More (VERY BASIC) Questions on Example on Wedge Products

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I am reading Barrett O'Neil's book: Elementary Differential Geometry ...

I need help with some more issues/problems with the example on wedge products of differential forms in O'Neill's text on page 31 in Section 1.6 ..The example reads as follows:
Printed page of a differential forms example: d(φ∧ψ) = d(fg dx dy) = ∂(fg)/∂z dz dx dy = (f ∂g/∂z + g ∂f/∂z) dx dy dz, labelled (*), then dφ∧ψ and φ∧dψ computed and subtracted.
In my working of the example above ... trying to understand (*) ... I proceeded as follows ... but had some questions almost immediately ...

We have [itex]\phi = f \ dx[/itex] and [itex]\psi = g \ dy[/itex]so we can write (can we?)[itex]d ( \phi \wedge \psi ) = d( f \ dx \wedge \ g \ dy )[/itex][itex]= d (f \ dx \ g \ dy )[/itex]But is this last step correct? That is is the wedge product of these elements just a concatenation? If so, why?
Continuing ...

[itex]d (f \ dx \ g \ dy ) = d( fg \ dx \ dy )[/itex]

Is this correct? Why?
It assumes all these type of elements commute ... is that correct? Why?... I am also perplexed by what is happening (and the justification) in O'Neill's step in (*) where he writes:

[itex]d(fg \ dx \ dy ) = \partial (fg) / \partial z \ dz \ dx \ dy[/itex]Can someone explain and justify this step please ... preferably by providing all the intervening steps and their justification ...
Hope someone can help with these questions/issues/problems ...

Peter
 
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  • Printed textbook page with worked wedge product example: d(φ∧ψ) = d(fg dx dy) = ∂(fg)/∂z dz dx dy = (f ∂g/∂z + g ∂f/∂z) dx dy dz (*), then dφ∧ψ and φ∧dψ computed and subtracted to get (*).
    O'Neill - EXAMPLE on page 31 - Wedge Products of Differential Forms.png
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Math Amateur said:
d(f dx g dy)=d(fg dx dy)

Is this correct? Why?
It assumes all these type of elements commute ... is that correct? Why?
I can do the middle one, as it's very quick, and I only have a minute right now.
f and g are scalars, whereas dx and dy are one-forms, which are elements of a vector space. As is usual with vector spaces, we can move scalars around as much as we like, and that's what's done above. However the vectors / one-forms dx and dy do not commute with each other (in fact if you change the order you change the sign - they anti-commute), so their order cannot be changed.
Hope that helps.
 
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andrewkirk said:
I can do the middle one, as it's very quick, and I only have a minute right now.
f and g are scalars, whereas dx and dy are one-forms, which are elements of a vector space. As is usual with vector spaces, we can move scalars around as much as we like, and that's what's done above. However the vectors / one-forms dx and dy do not commute with each other (in fact if you change the order you change the sign - they anti-commute), so their order cannot be changed.
Hope that helps.
Thanks Andrew ... yes, most helpful

Peter