# 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:

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

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

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

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:

$$$$F = \partial_\theta A_{ \varphi} \; \mathrm{d}\theta \wedge \mathrm{d} \varphi$$$$

Applying the Hodge operator according to the formula above gives:

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

such that:

$$$$*F = (\partial_\theta A_{ \varphi}) \mathrm{d} r$$$$

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

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

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.

Baez and Muniain define the Hodge operator as:

$$$$\omega \wedge * \mu = \langle \omega , \mu \rangle \mathrm{vol}$$$$

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?

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

1. the Levi-Civita symbol $$[\mu_1,\ldots,\mu_d]$$, which is not a tensor (but rather a $$(d,0)$$ tensor density) and whose values are only $$0$$ and $$\pm 1$$; and

2. the Levi-Civita tensor$$^1$$ $$\varepsilon^{\mu_1,\ldots,\mu_d}~=~\pm \frac{[\mu_1,\ldots,\mu_d]}{\sqrt{|g|}} , \qquad \varepsilon_{\mu_1,\ldots,\mu_d}~=~\pm \sqrt{|g|}[\mu_1,\ldots,\mu_d] ,$$ whose definition differs from the Levi-Civita symbol by a factor of $$\sqrt{|g|}\equiv \sqrt{|\det(g_{\mu\nu})|}$$.

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$$^1$$ The $$\pm$$ is included to acknowledge that different authors have different conventions.

When working in index notation, we should keep in mind that we should get a formulae that is appropriately covariant under general co-ordinate transformations. Hence, we should use this general formulae when working with index notation:

You can see how the inverse metric and the determinant factor of the metric conspire to give you the correct factor of $$\frac{1}{r^2 sin(\theta)}$$.

Hodge star operator on a vector space V is a linear operator on the exterior algebra of V mapping k vectors to (n-k) vectors with n = dim V . So, given two k vectors a, b; a /\ *b = < a b> w. So it bringing k vector outside of k dim of k vector space. So, in your curvature vector space between theta and phi , you took inner product and then projected onto outside space by multiplying by appropriate unit vector.