exterior derivative

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exterior derivative

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exterior derivative (plural exterior derivatives)

1. (calculus) 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 ${\displaystyle f=f(x^{1},x^{2},...,x^{n})}$ where the ${\displaystyle x^{i}}$’s are coordinates of ${\displaystyle \mathbb {R} ^{n}}$, is ${\displaystyle df={\partial f \over \partial x^{1}}dx^{1}+{\partial f \over \partial x^{2}}dx^{2}+...+{\partial f \over \partial x^{n}}dx^{n}}$.

   The exterior derivative of a k-blade ${\displaystyle f\,dx^{i_{1}}\wedge dx^{i_{2}}\wedge ...\wedge dx^{i_{k}}}$ is ${\displaystyle df\wedge dx^{i_{1}}\wedge dx^{i_{2}}\wedge ...\wedge dx^{i_{k}}}$.

   The exterior derivative ${\displaystyle d}$ may be though of as a differential operator del wedge: ${\displaystyle \nabla \wedge }$, where ${\displaystyle \nabla ={\partial \over \partial x^{1}}dx^{1}+{\partial \over \partial x_{2}}dx^{2}+...+{\partial \over \partial x^{n}}dx^{n}}$. Then the square of the exterior derivative is ${\displaystyle d^{2}=\nabla \wedge \nabla \wedge =(\nabla \wedge \nabla )\wedge =0\wedge =0}$ because the wedge product is alternating. (If u is a blade and f a scalar (function), then ${\displaystyle fu\equiv f\wedge u}$, so ${\displaystyle d(fu)=\nabla \wedge (fu)=\nabla \wedge (f\wedge u)=(\nabla \wedge f)\wedge u=df\wedge u}$.) Another way to show that ${\displaystyle d^{2}=0}$ is that partial derivatives commute and wedge products of 1-forms anti-commute (so when ${\displaystyle d^{2}}$ is applied to a blade then the distributed parts end up canceling to zero.)