## Differential Forms Part 1: Dimensions and Notation

Differential forms is a topic that, in some sense, extends ideas presented in vector calculus with more suggestive notation and geometric intuition into higher dimensions. The distinction may seem small and insignificant especially in the third dimension that we live in but its results and implications are quite elegant and can lead to nice formalization of certain results such as Stokes’ Theorem. Perhaps the greatest theorem in mathematics, within it resides the divergence theorem, Green’s theorem, the fundamental theorem of calculus, Cauchy’s integral formula, and of course the less generalized Stokes’ Theorem. It is presented below but its meaning will become more clear upon completion of this two part post.

$\int_{\partial\Omega}\omega=\int_\Omega&space;\textup{d}\omega$

This area of mathematics studies objects called $\inline&space;k$-forms. In some sense, $\inline&space;k$ is a representation of dimension. A $\inline&space;k$-form is a single point in space or a function with a singular output. $\inline&space;\phi=\frac{x^2}{y}$ is a $\inline&space;0$-form. This is simply all there is to $\inline&space;0$-forms.

A $\inline&space;1$-form represents a one-dimensional structure that usually requires some basis directions to describe it. It can be thought of as similar to a vector. The difference however is in notation. Notice how the same idea is represented by both a vector $\inline&space;v$ and $\inline&space;1$-form $\inline&space;\alpha$.

$v=(3,1,2)=3\hat{i}+\hat{j}+2\hat{k}$

$\alpha=3\textup{d}x+\textup{d}y+2\textup{d}z$

The transformation between vectors to $\inline&space;1$-forms and the other way around are called musical isomorphisms because of the symbols that represent them: $\inline&space;\alpha^\sharp=v$ and $\inline&space;v^\flat=\alpha$. In some sense, both of these indicate the same idea but the latter has slightly more versatility because the coefficients can be functions. Consider the following $\inline&space;1$-form $\inline&space;\beta$.

$\beta=y\textup{d}x+zx\textup{d}y+x^2\textup{d}z$

In fact, this is simply a concise way to describe a vector field. If these don’t convince you of the convenience of this notation, the consideration of $\inline&space;2$-forms will definitely show the merit of this approach.

Predictably, $\inline&space;2$-forms represent some sort of two-dimensional structure like a parallelogram. Parallelograms are defined by the two vectors that make them up or in this case $\inline&space;1$-forms. As a result, the parallelogram made up of the $\inline&space;1$-forms $\inline&space;\alpha$ and $\inline&space;\beta$ will be denoted $\inline&space;\alpha\wedge\beta$. We call $\inline&space;\wedge$ a wedge product. A $\inline&space;2$-form then is simply the wedge of two $\inline&space;1$-forms. We can deduce a few properties about this product.

$\inline&space;\alpha&space;\wedge&space;\alpha=0$ because a line cannot form a parallelogram with itself.

$\inline&space;\alpha&space;\wedge&space;\beta&space;=&space;-\beta&space;\wedge&space;\alpha$ so orientation can be accounted for as well.

$\inline&space;\alpha&space;\wedge&space;(\beta&space;+&space;\gamma)&space;=&space;\alpha&space;\wedge&space;\beta&space;+&space;\alpha&space;\wedge&space;\gamma$ We will simply assume this for now.

Using these properties alone, we can take the wedge product of two random $\inline&space;1$-forms. Let’s try this with the $\inline&space;1$-forms $\inline&space;\alpha=\textup{d}x+\textup{d}y+\textup{d}z$ and $\inline&space;\beta=\textup{d}x-\textup{d}y$. Observe the work below.

$\alpha\wedge\beta=(\textup{d}x+\textup{d}y+\textup{d}z)\wedge({\textup{d}x-\textup{d}y})$

$=\textup{d}x\wedge\textup{d}x-\textup{d}x\wedge\textup{d}y+\textup{d}y\wedge\textup{d}x-\textup{d}y\wedge\textup{d}y+\textup{d}z\wedge\textup{d}x-\textup{d}z\wedge\textup{d}y$

$=0-2\textup{d}x\wedge\textup{d}y-0+\textup{d}z\wedge\textup{d}x+\textup{d}y\wedge\textup{d}z$

$=-2\textup{d}x\wedge\textup{d}y+\textup{d}y\wedge\textup{d}z+\textup{d}z\wedge\textup{d}x$

We see that at the end there the 3 basis planes in $\inline&space;\mathbb{R}^3$ that are formed. To help visualize it better, look at the below image.

To convince you that this is well defined, consider the following. The cross product of the vector forms of $\inline&space;\alpha$ and $\inline&space;\beta$ is $\inline&space;(1,1,-2)$. The cross product is defined as being the vector with the same area as and is perpendicular to the parallelogram formed by the cross of two vectors. Taking the wedge product’s basis planes that formed and converting each to the $\inline&space;1$-form that is perpendicular to it, we get $\inline&space;-2\textup{d}z+\textup{d}x+\textup{d}y$. This is the exact same as the cross product of the vectors.

The reason this version is, in some sense, superior is because the wedge product fundamentally describes planes while cross products are a roundabout way of describing planes taking advantage of some nice properties of the third dimension. In a four-dimensional world, one could not create a single vector that would be perpendicular to a plane. A single vector in that space would be perpendicular to a volume. In that scenario, a form created by a wedge product is superior because it describes fundamentally what the form is made of: basis planes not normal vectors.

This is so important to consider. For example, in classical mechanics, we are taught that torque points perpendicular to the plane of rotation. This is simply not true! There is nothing in that direction. Torque is a two form that is defined in the plane. It is a planar quantity and one should not be crossing the force and radial vector, they should be wedging it. The distinction is subtle but important to really understand what is mathematically occurring on a fundamental level.

Again, this form is superior to just regular planes because the coefficients can also be functions. If the coefficient of one basis plane becomes the coefficient of a Jacobian matrix, it would create the surface described by that Jacobian. Otherwise, one is essentially creating a plane field similar to how a $\inline&space;1$-form creates a vector field.

The last form that will be discussed is the $\inline&space;3$-form. From the earlier discussion, one may be able to predict what a $\inline&space;3$-form would consists of: basis volumes. However, the only basis volume is $\inline&space;\textup{d}x\wedge\textup{d}y\wedge\textup{d}z$. Changing the permutation of the coordinates within the wedge would simply negate it. A volume field in this case would just be a density function across the space.

There are some important conclusions that can be made from these discussions of forms. We saw that all $\inline&space;0$-forms are constructed from numbers or more fundamentally $\inline&space;1$, $\inline&space;1$-forms are constructed from 3 basis vectors, $\inline&space;2$-forms also have 3 basis planes, and $\inline&space;3$-forms have 1 basis volume. In fact, these are simply the binomial coefficients $\inline&space;\binom{n}{k}$ for a cubic because there are 3 dimensions or directions to choose from and each $\inline&space;k$-form has $\inline&space;k$ directions that make it up. The fact that $\inline&space;\binom{2}{1}=\binom{2}{2}$ is what makes it possible for us to describe planes as vectors in our 3-dimensional world because they are simply dual to each other. In differential forms, this concept is called the Hodge dual. The Hodge dual or Hodge star denoted $\inline&space;\star$ maps a $\inline&space;k$-form to an $\inline&space;(n-k)$-form. The following are the relationships in 3 dimensions.

$\star&space;1=&space;\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z$

$\star&space;\mathrm{d}x=&space;\mathrm{d}y\wedge\mathrm{d}z,\star&space;\mathrm{d}y=&space;\mathrm{d}z\wedge\mathrm{d}x,\star&space;\mathrm{d}z=&space;\mathrm{d}x\wedge\mathrm{d}y$

The fact that binomial coefficients are symmetric makes this sort of map possible. Applying the Hodge star twice would simply yield the identity for three dimensional Euclidean metric spaces. In other dimensions and metrics, it gets slightly more complicated but I will not really cover it but it is important to understand that the Hodge star converts a form to one, in some sense, orthogonal to it. In fact, combining this with the wedge product earlier gives us a nice definition for the cross product.

$\star(\alpha\wedge\beta)=(\alpha^\sharp\times\beta^\sharp)^\flat$

A huge limitation with the cross product is that it was limited to 3 dimensions. One cannot take the cross product of three vectors in four dimensions because it is a binary operation but that does not mean a perpendicular vector does not exist in that space. To find it, you would just wedge all three together and create a volume form and take the Hodge dual of that getting back a $\inline&space;1$-form which is the vector you were searching for.

Now that we have established a basis for this way of looking at classical concepts in vector calculus, we can move on to describe differential operators and the almighty Stokes’ theorem. This however will be done in the next post. For now, it is important to digest entirely this idea of forms and how it leads to a more fundamental and holistic description of structures in space and how dimensions are formed.

If you want to know more or see where I learned it from, read the Quick and Dirty Introduction to Exterior Calculus.