The exterior derivative \(d\) is a map which maps smooth \(k\)-forms to smooth \(k+1\)-forms. It is convenient to define it in local coordinates. Let \(\omega\in \Omega^k(M)\) then
\begin{equation*} d\Bigl(\sum_{J}' \omega_Jdx^J\Bigr)=\sum_{J}' d\omega_J\wedge dx^J=\sum_{J}' \frac{\partial \omega_J}{\partial x^i}dx^i\wedge dx^{j_1}\wedge \cdots \wedge dx^{j_k}, \end{equation*}where we are using the primed sum sign defined in (0x66d1b59f) .
Remarks
- For 0-form \(f\) the exterior derivative \(df\) coincide with the differential of \(f\).
- The exterior differential of a 1-form motivates the exterior derivative.
Properties Link to heading
- \(d\) is linear
- \(d\circ d=0\)
- Product rule for wedge products : Let \(\omega\in \Omega^k(M)\) and \(\eta\in \Omega^l(M)\), then \begin{equation*} d(\omega\wedge \eta)=d\omega\wedge \eta + (-1)^k\omega\wedge d\eta. \end{equation*}
- Commutes with pullbacks: Let \(F:M\to N\) smooth and \(\omega\) be a smooth \(k\)-form on \(N\). Then \begin{equation*} F^*(d\omega)=d(F^*\omega). \end{equation*}