Exterior derivative

This article or section needs to be written in plain English, using plain English that most of our readers can understand. Articles that depend excessively on technical terms accessible only to specialists are useless for our purposes, so writers are admonished to avoid jargon

Let <math>f:M\rightarrow \mathbb{R}</math> be a smooth function on a manifold. The differential (or exterior derivative), <math>df</math>, is a covector field on M defined as follows: for v a tangent vector at a point <math>p</math>

<math> df(v) = D_{v}(f) </math>

i.e., <math>df(v)</math> is the directional derivative of f in the direction v.

Note that if <math>x_1,...,x_n</math> are a local coordinate system for M at p, then <math>dx_1,...,dx_n</math> define a local co-frame near p. Thus, near p, we may write the differential of f as a linear combination:

<math> df = g_1 dx_1 +...+ g_n dx_n </math>

In fact, since <math>dx_i(\frac{\partial}{\partial x_j}) = \delta^i_j</math>, we get that:

<math> df = \frac{\partial f}{\partial x_1} dx_1 + ... + \frac{\partial f}{\partial x_n} dx_n </math>

Exterior derivative of differential forms

If <math>\omega</math> is a differential k-form (i.e., a smooth section of <math>\Lambda^k T^*M</math>), the exterior derivative <math>d\omega</math> is a differential (k+1)-form defined as follows:

If we can write <math>\omega</math> in local coordinates as

<math> f_{i_1\cdots i_k} dx_{i_1}\wedge\cdots\wedge dx_{i_k} </math>

then in this coordinate system, <math>d\omega</math> equals

<math> df_{i_1\cdots i_k}\wedge dx_{i_1}\wedge\cdots\wedge dx_{i_k} </math>

More generally, we define the differential <math>d\omega</math> by extending the above definition by linearity.

Cohomological properties of the differential

The operator d has the important property that <math>d\circ d = 0</math>. This essentially follows from the equality of mixed partial derivatives. The following simplest example illustrates the general proof: Let <math>f(x,y)</math> be a smooth function in two variables. Then

<math> df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy </math>

Thus

<math> d^2 f = \frac{\partial^2 f}{\partial y\partial x}dy\wedge dx + \frac{\partial^2 f}{\partial x\partial y}dx\wedge dy </math>

Since <math>dx\wedge dy = -dy\wedge dx</math>, the equality of mixed partials shows that <math>d^2 f = 0</math>.