How to understand the Second covariant derivative? I am reading P. Li's lectures on Geometric analysis. On page 14, the author defines the second covariant derivative as follows:
Let $f$ be a smooth function on $M$. $\omega_1, \cdots, \omega_n$ be a local orthonormal basis of $T^*M$ around a fixed point $p$. And $d\omega_i=\sum_j\omega_{ij}\wedge\omega_j$. Then $$df=\sum_i f_i\omega_i$$
THen the author give the definition:
$$
f_{ij}\omega_j=df_i+f_j\omega_{ji}.
$$
My question is how to understand this definition compare with the usual one, i.e.
$$
\nabla^2_{X,Y}f=XY(f)-\nabla_XY (f)
$$
Also the third order covaiant derivative defines in the similar manner:
$$
f_{ijk}\omega_k=df_{ij}+f_{kj}\omega_{ki}+f_{ik}\omega_{kj}
$$
The similar question for this expression, how to understand these indices?
Thank you for any detailed answer, I am kind of afraid of this local calculations compare with the ususal global definition. However, in most papers, the local calculations are more common. So I want to figure it out clearly. 
 A: They are actually the same. I am used to a slightly different notations and I will stick to mine. Let $\{\omega^i\}_{i=1}^n$ be a local orthonormal coframe dual to the orthonormal frame $\lbrace e_i\rbrace_{i=1}^n$ and we define the connection $1$-form $\omega_i^j$ by $\nabla _X e_i = \omega_i^j(X)e_j$ (Einstein notation). Then by differentiating $\delta_i^j=\omega^j(e_i)$, we have 
$\nabla _X \omega^j =-\omega_i^j (X)\omega^i$. 
The equation $d\omega^i=\omega^j\wedge \omega_j^i$ (note that $\omega_i^j$ is $\omega_{ij}$ in Peter Li's note, and also $\omega_i^j=-\omega_j^i$ due to orthogonality), known as the (first) Cartan's structure equation, can be obtained by
$$d\omega^i(X, Y)= (\nabla _X \omega^i)(Y)- (\nabla _Y \omega^i)(X)=-\omega^i_j(X)\omega^j(Y)+\omega^i_j(Y)\omega^j(X)=(\omega^j\wedge \omega ^i_j)(X, Y).$$
Alternatively, from the above it's easy to see that if $\omega^i_j $ satisfies the Cartan's equation, then $\nabla _X \omega^j =-\omega_i^j (X)\omega^i$. The connection 1-forms are exactly the analogue of the Christoffel symbols in a coordinates system.
Recall that for a function $f$, the covariant derivative $\nabla f$ coincides with $df$, i.e. $\nabla f(X)= df(X)(=X(f))$. The second covariant derivative (i.e. Hessian) is given by $\nabla ^2 f :=\nabla (\nabla f)= \nabla (df)$. In an orthonormal frame, we have
\begin{equation*}
  \begin{split}
    f_{ij}:=(\nabla ^2 f)(e_i , e_j)= (\nabla (\nabla f))(e_i, e_j):= (\nabla _{e_i}\nabla f)(e_j)= &\nabla _{e_i}(\nabla _{e_j}f)- (\nabla f)(\nabla _{e_i}e_j)\\
    =& e_i(e_j(f))-(\nabla _{e_i}e_j)f\\
    =& e_i(f_j)-(\nabla _{e_i}e_j)f.
  \end{split}
\end{equation*}
where $f_j:= e_j(f)$. 
Using the language of moving frames, this computation becomes
\begin{equation*}
  \begin{split}
    f_{ij}= e_i(f_j)-f_k \omega_j^k(e_i).
  \end{split}
\end{equation*}
As $dh=e_i (h) \omega^i$ for a function $h$, this means that 
$$ f_{ij}\omega^i= df_j- f_k \omega^k_j= df_ j + f_k \omega_k^j.$$
It turns out that it doesn't matter if we define $f_{ij}$ as $\nabla ^2 f(e_i, e_j)$ or $\nabla ^2 f (e_j, e_i)$. However, for the third derivative, the order of $i, j, k$ does matter. In Peter Li's note, $f_{ijk}$ is defined as $(\nabla _k (\nabla ^2 f))(e_j, e_i)$, which in my convention is $(\nabla (\nabla ^2 f))(e_k, e_j, e_i)= :\nabla ^3 f(e_k , e_j, e_i)$. With this in mind, we now compute $f_{ijk}$. 
As $\nabla ^2 f= f_{lm}\omega^l \omega^m$, we have
\begin{equation*}
  \begin{split}
    f_{ijk}= (\nabla _k (\nabla ^2f))(e_j, e_i)
    =& \nabla _k (\nabla ^2 f (e_j , e_i))- \nabla ^2 f (\nabla _k e_j , e_i)-\nabla ^2 f ( e_j, \nabla _k e_i)\\
    =&e_k(f_{ij})- \omega_j^l(e_k)f_{li}-\omega_i^l (e_k)f_{jl}\\
    =& (df_{ij}- f_{li}\omega^l_k- f_{jl}\omega^l_i)(e_k)
  \end{split}
\end{equation*}
Therefore 
$$ f_{ijk}\omega^k=df_{ij}- f_{li}\omega^l_k- f_{jl}\omega^l_i=df_{ij}+f_{li}\omega^k_l+f_{jl}\omega^i_l.$$
