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 with $\omega$ 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.

  • 1
    $\begingroup$ Can you elaborate a little on your motivation? -- it's interesting and so am wondering if there is a reference. $\endgroup$ Oct 9, 2012 at 22:07
  • $\begingroup$ I edited the problem statement to give a bit more explanation of the connection between magnetic fields and contact geometry. $\endgroup$
    – Josh Burby
    Oct 10, 2012 at 21:42
  • $\begingroup$ 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? $\endgroup$ Oct 12, 2012 at 23:59
  • $\begingroup$ 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 $B$. $\endgroup$
    – Josh Burby
    Oct 13, 2012 at 23:13

2 Answers 2


I think your question is equivalent to asking whether the gauge choice affects the total helicity. The total helicity changes by $$\int_D H(A+\mathbf{s})-H(A)=\int_D \mathbf{s}\wedge dA = -\int_D d( \mathbf{s}\wedge A) = -\int_{\partial D} (\mathbf{s}\wedge A)$$ by Stokes theorem.

On a manifold without boundary, the total helicity is thus gauge invariant (in particular, if the total helicity is zero it remains zero in all gauges, excluding the helicity density to be non-vanishing everywhere).

On a manifold with boundary, the situation may be richer. Hodge theory on manifolds with boundary suggests that all closed one-form can be expressed as the sum of an exact and a harmonic form tangential to the boundary, i.e $s = d f + \lambda$, where $f$ is a smooth function and $\lambda$ is such that $d\lambda=0$, $\delta \lambda=0$ and $\mathbf{n}\cdot \lambda\big|_{\partial D}=0$. The change in total helicity is then $$ -\int_{\partial D} df\wedge A - \int_{\partial D} \lambda\wedge A = \int_{\partial D} f B - \int_{\partial D}\lambda\wedge A$$

These boundary terms vanish in both of your examples; in the first, $\mathbf{B}$ is tangential to the boundary and $\lambda=0$ (simply connected domain where closed one-forms are exact), in the second, $B$ is tangential, $\lambda= I d\theta + Jd\zeta$ ($I,J$ are numbers) and $A(\theta,\zeta,0) = A(\theta,\zeta,2\pi) = - d\zeta$ is periodic.


Here is another point of view. If $A$ is a contact form, then, by Darboux theorem, there is a local system of coordinates where $A=p dq - dS$ and $dA=dp\wedge dq$ (sometimes referred to as Clebsch coordinates). In this chart, the helicity density reads $H(A)=dS \wedge dq\wedge dp$, but by adding the "gauge term" $dS$, it can be made to vanish entirely $H(A+dS)=0$. Conversely, if you're given Clebsch potentials $dp$ and $dq$, you shift $A$ by a multiple of $dS$ to get a nowhere vanishing helicity.

However, it is important to remember that Darboux theorem is local. There will be topological (homological) obstructions to porting this result to the whole manifold. So, while the helicity density can be made to vanish locally, its integral of the whole manifold may be finite (see previous answer).


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