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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: Proof from Gockeler's book

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.

Thanks in advance.

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    $\begingroup$ Your second line seems to have a typo: shouldn't it be $\eta^{i_1i_1}\cdots\eta^{i_pi_p}$ instead of $\eta^{i_1i_1}\cdots\eta^{i_ni_n}$? $\endgroup$
    – Kajelad
    Feb 16 at 21:33
  • $\begingroup$ @Kajelad You're right, I fixed it. $\endgroup$
    – Muhammad20
    Feb 17 at 9:35
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If you're familiar with index notation, it possible, though tedious, to verify this using a direct computation. Here's an alternate route which uses the basis-independent definition.

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 $$ \langle e^{i_1}\wedge\cdots\wedge e^{i_p},e^{j_1}\wedge\cdots\wedge e^{j_p}\rangle=\eta^{i_1j_1}\cdots\eta^{i_pj_p} \\ i_1<\cdots<i_p,\ \ \ j_1<\cdots<j_p $$ Where $e^i$ is any orthonormal basis i.e. we define the basis $\{e^{i_1}\wedge\cdots\wedge e^{i_p}:i_1<\cdots<i_p\}$ 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.

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  • $\begingroup$ This is such a cool answer. In grad school I would handle all of this purely via index calculus. Wish I had seen this. Thanks $\endgroup$ Feb 17 at 15:55

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