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so let $\vec{v}: \Omega \subset \mathbb R^n \to \mathbb R^n$ be a vector-field.

Now we know:

(I) $\vec{v}$ conservative $\quad\Rightarrow\quad \nabla\times\vec{v}=0$

If further, $\Omega$ is simply-connected i.e. path-connected and null-homotopic, we can also state:

(II) $\nabla\times\vec{v}=0 \quad \Rightarrow \quad \vec{v}$ is conservative.

Question: Why do we need null-homotopy for (II)?

Davide Morgante
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xotix
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    Otherwise there are counterexamples, see for instance https://math.stackexchange.com/questions/774504/non-conservative-field-with-zero-curl. The sufficiency of the null-homotopy condition is due to Stokes' theorem. – Gal Porat Jul 20 '18 at 15:36
  • @GalPorat ah true! Exactly! Jesus, that took me ages. Thanks, solved :D – xotix Jul 20 '18 at 15:52

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