This is an old revision of the document!
On an oriented Riemannian manifold $M^n$, the Hodge star operator $\star \colon \Lambda^k(M) \to \Lambda^{n-k}$ is defined by the relation $$\alpha \wedge \star \beta = (\alpha, \beta) \mathrm{vol}$$ for all $\alpha \in \Lambda^{n-k}(M), \beta \in \Lambda^k(M)$.
Properties
The Hodge star is linear: $\star (a\alpha + \beta) = a\star \alpha + \star \beta$.
The Hodge star is compatible with musical isomorphisms, hence wedge products: $\star (\iota_u \alpha) = (-1)^{|\alpha| - 1} u^\flat \wedge \star \alpha$ for a form $\alpha$, where $\iota_u$ denotes the contraction (interior multiplication) by $u$ and $u^\flat := \langle u, \cdot \rangle$. In the usual notation, for a 1-form $\omega$ we have $\star e^*(\omega) = (-1)^{k-1} e(\omega) \star$ when acting on $k$-forms.
The Hodge star satisfies $\star^2 = (-1)^{k(n-k)}$ where $k$ is the degree of the form on which it acts. Thus $(-1)^{k(n-k)} \alpha \wedge \beta = (\alpha, \star \beta) \mathrm{vol}$. On the other hand, $(-1)^{k(n-k)} \alpha \wedge \beta = \beta \wedge \alpha = (-1)^{k(n-k)} (\star \alpha, \beta) \mathrm{vol}$, so the Hodge star is self-adjoint.