# Commutator of Lie derivative and Hodge star operator

I want to derive and expression for the commutator $[\mathcal{L}_Z,\star]\omega$. I found this post of mathoverflw that answers this question, but I have a few questions about Willie Wong's proof. How is the inverse metric $g^{-1}$ defined? I would guess that $g^{-1}$ is the metric defined on pairs of $1$-forms $g^{-1}(\alpha,\beta)=g(\alpha^\sharp,\beta^\sharp)$ where $\sharp$ denotes the musical isomorphism. He then states that $\star\omega = \text{Vol}_g\cdot(g^{-1})\cdot\omega$. I don't understand where this expression comes from, or what the $\cdot$ operator represents. Most likely it represents tensor contraction. I also don't understand where the expression $\mathcal{L}_Zg^{-1} = g^{-1}(\mathcal{L}_Zg)g^{-1}$ comes from. I would be very appreciative if someone could explain what exactly is going on here.

Yes, you can define $g^{-1}$ as you say, by $g^{-1}(\alpha, \beta) = g(\alpha^{\sharp}, \beta^{\sharp})$, and you can also define it as the unique $(0, 2)$-tensor that satisfies $\operatorname{tr}(g^{-1} \otimes g) = \operatorname{id}$, where $\operatorname{id}$ is the identity endomorphism $\operatorname{id}(X) = X$ and $\operatorname{tr}$ is any trace. Either way, this means that if we choose a frame $(E_a)$, we can define the matrix function $[g] = (g_{ab})$ whose components components $g_{ab} := g(E_a, E_b)$. Then, matrix representation of $g^{-1}$ with respect to the dual coframe is $[g^{-1}] = [g]^{-1}$ (hence the notation $g^{-1}$).
In the equation $\star \omega = \operatorname{Vol}_g \cdot (g^{-1}) \cdot \omega$:
• The expression $(g^{-1}) \cdot \omega$ just means that we use the musical isomorphism $\sharp$ to convert $\omega$ to a totally contravariant tensor. It is defined by declaring it to be linear and by declaring, for decomposable tensors, $(g^{-1}) \cdot (\alpha^1 \otimes \cdots \otimes \alpha^k) = (\alpha^1)^{\sharp} \otimes \cdots \otimes (\alpha^k)^{\sharp}$.
• The expression $\operatorname{Vol}_g \cdot \psi$ means, as you suggest, that we fully contract $\psi$ into $\operatorname{Vol}_g$: Again, we declare this to be linear, after which it is characterized by its behavior on decomposable tensors: $\operatorname{Vol}_g \cdot (X^1 \otimes \cdots \otimes X^{\ell}) := \operatorname{Vol}_g(X^1, \cdots, X^{\ell}, \,\cdot\,, \cdots ,\,\cdot\,)$.
Finally, the derivation of the last identity (NB as written there is a sign error) is standard. If we apply $\mathcal L_Z$ to both sides of the identity $\operatorname{tr}(g^{-1} \otimes g) = \operatorname{id}$ and use the fact that Lie derivatives commute with traces, we get $$\operatorname{tr}(\mathcal{L}_Z (g^{-1}) \otimes g) + \operatorname{tr}((g^{-1}) \otimes \mathcal{L}_Z g) = 0 .$$ Tensoring both sides with $g^{-1}$, taking a trace, using the above identity (equivalently, applying $\cdot^{\sharp}$ to both sides), and then rearranging gives the result. (If this isn't clear, it might be easier to think about this manipulation by writing all of this in terms of matrix representations.)
• It's just tensor contraction over two specified indices. For an endomorphism, it's just the usual trace, and for a $(k, \ell)$-tensor, $k, \ell > 0$, you trace over the relevant indices, and the other arguments of the tensor just come for the ride. NB I didn't specify the particular traces in my answer---this is because in each case the symmetry of the involved tensors means that all of the traces are the same, i.e., that it doesn't matter which indices one picks. Aug 25 '18 at 13:50