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Quickly: Multlinearity of exterior derivative, and proof of invariant formula

  1. Jun 4, 2012 #1
    Hi all,

    I am trying to prove the invariant form for the exterior derivative http://en.wikipedia.org/wiki/Exteri...tions_of_grad.2C_curl.2C_div.2C_and_Laplacian by following these notes http://idv.sinica.edu.tw/ftliang/diff_geom/*diff_geometry(II)/3.11/exterior_derivative_2.pdf.

    I am confused however, as the beginning of the proof is to show that the RHS form is "multilinear", in the sense that, for arbitrary vectors [itex]V_i[/itex]

    [tex] RHS(V_1, \ldots, fV_p, \ldots, V_{k+1}) = (-1)^{p-1} f RHS(V_1, \ldots, V_p, \ldots, V_{k+1}) [/tex]

    Given that the right hand side is equal to the exterior derivative of something, does this not imply that the exterior derivative of something has the same weird symmetry. I would have thought that since the exterior derivative is a tensor, it is the usual linear
    [tex] d \Omega (V_1, \ldots, fV_p, \ldots, V_{k+1}) = f d \Omega(V_1, \ldots, V_p, \ldots, V_{k+1}). [/tex]
    Could someone please clarify this?

    As for the general proof, after proving that things are multilinear in that unusual sense, the author states that [itex] \omega = f dx ^I[/itex] for some index set [itex] I [/itex]. Is this meant to be [itex] \omega = f_I dx ^I[/itex]?

    Cheers
     
  2. jcsd
  3. Jun 5, 2012 #2

    quasar987

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    If you read the proof, you will see that they actually prove that RHS is multilinear in the usual sense of the word, and so that (-1)^{p-1} factor is some kind of typo.

    Next, the authors make the remark that since a general k-form is (locally) a sum of terms of the form [itex]f dx ^I[/itex] (for some multiindex I), and since both RHS and d are linear, it suffices to check that RHS = d on forms [itex]\omega[/itex] of the type [itex]\omega = f dx ^I[/itex]. It is completely unnecessary to clog up the notation by giving f a multiindex, but you may do so if its absence makes you uneasy.
     
  4. Jun 5, 2012 #3
    Thanks. I soon figured out that it was multilinear in the usual sense, though the initial error was confusing.

    As for the next point, I just wanted to clarify that they were not assuming that all the coefficients were equal, i.e. letting [itex]f = f_{a_1\ldots a_{p}}[/itex] for all [itex]a_i[/itex].

    Ah, I see --- maybe they can assume that, as long as they show that the equation is linear in [itex]\omega[/itex]. I think they state this, but do not prove it.

    Cheers
     
  5. Jun 5, 2012 #4

    quasar987

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    I suppose that the complete argument would be this: pick a multiindex I and a function f, then verify the equality for [itex]\omega=fdx^I[/itex]. Then, observe that nowhere in this is verification did we explicitely use the value of I or any special property about the function f. So the equality is valid for any multiindix I and associated function f_I and hence for any k-form by linearity.
     
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