Consider an inverse square vector field $\vec{F}$ and it's curl and divergence shown below
$$\vec{F} = \hat{r}/r^2\\ \nabla \times \vec{F} = \vec{0}\\ \nabla \cdot \vec{F} = 0 $$
$\vec{F}$ is defined over all space except the origin. This is important because it says that the domain of $\vec{F}$ is not simply connected. Let's apply Helmholtz decomposition theorem over a simply connected domain $V \subset \mathbb{R}^3$ (I just want to avoid the trouble point at the origin. In order to apply Helmholtz, I don't know if the domain has to be simply connected, but let's just assume $V$ is simply connected anyways - Wikipedia says that $V$ just needs to be a domain over which $\vec{F}$ is a class $C^2$ function). Anyways, I decompose $\vec{F}$ into
$$\vec{F} = \nabla{\Phi} + \nabla \times \vec{A} $$
The potential $\Phi$ and the vector potential $\vec{A}$ are made out of two terms each. In the context of $\vec{F} = \hat{r}/r^2$, we are left with
$$ \Phi = \oint_{\text{surface of V}} \cdots \ dS \\ \vec{A} = \oint_{\text{surface of V}} \cdots \ dS $$
Question Does Helmholtz theorem make sense or did I apply it incorrectly? I already know that $\Phi$ should be $-1/r$ since
$$ \nabla \Big(-\frac{1}{r}\Big) = \frac{\hat{r}}{r^2}$$
The only way I can make sense of Helmholtz theorem is if $\vec{A}$ happens to be $\vec{0}$ so that the curl of $\vec{0}$ is zero. Or maybe $\vec{A}$ is a constant, so that the curl of a constant is zero. Or just in general $\vec{A}$ has to be conservative (say $\nabla \lambda$) so that $\nabla \times \nabla\lambda = \vec{0}$. Or maybe $\vec{A} = \text{co}\vec{\text{ns}}\text{t} + \nabla\lambda$ so that the curl is $0$. Is this correct?
Question If you have a vector field that is completely specified by a vector potential $\vec{F} = \nabla \times \vec{A}$, then it follows that the divergence of $\vec{F}$ is $0$. I know the logic doesn't follow the other way around. But I thought that if you know a vector field is divergence free and the domain is simply connected, then $\vec{F}$ is completely specified by a vector potential. Is this not true? I know that the inverse square field has a divergence of $0$. Over simply connected domains not containing the origin, does it not follow that $\hat{r}/r^2$ is completely specified by some vector potential $\vec{A}$, in which case $\Phi$ has to be a constant so that $\nabla \Phi = 0$? In my first question, I found $\Phi$ and concluded the form of $\vec{A}$ to make the curl $0$. Now I'm asking, couldn't $\vec{F}$ over simply connected domains be specified via the curl of $\vec{A}$, in which case I would have to conclude that $\Phi$ is a constant so the gradient is $0$?