General gauge algebra identity

Question

In https://arxiv.org/abs/1011.1145 the following rather general gauge algebra identity (2.4) is used

$$\delta_{gct}(\xi) B_\mu^{\>\>A} + \xi^\lambda R_{\mu\lambda}^{\quad\! A} -\sum_{\{C\}}\delta(\xi^\lambda B_\lambda^{\>\>C})B_\mu^{\>\>A} = 0 $$

The free $A$ index references any chosen gauge transformation in the gauge algebra, $B_\mu^{\>\>A}$ its corresponding gauge field / connection, $R_{\mu\nu}^{\quad\! A}$ its corresponding field strength / curvature, and $\xi = \xi^\lambda \partial_\lambda$ the generator of the general coordinate transformations $\delta_{gct}(\xi)$. The last sum over $C$ is the sum over all gauge transformations $\delta_C$ in the gauge algebra.

What follows is clear (and is not relevant to my post) but the author provides no explanation of this identity nor cites a reference where to look it up. Presumably it is valid for any gauge algebra that closes off-shell? Anyway, this relation is not clear to me. Can anyone please provide a derivation of this identity or a reference where it is derived? Thank you.

Answer 1

OK, I figured it out myself.

$$\begin{eqnarray}\delta_{gct}(\xi) B_\mu^{\>\>A} &\;=\;& \mathcal L_\xi B_\mu^{\>\>A} \\ &=& \xi^\lambda\partial_\lambda B_\mu^{\>\>A} + B_\lambda^{\>\>A}\partial_\mu \xi^\lambda \\ &=& \xi^\lambda\partial_{[\lambda} B_{\mu]}^{\>\>A} + \partial_\mu(\xi^\lambda B_\lambda^{\>\>A})\\ &=& \xi^\lambda \big( \partial_{[\lambda} B_{\mu]}^{\>\>A} + f_{BC}^{\quad A} B_\lambda^{\>\>B}B_\mu^{\>\>C} \big) + \partial_\mu(\xi^\lambda B_\lambda^{\>\>A}) + f_{BC}^{\quad A} B_\mu^{\>\>B}(\xi^\lambda B_\lambda^{\>\>C})\\ &=& \xi^\lambda R_{\lambda\mu}^{\quad \!A} + \delta \big(\epsilon = {\textstyle\sum}_C \xi^\lambda B_\lambda^{\>\>C}) B_\mu^{\>\>A} \end{eqnarray}$$ Notation: Here $\mathcal L$ is the Lie derivative, index antisymetrization is normalized as $T_{[ab]} \equiv T_{ab}-T_{ba}$, and summation on a contracted internal gauge index $A, B, C,...$ means over the entire gauge algebra. So for gauged Poincare the sum on $B^C$ runs over the vielbein $e^a$ and Lorentz connection $\omega_a^{\>\,b}$. The last term happens to be the action of gauge algebra transformation $\delta(\epsilon)$ acting on the fixed connection $B_\mu^{\>\>A}$, but with the usual gauge parameter $\epsilon$ replaced by $\sum_C\xi^\lambda B_\lambda^{\>\>C}$.