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Unformatted text preview: 3.5. VECTOR ANALYSIS IN R3 87 • Therefore
(∗dα)i = εi jk ∂ j αk ,
so that
∗dα = (∂2 α3 − ∂3 α2 )dx + (∂3 α1 − ∂1 α3 )dy + (∂1 α2 − ∂2 α1 )dz .
• We see that
∗dα = curl α .
• TwoForms.
For a 2form β there holds
(dβ)i jk = ∂i β jk + ∂ j βki + ∂k βi j ,
or
dβ = (∂1 β23 + ∂2 β31 + ∂3 β12 )dx ∧ dy ∧ dz .
• Hence, 1
∗dβ = εi jk ∂i β jk = ∂1 β23 + ∂2 β31 + ∂3 β12 .
2 • Now let α be a 1form
α = α1 dx + α2 dy + α3 dz .
Then
∗α = α1 dy ∧ dz − α2 dx ∧ dz + α3 dx ∧ dy ,
and
d ∗ α = (∂1 α1 + ∂2 α2 + ∂3 α3 )dx ∧ dy ∧ dz ,
or
∗d ∗ α = ∂1 α1 + ∂2 α2 + ∂3 α3 .
So,
∗d ∗ α = div α . diﬀgeom.tex; April 12, 2006; 17:59; p. 89 88 CHAPTER 3. DIFFERENTIAL FORMS diﬀgeom.tex; April 12, 2006; 17:59; p. 90 ...
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 Spring '10
 Wong
 Geometry

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