# A simple question about the Hodge star

The usual definition of the Hodge star says that it maps $$\Lambda^k(V)$$ to $$\Lambda^{n-k}(V)$$ in such a way that for each pair $$\omega, \eta \in \Lambda^k(V)$$ holds $$\omega \wedge *\eta = \langle \omega, \eta \rangle \operatorname{vol}$$. I was curious whether this definition is equivalent to saying that $$\omega \wedge *\omega = \operatorname{vol}$$ for each $$\omega$$ of unit norm.

Initially I thought that it should follow immediately from the polarization identity as follows: \begin{aligned} 2 \, \langle \omega, \eta \rangle \operatorname{vol} & = \langle \omega, \omega \rangle \operatorname{vol} + \langle \eta, \eta \rangle \operatorname{vol} -\langle \omega-\eta, \omega-\eta\rangle \operatorname{vol} \\[3pt] & = \omega \wedge *\omega + \eta \wedge *\eta - (\omega-\eta) \wedge *(\omega-\eta) \\[3pt] & = \omega \wedge *\eta + \eta \wedge *\omega. \end{aligned} This would give a proof if $$\omega \wedge * \eta$$ were equal to $$\eta \wedge *\omega$$, but this doesn't seem to follow from the definition of $$*\omega$$ by $$\omega \wedge *\omega = \operatorname{vol}$$. So my question is what am I missing? Or does the definition with one $$\omega$$ instead of two $$\omega, \eta$$ allow for a different choice of $$*\omega$$?

You don't get uniqueness this way. Take $$V=\langle e_1,e_2\rangle$$, so $$n=2$$ and take $$k=1$$. Then $$\text{vol} = e_1\wedge e_2$$ and if $$\ast e_1 = ae_1 + be_2$$ and $$\ast e_2 = ce_1 + de_2$$, all we obtain is that $$e_1 \wedge e_2 = e_1 \wedge \ast e_1 = b e_1 \wedge e_2,$$ so $$b=1$$ and similarly $$c=-1$$. We don't obtain information about $$a$$ or $$d$$.
• Thanks! I understood this right before you wrote the answer :) So assuming $\omega \wedge *\eta = \langle \omega, \eta \rangle \operatorname{vol}$ is indeed necessary. Apr 17 at 10:59
• In other word, the condition $\omega \wedge *\omega = \operatorname{vol}$ identifies $\omega$ up to an element of $\{ \eta \colon \omega \wedge \eta = 0 \}$. Apr 17 at 11:04