Hodge star is conformally invariant on $\Lambda^{n/2}(V)$, for $n$ even

Question

I am studying the Hodge star operator for the first time. I am trying to prove that for $n$ even, then for any $\omega \in \Lambda^{n/2}(V)$ $\star_g \omega= \star_{\tilde{g}} \omega$, where $g$ and $\tilde{g}$ are conformal ($g=\lambda \tilde{g}$ for $\lambda >0$), it seems obvious but I don't really know where to start.
My definition of Hodge star is: for any $\omega, \mu \in \Lambda^k(V)$, $e_1, \dots , e_n$ a positively oriented orthonormal basis of $V$ wrt the metric $g$ $$ \omega \wedge \star \mu = g(\omega,\mu) e_1 \wedge \dots \wedge e_n $$ Thanks!

Answer 1

It suffices to check just on an orthonormal basis. If $e_1,\dots,e_n$ is an oriented $\tilde g$-orthonormal basis, then $\bar e_1=e_1/\sqrt{\lambda}, \dots, \bar e_n=e_n/\sqrt{\lambda}$ will be an oriented $g$-orthonormal basis. With $I$ an ordered multiindex of length $n/2$, let $I'$ denote a complementary ordered multiindex so that $\tilde\star e_I = e_{I'}$. It follows, then, that $\star\bar e_I = \bar e_{I'}$; in other words, $\dfrac1{(\sqrt\lambda)^{n/2}}{\star} e_I = \dfrac1{(\sqrt\lambda)^{n/2}}e_{I'}$, which means that $\star e_I = e_{I'}$, as before.