exterior derivative

Noun

 * 1)  A differential operator which acts on a differential k-form to yield a differential (k+1)-form, unless the k-form is a pseudoscalar, in which case it yields 0.
 * The exterior derivative of a &ldquo;scalar&rdquo;, i.e., a function $$f = f(x^1, x^2, ..., x^n)$$ where the $$x^i$$&rsquo;s are coordinates of $$\mathbb{R}^n$$, is $$df = {\partial f \over \partial x^1} dx^1 + {\partial f \over \partial x^2} dx^2 + ... + {\partial f \over \partial x^n} dx^n$$.
 * The exterior derivative of a k-blade $$f \, dx^{i_1} \wedge dx^{i_2} \wedge ... \wedge dx^{i_k}$$ is $$df \wedge dx^{i_1} \wedge dx^{i_2} \wedge ... \wedge dx^{i_k}$$.
 * The exterior derivative $$d$$ may be though of as a differential operator del wedge: $$\nabla \wedge$$, where $$\nabla = {\partial \over \partial x^1} dx^1 + {\partial \over \partial x_2} dx^2 + ... + {\partial \over \partial x^n} dx^n$$. Then the square of the exterior derivative is $$d^2 = \nabla \wedge \nabla \wedge = (\nabla \wedge \nabla) \wedge = 0 \wedge = 0$$ because the wedge product is alternating. (If u is a blade and f a scalar (function), then $$f u \equiv f \wedge u $$, so $$d (f u) = \nabla \wedge (f u) = \nabla \wedge (f \wedge u) = (\nabla \wedge f) \wedge u = df \wedge u$$.) Another way to show that $$d^2 = 0$$ is that partial derivatives commute and wedge products of 1-forms anti-commute (so when $$d^2$$ is applied to a blade then the distributed parts end up canceling to zero.)