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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 oneform 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[av1  · · · 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 oneforms) 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 mtuple 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[ae0  · · · 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.
 Spring '09
 Robinson

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