In Sean Carroll's GR book, pg 84, the exterior derivative $d$ is defined as $$(dA)_{\mu_1\mu_2...\mu_{p+1}} = (p+1) \partial_{[\mu_1}A_{\mu_2...\mu_{p+1}]}\tag{2.76}$$ where $A$ is a $p$-form and the RHS is the appropiately normalized and antisymmetrized partial derivative.

But on pg. 454, equation E.5, he replaced the partial derivative with a covariant derivative:

$(dA)_{\mu_1\mu_2...\mu_{p+1}} = (p+1) \nabla_{[\mu_1}A_{\mu_2...\mu_{p+1}]}\tag{2.76}$

Are these two definitions the same?


2 Answers 2


Provided that the connection is torsion-free, $\Gamma^{\sigma}_{[\mu\nu]}=0$, there's a neat cancellation that happens. I'll work out a few examples to get a feel for this: $$ \require{cancel} \newcommand\ccancel[2][black]{\color{#1}{\cancel{\color{black}{#2}}}} (dA)_{\mu_1\mu_2\dots\mu_{p+1}}\sim\nabla_{[\mu_1}A_{\mu_2\dots\mu_{p+1}]} $$

$$ \\(dA)_{\mu\nu}\sim\nabla_{[\mu}A_{\nu]} \tag{p = 1} \\\sim\nabla_\mu A_\nu-\nabla_\nu A_\mu \\=\partial_\mu A_\nu-\ccancel[red]{\Gamma^{\sigma}{}_{\mu\nu}A_\sigma} \\-\partial_\nu A_\mu+\ccancel[red]{\Gamma^{\sigma}{}_{\nu\mu}A_\sigma} \\=\partial_{[\mu}A_{\nu]} $$

$$ (dA)_{\mu\nu\sigma}\sim\nabla_{[\mu}A_{\nu\sigma]} \tag{p = 2} \\\sim\nabla_\mu A_{\nu\sigma}-\nabla_\nu A_{\sigma\mu}+\nabla_\sigma A_{\mu\nu} \\=\partial_\mu A_{\nu\sigma}-\ccancel[red]{\Gamma^\rho{}_{\mu\nu}A_{\rho\sigma}}-\ccancel[blue]{\Gamma^\rho{}_{\mu\sigma}A_{\rho\nu}} \\-\partial_\nu A_{\sigma\mu}+\ccancel[green]{\Gamma^\rho{}_{\nu\sigma}A_{\rho\mu}}+\ccancel[red]{\Gamma^\rho{}_{\nu\mu}A_{\rho\sigma}} \\+\partial_\sigma A_{\mu\nu}-\ccancel[blue]{\Gamma^\rho{}_{\sigma\mu}A_{\rho\nu}}-\ccancel[green]{\Gamma^\rho{}_{\sigma\nu}A_{\rho\mu}} \\=\partial_{[\mu}A_{\nu\sigma]} $$

since the Christoffel symbols cancel out diagonally due to the absence of torsion. It is not the greatest leap of faith to assume that this result generalises for all $p$-forms. However, we have not yet made use of the crucial point - the antisymmetry of the form! (although we didn't actually need to for the $p=1$ and $p=2$ cases)

There are prima facie three antisymmetric "factors" here: the torsion-free Christoffel symbols, the antisymmetrisation over all the indices, and the antisymmetric nature of the forms.

Let $(a,b,c,...n)$ be a permutation of $(1,2,...p+1)$. For the exterior derivative of a $p$-form, $(dA)_{\mu_1\mu_2\dots\mu_{p+1}}$, compare the terms $\Gamma^\rho{}_{\mu_a\mu_b}A_{\rho\mu_c...\mu_n}$ and $\Gamma^\rho{}_{\mu_b\mu_a}A_{\rho\,\sigma(\mu_c...\mu_n)}$ where $\sigma$ is some permutation (remember, $a$ and $b$ need not be adjacent numbers). Both terms are unique, and are specified up to a sign, which we would like to be opposite in order to get them to cancel out. There are two cases:

  • $\sigma$ is even: the Christoffel symbols give a minus sign, the antisymmetry of the form does nothing and the overall antisymmetrisation does nothing - in total, a relative minus sign

  • $\sigma$ is odd: the Christoffel symbols give us a minus sign, the antisymmetry of the form gives us a minus sign and the overall antisymmetrisation also yields a minus sign - again in total, a relative minus sign

We conclude that $$\Gamma^\rho{}_{\mu_b\mu_a}A_{\rho\,\sigma(\mu_c...\mu_n)}=-\Gamma^\rho{}_{\mu_a\mu_b}A_{\rho\mu_c...\mu_n},$$

for the specific permutation $\sigma$ of the indices that appears in the exterior derivative. So all the Christoffel symbol terms vanish. Thus, depending on the normalisation conventions of the author, $(dA)_{\mu_1\mu_2\dots\mu_{p+1}}\sim\nabla_{[\mu_1}A_{\mu_2\dots\mu_{p+1}]}$ up to a prefactor.


Nihar Karve provided an explanation why the two formulas which are denoted by $(\text{2.76})$ in the original post are identical in a curved spacetime with no torsion, i.e. with a symmetric connection. Here is the missing part to his post (which is useful in GR extensions with torsion):

Let $\nabla$ be an operator which generalizes the $d$ from curved spacetime with no torsion. This acts on a p-form on a (cotangent bundle of a) manifold with torsion $T^{\mu}_{~~\nu\rho} :=\Gamma^{\mu}_{~~\nu\rho} - \Gamma^{\mu}_{~~\rho\nu}$.

Then: $$ \displaystyle{\left(\nabla\omega\right)_{\sigma\mu_1 \mu_2...\mu_p} :=\left(d\omega\right)_{\sigma\mu_1 \mu_2...\mu_p} + (-)^p \frac{p(p+1)}{2}T^{\nu}_{~~[\sigma\mu_1} \omega_{\mu_2...\mu_p]\nu}} $$

The first term is given by $(\text{2.76})$ in the original post with $\nabla_s$ a symmetric connection. Of course, since $d^2 =0$, we have that $\nabla\nabla\omega \neq 0$.


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