I am reading Barrett O'Neil's book: Elementary Differential Geometry ...(adsbygoogle = window.adsbygoogle || []).push({});

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:

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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# More (VERY BASIC) Questions on Example on Wedge Products

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