Let $M$ be a closed Riemannian manifold and $\omega$ and $\eta$ two differential forms of the same degree. Then one can consider $\int_M \omega \wedge *\eta$, where $*$ denotes the Hodge star operator. Can you tell me, why this defines a scalar product or at least where I can find a proof of this fact? In particular I would be very interested why this expression is symmetric in $\omega$ and $\eta$ and why it is positive definite.
[Math] Scalar product on manifold.
differential-geometry
Best Answer
The Riemannian metric extends to a metric on all tensors of rank $(k,l)$, and in particular it extends to $p$-forms. The very definition of the Hodge star operator is that $$ \omega \wedge \ast \eta = \langle \omega, \eta \rangle d\mathrm{vol} $$ where $\langle \omega, \eta \rangle$ is the pairing on $p$-forms induced by the metric, and $d\mathrm{vol}$ is the volume form (or density) induced by the metric. So $$ \int_M \omega \wedge \ast \eta = \int_M \langle \omega, \eta \rangle d\mathrm{vol} $$ which is very obviously symmetric and positive definite.
If you are using some other definition of the Hodge star (for example using te $\epsilon$ symbol), then all you have to do is to check that it is equivalent to the condition $\omega \wedge \ast \eta = \langle \omega, \eta \rangle d\mathrm{vol}$.