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.


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.


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