Let $\Omega^p(M)$ be the smooth degree $p$ differential forms on an $n$-dimensional manifold $M$. The Hodge $\ast$ operator maps $\ast : \Omega^p(M) \to \Omega^{n-p}(M)$. Using the Hodge dual we can dualize some operations. For example,
\begin{array}{ccc}
 \Omega^p(M) & \xrightarrow{d} & \Omega^{p+1}(M)\\
 \downarrow \ast & & \downarrow \ast \\
 \Omega^{n-p}(M) & \xrightarrow{\delta} & \Omega^{n-p-1}(M)
\end{array}
where $\delta$ is the codifferential.


If $X$ is a vector field, then
\begin{array}{ccc}
 \Omega^p(M) & \xrightarrow{i\_X} & \Omega^{p-1}(M)\\
 \downarrow \ast & & \downarrow \ast \\
 \Omega^{n-p}(M) & \xrightarrow{X^\flat \wedge} & \Omega^{n-p+1}(M)
\end{array}
where $X^\flat$ is the differential form obtained by musical isomorphism.


In the case of a Lie derivative, we have the diagram
\begin{array}{ccc}
 \Omega^p(M) & \xrightarrow{\mathcal{L}\_X} & \Omega^{p}(M)\\
 \downarrow \ast & & \downarrow \ast \\
 \Omega^{n-p}(M) & \xrightarrow{\mathcal{L}^\ast\_X} & \Omega^{n-p}(M)
\end{array}
where $\mathcal{L}^\ast\_X = X^\flat \wedge \delta + \delta X^\flat \wedge$.


My questions is first if this operator $\mathcal{L}^\ast\_X$ has a more standard name and notation. The second question is how should I think about it? The usual Lie derivative can be thought as the transformation under a diffeomorphism generated by the vector field $X$. Is there a similar interpretation for $\mathcal{L}^\ast\_X$?


I know a term like this appears in the definition of the Killing-Yano tensor:
$$
 \nabla\_{X} f = \frac 1 {p+1} i\_{X} d f - \frac 1 {n-p+1} X^{\flat} \wedge \delta f,
$$
but again I don't have a good intuition about this formula.



dual of the Lie derivative
$\mathcal L^\*\_X\omega = \* \mathcal L\_X (\*^{-1}\omega) = \*(\mathcal L\_X \*^{-1}) \omega + \mathcal L\_X(\omega)$. 


So $\mathcal L\_X^\*$ is the infinitesimal deformation along $X$ of $\*^{-1}= \pm \*$ combined with the infinitesimal deformation of the form. 


