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If we have a covariant derivative for vector fields given by an affine connection on a manifold, can we extend that to a covariant derivative for k-vectors by assuming that the product rule hold for wedge products of vectors? In other words, does it make sense to assume that:

$$\nabla_v(a\wedge b) = (\nabla_va)\wedge b+a\wedge (\nabla_vb) $$

where $v$ is a vector field, $a$ and $b$ are $p$-vector and $q$-vector fields, and $\nabla_va$ agrees with the covariant derivative of vectors when $a$ is a vector field?

(I am wondering since the product rule for the exterior derivative has a grade-dependent sign in it, but I think that should not be the case here.)

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Yes, $\nabla_v (a\wedge b)=(\nabla_v a)\wedge b+a\wedge(\nabla_v b)$ for any $v\in TM$ and $a,b\in\mathcal{T}^{\bullet,0}M$. This is the Leibniz rule for tensor product, projected down to the alternating parts. See, for example, this question.

(The $(-1)^k$ sign for exterior derivative comes from "moving $v$ pass other vectors" when impose the constraint $v$ also participate in the antisymmetry.)

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