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More (VERY BASIC) Questions on Example on Wedge Products

  1. Feb 4, 2016 #1
    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:


    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 ...

  2. jcsd
  3. Feb 4, 2016 #2


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    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.
  4. Feb 5, 2016 #3

    Thanks Andrew ... yes, most helpful

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