5
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.