Simons' identity

Question

I'm trying understand the proof of the Simons' identity:

$\begin{align*} \nabla_k \nabla_l h_{ij} &= \nabla_i \nabla_j h_{kl} + h_{kl}h_{ip}g^{pq}h_{qj} - h_{il} h_{kp} g^{pq} h_{qj}\\ &+ h_{kj} h_{ip} g^{pq} h_{ql} - h_{ij} h_{kp} g^{pq} h_{qj} + \overline{R}_{kilp} g^{pq} h_{qj}\\ &+ \overline{R}_{kijp} g^{pq} h_{ql} + \overline{R}_{pjil} g^{pq} h_{kq} + \overline{R}_{0i0j} h_{kl} - \overline{R}_{0k0l} h_{ij}\\ &+ \overline{R}_{pljk} g^{pq} h_{iq} + \overline{\nabla}_k \overline{R}_{0jil} + \overline{\nabla}_i \overline{R}_{0ljk}. \end{align*}$

I'm reading Lectures on Mean Curvature Flows by Xi-Ping-Zhu and I'm trying understand the following identity stated in his proof:

Then compute from the definition of $h_{ij}$ $$\nabla_k (\overline{R}_{0jil}) = \nabla_k \overline{R}_{0jil} + h_{ip} g^{pq} \overline{R}_{qjil} - h_{ik} \overline{R}_{0j0l} - h_{lk} \overline{R}_{0ji0}.$$

The local field of frames is the same as in the beginning of the section $2$ of this paper.

$\textbf{My attempt:}$

$\nabla_k \overline{R}_{0jil} = \nabla_k (\overline{R}_{0jil}) - \overline{R}(\overline{\nabla}_k \nu,e_j,e_i,e_l) - \overline{R}(\nu,\overline{\nabla}_k e_j,e_i,e_l) - \overline{R}(\nu,e_j,\overline{\nabla}_k e_i,e_l) - \overline{R}(\nu,e_j,e_i,\overline{\nabla}_k e_l)$

Computing the terms on the right separately,

$\begin{align*} \overline{R}(\overline{\nabla}_k \nu,e_j,e_i,e_l) &= \overline{R}(h^q_k e_q,e_j,e_i,e_l)\\ &= g^{qp} h_{pk} \overline{R}(e_q,e_j,e_i,e_l)\\ \end{align*}$

The problem above is that I obtained the term $h_{pk}$ instead of $h_{ip}$.

The problem below is that I don't know what to do with the terms involving Christoffel symbols. I thought use geodesic normal coordinates, but the expression for $\nabla_k (\overline{R}_{0jil})$ involves the term $g^{pq}$, then I don't need geodesic normal coordinates, but I don't know what to do with these terms. I thought find a relation between Christoffel's symbols in order to use the Bianchi's indentity, but I couldn't find a relation.

$\begin{align*} \overline{R}(\nu,\overline{\nabla}_k e_j,e_i,e_l) &= \overline{R}(e_i,e_l,\nu,\overline{\nabla}_k e_j)\\ &= \overline{R}(e_i,e_l,\nu,\overline{\Gamma}^m_{kj} e_m - h_{kj} \nu)\\ &= \overline{\Gamma}^m_{kj} \overline{R}(e_i,e_l,\nu,e_m)\\ &= \overline{\Gamma}^m_{kj} \overline{g}(\overline{R}(e_i,e_l)\nu,e_m)\\ &\overset{(1)}{=} - \overline{\Gamma}^m_{kj} \overline{g}(\nu,\overline{R}(e_i,e_l)e_m) \end{align*}$

$\begin{align*} \overline{R}(\nu,e_j,\overline{\nabla}_k e_i,e_l) &= \overline{R}(\nu, e_j,\overline{\Gamma}^m_{ki} e_m - h_{ki} \nu,e_l)\\ &= \overline{\Gamma}^m_{ki} \overline{R}(\nu,e_j,e_m,e_l) - h_{ki} \overline{R}(\nu, e_j,\nu,e_l)\\ &= - \overline{\Gamma}^m_{ki} \overline{R}(e_m,e_l,e_j,\nu) - h_{ki} \overline{R}(\nu, e_j,\nu,e_l)\\ &= - \overline{\Gamma}^m_{ki} \overline{g}(\nu,\overline{R}(e_m,e_l)e_j) - h_{ki} \overline{R}(\nu, e_j,\nu,e_l) \end{align*}$

$\begin{align*} \overline{R}(\nu,e_j,e_i,\overline{\nabla}_k e_l) &= \overline{R}(\nu, e_j,e_i,\overline{\Gamma}^m_{kl} e_m - h_{kl} \nu)\\ &= \overline{\Gamma}^m_{kl} \overline{R}(\nu,e_j,e_i,e_m) - h_{kl} \overline{R}(\nu, e_j,e_i,\nu)\\ &= - \overline{\Gamma}^m_{kl} \overline{R}(e_i,e_m,e_j,\nu) - h_{kl} \overline{R}(\nu, e_j,e_i,\nu)\\ &= - \overline{\Gamma}^m_{kl} \overline{g}(\nu,\overline{R}(e_i,e_m)e_j) - h_{kl} \overline{R}(\nu, e_j,e_i,\nu) \end{align*}$

$(1)$: We use here the fact that $\overline{\nabla} g \equiv 0$ and the lemma of this lecture notes on page $4$ applied for $T \equiv g$.

Thanks in advance!

Answer 1

I finally understood how derived $\nabla_k \overline{R}_{0jil}$ correctly. I will leave in my OP my original attempt to call attention for mistakes that I was making because some details.

Firstly, observe that the definition of curvature tensor given by Do Carmo's book, which is widely used in graduate programs, is a bit different of the definition of curvature tensor used in Huisken's paper, where I read the Simons' identity and my motivation to try to prove this identity. Indeed, the definition of the curvature tensor used by Huisken can be found on page $6$ of this lecture notes written by a former Huisken's student. That said, I will proceed using this last definition as follows:

$\nabla_k \overline{R}_{0jil} = \nabla_k (\overline{R}_{0jil}) - \overline{R}(\overline{\nabla}_k \nu,e_j,e_i,e_l) - \overline{R}(\nu,\overline{\nabla}_k e_j,e_i,e_l) - \overline{R}(\nu,e_j,\overline{\nabla}_k e_i,e_l) - \overline{R}(\nu,e_j,e_i,\overline{\nabla}_k e_l)$

Computing the terms on the right separately,

$\begin{align*} \overline{R}(\overline{\nabla}_k \nu,e_j,e_i,e_l) &= \overline{R}(h^q_k e_q,e_j,e_i,e_l)\\ &= g^{qp} h_{pk} \overline{R}(e_q,e_j,e_i,e_l)\\ \end{align*}$

As pointed out on the comments, the computation above isn't agree because a possible typo.

$\begin{align*} \overline{R}(\nu,\overline{\nabla}_k e_j,e_i,e_l) &= \overline{R}(e_i,e_l,\nu,\overline{\nabla}_k e_j)\\ &= \overline{R}(e_i,e_l,\nu,\overline{\Gamma}^m_{kj} e_m - h_{kj} \nu)\\ &= \overline{\Gamma}^m_{kj} \overline{R}(e_i,e_l,\nu,e_m)\\ &= \overline{\Gamma}^m_{kj} \overline{g}(\overline{R}(e_i,e_l)e_m,\nu) \end{align*}$

$\begin{align*} \overline{R}(\nu,e_j,\overline{\nabla}_k e_i,e_l) &= \overline{R}(\nu, e_j,\overline{\Gamma}^m_{ki} e_m - h_{ki} \nu,e_l)\\ &= \overline{\Gamma}^m_{ki} \overline{R}(\nu,e_j,e_m,e_l) - h_{ki} \overline{R}(\nu, e_j,\nu,e_l)\\ &= \overline{\Gamma}^m_{ki} \overline{R}(e_m,e_l,\nu,e_j) - h_{ki} \overline{R}(\nu, e_j,\nu,e_l)\\ &= \overline{\Gamma}^m_{ki} \overline{g}(\overline{R}(e_m,e_l)e_j,\nu) - h_{ki} \overline{R}(\nu,e_j,\nu,e_l) \end{align*}$

$\begin{align*} \overline{R}(\nu,e_j,e_i,\overline{\nabla}_k e_l) &= \overline{R}(\nu, e_j,e_i,\overline{\Gamma}^m_{kl} e_m - h_{kl} \nu)\\ &= \overline{\Gamma}^m_{kl} \overline{R}(\nu,e_j,e_i,e_m) - h_{kl} \overline{R}(\nu, e_j,e_i,\nu)\\ &= \overline{\Gamma}^m_{kl} \overline{R}(e_i,e_m,\nu,e_j) - h_{kl} \overline{R}(\nu, e_j,e_i,\nu)\\ &= \overline{\Gamma}^m_{kl} \overline{g}(\overline{R}(e_i,e_m)e_j,\nu) - h_{kl} \overline{R}(\nu,e_j,e_i,\nu) \end{align*}$

Observe that $\overline{g}(\overline{R}(e_i,e_l)e_m,\nu) = \overline{g}(\overline{R}(e_m,e_l)e_j,\nu) = \overline{g}(\overline{R}(e_i,e_m)e_j,\nu) = 0$ because the tangential component of the connection of the ambient space coincides with the connection of the hypersurface, therefore $\overline{R}$ coincides with $R$ when restrict to smooth vectorfields tangent to $T_pM \cong T_{X(p)}X(M)$ (I'm denoting by $X$ the immersion of $M^n$ into $N^{n+1}$) and $\overline{R}(e_i,e_l)e_m, \overline{R}(e_m,e_l)e_j$ and $\overline{R}(e_i,e_m)e_j$ are smooth vectorfields tangent to $T_pM \cong T_{X(p)}X(M)$ (I'm considering $e_i$ as local extensions to smooth vectorfields on $T_pM \cong T_{X(p)}X(M)$ to my argument be true), then the identity for $\nabla_k \overline{R}_{0jil}$ follows.