Relation between exterior (second) derivative $d^2=0$ and second derivative in multi-variable calculus.

What does an exterior (second) derivative such as in $d^2=0$ have to do with second derivatives as in single- or multi-variable calculus? Is this a correct start:

Calculus derivatives are good for Taylor expansions (and thus optimization), and curvature.

Exterior derivatives are needed for integration -- and are essential for generalizing the Fundamental Theorem of Calculus (i.e., Stokes theorem).

Good question! Here's a start. The ordinary derivative in one-variable calculus is a Lie derivative along a special vector field on $\mathbb{R}$; in particular, it is not a special case of the exterior derivative. The exterior derivative is instead some kind of "universal derivative": it records all of the information you would need to determine the derivative of a function along any vector field, for example. In particular, unlike the ordinary derivative, the exterior derivative of a function is a different kind of object, namely a $1$-form. Roughly speaking, a $1$-form is "the kind of thing that pairs with a vector field to return a number," so you can see the relationship there to what I said above.

• Thank you for the starters -- but isn't it the opposite? The calculus derivative contains all information, while exterior derivatives are projections that extract interesting info? Think of the second derivative of a function f: R^2 -> R^3 (and the associated area form). I like to think of a derivative as a (multi-)linear functional valued map (with many symmetries). My best intuit is that the reduction (projection) onto the alternating part is exactly what makes a generalization of the fundamental theorem (to Stokes' theorem) natural. How far off is this? I am thinking of the next time I am t – user128171 Feb 12 '14 at 7:17
• What is the relation between 2nd derivative of a 0-form in R and lie derivative of exterier derivative of the same 0-form? Are they somehow related? – tantuni May 25 '18 at 7:52

That $d^2 = 0$ is probably something you were already taught in vector calculus.

For instance, you probably remember that $\nabla \times \nabla \phi = 0$ for $\phi$ a scalar field, or that $\nabla \cdot (\nabla \times E) = 0$ for $E$ a vector field. It's a good exercise to show that both of these can be written as $d^2 f = 0$ and $d^2 E = 0$.

Of course, you know that $\nabla$ can be written in terms of partial derivatives:

$$\nabla = (dx) \frac{\partial}{\partial x} + (dy) \frac{\partial}{\partial y} + dz \frac{\partial}{\partial z}$$

You should also know that the above vector calculus identities rely upon the equality of mixed partial derivatives:

$$\nabla \times \nabla f = \frac{\partial^2 f}{\partial x \partial y} dx \times dy + \frac{\partial^2 f}{\partial y \partial x} dy \times dx + \ldots$$

But since $dx \times dy = -dy \times dx$, the equality of mixed partial derivatives reduces this to zero. The logic is similar for the vector field case.