I was wondering whether every vector field is the gradient of a certain scalar field. If not, could you provide a counterexample? Thanks.
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$\begingroup$No ! This is the beginning of cohomology. Take simply the vector field $1/z$ in $\mathbb C^*$. If it was the gradient of a function, this would implies that we have a well-defined logarithm on $\mathbb C^*$ everywhere which is known to be false.
Here is the real version of the argument. You can take $V(x,y) = (\frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2})$. This is well defined on $\mathbb R^2 \backslash \{(0,0)\}$. But since this is the derivative of the argument function $\theta(x,y)$, there is no functions $f$ with $grad(f) = V$ as it would implies that there is a globally defined angle function on $\mathbb R^2 \backslash 0$ !
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