Solving Equation 15.43 Line 2 to 3 in Tevian Dray's Differential Forms

In summary, the conversation discusses the equality implied in the move from equation 15.43 line 2 to line 3. The speaker struggles to understand Dray's book, but writing about their mistake helps them see it. They also mention a potential minus sign that can arise when applying the product rule for the exterior derivative of a product of two forms.
  • #1
gnnmartin
71
5
TL;DR Summary
In 15.43 he implies dα ∧ β = β ∧ dα where α,β are one forms.
I expected dα ∧ β = −β ∧ dα. Am I misreading the text, or have I simply lost the plot?
The equality is implied in the move from equation 15.43 line 2 to line 3.

I do find Dray's book is admirably clear and absolutely says something I wish to understand, but my 78 year old brain has difficulty. However, in this case I can be precise about where I fail to follow.

Oh! I find after all, writing this has enabled me to see my mistake, but I'll post the question all the same so that some kind person can confirm where I went wrong. If α is a one form, dα is a two form, so dα ∧ β = −−β ∧ dα = β ∧ dα.
 
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  • #2
Note that ##d\alpha## is a 2-form if ##\alpha## is a 1-form. In general, if ##\omega## and ##\eta## are ##p##- and ##q##-forms, respectively, then
$$
\omega\wedge\eta = (-1)^{p q} \eta \wedge\omega.
$$

Here you have ##p=2## and ##q=1## so ##(-1)^{p q} = (-1)^2 = +1##.

gnnmartin said:
Oh! I find after all, writing this has enabled me to see my mistake, but I'll post the question all the same so that some kind person can confirm where I went wrong. If α is a one form, dα is a two form, so dα ∧ β = −−β ∧ dα = β ∧ dα.
Indeed.
 
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  • #3
Thanks.
 
  • #4
Comment regarding a somewhat different but related issue that someone reading this in the future might also encounter:

Note that the exterior derivative of the product ##\omega \wedge \eta## also has a potential minus sign popping up when applying the product rule:
$$
d(\omega\wedge\eta) = (d\omega) \wedge \eta + (-1)^p \omega \wedge d\eta
$$
 
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  • #5
Orodruin said:
Comment regarding a somewhat different but related issue that someone reading this in the future might also encounter:

Note that the exterior derivative of the product ##\omega \wedge \eta## also has a potential minus sign popping up when applying the product rule:
$$
d(\omega\wedge\eta) = (d\omega) \wedge \eta + (-1)^p \omega \wedge d\eta
$$
Thanks, yes, it was not immediately obvious to me, but given the prompt I can see it.
 

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