Some intuition for the wedge product $dx^\mu\wedge dx^\nu$. What kind of object is this?

Question

In electromagnetism, we encounter a 2-form $F=\frac{1}{2!}F_{\mu\nu}~dx^\mu\wedge dx^\nu$ where $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$, and $dx^\mu\wedge dx^\nu$, called the wedge product, has the property $dx^\mu\wedge dx^\nu=-dx^\nu\wedge dx^\mu$.

Is there a definition of the $\wedge$ operation in $dx^\mu\wedge dx^\nu$ other than the property that a wedge product is antisymmetric? Let me explain a bit. The cross product of two 3-vectors ${\vec A}$ and ${\vec B}$ written as ${\vec A}\times{\vec B}$ is not only antisymmetric but also has a meaning and a way of computing it. ${\vec A}\times{\vec B}$ itself is a vector given by ${\vec A}\times{\vec B}=AB\sin\theta\hat{n}$ where $\theta$ is the acute angle between them and $\hat{n}$ is a vector $\perp^r$ to the plane containing ${\vec A}$ and ${\vec B}$ and determined by right-hand thumb rule.

So my question is there a detailed formula for $dx^\mu\wedge dx^\nu$ and what kind of object is it?

Answer 1

It's 2-form. This is an object into which you can plug 2 vector fields to get a number: $$ dx^\mu\wedge dx^\nu (X,Y)= (X^\mu Y^\nu- X^\nu Y^\mu) $$ If you are doing an integral over a surface $\Omega$, tile the surface with small parallelograms defined by their sides being small displacements vectors $\delta x^\alpha $, $\delta y^\beta$ then each parallegram gives $$ \delta x^\mu \delta y^\nu - \delta x^\nu \delta y^\mu $$ Sum over all paralellograms (and take a limit that their size goes to zero) to get $$ \int_ \Omega dx^\mu\wedge dx^\nu $$

Answer 2

First off I want to say that the $\mathrm{d}x^{\mu}$ is just notation, it simply means 'give me the $\mu$-th component of a vector' and is the dual basis of the cartesian basis. In a formula: $\mathrm{d}x^{\mu}(\mathbf{x}) = x^{\mu}$. The wedge product is a way to create alternating linear maps from these basis components. From this antisymmetry it follows/is defined that a general k-form can be calculated with as determinant (remember swapping rows/columns gives you a minus sign): $$ (\mathrm{d}x_{i_1}\wedge\dots\wedge\mathrm{d}x_{i_k})(\mathbf{v}^{(1)}, \dots, \mathbf{v}^{(k)}) = \begin{vmatrix} v^{(1)}_{i_1} & v^{(2)}_{i_1} & \dots & v^{(k)}_{i_1} \\ v^{(1)}_{i_2} & \ddots & & \vdots \\ \vdots & & \ddots & \vdots \\ v^{(1)}_{i_k} &\dots & \dots & v^{(k)}_{i_k}\end{vmatrix} $$ so a k-form $\omega: \underbrace{V\times V \times \dots\times V}_{k} \to \mathbb{R}$ eats k vectors and gives you a number. Determinants can be thought of as signed volumes, so a two-form is like the signed area of the parallelogram spanned by the two input vectors.

So a two-form would act like $$ \mathrm{d}x_i \wedge\mathrm{d}x_j(\mathbf{u}, \mathbf{w}) = \begin{vmatrix} u_i & w_i \\ u_j & w_j\end{vmatrix} $$

Since you're bringing up the crossproduct I also want to mention that the crossproduct is actually more of an accident of working with $\mathbb{R}^3$. In $\mathbb{R}^3$ the basis two-forms are $\{\mathrm{d}x\wedge\mathbf{d}y, \mathrm{d}x\wedge\mathrm{d}z, \mathrm{d}y\wedge\mathrm{d}z\}$, since all other combinations are covered by the antisymmetry of the wedge product. Therefore this vectorspace has the same dimension as the vector vectorspace formed by $\{\mathbf{e}_x, \mathbf{e}_y, \mathbf{e}_z\}$. Which means that there's a mapping from two-forms to one-form(vectors). So really the signed area is the underlying thing of the crossproduct and not the vector, whose length is equal to the area, we typically associate with it.

Answer 3

First, there are several ways to view $\mathrm{d}x^\mu.$ Here we define it to be the function that takes a vector $\mathbf{v}=v^\mu\mathbf{e}_\mu,$ where $\mathbf{e}_\mu$ is the basis induced by the coordinate system $x^\mu,$ and gives its $\mu$-component: $\mathrm{d}x^\mu(\mathbf{v})=v^\mu$.

Then we define the tensor product $\mathrm{d}x^\mu \otimes \mathrm{d}x^\nu$ as $$ (\mathrm{d}x^\mu \otimes \mathrm{d}x^\nu)(\mathbf{u}, \mathbf{v}) = \mathrm{d}x^\mu(\mathbf{u}) \, \mathrm{d}x^\nu(\mathbf{v}) = u^\mu v^\nu. $$ Thus, $\mathrm{d}x^\mu \otimes \mathrm{d}x^\nu$ is a function taking two vectors.

Finally we define $\mathrm{d}x^\mu \wedge \mathrm{d}x^\nu$ as $$ \mathrm{d}x^\mu \wedge \mathrm{d}x^\nu = \mathrm{d}x^\mu \otimes \mathrm{d}x^\nu - \mathrm{d}x^\nu \otimes \mathrm{d}x^\mu, $$ i.e. $$ (\mathrm{d}x^\mu \wedge \mathrm{d}x^\nu)(\mathbf{u}, \mathbf{v}) = u^\mu v^\nu - u^\nu v^\mu. $$ So with this view of $\mathrm{d}x^\mu$ the wedge product is a function.

If you want a more geometric view, you can think of $\mathrm{d}x^\mu$ as a vector, as the gradient of $x^\mu,$ and of $\mathrm{d}x^\mu \wedge \mathrm{d}x^\nu$ as the parallelogram spanned by $\mathrm{d}x^\mu$ and $\mathrm{d}x^\nu$ together with its size and orientation (a unit normal).

Read more on the Wikipedia page about exterior algebra.