Exterior covariant derivative and Lie derivative in Penrose abstract index notation?

How does one express the Lie derivative of tensors, and exterior covariant derivative for forms with values in a vector bundle in Penrose abstract index notation? I've tried looking through Penrose's negative dimensional tensors article but didn't see it written down.

For say a vector-bundle valued 3-form $\omega_{bcd}^A$ where upper case indices correspond to the vector bundle and lower case indices are form indices, would the exterior covariant derivative just be $\nabla_{[a}\omega_{bcd]}^A$, where square brackets indicate antisymmetrization?

For the exterior covariant derivative, your formula is correct, possibly up to a normalization constant depending on your convention for antisymmetrization. By linearity it suffices to prove it for a decomposable form $\omega = \theta \otimes s$ where $\theta \in \Omega^k(M), s \in \Gamma(E).$ Recalling the definition $$d^\nabla (\theta \otimes s) = d \theta \otimes s +(-1)^k \theta \wedge \nabla s,$$ note that in index notation this becomes $$d^\nabla_{j_0} (\theta_{j_1\cdots j_k} s^A)=\nabla_{[j_0} \theta_{j_1\cdots j_k]}s^A+(-1)^k\theta_{[j_0\ldots j_{k-1}}\nabla_{j_k]}s^A.$$ Here we used the formula $d\theta = \mathrm{Alt}(\nabla \theta)$ for the standard exterior derivative, so we're also assuming $\nabla$ is torsion-free here.
Shuffling $j_1$ up to the final slot of the second term requires $k$ transpositions, so using the Leibniz rule this becomes \begin{align} d^\nabla_{j_0} (\theta_{j_1\cdots j_k} s^A)& =\nabla_{[j_0} \theta_{j_1\cdots j_k]}s^A+\theta_{[j_1\ldots j_k}\nabla_{j_0]}s^A\\ &= \nabla_{[j_0}\left(\theta_{j_1\cdots j_k]} s^A\right) = \nabla_{[j_0} \omega_{j_1\ldots j_k]}^A \end{align} as desired. In other words, the exterior covariant derivative is just the antisymmetrization (in the $TM$ indices) of the full covariant derivative (taken with respect to the tensor product connection on $(T^*M)^{\otimes k} \otimes E$).
• By the way, I think maybe the indices should be numbered starting from $0$, since you assumed $\theta$ was a $k$-form, so we should have $\nabla_{[j_0}\theta_{j_1 \ldots j_k]}s^A$. – ಠ_ಠ Dec 7 '17 at 3:47