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Chern Simons form: calculation

  1. Jul 31, 2010 #1
    Hey folks, I'm trying to understand a couple of calculations in Chern-Simons theory and I'm stuck.

    I want to prove the following in four spacetime dimensions:

    Let [tex]G[/tex] be a Lie Group with generators [tex]T_{a}[/tex]. Further let
    [tex]A=T_{a} A^{a}_{\mu} dx^{\mu}[/tex]
    be a connection and
    [tex]F=dA+A \wedge A[/tex]
    its field strength. Now I want to prove
    [tex]Tr F \wedge F= d Tr(A \wedge dA+\frac{2}{3}A\wedge A\wedge A)[/tex]

    First I tried the following
    [tex]F\wedge F=dA\wedge dA+dA \wedge A\wedge A+A\wedge A\wedge dA+A\wedge A\wedge A\wedge A[/tex]

    [tex]d (A \wedge dA+\frac{2}{3}A\wedge A\wedge A)= dA\wedge dA+\frac{2}{3}dA\wedge A\wedge A +\frac{2}{3} A\wedge dA\wedge A+\frac{2}{3} A\wedge A \wedge dA [/tex]

    When I take the trace, I can make a [tex]A\wedge A\wedge dA[/tex] into [tex]dA \wedge A\wedge A[/tex], but that leaves the questions of why the trace of [tex]A^4[/tex] has to vanish and where the factors of 2/3 come from.

    Alternatively, I tried the following:

    Let [tex]A_t=tA[/tex] and
    [tex]F_t=dA_t+A_t\wedge A_t=dt\wedgeA+t dA+t^2 A\wedge A[/tex]

    Now one can evaluate [tex]Tr F_t \wedge F_t [/tex] and afterwards set t=1. This works out fine, but I still don't understand why the trace of [tex]A^4[/tex] has to vanish.

    So for the moment, it all boils down to: why is [tex]Tr A^4=0[/tex]?

    Thank you for your help!
  2. jcsd
  3. Jul 31, 2010 #2
    The sign of the third term on the right hand side should be minus, due to antisymmetry of the exterior derivative. We also have
    [tex]Tr\left[dA\wedge A\wedge A \right]= \epsilon^{ijkl}Tr\left[(\partial_{i}A_{j})A_{k}A_{l} \right]= \epsilon^{ijkl}Tr\left[A_{l}(\partial_{i}A_{j})A_{k} \right] = -\epsilon^{ijkl}Tr\left[A_{i}(\partial_{j}A_{k})A_{l} \right]=-Tr\left[A\wedge dA\wedge A \right][/tex],
    in which the cyclic property of trace and the antisymmetry of the [tex]\displaystyle{\epsilon}[/tex] tensor is used.

    [tex]Tr\left[A^{4}\right]=0[/tex] can be proved similarly.
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