Exterior derivative

Definition

Let $U \subseteq \mathbb{R}^n$ be open, and let $\omega = \sum_I f_I \, dx_I$ be a differential k-form on $U$ whose coefficients $f_I$ are of class $C^1$. The exterior derivative of $\omega$ is the $(k + 1)$-form $$d\omega = \sum_I \sum_{j=1}^n \frac{\partial f_I}{\partial x_j} \, dx_j \wedge dx_I,$$ where $\wedge$ is the wedge product.

Remarks

For a function $f$, $df = \sum_j \partial_j f \, dx_j$ is its differential, the one-form that corresponds to the gradient. The operator $d$ is linear and satisfies $d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^k \alpha \wedge d\beta$ for a $k$-form $\alpha$. For $C^2$ forms, d d = 0. In the plane, $d(L \, dx + M \, dy) = (\partial_x M - \partial_y L) \, dx \wedge dy$. In $\mathbb{R}^3$, $d(F_1 \, dx + F_2 \, dy + F_3 \, dz)$ has the components of the curl of $F$ as coefficients of $dy \wedge dz$, $dz \wedge dx$ and $dx \wedge dy$. Also $d(F_1 \, dy \wedge dz + F_2 \, dz \wedge dx + F_3 \, dx \wedge dy) = (\nabla \cdot F) \, dx \wedge dy \wedge dz$, which is the divergence.

Explanations

L1 human

The exterior derivative is one operation that does the work of gradient, curl and divergence together. You differentiate each coefficient in each direction and attach the new direction with a wedge product.