# Exterior Algebra I: Revenge of the Line Integrals

## Introduction

Note: this post assumes some knowledge of multivariable calculus and linear algebra

If you’ve spent a reasonable amount of time learning math, you’ve seen them. The $\mathrm{d}$ in $\frac{\mathrm{d}f}{\mathrm{d}x}$. Actually, the $\mathrm{d}$ in integrals, too. You probably asked your teacher what $\mathrm{d}$ actually meant, and you were probably brushed off with an excuse about it representing an infinitesimal number. Yet, it seemed that when it came down to algebraically manipulating these symbols, you were expected to simply accept that such an algebra exists, without anything like a rigorous proof given. Later, in multivariable calculus, you probably saw something like this:

$$\vec{F} = P(x, y, z) \hat{\imath} + Q(x, y, z) \hat{\jmath} + R(x, y, z) \hat{k}$$ $$I = \int_C (P(\vec{r}(t)) \frac{\partial r_x}{\partial t} + Q(\vec{r}(t)) \frac{\partial r_y}{\partial t} + R(\vec{r}(t)) \frac{\partial r_z}{\partial t}) \, \mathrm{d}t$$

And then the teacher “cancelled out” the derivatives:

$$I = \int_C P(\vec{r}) \, \mathrm{d}r_x + Q(\vec{r}) \, \mathrm{d}r_y + R(\vec{r}) \, \mathrm{d}r_z$$

Or, to be less pedantic:

$$I = \int_C P \, \mathrm{d}x + Q \, \mathrm{d}y + R \, \mathrm{d}z$$

As it turns out, there is a theoretically consistent way to express the algebra of $\mathrm{d}x$-type expressions. It’s known as exterior algebra.

## 1-forms

The expression shown before ($P \, \mathrm{d}x + Q \, \mathrm{d}y + R \, \mathrm{d}z$) is known as a “differential 1-form”. You can imagine that $\mathrm{d}x$ is actually a function $\mathrm{d}$ acting on a vector field $\vec{r}(x, y, z) = x$, creating an infinitesimal vector in the $\hat{\imath}$ direction. Thus, a differential 1-form is kind of like a vector field, in that they carry essentially the same information. We now define an operator:

For some vector field $$\vec{F}(x_1, x_2, \dots, x_k) = P_1 \hat{x_1} + P_2 \hat{x_2} + \dots + P_k \hat{x_k}$$ there exists a differential 1-form $\widetilde{F}$ such that $$\widetilde{F} = P_1\,\mathrm{d}x_1 + P_2\,\mathrm{d}x_2 + \dots + P_k\,\mathrm{d}x_k$$

In $\mathbb{R}^n$, there are $n$ basis 1-forms, i.e.: $\mathrm{d}x_1, \dots, \mathrm{d}x_n$

Since there are 1-forms, you might wonder what a 0-form is. They are just scalar functions on a manifold, e.g.: $P_1 : \mathbb{R}^n \to \mathbb{R}$ is a 0-form.

The “degree” of a $p$-form $\alpha$ is $p$, and is denoted $|\alpha| = p$

## 2-forms and $n$-forms

To create 2-forms, we introduce the notion of an exterior product. This product is denoted by $\wedge$, and is fittingly known as the wedge product.

$$(\wedge) : \mathrm{Form}^p \to \mathrm{Form}^q \to \mathrm{Form}^{p + q}$$

The wedge product conforms to most of the properties of multiplication, except it is not commutative:

$$\alpha \wedge (\beta \wedge \gamma) = (\alpha \wedge \beta) \wedge \gamma = \alpha \wedge \beta \wedge \gamma$$ $$\alpha \wedge (\beta + \gamma) = (\alpha \wedge \beta) + (\alpha \wedge \gamma)$$ $$\alpha \wedge \beta = (-1)^{|\alpha||\beta|} (\beta \wedge \alpha)$$

In $\mathbb{R}^n$ we have $k$ different basis $p$-forms, where $k = C(p, n)$ and $C(n, r)$ is the binomial function.

In $\mathbb{R}^2$, this means that there is one basis 0-form (the number 1), two basis 1-forms ($\mathrm{d}x$ and $\mathrm{d}y$), and one basis 2-form ($\mathrm{d}x \wedge \mathrm{d}y$)

In $\mathbb{R}^3$, there is one basis 0-form (1), three basis 1-forms ($\mathrm{d}x$, $\mathrm{d}y$ and $\mathrm{d}z$), three basis 2-forms ($\mathrm{d}y \wedge \mathrm{d}z$, $\mathrm{d}z \wedge \mathrm{d}x$, $\mathrm{d}x \wedge \mathrm{d}y$), and one basis 3-form ($\mathrm{d}x \wedge \mathrm{d}y \wedge \mathrm{d}z$).

In $\mathbb{R}^n$ we define $$\mathrm{d}V = \mathrm{d}x_1 \wedge \mathrm{d}x_2 \wedge \dots \wedge \mathrm{d}x_n$$

Multiplying a scalar by this “volume form” creates a “pseudoscalar”.

## Exterior derivatives

Now, let’s unpack the algebraic properties of the $\mathrm{d}$ operator. Here are its properties:

$$\forall p \in \mathbb{N}, \forall \alpha, \beta \in \mathrm{Form}^p, \forall f : \mathbb{R}^n \to \mathbb{R}$$ $$\mathrm{d}(\mathrm{d}(\alpha)) = 0$$ $$\mathrm{d}(f) = \widetilde{\nabla f}$$ $$\mathrm{d}(a f + b g) = a (\mathrm{d}(f)) + b (\mathrm{d}(g))$$ $$\mathrm{d}(\alpha \wedge \beta) = (\mathrm{d}(\alpha) \wedge \beta) + ((-1)^{|\alpha|} (\alpha \wedge \mathrm{d}(\beta)))$$

Also, we will introduce some terminology: a form $\alpha$ is closed if $\mathrm{d}(\alpha) = 0$, and a closed form $\alpha$ is exact if there exists some $\beta$ such that $\alpha = \mathrm{d}(\beta)$. Note the similarity between the idea of an exact 1-form and a conservative vector field.

## Dot products

We define the dot product on differential forms in a fairly natural way. Firstly, the dot product of a $p$-form and a $q$-form where $p \neq q$ is zero. Secondly, assuming we have two $p$-forms that are the linear combination of basis $k$-forms and 0-forms, the dot product is as follows: $$\alpha = P_1 \mu_1 + P_2 \mu_2 + \dots$$ $$\beta = Q_1 \mu_1 + Q_2 \mu_2 + \dots$$ $$\alpha \cdot \beta = P_1 Q_1 + P_2 Q_2 + \dots$$ Where $\mu_i$ are the basis $k$-forms.

## Hodge dual

The Hodge dual is the last operator in exterior algebra we will consider for now. It is a unary operator denoted by a $\star$.

In $\mathbb{R}^n$: $$(\star) : \mathrm{Form}^p \to \mathrm{Form}^{n - p}$$

The Hodge dual is an involution, and is linear:

$$\star \star \alpha = \alpha$$ $$\star (a \alpha + b \beta) = a (\star \alpha) + b (\star \beta)$$

The Hodge dual is defined by the dot product as follows:

$$\star (\alpha \wedge (\star \beta)) = \alpha \cdot \beta$$

I will let you work out the consequences of this operator yourself, but here are some examples, in $\mathbb{R}^3$:

$$\star \mathrm{d}x = \mathrm{d}y \wedge \mathrm{d}z$$ $$\star \mathrm{d}y = \mathrm{d}z \wedge \mathrm{d}x$$ $$\star \mathrm{d}z = \mathrm{d}x \wedge \mathrm{d}y$$ $$K = A\,(\mathrm{d}y \wedge \mathrm{d}z) + B\,(\mathrm{d}z \wedge \mathrm{d}x) + C\,(\mathrm{d}x \wedge \mathrm{d}y)$$ $$\star K = A\,\mathrm{d}x + B\,\mathrm{d}y + C\,\mathrm{d}z$$

## Some motivation

We will now examine the properties of this basic algebra in more detail. First, let’s try taking the wedge product of two arbitrary 1-forms in $\mathbb{R}^3$

$$M = (P\,\mathrm{d}x + Q\,\mathrm{d}y + R\,\mathrm{d}z) \wedge (A\,\mathrm{d}x + B\,\mathrm{d}y + C\,\mathrm{d}z)$$

We distribute: $$M = J + K + L$$ $$J = (P\,\mathrm{d}x + Q\,\mathrm{d}y + R\,\mathrm{d}z) \wedge (A\,\mathrm{d}x)$$ $$K = (P\,\mathrm{d}x + Q\,\mathrm{d}y + R\,\mathrm{d}z) \wedge (B\,\mathrm{d}y)$$ $$L = (P\,\mathrm{d}x + Q\,\mathrm{d}y + R\,\mathrm{d}z) \wedge (C\,\mathrm{d}z)$$

We distribute again and remove constant factors: $$J = P A\,(\mathrm{d}x \wedge \mathrm{d}x) + Q A\,(\mathrm{d}y \wedge \mathrm{d}x) + R A\,(\mathrm{d}z \wedge \mathrm{d}x)$$ $$K = P B\,(\mathrm{d}x \wedge \mathrm{d}y) + Q B\,(\mathrm{d}y \wedge \mathrm{d}y) + R B\,(\mathrm{d}z \wedge \mathrm{d}y)$$ $$L = P C\,(\mathrm{d}x \wedge \mathrm{d}z) + Q C\,(\mathrm{d}y \wedge \mathrm{d}z) + R C\,(\mathrm{d}z \wedge \mathrm{d}z)$$

We combine like terms, flipping wedges when necessary $$J = P A\,(\mathrm{d}x \wedge \mathrm{d}x) + (Q C - R B)\,(\mathrm{d}y \wedge \mathrm{d}z)$$ $$K = Q B\,(\mathrm{d}y \wedge \mathrm{d}y) + (R A - P C)\,(\mathrm{d}z \wedge \mathrm{d}x)$$ $$L = R C\,(\mathrm{d}z \wedge \mathrm{d}z) + (Q A - P B)\,(\mathrm{d}x \wedge \mathrm{d}y)$$

Since $\alpha \wedge \alpha = -\alpha \wedge \alpha$ if $\alpha$ is a 1-form, we can conclude that $\alpha \wedge \alpha = 0$ $$J = (Q C - R B)\,(\mathrm{d}y \wedge \mathrm{d}z)$$ $$K = (R A - P C)\,(\mathrm{d}z \wedge \mathrm{d}x)$$ $$L = (Q A - P B)\,(\mathrm{d}x \wedge \mathrm{d}y)$$

$$\therefore M = (Q C - R B)\,(\mathrm{d}y \wedge \mathrm{d}z) + (R A - P C)\,(\mathrm{d}z \wedge \mathrm{d}x) + (Q A - P B)\,(\mathrm{d}x \wedge \mathrm{d}y)$$

Now let’s calculate $\star M$ $$\star M = (Q C - R B)\,\mathrm{d}x + (R A - P C)\,\mathrm{d}y + (Q A - P B)\,\mathrm{d}z$$

Let’s define two vector fields, $\vec{F}$ and $\vec{G}$ such that: $$\widetilde{F} = P\,\mathrm{d}x + Q\,\mathrm{d}y + R\,\mathrm{d}z$$ $$\widetilde{G} = A\,\mathrm{d}x + B\,\mathrm{d}y + C\,\mathrm{d}z$$

Rephrasing $\star M$ in this light, we get: $$\star M = (F_y G_z - F_z G_y)\,\mathrm{d}x + (F_z G_x - F_x G_z)\,\mathrm{d}y + (F_y G_x - F_x G_y)\,\mathrm{d}z$$

If we define a third vector field $\vec{H} = \vec{F} \times \vec{G}$, we find that $\widetilde{H} = \star M$

We just discovered the cross product! This shows the generality of these methods, and how the classical tools of vector calculus simply fall out as consequences of exterior algebra.

Hopefully next time we can see some more interesting consequences of exterior algebra, including an application to electrodynamics.