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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 ?

Timofei
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2 Answers2

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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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    If the connection does have torsion, you can still get an expression using the torsion tensor.

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

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

    and so on

     – Lawrence D'Anna Apr 10 '18 at 00:12
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    This can be derived from the expression given for $dΩ$ 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

     – Lawrence D'Anna Apr 10 '18 at 00:18
  • This means, in the torsionless case, that $(\mathrm{d}\omega)(X,Y,Z,\dotsc) = (\nabla\omega)(X \wedge Y \wedge Z \wedge \dotsb)$, provided $X \wedge Y \wedge \dotsb = X\otimes Y \otimes \dotsb - Y\otimes X \otimes \dotsb + \dotsb $ ? – pglpm Oct 22 '20 at 05:52
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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.

Hans Lundmark
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  • The wikipedia formula does not involve covariant derivative. Actually you get the one I'm talking about by replacing $[X,Y]$ with $D_X Y - D_Y X$. – Timofei Mar 31 '11 at 10:40
  • @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. – Hans Lundmark Mar 31 '11 at 11:15