# Formal construction of Hodge star operator.

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?

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

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.