Convex Manifold with constant mean curvature is hypersphere I wish to understand the proof the following theorem:
Assume $M$ is an unbounded, compact, connected, strictly convex $\mathbb{R}^{n+1}$ hypersurface.
If the mean curvature of $M$ is constant, $M$ is a hypersphere.
The proof starts like this:
With regard to $\mathbb{R}^{n+1}$ hypersurfaces, the equations become
\begin{align*}
R_{ijkl} &=h_{ik}h_{jl}-h_{il}h_{jk}\quad\quad &&(5.2.40)\\
h_{ijk}-h_{ikj}&=0\quad\quad&&(5.2.41)
\end{align*}
Using the $\Delta$ expression for the Laplacian $\Delta = \left ( \frac{\partial}{\partial x_1} , \dots , \frac{\partial}{\partial x_n} \right )$ on $M$, then
$$\frac{1}{2} \Delta \big(\|{B}\|^2\big)= \frac{1}{2} \Delta\Big( \sum_{i,j} h_{ij}^2 \Big) = \sum_{i,j,k} h_{ijk}^2+\sum_{i,j}h_{ij}\Delta h_{ij}\quad (5.2.42)$$
with
\begin{align*}
\Delta h_{ij} &=\sum_{k}h_{ijkk},\\
\sum_l h_{ijkl}\omega^l &= dh_{ijk} - \sum_l h_{ljk}\omega^l_i - \sum_l h_{ilk}\omega^l_j -\sum_{l}h_{ijl}\omega_k^l.
\end{align*}
This is, where my first questions come up. I don't understand the usage of the second fundamental form here. 
How does its indexes work? 
And where does the second equality in (5.2.42) come from? 
Do I already know at this point that $\|B\|^2$ is a harmonic function?
and continues with more facts about $h_{ijkk}$:
Using (5.2.41) as well as the Ricci equation, we have
\begin{align*}
\sum_{k}h_{ijkk}  &=\sum_{k}h_{kijk}\\
      &= \sum_{k}h_{kikj}+\sum_{k,l}h_{li}R_{lkjk}+\sum_{k,l}h_{kl}R_{lijk}\\
      &=\sum_{k}h_{kkij}+\sum_{k,l}\Big(h_{li}R_{lkjk}+h_{kl}R_{lijk}\Big)\\
      &=\sum_{k,l}\Big(h_{li}R_{lkjk}+h_{kl}R_{lijk}\Big)
\end{align*}
The last equality is due to $M$s mean curvature being constant.
HERE I don't understand the second and the last equality.
Thanks for your help!
 A: I found the answer to most my question.
In fact, my whole problem was the notation and the follow-up misunderstanding of $h_{ijkl}$
Here is the solution with the following notation $h_{ijkk}=h_{ij,kk}$ and 
$$\Delta h_{ij} \overset{(1)}{=}\sum_{k}h_{ijkk}$$
\begin{align*}
\frac{1}{2} \Delta \big(\|{B}\|^2\big) = &\frac{1}{2} \Delta\Big( \sum_{i,j} h_{ij}^2 \Big) \\
       = &\frac{1}{2} \sum_k \Big( \sum_{i,j} h_{ij}^2 \Big) _{,kk}\\
       = &\frac{1}{2} \sum_k \Big( 2\sum_{i,j} h_{ij}h_{ij,k} \Big) _{,k}\\
       = & \sum_k \Big( \sum_{i,j} h_{ij,k}^2+h_{ij}h_{ij,kk} \Big) \\
       \overset{using (1)}{=} &\sum_{i,j,k} h_{ijk}^2+\sum_{i,j}h_{ij}\Delta h_{ij}\quad (5.2.42)
\end{align*}
The answer to the second question lies in the Ricci equation being:
$$h_{ki,jk}-h_{ki,kj}=h_{li}R_{lkjk}+h_{kl}R_{lijk}.$$
Inserting it in $h_{ij,kk}$ which is equal to $h_{ki,jk}$ due to symmetry, solves the problem.
If the mean curvature is constant, $\frac{1}{n}\sum_k h_{kk}$ is constant, moreover $\sum_k h_{kk}$ is constant. Thus $\sum_k h_{kk,ij}=0.$
Still unanswered stays:
Do I already know at this point that $∥B∥^2$ is a harmonic function?
