By defintion, the Hodge star of a $p$-form $\omega_{a_1\cdots a_p}$ on a $n$-dimensional manifold is given by

$*\omega_{b_1\cdots b_{n-p}}=\frac{1}{p!}\omega^{a_1\cdots a_p}\epsilon_{a_1\cdots a_pb_1\cdots b_{n-p}}$

Here, I assume the manifold is equipped with a metric $g_{ab}$ and the volume element is compatible with the metric, ie., $\epsilon_{a_1\cdots a_n}=\sqrt{|g|}\tilde\epsilon_{a_1\cdots a_n}$ with $\tilde\epsilon_{a_1\cdots a_n}=+1$ when $a_1,\cdots,a_n$ is an even permutation of $1,2,\cdots,n$; $=-1$ when old permutation; $=0$ when there are repeated indices.

From this formula, I guess we can calculate the Hodge dual by first take the tensor product of $\omega_{a_1\cdots a_p}$ with $\epsilon_{a_1\cdots a_n}$ and then contract the first $p$ indices. Am I correct?


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  • $\begingroup$ My definition of the Hodge star is that $\alpha\wedge *\beta=g(\alpha,\beta)\epsilon$. Here $\alpha$ and $\beta$ are $p$-forms and $\epsilon$ is the volume form. So in particular $*\beta$ has order $n-p$, but it looks to me like your proposal would have order $n$. $\endgroup$ – Kevin Carlson Aug 28 '14 at 17:15
  • $\begingroup$ @KevinCarlson I did the calculation and found out that your definition is equivalent to the expression I gave. The expression contains tensor product and contraction, so my guessing should be correct. And I cannot see why tensor product and contraction will make the order of $*\omega$ be n. $*\omega$ has $n-p$ indices, so the order is at most $n-p$, isn't it? $\endgroup$ – Drake Marquis Aug 28 '14 at 20:47

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