Hodge Laplacian and scalar

Question

I'm reading Nakahara GEOMETRY, TOPOLOGY AND PHYSICS now.Then, Hodge Laplacian is given by \begin{align} \Delta=(d+d^{\dagger})^2=dd^{\dagger}+d^{\dagger}d \end{align} For example, we consider 0-form $f$, then $d^{\dagger}f=0$ \begin{align} \Delta f&=d^{\dagger}df=d^{\dagger}(\partial_{\mu}f dx^{\mu})\\ &=-\ast d \ast(\partial_{\mu}f dx^{\mu})=-\ast d(\frac{\sqrt{|g|}}{(m-1)!}\partial_{\mu}f g^{\mu \lambda}\epsilon_{\lambda \nu_2 \cdot \nu_m} dx^{\nu_2}\wedge \cdots \wedge dx^{\nu_m})\\ &=-\ast \frac{1}{(m-1)!}\partial_{\nu}[\sqrt{|g|}g^{\mu \lambda} \partial_{\mu}f] \epsilon_{\lambda \nu_2 \cdot \nu_m} dx^{\nu}\wedge dx^{\nu_2}\wedge \cdots \wedge dx^{\nu_m} \end{align} So far, so good, but I don't understand the next transformation. \begin{align} \Delta f&=-\ast \partial_{\nu}[\sqrt{|g|}g^{\mu \lambda} \partial_{\mu}f] g^{-1} dx^1\wedge \cdots \wedge dx^m\\ &=-\frac{1}{\sqrt{|g|}}\partial_{\nu}[\sqrt{|g|}g^{\mu \lambda} \partial_{\mu}f] \end{align} Why did $g^{-1}$ come up? I would appreciate it if you could tell me. For your information, $\ast$ is defined as follows. \begin{align} \ast (dx^{\mu_1}\wedge dx^{\mu_2}\wedge \cdots \wedge dx^{\mu_r})=\frac{\sqrt{|g|}}{(m-r)!}\epsilon^{\mu_1\mu_2 \cdots \mu_r}\ _{\nu_{r+1}\cdots \nu_{m}}dx^{\nu_{r+1}}\wedge \cdots \wedge dx^{\nu_{m}} \end{align} where $m$ is just dimension of manifold.

Answer 1

Let's start with this expression,

$-\ast \frac{1}{(m-1)!}\partial_{\nu}[\sqrt{|g|}g^{\mu \lambda} \partial_{\mu}f] \epsilon_{\lambda \nu_2 \cdots\nu_m} dx^{\nu}\wedge dx^{\nu_2}\wedge \cdots \wedge dx^{\nu_m}.$

Now,

\begin{equation} \ast dx^{\nu}\wedge dx^{\nu_2}\wedge \cdots \wedge dx^{\nu_m}=\sqrt{|g|}\epsilon^{\nu\nu_2\cdots\nu_m}. \end{equation}

Therefore, the first expression becomes,

\begin{equation} \frac{1}{(m-1)!}\partial_{\nu}[\sqrt{|g|}g^{\mu \lambda} \partial_{\mu}f] \epsilon_{\lambda \nu_2 \cdots\nu_m}\sqrt{|g|}\epsilon^{\nu\nu_2\cdots\nu_m} \end{equation}

Using equation (7.171b) of the book, we can express the Levi Civita tensor with all contravariant indices in terms of the tensor with all covariant indices, \begin{equation} \frac{1}{(m-1)!}\partial_{\nu}[\sqrt{|g|}g^{\mu \lambda} \partial_{\mu}f] \epsilon_{\lambda \nu_2 \cdots\nu_m}\sqrt{|g|}g^{-1}\epsilon_{\nu\nu_2\cdots\nu_m}=\frac{1}{(m-1)!}\partial_{\nu}[\sqrt{|g|}g^{\mu \lambda} \partial_{\mu}f] \frac{1}{\sqrt{|g|}}(m-1)!\delta_{\lambda}^{\nu}. \end{equation} The second expression follows from a property of the Levi Civita tensor (see Wikipedia). This gives the desired expression.