Nowhere vanishing magnetic helicity Suppose you are given a nowhere-vanishing exact 2-form $B=dA$ on an open, connected domain $D\subset\mathbb{R}^3$. I'd like to think of $B$ as a magnetic field.
Consider the product $H(A)=A\wedge dA$. At least in the plasma physics literature, $H(A)$ is known as the magnetic helicity density. 
How can one determine if there is a closed one-form $\mathbf{s}$ such that $H(A+\mathbf{s})$ is non-zero at all points in $D$? 
The reason I am interested in this question is that if you can find such an $\mathbf{s}$, then  $A+\mathbf{s}$ will define a contact structure on $D$ whose Reeb vector field gives the magnetic field lines. Thus, the question is closely related to the Hamiltonian structure of magnetic field line dynamics. 
I'll elaborate on this last point a bit. If there is a vector potential $A$ such that $A\wedge dA$ is non-zero everywhere, then the distribution $\xi=\text{ker}(A)$ is nowhere integrable, meaning $\xi$ defines a contact structure on $D$ with a global contact 1-form $A$. The Reeb vector field of this contact structure relative to the contact form $A$ is the unique vector field $X$ that satisfies $A(X)=1$ and $\text{i}_XdA=0$. Using the standard volume form $\mu_o$, $dA$ can be expressed as $\text{i}_{\mathbf{B}}\mu_o$ for a unique divergence-free vector field $\mathbf{B}$. Thus, the second condition on the Reeb vector field can be expressed as $\mathbf{B}\times X=0$, which implies the integral curves of $X$ coincide with the magnetic field lines. 
An example where $D=$3-ball and no $\mathbf{s}$ can exist:
Let $D$ consist of those points in $\mathbb{R}^3$ with $x^2+y^2 < a^2$ for a real number $a>1$. Note that all closed 1-forms are exact in this case. Let $f:[0,\infty)\rightarrow\mathbb{R}$ be a smooth, non-decreasing function such that $f(r)=0$ for $r<1/10$ and $f(r)=1$ for $r\ge1/2$. Let $g:\mathbb{R}\rightarrow \mathbb{R}$ be the polynomial $g(r)=1-3r+2r^2$. Define the 2-form $B$ using the divergence free vector field $\mathbf{B}(x,y,z)=f(\sqrt{x^2+y^2})e_\phi(x,y,z)+g(\sqrt{x^2+y^2})e_z$. Here $e_\phi$ is the azimuthal unit vector and $e_z$ is the $z$-directed unit vector. It is easy to verify that $B$, thus defined, is an exact 2-form that is nowhere vanishing. 
Because $g(1)=0$ and $f(1)=1$, the circle, $C$, in the $z=0$-plane, $x^2+y^2=1$, is an integral curve for the vector field $\mathbf{B}$. I will use this fact to prove that the helicity density must have a zero for any choice of gauge. Let $A$ satisfy $dA=B$ and suppose $A\wedge B$ is non-zero at all points in $D$. Note that $A\wedge B=A(\mathbf{B})\mu_o$, meaning $h=A(\mathbf{B})$ is a nowhere vanishing function. Without loss of generality, I will assume $h>0$. Thus, the line integral $I=\oint_C h\frac{dl}{|\mathbf{B}|}$ satisfies $I>0$. But, by Stoke's theorem, $I=2\pi\int_0^1g(r)rdr=0$, as is readily verified by directly evaluating the integral. Thus, there can be no such $A$.
 A: I believe that the answer, in generality, is that you cannot find $\mathbf s$. As an example, consider a field configuration that is translationally symmetric in, say, the $z$ direction (for example, if there is a smooth current density that directed parallel to the $z$ axis). Speaking in physics terms, if we assume that the helicity is non-zero in some gauge $\mathbf s=\boldsymbol\nabla \phi$, then without loss of generality, we can assume that $H > 0$ everywhere in the domain $D$. However, consider the line integral of $H$ over any closed field loop:
$$0<\oint H\frac{\,dl}{|\mathbf B|}=\oint (\mathbf A + \boldsymbol\nabla\phi) \cdot\,d\mathbf l=\iint (\boldsymbol\nabla\times \mathbf A)\cdot\,d\boldsymbol\sigma= \iint \mathbf B\cdot \,d\boldsymbol\sigma=0$$
Here, $\,dl$ is a line measure, and $\,d\boldsymbol\sigma$ is the surface measure. The first integral is well defined since $\mathbf B$ is nowhere vanishing. 
A: Since $s$ is closed:
\begin{eqnarray}H(A+s) &=& (A+s) \wedge d(A+s)\\
& =& (A+s) \wedge (dA + ds) = (A+s) \wedge dA \\
&=& (A \wedge dA) + (s \wedge dA)\end{eqnarray}
If $dA$ is nowhere-vanishing on your domain, $|dA|$ should have a maximum value on the closure $\overline{D}$.  
So you should be able to add a sufficiently large constant 1-form, so that $H$ is also nowhere vanishing.
