First, the following question is highly related to this. I read that several times, but I can't solve my problem so please let me submit a similar question.
Second, I major in Physics so I guess that my writing lacks mathematical rigorousness, sorry.
Let $X$ be a real manifold whose dimension is $d= \dim X$ and $A_p, B_p$ be $p$-forms on $X$. Then I define the component or coordinate expression of $p$-form as \begin{align} A_p = \frac{1}{p!} A_{m^1 \dots m^p} dx^{m^1} \wedge \cdots \wedge dx^{m^p}. \end{align} The wedge product of two forms becomes
\begin{align} A_p \wedge B_p = \frac{1}{p!q!} A_{m^1 \dots m^p} B_{n^1 \dots n^p} dx^{m^1} \wedge \cdots \wedge dx^{m^p} \wedge dx^{n^1} \wedge \cdots \wedge dx^{n^q}. \end{align}
The hodge star operator $\star$ is a map from $\Omega^p$ to $\Omega^{d-p}$ and acts on $p$-form as \begin{align} \star (dx^{m^1} \wedge \cdots \wedge dx^{m^p} ) = \frac{1}{(d-p)!} |g|^{1/2} g^{m^1 n^1} \cdots g^{m^{p}n^{p}}\epsilon_{n^1 \dots n^p n^{p+1} \dots n^d} dx^{ n^{p+1}} \cdots dx^{n^d}. \end{align}
Here $g$ is the metric and $|g|$ is the absolute value of $\mathrm{det}g$ .
Then I can define the inner product of two $p$-form:
\begin{align} \langle A_p, B_p \rangle \mathrm{vol}_X &\equiv \frac{1}{p!}A_p\wedge \star B_p \\ &= |g|^{1/2} A_{m^1 \dots m^p}B^{m^1 \dots m^p} \mathrm{vol}_X. \end{align} The last equality is written on a physics textbook (page 584) which I am now reading, so I have to prove that. Since normalisation factors may be incorrect, please treat them roughly.
I tried as follows: Since \begin{align} \star B_p &= \frac{|g|^{1/2}}{p! (d-p)!} B_{m^1 \dots m^p} g^{m^1 n^1} \cdots g^{m^{p}n^{p}}\epsilon_{n^1 \dots n^p n^{p+1} \dots n^d} dx^{ n^{p+1}} \cdots dx^{n^d} \\ &= \frac{|g|^{1/2}}{p! (d-p)!} B^{n^1 \dots n^p} \epsilon_{n^1 \dots n^p n^{p+1} \dots n^d} dx^{ n^{p+1}} \cdots dx^{n^d}, \end{align} $A_p \wedge \star B_p$ becomes
\begin{align} A_p \wedge \star B_p &= \frac{|g|^{1/2}}{(p!)^2 (d-p)!} A_{m^1 \dots m^p}B^{n^1 \dots n^p} \epsilon_{n^1 \dots n^p n^{p+1} \dots n^d} \underbrace{dx^{m^1}\wedge \cdots \wedge dx^{m^{p}} \wedge dx^{ n^{p+1}} \cdots dx^{n^d}}_{= \epsilon^{m^1 \dots m^p n^{p+1} \dots n^{d}} \mathrm{vol} X} \\ &=\frac{|g|^{1/2}}{(p!)^2 (d-p)!} A_{m^1 \dots m^p}B^{n^1 \dots n^p} \epsilon_{n^1 \dots n^p n^{p+1} \dots n^d}\epsilon^{m^1 \dots m^p n^{p+1} \dots n^{d}} \mathrm{vol} X \end{align} Now the normalisation factors are clearly strange, but I would like to shut my eyes to that, but I think \begin{align} \epsilon_{n^1 \dots n^p n^{p+1} \dots n^d}\epsilon^{m^1 \dots m^p n^{p+1} \dots n^{d}} = \delta^{m^1 \dots m^p}_{n^1 \dots n^p}, \end{align} here $\delta$ is the generalised Kronecker delta. What I want to say here is that there is no necessity for $m^i$ and $n^i$ $(i = 1, \dots , p)$ to match each other. If so, I can't extract the structure of the inner product from this... where I have missed?
Thank you.