When can I use $\wedge$ instead of curl?

It seems in some circles the wedge product is used in preference to curl. I have a basic understanding of Green and Stokes' formula, I wish to use the $\wedge$ notation from now on.

Can someone tell me if this is commonly done, and if so what is the underlying assumption of the surface. If it is not too much to ask, can someone show me how to write say Maxwell's equations using $\wedge$ instead of Curl

• AFAIK, in 3D, $\wedge\equiv\times$. – Kyle Kanos Aug 29 '14 at 2:01
• I only use $\wedge$ for $\times$ when I'm writing by hand, to avoid confusing $\times$ with $x$. But to be precise, wedge and cross products truly are distinct operations. One gives you a rank-2 tensor, the other a pseudovector. – Brian Bi Aug 29 '14 at 2:56

The wedge product has its roots in exterior algebra. Exterior algebra lets you talk about objects like planes or volumes as algebraic elements of their own, separate from ordinary vectors, but still obeying the same notions of being "vectors" in their own vector spaces. The wedge product of two vectors is a bivector, and many concepts you may have been taught in vector calculus can be thought of in terms of bivectors instead. Normal vectors are just the unique vectors perpendicular to bivectors that are tangent to some surface instead. Rotations can be thought of as rotating within a bivector, instead of around a rotation axis (and incidentally, rotating within a bivector is a concept that still works in Minkowski space, unlike rotation axes).

Exterior algebra by itself is not enough, though: you need to have a way for vectors and bivectors to interact. There are a couple formalisms that give all the tools to do this:

1. Differential forms. Forms are very common and extensively used in high-level mathematics and for advanced electromagnetism and general relativity. Differential forms contributes the machinery to do calculus on bivectors and the like, using the "exterior derivative" $d$. It also has the concept of Hodge duality, with the Hodge star operator $\star$ that turns bivectors in 3d to their normal vectors and vice versa. This ultimately gives you the power to use forms effectively in metrical contexts: like special and general relativity.

2. Clifford algebra and geometric calculus. Clifford algebra is extensively used in quantum mechanics, through the so-called gamma or pauli matrices. The algebra of these objects, however, is worthy of study independent of the idea that these are matrices using matrix multiplication. In this mindset, you can use clifford algebra for run-of-the-mill 3d vector geometry and calculus. Clifford algebra introduces a "geometric product" of vectors instead, which incorporates both the metric and the exterior algebra into one nifty operation. Many of the core concepts are the same as with differential forms, but the notation is often a little closer to traditional vector calculus in look.

You should use wedge products anytime you're not in 3d space, as the cross product is only used in 3d or 7d and not an immediately generalizable concept.

A traditional differential forms writing of Maxwell's equations might be like this:

$$\star d (\star E) = \rho, \quad dE = -\partial_t B, \quad dB = 0, \quad \star d (\star B) = j + \partial_t E$$

This form takes $B$ as a 2-form (a bi-covector), but it's not particularly common, as forms are more often used in the context of spacetime, where $E$ is also a 2-form and the two fields come together in the Faraday 2-form $F$. Maxwell's equation in vacuum then take on the simpler form

$$d(\star F) = J, \quad dF = 0$$

In geometric calculus, the derivative operator $\nabla$ has the capability to do both divergences and curls in one operation, so the last equation is typically written

$$\nabla F = \nabla \cdot F + \nabla \wedge F = J + 0 = J$$

The relationship between wedges and the cross product, is as follows: In differential forms,

$$a \times b = \star (a \wedge b)$$

While in clifford algebra, we typically keep explicit the unit trivector $i$:

$$a \times b = i^{-1} (a \wedge b)$$

Canonically, the wedge product is distinct from the cross product and should not be confused, however in three dimensions they are inextricably linked.

The wedge product (or outer product) comes from exterior algebra, first due to Grassmann who generalized vector products to arbitrary dimensions. This would later be extended by Clifford into the Clifford Algebra, which would unite the inner and outer products for arbitrary dimensions.

The cross product itself would come from a term in the quaterion's first described by Hamilton. It only exists in three dimensions, and is defined as

$$\vec c = \vec a \times \vec b$$ $$c_i = \epsilon_{ijk} a_j b_k$$

Where $\epsilon$ is the Levi-Cevita symbol, a completely anti-symmetric "tensor"

$$\epsilon_{ijk} = \begin{cases} 1 & \text{ijk an even permutation of 123} \\ -1 & \text{ijk an odd permutation of 123} \\ 0 & \text{otherwise} \end{cases}$$

This defines a vector that is perpendicular to the other two vectors. And there we can immediately see the problem. In dimensions higher than 3 there is no longer a unique direction perpendicular to two vectors, there is an entire subspace.

To get around this difficulty, Grassmann had to change the geometric interpretation of the vector product. His wedge product (or outer product) takes two vectors and returns an object that represents the oriented, scaled plane that the two vectors span, and axiomatically is defined to be antisymmetric.

$$\vec a \wedge \vec b = - \vec b \wedge \vec a$$ From wikimedia

This can naturally be extended to further geometric objects by wedging by still further vectors to build up a fully complete algebra.

Now we can write the correspondence between the cross product and the wedge product:

$$\vec a \times \vec b = \star ( \vec a \wedge \vec b)$$

This is pretty unsatisfactory as an equation, for I've hidden all of the relevant bits into a new symbol: ($\star$), which represents the Hodge dual. You see, one of the consequences of this geometric algebra of Clifford is that you can only wedge things against each other so far, eventually you run out of space to wedge against. Additionally, the dimensionality of any particular rank of objects in this algebra happen to correspond to rows of Pascal's triangle. In particular, in three dimensions we have:

$$1 \quad 3 \quad 3 \quad 1$$

Which is to say, we have 1 linearly independent scalar, 3 independent vectors, 3 independent planes and only 1 independent volume element. For four dimensions you would have

$$1 \quad 4 \quad 6 \quad 4 \quad 1$$

1 scalar, 4 vectors, 6 plane elements, 4 independent (3D) volumes and only 1 4D volume. It is always true that these sequences are symmetric about the middle. This allows us to define an operator: ($\star$) the hodge dual, that maps from an object on the left half of one of these sequences to the right half and vice versa. In particular, this allows us to associate any plane in 3D with a corresponding vector. This is entirely natural, we do this all of the time in 3D, the natural vector associated with a plane is just the vector perpendicular to the plane. In 4D, every vector is associated with a 3D volume element and vice versa.

At this point, we notice that 3 dimensions is quite special. The natural vector product ($\wedge$) takes two vectors and creates a plane element, but in 3D we can always uniquely assign a vector to this (orientated) plane, this associated vector is the cross product. The wedge of two vectors in any higher dimension cannot be uniquely associated with a vector, as I have hopefully made clear.

It all seems nice a tidy, but the fact that the vector generated by the cross product isn't the natural result of vector multiplication leaves a sort of smudge on the vectors created by cross products: they transform differently under reflections. This creates the whole confusion over pseudovectors. The product of vectors is naturally a plane, and planes don't switch sign under reflection, while vectors do. For more details see this answer by ACuriousMind to another question.