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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?

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For the Lie derivative, assuming that you're working with a torsion-free connection: just take the usual coordinate formula (see e.g. here) and replace the partial derivatives with covariant derivatives. (The antisymmetric nature of the Lie derivative combined with the symmetry of the Christoffel symbols means all the Christoffel terms vanish.)

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 $(T^*M)^{\otimes k} \otimes E$).

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  • $\begingroup$ Awesome, thanks for your help! $\endgroup$ – ಠ_ಠ Dec 3 '17 at 6:35
  • $\begingroup$ 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 $\nabla_{[j_0}\theta_{j_1 \ldots j_k]}s^A$. $\endgroup$ – ಠ_ಠ Dec 7 '17 at 3:47
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    $\begingroup$ @ಠ_ಠ: yeah, you're right. $\endgroup$ – Anthony Carapetis Dec 7 '17 at 3:58

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