Computation of exterior derivative of an $(n-1)$ form

Question

Suppose $(M,g)$ is a Lorentzian manifold of dimension $n$. Let $V$ be a one-form on $M$ and define the $(n-1)$ form $\omega = \ast V$ where $\ast$ is the Hodge dual.

In a chart $(U,x)$, if we have

$$\epsilon_{\mu_1\dots \mu_n}=\sqrt{|g|} \varepsilon_{\mu_1\dots \mu_n},$$

where $\varepsilon$ is the Levi civita symbol. Then the Hodge dual in components will be

$$\omega_{\mu_1\dots \mu_{n-1}}=\epsilon_{\nu \mu_1\dots \mu_{n-1}}V^\nu$$

where $V^\nu = g^{\mu\nu} V_\mu$ and $V = V_\mu dx^\mu$. That is all fine, now comes the thing. In the book Spacetime and Geometry by Sean Carroll the author performs the following computation:

$$(d\omega)_{\lambda \mu_1\dots \mu_{n-1}}=(d\ast V)_{\lambda \mu\dots \mu_{n-1}}=n\nabla_{[\lambda}(\epsilon_{|\nu| \mu_1\dots \mu_{n-1}]}V^\nu)=n\epsilon_{\nu[\mu_1\dots \mu_{n-1}}\nabla_{\lambda]}V^\nu.$$

I can't understand this computation. I have several issues:

1. The first is the notation. I can't really understand this bracket notation. All I understand is that if we have something like $A_{\mu_1\dots \mu_k}$ then $A_{[\mu_1\dots \mu_k]}=\frac{1}{k!}\sum_{\sigma \in S_k}\operatorname{sgn}(\sigma) A_{\mu_{\sigma (1)}\dots \mu_{\sigma (k)}}$ and that whenever an index is enclosed in bars it is not included in the permutations. I believe the author is actually using the fact that $d\omega = \operatorname{alt}(\nabla \omega)$.

2. I can't understand where the $n$ factor comes from. I have some handwavying understanding that it comes from not including one index of $\epsilon$ on the antisymetrization, but I can't rigorously show that this $n$ is there.

3. I also can't understand the manipulation with the covariant derivative, where it was brought inside the antisymetrization bracket. I believe it is one issue with notation for the covariant derivative. The notation I'm used to doesn't include indices. So it is like $\nabla_X \omega$ and thus I need to write $\omega$ explicitly first and differentiate it.

Anyway, what is behind this computation and how to address these points?

Answer 1

1. The first is the notation. I can't really understand this bracket notation. All I understand is that if we have something like $A_{\mu_1\dots \mu_k}$ then $A_{[\mu_1\dots \mu_k]}=\frac{1}{k!}\sum_{\sigma \in S_k}\operatorname{sgn}(\sigma) A_{\mu_{\sigma (1)}\dots \mu_{\sigma (k)}}$ and that whenever an index is enclosed in bars it is not included in the permutations. I believe the author is actually using the fact that $d\omega = \operatorname{alt}(\nabla \omega)$.

The bracket notation is exactly what you say it is. Do consider that different sources use different conventions when embedding the exterior algebra into the tensor algebra. The convention used by Carroll is the same that is used by most differential geometers (but not Kobayashi & Nomizu for example) is that $$ (\omega\wedge\mu)_{\mu_1...\mu_k\nu_1...\nu_l}=\frac{(k+l)!}{k!l!}\omega_{[\mu_1...\mu_k}\mu_{\nu_1...\nu_l]} \\ (d\omega)_{\mu_1...\mu_k\mu_{k+1}}=(k+1)\partial_{[\mu_1}\omega_{\mu_2...\mu_{k+1}]}. $$ The second follows from the first via some annoying combinatorics.

This also explains your second question - he exterior differentiated an $n-1$-th order form, so he got a factor of $n-1+1=n$ into the antisymmetrization brackets.

3. I also can't understand the manipulation with the covariant derivative, where it was brought inside the antisymetrization bracket. I believe it is one issue with notation for the covariant derivative. The notation I'm used to doesn't include indices. So it is like $\nabla_X \omega$ and thus I need to write $\omega$ explicitly first and differentiate it.

There are several things to note here.

1) The covariant derivative was also part of the antisymmetrization from the get-go. If you check the first formula involving the covariant derivative, the antisymmetrization bracket starts on $\nabla$'s index.

2) The covariant derivative ends up there, because the expression $$ (d\omega)_{\mu\mu_1...\mu_k}=(k+1)\nabla_{[\mu}\omega_{\mu_1...\mu_k]} $$ is insensitive to which connection $\nabla$ is put in there as long as it is torsionless. Because the exterior derivative is a local operation, we can also use a local chart's "trivial" connection, $\partial_\mu$, which acts by partially differentiating the tensor components to obtain the same result. So essentially, covariants can be used to calculate $d$ instead of partials.

3) If the connection is the Levi-Civita connection, then $\epsilon$ is constant with respect to it, the same way $g_{\mu\nu}$ is. So $\nabla$ is brought into the bracket because $\epsilon$ is a constant.

Appendix:

Here is more info on the factor of $(k+1)$. A differential k-form $\omega$ can be represented via tensor components as $$ \omega=\omega_{\mu_1...\mu_k}dx^{\mu_1}\otimes...\otimes dx^{\mu_k}, $$ where the array $\omega_{\mu_1...\mu_k}$ is totally antisymmetric, or via differential form components as $$ \omega=\sum_{\mu_1<...<\mu_k}\omega_{\mu_1...\mu_k}dx^{\mu_1}\wedge...\wedge dx^{\mu_k}, $$ where it doesn't make sense to talk about antisymmetry, because the indices have restricted values, or as a mixture by $$ \omega=\frac{1}{k!}\omega_{\mu_1...\mu_k}dx^{\mu_1}\wedge...\wedge dx^{\mu_k}, $$ here the components are tensor components, thus antisymmetric, and the sum is not restricted whatsoever.

What happens if we take a not necessarily antisymmetric array $\omega_{a_1...a_k}$ and form $\omega_{a_1...a_k}dx^{a_1}\wedge...\wedge dx^{a_k}$? (here I switched to latin indices because my fingers are falling off from the typing...)

Well, we have $dx^{a_1}\wedge...\wedge dx^{a_k}=dx^{[a_1}\wedge...\wedge dx^{a_k]}$, because the expression was already antisymmetric, but it is also true that if $A$ and $B$ are arbitrary arrays, then $A_{a_1...a_k}B^{[a_1...a_k]}=A_{[a_1...a_k]}B^{a_1...a_k}$, so $$ \omega_{a_1...a_k}dx^{a_1}\wedge...\wedge dx^{a_k}=\omega_{[a_1...a_k]}dx^{a_1}\wedge...\wedge dx^{a_k} $$ even if $\omega_{a_1...}$ was not antisymmetric.

Now let $\omega=\sum_{a_1<...<a_k}\omega_{a_1...a_k}dx^{a_1}\wedge...\wedge dx^{a_k}$ be a $k$-form. We have then $\omega=\frac{1}{k!}\omega_{a_1...a_k}dx^{a_1}\wedge...\wedge dx^{a_k}$ and $$ d\omega=\frac{1}{k!}\partial_a\omega_{a_1...a_k}dx^a\wedge dx^{a^1}\wedge...\wedge dx^{a_k}=\frac{1}{k!}\partial_{[a}\omega_{a_1...a_k]}dx^a\wedge dx^{a^1}\wedge...\wedge dx^{a_k}=\frac{(k+1)!}{k!}\partial_{[a}\omega_{a_1...a_k]} dx^{a}\otimes dx^{a_1}\otimes...\otimes dx^{a_k}, $$ where the factor of $(k+1)!$ appeared because $d\omega$ is a $k+1$-form and we switched to tensor component notation from differential form component notation, and we have $\frac{(k+1)!}{k!}=k+1$. So in tensor components we have $$ (d\omega)_{\mu\mu_1...\mu_k}=(k+1)\partial_{[\mu}\omega_{\mu_1...\mu_k]} $$