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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 $A∧dA$ is non-zero everywhere, then the distribution $ξ=\text{ker}(A)$ is nowhere integrable, meaning $ξ$ 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 $i_XdA=0$. Using the standard volume form $μ_o$, $dA$ can be expressed as $i_B\mu_o$ for a unique divergence-free vector field $\mathbf{B}$ (I'm having trouble typing $\mathbf{B}$ as a subscript). Thus, the second condition on the Reeb vector field can be expressed as $\mathbf{B}×X=0$, which implies the integral curves of X coincide with the magnetic field lines.

More generally, suppose $M$ is an orientable odd-dimensional manifold equipped with an exact 2-form $\omega$ of maximal rank. Also assume that the characteristic line bundle associated to $\omega$ admits a non-vanishing section $b:M\rightarrow \text{ker}(\omega)$. What is the obstruction to the existence of a 1-form $\vartheta$ with $d\vartheta=\omega$ and $\vartheta(b)>0$?

some observations/comments:

1) If $A(\mathbf{B})$ is bounded above and below on $D$, then a sufficient condition for there to be an $\mathbf{s}$ that gives a nowhere-vanishing helicity density is the existence of a closed one-form $\alpha$ with $\alpha(\mathbf{B})$ nowhere vanishing. In that case, $\mathbf{s}=\lambda \alpha$, where $\lambda$ is some large real number (with appropriate sign), would work.

If there is such an $\alpha$, then, being closed, it defines a foliation whose leaves are transverse to the divergence-free field $\mathbf{B}$. I suspect the question that asks whether a given non-vanishing divergence-free vector field admits a transverse co-dimension one foliation has been studied before, but I am not familiar with any work of this type.

An example where $D=$3-ball and helicity density must have a zero:

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 $A$.

An example where $D=T^2\times (0,2\pi)$ and helicity density must have a zero:

Set $D=S^1\times S^1\times(0,2\pi)$ and let $(\theta,\zeta,r)$ be the obvious coordinate system. Set $B=f(r) dr\wedge d\theta+g(r) dr\wedge d\zeta$ where $$f(r)=\cos(2r),$$ and $$g(r)=\sin(r). $$ Clearly, $A=\frac{1}{2}\sin(2r)d\theta-\cos(r)d\zeta$ satisfies $B=dA$ and $B$ is nowhere vanishing. A quick calculation shows that $\int_D A\wedge B=0$.

Now suppose that $\mathbf{s}$ is an arbitrary closed 1-form. Either by using Stoke's theorem or by direct calculation, the fact that the total toroidal and poloidal fluxes, $2\pi\int_0^{2\pi}f(r)dr$ and $2\pi\int_0^{2\pi}g(r)dr$, are zero implies that $\int_D(A+\mathbf{s})\wedge B=0$. Thus, the helicity density must always have a zero.

share|improve this question
Can you elaborate a little on your motivation? -- it's interesting and so am wondering if there is a reference. –  Chris Gerig Oct 9 '12 at 22:07
I edited the problem statement to give a bit more explanation of the connection between magnetic fields and contact geometry. –  Josh Burby Oct 10 '12 at 21:42
I have some questions. 1) What is $A(B)$ ? 2) Do you have an example when such form $s$ does not exist? 3) Can you prove that such $s$ exists if $D$ is a 3-ball? 4) What role plays for you that $B$ can be unbounded? –  Dmitri Oct 12 '12 at 23:59
1) $A(\mathbf{B})$ is just the contraction of the 1-form $A$ with the vector field $\mathbf{B}$, $A(B)=\text{i}_{\mathbf{B}} A$. 2+3) I have an example where $D$ is a 3-ball and such an $s$ cannot exist. I've added this above. 4) I am not terribly interested in unbounded $B$. –  Josh Burby Oct 13 '12 at 23:13
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