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Assume that $V$ is oriented finite dimensional vectorspace with dimension $n$, $g \in T^0_2(V)$ a given symmetric and nondegenerate tensor. Let $\mu$ be the corresponding volume element of $V$. Assume that $\{e_1, \dotsc,e_n\}$ is a oriented arbitrary (not necessarily $g$-orthonormal) basis of $V$ and $\{e^1, \dotsc,e^n\}$ its dual basis. Let $\ast \colon \Lambda^k(V) \to \Lambda^{n-k}(V)$ be the unique isomorphism satisfying $$\alpha \wedge \ast \beta = \langle\!\langle \alpha, \beta \rangle\!\rangle \mu \quad \text{for } \alpha, \beta \in \Lambda^k(V),$$ where it is known that $\langle\!\langle,\rangle\!\rangle$ fulfills the following $$\langle\!\langle e^{i_1} \wedge \dotsb \wedge e^{i_k}, e^{j_1} \wedge \dotsb \wedge e^{j_k} \rangle\!\rangle = \det (g^{i_l j_{l^\prime}})_{l,l^\prime \in \{1,\dotsc,k\}} = \det \left(g\left((e^{i_l})^\sharp,(e^{j_{l^\prime}})^\sharp\right)\right)_{l,l^\prime \in \{1,\dotsc,k\}}$$ and $\mu$ is given by $$\mu = \sqrt{|\det [g(e_i,e_j)]|} e^1 \wedge \dotsb \wedge e^n.$$

Then $$\ast (e^{i_1} \wedge \dots \wedge e^{i_k}) = \sum_{j_{k+1} < \dotsb < j_n}c^{i_1 \dotsm i_k}_{j_{k+1} \dotsm j_n} e^{j_{k+1}} \wedge \dotsb \wedge e^{j_n},$$ where $$c^{i_1 \dotsm i_k}_{j_{k+1} \dotsm j_n} = \sqrt{|\det [g(e_i,e_j)]|} \mathrm{sign} \binom{1 \dotsm n}{j_1 \dotsm j_n}g^{i_1j_1} \dotsm g^{i_kj_k} .$$

Result has been stated in Abraham, R. and Marsden, J.E. and Ratiu, T."Manifolds, Tensor Analysis, and Applications (3rd Ed)" 7.2.14 Examples (2)

Proof: Set $\alpha = e^{j_1} \wedge \dotsb \wedge e^{j_k}$ with $j_1 < \dotsb < j_k$ and $\beta = e^{i_1} \wedge \dotsb \wedge e^{i_k}$ One can compute \begin{align} \alpha \wedge \ast \beta &= e^{j_1} \wedge \dotsb \wedge e^{j_k }\wedge \sum_{l_{k+1} < \dotsb < l_n} c^{i_1 \dotsm i_k}_{l_{k+1} \dotsm l_n} e^{l_{k+1}} \wedge \dotsb \wedge e^{l_n} \\ &= c^{i_1 \dotsm i_k}_{j_{k+1} \dotsm j_n} e^{j_1} \wedge \dotsb \wedge e^{j_k } \wedge e^{j_{k+1}} \wedge \dotsb \wedge e^{j_n} \\ &= c^{i_1 \dotsm i_k}_{j_{k+1} \dotsm j_n} \mathrm{sign} \binom{1 \dotsm n}{j_1 \dotsm j_n} e^1 \wedge \dotsb \wedge e^n. \end{align} On the other hand it holds that \begin{align} \langle\!\langle \alpha, \beta \rangle\!\rangle \mu &= \det (g^{j_l i_{l^\prime}})_{l,l^\prime \in \{1,\dotsc,k\}} \sqrt{|\det [g(e_i,e_j)]|} e^1 \wedge \dotsb \wedge e^n. \end{align}

Since $\alpha \wedge \ast \beta = \langle\!\langle \alpha, \beta \rangle\!\rangle \mu$, I get $$c^{i_1 \dotsm i_k}_{j_{k+1} \dotsm j_n} = \sqrt{|\det [g(e_i,e_j)]|} \mathrm{sign} \binom{1 \dotsm n}{j_1 \dotsm j_n} \det (g^{j_l i_{l^\prime}})_{l,l^\prime \in \{1,\dotsc,k\}}.$$

But this is obviously not the full truth, what am I missing, to get the desired formula? What are my mistakes in argumentation so far? I appreciate your help.

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  • $\begingroup$ Nice, I was looking for a formula for the Hodge star in an arbitrary basis! I am wondering if maybe you have the wrong formula there? Is it equivalent to a sum over $(k,n-k)$ shuffles as the book states? $\endgroup$ – Emil Oct 11 '13 at 0:19
  • $\begingroup$ The sum $\sum_{j_{k+1} < \dotsb < j_n} c^{i_1 \dotso i_k}_{j_{k+1} \dotso j_n} \, e^{j_{k+1}} \wedge \dotsb \wedge e^{j_n}$ is equivalent to the sum of all $(k,n-k)$-shuffles of $$\binom{1 \dotso n}{j_1 \dotso j_n}$$. I really don't know, where my mistake could be. $\endgroup$ – varsop Oct 12 '13 at 19:44
  • $\begingroup$ I treat the Hodge star in an arbitrary basis in section 10 of the arXiv.org submission 1110.3350v1. $\endgroup$ – Richard A. Smith Aug 8 '14 at 1:52

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