# What is the name for a vector field that is both divergence-free and curl-free?

Consider a smooth vector field $\mathbf u\colon\Omega\to\mathbb R^3$ defined on an open domain $\Omega\subseteq\mathbb R^3$ such that $\mathbf u$ has zero divergence and zero curl on $\Omega$, that is, \begin{align}\nabla\cdot\mathbf u&=0,\\\nabla\times\mathbf u&=0.\end{align} Is there a specific technical name for such a vector field?

Wikipedia calls it a Laplacian vector field, but

1. it does not cite any references, and
2. it asserts that any such vector field is the gradient of a harmonic function, but this is only true if $\Omega$ is simply connected (counterexample: $\big(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2},0\big)$ on the region $x^2+y^2>0$),

so I'm disinclined to trust it. Can anyone provide references supporting this or any other name?

• P.S. The counterexample above is the magnetic field due to an infinite current-carrying wire, in the region outside the wire. – user856 May 24 '13 at 19:00
• Well, even the curl 0 implying that it's a gradient holds only for simply connected regions, right? Or at least the proof I know does. – ronno May 24 '13 at 19:13
• @ronno: Yes, Wikipedia's logic goes like, zero curl implies it's the gradient of a scalar potential, then zero divergence implies the potential is harmonic. But the first implication requires a simply connected domain. I guess they're taking $\Omega=\mathbb R^3$, but it would be nice if they said so. – user856 May 24 '13 at 20:40

Put musical isomorphism aside, I believe what Rahul Narain refers to is just harmonic $1$-form.

In Hodge decomposition for $k$-forms $\omega$: $$\omega =\mathrm{d}\alpha +\delta \beta + \gamma$$ where $\gamma$ is harmonic in that $(\mathrm{d}\delta + \delta\mathrm{d})\omega = 0$, and $\delta = (-1)^{nk+n+1} \star^{n-k+1}\mathrm{d}^{n-k}\star^k$.

In the 3-dimensional case. We have the cochain complex: $$\Lambda^0\ \stackrel{\mathrm{d}^0}{\longrightarrow}\ \Lambda^1 \ \stackrel{\mathrm{d}^1}{\longrightarrow}\ \Lambda^2\ \stackrel{\mathrm{d}^2}{\longrightarrow}\ \Lambda^3.$$ Define $\delta = \mathrm{d}^*_{k}: \Lambda^*_k\to\Lambda^*_{k-1}$ as the adjoint of $\mathrm{d}^{k-1}: \Lambda^{k-1}\to\Lambda^{k}$ with respect to the inner product. We can have somewhat a dual complex: $$\Lambda^*_3\ \stackrel{\mathrm{d}^*_3}{\longrightarrow}\ \Lambda^*_2 \ \stackrel{\mathrm{d}^*_2}{\longrightarrow}\ \Lambda^*_1\ \stackrel{\mathrm{d}^*_1}{\longrightarrow}\ \Lambda^*_0.$$ For a harmonic 1-form $\gamma$: $$(\mathrm{d}\delta + \delta\mathrm{d}) \gamma = (\mathrm{d}^0\mathrm{d}^*_1 + \mathrm{d}^*_2 \mathrm{d}^1)\gamma = 0. \tag{1}$$ Note: $$\mathrm{d}^1 = \nabla \times, \quad\mathrm{d}^*_1 = (-1)\star^{0}\mathrm{d}^{2}\star^1 = -\nabla \cdot$$ (1) is: $$\nabla \times (\nabla \times \gamma) - \nabla (\nabla \cdot \gamma) = 0.\tag{2}$$ We can say a curl-free and a divergence-free vector field is harmonic under musical isomorphism for $$\nabla \times \gamma = 0\;\text{ and } \nabla \cdot \gamma = 0\Longrightarrow\nabla \times (\nabla \times \gamma) - \nabla (\nabla \cdot \gamma) = 0 .$$

I am guessing the wikipedia page was using "Laplacian vector fields" in that (2)'s left side is actually the vector Laplace operator (or Laplace-Beltrami) acting on a vector field.

For references, we use this term a lot in computational geometry, a field which inherits a lot of terminologies from vector calculus, it is like almost a tradition that saying a vector field is harmonic means it is curl-free and divergence-free w/o citing anyone's book.

• "$\nabla \times (\nabla \times \gamma) - \nabla (\nabla \cdot \gamma) = 0 \Longleftrightarrow \nabla \times \gamma = 0\;\text{ and } \nabla \cdot \gamma = 0.$" I don't think that's right. The vector field $(y,0,0)$ has zero vector Laplacian but its curl is nonzero. – user856 May 24 '13 at 22:08
• Nevertheless, I'm very happy to know that the term "harmonic vector field" has an established usage and means exactly what I want. – user856 May 24 '13 at 22:13
• @RahulNarain Thanks for correcting the mistakes. – Shuhao Cao May 24 '13 at 22:27
• In order to make it an equivalence, you need to assert that the vector field does decay at infinity. The example by Rahul does not vanish at infinity, and does not satisfy the equivalence. – shuhalo May 17 '16 at 22:42

In geometric calculus literature (see, for example, Doran and Lasenby), such a function is called monogenic. Monogenic functions are generalizations of complex analytic (or holomorphic) functions. This condition is strictly stronger than being harmonic--all monogenic functions are harmonic, but not all harmonic functions are monogenic.

The term monogenic is not restricted to vector fields, as well; a scalar field with zero gradient would also be referred to as monogenic.

Edit: Phrased in the language of geometric calculus, we define the vector derivative of a vector field $u$ as $\nabla u$, given by
$$\nabla u = \nabla \cdot u + \nabla \wedge u$$
When $\nabla u = 0$, then $\nabla^2 u = \nabla \wedge (\nabla \cdot u) + \nabla \cdot (\nabla \wedge u) = 0$ as well, fulfilling the harmonic condition.