How to Derive Yang-Mills Theories from a Kaluza-Klein Perspective?

[BVM]

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.

 

[fzero]

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.

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.

 

[BVM]

Oh my god how did I miss this.

Thanks a lot!