TITLE: Explicit formula for the curvaure of a connection
QUESTION [10 upvotes]: Let $E$ be a vector bundle over $M$ and denote by $\mathcal{A}^k(E)$ the space of sections of $\Lambda^k (TM)^* \otimes E$, i.e. the space $k$-forms with values in $E$.
A connection $\nabla:\mathcal{A}^0(E) \to \mathcal{A}^1(E)$ extends to a map $\nabla:\mathcal{A}^k(E) \to \mathcal{A}^{k+1}(E)$ by setting $\nabla(\alpha \otimes s)= d \alpha \otimes s + (-1)^k \alpha \wedge\nabla s$ and we can define the curvature of $\nabla$ as the composition $F_\nabla=\nabla \circ \nabla: \mathcal{A}^0(E) \to \mathcal{A}^2(E)$.
From the Leibniz rule one can see that $F_{\nabla}(fs) = f F_{\nabla}(s)$ and therefore we can view $F_\nabla$ as an element of $\mathcal{A}^2(\text{End}(E))$ ($F_\nabla$ evaluated at a pair of tangent vectors is the endomorphism $s \mapsto F_\nabla(X,Y)(s)$.
My question is whether the formula
$F_{\nabla}(X,Y) = \nabla_X \nabla_Y  - \nabla_Y \nabla_Y - \nabla_{[X,Y]}$ holds true.
I could prove it when $E=TM$ and $\nabla$ is a torsion free connection (because in that case we have a formula relating the exterior derivative with $\nabla$) but I don't know how to deal with the general case.

REPLY [3 votes]: I guess I've  found an alternative computation. Can anyone tell me if this is right?
Let $s\in \mathcal A^0(E)$ a section and consider $\nabla s \in \mathcal A^1(E)$. By linearity we may assume that $\nabla s = \alpha \otimes t$ with $\alpha$ a 1-form and therefore $\nabla_X s = \alpha(X)t$.
If $X,Y$ are vector fields, the invariant formula for the exterior derivative gives $d \alpha (X,Y)= X \alpha(Y) - Y\alpha(X) - \alpha([X,Y])$ and therefore
$F_\nabla(X,Y)(s) = \nabla(\alpha \otimes t)(X,Y) = (d \alpha \otimes t - \alpha \wedge \nabla t)(X,Y)$
$=(d \alpha)(X,Y)t - \alpha(X) \nabla_Y t + \alpha(Y) \nabla_X t$
$= [X \alpha(Y) - Y\alpha(X) - \alpha([X,Y])]t - \alpha(X) \nabla_Y t + \alpha(Y) \nabla_X t$
$= (X \alpha(Y))t + \alpha(Y)\nabla_X t - (Y \alpha(X))t - \alpha(X)\nabla_Y t - \alpha([X,Y])t $
$= \nabla_X (\alpha(Y) t) - \nabla_Y(\alpha(X) t) - \alpha([X,Y])t$
$=\nabla_X \nabla_Y s - \nabla_Y \nabla_X s- \nabla_{[X,Y]}s.$