Doubt on Double Covariant Derivatives I simply want to calculate the Components of Riemann Tensor for a general connection from the definition given by:
\begin{equation}
    \mathrm{Riem}(V,W,U) := \nabla_{V}\nabla_{W}U - \nabla_{W}\nabla_{V}U - \nabla_{[V,W]}U
\end{equation}
But I'm struggling a little with the double covariant derivatives.
My calculation is given by:

$$\nabla_{V}\nabla_{W}U = \nabla_{V}(\nabla_{W}U):=\nabla_{V}(A) \tag{1}$$
In a local chart we have:
$$\nabla_{V}(A) = (V^{a}\nabla_{a}A^{c})\partial_{c} \tag{2}$$
But, note that $A^{c}$ are the components of $\nabla_{W}U$,
$$A^{c} = (\nabla_{W}U)^{c} = W^{b}\nabla_{b}U^{c} \tag{3}$$
Therefore,
$$\nabla_{V}\nabla_{W}U = (V^{a}\nabla_{a}(W^{b}\nabla_{b}U^{c}))\partial_{c} \tag{4}$$

So, my doubt is: the components of $\nabla_{V}\nabla_{W}U$ are, in fact, that of $(4)$?
 A: I find it pretty dangerous to ask for the component of $\nabla_V\nabla _W U$, while $$(U, V, W)\mapsto \nabla_V\nabla _W U$$
is not a tensor: It is not $C^\infty(M)$-linear in $U, W$. On the other hand, it makes sense to ask for the component of $\operatorname{Riem}$, since it is a tensor (some checking is needed).
Anyway, if we know that $\operatorname{Riem}$ is a tensor, then
$$ \operatorname{Riem} = \sum_{i,j,k,l=1}^n {R_{ijk}}^l dx^i \otimes dx^j \otimes dx^k \otimes \frac{\partial }{\partial x^l}, $$
where ${R_{ijk}}^l$ are the components of $\operatorname{Riem}$ given by
$$ \tag{1} \operatorname{Riem}\left( \frac{\partial }{\partial x^i},\frac{\partial }{\partial x^j},\frac{\partial }{\partial x^k}\right) = \sum_{l=1}^n  {R_{ijk}}^l\frac{\partial }{\partial x^l}. $$
Calculating the left hand side of (1) is easy:
\begin{align}
\operatorname{Riem}\left( \frac{\partial }{\partial x^i},\frac{\partial }{\partial x^j},\frac{\partial }{\partial x^k}\right) &= \nabla_{\partial_i }\nabla_{\partial_j} \frac{\partial}{\partial x^k} - \nabla_{\partial_j }\nabla_{\partial_i} \frac{\partial}{\partial x^k} \\
&= \sum_l \left( \nabla_{\partial_i } \left(\Gamma_{jk}^l \frac{\partial}{\partial x^l}\right)-\nabla_{\partial_j } \left(\Gamma_{ik}^l \frac{\partial}{\partial x^l}\right)\right) \\
&= \sum_l (\partial_i \Gamma_{jk}^l-\partial_j \Gamma_{ik}^l) \frac{\partial}{\partial x^l} + \sum_m \left( \Gamma_{jk}^m \nabla_{\partial_i} \frac{\partial}{\partial x^m} - \Gamma_{ik}^m \nabla_{\partial_j} \frac{\partial}{\partial x^m} \right) \\
&=\sum_l \left(\partial_i \Gamma_{jk}^l-\partial_j \Gamma_{ik}^l + \sum_m\left( \Gamma_{jk}^m \Gamma_{im}^l - \Gamma_{ik}^m \Gamma_{jm}^l\right) \right)\frac{\partial}{\partial x^l}
\end{align}
Thus
$${R_{ijk}}^l = \partial_i \Gamma_{jk}^l-\partial_j \Gamma_{ik}^l + \sum_m\left( \Gamma_{jk}^m \Gamma_{im}^l - \Gamma_{ik}^m \Gamma_{jm}^l\right).$$
