In three dimensions, the curl operator $\newcommand{curl}{\operatorname{curl}}\curl = \vec\nabla\times$ fulfils the equations
$$\curl^2 = \newcommand{grad}{\operatorname{grad}}\renewcommand{div}{\operatorname{div}}\grad\div-\Delta,\\
\curl\grad = 0,\\
\div\curl = 0 $$
where $\Delta$ denotes the (vector) Laplacian $\nabla^2$. Since none of these equations requires the cross-product, which is only defined in three dimensions, can they be used to generalize the curl operator to an arbitrary $d$-dimensional space?
I know taking the root of an operator is not exactly a funny thing, but Dirac has managed that before, even if it led to requiring anticommuting Grassman numbers and spinors…
So my questions are:
- In which dimensions $d$ is $\curl :=+\sqrt{\grad\div-\Delta}$ uniquely defined (by the additional constraints mentioned above, or maybe including other properties of the 3D $\curl$)?
- In which of these dimensions does this work for simple complex numbers without requiring the introduction of spinors?
- How to actually calculate it?
- Bonus question: Use this generalization to generalize the cross product
Note to answerers: I declared the operators \curl, \grad and \div for convenience, they should work everywhere below the question.
As an explicit example, observe $n=2$:
Claiming that $\curl = \begin{pmatrix}a&b\\c&d\end{pmatrix}$, use $\curl\grad=0$ and $\div\curl=0$ to obtain
$$\curl = \alpha\begin{pmatrix}\partial_y^2 & -\partial_x\partial_y \\ -\partial_x\partial_y & \partial_x^2\end{pmatrix}$$
the square of which is
$$\curl^2=\alpha^2\begin{pmatrix}\partial_y^2\Delta & -\partial_x\partial_y\Delta\\ -\partial_x\partial_y\Delta & \partial_x^2\Delta\end{pmatrix}
\stackrel!= \begin{pmatrix}-\partial_y^2 & \partial_x\partial_y\\\partial_x\partial_y & -\partial_x^2\end{pmatrix} = \grad\div-\Delta$$
So $\alpha^2 \stackrel!= -\Delta^{-1}$ and formally
$$\curl = \begin{pmatrix}\partial_y^2 & -\partial_x\partial_y\\ -\partial_x\partial_y & \partial_x^2\end{pmatrix}\otimes(\sqrt{-\Delta})^{-1}$$
$\sqrt\Delta$ requires Spinors, as feared, and I assume they will arise for all even dimensions. One interesting question is, are usual vectors enough in the odd dimensional extension? But doing this manually with even just a $5\times5$ matrix would be a bit too tedious…
Best Answer
div, grad, and curl are secretly just the three exterior derivatives in $\mathbb{R}^3$. Said another way, they are the three nontrivial differentials in the de Rham complex
$$0 \to \Omega^0(\mathbb{R}^3) \xrightarrow{d_0} \Omega^1(\mathbb{R}^3) \xrightarrow{d_1} \Omega^2(\mathbb{R}^3) \xrightarrow{d_2} \Omega^3(\mathbb{R}^3) \to 0$$
of $\mathbb{R}^3$. In particular, grad is secretly $d_0$, curl is secretly $d_1$, and div is secretly $d_2$. I say "secretly" because there is some additional funny business going on involving the Hodge star and the musical isomorphisms.
So in $n$ dimensions there are $n$ generalizations of div, grad, and curl $d_0, ... d_{n-1}$ which satisfy $d_{i+1} \circ d_i = 0$ (the defining equation of a chain complex).