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Wedge Product as an (r+s) form

  1. Oct 11, 2008 #1
    1. The problem statement, all variables and given/known data

    If [itex]\omega \in \Lambda^{r}(V^*)[/itex] and [itex]\nu \in \Lambda^{s}(V^*)[/itex] show that [itex]\omega \wedge \nu \in \Lambda^{r+s}(V^*)[/itex]

    2. Relevant equations

    [itex]\omega \wedge \nu (v_1,... ,v_r,v_{r+1},... ,v_{r+s})= \frac{1}{(r+s)!} \Sigma_{\sigma}(-1)^\sigma \omega(v_{\sigma(1)},v_{\sigma(2)},...,v_{\sigma(r)}) \nu(v_{\sigma(r+1)},v_{\sigma(r+2)},...,v_{\sigma(r+s)})[/itex]

    3. The attempt at a solution

    It would take really long to write out my solution thus far, I broke the sum on the right into even and odd sums. Then proceeded to show (I hope) that as [itex]\omega[/itex] and [itex]\nu[/itex] behave like [itex]\omega[/itex](odd interchange) [itex]= - \omega[/itex](no interchange) and the same for [itex]\nu[/itex].

    Then the "odd sum" will have three negatives, one from the [itex](-1)^\sigma[/itex] and two from the above relations. This leaves a net (-1) as per definition of an (r+s) form, similar reasoning is used for the even sum.

    Combining these two, we have, up to a scaling factor, that this new object behaves just as an (r+s) form.

    Cheers, it's mainly the permutation thing is bugging me.
  2. jcsd
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