# relation between differential forms and standard calculus?

I have been studying calculus for a long time, including multivariate calculus.

Now, suddenly, I'm reading about the concept of a "differential form", referring to $dx$ and $dy$ and so forth.

I've been seeing these symbols all the time in formulas like $\int f(x)dx$ and $dy=\frac {\delta f}{\delta x}dx + \frac {\delta f }{\delta y} dy$

My question is: is this term "differential form" an extension/advanced-version of the $dx$'s and $dy$'s that I've been working with, or have I been working with differential forms all this time without knowing it?

Yep, you've been working with differential forms this whole time without knowing it. In fact, the things which you would call vector field were actually differential forms (although vector fields are still a thing). Scalar functions are $0$-forms, the "vector fields" that you would integrate over curves (often written as $f\,dx+g\,dy+h\,dz$) were $1$-forms, the ones you integrated over surfaces were $2$-forms, and the scalar functions you integrated over volumes can be thought of as $3$-forms (recall that the integral of a $3$-form $f\, dx\wedge dy\wedge dz$ defined on a $3$-manifold in $\mathbb R^3$ is equal to the ordinary integral of $f$ over the volume).
Notice how the "vector fields" you integrated over curves are a different object than the "vector fields" you integrated over surfaces; the former is a $1$-form, the latter is a $2$-form. This would be more clear if we lived in a four-dimensional universe, for in that case, a $1$-form would have four components whereas a $2$-form would have ${4\choose 2}=6$ components.
Not only that, but the gradient, curl, and divergence are all special cases of the exterior derivative. For example, let $\omega = f\, dx\wedge dy+g\,dy\wedge dz+h\, dz\wedge dx$ be a $2$-form. Then $$d\omega = df\wedge dx\wedge dy+dg\wedge dy\wedge dz+dh\wedge dz\wedge dx$$ $$= \left(\frac{\partial f}{\partial x}+\frac{\partial g}{\partial y}+\frac{\partial h}{\partial z} \right)dx\wedge dy\wedge dz$$ If you carry out the computation, the exterior derivative of a $1$-form gives the formula for the "curl."
• What still buggs me is that I've always been taught that $dx$ and $dy$ are only "informal" notations, without a rigorous definition. For example $\frac {dy}{dx}$ is a single object, the derivative, and we cannot speak of $dx$ as some separately existing object. Similarly, $\int f(x)dx$ is a single object, and we cannot treat the $dx$ as separate in itself. So when you call $dx$ a "differential form", this suggests to me that it has a rigorous meaning as a separately existing object. – user56834 Jun 5 '17 at 6:28
• So this theory of differential forms must be a fundamentally different way of thinking about calculus from the traditional way derivatives and integrals are defined? For this reason, I find it hard to grasp the exact formal relation between this theory of differential forms, and the more traditional way in which the derivative and integral are defined, namely without specific mention to any formal meaning of $dx$ or $dy$: $\frac {dy}{dx}=\lim_{$\Delta x\to 0}(\frac{f(x+\Delta x)-f(x)}{\Delta x}$. – user56834 Jun 5 '17 at 6:31 • First, note that the$dx$'s in$\frac{dy}{dx}$and$\int f(x)\, dx$are not differential forms. The definitions of derivatives and (scalar) integrals come way before differential forms, in terms of the order in which things are constructed. Derivatives are necessary to define what a smooth manifold is, and integrals of differential forms are defined in terms of ordinary integrals. So, differential forms aren't a "different approach" than the standard definitions in single-variable calculus; the$dx$'s that appear there are just notation. – florence Jun 5 '17 at 6:35 • (1/2) As for what$dx$actually is in the context of differential forms, it is the exterior derivative (the gradient, in this case) of the function$f:\mathbb R^3\to \mathbb R$given by$f(x,y,z)=x$. More formally, if$M$is an$n$dimensional manifold, then a$k$-form$\omega$is a function which assigns (smoothly) to each point in$M$an alternating multilinear$k$-form on the tangent space of$M$at that point. – florence Jun 5 '17 at 6:40 • (2/2) The tangent space of$M$at$x$is the vector space of all tangent vectors to$M$at$x$, and a multilinear$k$-form over an$\mathbb{R}$-vector space$V$is a map$g: V^k\to \mathbb{R}$which is linear in each of its arguments. A multilinear$k$-form is alternating provided that if two of its arguments are the same, then it evaluates to zero.$dx$is a$1\$-form which is, in a sense, "induced" by the map above. – florence Jun 5 '17 at 6:40