Manifolds 50 | Example of Exterior Derivative

TL;DR

On a chart U≅R^n, the exterior derivative of a term f times a wedge of coordinate 1-forms is df wedged with the remaining factors, where df=Σ_j (∂f/∂x_j) dx_j.

Briefing Cornell Notes

Briefing

Exterior (Cartan) derivative sends a k-form to a (k+1)-form in a way that behaves like differentiation while respecting the wedge product’s antisymmetry. Locally, on a chart U identified with R^n, a k-form can be written as a sum of component functions times wedge products of coordinate 1-forms. Because the exterior derivative is linear, it suffices to compute it on one term: for a component function f, the exterior derivative produces df wedged with the remaining 1-forms. In coordinates, df becomes a sum of partial derivatives, so the local rule is essentially “take ordinary derivatives of the coefficients and wedge them into the form.” This local formula is the engine behind the key algebraic properties.

Using that chart-level definition, the exterior derivative inherits a product rule for wedge products. For a k-form ω and an ℓ-form η, applying d to ω∧η splits into two terms: d(ω∧η)=dω∧η plus a sign-adjusted term involving ω∧dη. The sign comes from moving dη past ω’s k 1-form factors, using the wedge product’s graded antisymmetry. This mirrors the familiar Leibniz rule from calculus, but the bookkeeping depends on k.

The same local computation proves the “complex property” d∘d=0. Starting with a k-form ω, applying d twice yields a (k+2)-form whose coefficient contains second derivatives. Those second derivatives are symmetric in the indices by Schwartz’s theorem (mixed partials commute for smooth functions), while the wedge part is antisymmetric in the same indices. When the symmetric and antisymmetric contributions are summed over all index placements, the terms cancel pairwise, forcing the result to be zero on each chart—and therefore everywhere.

A concrete example on R^3 ties the abstract machinery to classical vector calculus. Take a 1-form ω=v1 dx1+v2 dx2+v3 dx3, where the coefficients v1,v2,v3 form a vector field V. Computing dω using partial derivatives produces a 2-form whose components match the curl of V. The wedge-product structure automatically enforces the correct antisymmetry, so terms with dx_i∧dx_i vanish and the remaining cross terms combine with the appropriate minus signs. The resulting 2-form corresponds to curl(V) in the standard identification between 2-forms and pseudovectors in three dimensions.

Finally, the familiar identity “div(curl(V))=0” appears as a special case of d∘d=0: in three dimensions, applying exterior derivative twice translates to divergence of curl vanishing. The calculation shows that the higher-dimensional statement is the underlying reason, while the three-dimensional curl/divergence formulas are recovered once the differential forms are interpreted with the correct ordering conventions.

Cornell Notes

The exterior (Cartan) derivative d maps k-forms to (k+1)-forms by differentiating the coefficient functions and wedging the result into the form. On a chart U≅R^n, if a k-form term looks like f times a wedge of coordinate 1-forms, then d(f·(…)) becomes df∧(…), with df expressed as a sum of partial derivatives times dxj. This leads to a wedge-product Leibniz rule: d(ω∧η)=dω∧η+(-1)^k ω∧dη, where the sign (-1)^k comes from moving dη past k one-form factors. A second application gives d(dω)=0 because mixed partial derivatives are symmetric (Schwartz’s theorem) while the wedge part is antisymmetric, so contributions cancel. In R^3, d applied to ω=v1 dx1+v2 dx2+v3 dx3 yields a 2-form whose components match curl(V), and the identity div(curl(V))=0 follows as a special case of d∘d=0.

What is the local coordinate rule for the exterior derivative of a k-form term?

On a chart U identified with R^n, a k-form term can be written as f times a wedge product of coordinate 1-forms. Linearity reduces the computation to one term: d(f·(…)) = df ∧ (…). The differential df is computed like ordinary calculus: df = Σ_{j=1}^n (∂f/∂x_j) dx_j. The resulting object is a (k+1)-form because df contributes one extra 1-form factor via the wedge product.

How does the Leibniz (product) rule for d work with wedge products, and where does the sign come from?

For a k-form ω and an ℓ-form η, the rule is d(ω∧η)=dω∧η+(-1)^k ω∧dη. The first term comes from differentiating the coefficient part of ω. For the second term, dη must be moved past the k one-form factors already present in ω. Because the wedge product is antisymmetric, each swap of one-forms introduces a minus sign; moving past k factors produces the overall factor (-1)^k.

Why does applying d twice always give zero (d∘d=0)?

After two applications, the coefficients involve second derivatives like ∂^2f/(∂x_i∂x_j). For smooth functions, Schwartz’s theorem makes these second derivatives symmetric in i and j. But the wedge part of a (k+2)-form is antisymmetric in i and j. When the symmetric coefficient multiplies an antisymmetric wedge basis and the sum over index placements is carried out, the plus-signed and minus-signed contributions cancel, forcing d(dω)=0 on the chart, hence everywhere.

How does d act on a 1-form in R^3 of the form ω=v1 dx1+v2 dx2+v3 dx3?

Using the coordinate rule, dω is built from partial derivatives of v1,v2,v3. Terms where the same coordinate 1-form wedges with itself vanish because dx_i∧dx_i=0. The remaining cross terms combine with signs determined by the antisymmetry of the wedge product, yielding a 2-form whose coefficients match the components of curl(V). Concretely, the coefficient of dx2∧dx3 involves ∂v3/∂x2−∂v2/∂x3, and similarly for the other pairs.

How does the identity div(curl(V))=0 relate to d∘d=0?

In three dimensions, the curl of a vector field corresponds to a 2-form obtained from dω, and taking divergence corresponds to applying d again in the appropriate identification between forms and vector calculus operators. Since d∘d=0 holds generally, the three-dimensional statement div(curl(V))=0 is recovered as a special case of the exterior-derivative complex property.

Review Questions

What is the coordinate expression for df in terms of partial derivatives and dx_j, and how does it increase the degree of a form?
In the formula d(ω∧η)=dω∧η+(-1)^k ω∧dη, why exactly does the exponent k appear in the sign?
In the proof that d∘d=0, which part is symmetric (Schwartz’s theorem) and which part is antisymmetric (wedge product), and how does that force cancellation?

Key Points

1
On a chart U≅R^n, the exterior derivative of a term f times a wedge of coordinate 1-forms is df wedged with the remaining factors, where df=Σ_j (∂f/∂x_j) dx_j.
2
The exterior derivative is linear, so computations reduce to checking it on one coefficient function term at a time.
3
The wedge-product Leibniz rule is d(ω∧η)=dω∧η+(-1)^k ω∧dη for a k-form ω, with the sign determined by moving dη past k one-form factors.
4
The complex property d∘d=0 follows because mixed partial derivatives are symmetric (Schwartz’s theorem) while the wedge basis is antisymmetric, making terms cancel.
5
In R^3, applying d to ω=v1 dx1+v2 dx2+v3 dx3 produces a 2-form whose components match curl(V).
6
The familiar vector identity div(curl(V))=0 is a three-dimensional instance of the general rule d∘d=0, once the form/vector identifications are used.

Highlights

On a chart, d acts like “differentiate the coefficients and wedge in the result,” turning k-forms into (k+1)-forms.

The product rule for wedge products includes a graded sign: d(ω∧η)=dω∧η+(-1)^k ω∧dη.

The cancellation behind d∘d=0 comes from symmetry of second derivatives versus antisymmetry of wedge products.

For ω=v1 dx1+v2 dx2+v3 dx3 on R^3, dω reproduces curl(V) as a 2-form.

div(curl(V))=0 drops out as a special case of the universal identity d∘d=0.

Topics

Exterior Derivative
Wedge Product
Leibniz Rule
Complex Property
Curl in R3