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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?

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

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    $\begingroup$ $\uparrow$ Yes. $\endgroup$ – Qmechanic Aug 21 '18 at 17:50
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    $\begingroup$ You can prove it! Assuming there is no torsion means you have symmetric Christoffel symbols, and you know how to write the covariant derivative in terms of Christoffel symbols, so take the derivatives and simplify until you get only that product rule. $\endgroup$ – N. Steinle Aug 21 '18 at 18:03
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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.

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  • $\begingroup$ I added an extra comment to me answer. $\endgroup$ – mike stone Aug 22 '18 at 18:44

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