More (VERY BASIC) Questions on Example on Wedge Products

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The discussion centers on understanding the wedge product of differential forms as presented in Barrett O'Neil's "Elementary Differential Geometry." Participants seek clarification on whether the expression d(φ ∧ ψ) simplifies to d(fg dx dy) and the implications of commuting scalars and one-forms. It is confirmed that while scalars f and g can be rearranged freely, the one-forms dx and dy do not commute and instead anti-commute, affecting the order of operations. The justification for the transformation in O'Neill's example is also requested, emphasizing the need for clear steps in the derivation. Overall, the thread highlights the complexities of working with wedge products in differential geometry.
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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:
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In my working of the example above ... trying to understand (*) ... I proceeded as follows ... but had some questions almost immediately ...

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

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 am also perplexed by what is happening (and the justification) in O'Neill's step in (*) where he writes:

d(fg \ dx \ dy ) = \partial (fg) / \partial z \ dz \ dx \ dyCan 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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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
 

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