External derivative of a product of forms REVIEWED
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A short working note by Phil dated 5.17.16, marked as reviewed and already folded into his wedge document. It proves the product rule for the exterior derivative using coordinate expansions of k-forms and k'-forms. It then checks that the rule agrees with the graded commutativity α^β = (-1)^(kk')β^α, and gives a shorter proof using df terms. It cites Sjamaar and equation numbers 10.3.27 and 10.4.1.
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Extracted text (machine-read; may contain errors)
External derivative of a product of forms PhL 5.17.16
A section on this was added to wedge doc, do not edit here.
Pause; What about d(α^β)? This appears page 21 old Sjamaar but I have omitted it! Let's give it a try right here:
α = Σ'I fI(x) λ^I // a k-form
dα = Σ'I Σs=1n [∂sfI(x)] λs ^ λ^I
β = Σ'J gJ(x) λ^J // a k'-form
dβ = Σ'J Σs=1n [∂sgJ(x)] λs ^ λ^J
α ^ β = (Σ'I fI(x) λ^I) ^ Σ'J gJ(x) λ^J
= Σ'I Σ'J fI(x)gJ(x) λ^I ^ λ^J
d(α ^ β) = Σ'I Σ'J Σs ∂s[fI(x)gJ(x)] λs ^ λ^I ^ λ^J
Desired answer is
d(α ^ β) = (dα)^β + (sign) α ^ (dβ)
Now
(dα)^β = ( Σ'I Σs=1n [∂sfI(x)] λs ^ λ^I) ^ (Σ'J gJ(x) λ^J)
= Σ'I Σ'J Σs=1n [∂jfI(x)] gJ(x) λs ^ λ^I ^ λ^J
α ^ (dβ) = (Σ'I fI(x) λ^I) ^ Σ'I Σs=1n [∂sgI(x)] λs ^ λ^I
= Σ'I Σ'J Σs=1n fI(x)[∂sgI(x)] λI ^ λ^s ^ λ^J
= Σ'I Σ'J Σs=1n fI(x)[∂sgI(x)] (-1)k λs ^ λ^I ^ λ^J
Then
(dα)^β + (-1)k α ^ (dβ) = Σ'I Σ'J Σs=1n [∂sfI(x)] gJ(x) λs ^ λ^I ^ λ^J
+ Σ'I Σ'J Σs=1n fI(x) [∂sgI(x)] λs ^ λ^I ^ λ^J
= Σ'I Σ'J Σs=1n { [∂sfI(x)] gJ(x) + fI(x) [∂sgI(x)] } λs ^ λ^I ^ λ^J
= Σ'I Σ'J Σs=1n ∂s[fI(x)gJ(x)] λs ^ λ^I ^ λ^J
= d(α ^ β)
OK, but I am confused by this:
It seems clear that d(β^α) = (-1)kk' d(α^β) since d is linear, using 10.4.1
so d(β^α) =(-1)kk'[(dα)^β + (-1)k α ^ (dβ) ] **
On the other hand, (10.3.27) says d(β ^ α) = (dβ) ^ α + (-1)k' β ^ (dα). So we are claiming
d(β^α) =(-1)kk'[(dα)^β + (-1)k α ^ (dβ) ] **
d(β^α) = (dβ) ^ α + (-1)k' β ^ (dα) ***
Show these are consistent with this rule
α ^ β = (-1)kk'β ^ α
OK, here goes:
α^(dβ) = (-1)k(k'+1)(dβ)^α
(dα)^β = (-1)(k+1)k'β^(dα)
Then ** says
d(β^α) = (-1)kk'[(dα)^β + (-1)k α ^ (dβ) ]
= (-1)kk'[ (-1)(k+1)k'β^(dα) + (-1)k (-1)k(k'+1)(dβ)^α ]
Sign on first term is (-1)k. Sign on second term is +1, then
= (-1)k.β^(dα) + )(dβ)^α = *** QED
Simpler proof of (10.3.27),
α = Σ'I fI(x) λ^I // a k-form dα = Σ'I dfI(x) ^ λ^I
β = Σ'J gJ(x) λ^J // a k'-form dβ = Σ'J dgJ(x) ^ λ^J
α ^ β = ( Σ'I fI(x) λ^I) ^ (Σ'J gJ(x) λ^J) = Σ'I Σ'J fI(x)gJ(x) λ^I ^ λ^J
d(α ^ β) = Σ'I Σ'J d[fI(x)gJ(x)] ^ λ^I ^ λ^J
= Σ'I Σ'J gJ(x) dfI(x) ^ λ^I ^ λ^J + Σ'I Σ'J fI(x) {dgJ(x) ^ λ^I} ^ λ^J
= Σ'I Σ'J gJ(x) dfI(x) ^ λ^I ^ λ^J + Σ'I Σ'J fI(x) {(-1)k λ^I ^ dgJ(x)} ^ λ^J †
= ( Σ'IdfI(x)λ^I) ^ (Σ'JgJ(x)λ^J) + (-1)k ( Σ'I fI(x)λ^I ) ^ (Σ'JdgJ(x)λ^J)
= dα ^ β + (-1)k α ^ dβ
But here you have to use (10.4.1) which I have not yet stated.