Hodge star operator independent of orthonormal basis

I'm reading the book on differential geometry by Gockeler and Schucker and I came across a proof which I don't understand regarding the Hodge star operator. In the book, they define the Hodge star operator on an $$n$$-dimensional oriented vector space by picking some positively oriented orthonormal basis $$\{e_i\}_{i=1}^n$$ and defining the map on the forms induced by the duals: $$*:\Lambda^p V\rightarrow \Lambda^{n-p}V$$ $$*(e^{i_1}\wedge\cdots\wedge e^{i_p})=\epsilon_{i_1\ldots i_n}\eta^{i_1i_1}\cdots\eta^{i_pi_p}e^{i_{p+1}}\wedge\cdots\wedge e^{i_n}$$ Where $$\epsilon$$ is the Levi-Civita symbol and $$\eta$$ is a pseudo-Riemannian metric, which is diagonal with $$r$$ diagonal entries equal to $$1$$ and $$s$$ equal to $$-1$$.

They then claim that this definition does not depend on the choice of oriented orthonormal basis, and they explain it as follows: Where equation $$3.16$$ is the equation written above and equation $$3.9$$ is the equation which defines the elements of $$O(r,s)$$: $$(\Lambda ^{-1})^t\eta \Lambda=\eta$$ I don't really understand their argument, and I'd appreciate any help in understanding it, even for the simple case of $$\eta$$ being the identity matrix (so just the usual Euclidean metric). I came across this related post, but it didn't contain a complete solution I could understand.

I tried proving it myself, by taking another positive orthonormal basis $$\{x_i\}_{i=1}^n$$, and writing $$e_i=Ax_i$$ for some matrix $$A\in SO(r,s)$$ and plugging this into the formula above, but I'm not sure on how to proceed from here.

• Your second line seems to have a typo: shouldn't it be $\eta^{i_1i_1}\cdots\eta^{i_pi_p}$ instead of $\eta^{i_1i_1}\cdots\eta^{i_ni_n}$? Feb 16 '21 at 21:33
• @Kajelad You're right, I fixed it. Feb 17 '21 at 9:35

First, we note that there exists a canonical top form $$\omega\in\Lambda^nV$$ defined by $$\omega=e^1\wedge\cdots\wedge e^n$$ where $$e^i$$ is any oriented orthonormal basis. This is independent of the choice of basis, since $$Ae^1\wedge\cdots\wedge Ae^n=\det(A)e^1\wedge\cdots\wedge e^n$$ and every element of $$SO(V)$$ has unit determinant.
Additionally, since $$\eta$$ defines an inner product $$\langle\ ,\ \rangle$$ on $$V$$, we can define an inner product of $$\Lambda^pV$$ by $$\langle e^{i_1}\wedge\cdots\wedge e^{i_p},e^{j_1}\wedge\cdots\wedge e^{j_p}\rangle=\eta^{i_1j_1}\cdots\eta^{i_pj_p} \\ i_1<\cdots Where $$e^i$$ is any orthonormal basis i.e. we define the basis $$\{e^{i_1}\wedge\cdots\wedge e^{i_p}:i_1<\cdots to be orthonormal. This definition is independent of the choice of basis, since $$\langle Au,Av\rangle=\langle u,v\rangle$$ for $$A\in SO(V)$$.
The Hodge star can then be defined by $$\mu\wedge(\star\nu)=\langle\mu,\nu\rangle\omega\ \ \ \ \ \forall\mu,\nu\in\Lambda^pV$$ Since everything in this expression is basis-independent, it suffices to show that $$\star$$ is well-defined. One way of doing this is by showing that this expression is uniquely satisfied by your expression for $$\star(e^{i_1}\wedge\cdots\wedge e^{i_p})$$ and the rest follows from linearity.