Introduction to differential forms

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If C is piecewise C 1 , then we simply add up the integrals over the C 1 pieces. Although we’ve done everything at once, it is often easier, in practice, to do this in steps. First change the variables from x and y to expresions in t, then replace dx by dx dt dt etc. Then integrate with respect to t. For example, if we parameterize the unit circle c by x = cos θ, y = sin θ, 0 ≤ θ ≤ 2π, we see − y x 2 + y 2 dx + x x 2 + y 2 dy = − sin θ(cos θ)′ dθ + cos θ(sin θ) ′ dθ = dθ and therefore From the chain rule, we get LEMMA 4.3 ∫ C − y ∫ x 2 + y 2 dx + x 2π x 2 + y 2 dy = dθ = 1 ∫ −C ∫ F dx + Gdy = − F dx + Gdy C If C and C ′ are equivalent, then ∫ ∫ F dx + Gdy = C C ′ 0 F dx + Gdy While we’re at it, we can also define a line integral in R 3 . Suppose that F dx + Gdy + Hdz is a differential form with C 1 coeffients. Let C : [a, b] → R 3 be a piecewise C 1 parametric curve, then DEFINITION 4.4 ∫ b a [F (x(t), y(t), z(t)) dx dt ∫ C F dx + Gdy + Hdz = dy dz + G(x(t), y(t), z(t)) + H(x(t), y(t), z(t)) dt dt ]dt The notion of exactness extends to R 3 automatically: a form is exact if it equals df for a C 2 function. One of the most important properties of exactness is its path independence: C 1 PROPOSITION 4.5 If ω is exact and C 1 and C 2 are two parametrized curves with the same endpoints (or more acurately the same starting point and ending point), then ∫ ∫ ω = ω It’s quite easy to see why this works. If ω = df and C 1 : [a, b] → R 3 then ∫ ∫ b df df = C 1 dt dt by the chain rule. Now the fundamental theorem of calculus shows that the last integral equals f(C 1 (b)) − f(C 1 (a)), which is to say the value of f at the a C 2 4
endpoint minus its value at the starting point. A similar calculation shows that the integral over C 2 gives same answer. If the C is closed, which means that the starting point is the endpoint, then this argument gives COROLLARY 4.6 If ω is exact and Cis closed, then ∫ C ω = 0. Now we can prove theorem 4.1. If F dx + Gdy is a closed form on R 2 , set ∫ f(x, y) = F dx + Gdy where the curve is indicated below: C (x,y) (0,0) (x,0) We parameterize both line segments seperately by x = t, y = 0 and x = x(constant), y = t, and sum to get f(x, y) = ∫ x F (t, 0)dt + ∫ y 0 0 G(x, t)dt Then we claim that df = F dx + Gdy. To see this, we differentiate using the fundamental theorem of calculus. The easy calcutation is ∂f ∂y ∫ y = ∂ G(x, t)dt ∂y 0 = G(x, y) Slightly trickier is ∂f ∂x ∫ x ∫ y = ∂ F (x, 0)dt + ∂ G(x, t)dt ∂x 0 ∂x 0 ∫ y ∂G(x, t) = F (x, 0) + dt 0 ∂x ∫ y ∂F (x, t) = F (x, 0) + dt 0 ∂t = F (x, 0) + F (x, y) − F (x, 0) = F (x, y) 5
Page 1 and 2: Introduction to differential forms
Page 3: will be called −C. This represent
Page 7 and 8: A typical 2-form looks like this:
Page 9 and 10: 9 “d” of a 2-form and divergenc
Page 11 and 12: n T v T u S v=constant u=constant F
Page 13 and 14: = ( ∂x ∂y ∂u ∂v − ∂x
Page 15 and 16: 13 Flux In many situations arising
Page 17 and 18: THEOREM 14.3 (Stokes’ theorem II)
Page 19 and 20: We are subtracting the second surfa

Introduction to differential forms

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