HW04sol - Differential Geometry 1 Homework 04 1. Let ζ...

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Unformatted text preview: Differential Geometry 1 Homework 04 1. Let ζ ∈ Vec(S 1 ) be the standard vector field given by (a, b) ∈ S 1 ⇒ ζ(a,b) = τ(a,b) (−b, a) and let Θ ∈ Ω1 (S 1 ) be the standard one-form given by Θ(ζ ) ≡ 1. (i) Find explicitly the integral curve γ of ζ through (a, b) ∈ S 1 . (ii) Show that if g ∈ C ∞ (S 1 ) and g Θ = df for some f ∈ C ∞ (S 1 ) then 2π g (γ (t))dt = 0. 0 2. Let ω ∈ Ωm (Rm+1 ) be defined by ωa (τa (v1 ), . . . , τa (vm )) = Det[a|v1 | · · · |vm ] for a, v1 , . . . , vm ∈ Rm+1 . Show that m (−1)i xi dx0 ∧ · · · ∧ dxi ∧ · · · ∧ dxm ω= i=0 in terms of the standard coordinates (x0 , · · · , xm ) on Rm+1 . Solutions (1) (i) In standard coordinates, let γ (t) = (x(t), y (t)) for t real. The differential equation to be solved for γ is γ • (t) = ζγ (t) or (x (t), y (t))(x(t),y(t)) = (−y (t), x(t))(x(t),y(t)) . With initial conditions x(0) = a and y (0) = b, the resulting system x = −y, 1 y =x has solution given by x(t) = a cos t − b sin t, y (t) = b cos t + a sin t. (1) (ii) Evaluate both sides of the supposed equality (of one-forms) at γ (t) on the tangent vector γ • (t) to obtain (f ◦ γ ) (t) = γ • (t)f = dfγ (t) (γ • (t)) = (g Θ)γ (t) (ζγ (t) ) = g (γ (t)) and then integrate over [0, 2π ] to deduce (FTOC) that 2π (g ◦ γ ) = f (γ (2π )) − f (γ (0)) = 0. 0 (2) Recall that ω (∂I )dxI ω= I↑ where ∂I = ( ∂ ∂ , . . . im ). ∂xi1 ∂x Of course, the m-tuple I is simply (0, . . . , i, . . . , m) for some i and then ∂ ∂ ∂ ,..., i ,..., m ) 0a ∂x ∂x a ∂x a ωa (τa (e0 ), . . . , τa (ei ), . . . , τa (em )) Det[a|e0 | · · · |ei | · · · |em ] (−1)i Det[e0 | · · · |a| · · · |em ] (−1)i ai = (−1)i xi (a) ω (∂I )(a) = ωa ( = = = = whence m (−1)i xi dx0 ∧ · · · ∧ dxi ∧ · · · ∧ dxm . ω= i=0 2 ...
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This note was uploaded on 07/08/2011 for the course MTG 6256 taught by Professor Robinson during the Spring '09 term at University of Florida.

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HW04sol - Differential Geometry 1 Homework 04 1. Let ζ...

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