How to derive the coordinate expression of the Hodge dual?

Question

I'm trying to obtain the coordinate expression of the Hodge dual.

A possible definition of the Hodge dual of a $r$-form $w$, given a metric $g$, is the unique $n-r$-form such that

$$ v \wedge \star w = \langle v,w\rangle \omega \tag1$$

for any $r$-form $v$.

I should obtain that the components of the Hodge dual are

$$\star w_{\mu_{r+1}\ \ ...\mu_n}=g_{\mu_{r+1}\ \ \nu_{r+1}}\ \dots g_{\mu_{n}\nu_{n}} \frac{\epsilon^{\nu_{r+1}\ \ ...\nu_n \sigma_1...\sigma_r}}{\sqrt g} w_{\sigma_1...\sigma_r} \tag2$$

I tried substituting $\omega=\sqrt g dx^1\wedge...\wedge dx^n$ and $$v\wedge \star w = v_{\alpha_1...\alpha_r} \star \omega_{\mu_{r+1}\ \ ...\mu_n} \epsilon^{\alpha_1...\alpha_r\mu_{r+1}\ \ ...\mu_n} \ dx^1\wedge ...\wedge dx^n \tag3$$

but I don't know how to expand $\langle v,w\rangle$. I only know it involves the scalar product and the determinant. So, what is the precise expression of $\langle v,w\rangle$?

Answer 1

So, what is the precise expression of $⟨v,w⟩$?

Let $(M,g)$ be an orientable Riemannian manifold of dimension $n$. For any $0 \le k \le n$, there is a unique extension of the metric $g$ to the metric $⟨ \, ,\, ⟩: \Omega^k(M) \times \Omega^k(M) \to \mathbb{R}$ such that for any O.N.B. $ \{e^1, ...,e^n \} $ of one-forms, the basis of $\Omega^k(M)$ is an O.N.B. with respect to $⟨ \, ,\, ⟩$

$$\text{i.e. } ⟨ e^{i_1} \wedge...\wedge e^{i_k} \, ,\, e^{j_1} \wedge...\wedge e^{j_k}⟩ = \delta^{i_1 j_1}...\delta^{i_k j_k} \, .\tag4$$

More generally, for $v^1,...,v^k,w^1,...,w^k \in \Omega(M)$,

$$ ⟨v^1\wedge...\wedge v^k,w^1\wedge...\wedge w^k⟩ := \mathrm{det}(g(v^i,w^j)) \,. \tag5$$

It is easy to check that this form is bilinear, symmetric and positive-definite.

Now suppose $ v := v_{i_1...i_k} e^{i_1} \wedge...\wedge e^{i_k}$ and $w:=w_{j_1...j_k}e^{j_1} \wedge...\wedge e^{j_k}$ are the coordinate representations of the $k$-forms $v$ and $w$. Therefore,

\begin{align*} ⟨v,w⟩ &= v_{i_1...i_k} \ w_{j_1...j_k} \,\delta^{i_1 j_1}...\delta^{i_k j_k} \\ &= v_{i_1...i_k} \ w^{i_1...i_k} \,. \tag8 \\ \end{align*}

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How to derive the coordinate expression of the Hodge dual?

To begin with, I will take your definition and write it down as follows.

The Hodge dual is the unique isomorphism

\begin{align*} \star:\Omega^k(M) &\to \Omega^{n-k}(M), \\ \omega &\mapsto \star \omega \\ \end{align*}

such that the following holds:

$$ \forall \omega, \eta \in \Omega^k(M): \omega \wedge \star \eta = ⟨\omega,\eta⟩ \ vol$$

where $vol := \sqrt{g} \ \mathrm{d} x^1 \wedge ... \wedge \mathrm{d}x^n$ is the volume form.

I have purposefully deviated from the notation used in (1) to avoid confusion between $w$ and $\omega$. Moreover, I believe that (2) is not correct at all. I will derive the correct coordinate representation below. Since the Hodge star operator is a $C^\infty (M)$-linear map, it suffices to evaluate it on the basis elements.

Proposition:

$$ \star (\mathrm{d}x^{i_1}\wedge...\wedge \mathrm{d}x^{i_k}) = \frac{\sqrt{g}}{(n-k)!} g^{i_1 j_1} ... g^{i_k j_k} \ \varepsilon_{j_1 ... j_n} \ \mathrm{d}x^{j_{k+1}} \wedge ... \wedge \mathrm{d}x^{j_n}\,.\tag9$$

This immediately implies

$$ \boxed{(\star \omega)_{j_{k+1} ... j_n} = \frac{\sqrt{g}}{(n-k)!} g^{i_1 j_1} ... g^{i_k j_k} \ \varepsilon_{j_1 ... j_n} \ \omega_{i_1 ... i_k}} \,. \tag{10}$$

Note the structural difference between this (10) and your claim (2).

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Lemma: The permutation symbol obeys

$$ \varepsilon^{i_1 ... i_k} = \frac{1}{(n-k)!} g^{i_1 j_1} ... g^{i_k j_k} \ \varepsilon_{j_1 ... j_n} \ \varepsilon^{j_{k+1} ... j_n} \,.\tag{11}$$

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Proof of Prop.: Let $\eta := \mathrm{d}x^{i_1}\wedge...\wedge \mathrm{d}x^{i_k}$ and $\widetilde{\eta} := \frac{\sqrt{g}}{(n-k)!} g^{i_1 j_1} ... g^{i_k j_k} \ \varepsilon_{j_1 ... j_n} \ \mathrm{d}x^{j_{k+1}} \wedge ... \wedge \mathrm{d}x^{j_n}$. In order that $\widetilde{\eta} = \star \eta$, we have to check if

$$ \forall \omega \in \Omega^k(M): \omega \wedge \widetilde{\eta} = ⟨\omega,\eta⟩ \ vol$$

Since both sides are linear in $\omega$, it suffices to check on a basis. Let $\omega := \mathrm{d}x^{1}\wedge...\wedge \mathrm{d}x^{k}$. Then, one immediately notices that $\eta = \varepsilon^{i_1 ... i_k} \omega$.

Observe that \begin{align*} ⟨\omega,\eta⟩ \ vol &= \varepsilon^{i_1 ... i_k} \ ⟨\omega,\omega⟩ \sqrt{g} \ \mathrm{d} x^1 \wedge ... \wedge \mathrm{d}x^n\\ &= \sqrt{g} \ \varepsilon^{i_1 ... i_k} \ g^{1 j_1} ... g^{k j_k} \ \varepsilon_{j_1 ... j_k} \mathrm{d} x^1 \wedge ... \wedge \mathrm{d}x^n\\ &= \sqrt{g} \ \varepsilon^{i_1 ... i_k} \ \mathrm{d} x^1 \wedge ... \wedge \mathrm{d}x^n\\ &= \sqrt{g} \ \Big( \frac{1}{(n-k)!} g^{i_1 j_1} ... g^{i_k j_k} \ \varepsilon_{j_1 ... j_n} \ \varepsilon^{j_{k+1} ... j_n} \Big) \ \mathrm{d} x^1 \wedge ... \wedge \mathrm{d}x^n\\ &= \mathrm{d} x^1 \wedge ... \wedge \mathrm{d} x^k \wedge \Big( \frac{\sqrt{g}}{(n-k)!} g^{i_1 j_1} ... g^{i_k j_k} \ \varepsilon_{j_1 ... j_n} \Big) \ \mathrm{d} x^{j_{k+1}} \wedge ... \wedge \mathrm{d}x^{j_n}\\ &= \omega \wedge \widetilde{\eta} \end{align*}

This concludes our proof. $\tag{Q.E.D.}$

Answer 2

I wanted to complement Nanashi's answer with a different approach. It has the advantage that it is constructive.

First, let me for notational convenience skip the wedges. I will also use the notation that every $k$-form is expressed uniquely as $$\mu=\frac{1}{k!}\mu_{a_1\cdots a_k}dx^{a_1}\cdots dx^{a_k},$$ with $\mu_{a_1\cdots a_k}$ antisymmetric.

With this notation we have $$dx^{a_1}\cdots dx^{a_k}\star\mu=\frac{1}{(d-k)!}\epsilon^{a_1\cdots a_d}(\star\mu)_{a_{k+1}\cdots a_d}d^dx.$$ In here I have used the notation $$d^dx=dx^1\cdots dx^d,$$ and the simple realization that since $\Omega^d(M)$ is one-dimensional and the product of forms is antisymmetric $$dx^{a_1}\cdots dx^{a_d}=\epsilon^{a_1\cdots a_d}d^dx.$$

On the other hand, by definition of the $\star$, we have $$dx^{a_1}\cdots dx^{a_k}\star\mu=\langle dx^{a_1}\cdots dx^{a_k},\mu\rangle dvol_g.$$ We can also expand the inner product into components $$dx^{a_1}\cdots dx^{a_k}\star\mu=\frac{1}{k!}\mu_{b_1\cdots b_k}\langle dx^{a_1}\cdots dx^{a_k},dx^{b_1}\cdots dx^{b_k}\rangle dvol_g.$$ Using the definition of the determinant, as well as the fact that $\langle dx^a,dx^b\rangle=g^{ab}$ we further have $$dx^{a_1}\cdots dx^{a_k}\star\mu=\frac{1}{k!}\sum_{\sigma\in S_k}sign(\sigma)\mu_{b_1\cdots b_k}g^{a_{\sigma(1)} b_{1}}\cdots g^{a_{\sigma(k)} b_{k}} dvol_g.$$ Now, by symmetry we can reorder $$g^{a_{\sigma(1)} b_{1}}\cdots g^{a_{\sigma(k)} b_{k}}=g^{a_{1} b_{\sigma^{-1}(1)}}\cdots g^{a_{k} b_{\sigma^{-1}(k)}}.$$ Doing this reordering on the differential form we obtain another sign of the permutation $$\mu_{b_1\cdots b_k}=sign(\sigma)\mu_{b_{\sigma^{-1}(1)}\cdots b_{\sigma^{-1}(k)}}.$$ Given that $sign(\sigma)^2=1$ and relabelling the dummy indices $b_{\sigma^{-1}(j)}\mapsto b_j$, we obtain $$dx^{a_1}\cdots dx^{a_k}\star\mu=\frac{1}{k!}\sum_{\sigma\in S_k}\mu_{b_1\cdots b_k}g^{a_1 b_{1}}\cdots g^{a_k b_{k}} dvol_g.$$ Now, this is independent of the permutation, so that the sum cancels the $k!$ term $$dx^{a_1}\cdots dx^{a_k}\star\mu=\mu_{b_1\cdots b_k}g^{a_1 b_{1}}\cdots g^{a_k b_{k}} dvol_g.$$ Finally, using the component form $dvol_g=\sqrt{g}d^dx$, we obtain $$\frac{1}{(d-k)!}\epsilon^{a_1\cdots a_d}(\star\mu)_{a_{k+1}\cdots a_d}=\mu_{b_1\cdots b_k}g^{a_1 b_{1}}\cdots g^{a_k b_{k}} \sqrt{g}.$$

We can proceed to multiply by a Levi-Civita symbol $$\frac{1}{(d-k)!}\epsilon_{a_1\cdots a_k b_{k+1}\cdots b_d}\epsilon^{a_1\cdots a_d}(\star\mu)_{a_{k+1}\cdots a_d}=\epsilon_{a_1\cdots a_k b_{k+1}\cdots b_d}\mu_{b_1\cdots b_k}g^{a_1 b_{1}}\cdots g^{a_k b_{k}} \sqrt{g}.$$ Only the $(a_1,\dots,a_k)$ for which $(a_1,\dots,a_k,b_{k+1},\dots,b_d)$ are a permutation of $(1,\dots,d)$ will contribute. There are $k!$ of these (assuming of course, that $b_{k+1},\dots,b_d$ are all different, otherwise we just have $0=0$). For each, only the $(a_{k+1},\dots,a_k)$ which are permutations of $(b_{k+1},\dots,b_d)$ will contribute. There are $(d-k)!$ of these. For each of these, we can reorder $(a_{k+1},\dots,a_k)$ into $(b_{k+1},\dots,b_d)$ without obtaining a sign as long as we do the same on the components of $\star\mu$. After doing this, both Levi-Civita symbols contribute the same sign and thus cancel. We conclude that all choices give $(\star\mu)_{b_{k+1}\cdots b_d}$ and there are $k!(d-k)!$ of them, so that $$\frac{k!(d-k)!}{(d-k)!}(\star\mu)_{b_{k+1}\cdots b_d}=\epsilon_{a_1\cdots a_k b_{k+1}\cdots b_d}\mu_{b_1\cdots b_k}g^{a_1 b_{1}}\cdots g^{a_k b_{k}} \sqrt{g}.$$ We conclude that, after a trivial relabeling $$(\star\mu)_{a_{k+1}\cdots a_d}=\frac{1}{k!}\epsilon_{a_1\cdots a_d }\mu_{b_1\cdots b_k}g^{a_1 b_{1}}\cdots g^{a_k b_{k}} \sqrt{g}.$$

This differs from Nanashi's answer because they used the convention $$\mu=\mu_{a_1\cdots a_k}dx^{a_1}\cdots dx^{a_k}.$$