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I'm studying differential geometry, and I'm looking for a formal construction of the Hodge star operator. For example, in the Baez and Muniain's book, the Hodge operator is defined as the unique linear operator $\star:\Omega^p(M)\rightarrow\Omega^{n-p}(M)$ such that, for all $\mu$, $\nu\in\Omega^p(M)$: $$\omega\wedge\nu=\langle\omega,\nu\rangle dV $$ where $dV$ is the volume form. What I'm looking for is a statement like this one:

Proposition: There exist an unique linear operator $\star:\Omega^p(M)\rightarrow\Omega^{n-p}(M)$ such that, for all $\mu$, $\nu\in\Omega^p(M)$: $$\omega\wedge\nu=\langle\omega,\nu\rangle dV $$

And a proof that involves the construction of such operator, and the proof of its uniqueness as a mathematician would do it.

I've searched in many references, but none of them offer a proof of the statement, or a proof just involving few mathematical tools, like vector bundle orientations and differential forms. Could anybody help me?

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  • $\begingroup$ Check Paul Renteln's Manifolds, Tensors and Forms. $\endgroup$
    – Ivo Terek
    Commented Mar 1, 2020 at 22:46
  • $\begingroup$ You need to do some proofreading. $\endgroup$ Commented Mar 2, 2020 at 7:16
  • $\begingroup$ In Lee's Introduction to Smooth Manifolds (2nd Edition) it was outlined in Problem 16-18. Perhaps it would provide some help. $\endgroup$ Commented Feb 3, 2021 at 16:59

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Think of this in the level of vector spaces, to make it easier. Let $(V,\langle\cdot,\cdot\rangle)$ be a space with (pseudo-)Euclidean scalar product, $\dim V = n$, and fix a volume form $\sigma \in V^{\wedge n}$. A choice of such $\sigma$ determines an isomorphism $V^{\wedge n} \cong \Bbb R$: any top form is a multiple of $\sigma$, and we map the form to such multiple. Given $\omega \in V^{\wedge k}$, look at the functional $(V^{\wedge (n-k)})^*$ given by multiplication by $\omega$. That is, we map $\nu \in V^{\wedge(n-k)}$ to $\omega \wedge \nu \in V^{\wedge n} \cong \Bbb R$. In particular, $V^{\wedge k}$ is equipped with a non-degenerate scalar product induced by $\langle \cdot,\cdot\rangle$ (for which we'll use the same notation), and so this functional has the form $\langle \text{some $(n-k)$-form}, \cdot\rangle$. This $(n-k)$-form is then called $\star \omega$. This establishes that $$\omega \wedge \nu = \langle \star \omega, \nu\rangle \sigma,$$as wanted. Note that $\star \omega$ depends both on $\langle \cdot,\cdot\rangle$ and $\sigma$. On the manifold level, this is done pointwise for all the tangent spaces at the same time.

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