I came across one very interesting, but old topic Anti-curl operator. I do not know the completeness of the solution of this problem, but I was interested in some aspect of it: is it possible to create an anti-curl operator for a vector field of arbitrary dimension?

I'll make a short note. Let's say we have an equation with $\text{Curl}$:


$\textbf{V}$ is known, and need to find $\textbf{F}$.

In practice, special cases of $\text{Curl}$ for 2- and 3-dimensional vector fields are common. In both cases, a system of partial differential equations is compiled, the structure of which shows that $\textbf{F}$ is far from unique. Therefore, arbitrary boundary/initial conditions are used.

$\text{Curl}$ of a vector field with dimension $N>3$ is a tensor, and the system of equations compiled on its basis will obviously be strongly overdetermined.

Question: Will the formulas from the Helmholtz theorem work for $N$-dimensional vector fields? Are there any developments on the issue under consideration? Are there any nuances associated with formula for high dimensions?


In the structure of the tensor, after $\text{Curl}$ is applied to a vector field $\textbf{F}$, there are (I don't know what to call them correctly) "symmetric" pairs of equations. This can help in solving a reduced-dimensional PDE system. Below - example for 4-dimensional vector field $\textbf{F}$. Circles mark symmetric pairs of equations.

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1 Answer 1


$ \newcommand\R{\mathbb R} \newcommand\dd{\mathrm d} $

I think what you're looking for is given by geometric calculus. This is essentially calculus done over the Clifford algebra associated to the standard inner product of $\R^N$. We can formulate your "$N$-dimensional curl" as $\nabla\wedge F$ for $F$ a vector field, and together with the divergence-free condition $\nabla\cdot F = 0$ the turn into one equation $$ \nabla F = \nabla\cdot F + \nabla\wedge F = 0 + V = V $$ where $\nabla\wedge F = V$ is what's called a bivector. The LHS is the geometric or Clifford product of $\nabla$ with $F$. This is advantageous because the operator $F \mapsto \nabla F$ has a Green's function; we can directly solve to get $$ F(x) = \frac1{S_N}\int_{\partial R}|\dd^{N-1}x'|\,\frac{x'-x}{|x'-x|^N}\hat n(x')F(x') - \frac1{S_N}\int_R|\dd^Nx'|\,\frac{x'-x}{|x'-x|^N}V(x') $$ where $R$ is some region of space with non-zero volume containing $x$, $|\dd^{N-1}x'|$ is the surface area measure on the boundary $\partial R$, $|dd^Nx'|$ is the volume measure on $R$, $S_N$ is the surface area of a unit $N$-ball (i.e. the $(N-1)$-volume of an $(N-1)$-sphere), $n(x')$ is the outward-pointing unit normal to $\partial R$ at $x'$, and $x'$ is the variable of integration. The first term is a boundary term, and requires specifying boundary conditions on $F$.

The above formula is (apart from some differences in notation) equation 4.16 of section 7-4 in Clifford Algebra to Geometric Calculus (1984) by David Hestenes and Garret Sobczyk. I will warn you that this book is difficult to read. I would instead recommend Geometric Algebra for Physicists (2003) by Chris Doran and Anthony Lasenby as an introduction geometric algebra (a particular way of thinking about Clifford algebras) and geometric calculus, though I believe the above equation is stated only for the (important) case $\nabla F = 0$ (i.e. zero divergence and zero curl).

  • $\begingroup$ Thank you for your answer and links to literature! I would check with interest the formula that is proposed here, but I do not know how to calculate such integrals. Could you show how this is done in Mathematica for example for V = {x, -y, z} $\endgroup$
    – ayr
    Commented Nov 27, 2022 at 6:21
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    $\begingroup$ @dtn I don't know Mathematica. All you need is a way to do the Clifford products, and then the integrals are just the usual volume and surface integrals. A quick search gave me this package. Alternatively, the first integrand is a product of three vectors $XnF$; since the result has to be a vector this can be expanded as $$ (X\cdot n)F + (n\cdot F)X - (X\cdot F)n. $$ If the bivector $V = v_1\wedge v_2$ for vectors $v_1, v_2$ then the second integrand $XV$ can be expanded as $$ (X\cdot v_1)v_2 - (X\cdot v_2)v_1. $$ (cont.) $\endgroup$ Commented Nov 27, 2022 at 8:18
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    $\begingroup$ (cont.) In this case $V$ can be thought of as representing the plane spanned by $v_1, v_2$; however, not all bivectors are of this form. $\endgroup$ Commented Nov 27, 2022 at 8:21

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