Suppose $(M,g)$ is a compact Riemannian manifold, and $*_g$ is the Hodge star operator defined on the de Rham algebra $\Omega^*(M)$ with respect to the metric $g$. Let $\phi:M\to M$ be a diffeomorphism. Then, can we find another metric $h$ (which should be related to $\phi$) so that
$$
(\phi^{-1})^* \circ *_g\circ \phi^*
$$
agree with the new Hodge star operator $*_h$?
EDIT: I guess this should be the Hodge star operator for the pullback metric. On the other hand, we know $\phi$ induces a map $$
\phi^*:H^*(M)\to H^*(M)$$ on de Rham cohomology groups, and meanwhile we also know for any metric $h$ there is the so-called Hodge isomorphism $\mathcal H^*_h(M) \to H^*(M)$ where $\mathcal H^*_h(M)$ denotes the $h$-harmonic forms. Now another interesting question is that can we find a metric $h$ so that we can induce a map
$$ \phi^*: \mathcal H_g(M) \to \mathcal H_h(M)$$?
Best Answer
Yes, here is the proof.
$\alpha\wedge{*}\beta=<\alpha,\beta>\Omega_{g}$, this is the definition of hodge star.
Pullback by $\varphi$, we get:
$LHS=\varphi^{*}(\alpha\wedge{*_{g}}\beta)=\varphi^{*}\alpha\wedge\varphi^{*}{*_{g}}\beta$
$RHS=\varphi^{*}(<\alpha,\beta>\Omega_{g})=\varphi^{*}<\alpha,\beta>\Omega_{\varphi^{*}g})=<\varphi^{*}\alpha,\varphi^{*}\beta>\Omega_{\varphi^{*}g}=\varphi^{*}\alpha\wedge*_{\varphi^{*}g}\varphi^{*}\beta$.
Since $\alpha,\beta$ are arbitrarily given, compare both sides we get the conclusion.