Differential Forms

We will start by discussing what a differential is. In some ways this is familiar, but in other ways it might not be.

Let us start by revisiting an example in the one-dimensional case. We proceed informally for now. Intuitively, a differential is just a very small change in some function $\varphi$ caused by some very small change in its input. That is, when $\varphi$ is a function evaluated at $x$ and when $dx$ is very small, we have that:

$$\varphi(x + dx) - \varphi(x) \approx d\varphi_x$$

Where the subscript $x$ specifies the point at which we are differentiating. Let us work out a simple example. Suppose we have the function $\varphi(x) = x^3$. It is easy to compute the differential using the binomial theorem. Since $dx \approx 0$ and the squares of small numbers, less than 1, are much smaller than the original number, higher order $dx$ terms are negligible, so:

$$ \begin{align*} \varphi(x + dx) - \varphi(x) &= (x + dx)^3 - x^3\\ &= x^3 + 3x^2\,dx + 3x\,dx^2 + dx^3 - x^3\\ &\approx 3x^2\,dx \end{align*} $$

Therefore $d\varphi_x = 3x^2\,dx$. What it means for a function to be differentiable is that when we do this sort of thing, we always end up with a sensible result like this one for the differential: it is just some number, the derivative $\varphi'(x)$, multiplied by the size of the tiny nudge $dx$, or $d\varphi_x = \varphi'(x)\,dx$

To understand the generalization into higher dimensions, it is crucial to understand a few details:

1. Although the derivative is a function of $x$, the point at which we are differentiating, we consider $d\varphi_x$ to be obtained after fixing a particular x. It is therefore only a function of the little nudge, $dx$.
2. In one-dimension, it is a little bit weird, but imagine that $dx$ is actually a very small nudge vector, rather than a very small nudge amount, and imagine that the derivative at $x$, $\varphi'(x)$, is actually a matrix. In this case, it would just be a 1x1 matrix. $d\varphi_x$ is then a linear map that eats nudge vectors and spits out approximations of the change in $\varphi$ over a nudge of that size. Its matrix representation is the derivative at $x$.
3. For some reason, although in one dimension everyone agrees that the linear map $d\varphi_x$ is a differential and its matrix representation $\varphi'(x)$ is the derivative, for higher dimensions, the "convention" is to flip-flop between calling $d\varphi_x$ the differential or derivative and then agreeing to call the matrix representation the Jacobian. I will not be using this standard convention. I will simply stick to the same terminology that is standard in one dimension and naively applying it to higher dimensions.

With that out of the way, we can finally ask: what does it take to generalize the derivative and differential to higher dimensions? The answer is basically nothing. A computational example should help illustrate this. Consider a function $\varphi: \mathbb{R}^3 \rightarrow \mathbb{R}^2$ defined by:

$$\varphi\begin{pmatrix}x \\ y \\ z \\ \end{pmatrix} = \begin{pmatrix}x^2y\\ y^2 + z\end{pmatrix}$$

We can find the differential by fixing $a = (x, y, z)$ and considering a nudge vector $da = (dx, dy, dx)$, then doing the same thing we did last time.

$$ \begin{align*} \varphi(a + da) - \varphi(a) &= \begin{pmatrix} (x + dx)^2(y + dy) \\ (y + dy)^2 + (z + dz) \end{pmatrix} - \begin{pmatrix}x^2y\\ y^2 + z\end{pmatrix}\\ &= \begin{pmatrix} (x^2 + 2x\,dx + dx^2)(y + dy) \\ y^2 + 2y\,dy + dy^2 + z + dz \end{pmatrix} - \begin{pmatrix}x^2y\\ y^2 + z\end{pmatrix}\\ &= \begin{pmatrix} 2xy\,dx + y\,dx^2 + x^2\,dy + 2x\,dx\,dy + dx^2\,dy \\ 2y\,dy + dy^2 + dz \end{pmatrix}\\ &\approx \begin{pmatrix} 2xy\,dx + x^2\,dy \\ 2y\,dy + dz \end{pmatrix}\\ &= \begin{pmatrix} 2xy & x^2 & 0 \\ 0 & 2y & 1 \end{pmatrix}\begin{pmatrix} dx \\ dy \\ dx \end{pmatrix}\\ &= \varphi'(a)\,da \end{align*} $$

Once again, the same sort of thing happens. How can we make this more formal? [TODO: Add detailed motivation]

Definition 1.1: We define the derivative $\varphi'(a)$ of a function $\varphi: \mathbb{R^n} \rightarrow \mathbb{R^m}$ to be the matrix such that $\varphi(a + h) - \varphi(a) - \varphi'(a)h \in o(\| h \|)$, and the differential $d\varphi_a$ to be the linear map it represents, where: $$o(\| h \|) = \{f: \mathbb{R^n} \rightarrow \mathbb{R^m} : \lim_{{h}\rightarrow{0}} \frac{\| f(h) \|}{\| h \|}\}$$ If such a matrix exists, we say $\varphi$ is differentiable.

Exercise 1.1: Verify that the set $o(\| h \|)$ is closed under addition and scalar multiplication of functions.

Theorem 1.1: If $\varphi: \mathbb{R^n} \rightarrow \mathbb{R^m}$ is differentiable, then the derivative is unique. Proof: Suppose $\varphi$ has the two derivative matrices $A$ and $B$. It follows that: $$\begin{align*} Ah - (\varphi(a + h) - \varphi(a)) \in o(\| h \|)\\ \varphi(a + h) - \varphi(a) - Bh \in o(\| h \|) \end{align*}$$ So, by Exercise 1.1, $Ah - Bh = (A - B)h \in o(\| h \|)$. Let $v \in \mathbb{R^n}$ be a non-zero vector. By the definition of $o(\| h \|)$ and linearity, this means that: $$\begin{align*} 0 &= \lim_{{c}\rightarrow{0}} \frac{\|(B - A)(cv)\|}{\|cv\|}\\ &= \lim_{{c}\rightarrow{0}} \frac{\|c\|\cdot\|(B - A)(v)\|}{\|c\|\cdot\|v\|}\\ &= \lim_{{c}\rightarrow{0}} \frac{\|(B - A)(v)\|}{\|v\|}\\ &= \frac{\|(B - A)(v)\|}{\|v\|} \end{align*}$$ So $\|(B - A)(v)\| = 0$ for all non-zero $v$. This is only possible if $B - A$ is the zero matrix, meaning $A = B$. $\square$