- #1

BVM

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## Homework Statement

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.

## Homework 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.