BIGpedia - Exterior derivative - Encyclopedia and Dictionary Online

In mathematics, the exterior derivative operator of differential topology, extends the concept of the differential of a function to differential forms of higher degree. It is important in the theory of integration on manifolds, and is the differential used to define de Rham and Alexander-Spanier cohomology. Its current form was invented by Élie Cartan.

Contents

1 Definition

2 Properties

3 Invariant formula

4 Connection with vector calculus

4.1 Gradient
4.2 Curl
4.3 Divergence

5 Examples

6 See also

Definition

The exterior derivative of a differential form of degree k is a differential form of degree k + 1.

For a k-form ω = f_I dx_I over Rⁿ, the definition is as follows:

$d{\omega} = \sum_{i=1}^n \frac{\partial f_I}{\partial x_i} dx_i \wedge dx_I.$

For general k-forms Σ_I f_I dx_I (where the multi-index I runs over all ordered subsets of {1, ..., n} of cardinality k), we just extend linearly. Note that if $i = I$ above then $dx_i \wedge dx_I = 0$ (see wedge product).

Properties

Exterior differentiation satisfies three important properties:

linearity

the wedge product rule (see antiderivation)

$d(\omega \wedge \eta) = d\omega \wedge \eta+(-1)^{{\rm deg\,}\omega}(\omega \wedge d\eta)$

and d² = 0, a formula encoding the equality of mixed partial derivatives, so that always

$d(d\omega)=0 \, \!$

It can be shown that exterior derivative is uniquely determined by these properties and its agreement with the differential on 0-forms (functions).

The kernel of d consists of the closed forms, and the image of the exact forms (cf. exact differentials).

Invariant formula

Given a k-form ω and arbitrary smooth vector fields V₀,V₁, …, V_k we have

$d\omega(V_0,V_1,...V_k)=\sum_i(-1)^i V_i\omega(V_0,...,\hat V_i,...,V_k)$

$+\sum_{i<j}(-1)^{i+j}\omega([V_i,V_j],V_0,...,\hat V_i,...,\hat V_j,...,V_k)$

where $[V_i,V_j]\,\!$ denotes Lie bracket and $\omega(V_0,...,\hat V_i,...,V_k)=\omega(V_0,..., V_{i-1},V_{i+1}...,V_k).$

In particular, for 1-forms we have:

d ω(X, Y) = X (ω(Y)) - Y (ω(X)) - ω([X, Y]).

Connection with vector calculus

The following correspondence reveals about a dozen formulas from vector calculus as merely special cases of the above three rules of exterior differentiation.

Gradient

For a 0-form, that is a smooth function f: Rⁿ→R, we have

$df = \sum_{i=1}^n \frac{\partial f}{\partial x_i}\, dx_i.$

Therefore

$df(V) = \langle \mbox{grad }f,V\rangle,$

where grad f denotes gradient of f and <•, •> is the scalar product.

Curl

For a 1-form $\omega=\sum_{i} f_i\,dx_i$ on R³,

$d \omega=\sum_{i,j}\frac{\partial f_i}{\partial x_j} dx_j\wedge dx_i,$

which restricted to the three-dimensional case $\omega= u\,dx+v\,dy+w\,dz$ is

$d \omega = \left(\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right) dx \wedge dy + \left(\frac{\partial w}{\partial y} - \frac{\partial v}{\partial z} \right) dy \wedge dz + \left(\frac{\partial u}{\partial z} - \frac{\partial w}{\partial x} \right) dz \wedge dx.$

Therefore, for vector field V=[u,v,w] we have $d \omega(U,W)=\langle\mbox{curl}\, V \times U,W\rangle$ where curl V denotes the curl of V, × is the vector product, and <•, •> is the scalar product.

(what are U and W here? this assertion needs clarification - Gauge 23:37, 7 Apr 2005 (UTC))

Divergence

For a 2-form $\omega = \sum_{i,j} h_{i,j}\,dx_i\,dx_j,$

$d \omega = \sum_{i,j,k} \frac{\partial h_{i,j}}{\partial x_k} dx_k \wedge dx_i \wedge dx_j.$

For three dimensions, with $\omega = p\,dy\wedge dz+q\,dz\wedge dx+r\,dx\wedge dy$ we get

$d \omega = \left( \frac{\partial p}{\partial x} + \frac{\partial q}{\partial y} + \frac{\partial r}{\partial z} \right) dx \wedge dy \wedge dz = \mbox{div}V dx \wedge dy \wedge dz,$

where V is a vector field defined by $V = [p, q, r].$

Examples

For a 1-form $\sigma = u\, dx + v\, dy$ on R² we have

$d \sigma = \left(\frac{\partial{v}}{\partial{x}} - \frac{\partial{u}}{\partial{y}}\right) dx \wedge dy$

which is exactly the 2-form being integrated in Green's theorem.