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?

  • $\begingroup$ 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

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .