Exterior derivative

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>.