3
$\begingroup$

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 $*\omega$?

$\endgroup$

1 Answer 1

4
$\begingroup$

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 $d$.

$\endgroup$
2
  • $\begingroup$ Thanks! I understood this right before you wrote the answer :) So assuming $\omega \wedge *\eta = \langle \omega, \eta \rangle \operatorname{vol}$ is indeed necessary. $\endgroup$
    – tsnao
    Apr 17 at 10:59
  • 3
    $\begingroup$ In other word, the condition $\omega \wedge *\omega = \operatorname{vol}$ identifies $\omega$ up to an element of $\{ \eta \colon \omega \wedge \eta = 0 \}$. $\endgroup$
    – tsnao
    Apr 17 at 11:04

You must log in to answer this question.

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