I am currently working on extrinsic riemannian geometry and I am looking for a sort of commutation relation between the covariant and Lie derivatives.


To be more precise : considering an hypersurface $H \subset M$ of a riemannian manifold, $\nu$ a vector field normal to $H$ and $S$ its *shape operator* (or *Wiengarten operator*) defined by $SX = \nabla\_X \nu$, you can consider normal geodesics emanating from $H$ as geodesics veryfing $\gamma(0) \in H$, $\dot\gamma(0) = \nu$. Writing the parameters of these geodesics $r$, you get a vector field $\partial\_r = \dot\gamma$. If $(x^1,\ldots,x^n)$ are local coordinates on $H$, then you have *Fermi coordinates* $(r,x^1,\ldots,x^n)$ on $M$.


We have the Ricatti equation, where $R\_{\partial\_r} = R(\partial\_r,\cdot)\partial\_r$ :
\begin{align\*}
\mathcal{L}\_{\partial\_r}S=\partial\_r S = -S^2 - R\_{\partial\_r}
\end{align\*}


(in fact, the equation is still true while replacing $\mathcal{L}\_{\partial\_r}$ by $\nabla\_{\partial\_r}$, it's a property of the shape operator).


I want to find a differential equation for $\nabla\_{\partial\_j}S$ where $\partial\_j = \frac{\partial}{\partial x^j}$. My idea is to differentiate the Ricatti equation with respect to $\nabla\_{\partial\_j}$ and use a sort of commutation relation to get a differential equation involving $S$, $\nabla\_{\partial\_j}S$, $R\_{\partial\_r}$, *etc*. with variable $r$.


So, my question is : do we have a nice relation between $\nabla\_{\partial\_j} \mathcal{L}\_{\partial\_r} S$ and $\mathcal{L}\_{\partial\_r}\nabla\_{\partial\_j}S$ ?


Thank you for reading me.


**Edit**


I recently tried something : expanding the lie derivative to the connexion itself. That is :
\begin{align}
\mathcal{L}\_{\partial\_r} \left( \nabla\_j S) \right) &= \left(\mathcal{L}\_{\partial\_r}\nabla\_j\right) S + \nabla\_j \left( \mathcal{L}\_{\partial\_r}S\right)
\end{align}
In *Einstein Manifolds*, Besse, there is a formula for the derivative of the connection with respect to the metrics, in the direction of a symmetric tensor, that is :
\begin{align}
g\left((\nabla'(g)\cdot h)(X,Y),Z\right) &= \dfrac{1}{2}\left(\nabla\_Xh (Y,Z) + \nabla\_Yh(X,Z) - \nabla\_Zh (X,Y) \right)
\end{align}
With that in mind, and recalling that $\mathcal{L}\_{\partial\_r}g = 2g\left(S\cdot,\cdot\right)$, something is appearing. I would post somthing if this answers the original question.



Commutation relations between covariant and Lie derivatives
I recently answered my question by finding a formula I wasn't aware of.
Let $\nabla$ be a connexion and $X$ a vector field. Then $\mathcal{L}\_X\nabla$ is a tensor and
\begin{align}
\mathcal{L}\_X\nabla &= -i\_X\circ R^{\nabla} + \nabla^2X
\end{align}


where $R^{\nabla}(U,V) = \nabla\_{[U,V]} - [\nabla\_U,\nabla\_V]$ is the curvature tensor of $\nabla$, and $\nabla\_{U,V}^2X = \nabla\_U\nabla\_VX - \nabla\_{\nabla\_UV}X$. Applying this to $\nabla\_{\partial\_j}S$ we get


\begin{align}
\mathcal{L}\_{\partial\_r}\left(\nabla\_{\partial\_j}S\right) &= \mathcal{L}\_{\partial\_r}(\nabla)(\partial\_j,S) + \nabla\_{[\partial\_r,\partial\_j]}S + \nabla\_{\partial\_j}(\mathcal{L}\_{\partial\_r}S)
\end{align}


and using the above formula and the Riccati equation for $S$ leads to the wanted linear differential equation.


