# Commutation relations between covariant and Lie derivatives

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 $$\mathcal{L}_{\partial_r}\nabla_{\partial_j}S$$ ?

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.

I recently answer my question by finding a formula I didn't know. 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.