Proof of $\nabla\times\vec{E}=0\Rightarrow\exists\Phi:\vec{E}=\nabla\Phi$ Would anyone be able to help me prove these two statements involving vector fields?
For a suitable region
$$\nabla\times\vec{E}=0\Rightarrow\exists\Phi:\vec{E}=\nabla\Phi$$
and
$$\nabla\cdot\vec{B}=0\Rightarrow\exists\vec{A}:\vec{B}=\nabla\times\vec{A}$$
Many thanks
 A: The existence of the potentials depend on the shape of your region. More explicitly, they depend on the first and second de Rham cohomology groups of the region vanishing. This is not always the case. It is true for convex sets, and more generally $p$-convex sets, defined below. However, proving it for more general spaces requires more advanced machinery, like homotopy invariance et cetera.
If your region $U\subseteq \mathbb{R}^3$ is $p$-convex for a certain point $p\in U$, meaning that it contains the line segment connecting $p$ to any other point in $U$, then the scalar and vector potentials can be explicitly constructed as follows:
Let $f=(f_1,f_2,f_3)$ be a $C^2$ vector field on a $p$-convex region $U$, where we choose $p$ to be the origin (We can always translate the coordinates to make it so).
1) If $\text{curl}(f)=0$, then there exists a function $F:U\rightarrow \mathbb{R}$ such that $\frac{\partial F}{\partial x_i} = f_i$ for $i=1,2,3$, given by
$$F(x_1,x_2,x_3) = \int_0^1 x_1 f(t x)+x_2f_2(tx) + x_3 f_3(tx) \text{d}t$$ where $x=(x_1,x_2,x_3)$.
2) If $\text{div}(f)=0$, then there exists a $F:U\rightarrow \mathbb{R}^3$ with $\text{curl}(F)=f$, given by
$$F(x) = \int_0^1 f(tx)\times (tx) \text{d}t$$
I'll let you check that these $F$'s really do give the desired results. The calculations are not very difficult.
