# Exterior Derivative

## Main Question or Discussion Point

I have found the exterior derivative very difficult to visualize. Does it have anything to do with the ordinary derivative of a scalar function? What I mean is that the ordinary differentiation is the rate of change of the scalar function with respect to the variable. So the exterior derivative is also the rate of change of something? Or it is entirely something else? It simply doesnt seem to me that way. How can I visualize taking exterior derivative of a 1-form is a 2-form?

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Hurkyl
Staff Emeritus
Gold Member
Forms are functions. You don't visualize them: you visualize what they do.

A k-form is simply a way of measuring k-dimensional surfaces in your manifold.

By Stokes' theorem, $\int_S d\omega = \int_{\delta S} \omega$.

This tells you exactly what the exterior derivative does: dw measures a region by applying w to the boundary.

If you want to picture it locally, then just imagine little tiny regions. For example, let's work in 3-space.

Let w be the 2-form that measures how much of your surface is perpendicular to a certain vector field F. By picturing little tiny spheres, what do you think dw is?

Hint below:

Note that

$$\int_S \omega = \\int_S \vec{F} \cdot \hat{n} \, dA$$

where n is the unit normal to your surface.

Last edited:
mathwonk