differential geometry - Hodge star operator on curvature?

I've a question regarding the Hodge star operator. I'm completely new to the notion of exterior derivatives and wedge products. I had to teach it to myself over the past couple of days, so I hope my question isn't trivial.

I've found the following formulas on the internet, which seem to match the definitions of the two books (Carroll and Baez & Muniain) that I own. For a general $p$-form on a $n$-dimensional manifold:

\begin{equation} v=\frac{1}{p!} v_{i_1 \ldots i_p} \mathrm{d}x^{i_1} \wedge \cdots \wedge \mathrm{d}x^{i_p} \end{equation}

the Hodge operator is defined to act on the basis of the $p$-form as follows:

\begin{equation} *\left( \mathrm{d}x^{i_1} \wedge \cdots \wedge \mathrm{d}x^{i_p} \right) = \frac{1}{q!} \tilde{ \varepsilon}_{j_1,\ldots,j_q}^{i_1,\ldots,i_p} \mathrm{d} x^{j_1} \wedge \cdots \wedge \mathrm{d} x^{j_q} \end{equation}

where $q=n-p$ and $\tilde{ \varepsilon}$ is the Levi-Civita tensor. Up until here everything is fine, I managed to do some exercises and get the right answers. However, actually trying to calculate the curvature does cause some problems with me.

To give a bit of background. I'm working with a curvature in a Yang-Mills theory in spherical coordinates $(r, \theta, \varphi)$. Using gauge transformation, I've gotten rid of time-dependence, $r$ dependence and $\theta$ dependence. Therefore, the curvature is given by:

\begin{equation} F = \partial_\theta A_{ \varphi} \; \mathrm{d}\theta \wedge \mathrm{d} \varphi \end{equation}

Applying the Hodge operator according to the formula above gives:

\begin{equation} * \left(\mathrm{d} \theta \wedge \mathrm{d} \varphi\right) = \frac{1}{(3-2)!} \tilde \varepsilon^{\theta \varphi}_r \mathrm{d}r=\mathrm{d}r \end{equation}

such that:

\begin{equation} *F = (\partial_\theta A_{ \varphi}) \mathrm{d} r \end{equation}

However, three different sources give a different formula. Specifically they give:

\begin{equation} *F = (\partial_\theta A_{ \varphi}) \frac{1}{r^2 \sin \theta} \mathrm{d} r \end{equation}

It is not clear to me where they get this from. Something is being mentioned about the fact that the natural volume form is $\sqrt{g} \; \mathrm{d} r \wedge \mathrm{d} \varphi \wedge \mathrm{d} \theta$ with $\sqrt{g}=r^2 \sin \theta$, which I agree with. However, I do not understand why that term is incorporated in the Hodge operator.

Boaz and Muniain define the Hodge operator as:

\begin{equation} \omega \wedge * \mu = \langle \omega , \mu \rangle \mathrm{vol} \end{equation}

But I don't see how that formula is applicable to calculating the Hodge operator on the curvature. Could anybody tell me where I am going wrong or provide me a source where they explain this?

Answer

It seems the resolution to OP's question lies in the difference between

1. 
   

   the Levi-Civita symbol, which is not a tensor and whose values are only $0$ and $\pm 1$; and

   
   


2. 
   

   the Levi-Civita tensor, whose definition differs from the Levi-Civita symbol by a factor of $\sqrt{|\det(g_{\mu\nu})|}$.