Hodge double star operator I want to prove that $**\omega=\left(-1\right)^{k\left(n-k\right)}\omega$, where $*$ is the Hodge star operator acting on differential $k$-forms $\omega$ on $\mathbb{R}^n$. Where can I find the proof of this? 
 A: The proof follows directly from applying the definition of the Hodge star twice. The most annoying thing is that you usually need some identity for the contraction of totally antisymmetric tensors. It's spelled out in Nakahara's "Geometry, Topology and Physics", page 291. He defines the dual as
$$
\star \omega = \frac{\sqrt{\vert g \vert}}{r!(m-r)!} \omega_{m_1 \dots m_r} \epsilon^{m_1 \dots m_r}_{\phantom{m_1\dots m_r} n_{r+1} \dots n_m} dx^{n_{r+1}} \wedge \dots \wedge dx^{n_m}.
$$
Using this twice you'll need some identity for contracting the $\epsilon$'s, which you can calculate quickly using induction (or you can just guess it). If you don't want to get your hands dirty by shifting indices around, I found another version of the proof in Voisin's "Hodge Theory and Complex Algebraic Geometry, volume 1" around page 120. The key to that is that $\star$ preserves metrics, so
$$
\langle \alpha, \beta \rangle \text{vol} = \langle \star \alpha, \star \beta \rangle \text{vol}.
$$
A: Starting from
\begin{align}
  &w_p = \frac{1}{p!} w_{m_1 m_2 \cdots m_p} dx^{m_1} \wedge dx^{m_2} \wedge \cdots \wedge dx^{m_p} \\
  &*w_p = \frac{\sqrt{g}}{p!(n-p)!} w_{m_1 m_2 \cdots m_p} \epsilon^{m_1 m_2 \cdots m_p}_{\phantom{m_1 m_2 \cdots m_p} m_{p+1} \cdots m_n  }
    dx^{m_{p+1}}\wedge \cdots \wedge dx^{m_n} 
    \end{align}
since $*w_p$ is $n-p$ forms we can write
\begin{align}
   *w_p  = \frac{1}{(n-p)!} \tilde{w}_{m_{p+1} \cdots m_n} dx^{m_{p+1}} \wedge \cdots \wedge dx^{m_n}
\end{align}  where $\tilde{w}_{m_{p+1} \cdots m_n} = \frac{\sqrt{g}}{p!}  w_{m_1 m_2 \cdots m_p} \epsilon^{m_1 m_2 \cdots m_p}_{\phantom{m_1 m_2 \cdots m_p} m_{p+1} \cdots m_n  }$
Then
\begin{align}
 ** w_p &= \frac{\sqrt{g}}{p! (n-p)!}  \tilde{w}_{m_{p+1} \cdots m_n} \epsilon^{m_{p+1} \cdots m_n  }_{\phantom{m_{p+1} \cdots m_n }  m_1 \cdots m_p}
 dx^{m_{1}} \wedge \cdots \wedge dx^{m_p} \\
 & = \frac{\sqrt{g}}{p! (n-p)!} \frac{\sqrt{g}}{p!}  w_{m_1 m_2 \cdots m_p} \epsilon^{m_1 m_2 \cdots m_p}_{\phantom{m_1 m_2 \cdots m_p} m_{p+1} \cdots m_n  }   \epsilon^{m_{p+1} \cdots m_n  }_{\phantom{m_{p+1} \cdots m_n }  m_1 \cdots m_p}
 dx^{m_{1}} \wedge \cdots \wedge dx^{m_p}   \\
 & = \frac{|g|}{p! (n-p)!}  \epsilon^{m_1 m_2 \cdots m_p}_{\phantom{m_1 m_2 \cdots m_p} m_{p+1} \cdots m_n  }   \epsilon^{m_{p+1} \cdots m_n  }_{\phantom{m_{p+1} \cdots m_n }  m_1 \cdots m_p} w_p \\
& = \frac{1}{p! (n-p)!}  \varepsilon^{m_1 m_2 \cdots m_p}_{\phantom{m_1 m_2 \cdots m_p} m_{p+1} \cdots m_n  }   \varepsilon^{m_{p+1} \cdots m_n  }_{\phantom{m_{p+1} \cdots m_n }  m_1 \cdots m_p} w_p \\
 & =  (-1)^{p(n-p)} w_p
\end{align}
where we have been used contraction of the Levi-Civita tensor, denoted $\varepsilon$, while $\epsilon$ is the Levi-Civita symbol. The relation between the two is
$$\varepsilon = \sqrt{g}\, \epsilon$$
