Exterior and Covariant Derivatives

Question

Is the following guaranteed to be true for any covariant vector $f_\mu$ (1-form $\boldsymbol{f}$) in the absence of torsion? $$\nabla_{[\alpha}\nabla_{\beta}f_{\mu]}=\partial_{[\alpha}\partial_{\beta}f_{\mu]}=\boldsymbol{d}\boldsymbol{d}\boldsymbol{f}=0,$$ where $\nabla_{\alpha}$ is the covariant derivative, $\partial_{\beta}$ is the partial derivative, and $\boldsymbol{d}$ is the exterior derivative, and brackets in the subscript means antisymmetrization.

$\partial_{[\alpha}\partial_{\beta}f_{\mu]}=\boldsymbol{d}\boldsymbol{d}\boldsymbol{f}$ is just the definition of $\boldsymbol{d}$ and is everywhere in textbooks, and the fact that it is zero is also all over the place. So my real question is:

Is the following always true?

$\nabla_{[\alpha}\nabla_{\beta}f_{\mu]}=\partial_{[\alpha}\partial_{\beta}f_{\mu]}$

Answer 1

I think you need the first Bianchi identity (which holds for torsion free). Your antisymmetrized expression is the LHS of

$$ [\nabla_a, \nabla_b] f_c+ [\nabla_b, \nabla_c] f_a+[\nabla_c, \nabla_a] f_b =- f_d({R^d}_{cab}+ {R^d}_{abc}+{R^d}_{cab})\\\qquad\qquad =0 $$ I think my minus sign is correct for the covariant case of the commutator/curvature expression.

Added comment: Actually your derivation using $d^2=0$ is also quite correct --- so my first Bianchi identity route can be turned around to give a neat (and previously unknown to me) proof of the first Bianchi identity for torsion-free connections.