# Can the curl operator be generalized to non-3D?

In three dimensions, the curl operator $\newcommand{curl}{\operatorname{curl}}\curl = \vec\nabla\times$ fulfils the equations


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...

• Related question: What are the Eigenvectors of the curl operator? which might be used to construct $\curl$ from its Eigenvectors. Mar 22, 2013 at 15:30
• Do you know about differential forms? All of these strange operators are explained once you know how these work. Mar 22, 2013 at 15:32
• @muzzlator Everytime I re-read about them, for a few minutes I'm convinced I finally understood them. Then those few minutes are over :-/ But that doesn't mean I would oppose to an answer using them Mar 22, 2013 at 15:33

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).

• Even more comes out of this picture. Physicists' pseudo-vectors are precisely vector fields arising from the isomorphism $$\Omega^2(\mathbb{R}^3) \xrightarrow{\ast} \Omega^1(\mathbb{R}^3) \xrightarrow{\sharp} \Gamma(\mathbb{R}^3),$$ where $\ast: \Omega^2(\mathbb{R}^3) \to \Omega^1(\mathbb{R}^3)$ is the Hodge star, and $\sharp : \Omega^1(\mathbb{R}^3) \to \Gamma(\mathbb{R}^3)$ is the musical isomorphism (viz, index raising) defined by the canonical flat Riemannian metric on $\mathbb{R}^3$. Mar 22, 2013 at 21:55
• Similarly, physicists' pseudoscalars are precisely the scalar-valued functions arising from the isomorphism defined by the Hodge star $$\Omega^3(\mathbb{R}^3) \xrightarrow{\ast} \Omega^0(\mathbb{R}^3) = C^\infty(\mathbb{R}^3).$$ The precise reason why pseudovectors (like the magnetic field) and pseudoscalars (like the magnetic flux) flip signs when the orientation is flipped (i.e., an an "improper" rotation $R$ with $\det R = -1$ is applied to the coordinate system) is that the Hodge star itself flips signs when the orientation is flipped. Mar 22, 2013 at 22:01
• @BranimirĆaćić Point of pedantry: you seem to be talking about pseudovector fields and pseudoscalar fields, when the unit pseudoscalar and pseudovectors are objects in unto themselves. I feel this is partly why people don't recognize how the Hodge star is really contraction with the unit pseudoscalar. Mar 22, 2013 at 22:22
• Some prefer to work with the exterior calculus perspective, others with the geometric algebra perspective. De gustibus non disputandum est... Mar 22, 2013 at 22:59
• Indeed, $\Gamma(\mathbb{R}^3)$ is the space of vector fields, and $\Omega^0(\mathbb{R}^3) = C^\infty(\mathbb{R}^3)$ is the space of scalar fields. However, $$\operatorname{curl} = \sharp \circ \ast \circ d_1 \circ \flat, \quad \operatorname{div} = \ast \circ d_2 \circ \ast \circ \flat,$$ where $\flat = \sharp^{-1}$. Moreover, it might also be worth observing that for $\xi$, $\eta \in \Gamma(\mathbb{R}^3)$, $$\xi \times \eta = (\sharp \circ \ast)\left(\flat(\xi) \wedge \flat(\eta)\right).$$ From this perspective, then, cross products are pseudo-vectors because they're $2$-forms in disguise. Mar 25, 2013 at 10:13

In geometric algebra and calculus, the cross product and curl are generalized by the wedge product and exterior derivative.

Let's start with the cross product. Let $$u = a \times b$$. We usually say $$u$$ is orthogonal to both $$a$$ and $$b$$, or normal to the plane spanned by $$a$$ and $$b$$. Geometric algebra works with that plane directly, as a "2-vector" or bivector that we call $$a \wedge b$$. This is something that can be in any number of dimensions, unlike the cross product.

We compute the exterior derivative of a field $$A$$ as $$\nabla \wedge A$$ in the same manner. Together, the exterior derivative with the interior derivative $$\nabla \cdot A$$ constitute the full, geometric derivative $$\nabla A$$. This operator has all of the information needed to reconstruct the original $$A$$ through integration.

Thus, if one knows $$\nabla A$$ and $$\nabla \cdot A$$, then one can always compute the curl through $$\nabla \wedge A = \nabla A - \nabla \cdot A$$.

The relationship to the Laplacian: the exterior derivative and interior derivative obey the following relationships in flat space.

$$\nabla \wedge \nabla \wedge A = \nabla \cdot (\nabla \cdot A) = 0$$

This follows as a consequence of the equality of mixed partial derivatives. This means that the Laplacian can always be written as

$$\nabla (\nabla A) = \nabla^2 A = \nabla \cdot (\nabla \wedge A) + \nabla \wedge (\nabla \cdot A)$$

This is true for both the scalar and vector Laplacian, although when $$A$$ is a scalar field, $$\nabla \cdot A = 0$$ always. The dot should be interpreted as "contracting" or "grade-lowering", and clearly you can't reduce a scalar to something lower-dimensional.

Edit: to build off of what Qiaochu Yuan is saying, let's look at various fields in 3d to understand how this theory captures the usual vector calculus derivatives. Let $$E$$ be a vector field and $$\varphi$$ a scalar field. We usually have the following derivatives:

Divergence: $$\nabla \cdot E$$

Gradient: $$\nabla \varphi$$

Curl: $$\nabla \times E$$

The relationship between curl and the wedge derivative is through Hodge duality. This is denoted by multiplication with a pseudoscalar, called $$i$$. $$i$$ converts vectors to 2-vectors ("pseudovectors") and vice versa. The canonical relationship between the curl and this derivative is

$$\nabla \times E = -i \nabla \wedge E$$

$$\nabla \wedge E$$ is what differential forms people call the curl.

What's interesting to note is when the derivative acts on a 2-vector instead. Let such an object be $$B$$:

$$\nabla \cdot B = \text{vector}, \quad \nabla \wedge B = \text{pseudoscalar (0d vector space)}$$

This latter is what differential forms people tend to call the divergence. To me, this is deceptive: $$\nabla \cdot E$$ is just as equally the divergence.

Edit edit: you mention complex numbers. Let's talk about them and spinors. You can use this definition of the vector derivative $$\nabla$$ and its associated operations to work on spinors also. In GA, spinors are just linear combinations of scalars and 2-vectors. Let $$\psi(z)$$ be such a "spinor" field, or a function of a so-called complex variable. If $$\psi = u + iv$$ is holomorphic, then it obeys

$$\nabla \psi = \nabla u + (\nabla v) i = 0$$

This may seem somewhat backwards from complex analysis, but this is because $$\nabla$$ can be identified with $$\partial/\partial \bar z$$. The benefit here is, of course, that $$\nabla$$ is valid in all dimensions. You shouldn't be afraid of spinors. They're very useful, and we can treat derivatives of them in the same way we treat fields of other kinds. The spinors of 3d are quaternions, and their usefulness for 3d rotations is well-known.

Finally, let's calculate some derivatives of some sample fields. Let $$\varphi$$ be a scalar field, $$E$$ be a vector field, $$B$$ a bivector field, and $$\gamma$$ a pseudoscalar field. Let $$e_x, e_y, e_z$$ be the unit cartesian vectors. Let $$e_{zx} = e_z \wedge e_z$$, and so on. Let $$e^x$$ be the unit cotangent vector perpendicular to the $$yz$$ plane. Note that for Cartesian coordinates, this is equal to $$e_x$$, but vector derivatives always depend on these one-forms, so this is more general. A quick summary:


Interior derivatives: \begin{align*} \nabla \cdot \varphi &= 0 \\ \nabla \cdot E &= \partial_x E^x + \partial_y E^y + \partial_z E^z \\ \nabla \cdot B &= (\partial_z B^{zx} - \partial_y B^{xy}) e_x + (\partial_x B^{xy} - \partial_z B^{yz}) e_y + (\partial_y B^{yz} - \partial_x B^{zx}) e_z \\ \nabla \cdot \gamma &= \partial_x \gamma^{xyz} e_{yz} + \partial_y \gamma^{xyz} e_{zx} + \partial_z \gamma^{xyz} e_{xy}\end{align*}

Exterior derivatives:

\begin{align*} \nabla \wedge \gamma &= 0 \\ \nabla \wedge B &=( \partial_x B^{yz} + \partial_y B^{zx}+ \partial_z B^{xy}) e_{xyz} \\ \nabla \wedge E &= (\partial_z E^y - \partial_y E^z) e_{yz} + (\partial_x E^z - \partial_z E^x) e_{zx} + (\partial_y E^x - \partial_x E^y) e_{xy} \\ \nabla \wedge \varphi &= \partial_x \varphi e^x + \partial_y \varphi e^y + \partial_z \varphi e^z\end{align*}

Edit$$^3$$: let's talk about Hodge duality. The unit pseudoscalar $$i$$ turns dots to wedge and wedges to dots. This is true outside of the calculus, too, in a generalized sense: let $$b$$ be a vector. Then $$b \cdot b = b \wedge (bi) i^{-1}$$. This is how dot products are often calculated in differential forms (I personally find it unduly circuitous). Applying this to the calculus yields various relations. Let $$\varphi, E, B, \gamma$$ be defined as before.

$$\nabla \cdot \gamma = [\nabla \wedge (\gamma i)] i^{-1}$$: this converts the interior derivative of a pseudoscalar field to a gradient of a scalar field, which is then dualized back to a pseudoscalar. This is one way traditional vector calculus converts back and forth between scalars and pseudoscalars, avoiding the difference between the two.

$$\nabla \wedge B = [\nabla \cdot (Bi)] i^{-1}$$: this is traditionally how differential forms treads the divergence, by considering the exterior derivative of a 2-form.

Edit$$^4$$: it may be I breezed right past what is of interest here. Let me translate a few things to make apparent a broader point.

\begin{align*} \text{Vector calculus:} &\curl \grad \psi = 0 \\ \text{Geometric calculus:} & \nabla \wedge \nabla \wedge \psi = 0 \end{align*}

Here, $$\nabla \wedge \psi$$ is the gradient, and the second $$\nabla \wedge$$ on the left plays the role of curl.

\begin{align*} \text{Vector calculus:} &\div \curl E = 0 \\ \text{Geometric calculus:} & \nabla \wedge \nabla \wedge E = 0 \end{align*}

On the other hand, here the divergence and curl both take on the meaning of $$\nabla \wedge$$. This is why differential forms likes to call them all aspects of $$\nabla \wedge$$. But both of these identities for divergence, curl, and gradient have exactly the same expression in both differential forms and geometric calculus.

Again, remembering that $$\nabla \wedge \nabla \wedge X = 0$$ for any kind of $$X$$ is what allows us to write

\begin{align*} \text{Vector calculus:} &\curl \curl E = \grad \div E - \Delta E \\ \text{Geometric calculus:} & -\nabla \cdot (\nabla \wedge E) = \nabla \wedge (\nabla \cdot E) - \nabla^2 E\end{align*}

This is an important point: the second applied curl ends up taking the role of $$\nabla \cdot$$ on the 2-form $$\nabla \wedge E$$. This is why, ultimately, I think your search for a generalized curl specifically may be somewhat misguided, for curl, divergence, and even gradient take on wildly different meanings depending on what they're acting on, while $$\nabla \wedge$$ and $$\nabla \cdot$$ have fixed meanings in geometric calculus as raising (for $$\nabla \wedge$$) or lowering ($$\nabla \cdot$$) the dimensionality of the objects they differentiate. It's for this reason that, in geometric calculus, $$\nabla \cdot$$ is sometimes referred to as the divergence and $$\nabla \wedge$$ is referred to as the curl, but this lazy (yet useful) terminology doesn't always coincide with vector calculus. Again, the reason that the meanings of divergence, curl, and gradient fluctuate in vector calculus is because of the use of Hodge duality to simplify all objects to scalars and vectors.

• I assume your choice of $E$ / $B$ is on purpose, relating them to the electric / magnetic fields in physics? I'm not entirely done reading your answer, but - on the risk of sounding very stupid - is the pseudoscalar $i$ in any way related to complex numbers or the $\partial_t$ of the Maxwell equations? Mar 25, 2013 at 10:30
• @TobiasKienzler Yes, the use of $E$ and $B$ is meant to be suggestive that in 3d, the magnetic field is better described as a 2-form field than as a vector field. Here, $\gamma$ could be a magnetic scalar potential (which we see is really pseudoscalar). The pseudoscalar is often called $i$ because in 2d, $ii = e_{xy} e_{xy} = -1$, so it squares to $-1$, just the way the complex imaginary does. So instead of a purely algebraic object like the complex imaginary, we have a geometric notion of multiplication and a geometric object that plays the same role. Mar 25, 2013 at 14:13
• Thanks for your extended answer, it's very interesting - I wish Theoretical Physics lectures would emphasize a bit more on this, we basically just treated the magnetic field as a usual vector field and probably didn't even mention the pseudo-bit... Mar 26, 2013 at 6:37
• @TobiasKienzler I've added one last section on the translation of your identities for the curl to geometric calculus. I think this bears specifically on your search for a generalized curl. Mar 26, 2013 at 18:25
• Thanks, I think your edit$^4$ summarizes it nicely. But somehow I have contradicting thoughts about this - on the one hand, it's pretty obvious from your answer (plus the deRahm complex mentioned in Qiaochu's answer) that only in 3D $\nabla\wedge$ acting on a vector can be "hodged" back to an actual vector, while in all other dimensions this yields a higher-dimensional object. But on the other hand, I'm not looking for the meaning of $\nabla\wedge$ in other dimensions, but for a operator that, when squared, behaves like the well-defined $\curl^2$... Mar 27, 2013 at 8:15

Based on Qiaochu Yuan's answer and Branimir Ćaćić's comments to it, I propose the following generalizations for $$n\ge2$$ dimensions:


using the path $$\Gamma(\mathbb R^n)\xrightarrow{\flat}\Omega^1(\mathbb R^n)\xrightarrow{d_1}\Omega^2(\mathbb R^n)\xrightarrow{\ast}\Omega^{n-2}(\mathbb R^n)\xrightarrow{d_{n-2}}\Omega^{n-1}(\mathbb R^n)\xrightarrow{\ast}\Omega^1(\mathbb R^n)\xrightarrow{\sharp}\Gamma(\mathbb R^n)$$

I am not yet sure how to express $$\Delta$$ and therefore can't check whether $$\curl^2=\grad\div-\Delta$$. Since $$d_{i+1}\circ d_i = 0$$, one obtains $$\curl^2\grad=0$$ and $$\div\curl^2=0$$, which is necessary (but not sufficient) for the criteria $$\curl\grad=0$$ and $$\div\curl=0$$ from the question. These require

$$\curl = \sharp\circ\ast\circ d_{n-2}\circ A\circ d_1\circ\flat$$ for some $$A$$ which, by comparing $$\curl^2$$, must obey $$A\circ d_1\circ\ast\circ d_{n-2}\circ A = \ast$$ or $$d_1\circ\ast\circ d_{n-2} = A^{-1}\circ\ast\circ A^{-1}.$$ So, can $$A$$ be $$d_1^{-1}$$ and $$d_{n-2}^{-1}$$ at the same time for $$n\neq3$$?

• Let $* d * = \delta : \Omega^n \to \Omega^{n-1}$, the coexterior derivative. Then curl$^2:=\sharp \,\delta\, d\, \flat$. The Laplacian is defined as $\Delta \, := \sharp\, (d + \delta)^2\, \flat$, or $\Delta \, := \sharp\, (d_0\, \delta_1 + \delta_2\, d_1)\, \flat \,:\, \Gamma \to \Gamma$. Interpreting grad $:= \sharp \, d_0 \,\flat$ and div $:= \sharp\,\delta_1\,\flat$, then by definition curl$^2 = \sharp\, (\delta_2\, d_1)\, \flat = \sharp\, (d_0\, \delta_1)\, \flat - \Delta$ = grad div - $\Delta$.
– Kai
Feb 13 at 23:13
• really all I'm saying is that (dropping the musical isomorphisms since they're isomorphisms) $\Delta_1 = d_0 \, \delta_1 + \delta_2 \, d_1$ = grad div + curl curl, if we interpret both $d_1$ and $\delta_2$ as being a "curl".
– Kai
Feb 13 at 23:20