# Prove $\nabla \cdot \vec{B}=0 \Rightarrow \exists \vec{A}:\vec{B} = \nabla \wedge \vec{A}$

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 $\vec{B} = \nabla \wedge \vec{A}$?

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Could you please give a little more explanation? For example, does $\nabla \cdot \vec{B}$ mean that $\vec{B}$ is incompressible? Moreover, what does $\wedge$ mean in $\nabla \wedge \vec{A}$? Finally, what type of manifold are your vector fields defined on? Closed? Compact? Connected? Genus one or two? Etc... It is also suggested that you explain what research you have done, and any attempts you have made to solve the problem. – Fly by Night Oct 8 '12 at 21:55
@LHS: Do you mean $\vec{B} = \nabla\times\vec{A}$ instead of $\vec{B} = \nabla\wedge\vec{A}$? – Michael Albanese Oct 8 '12 at 22:06
@Fly by night: $\wedge$ is the notation my university uses for cross product. $\vec{B}$ is a vector function, definded on a simple closed space R, the question I'm working with says 'suitable region'. I have solved the previous question which was involving $\nabla \wedge \vec{B}=0 \Rightarrow \exists \vec{A} : \vec{B}=\nabla \vec{A}$, I have been trying to use this – Freeman Oct 8 '12 at 22:46
@MichaelAlbanese: yes, but my university uses the notation I used – Freeman Oct 8 '12 at 22:53
@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. – Fly by Night Oct 9 '12 at 15:51

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.