Question about identities regarding second covariant derivatives For a vector field $X$ and a covariant derivative(Levi-Civita connection for a given metric, for simplicity) $\nabla$ let us suppose that $\nabla_a X_b+\nabla_b X_a=0$ holds. Then it is said that $\nabla_c\nabla_a X_b+\nabla_c\nabla_b X_a=0$ holds trivially. But I cannot understand how this second derivative identity holds.... Isn't $\nabla_c\nabla_bX_a=(\nabla_c\nabla_bX)_a$? Then from by calculations regarding Christoffel symbols, 
\begin{align}
\partial_c(\nabla_aX_b)=\nabla_c\nabla_aX_b+\Gamma^f_{cb}\nabla_aX_f\\\partial_c(\nabla_bX_a)=\nabla_c\nabla_bX_a+\Gamma^f_{ca}\nabla_bX_f\\
\end{align}
Since $\partial_c(\nabla_aX_b)=-\partial_c(\nabla_bX_a)$ we can equate the above two equations by - sign. However, $\Gamma^f_{cb}\nabla_aX_f+\Gamma^f_{ca}\nabla_bX_f$ simply doesn't vanish as should be....As a result, $\nabla_c\nabla_aX_b=\nabla_c\nabla_bX_a$ cannot be obtained.... It seems that I am terribly misunderstanding something and it is extremely frustrating.... Could anyone please help me?
 A: The traditional way of using indexes in the notation for covariant derivatives is notoriously confusing.
If you have a vector field $X$, you would write its components as $X^a$, but in presence of a Riemaniinan metric it is customary to identify vectors with co-vectors (= 1-forms), so you can write a condition such as $\nabla_a X_b+\nabla_b X_a=0$ still thinking of $X$ as of a vector field.
The covariant derivative of $X$ with respect to some connection $\nabla$ is denoted by $\nabla X$, and this is a tensor field with two indexes. Up to the identification mentioned above ("the musical isomorphisms"), you can write the components of $\nabla X$ as $\nabla_a X_b$, as you do. The formulas using the Christoffel symbols allow to express these components in terms of a local coordinate patch.
But then the covariant derivative of $\nabla X$ in turn is denoted as $\nabla \nabla X$, and $\nabla_a \nabla_b X_c$ is a confusing but wide-spread manner of denoting the components of this new tensor with three indexes. Really, it would be better to write $(\nabla \nabla X)_{a b c}$, but who would bother :)
Now, $\nabla_a X_b+\nabla_b X_a=0$ means, that you have a sum of two tensor fields $\nabla X$ and $\text{[Swap a and b]}\nabla X$, where the components of the latter field are given by 
$$
(\text{[Swap a and b]}\nabla X)_{a b} = (\nabla X)_{b a}
$$
For any two tensor fields $T$ and $S$, the covariant derivative enjoys the linearity property:
$$
\nabla (T + S) = \nabla T + \nabla S
$$
Write it in terms of components, and you get:
$$
(\nabla (T + S))_{abc\dots} = (\nabla T)_{abc\dots} + (\nabla S)_{abc\dots}
$$
Applying this to our situation, we see
$$
\begin{align}
(\nabla (\nabla X + \text{[Swap a and b]}\nabla X))_{c a b} 
& = (\nabla \nabla X)_{c a b} + (\nabla \text{[Swap a and b]} \nabla X )_{c a b} \\
& = (\nabla \nabla X)_{c a b} + (\nabla  \nabla X)_{c b a}
\end{align}
$$
Sorry for this ad-hoc $\text{[Swap a and b]}$ notation. It is only used to convey the idea.
Appendix: the second covariant derivative of a vector field in terms of the Christoffel symbols.
As the OP requested in chat, I put here a few lines regarding an expression for the second covariant derivative of a vector field in terms of the Christoffel symbols.
I will assume that the following two identities for the components of the covariant derivatives of a co-vector field (1-form) $\omega$ and a vector field $X$  are known:
$$
(\nabla \omega)_{a b} = \nabla_a \omega_b = \partial_a \omega_b - \Gamma_{a b}{}^{c} \omega_c \tag{I}
$$
and 
$$
(\nabla X)_a{}^b = \nabla_a X^b = \partial_a X^b + \Gamma_{a c}{}^{b} X^c \tag{II}
$$
where the Einstein summation convention is used.
Now, let us express the second covariant derivative $\nabla \nabla X$, using these formulas. 
Since $\nabla X$ is a tensor field, we may think of it as of a finite sum of tensor products of co-vectors and vectors. Thus, we may run our calculation for a tensor of the form $\omega \otimes X$, and then extend our results by linearity.
So, let us imagine for a moment, that $\nabla X \equiv \omega X$.
Notice, that for a tensor product $(\omega X)_a{}^b \equiv \omega_a X^b$.
Then we can write
$$
(\nabla \nabla X)_{a b}{}^{c} = (\nabla (\omega X))_{a b}{}^{c} = ((\nabla \omega)_{a b} X^c + \omega_b (\nabla X)_{a}{}^{c})
$$
where we have used the Leibniz rule.
Substituting the identities $(I)$ and $(II)$ into the above result, we obtain:
$$
\begin{align}
(\nabla \nabla X)_{a b}{}^{c} & = (\partial_a \omega_b - \Gamma_{a b}{}^{d} \omega_d) X^c + \omega_b ( \partial_a X^c + \Gamma_{a d}{}^{c} X^d) \\
& = (\partial_a \omega_b) X^c - \Gamma_{a b}{}^{d} (\omega_d X^c) +  \omega_b ( \partial_a X^c) + \Gamma_{a d}{}^{c} (\omega_b X^d) \\
& = \partial_a(\omega_b X^c) - \Gamma_{a b}{}^{d}(\omega_d X^c) + \Gamma_{a d}{}^{c} (\omega_b X^d)
\end{align}
$$
Now, watch the hands: $\omega X$ was representing $\nabla X$ (this is where we "extend by linearity"), so, we can write:
$$
\begin{align}
(\nabla \nabla X)_{a b}{}^{c} & = \partial_a(\nabla_b X^c) - \Gamma_{a b}{}^{d}(\nabla_d X^c) + \Gamma_{a d}{}^{c} (\nabla_b X^d)
\end{align}
$$
But for $\nabla X$ we can substitute the identity $(II)$ again, and continue the calculation!
So,
$$
\begin{align}
(\nabla \nabla X)_{a b}{}^{c} & = \partial_a(\partial_b X^c + \Gamma_{b d}{}^{c} X^d) - \Gamma_{a b}{}^{d}(\partial_d X^c + \Gamma_{d e}{}^{c} X^e) + \Gamma_{a d}{}^{c} (\partial_b X^d + \Gamma_{b e}{}^{d} X^e) \\
& = \partial_a \partial_b X^c + (\partial_a \Gamma_{b d}{}^{c}) X^d + \Gamma_{b d}{}^{c} \partial_a X^d \\
& - \Gamma_{a b}{}^{d}(\partial_d X^c) - \Gamma_{a b}{}^{d} \Gamma_{d e}{}^{c} X^e + \Gamma_{a d}{}^{c} (\partial_b X^d) +  \Gamma_{a d}{}^{c} \Gamma_{b e}{}^{d} X^e
\end{align}
$$
Finally, we can present the above result as
$$
\begin{align}
(\nabla \nabla X)_{a b}{}^{c} & = \partial_a \partial_b X^c - \Gamma_{a b}{}^{d} \Gamma_{d e}{}^{c} X^e +  \Gamma_{a d}{}^{c} \Gamma_{b e}{}^{d} X^e  \\
& + (\partial_a \Gamma_{b d}{}^{c}) X^d + \Gamma_{b d}{}^{c} \partial_a X^d - \Gamma_{a b}{}^{d}(\partial_d X^c)  X^e + \Gamma_{a d}{}^{c} (\partial_b X^d) 
\end{align}
$$
where one can clearly recognize the "tensorial" and "non-tensorial" parts.
I hope that this helps.
