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

[gnnmartin]

TL;DR

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.

 

[Orodruin]

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 $(-1)^{p q} = (-1)^2 = +1$.

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.

 

[gnnmartin]

Thanks.

 

[Orodruin]

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:
$$
d(\omega\wedge\eta) = (d\omega) \wedge \eta + (-1)^p \omega \wedge d\eta
$$

 

[gnnmartin]

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.