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Exactness of a differential form

  1. Dec 7, 2005 #1
    Suppose your manifold is just M = R^2 with the standard differential structure (so the atlas is {(R^2, id, R^2)}). Suppose we have a 1-form [tex]\omega[/tex] on M. Then ofcourse [tex]\omega = a_1 dx_1 + a_2 dx_2[/tex], where the a_i are just c-infinity functions from R^2 to R. Suppose we have a function f on M defined as follows:

    [tex]f(x) = x_1 \int_0^1 a_1(tx) dt + x_2 \int_0^1 a_2(tx) dt[/tex]

    Show that df = w when dw = 0. Use the fact that:

    [tex]\frac{d}{dt}(t a(tx)) = a(tx) + tx_1 \frac{\partial a}{\partial x_1}(tx) + tx_2 \frac{\partial a}{\partial x_2}(tx)[/tex]

    Using this hint i rewrote f(x):

    [tex]f(x) = a_1 (x)x_1 + a_2 (x)x_2 - x_1^2 \int_0^1 t\frac{\partial a_1}{\partial x_1}dt - x_2^2 \int_0^1 t \frac{\partial a_2}{\partial x_2}dt - x_1 x_2 \int_0^1 t\left(\frac{\partial a_1}{\partial x_2} + \frac{\partial a_2}{\partial x_1}\right) dt[/tex]

    Now I want to use the fact that [tex]d\omega = 0[/tex]
    This means that:

    [tex]d\omega = d(a_1 dx_1 + a_2 dx_2) = d(a_1)\wedge dx_1 + d(a_2)\wedge dx_2 = \left(\frac{\partial a_2}{\partial x_1} - \frac{\partial a_1}{\partial x_2}\right) dx_1 \wedge dx_2
    [/tex]

    so [tex]\frac{\partial a_2}{\partial x_1} = \frac{\partial a_1}{\partial x_2}[/tex]

    If somehow df = w, the differential of the integral terms in f(x) should be zero. But I don't see how i could show that.

    Could someone give me a hint? :smile:
     
    Last edited: Dec 8, 2005
  2. jcsd
  3. Dec 7, 2005 #2

    matt grime

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    I think you just need to apply d to f and see that df=w+dw, or perhpas w+Cdw for some object C, then under the hypotheses that dw=0 you're done.

    ie take this line

    [tex]f(x) = a_1 (x)x_1 + a_2 (x)x_2 - x_1^2 \int_0^1 t\frac{\partial a_1}{\partial x_1}dt - x_2^2 \int_0^1 t \frac{\partial a_2}{\partial x_2}dt - x_1 x_2 \int_0^1 t\left(\frac{\partial a_1}{\partial x_2} + \frac{\partial a_2}{\partial x_1}\right) dt[/tex]

    and apply d to it, notice that d kills the first two terms on the RHS (since that is w) by assumption, in theory what should pop out is w, but i'm not quick enough to see that, and too lazy to check. It just strikes me that you're yet to work out what df is.
     
  4. Dec 7, 2005 #3
    I don't completely agree with you. You said the first two terms, [tex]a_1(x)x_1 + a_2(x)x_2 = \omega[/tex]. But this isn't true. The differential, [tex]d(a_1(x)x_1 + a_2(x)x_2) = a_1 dx_1 + a_2 dx_2 = \omega[/tex] of these two terms equals [tex]\omega[/tex].
    Another reason why your statement can't be true is that [tex]a_1(x)x_1 + a_2(x)x_2[/tex] is just a function so a 0-form, and [tex]\omega[/tex] is a 1-form.
     
  5. Dec 7, 2005 #4

    matt grime

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    sorry, read it too quickly,

    In any case, applying d to f would seem to be the best place to start wouldn't you agree?

    I think you overly focused on the wrong part of my answer there given by your length of response to something that was simply wrong; you didn't have to prove it was wrong, just point out that i misread it.

    in this case d should kill the other terms, shuldn't it?
     
    Last edited: Dec 7, 2005
  6. Dec 7, 2005 #5
    Yeah d should kill the other terms but the question is how! :smile:

    When i apply d for example I get a bunch of messy derivatives and stuff, and i really don't see how this could be zero:mad:
     
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