Exterior derivative: Information from Answers.com

In differential geometry, the exterior derivative extends the concept of the differential of a function, which is a form of degree zero, to differential forms of higher degree. Its current form was invented by Élie Cartan.

The exterior derivative d has the property that d² = 0 and is the differential (coboundary) used to define de Rham (and Alexander-Spanier) cohomology on forms. Integration of forms gives a natural homomorphism from the de Rham cohomology to the singular cohomology of a smooth manifold. The theorem of de Rham shows that this map is actually an isomorphism. In this sense, the exterior derivative is the "dual" of the boundary map on singular simplices.

1 Definition
2 Examples
3 Further properties
- 3.1 Closed and exact forms
- 3.2 Naturality
4 The exterior derivative in calculus
5 See also
6 References

Definition

The exterior derivative of a differential form of degree k is a differential form of degree k + 1. There are a variety of equivalent definitions of the exterior derivative.

Exterior derivative of a function

If ƒ is a smooth function, then the exterior derivative of ƒ is its differential of ƒ. That is, dƒ is the unique one form such that for every smooth vector field X,

d f (X) = X (f),

where $X (f)$ is the directional derivative of ƒ in the direction of X. Thus the exterior derivative of a function (or 0-form) is a one-form.

Exterior derivative of a k-form

The exterior derivative is defined to be the unique R-linear mapping from k-forms to (k+1)-forms satisfying the following properties:

dƒ is the differential of ƒ for smooth functions ƒ.
$d\,df = 0$ for any smooth function ƒ.
$d(\alpha\wedge\beta) = d\alpha\wedge\beta + (-1)^{p} \alpha \wedge d\beta$ where α is a p-form. That is to say, d is a derivation of degree 1 on the exterior algebra of differential forms.

The second defining property holds in more generality: in fact, $d\,d\alpha=0$ for any k-form α. This is part of the Poincaré lemma. The third defining property implies as a special case that if ƒ is a function and α a k-form, then

$d(f\alpha) = df\wedge\alpha + fd\alpha$

because functions are forms of degree 0.

Exterior derivative in local coordinates

Alternatively, one can work entirely in a local coordinate system (x₁,…,x_n). First, the coordinate differentials dx₁,…,dx_n form a basic set of one-forms within the coordinate chart. Given a multi-index $I=(i_1, i_2, \dots, i_k)$ with $1\le i_1, i_2, \dots, i_k \le n,$ the exterior derivative of a k-form

$\omega = f_Idx_I=f_{i_1i_2\cdots i_k}dx_{i_1}\wedge dx_{i_2}\wedge\cdots\wedge dx_{i_k}$

where

$dx_I = dx_{i_1}\wedge dx_{i_2}\wedge\cdots\wedge dx_{i_k}$

$f_I = f_{i_1i_2\cdots i_k}$

over Rⁿ is defined as

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

For general k-forms ω = Σ_I f_I dx_I (where the components of the multi-index I run over all the values in {1, ..., n}), the definition of the exterior derivative is extended linearly. Note that whenever $i$ is one of the components of the multi-index $I,$ then $dx_i \wedge dx_I = 0$ (see wedge product).

The definition of the exterior derivative in local coordinates follows from the preceding definition. Indeed, if $\omega = f_I dx_{i_1} \wedge \cdots \wedge dx_{i_k}$ , then

$d{\omega} = d ( f_I dx_{i_1} \wedge \cdots \wedge dx_{i_k} ) = d \left ( f_I \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k} \right)$

$= d(f_I) \wedge (dx_{i_1} \wedge \cdots \wedge dx_{i_k}) + f_I (-1)^{p} d(dx_{i_1} \wedge \cdots \wedge dx_{i_k})$

$= d(f_I) \wedge (dx_{i_1} \wedge \cdots \wedge dx_{i_k})$

$= \frac{\partial(f_I)}{\partial x^i} dx_i \wedge (dx_{i_1} \wedge \cdots \wedge dx_{i_k})$

we have here interpreted f_I as a zero-form, and then applied the properties of the exterior derivative.

Invariant formula

Alternatively, an explicit formula can be given for the exterior derivative of a k-form ω, when paired with k+1 arbitrary smooth vector fields V₀,V₁, …, V_k:

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

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

where $[V i, V j]$ denotes Lie bracket and the hat denotes the omission of that element: $\omega(V_0, \ldots, \hat V_i, \ldots,V_k) = \omega(V_0, \ldots, V_{i-1}, V_{i+1}, \ldots, V_k).$

In particular, for 1-forms we have:

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

Examples

1

Consider $\sigma = u\, dx_1 \wedge dx_2$ over a 1-form basis $dx_1, dx_2, \ldots, dx_n$ . The exterior derivative is:

$d \sigma = \sum_{i=1}^n \frac{\partial u}{\partial x_i} dx_i \wedge dx_1 \wedge dx_2$

$= \sum_{i=3}^n \frac{\partial u}{\partial x_i} dx_i \wedge dx_1 \wedge dx_2$

The last formula follows easily from the properties of the wedge product.

2

For a 1-form $\sigma = u\, dx + v\, dy$ on R² we have, by applying the above formula to each term (consider $x 1 = x$ and $x 2 = y$ in the following sum),

$d \sigma = \left( \sum_{i=1}^2 \frac{\partial u}{\partial x_i} dx_i \wedge dx \right) + \left( \sum_{i=1}^2 \frac{\partial v}{\partial x_i} dx_i \wedge dy \right)$

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

$= 0 -\frac{\partial{u}}{\partial{y}} dx \wedge dy + \frac{\partial{v}}{\partial{x}} dx \wedge dy + 0$

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

Further properties

Closed and exact forms

Main article: Closed and exact forms

Differential forms in the kernel of d are said to be closed forms. The image of d is said to consist of exact forms (cf. exact differentials). Closed and exact forms are related, because of the identity $d\,d\alpha=0$ for any k-form α. This implies that every exact form is closed. The converse is true in contractable regions, by the Poincaré lemma.

Naturality

The exterior derivative is natural. If f: M → N is a smooth map and Ω^k is the contravariant smooth functor that assigns to each manifold the space of k-forms on the manifold, then the following diagram commutes

so d(f*ω) = f*dω, where f* denotes the pullback of f. This follows from that f*ω(·), by definition, is ω(f_*(·)), f_* being the pushforward of f. Thus d is a natural transformation from Ω^k to Ω^k+1.

The exterior derivative in calculus

Most vector calculus operators are special cases of, or have close relationships to, the notion of exterior differentiation.

Gradient

A smooth function f: Rⁿ → R is a 0-form. The exterior derivative of this 0-form is the 1-form

$\mathrm{d}f = \sum_{i=1}^n \frac{\partial f}{\partial x^i}\, \mathrm{d}x^i = \langle \nabla f ,\cdot \rangle.$

That is, the form df acts on any vector field V by outputting, at each point, the scalar product 〈 , 〉 of V with the gradient $\nabla f$ of f.

The 1-form df is a section of the cotangent bundle, that gives a local linear approximation to f in the cotangent space at each point.

Divergence

A vector field V = (v₁, v₂, . . . v_n) on Rⁿ has a corresponding (n-1)-form

$\omega _V = v_1 \; (\mathrm{d}x^2 \wedge \mathrm{d}x^3 \wedge \cdots \wedge \mathrm{d}x^n) - v_2 \; (\mathrm{d}x^1 \wedge \mathrm{d}x^3 \cdots \wedge \mathrm{d}x^n) + \cdots + (-1)^{n-1}v_n \; (\mathrm{d}x^1 \wedge \cdots \wedge \mathrm{d}x^{n-1}).$

(For instance, when n = 3, in three-dimensional space, the 2-form ω_V is locally the scalar triple product with V.) The integral of ω_V over a hypersurface is the flux of V over that hypersurface.

The exterior derivative of this (n-1)-form is the n-form

$\mathrm{d} \omega _V = \operatorname{div}(V) \; (\mathrm{d}x^1 \wedge \mathrm{d}x^2 \wedge \cdots \wedge \mathrm{d}x^n).$

Curl

A vector field V on Rⁿ also has a corresponding 1-form

$\eta_V = v_1 \; \mathrm{d}x^1 + v_2 \; \mathrm{d}x^2 + \cdots + v_n \; \mathrm{d}x^n.$ ,

Locally, η_V is the dot product with V. The integral of η_V along a path is the work done against -V along that path.

When n = 3, in three-dimensional space, the exterior derivative of the 1-form η_V is the 2-form

$\mathrm{d} \eta_V = \omega _{\operatorname{curl}(V)}.$

Invariant formulations of grad, curl, and div

The three operators above can be written in coordinate-free notation as follows:

$\begin{array}{rcccl} && \nabla f &=& \left( {\mathbf d} f \right)^\sharp \\ \operatorname{div}(F) &=& \nabla \cdot F &=& \star {\mathbf d} \left( \star F^\flat \right) \\ \operatorname{curl}(F) &=& \nabla \times F &=& \left[ \star \left( {\mathbf d} F^\flat \right) \right]^\sharp, \\ \end{array}$

where $\star$ is the Hodge star operator and $\flat$ and $\sharp$ are the musical isomorphisms.

References

Flanders, Harley (1989). Differential forms with applications to the physical sciences. New York: Dover Publications. pp. 20. ISBN 0-486-66169-5.
Ramanan, S. (2005). Global calculus. Providence, Rhode Island: American Mathematical Society. pp. 54. ISBN 0-8218-3702-8.
Conlon, Lawrence (2001). Differentiable manifolds. Basel, Switzerland: Birkhäuser. pp. 239. ISBN 0-8176-4134-3.
Darling, R. W. R. (1994). Differential forms and connections. Cambridge, UK: Cambridge University Press. pp. 35. ISBN 0-521-46800-0.

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Exterior derivative

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