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The proofs I've found of the fact that $**\alpha=(-1)^{k(n-k)}\alpha$, or, equivalently, the fact that the Hodge star is an isometry, all use an orthonormal basis. Is there a basis-free proof of either of these facts directly from the definition of the Hodge star by

$$\alpha\wedge*\beta=\langle\alpha,\beta\rangle\mathrm{vol}?$$

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    $\begingroup$ I don't agree with the "equivalently." The definition I prefer to use for $\star$ uses an orthonormal basis, so I'm not sure what definition you mean. $\endgroup$ – Ted Shifrin Jun 7 '18 at 1:01
  • $\begingroup$ @TedShifrin I think that he is right claiming that the facts are equivalent. Please see my partial answer below. $\endgroup$ – Yuri Vyatkin Jun 7 '18 at 7:25
  • $\begingroup$ @YuriVyatkin, thanks for adding those details; that's exactly the question. I'm not sure which of the two facts ($*$ is an isometry; $**=(-1)^{k(n-k)}$) would be easier to prove without coordinates, but either fact is enough. $\endgroup$ – Yasha Berchenko-Kogan Jun 9 '18 at 17:21
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Let me make just a partial contribution as an answer, as it is too long for a comment.

Assuming we know that $\star$ is an isometry, and taking the identity in the question as a definition,

$$ \begin{align} \langle \alpha, \beta \rangle \mathrm{vol} & = \langle \star \alpha, \star \beta \rangle \mathrm{vol} = \star \alpha \wedge \star \star \beta = (- 1)^{k(n-k)} \star \star \beta \wedge \star \alpha \\ & = (- 1)^{k(n-k)} \langle \star \star \beta , \alpha \rangle \mathrm{vol} = (- 1)^{k(n-k)} \langle \alpha , \star \star \beta \rangle \mathrm{vol} \end{align} $$

so, we get the desired fact from the non-degeneracy of the metric.

Conversely, if we know, that $**\alpha=(-1)^{k(n-k)}\alpha$ holds, then, considering that

$$ \langle \star \alpha, \star \alpha \rangle \mathrm{vol} = \star \alpha \wedge \star \star \alpha = (-1)^{k(n-k)} \star \alpha \wedge \alpha = \alpha \wedge \star \alpha = \langle \alpha, \alpha \rangle \mathrm{vol} $$

which implies that $\star$ is an isometry.

Now, we probably want to prove in a coordinate-free manner, that $\star$ is an isometry. However, it looks like a delicate matter, as we can see from the discussion around this question, so I feel that I need to take a timeout :)

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