncture at $0$. Take $f(x, y) = 1 / (x^2 + y^2)$, for instance. Then the antiderivative $y^{-1}\arctan(x / y)$ is not defined everywhere. – Frank Nov 07 '22 at 03:46