exterior derivative

exterior derivative

English

Noun

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: ${\\displaystyle \abla \\wedge }$, where ${\\displaystyle \abla ={\\partial \\over \\partial x^{1}}dx^{1}+{\\partial \\over \\partial x_{2}}dx^{2}+...+{\\partial \\over \\partial x^{n}}dx^{n}}$. Then the square of the exterior derivative is ${\\displaystyle d^{2}=\abla \\wedge \abla \\wedge =(\abla \\wedge \abla )\\wedge =0\\wedge =0}$ because the wedge product is alternating. (If u is a blade and f a scalar (function), then ${\\displaystyle fu\\equiv f\\wedge u}$, so ${\\displaystyle d(fu)=\abla \\wedge (fu)=\abla \\wedge (f\\wedge u)=(\abla \\wedge f)\\wedge u=df\\wedge u}$.) 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.)