More (VERY BASIC) Questions on Example on Wedge Products

1. Feb 4, 2016

Math Amateur

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 ..

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 $\psi = g \ dy$

so 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:

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

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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2. Feb 4, 2016

andrewkirk

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.

3. Feb 5, 2016

Math Amateur

Thanks Andrew ... yes, most helpful

Peter