Exterior derivative

English

Pronunciation

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Noun

1. Template:Lb 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 “scalar”, i.e., a function $f=f(x^1,x^2,...,x^n)$ where the $x^i$’s are coordinates of ${ℝ}^n$, is $df=\frac{\partial f}{\partial x^1}d{x}^1+\frac{\partial f}{\partial x^2}d{x}^2+...+\frac{\partial f}{\partial x^n}d{x}^n$.

   The exterior derivative of a k-blade $fd{x}^{i_1}\wedge d{x}^{i_2}\wedge ...\wedge d{x}^{i_k}$ is $df\wedge d{x}^{i_1}\wedge d{x}^{i_2}\wedge ...\wedge d{x}^{i_k}$.

   The exterior derivative $d$ may be though of as a differential operator del wedge: $\mathrm{\nabla }\wedge$, where $\mathrm{\nabla }=\frac{\partial }{\partial x^1}d{x}^1+\frac{\partial }{\partial x_2}d{x}^2+...+\frac{\partial }{\partial x^n}d{x}^n$. Then the square of the exterior derivative is $d^2=\mathrm{\nabla }\wedge \mathrm{\nabla }\wedge =(\mathrm{\nabla }\wedge \mathrm{\nabla })\wedge =0\wedge =0$ because the wedge product is alternating. (If u is a blade and f a scalar (function), then $fu\equiv f\wedge u$, so $d(fu)=\mathrm{\nabla }\wedge (fu)=\mathrm{\nabla }\wedge (f\wedge u)=(\mathrm{\nabla }\wedge f)\wedge u=df\wedge u$.) 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.)