Exterior Derivative Sample Clauses

Exterior Derivative. The intention of introducing the exterior derivative is to capture all of the classical theorems of “vector analysis” into one unified Xxxxxx’x Theorem, which asserts that the integral of a form on the boundary of a manifold is the integral of the exterior derivative of the form on the interior of the manifold:4 ∫ ∫ ω = ∂M M dω. (5.22) As we have seen in equation (3.34), the differential of a function on a manifold is a one-form field. If a function on a manifold is considered to be a form field of rank zero,5 then the differential operator increases the rank of the form by one. We can generalize this to k-form fields with the exterior derivative operation. Consider a one-form ω. We define6 dω(v1, v2) = v1(ω(v2)) − v2(ω(v1)) − ω([v1, v2]). (5.23) More generally, the exterior derivative of a k-form field is a k + 1- form field, given by:7 dω(v0,... , vk) = (5.24) Σk i=0 Σk . i ((−1) vi(ω(v0,... , vi−1, vi+1,... , vk))+ (−1)i+j ω([vi, vj], v0,... , vi−1, vi+1,... , vj−1, vj+1,... , vk))} . j=i+1 This formula is coordinate-system independent. This is the way we compute the exterior derivative in our software. 4This is a generalization of the Fundamental Theorem of Calculus. 5A manifold function f induces a form field ˆf of rank 0 as follows: ˆf()(m) = f(m). 6The definition is chosen to make Xxxxxx’x Theorem pretty. 7See Xxxxxx, Differential Geometry, Volume 1, p.289.
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Exterior Derivative. 63 If the form field ω is represented in a coordinate basis nΣ−1 ω = ai ,...,i dxi0 ∧ ··· ∧ dxik−1 (5.25) 0 i0=0,...,ik−1=0 k−1 then the exterior derivative can be expressed as nΣ−1 dω = dai ,...,i ∧ dxi0 ∧ ··· ∧ dxik−1 . (5.26) i0=0,...,ik−1=0 0 k−1 Though this formula is expressed in terms of a coordinate basis, the result is independent of the choice of coordinate system. Computing Exterior Derivatives We can test that the computation indicated by equation (5.24) is equivalent to the computation indicated by equation (5.26) in three dimensions with a general one-form field: (define a (literal-manifold-function ’alpha R3-rect)) (define b (literal-manifold-function ’beta R3-rect)) (define c (literal-manifold-function ’gamma R3-rect)) (define theta (+ (* a dx) (* b dy) (* c dz))) The test will require two arbitrary vector fields (define X (literal-vector-field ’X-rect R3-rect)) (define Y (literal-vector-field ’Y-rect R3-rect)) (((- (d theta) (+ (wedge (d a) dx) (wedge (d b) dy) (wedge (d c) dz)))

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