# dual of the Lie derivative

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.

• $\mathcal L^*_X$ is depends heavily on the choice of the (pseudo) Riemann metric which is built into the Hodge star. – Peter Michor Feb 14 '15 at 19:05
• The Killing-Yano definition you write comes from the decomposition of the covariant derivative on forms into irreducibles. It has three parts. A form is Killing-Yano if the third part vanishes, leaving you with the equation above. – Paul Reynolds Feb 14 '15 at 20:59

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