Exterior Derivative

Consider a Differential k-Form

$$\omega^1 = b_1\,dx_1+b_2\,dx_2.$$ (1)

Then its exterior derivative is

$$d\omega^1 = db_1\wedge dx_1+db_2\wedge dx_2,$$ (2)

where $\wedge$ is the Wedge Product. Similarly, consider

$$\omega^1 =b_1(x_1,x_2)\,dx_1+b_2(x_1,x_2)\,dx_2.$$ (3)

Then

$$d\omega^1 = db_1\wedge dx_1+db_2\wedge dx_2$$

$$= \left({{\partial b_1\over\partial x_1}dx_1+{\partial b_1\over\par... ...2\over\partial x_1}dx_1+{\partial b_2\over\partial x_2}dx_2}\right)\wedge dx_2.$$ (4)

Denote the exterior derivative by

$$Dt\equiv {\partial \over \partial x} \wedge t.$$ (5)

Then for a 0-form $t$,

$$(Dt)_\mu \equiv {\partial t\over \partial x^\mu},$$ (6)

for a 1-form $t$,

$$(Dt)_{\mu\nu} \equiv {1\over 2}\left({{\partial t_\nu\over \partial x^\mu} - {\partial t_\mu\over \partial x^\nu}}\right),$$ (7)

and for a 2-form $t$,

$$(Dt)_{ijk} \equiv {\textstyle{1\over 3}}\epsilon_{ijk}\left(... ...ver \partial x^2}+{\partial t_{12}\over \partial x^3}}\right),$$ (8)

where $\epsilon_{ijk}$ is the Permutation Tensor.

The second exterior derivative is

$$D^2t = {\partial\over\partial x}\wedge \left({{\partial\over... ...\partial x}\wedge {\partial\over\partial x}}\right)\wedge t=0,$$ (9)

which is known as Poincaré's Lemma.

See also Differential k-Form, Poincaré's Lemma, Wedge Product