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Yang Mills Fields from Isometries

  1. Jun 6, 2015 #1

    BVM

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    1. The problem statement, all variables and given/known data
    I am going through the book "Symmetries in Fundamental Physics" by Kurt Sundermeyer and in his part on deriving Yang-Mills theories from a Kaluza-Klein perspective I seem to be stuck on a small step in the derivation.

    2. Relevant equations
    He expands the metric in a typical Kaluza-Klein-like style
    ## \begin{cases}
    &\hat{g}^{0}_{\mu \nu} (x) = W(\phi) g_{\mu \nu} + f(\phi)h_{\iota \kappa} \mathcal{B}^{\iota}_{\mu}\mathcal{B}^{\kappa}_{\nu}\\
    &\hat{g}^{0}_{\mu \kappa} (x) = f(\phi) h_{\iota \kappa}\mathcal{B}^{\iota}_{\mu}\\
    &\hat{g}^{0}_{\iota \kappa} (x) = f(\phi)h_{\iota \kappa}
    \end{cases}##
    and derives from it that under diffeomorphism, the B-field should transform as
    ##\mathcal{B}'^{\iota}_{\mu} = \frac{\partial x^{\nu}}{\partial x'^{\mu}}\left( \frac{\partial \theta'^{\iota}}{\partial \theta^{\kappa}} \mathcal{B}^{\kappa}_{\nu} - \frac{\partial \theta'^{\iota}}{\partial x^{\nu}} \right).##
    Now, by considering the isometries infinitesimally
    ##\theta'^{\iota} = \theta^{\iota} + \epsilon^a(x)K^{\iota}_a(\theta)##
    and expanding the B-fields in terms of the Killing vectors
    ##\mathcal{B}^{\iota}_{\mu} = g K^{\iota}_{a} \mathcal{A}^{a}_{\mu}##
    he gets
    ##g K^{\iota}_{d} \mathcal{A}^{d}_{\mu} = (\delta^{\iota}_{\kappa} + \epsilon^a \partial_{\kappa}K^{\iota}_a) g K^{\kappa}_{b} \mathcal{A}^{b}_{\mu} - K^{\iota}_{a}\partial_{\mu}\epsilon^a.##

    This i can follow, but now he claims that via
    ##K^{\kappa}_{a} \partial_{\kappa} K^{\iota}_{b} - K^{\kappa}_{b} \partial_{\kappa} K^{\iota}_{a} = f_{abc} K^{\iota}_{c}##
    he can get
    ##\mathcal{A}'^{a}_{\mu} = \mathcal{A}^{a}_{\mu} + \frac{1}{g}f^{bca}\mathcal{A}^b_{\mu}\epsilon^c - \frac{1}{g} \partial_{\mu}\epsilon^a.##

    3. The attempt at a solution
    By just plugging in the relation I can get as far as
    ##\mathcal{A}'^{a}_{\mu} = \mathcal{A}^{a}_{\mu} + f^{bca}\mathcal{A}^b_{\mu}\epsilon^c + \epsilon^c(K^{\iota}_a)^{-1}K^{\kappa}_{c}(\partial_{\kappa} K^{\iota}_b)\mathcal{A}^b_{\mu} - \frac{1}{g} \partial_{\mu}\epsilon^a##
    but I don't see how the extra term cancels or how the ##\frac{1}{g}## appears in the second term.
     
  2. jcsd
  3. Jun 8, 2015 #2

    fzero

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    Gold Member

    Performing the contraction, we find

    $$ \epsilon^c(K^{\iota}_a)^{-1}K^{\kappa}_{c}(\partial_{\kappa} K^{\iota}_b)\mathcal{A}^b_{\mu} = \epsilon^a(\partial_{\kappa} K^{\kappa}_b)\mathcal{A}^b_{\mu},$$

    but the divergence of a Killing vector vanishes, so this is zero.
     
  4. Jun 8, 2015 #3

    BVM

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    Oh my god how did I miss this.

    Thanks a lot!
     
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