# How do we compute Hodge duals?

The motivation for this question is to try to come up with a general expression for $$(\star F)_{\mu\nu}$$, the $$\mu,\nu$$ component of the Hodge dual of the Field strength tensor, which is of great importance in Electrodynamics. In the course I am taking, the instructor has defined it contravariantly, as $$(\star F)^{\mu\nu}=\frac{1}{2}\varepsilon^{\mu\nu\alpha\beta}F_{\alpha\beta}$$ Where $$F_{\alpha\beta}$$ are the components of the normal field strength tensor, $$F_{\alpha\beta}=\nabla_\alpha A_\beta-\nabla_\beta A_\alpha$$ And we have defined the raised Levi-Civita symbol using the generalized Kronecker delta: $$\varepsilon^{ijkl}=\delta^{i~j~k~l}_{1~2~3~4}$$ However, there are some problems with this equation. As far as I know, it only holds when the metric is simple as possible, i.e $$\boldsymbol{\eta}=\operatorname{diag}(-1,1,1,1)$$ the standard Minkowski metric. I don't think it holds in other coordinate systems e.g polar coordinates. Also, I thought the Hodge star was a map that took $$k$$ forms to $$d-k$$ forms, where $$d$$ is the dimension of the space. But my instructor's equation seems to have the Hodge dual being a fully contravariant tensor... are the components he is listing actually the components of $$(\star\mathbf{F})^{\sharp}$$, its index-raised counterpart?

With this motivation, I searched the internet for answers and stumbled across this post on physics SE. Given a $$d$$ dimensional manifold, the author of the post defines the Hodge dual of a totally skew-symmetric $$k$$ form $$\omega$$ as having components

$$(\star \omega)_{\mu_1\dots\mu_{d-k}}=\frac{1}{k!}\epsilon^{\nu_1\dots\nu_k}{}_{\mu_1\dots\mu_{d-k}}\omega_{\nu_1\dots\nu_k}$$

However, the author does not make it very clear what the object $$\epsilon^{\nu_1\dots\nu_k}{}_{\mu_1\dots\mu_{d-k}}$$ is. He does write $$\epsilon_{\mu_1\dots\mu_d}=\sqrt{-g}~\varepsilon_{\mu_1\dots\mu_d}$$ But not only does this look very different to$$\epsilon^{\nu_1\dots\nu_k}{}_{\mu_1\dots\mu_{d-k}}$$ in terms of the number and structure of the indices, it also raises other concerns, like what if $$g=\det \mathbf{g}$$ is a positive number? Will we start getting imaginary components of the Hodge dual, even when we are working on a real manifold?

My request: Could someone please provide some other formula for the Hodge dual of a totally skew symmetric $$k$$ form on a $$d$$ dimensional manifold, in terms of well known objects, such as the metric, Christoffel symbols, Kronecker delta, Levi-Civita symbol, etc? I searched the Wikipedia page for answers but I found it very dense and difficult to understand.

• Does this help? It gives you an alternate perspective. mathoverflow.net/questions/162366/… Commented Dec 6, 2021 at 21:18
• Looks like the components are those of $(\star F)^\sharp$. For the generalized Kroneckar, did you mean something like $\delta_{1234}^{ijkl}$? The expression involving $g$ is $\sqrt{|\det(g)|}$, no imaginary units are present. The Levi-Civita symbol has the same components in any co-ordinate system. To construct the Levi-Civita tensor, note $\epsilon^{\text{tensor}}_{ij\dots}=\sqrt{|g|}\varepsilon^{\text{symbol}}_{ij\dots}$. The expression in your post for Hodge dual is correct, and includes a mixture of up/ down indices on the Levi-Civita tensor, which may be raised using the metric
– Sal
Commented Dec 6, 2021 at 22:00
• @Sal yes, I did mean 1234... i will correct it. If it is supposed to be $\sqrt{|g|}$, why do I see so many people write $\sqrt{-g}$? Commented Dec 6, 2021 at 22:10
• @Sal So are you suggesting that $$\epsilon^{\nu_1\dots\nu_k}{}_{\mu_1\dots\mu_{d-k}}=\sqrt{|g|}\varepsilon^{\nu_1\dots\nu_k}{}_{\mu_1\dots\mu_{d-k}} \\=\sqrt{|g|}g^{\alpha_1 \nu_1}\dots g^{\alpha_k\nu_k}\varepsilon_{\alpha_1\dots\alpha_k~\mu_1\dots\mu_{d-k}}$$ ? Commented Dec 6, 2021 at 22:14
• Yes, that is precisely what I'm suggesting. I guess it's often written as $\sqrt{-g}$ because in 3+1 dimensions the determinant is necessarily negative... and it makes equations look impressively complex?
– Sal
Commented Dec 6, 2021 at 22:41

Since the Hodge dual maps $$k$$ forms to $$n-k$$ forms, and covariant skew-symmetric tensors are forms, the contravariant expression for $$(\star F)$$ must be interpreted as $$(\star F)^\sharp$$.

The expression for Hodge dual in OP is correct. What appears on the RHS is the Levi-Civita tensor, with a mixture of up/ down indices. The Levi-Civita symbol, which has the same components in all co-ordinates, is related to it by

$$\epsilon^{\text{tensor}}_{ij\dots}=\sqrt{|g|}\varepsilon^{\text{symbol}}_{ij\dots}$$

Since this is a tensor, indices may be raised and lowered with the metric as necessary

$$\epsilon^{\nu_1\dots\nu_k}{}_{\mu_1\dots\mu_{d-k}}=\sqrt{|g|}g^{\alpha_1 \nu_1}\dots g^{\alpha_k\nu_k}\varepsilon_{\alpha_1\dots\alpha_k~\mu_1\dots\mu_{d-k}}$$

The expression $$\sqrt{-g}$$ is more generally $$\sqrt{|\det(g)|}$$. In Lorentz spacetime the determinant is necessarily negative so there is no harm (and no imaginary units) in writing $$\sqrt{-g}$$.

References: Spacetime and geometry by S. Carrol.