exterior derivative
Noun
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Noun
exterior derivative (plural exterior derivatives)
- (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 f = f(x^1, x^2, ..., x^n) where the x^i’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.)
This text is extracted from the Wiktionary and it is available under the CC BY-SA 3.0 license | Terms and conditions | Privacy policy 0.002