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According to Gallot-Hulin-Lafontaine one has $d\alpha (X_0,\cdots,X_q) = \sum_{i=0}^q (-1)^i D_{X_i} \alpha (X_1,\cdots,X_{i-1},X_0,X_{i+1},\cdots,X_q)$

It seems to me that it should be $d\alpha (X_0,\cdots,X_q) = \sum_{i=0}^q (-1)^i D_{X_i} \alpha (X_0,\cdots,\hat{X_i},\cdots,X_q)$

Is this right ?

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    I think you are right. – 2011-04-30

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If $\theta$ is a 1-form, then

$ d\theta(X,Y) = (\nabla_X\theta)(Y) - (\nabla_Y\theta)(X) $

If $\Omega$ is a 2-form, then

$ d\Omega(X,Y,Z) = (\nabla_X\Omega)(Y,Z)+(\nabla_Z\Omega)(X,Y)+ (\nabla_Y\Omega)(Z,X) $

and so on ... but you have to have a zero-torsion (symmetric) connection. These formulae will be useful when manipulating structure equations, for instance to obtain Bianchi identities. -- Salem

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    This can be derived from the expression given for $dĪ©$ in Lemma 1 of "A Definition of the Exterior Derivative in Terms of Lie Derivatives" by Richard Palais http://vmm.math.uci.edu/PalaisPapers/ADefinitionOfTheExteriorDerivativeInTermsOfLieDerivatives.pdf – 2018-04-10
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The correct formula is given on Wikipedia. If the vector fields commute (for example, if the $X_k$'s are the vector fields associated to a coordinate system), then it reduces to your formula.

It's not even clear to me how to interpret the terms for $i=0$ or $i=1$ in their formula, and in any case the factor $(-1)^i$ looks strange, since they would get the alternating signs from the moving of the argument $X_0$ anyway.

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    @Timofei: Ah, sorry. Sloppy reading on my part... I'm more used to $\nabla$ for covariant derivatives, so I subconsciously read the $D$'s as Lie derivatives. – 2011-03-31