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Let $(E,g)$ be a real oriented inner product space with orthonormal basis $(e_1, \dots, e_n)$ with corresponding dual basis $(e^1, \dots, e^n)$. Then, for any $\beta \in \Lambda^0(V) := \mathbb{R}$, how does one prove that $$ * \beta = \beta \omega_E $$ where $\omega_E := e^1 \wedge \cdots \wedge e^n$ denotes the induced volume form for $(V,g)$?

I welcome feedback about the correctness of my approach and any alternate approaches that might be more efficient.

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First, note that $1 \in \Lambda^0(E)$ is the only positively-oriented orthonormal basis for $\Lambda^0(E)$. For, suppose that a nonzero scalar $k$ is basis which we can do since $\Lambda^0(E)$ is one-dimensional. Then if $k$ is orthonormal with respect to $g$, $$ g(k, k) = 1 \implies |k|^2 = 1 \implies k = \pm 1 $$ But, since we require positive orientation this reduces to $k = 1$. So, without loss of generality, we take $1$ as a basis for $\Lambda^0(E)$. Also, to be clear, the $g$ referenced here is inner product on $\Lambda^k(E)$, $k=1, \dots, n$, that is induced by the inner product on $(V, g)$. Perhaps a better notation would be $g_0$.

In any event, the Hodge-$*$ is the unique isomorphism $* \colon \Lambda^0(E) \rightarrow \Lambda^n(E)$ such that $$ \alpha \wedge (* \beta) = g(\alpha, \beta)\omega_{E} $$ for all $\alpha \in \Lambda^0(E)$. If follows then from the bilinearity and orthonormality of $g$ that \begin{align*} \alpha \wedge (* \beta) &= g(\alpha, \beta)\omega_{E}\\ &= g(\alpha \cdot 1, \beta \cdot 1) \omega_{E}\\ &= \alpha \beta g(1,1) \omega_{E}\\ &= \alpha \beta \omega_{E}. \end{align*} But $\alpha$ is just a scalar so $$ \alpha \wedge (*\beta) = \alpha \cdot (* \beta) $$ and consequently $$ \alpha \cdot (* \beta) = \alpha \beta \omega_E \implies * \beta = \beta \omega_E. $$

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  • $\begingroup$ That's correct. The last step is unnecessary: you have already shown that $\alpha\wedge (\beta \omega_E)=g(\alpha,\beta)\omega_E$, so by the characterization/definition you conclude $\beta \omega_E=\star(\beta)$. $\endgroup$ Jul 8 '12 at 20:26

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