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Differential forms, help finding the exterior dervative in dimensions greater than 3

  1. Jan 6, 2013 #1
    So say I have a n-1 form

    [itex]\sum^{n}_{i=1}x^{2}_{i}dx_{1}...\widehat{dx_{i}}...dx_{n}[/itex]

    and I want to find the exterior derivative, how do I know where to put which partial derivative for each term,

    would it simply be??

    [itex]\sum^{n}_{i=1} \frac{∂x^{2}_{i}}{∂x_{i}}dx_{i}dx_{1}...\widehat{dx_{i}}...dx_{n}[/itex]

    hopefully this will clarify, for this 2-form

    [itex]\alpha=x_{1}x_{2}dx_{2}dx_{4} + x_{3}x_{4}dx_{1}dx_{2}[/itex]

    how would one go about finding the exterior derivative? I have no idea which partials to put where, this is simple for a normal ℝ3 2 form, but I have no idea here. let me know if I need to clarify.
     
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  3. Jan 7, 2013 #2

    tiny-tim

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    hi saminator910! :smile:

    i'll rewrite your question slightly, since i find it a little confusing :redface:

    how do i find the exterior derivative of an n-1 form [itex]\sum^{n}_{i=1}f_i(x_1,\cdots x_n)dx_{1}..\cdots\widehat{dx_{i}}\cdots dx_{n}[/itex] ?​

    i find it easier to write it as [itex](\sum^{n}_{l=1}d_j)\wedge\sum^{n}_{i=1}f_i(x_1,...x_n)dx_{1}\wedge\cdots \widehat{dx_{i}}\cdots\wedge dx_{n}[/itex]

    then everything except j = i is zero, and you get

    [itex](\sum^{n}_{i=1}\partial f_i(x_1,...x_n)/\partial x_i)\ \ dx_{1}\wedge\cdots \wedge dx_{n}[/itex] :wink:
    if it's in ℝ3, where does x4 come from? :confused:
     
  4. Jan 7, 2013 #3
    Re: Differential forms, help finding the exterior dervative in dimensions greater tha

    thanks a lot, this seems to make sense. That is actually in ℝ4, how would one solve that? I was saying it would be simple for a 2-form in ℝ3, but it's difficult in ℝ4.
     
  5. Jan 8, 2013 #4

    tiny-tim

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    ok, then eg [itex]d\wedge (x_1x_2\,dx_2\wedge dx_4)[/itex]

    = [itex]\partial (x_1x_2)/\partial x_1\ (dx_1\wedge dx_2\wedge dx_4)[/itex]

    [itex]+ \partial (x_1x_2)/\partial x_3\ (dx_3\wedge dx_2\wedge dx_4)[/itex]

    = [itex]x_2\ (dx_1\wedge dx_2\wedge dx_4)[/itex] :smile:
     
  6. Jan 8, 2013 #5
    Re: Differential forms, help finding the exterior dervative in dimensions greater tha

    Okay thanks alot :biggrin:, that really makes things clearer. So for the 2 form in [itex]ℝ^{4}[/itex]

    I'm going through this step by step, just in case I make a mistake...

    [itex]β=x_{1}x_{2}dx_{3}dx_{4}+x_{3}x_{4}dx_{1}dx_{2}[/itex]

    [itex]dβ=d(x_{1}x_{2})dx_{3}dx_{4}+d(x_{3}x_{4})dx_{1}dx_{2}[/itex]

    [itex]dβ=(x_{2}dx_{1}+x_{1}dx_{2})dx_{3}dx_{4}+(x_{4}dx_{3}+x_{3}dx_{4})dx_{1}dx_{2}[/itex]

    now from here is where I think I'm doing something wrong, I "distribute" if you will, the dx's outside the parenthesis and get

    [itex]dβ=x_{2}dx_{1}dx_{3}dx_{4}+x_{1}dx_{2}dx_{3}dx_{4}+x_{4}dx_{3}dx_{1}dx_{2}+x_{3}dx_{4}dx_{1}dx_{2}[/itex]

    but the supposed answer is this, notice the switched dx's in the last two terms, why do I need to do this?

    [itex]dβ=x_{2}dx_{1}dx_{3}dx_{4}+x_{1}dx_{2}dx_{3}dx_{4}+x_{4}dx_{1}dx_{2}dx_{3}+x_{3}dx_{1}dx_{2}dx_{4}[/itex]
     
  7. Jan 9, 2013 #6

    tiny-tim

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    hi saminator910! :wink:
    that's right :smile:
    you don't need to do it, it's just neater

    ##dx_{3}\wedge dx_{1}\wedge dx_{2}## is the same as ##dx_{1}\wedge dx_{2}\wedge dx_{3}## (it's an even number of exchanges, so there's no minus-one factor)

    but the latter looks better o:)
     
  8. Jan 9, 2013 #7
    Re: Differential forms, help finding the exterior dervative in dimensions greater tha

    Thanks alot!
     
  9. Jan 10, 2013 #8

    HallsofIvy

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    Re: Differential forms, help finding the exterior dervative in dimensions greater tha

    Its differential is [itex]d\alpha= d(x_{1}x_{2}dx_{2}dx_{4} + x_{3}x_{4}dx_{1}dx_{2})[/itex]
    [itex]= (x_2dx_1+ x_1dx_2)(dx_2dx_4)+ (x_4dx_3+ x_3dx_4)(dx_1dx_2)[/itex]
    [itex]= (x_2dx_1dx_2dx_4+ x_1dx_2dx_2dx_4)+ (x_4dx_3dx_1dx_2+ x_3dx_4dx_1dx_2)[/itex]
    Now use the fact that the multiplication is "anti-symmetric" (which immediately implies that [itex]dx_2dx_2= 0[/itex]) to write that as
    [itex]x_2dx_1dx_2dx_4+ x_4dx_1dx_2dx3+ x_3dx_1d_2d_3[/itex]

    The first term, [itex]x_2dx_1dx_2dx_4[/itex] already has the differentials in the "correct" order. The last two, [itex]x_4dx_3dx_1dx_2[/itex] and [itex]x_3dx_4dx_1dx_2[/itex], each require two transpositions, [itex]x_4dx_3dx_1dx_2[/itex] to [itex]-x_4d_1dx_3dx_2[/itex] to [itex]-(-x_4dx_1dx_2dx_3)[/itex] and [itex]x_3dx_4dx_1dx_2[/itex] to [itex]-x_3dx_1dx_4dx_2[/itex] to [itex]-(-x_3dx_1dx_2dx_4)[/itex] and so have no net change in sign.

    (What is the "correct" order is, of course, purely conventional.)
     
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