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Chern-Simons form of the Chern character

  1. Aug 10, 2015 #1
    So I'm working my way through Nakahara Geometry Topology and Physics. In exercise 11.4 of section 11.5.2 I'm tasked to show

    [tex]\text{tr}[\epsilon^{ijkl}\mathcal{F}_{ij}\mathcal{F}_{kl}]=\partial_{i}[2\epsilon^{ijkl}\text{tr}(\mathcal{A}_j\partial_{k}\mathcal{A}_l+\frac{2}{3}\mathcal{A}_j\mathcal{A}_k\mathcal{A}_l)][/tex]

    where the field strength is lie algebra valued
    [tex] \mathcal{F}_{ij}=\partial_{i}\mathcal{A}_j-\partial_{j}\mathcal{A}_i+[\mathcal{A}_i,\mathcal{A}_j] [/tex]

    this should be a straightforward bit of tensor algebra but I'm getting a factor of 2 discrepancy in the first term on the RHS

    [tex]2\epsilon^{ijkl}\partial_{i}\text{tr}(\mathcal{A}_j\partial_{k}\mathcal{A}_l)=2\epsilon^{ijkl}\text{tr}(\partial_{i}\mathcal{A}_j\partial_{k}\mathcal{A}_l)[/tex]

    whereas on the LHS

    [tex]\epsilon^{ijkl}(\partial_{i}\mathcal{A}_j-\partial_{j}\mathcal{A}_i+[\mathcal{A}_i,\mathcal{A}_j])(\partial_{k}\mathcal{A}_l-\partial_{l}\mathcal{A}_k+[\mathcal{A}_k,\mathcal{A}_l])=\epsilon^{ijkl}4(\partial_{i}\mathcal{A}_j+\mathcal{A}_i\mathcal{A}_j)(\partial_{k}\mathcal{A}_l+\mathcal{A}_k\mathcal{A}_l)=4\partial_{i}\mathcal{A}_j\partial_{k}\mathcal{A}_l+\ldots[/tex]

    what am I missing it's driving me crazy

    Thanks
     
  2. jcsd
  3. Aug 16, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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