Show that $a \wedge * b = g(a,b) \operatorname{vol}$

Question

$\newcommand{\vol}{\operatorname{vol}}$

Let $\omega$ be a $p$-form on a Riemannian manifold $M^n$ with metric $g$ and let $\vol_{i_1,\ldots,i_n}=\sqrt{\lvert g\rvert} \epsilon_{i_1,\ldots,i_n}$ be a volume form, where $\epsilon_{i_1,\ldots,i_n}$ is the Levi-Civita symbol with $\epsilon_{1,\ldots,n}=1$. By definition $*\omega$ is the $(n-p)$-form on $M^n$ given by the formula $$(*\omega)_{i_{p+1},\ldots,i_n} = \frac{1}{p!} \vol_{i_1,\ldots,i_n} \omega^{i_1,\ldots,i_p}.$$ I need to show the formula $\alpha \wedge *\beta = g(\alpha,\beta) \vol$ is valid, where $\alpha = \alpha_{i_1,\ldots,i_p}$, $\beta=\beta_{i_1,\ldots,i_p}$ are arbitrary $p$-forms, $g(\alpha,\beta) = \alpha_{i_1,\ldots,i_p}\beta^{i_1,\ldots,i_p}$.

By definition $$ (\alpha \wedge *\beta)_{i_1,\ldots,i_n} = \frac{1}{p!(n-p)!}\delta^{j_1,\ldots,j_n}_{i_1,\ldots,i_n} \alpha_{j_1,\ldots,j_p}(*\beta)_{j_{p+1},\ldots,j_n} = \\ \frac{1}{p!^2(n-p)!}\delta^{j_1,\ldots,j_n}_{i_1,\ldots,i_n} \alpha_{j_1,\ldots,j_p} \vol_{k_1,\ldots,k_p,j_{p+1},\ldots,j_n}g^{k_1,s_1}\ldots g^{k_p,s_p} \beta_{s_1,\ldots,s_p}. $$ And I can't see what can I do with this expression. Please, help me to finish the demonstration.

Answer 1

One method is:

1. To work with the whole form $\alpha \wedge *\beta$ rather than its components $(\alpha \wedge *\beta)_{i_1\ldots i_n}$.

2. To use $$\tag{1}\mathrm{d}x^{i_1} \wedge \ldots \wedge \mathrm{d}x^{i_n}~=~\varepsilon^{i_1\ldots i_n} \mathrm{d}x^{1} \wedge \ldots \wedge \mathrm{d}x^{n}~=~\frac{\varepsilon^{i_1\ldots i_n}}{\sqrt{|\det(g_{\cdot\cdot})|}}\Omega,$$ where$^1$ $\varepsilon^{i_1\ldots i_n}$ here denotes the Levi-Civita symbol (as opposed to the Levi-Civita tensor), i.e., $\varepsilon^{1\ldots n}=1$.

3. To use

$$ \tag{2}\delta_{j_1\ldots j_p}^{i_1\ldots i_p}~=~\frac{1}{(n-p)!}\varepsilon^{i_1\ldots i_pk_{p+1}\ldots k_n} ~\varepsilon_{j_1\ldots j_pk_{p+1}\ldots k_n}. $$

Sketched proof:

$$\alpha \wedge *\beta~=~\frac{1}{p!} \alpha_{i_1\ldots i_p} \mathrm{d}x^{i_1} \wedge \ldots \wedge \mathrm{d}x^{i_p} ~\wedge~\frac{\sqrt{|g|}}{p!(n-p)!} \beta^{j_1\ldots j_p}~\varepsilon_{j_1\ldots j_n}\mathrm{d}x^{j_{p+1}} \wedge \ldots \wedge \mathrm{d}x^{j_n} $$ $$~\stackrel{(1)}{=}~ \frac{1}{(p!)^2 (n-p)!} \alpha_{i_1\ldots i_p}~\beta^{j_1\ldots j_p}~\varepsilon_{j_1\ldots j_n}~ \varepsilon^{i_1\ldots i_pj_{p+1}\ldots j_n}~\Omega $$ $$ \tag{3}~\stackrel{(2)}{=}~\frac{1}{(p!)^2 } \alpha_{i_1\ldots i_p}~\beta^{j_1\ldots j_p}~\delta_{j_1\ldots j_p}^{i_1\ldots i_p}~\Omega ~=~\frac{1}{p! } \alpha_{i_1\ldots i_p}~\beta^{i_1\ldots i_p}~\Omega ~=~g( \alpha,\beta) ~\Omega .$$

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$^1$ In the last equality of eq. (1), we allow for simplicity only local coordinate transformations that preserve orientation.

Answer 2

This calculation in differential forms is quite trivial in geometric algebra. Let $\alpha$ and $\beta$ be $p$-vectors. Let $\epsilon$ be some pesudoscalar.

The geometric product $\alpha \beta \epsilon$ has as its grade-$n$ component

$$\langle (\alpha \beta )\epsilon\rangle_n = \langle \alpha \beta \rangle_0 \epsilon = (\alpha \cdot \beta) \epsilon$$

But associativity of the geometric product tells us we can group $\beta \epsilon$ instead to get

$$\langle \alpha (\beta \epsilon) \rangle_n = \alpha \wedge (\beta \epsilon)$$

Alas, while I'm sure that this solution can be translated into differential forms parlance, the key ingredient--the geometric product and the use of grade projection--is something that would be pretty onerous in itself to translate.