What is the operator $I_\zeta$, satisfying $ d I_\zeta + I_\zeta d = 1$ that allows us to define Noether-Wald charge?

Question

In these lectures on general relativity by Geoffrey Compère, the author, in section 1.3.3., equation 1.62, define an operator $I_\zeta$ which satisfies the identity

$$ d I_\zeta + I_\zeta d = 1.\tag{1.60}$$

Here, $\zeta^\mu$ is a vector field and $I_\zeta$ acts on differential forms. The is used to define the "Noether Wald surface charge" by

$$Q_\zeta[\Phi] = - I_\zeta \Theta[\delta_\zeta \Phi; \Phi]\tag{1.63}$$

where $\Phi$ is a stand in for a general field (although I'm really just interested in the metric here), $\delta_\zeta \Phi$ is the change in this field due to a diffeomorphism by $\zeta$, and $\Theta$ is the so-called "presymplectic form," given for gravity in the notes in equation 1.109

$$\Theta^\mu [\mathcal{L}_\zeta g; g] \approx \frac{\sqrt{-g}}{16 \pi G}\nabla_\nu( \nabla^\nu \zeta^\mu - \nabla^\mu \zeta^\nu)\tag{1.109}$$

where $\mathcal{L}_\zeta$ is the Lie derivative and $\approx$ means "equal to on shell."

Here is my question. The expression above for $\Theta^\mu$ clearly depends on two derivatives of $\zeta$. However, the expression given for $I_\zeta$ in the notes (eq. 1.62) is

$\forall \omega_\zeta \in \Omega^k(M),$

$$I_\zeta \omega_\zeta = \frac{1}{n-k} \zeta^\alpha \frac{\partial}{\partial \partial_\mu\ \zeta^\alpha} \frac{\partial}{\partial dx^\mu} \omega_\zeta + \text{ (Higher derivative terms)}.\tag{1.62}$$

This issue is that these higher derivative terms are actually relavent for the case at hand because $\Theta$ depends on two derivatives of $\zeta$.

My question is, what are these higher derivative terms? I would like to carry out the calculation presented in equation 1.110, where it is shown

$$ Q_\zeta^{\mu \nu} = -I_\zeta \Theta[\delta_\zeta g; g] = \frac{\sqrt{-g}}{8 \pi G}(\nabla^\mu \zeta^\nu - \nabla^\nu \zeta^\mu) $$ which is Komar's surface term.

So, to reiterate, does anyone know

1. What are the higher derivative terms?
2. What this operator $I_\zeta$ is called?

Answer 1

$I_\zeta$ is called a homotopy operator because of eq. (1.60). The explicit form to all orders can e.g. be found in P.J. Olver's book Applications of Lie Groups to Differential Equations, 2nd edition, 1993, eq. (5.134).