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Can anyone help me prove that if $\vec{B}$ is a differentiable vector field with $\nabla \cdot \vec{B}=0$, then there exists an $\vec{A}$ such that $\vec{B} = \nabla \wedge \vec{A}$?

Thank you in advance!

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    @LHS Interesting. The cross product is only defined in a three dimensional vector space. The symbol $\wedge$ often denotes the exterior product. The cross product is the Hodge dual of the exterior product.2012-10-09

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A high-level view is that you are asking whether there are any closed $2$-forms that are not exact. (See the Wikipedia article on closed and exact differential forms.) Since $\mathbb{R}^3$ is contractible, all its cohomology vanishes, so all closed forms are exact.

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    I encountered this as part of my electromagnetism course, I'm afraid I have not covered what you are referring to2012-10-08