Hodge star operator independent of orthonormal basis

Question

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.

Thanks in advance.

Answer 1

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.

Answer 2

Maybe even more "basisless" construction can be performed given:

the exterior multiplication $\Lambda^kV\otimes\Lambda^{n-k}V\to\Lambda^nV$;

isomorphisms $\Lambda^j(V^*)\cong(\Lambda^jV)^*$;

a choice of a nonzero element of $\Lambda^nV$ ("volume form"), inducing an isomorphism between $\Lambda^nV$ and the standard 1-dimensional vector space;

a choice of an isomorphism $V\cong V^*$ (amounts to a nonsingular bilinear form on $V$).

Indeed these data give maps $\Lambda^kV\to(\Lambda^{n-k}V)^*\to\Lambda^{n-k}(V^*)\to\Lambda^{n-k}V$.

I believe their composite is the Hodge star operator. While it is already an isomorphism, in order for it to be its own inverse up to sign the corresponding nonsingular bilinear form must be symmetric, I believe. (Maybe a skew-symmetric one would work too, with different signs?)