# Differential forms without derivatives?

Recently, I've drawn increasingly attracted to using differential forms for routine calculus computations - for instance, I've come to like the equation $$dy=2x\,dx$$ which clearly states that the rate of change of $$y$$ is $$2x$$ times the rate of change of $$x$$ far better than the equation $$\frac{dy}{dx}=2x$$ which states that "the derivative of $$y$$ with respect to $$x$$" (whatever that means) is $$2x$$. I like the former equation due to the way that the differential form notation easily generalizes to many dimensions (since rates of change can be written as sums of several other rates, if desired), to implicit equations (since no variable is prioritized), and to problems involving multiple variables with relations between them (since differential forms allow for substitution just the same as any other object we do algebra with).

While it's easy enough to explain the notation of differential forms intuitively - and indeed, that is why I like them - it seems very difficult to pin down what they are formally for someone who is unfamiliar with calculus and linear algebra, since the usual definition of "A differential 1-form on a smooth manifold is an a section of the cotangent space of the manifold" rests upon a solid foundation in the calculus of functions $$\mathbb R^n\rightarrow\mathbb R^n$$ - so is pretty useless for explanation!

In particular, suppose I wanted to set up an intuitive framework where we imagine $$y=x^2$$ as defining a parabola that we may freely move upon. It's not so hard to see that if $$x$$ is negative, then increasing $$x$$ would decrease $$y$$ and if $$x$$ is positive, increasing $$x$$ increases $$y$$ - and that as we get further from $$x=0$$, changing $$x$$ by a little by changes $$y$$ by an ever growing proportion - and that we could also, equally well, imagine changing $$y$$ and see what happens to $$x$$. Basically, we have some set of states that we could be in, and we recognize that, if we had some "velocity" and were changing the state in some smooth manner, the quantities $$y$$ and $$x$$ would also be changing at some rates - and the way that these changes responded to a change in state are somehow encapsulated in the symbols $$dy$$ and $$dx$$ - and finally, that these rates turn out to be related if $$y$$ always equals $$x^2$$.

This seems all well and good until you want to define $$dx$$ and $$dy$$ and to try to prove $$dy=2x\,dx$$. This formal side works out fine if you take the usual quotient $$\lim_{h\rightarrow 0}\frac{(y+h)^2 - y^2}{h}$$ and while this might be fine later to argue about a relation, it doesn't get us any closer to understanding what $$dx$$ and $$dy$$ are - and we'd be breaking an inherent symmetry of the differential forms by declaring that our theory only will work if $$y$$ is a function of $$x$$. We could think about parameterized curves on $$y=x^2$$ and define instantaneous velocity, and say that $$dy$$ and $$dx$$ are (linear) rules for assigning rates of change to these velocities, but now our differential forms look very abstract - and we still had to define calculus to define velocity. Maybe we could more explicitly try to think of differential forms as "local approximations of a function up to a linear term", but this seems rather abstract. The most promising idea I can think of would be to search of a way to satisfying describe a tangent space as some sort of space of allowed "velocities" at a point on a curve/surface and then to define differential forms on that space - but I don't have a good sense of how to do this.

Is there a good way to explain differential 1-forms as the fundamental objects of calculus, without relying upon pre-existing calculus knowledge?

• If you only want $1$-forms in Euclidean space, interpret $dx$, $dy$ as orthogonal projections of the unit coordinate vectors to the tangent line. This gives $dy=2x\,dx$. Of course, these are really vectors rather than forms, and standard metric is used to identify the two, but beginners need not know about that. For higher forms you'll need to define exterior forms (multivectors, really) and explain how to orthogonally project them first. Feb 8, 2020 at 7:50
• You can see 1-forms on M as smooth functions on MxM vanishing on the diagonal M, modulo the ideal (in $C^\infty(M\times M)$) generated by products of such functions. Say if f is a function on M, let f1 and f2 denote f on the 1st and on the second M respectively, then f1-f2 corresponds to df, and if g is another function then both g1(f1-f2) and g2(f1-f2) correspond to g df. Feb 8, 2020 at 8:05
• I see some contradictions here. 1. On one hand, you want to connect $y = x^2$ with $dy = 2xdx$, which has to require derivatives in one way or another. On the other, you suggest that you don't want to rely on calculus. 2. You prefer not to prioritize any variable, so more or less you consider some smooth structure on $\mathbb{R}$. At the same time, you don't like the available definition of the tangent bundle (which is what we need to define differential forms in a coordinate-free way). Or maybe 2 is the whole point - you're just seeking for a nice description of tangent bundle? Mar 17, 2021 at 13:35
• @MichałMiśkiewicz (2) is the point - I am asking whether one could define the tangent bundle for someone unfamiliar with calculus - so the question is whether you can start talking about differential forms in a rigorous way (at least, say, 1-forms on manifolds embedded in $\mathbb R^n$) without first defining derivatives of functions $\mathbb R^n\rightarrow \mathbb R^m$ (which I suppose I'm calling "calculus"/"derivatives"). Mar 17, 2021 at 15:54
• @MiloBrandt: I don't know if this is useful to you, but you might find this article— What a Tangent Line is when it isn't a Limit—interesting. The idea is to define the tangent as the "best linear approximation" of a function around a point; this definition is independent of derivatives. Then, you could define $dx$ and $dy$ as changes along the tangent line.
– Joe
Jul 2, 2021 at 20:53

tl; dr: We can make sense of vector fields and differential forms in a fixed coordinate system without calculus. The catch is, we can't say how vector fields and differential forms transform under smooth mappings (such as changes of coordinates) without differentiation.

$$\newcommand{\Reals}{\mathbf{R}}\newcommand{\dd}{\partial}$$If $$U$$ is an open subset of Cartesian $$n$$-space, the tangent bundle of $$U$$ may be defined to be the set $$U \times \Reals^{n}$$ equipped with some structure:

• The topology of the inclusion $$U \times \Reals^{n} \subset \Reals^{n} \times \Reals^{n}$$, for which projection to the first factor is continuous;
• The vector space structure on each fibre $$\{x\} \times \Reals^{n}$$ coming from the identification $$(x, v) \leftrightarrow v \in \Reals^{n}$$.

Incidentally, and this is a warning sign, the cotangent bundle of $$U$$ has precisely the same description as a set-with-structure. Concretely, let $$x = (x^{k})_{k=1}^{n}$$ denote Cartesian coordinates in $$U$$, and introduce standard bases in the fibres: $$(\dd_{k})_{k=1}^{n}$$ for the tangent bundle, and $$(dx^{k})_{k=1}^{n}$$ for the cotangent bundle.

• A vector field in $$U$$ is a mapping $$Y:U \to U \times \Reals^{n}$$ of the form $$Y(x) = (x, y(x))$$ for some mapping $$y = (y^{k}):U \to \Reals^{n}$$. In terms of our standard basis, we write $$Y(x) = \sum_{k} y^{k}(x) \dd_{k}$$.
• A $$1$$-form in $$U$$ is a mapping $$\Omega:U \to U \times \Reals^{n}$$ of the form $$\Omega(x) = (x, \omega(x))$$ for some mapping $$\omega = (\omega_{k}):U \to \Reals^{n}$$. In terms of our standard basis, we write $$\Omega(x) = \sum_{k} \omega_{k}(x)\, dx^{k}$$.

In other words, we can tell students that a $$1$$-form is an expression $$\Omega(x) = \sum_{k} \omega_{k}(x)\, dx^{k}$$, with the understanding that $$dx^{k}$$ (n.b., not $$\mathrm{d}x^{k}$$, i.e., not "a little bit of $$x^{k}$$") is just a new formal variable. We can alternatively tell students that a $$1$$-form is a row-matrix-valued function, while a vector field is a column-matrix-valued function.

The snag is, differential forms and vector fields become useful because of their transformation properties under smooth mappings. The missing ingredient is the behavior of each bundle under a smooth mapping $$f$$ defined on $$U$$, i.e., the induced behavior on the fibres. As the notation for the respective standard frames suggests, providing that ingredient to each bundle is equivalent to differentiation, specifically to the chain rule. We can't explain what that means without the chain rule or some equivalent.

(Using only algebra we can say how vector fields and differential forms transform under polynomial mappings, because we can differentiate polynomials by a formal algebraic process. Unfortunately, the inverse of a locally-invertible polynomial mapping is not a polynomial mapping except in the simplest case of degree-one polynomials.)

If you just want to explain things intuitively, without formal definitions and proofs, then it's certainly possible to treat differentials as the fundamental concept of Calculus. This is what I've done when teaching Applied Calculus courses where rigorous definitions aren't expected. You can see my latest explanation at http://tobybartels.name/MATH-1400/2015SU/summary/ (although after nearly 8 years I now see some things in this that I'll want to change if I teach the course again).

In the non-applied Calculus courses (which I still teach), I emphasize calculations with differentials such as in your question, much more than usual. But since this course requires rigorous definitions (if not proofs), I find that I need to define derivatives of functions before I can define differentials. But even there, the development (which you can follow in Chapter 3 of my one-variable course notes at http://tobybartels.name/calcbook/) runs $$f ' \to \mathrm d y \to \frac { \mathrm d y } { \mathrm d x }$$.

You might also be interested in the argument for the use of differentials in Calculus courses in Putting differentials back into calculus by Tevian Dray and Corinne A. Manogue, about halfway down the page at https://bridge.math.oregonstate.edu/papers/. (But I don't agree with everything they say; I think that differentials are even easier than they make them out to be!)