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Exterior Derivative

Consider a Differential k-Form

\omega^1 = b_1\,dx_1+b_2\,dx_2.
\end{displaymath} (1)

Then its exterior derivative is
d\omega^1 = db_1\wedge dx_1+db_2\wedge dx_2,
\end{displaymath} (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.
\end{displaymath} (3)


$\displaystyle d\omega^1$ $\textstyle =$ $\displaystyle db_1\wedge dx_1+db_2\wedge dx_2$  
  $\textstyle =$ $\displaystyle \left({{\partial b_1\over\partial x_1}dx_1+{\partial b_1\over\par...
...2\over\partial x_1}dx_1+{\partial b_2\over\partial x_2}dx_2}\right)\wedge dx_2.$ (4)

Denote the exterior derivative by

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

Then for a 0-form $t$,
(Dt)_\mu \equiv {\partial t\over \partial x^\mu},
\end{displaymath} (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),
\end{displaymath} (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),
\end{displaymath} (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,
\end{displaymath} (9)

which is known as Poincaré's Lemma.

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

© 1996-9 Eric W. Weisstein