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 $\langle v,w\rangle$?
Best Answer
So, what is the precise expression of $⟨v,w⟩$?
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*}
How to derive the coordinate expression of the Hodge dual?
To begin with, I will take your definition and write it down as follows.
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.
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).
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. $\tag{Q.E.D.}$