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Wedge product and change of variables

  1. Apr 21, 2013 #1
    1. The problem statement, all variables and given/known data

    The question is:

    Let [itex]\phi: \mathbb{R}^n\rightarrow\mathbb{R}^n[/itex] be a [itex]C^1[/itex] map and let [itex]y=\phi(x)[/itex] be the change of variables. Show that

    d[itex]y_1\wedge...\wedge [/itex]d[itex]y_n[/itex]=(detD[itex]\phi(x)[/itex])[itex]\cdot[/itex]d[itex]x_1\wedge...\wedge[/itex]d[itex]x_n[/itex].

    2. Relevant equations


    3. The attempt at a solution
    Take a look at here and the answer given by Michael Albanese:

    My question is can we prove it without using the fact "[itex]\det A = \sum_{\sigma\in S_n}\operatorname{sign}(\sigma)\prod_{i=1}^na_{i \sigma(j)}[/itex]"?
  2. jcsd
  3. Apr 21, 2013 #2


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    Science Advisor

    Do you know the definition of the pullback of a differential form ? This is a generalization to multilinear

    maps of the "induced map" L* , from W* to V*, given a linear map L:V-->W , both V,W vector spaces.

    I'm trying to avoid heavy machinery, but I think you need to understand this, unless you just want

    a quick-and-dirty answer ( I assume you don't since you would have accepted the answer from the link

    if you did.). You are basically doing a change of bases for multilinear maps, an extension of the idea of

    basis change for a linear map.
    Last edited: Apr 21, 2013
  4. Apr 21, 2013 #3


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    Science Advisor

    Or maybe you can tell us the approach you want to follow .
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