Conformal Hodge scaling in every degree

On the inherited smooth manifold background [A1], since $\tilde{g}={\mathrm{\Omega }}^{2}g$ and $\mathrm{\Omega }>0$, its inverse is ${\tilde{g}}^{-1}={\mathrm{\Omega }}^{-2}{g}^{-1}$. The exterior pairing of two $k$-forms contains exactly $k$ inverse-metric factors, so $⟨\alpha ,\beta {⟩}_{\tilde{g}}={\mathrm{\Omega }}^{-2k}⟨\alpha ,\beta {⟩}_g$. In a positively oriented chart, $\det ({\tilde{g}}_{ab})={\mathrm{\Omega }}^8\det (g_{ab})$; therefore the metric volume form scales as ${\mathrm{v}\mathrm{o}\mathrm{l}}_{\tilde{g}}={\mathrm{\Omega }}^4{\mathrm{v}\mathrm{o}\mathrm{l}}_g$.

Applying the defining identity in [F1] with $\tilde{g}$ in place of $g$ and substituting step 1.1 gives, for every $k$-form $\alpha$, $\alpha \wedge {\star }_{\tilde{g}}\beta ={\mathrm{\Omega }}^{-2k}⟨\alpha ,\beta {⟩}_g{\mathrm{\Omega }}^4{\mathrm{v}\mathrm{o}\mathrm{l}}_g=\alpha \wedge \left({\mathrm{\Omega }}^{4-2k}{\star }_g\beta \right)$. For $k=0,1,2,3,4$ the multipliers are respectively ${\mathrm{\Omega }}^4,{\mathrm{\Omega }}^2,1,{\mathrm{\Omega }}^{-2},{\mathrm{\Omega }}^{-4}$, all smooth because $\mathrm{\Omega }$ is smooth and positive. Thus the candidate is a smooth bundle map satisfying the defining identity for ${\star }_{\tilde{g}}\beta$; uniqueness in [F1] gives ${\star }_{\tilde{g}}\beta ={\mathrm{\Omega }}^{4-2k}{\star }_g\beta$.